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http://mathhelpforum.com/discrete-math/164800-towers-hanoi-problem-print.html	# Towers of Hanoi Problem Printable View • November 29th 2010, 06:28 PM SMAlvarez Towers of Hanoi Problem I need help getting started I believe bk=ak-1, but I'm not sure. Can someone show me how to get B1 and I will attempt the rest? Thanks. The problem is: Quote: 1. Tower of Hanoi with Adjacency Requirement. Suppose that in addition to the requirement that they neer move a larger disk on top of a smaller one, the priests who move the disks of the Tower Hanoi are also required to move disks one by one from one pole to an adjacent pole. Assume poles A and C are at the two ends of the row and pole B is in the middle. Let an = the minimum number of moves needed to transfer a tower of n disks from pole A to pole C. The recurrence relation for ak is then ak = 3ak-1+2. Now let bk=the minimum number of moves needed to transfer a tower of n disks from pole A to pole B. a. Find b1, b2, and b3. b. Show that bk=ak-1+1+bk-1 for all integers k>=2. c. Show that bk <= 3bk-1+1 for all integers k>=2. • November 30th 2010, 02:22 AM emakarov Quote: Can someone show me how to get B1 and I will attempt the rest? Are you really wondering how many moves it takes to move one disk from A to B? It is easy to see that $b_2=4$. In my opinion, it is easier to see why the general formula for $b_k$ holds than to calculate $b_3$ from scratch. This, of course, will allow you to figure $b_3$ provided you know $a_1$. • November 30th 2010, 07:35 AM SMAlvarez Yes how many steps does it take to move from a to b? Also, is there a equation for bk? • November 30th 2010, 07:59 AM SMAlvarez Also how did you get b2 = 4? The book says b4=40 also when I say equation I mean like ak = 3ak-1+2. THANKS!! • November 30th 2010, 08:42 AM emakarov Quote: Also, is there a equation for bk? There is a recurrence equation in your first post, though it expresses $b_k$ through both $b_{k-1}$ and $a_{k-1}$. Most likely, it is possible to express $b_k$ through $b_{k-1}$ only, but the problem does not ask for this. Quote: Yes how many steps does it take to move from a to b? Move what? Moving one disk from A to B is something you'll have to figure out for yourself. Also, be careful about capital/lowercase letters: A and B denote poles while $a_k$ and $b_k$ are numbers. Quote: Also how did you get b2 = 4? Move the smaller disk to C (2 moves because of the adjacency requirement), move the larger disk to B, move the smaller disk to B. Quote: The book says b4=40 I agree. All times are GMT -8. The time now is 11:42 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9372614026069641, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/208809/compact-way-to-predicting-percent-values	# Compact way to predicting percent values I'm looking for different ways to calculate payouts after I deposit some units. Idea: There is a raffle for total of 1 000 000 000(TP). Everyone can deposit their Tickets(OT) (everyone has different amount of owned tickets) and at the end, I'll get the percentage of total tickets(TT) to deposited tickets(DT) Example, someone deposited 1000 tickets, and I'm going to deposit 234: $$currentPricePerTicket = \frac{TP}{TT} = 1000000$$ $$TTafterMyDeposit = TT + DT = 1000 + 234 = 1234$$ $$pricePerTicketAfterDeposit = \frac{TP}{TTafterMyDeposit} = \frac{1000000000}{1234} = 810372.77$$ $$myPercentOfDeposited = DT \div \frac{TTafterMyDeposit}{100} = 234 \div \frac{1234}{100} = 18.96\%$$ $$iWillGet = TP \cdot myPercentOfDeposited = 1000000000 \cdot 18.96\% = 189 600 000$$ This is what I'll get if nobody after me deposited their tickets. This is pretty straight way to calculate. My main question, is there more "short" version to calculate the answer (189 600 000)? Also, my second question: how do I invert this this? For example, if I want to get 249 365 000, how many tickets i should deposit? - ## 1 Answer So that I will not have to type so many zeros, let the prize be $P$. If the current total is $t$, and I deposit $d$, then my fractional share of the prize is $\dfrac{d}{t+d}$. So the amount I get is $$\frac{Pd}{t+d}.$$ That is precisely what you did. There does not seem to be anything simpler. The question now is, if we want our share to be the amount $S$, what should $d$ be? So we have the equation $$S=\frac{Pd}{t+d},$$ and want to solve for $d$. Multiply both sides of the above equation by $t+d$. We get $$S(t+d)=Pd.$$ So $St+Sd=Pd$, and therefore $St=(P-S)d$. It follows that $$d=\frac{St}{P-S}.$$ In your question, we have $P=1000000000$, $t=1000$, and $S=249365000$. Then the above equation gives $d\approx 332.2$. - I think this is a simplification, as it avoids going through the percentages. – Ross Millikan Oct 7 '12 at 16:34 Thank you very much, that's awesome :) – Bubonic Pestilence Oct 7 '12 at 16:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9382906556129456, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/tagged/weighted-mean	# Tagged Questions The weighted-mean tag has no wiki summary. 2answers 66 views ### How to compute weighted means and SDs? I want to calculate volume weighted average price and standard deviation for the following dataset: ... 1answer 42 views ### Confusion regarding weighted average This is a pretty simple question and I'm wondering how to go about finding a solution. I have the following data. ... 1answer 32 views ### Given weighted mean and raw data, derive weights to minimize error Imagine I have a set of data from an experiment: Observed Measurement 1: 5 Observed Measurement 2: 6 Observed Measurement 3: 7 Observed Measurement 4: 8 Observed Measurement 5: 9 I have a target ... 0answers 60 views ### Weighted Variance, one more time Unbiased weighted variance was already addressed here and elsewhere but there still seems to be a surprising amount of confusion. 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A fair coin is flipped; if it is Heads, a ball is drawn from Urn 1, and if it is Tails, a ball is drawn from Urn ... 1answer 256 views ### Bias correction in weighted variance For unweighted variance $$\text{Var}(X):=\frac{1}{n}\sum_i(x_i - \mu)^2$$ there exists the bias corrected sample variance, when the mean was estimated from the same data: ... 1answer 80 views ### Optimize Weights used in a Weighted Average I want to calculate a weighted average, mostly for illustration purposes. However, I have an outcome variable that is Y/N and would like to optimize the weights relative to this outcome. What are ... 1answer 169 views ### Voting system that uses confidence for each voter Let's say, we have simple "yes/no" question that we want to know answer to. And there are N people "voting" for correct answer. Every voter has a history - list of 1's and 0's, showing whether they ... 1answer 590 views ### A simpler way to calculate Exponentially Weighted Moving Average? Proposed Method: Given a time series $x_i$, I want to compute a weighted moving average with an averaging window of $N$ points, where the weightings favour more recent values over older values. In ... 1answer 74 views ### Handling missing attributes when using weighted average I was trying this optimization where I have some predictors and I want to predict the real value using weighted average of the predictors. What I am trying to do is given the predictors' prediction ... 0answers 71 views ### Is it possible to map observed responses on five-point Likert scale into three-class conceptual model? I am currently making my thesis proposal about employees productivity and I had a questionnaire which consists of a five-point Likert scale (5-strongly agree, 4-agree, 3-agree, 2-disagree, and ... 2answers 108 views ### How do I calculate the ranking of some galleries based on the rankings of the artists represented by them? 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Because all measurements are assumed ... 1answer 218 views ### Standard error for the sum of weighted means I am looking at a two-stage process of diagnosis. Stage 1 is a screen that identifies people with the diagnosis with some error, Stage 2 is a detailed examination (a gold standard) In Stage 1, N ... 0answers 33 views ### How to symmetrically adjust data by removing certain value Say I have a data set in the following format: {a:200, b: 100; c:400} The sum of a, b and c equals 700. Now, I have another value which is supposed to be the actual total say 500. How do I adjust ... 2answers 479 views ### Weighted mean of two 2D Gaussian random variables I have two 2D Gaussian random variables. I'd like to find the weighted average of the two (based on their covariance matrix around the mean, meaning that the mean of the final Gaussian should be ... 0answers 63 views ### Weighting datapoints by uncertainty - asymmetric CIs I have several estimates obtained using different softwares, these programs give me an estimate as well as a 95% confidence interval. However, these CIs are not symmetric around my point estimates, ... 1answer 178 views ### Averages of averages (of averages, of averages…) Consider the following cell biology experiment. We are comparing $T$ different treatments of cultured cells. Each treatment $t$ is replicated in several (microtiter) wells, indexed by the variable ... 1answer 206 views ### Determining true mean from noisy observations I have a large set of data points of the form (mean, stdev). I wish to reduce this to a single (better) mean, and a (hopefully) smaller standard deviation. Clearly I could simply compute \$\frac{\sum ... 1answer 363 views ### Quality-price trade-off Say there are 3 companies A, B and C. Each company has a quality rating from 0 to 100 and a price in USD. ... 1answer 181 views ### How do I introduce features and their confidence values into classifiers I would appreciate some advice on how best to weight or give more importance to a percentage with a larger denominator. Eg A. 1 out of 2 = 50% B. 5 out of 10 = 50% C. 500 out of 1000 = 50% Some of ... 1answer 12k views ### How do I calculate a weighted standard deviation? In Excel? So, I have a data set of percentages like so: ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9400244355201721, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/213657/geometric-significance-of-the-addition-of-square-roots-of-two-numbers	# Geometric Significance of the Addition of Square Roots of Two Numbers In a calculation, I've come across a relation along the lines of this: $${a}^{1/2}+{b}^{1/2}$$ My presumption would be that this is somewhat related to the Pythagorean relation: $${a}^{2}+{b}^{2}$$ I can understand the Pythagorean relation, but not the importance of the square root relation. Is there a hidden geometric significance to the first relation, just as there is one for the Pythagorean formula? Note: the other side of the equation in my calculations could be anything, it's not limited to ${c}^{1/2}$. Actually, in my calculations the full relation is: $${c}={d}^{1/2}\bigg[{a}^{1/2}+{b}^{1/2}\bigg]$$ - Without knowing the details, that is probably more geometrically meaningful written as $c=\sqrt{ad}+\sqrt{bd}$. – Henning Makholm Oct 14 '12 at 14:59 where did you come across this relation? – john mangual Nov 7 '12 at 17:49 I was doing original research. But to answer it fully without revealing my work, I was relating two bodies of mass and this relation came about because of that. It's really as simple as that; relating two masses. – Zchpyvr Nov 7 '12 at 23:29 ## 1 Answer For positive $a$ and $b$, the generalized mean with exponent 1/2: $M_{1/2}\left(a,b\right)=\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2}$ is equal to $a$ if $a=b$, and lies between the geometric mean and the arithmetic mean (the "average"): $\sqrt{ab}\le\left(\frac{\sqrt{a}+\sqrt{b}}{2}\right)^{2}\le\frac{a+b}{2}$. If $a$ and $b$ are areas of squares, then $M_{1/2}\left(a,b\right)$ is the area of a square with side length equal to the average of the original side lengths, and your expression is the sum of the side lengths, but without more geometric context I don't know how to get a better geometric picture of these things. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9607792496681213, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/209581-multiple-root.html	# Thread: 1. ## Multiple root Let $P$ be an arbitrary field, while $C$ is the complex field. $f(x)$ is a polynomial in $P$ with degree $>0$. Suppose $f$ has no multiple factor in $P$, prove then $f$ has no multiple root in $C.$ Here, we mean by $g(x)$ is a multiple factor of $f(x)$, if $g(x)$ is irreducible, and for some $k>1$, $g^k(x)\mid f(x).$ 2. ## Re: Multiple root This doesn't make any sense as written? I'm going to assume that $P$ is a subfield of $\mathbb{C}$. Recall then that $f$ having no multiple roots is equivalent to $(f,f')=1$, which is trivially independent of what field you are discussing. Once again, this question dreally doesn't make sense.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9603958129882812, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/5421/using-cbc-with-a-fixed-iv-and-a-random-first-plaintext-block/5422	# Using CBC with a fixed IV and a random first plaintext block What if, instead of using CBC mode in the normal way with a random IV, I used this approach: 1. Use a fixed IV (like a block of 0's). 2. Before encrypting, generate a random block and prepend it to the plaintext. 3. After decrypting, ignore the first block. Is this exactly the same in terms of security as using a random IV? Also, would it make any difference if I used something like a message counter instead of a random block in step 2? - Why to do that?Is like CTR mode with 1 counting step, but in CTR the IV+the counter are encrypted with AES and this is XORed with the plaintext. If the IV is not random it must be unique, that's why we use a counter – curious Nov 19 '12 at 11:16 3 – Ilmari Karonen Jan 21 at 19:28 2 @IlmariKaronen Good point. If nobody protests here until tomorrow (about noon UTC), I'll do the merge, into the first question (as it is the oldest one and has an accepted answer). – Paŭlo Ebermann♦ Jan 21 at 23:14 2 Note: I merged the two questions CBC with fixed IV but changing data and Why not use CBC, not send IV, and assume first plaintext block is garbage? into this question (which means transferring all answers + comments). Please edit the question so it becomes a "canonical" question. (@IlmariKaronen and anyone else.) – Paŭlo Ebermann♦ Jan 23 at 9:35 1 @Paŭlo and others: I made a few changes, but I'm sure it could be improved further yet. – Ilmari Karonen Jan 23 at 15:31 show 4 more comments ## 5 Answers The security of that approach is equivalent to that of normal CBC. Your scheme with first plaintext block $IV^\prime$ is clearly identical to normal CBC with $IV=AES(IV^\prime)$. Since a block cipher is a permutation over a block, a uniformly random first plaintext block will lead to a uniformly random IV for normal CBC. A ciphertext produced with your scheme is of the same nature as with plain CBC. i.e. you can decrypt a normal CBC ciphertext with your algo, and you can decrypt a ciphertext produced by your algo with normal CBC. The only downside of your scheme is that it costs one additional AES invocation per message. No difference in ciphertext size, and no CPU cost beyond that single call. One interesting question is what happens with IVs that are unique, but not unpredictable. I suspect your scheme is secure in such a situation, unlike normal CBC. But I don't have a proof for that. - Yes. It think that any strategy breaking the proposed scheme breaks CBC with random IV, and vice versa, without need for any adjustment. If the IVs are known, or worse predictable (e.g. a counter), I can't think of such a proof. – fgrieu Nov 19 '12 at 11:47 I'd postulate that the scheme is secure with predictable IVs such as counters as long as the number of encrypted blocks is below the birthday bound. The idea is that the AES primitive will never encrypt the same block twice. The failure when a collision happens might be more severe though. – CodesInChaos Nov 19 '12 at 11:53 If you look at the CBC diagram, you'll see that having a fixed IV is equivalent to having the first ciphertext block become the IV. If your cipher is a good pseudorandom permutation, then what you are doing does work, if and only if all timestamps are unique such that the "new IV" is unique and unpredictable. And in fact, if you do not use the decrypted value of the first block (the "IV") this is provably just as secure as normal CBC in the ideal cipher model, as the IV requirements of CBC (uniqueness and unpredictability) happen to coincide with what a pseudorandom permutation does. If you do use that value, I am not sure. I believe it is still secure but cannot think of a convincing argument of why it is (or is not) right now, perhaps someone can complete this. You may also set the "fixed IV" to zero for convenience - its value is irrelevant. That said, it is probably best to fix your IV transmission, just to do things cleanly and not have to explain your weird IV trick. After all, if you can transmit the ciphertext, why can't the IV make it? - +1 to fixing IV transmission. If you can transmit the ciphertext, you can transmit the IV. Just append the IV to the ciphertext, and unpack it on the other end. – Stephen Touset Jan 20 at 1:12 1 Also, be very sure your timestamps are unique. If they're only accurate to a second, you're probably going to be generating messages with identical timestamps, which could be very bad. You can also insert a random, unused nonce into each message, but then you might as well just use a dynamic IV. – Stephen Touset Jan 20 at 1:15 The ciphertext is transmited in a data structure using a Web Service. So adding an IV force me to change the Web Service... and I am not the only one using this Web Service. – user687254 Jan 20 at 18:48 So append the IV to the beginning of the ciphertext entry. It will be a fixed length, so can later be stripped off the front unambiguously. – Stephen Touset Jan 21 at 20:05 Sure, that's fine, but you're really just using the first block of ciphertext as the IV. If you choose the first plaintext block to be a running message counter (which you might as well do; it's easier than generating a random block) and your "discarded IV" to be all zeros (or vice versa) then your method is equivalent to standard CBC mode combined with the "encrypted counter" method of IV generation as described in Appendix C of NIST SP 800-38A: "There are two recommended methods for generating unpredictable IVs. The first method is to apply the forward cipher function, under the same key that is used for the encryption of the plaintext, to a nonce. The nonce must be a data block that is unique to each execution of the encryption operation. For example, the nonce may be a counter, as described in Appendix B, or a message number. The second method is to generate a random data block using a FIPS-approved random number generator." - +1 for the NIST quote – CodesInChaos Jan 8 at 20:02 How is the IV unpredictable if I'm combining all zeros with a message counter? – Joshua Jan 9 at 16:34 @Joshua: It becomes unpredictable (to an adversary who does not know the key) when it passes through the block cipher. – Ilmari Karonen Jan 9 at 16:43 Note that that's assuming that the adversary does not have oracle access to the block cipher. Even if the adversary does have such access to the high-level CBC mode encryption function (i.e. they can request the encryption of chosen plaintexts), CBC mode prevents them from directly feeding chosen inputs to the block cipher as long as they cannot predict the IV. Thus, the ability to predict the IV and oracle access to the block cipher are essentially equivalent for an IND-CPA adversary attacking CBC mode using this IV generation method -- either one implies the other. – Ilmari Karonen Jan 9 at 16:51 So then I am to understand that when you refer to being unable to predict the IV, that you mean that the first ciphertext block, whose plaintext is xor'd with the incrementing IV (or whatever), is sufficiently unpredictable between encryption runs, not that the IV used is sufficiently unpredictable? As it stands, if I were to pass the (incrementing) IV used in clear text along with the ciphertext, an eve would be able to see that the IV is clearly predictable, but the resulting first block still is not. – Joshua Jan 9 at 21:51 show 1 more comment The proper precautions, this is an acceptable way to implement CBC (and yes, it interoperates with the more traditional implementation of CBC, at least, implementations of CBC that put the IV immediately in front of the ciphertext). The proper precaution is to make sure, in the encrypt direction, that the value of the iv exclusive-or'ed with the block of garbage isn't the value you used with for a previous block (or, at least, it's sufficiently improbable). If you use the same IV all the time, then you can make the garbage block an incrementing counter; if you use the last ciphertext block as the IV, well, just about any garbage block that isn't correlated to the previous encryption will work. Now, why would you implement CBC this way? Well, I've personally run into two distinct scenarios where this trick helps: • You're using encryption hardware that always insists on using the last ciphertext block from the previous message. Now, we know that using predictable IVs can be insecure; this trick allows us to use such hardware without inheriting the potential security problems. • You're using hardware that can vectorize CBC mode (and so encrypting a message of 101 blocks is not much more expensive than encrypting a message of 100 blocks). With standard CBC mode, we need to generate an unpredictable IV; cryptographically secure RNGs can be expensive, at least, more expensive than writing a nonce into the garbage block, and asking hardware to encrypt one more block. - Sure you can do that and it would work just fine, but why would you do it? It doesn't save any on the communications side of the house as you still have to communicate the garbage ciphertext block. On the computation side of things, if you decrypt that garbage block and throw it away, you have made one additional call to the block cipher operation that you don't really need to make. In today's fast computers, that isn't much of an issue, but if you are on a low computation power device, making an extra call to the block cipher will add some unnecessary delay. - I guess its supposed to save on sending the IV out the door, but that the unnecessary calls to the encryption/decryption routines may be unacceptably wasteful – Joshua Jan 8 at 19:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9285008907318115, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/37713/if-it-is-given-which-intervals-are-spacelike-can-be-determined-which-intervals?answertab=oldest	# If it is given which intervals are spacelike, can be determined which intervals are lightlike? Provided that the notion of "$\mbox{spacelike}$"-ness (of an interval) is symmetric: $$\text{spacelike}( \, x - y \, ) \Longleftrightarrow \text{spacelike}( \, y - x \, ),$$ then for any set $X$ (of sufficiently many elements) the set $X^2$ (of all pairs, regardless of order, of not necessarily distinct elements of $X$) may be partitioned into three disjoint and generally non-empty subsets • of pairs containing the same element twice: $I_X := \{ x \in X: (x x) \}$, • of pairs of distinct elements whose interval is (called) "$\text{spacelike}$": $S_X := \{ x \in X \, \& \, y \in X \, \& \, x \ne y \, \& \, \text{spacelike}( x - y ): (x y) \}$, and • of all remaining pairs $K_X := X^2 \backslash (I_X \cup S_X)$. Provided further that the notion of "$\mbox{lightlike}$"-ness (of an interval) is symmetric as well: $$\text{lightlike}( \, x - y \, ) \Longleftrightarrow \text{lightlike}( \, y - x \, ),$$ and given a suitable set $X$ (of sufficiently many elements) and a suitable set $S_X$ satisfying $X^2 \cap S_X = S_X$, how would the corresponding set $K_X$ be partitioned further into two disjoint and generally non-empty subsets • of pairs of distinct elements whose interval is (called) "$\text{lightlike}$": $L_X := \{ x \in X \, \& \, y \in X \, \& \, x \ne y \, \& \, \text{lightlike}( x - y ): (x y) \}$, and • of all remaining pairs $T_X := K_X \backslash L_X$ ? Edit The wording of this question (apart from formatting issues) as it presently stands appears not adequate to the title. (Helpful responses to it have been received nevertheless, which I try to incorporate in going forward.) In trying to improve the detailed wording, what would need to be considered first is (put roughly, for the time being, as I've come to consider it only recently): (1) Whether and how the topological notion of "boundary" can be suitably generalized to the context of sets of pairs such as $X^2$ and $S_X$, and (2) Whether, given a particular set $X$, the predicate "$spacelike()$" in the definition of set $S_X$ implies certain relations to a corresponding set $K_X$ which I did not state explicitly above; such as the absence of "impossible figures" wrt. membership of certain pairs in $S_X$ or $K_X$. I plan to defer editing (apart from possible formatting) until these preliminary questions have been expressed more adequatly elsewhere. - – Qmechanic♦ Sep 18 '12 at 20:55 In main body of your question you are effectively asking that if we are given two symmetric relations $S$ and $L$ on a set $X$, and if set of pairs of points which are $S$- related to each other are known, then is it possible to find set of pairs which are $L$-related to each other. – user10001 Sep 19 '12 at 1:32 – David Zaslavsky♦ Sep 19 '12 at 2:30 `@Qmechanic`: Based on the standard you recommend, how would you express for instance "set of pairs (regardless of order) of distinct elements of set $X$", please? – user12262 Sep 19 '12 at 15:08 1 @user12262: If you would like, I could edit your question with standard notation. You can always roll back if you don't like the edits. Or improve further yourself. – Qmechanic♦ Sep 19 '12 at 16:59 show 8 more comments ## 1 Answer I think you can do this only if the set $X$ is a topological space. For any point $x\in X$, construct the set $C(x)\subset X$ of points $y'\in X$ so that $(x,y')\in K_X$. For this construction to work, each $C(x)\cup\{x\}$ has to be a closed set in $X$. Denote by $\partial C(x)$ the boundary of $C(x)$. Then, $$L_X=\{(x,y)\in X\times X\|y\in \partial C(x)\}$$ and $$T_X=\{(x,y)\in X\times X\|y\in \text{int } C(x)\}.$$ You can do an equivalent construction starting with $S_X$ instead of $K_X$. The intuition behind this construction is that the lightcone at a point $x$ is the boundary of the points which are causally connected with $x$. - `@Cristi Stoica`: Very nice and efficient and intuitive! (Meanwhile I've been struggling through trying to cast the similar intuition into overlong logical formulas and constructions à la A.A.Robb ... &) However: my question did not presume set $X$ to be given with a topology! (Sorry for having set the "general-relativity" tag then -- is there a more appropriate one? But I'm really glad you understood the sense of my question anyways.) So: Is there perhaps a suitable topology being induced on set $X$, plainly by the given set $S_X$ ? (The sum of difficulties might come out the same.) – user12262 Sep 18 '12 at 22:00 – Cristi Stoica Sep 19 '12 at 2:22 `@Cristi Stoica`: thanks for your insight and suggestions, FWIW to me in trying to phrase my question more adequatly (cmp. the recent "Edit" of my question). I wish I had a better grasp of (standard) mathematical terminology; to use (or avoid) what's available, and to claim what's not ... Presuming distance (or at least quasi-distance) relations is of course not helpful to the experimentalist who rightly asks "How do we get that?". Also, if some "additional structure" itself is available, then it may not be mandatory to consider any topology which may (or may not) be thereby induced. – user12262 Sep 19 '12 at 17:14	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9430893063545227, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/6721/entropy-of-an-empty-universe?answertab=votes	# Entropy of an empty universe After watching the first episode of wonders of the solar system, one question came up which is not explained. Bryan Cox says that ultimately the universe will be devoid of matter, so not even a single particle will actually exist. This however, seems to be a state where entropy is zero. There is no entropy since there's no disorder. On the other hand the 2nd law of thermodynamics says that the disorder increases. I can imagine that the answer might be that an empty universe has an infinite amount of entropy but it feels pretty counter-intuitive for me. - ## 2 Answers Dear Paulo, the space will be empty in the sense that the density of particles will be tiny (the particles will include diluted gas as well as the Hawking radiation from the black holes that will have evaporated). However, the Universe will also be huge because its expansion will continue - and it will become the exponential expansion of the de Sitter space (because of the dark energy, also known as the cosmological constant, that has a constant positive energy density). It's only the total entropy of the system that is increasing and the growth of the total volume of the Universe beats the decrease of the entropy density. There's another method to look at the entropy in the future. One may only look inside the "cosmic horizon" of the de Sitter space that our Universe is already becoming. This cosmic horizon - the maximally distant place where you can still see, if you wish - has a fixed radius and indeed, the entropy inside the cosmic horizon will go zero. In this description, however, the cosmic horizon itself carries a huge entropy equal to $$S = A / 4A_0$$ where $A_0=\hbar G / c^3$ is the Planck area, about $10^{-70}$ squared meters. The formula is identical to the Bekenstein-Hawking formula for a black hole. You may imagine that this entropy $S$ of the event horizon remembers the entropy of everything that may exist behind the cosmic horizon. For the value of the radius of the cosmic horizon which is determined by the cosmological constant in our Universe, $S$ is approximately $10^{120}$ in natural units (essentially bits of information, also equal to $10^{97}$ J/K or so) that is about 20 orders of magnitude larger than the current entropy $10^{100}$ of the visible Universe - the latter figure is dominated by the entropy of the large black holes in the middle of galaxies. (Before people knew about the large black holes and their entropy, they thought that most of the entropy of the current Universe was carried by the cosmic microwave background - but the CMB entropy is roughly 10-20 orders of magnitude lower than the entropy of the black holes.) The entropy of the Universe will continue to increase but in some proper language, it will converge to $10^{120}$ which is the maximum allowed entropy of our Universe. The latter statement - that the entropy of a de Sitter can't exceed this bound equal to the inverse cosmological constant in the Planck units - is a bit of the lore of quantum gravity. I suppose that Brian Cox didn't want to explain this point because the TV program would become really confusing for his typical audiences, much like this comment inevitably will. Whoever is confused by this bound should keep on assuming that $10^{120}$ is infinity and the entropy will continue to grow indefinitely. - I can know understand why he didn't explain it further. Thanks! :) – Paulo J. Matos Mar 14 '11 at 12:38 The problem with volume answers to this question is that if the universe is ultimately going to be de-Sitter, the volume that one is talking about is outside the horizon, and so of dubious status considering logical positivism. A good answer to this can be obtained by considering the entropy stored in the cosmological horizon. In a deSitter space, the maximum area of the cosmological horizon is in an empty universe, and the gain in area of the horizon swamps the entropy of anything that fell through. That this is always true is the content of the Bekenstein bound on entropy. So as the universe dilutes, the entropy of the cosmological horizon always increases by more than the entropy of whatever is inside goes down. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9485909342765808, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/168904/solve-for-the-intersection-point-given-two-sets-of-data	# Solve for the intersection point, given two sets of data I need to (numerically, specifically in C++) solve for the intersection point of two curves, f(x) and g(x). I am given two sets of data, one for each curve (eg. {($x_1$,f($x_1$)),($x_2$,f($x_2$)),($x_3$,f($x_3$))....} and {($x_4$,g($x_4$)),($x_5$,g($x_5$)),($x_6$,g($x_6$))....}). The intersection point lies outside of the two sets, and none of the $x_i$ values in "f-set" are in "g-set" (excuse the wording). The solution therefore involves some iterative loop of (a) extrapolation to a guess $x_n$ and calculating $f(x_n)$ and $g(x_n)$, and (b) checking some relation between these two values to see if $x_n =$ or $>$ or $< x_{intersection}$. If you like, you can visualize this problem as follows: you are given points on two sides of a triangle, and you would like to extrapolate these points to find the corner point. However, in general, the curves $f$ and $g$ are not straight lines. Could you please walk me through the methods used to solve this problem? - 1 Where did these points you have come from? One could certainly use piecewise polynomials to represent your two curves, but depending on the provenance of your points, there might be a more natural way to interpolate them... – J. M. Jul 10 '12 at 4:05 The visualization I provided is actually quite close to where they came from, but as stated, the sides of the triangle are not straight. Consider laying the "triangle" on side, with the point up. Given points on the "bottom edge" of the triangle, I "evolve" them upwards and can solve for where the top left/right edges are. I need to use some extrapolation method from these edges points to solve for the top point. Thanks – Kurt Jul 10 '12 at 4:10 I suppose you could start with building an interpolating Bézier curve from your points. The convenience of this is that there are a number of efficient algorithms for finding the intersection points of two Bézier curves... – J. M. Jul 10 '12 at 4:16 1 Do you have some underlying model for the data that would suggest a form for the extrapolation? – copper.hat Jul 10 '12 at 4:28 Unfortunately, no. The shapes are generally smooth, but could really be anything. I guess I could just try a cubic spline, but then what is an efficient method to solve for the intersection of two cubic splines? – Kurt Jul 10 '12 at 5:36	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9437851309776306, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/182257/a-qualitative-yet-precise-statement-of-godels-incompleteness-theorem?answertab=active	# A qualitative, yet precise statement of Godel's incompleteness theorem? I read online a statement to the effect that (I'm paraphrasing): Goedel's incompleteness theorem shows that we cannot even have a complete and consistent theory for the natural numbers. I am under the (qualitative) impression that this statement is true within the axioms of natural numbers themselves, so that • if one expanded the of axioms one could prove all of the true statements that can be expressed solely in terms of natural numbers. Note that this larger system itself is not complete and consistent. Does Godel's incompleteness theorem have the feature that it shows that these larger systems are somehow representable with the axioms of the natural numbers? - ## 1 Answer For any recursively axiomatized (consistent) theory $T$ that extends first-order Peano arithmetic, there is a sentence in the language of Peano arithmetic that is neither provable nor refutable in $T$. Note that the theory $T$ can be over a language that extends the usual language of Peano arithmetic. This states a version of the First Incompleteness Theorem. One can strengthen this, and at the same time make it more informal, by replacing the condition "$T$ is recursively axiomatized" by "there is an algorithm for enumerating the axioms of $T$." Remark: One can weaken the result, by omitting reference to Peano Arithmetic. Let $L_0$ be a language whose non-logical symbols are $0$, $S$ (for successor), $+$, and $\times$, and let $L$ be an extension of $L$. Then there is no (consistent) recursively axiomatized theory $T$ over $L$ such that all sentences of $L_0$ that are (under the usual interpretation of the non-logical symbols) true in $\mathbb{N}$ are theorems of $T$. - I didn't know it was such a softball. – Dave Aug 14 '12 at 0:34 @Dave:The Second Incompleteness Theorem is much harder to state semi-formally without distortion. – André Nicolas Aug 14 '12 at 0:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9224523305892944, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/223728/solving-for-multiple-unknown-variables	# Solving for multiple unknown variables Assume for this scenario that: `a = b * c` `d = b - c` I only know the value for `a` and `d` (in this case 0.075 and 239, respectively), and I need to solve for both `b` and `c`. Is this possible to solve given the limited information, and if so what is the best approach to do so? - ## 3 Answers This is known as a system of nonlinear simultaneous equations. In this case, it is very easy to solve the system. First, write $b$ in terms of $d$ and $c$, where $d$ is known: $$b = d+c.$$ Substitute this into the first equation: $$a = bc = (d+c)c = cd+c^2$$ Solve this for $c$ using the quadratic formula: $$c^2+dc-a = 0 \implies c = -\frac{d}{2}\pm \frac{\sqrt{d^2+4a}}{2}$$ Substitute the solutions back into the equation $b = d+c$ to find values of $b$. - The problem I am running into is, after solving the quadratic equation I get the results `-0.00031` and `-239` and I know neither of those to be possible values. Even as positive values, neither of them make sense given the real world number I am trying to solve for. Any idea as to why this would be? – Moses Oct 29 '12 at 19:28 I do not know, because I do not have the context of the problem. All I know is that there are finitely many possible solutions given the input data. – Arkamis Oct 29 '12 at 19:32 My apologies, you were correct in your approach and I was incorrect in my assumptions. Once I adjusted my assumptions your approach produced correct results. Thanks. – Moses Oct 29 '12 at 23:03 Yes, there is a solution. I recommend checking the exact form (not approximate) as provided by wolfram alpha. Solution This supports Ed's answer - Very cool, I did not know wolfram alpha was capable of that kind of analysis. I'll be going back there next time I find myself sans graphic calculator! – Moses Oct 29 '12 at 23:05 Rewrite your two equations: $$\begin{cases} .075 = b\times c \\ 239 = b-c \end{cases}$$ Now solve the second equation for $b$ so that $b=239+c$. Plug this into the first equation in place of $b$: $$.075 = (239+c)\times c$$ Do you know how to solve quadratic equations? Once you have the possible values of $c$, plug those back into our equation for $b$ to get the final answer. This method is called eliminating a variable by substitution. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9441354870796204, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/7080/robust-bayesian-optimization-in-markowitz-framework	# Robust-Bayesian optimization in Markowitz framework Suppose we are in the mean-variance optimization setting with a vector of returns $\alpha$ and a vector of portfolio weights $\omega$. In a robust setting, the returns are assumed to lie in some uncertainty region. I came accross a paper which lets this region, call it $U$, be given by the sphere centered at $\alpha$ with radius $\chi\|\alpha\|$ where $\chi$ lies between 0 and 1. The authors then turn their attention to: $\min_U r_{p}$ and end up with the following solution: $\min_U r_{p}=\alpha^\intercal\omega-\chi\|\alpha\|\|\omega\|$ ... ... ... (1) They do not provide much detail as to how they arrive at this but mention the following: "this uncertainty region corresponds to a one-sigma neighborhood under a Bayesian prior of an uncertain $\alpha$ distributed normally about the estimated $\alpha$, with $\sigma=\chi\|\alpha\|$..." Does anyone know how they might have arrived at equation (1)??? Thanks in advance! - 1 Can you post the link to the paper? Reminds me of an Atillio Meucci paper – Quant Guy Jan 23 at 14:54 His book has appendices that show the derivation for robust Bayesian optimization (available at symmys.com) – John Jan 23 at 15:21 – Geraldine Bailey Jan 23 at 16:56 ## 1 Answer In robust optimization, the true return is not known, we just have a prior $\alpha$ and you have to take into account a possible misestimate which can lower the true return. This is done under the assumption that the posterior return will be within the prior return $\alpha$ plus minus the error being in some $\sigma$-interval. Now a try for a more formal answer: The posterior return vector is estimated as $\vec{\alpha} +\vec{\chi}\cdot\|\alpha\|$ (1) with $\|\vec{\chi}\|\leq 1$, or equivalently $\vec{\chi}^{2} \leq 1$. This exactly describes a sphere around $\vec{\alpha}$. Now the return is the product of the return vector $\vec{\alpha}+\vec{\chi}\cdot\|\alpha\|$ times the weight vector $\vec{\omega}$ : $r=(\vec{\alpha}+\vec{\chi}\cdot\|\alpha\|)\cdot\vec{\omega}=\alpha^{T}\omega+\chi^{T}\omega\|\alpha\|$. (2) Here, $\vec{\chi}$ can have any orientation. We want the minimum of the second term. $\alpha^{T}\omega$ is minimal if $\vec{\alpha}$ and $\vec{\omega}$ look in opposite direction (property of the dot product), therefore $\min_U r_P=\alpha^{T}\omega-\chi\|\alpha\| \|\omega\|$. (3) The first term is just the dot product of $\vec{\alpha}$ and $\vec{\omega}$, so it can be written as $\|\alpha\|\|\omega\|\cos(\phi)$ where $\phi$ is the angle between the two vectors (in n dimensions). This is the next equation in the Golts and Jones working paper: $\min_U r_P=\|\alpha\|\|\omega\|(\cos(\phi)-\chi)$. (4) - Wow thank you very much! That helps alot. Many many many thanks! :) – Geraldine Bailey Jan 24 at 9:48 @Phillipe: Can you recommend a good source or reference that will help or that you have used??? – Geraldine Bailey Jan 24 at 10:34 Can you please explain, a little more clearly, how you get to the final equation from the second one? – Geraldine Bailey Jan 24 at 20:05 Regarding the references, I just glimpsed over the nice working paper from Golts and Jones (2009) you cited and the pdf version of an article by Goldfarb and Iyengar (CORC Technical Report TR-2002-03 Robust portfolio selection problems). There, equations (4) and (15) seem to state the same result. Regarding the mathematical steps, I will edit my answer – philippe Jan 24 at 22:47 From eqn (2) to (3), we want to have the worst possible r. The first term on the right side of (2) is constant in our setting given our initial $\vec{\alpha}$ and $\vec{\omega}$. Now, we want to reduce that by as much as possible. In $\chi'\omega\|\alpha\|$ \|\alpha\| is again constant, so we are looking for the smallest value of $\chi'\omega$. This can be written as $\|\chi\|\|\omega\|\cos(\delta)$ with $\|\chi\|$ and $\|\omega\|$ the length of the vectors and $\delta$ the angle between them. This is minimal for $\delta=\pi$ (ie both vectors have 180 degrees between them and look in opposite directions) – philippe Jan 24 at 23:15 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9230387210845947, "perplexity_flag": "head"}
http://mathoverflow.net/questions/11192/when-a-banach-space-is-a-hilbert-space/20754	## When a Banach space is a Hilbert space? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $\mathcal{X}$ be a Banach space. It is a well known fact that $\mathcal{X}$ is a Hilbert space (i.e. the norm comes from an inner product) if the parallelogram identity holds. Question: Are there other (simple) characterizations for a Banach space to be a Hilbert space? - 6 re: Leonid's comment; Another isomorphic characterisation of Hilbert spaces is that a Banach space $X$ is isomorphic to a Hilbert space if and only if every closed linear subspace of $X$ is complemented (that is, is the range of a continuous linear projection on $X$). I believe this result is due to Lindenstrauss and Tzafriri. Another result along these lines is that a separable infinite dimensional Banach space $X$ is isomorphic to $\ell_2$ if and only if every infinite dimensional closed subspace of $X$ is isomorphic to $X$. I believe that this result is due to Tim Gowers. – Philip Brooker Jan 9 2010 at 2:02 7 Characterizing Hilbert spaces isomorphically is a very interesting topic in Banach space theory. Another one is that every nuclear operator on the space has absolutely summable eigenvalues. Open is whether a Banach space all of whose subspaces have an unconditional basis must be isomorphic to a Hilbert space. A non characterization is that there are Banach spaces non isomorphic to a Hilbert space all of whose subspaces have a Schauder basis. – Bill Johnson Jan 9 2010 at 8:14 1 I wonder if the algebra $\mathcal{B}(X)$ of all bounded linear operators on the Banach space $X$ is a $C$-algebra with the operator norm if and only if $X$ is isometrically isomorphic to a Hilbert space. There are many isomorphic variants one could ask in this direction too. On a related note, the Eidelheit theorem ($\mathcal{B}(X)$ and $\mathcal{B}(Y)$ are isomorphic as Banach algebras if and only if $X$ and $Y$ are isomorphic as Banach spaces) gives an isomorphic characterisation of Hilbert spaces, though admittedly it is probably not easy to check. – Philip Brooker Mar 31 2010 at 3:46 Have only just noted this question from Philip Brooker, since the original question was bumped by a new answer. I think I've seen a proof (in work of Daws) that if E and F are Banach spaces and we have a closed-range unital homomorphism from A(E) into B(F), then E is isomorphic to a weakly complemented subspace of F. If I have remembered this correctly, then it would answer the natural isomorphic variant of Philip's question. – Yemon Choi Mar 2 2011 at 10:14 Yemon, thanks for bringing that to my attention. – Philip Brooker Mar 22 2011 at 23:58 show 1 more comment ## 6 Answers a real Banach space (X, ‖ · ‖) is a Hilbert space if and only if for any three points A,B,C of this space not belonging to a line there are three altitudes in the triangle ABC intersecting at one point. Many other references show when Googling "is a hilbert space if" banach - 11 For the really lazy among us: google.com/… – Tom LaGatta Mar 31 2010 at 5:24 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Bessaga and Pelczynski wrote a survey on Banach spaces. The chapter 4 is devoted to this question. http://matwbn.icm.edu.pl/ksiazki/or/or2/or214.pdf - More characterisations are in the book of Haim Brezis (Analyse fonctionnelle), at the appendix of Chapter 5. I will copy two of these below, toghether with the references: 1. If $\dim(E)\geq 2$ and every subspace $X\subset E$ of dimension $2$ is the image of a bounded projector $P$ such that $\\|P\\| = 1$, then $E$ is isometric to a Hilbert space (Kakutani, Japanese Journal of Mathematics, 1939); 2. if $\dim(E)\geq 3$ and the map $T$, defined as the identity on the unit ball and as $u/\\|u\\|$ when $\\|u\\|\geq 1$, is lipschitzian with constant $1$, then $E$ is isometric to a Hilbert space (de Figueiredo; Karlovitz, Bulletin of the American Mathematical Society, 1967). Also, if $E$ is isomorphic to all its infinite-dimensional subspaces, then it is isomorphic to a separable Hilbert space (Gowers, Annals of Mathematics, 2002). - Just two isometric/isomorphic characterizations: A Banach space $X$ is [isometric to] a Hilbert space if and only if there exists a Banach space $Y$ and a symmetric bilinear mapping $f:X\times X\rightarrow Y$ satisfying $\|\|f(x,z)\|\|$ $=$ $\|\|x\|\|\cdot\|\|z\|$\| for all $x,z$ $\in$ $X$. [J. Becerra Guerrero & A. Rodriguez-Palacios] A Banach space is [isomorphic to] a Hilbert space iff it is uniformly homeomorphic to a Hilbert space. [Per Enflo] - Yes, there are many (simple) characterizations of when a normed space is an inner product space. Here are two book references, one with Google preview, the other you can hopefully get at your library. - In this simple note http://arxiv.org/abs/0907.1813 (to appear in Colloq. Math.), Rossi and I proved a characterization in terms of "inversion of Riesz representation theorem". Here is the result: let $X$ be a normed space and recall Birkhoff-James ortogonality: $x\in X$ is orthogonal to $y\in X$ iff for all scalars $\lambda$, one has $\|\|x\|\|\leq\|\|x+\lambda y\|\|$. Let $H$ be a Hilbert space and $x\rightarrow f_x$ be the Riesz representation. Observe that $x\in Ker(f_x)^\perp$, which can be required using Birkhoff-James orthogonality: Theorem: Let $X$ be a normed (resp. Banach) space and $x\rightarrow f_x$ be an isometric mapping from $X$ to $X^$ such that 1) $f_x(y)=\overline{f_y(x)}$ 2) $x\in Ker(f_x)^\perp$ (in the sense of Birkhoff and James) Then $X$ is a pre-Hilbert (resp. Hilbert) space and the mapping $x\rightarrow f_x$ is the Riesz representation. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 55, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8767164349555969, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/35284/which-speed-can-an-electric-scooter-reach-on-a-given-slope	# Which speed can an electric scooter reach on a given slope? Electric scooters are always "given" as "working with slopes up to xx%", but what does it mean? Given motor torque and power, scooter+driver weight and wheels diameter, how can I determine at which speed it will run on a given slope, regardless of air friction? - ## 1 Answer Suppose you have your scooter on some slope: The mass of the scooter and driver is $m$ and the angle of the slope is $\theta$. The force $F$ down the slope is given by: $$F = mg\space sin\theta$$ And if your wheels have a radius of $r$ then this force creates a torque: $$\tau = r \space mg\space sin\theta$$ The scooter will stop when this torque equals the torque generated by the electric motor in your scooter. Effect of the slope on maximum speed When the scooter is moving at some speed $v$ the drag on it will come partly from mechanical friction and partly from air resistance. in general the drag will be a complicated function of speed, so I'll just write the drag as a function of velocity $D(v)$. If you know the torque of the electric motor, $\tau$, then the maximum velocity will be given by: $$\frac{\tau}{r} = D(v_{max})$$ where $r$ is the radius of the wheels. If you know the function $D(v)$ you can solve for $v_{max}$. Now suppose you are on a slope of angle $\theta$ as shown above, then there will be a force due to gravity of $mg\space sin\theta$. The maximum velocity is now given by: $$\frac{\tau}{r} - mg\space sin\theta = D(v_{max})$$ As above, if you know the form of the function $D(v)$ you can solve for $v_{max}$ to find the maximum speed on the slope. The maximum velocity on the slope will obviously be less than the maximum velocity on the flat. - So speed is only affected by air friction? As long as motor torque is greater than "slope torque", scooter can reach any speed it is capable of? But in real world I can reach 10 mph using P1 power on θ1 slope, and 15 mph on same P1 power and θ2= 0 (flat road), so I think there's something missing in this formula: I need a formula linking speed to P and θ, or to torque and θ. – jumpjack Sep 1 '12 at 10:48 /ot How do I add pictures and formulas to my posts? – jumpjack Sep 1 '12 at 10:49 The equation I've given tells you the maximum slope the scooter can climb i.e. above this slope the motor isn't powerful enough to move the scooter at all. Calculating the maximum speed as a function of slope would be a lot harder as there isn't a simple formula that links maximum speed to the torque - it's a complicated function of air resistance and mechanical friction. – John Rennie Sep 1 '12 at 11:12 – John Rennie Sep 1 '12 at 11:15 So ignoring frictions a vehicle can reach same speed on flat and on a slope?!? – jumpjack Sep 2 '12 at 12:14 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9197086095809937, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/265795/probability-that-a-random-graph-is-an-expander?answertab=votes	# Probability that a random graph is an expander I have a random graph $G = (V, E)$ and each edge is in the graph with probability $p$. I need to show that the probability that $G$ is $\delta$-edge-expander* when $\delta= \frac{np}{4}$ goes to $1$ as $n=\|V\|\to\infty$. *A graph is $\delta$-edge-expander if for every set of nodes $S$ of size at most $\frac{n}{2}$ it holds that the number of edges from $S$ to $S'$ (where $S'$ is all the nodes that don't appear in $S$) is at least $\delta$. You can see the definition in this wiki page. - 3 please elaborate on what is a δ−edge−expander – K.Ghosh Dec 27 '12 at 8:16 1 It's unclear what you mean by the event that $G$ is "$\delta--edge--expander$. And you should mention that $n$ is the number of vertices of the graph $G$, if that's what $n$ is here. – coffeemath Dec 27 '12 at 8:18 1 The definition on the wiki page requires the edge boundary to be at least $\mathrm{const}\,\|S\|$, and you wrote "at least $\delta$", without $\|S\|$. – user53153 Dec 27 '12 at 17:29	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9586965441703796, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=533843	Physics Forums Recognitions: Gold Member ## What is the nature of zero? With the last year or so I came across a great discussion in these forums about the different types of infinity. It had never occurred to me there would necessarily be different types with different "values". How about zero? Are there different mathematical concepts regarding it or are all zeros "equal" ? Any different types of zero with different mathematical meanings?? Any difference between approaching from plus, or minus, or "i" ?? Is a 1 x 1 matrix with zero the same as, say, an n x n matrix with all zeros? [Physically, "zero" for physicsts is not always so clear...like absolute zero; zero energy in a volume of spacetime isn't...it's zero point energy with virtual particles,quantum activity; zero time in cosmology (less than 1 Planck time) etc,etc. ] Thank you. PS: for those interested, Wikipedia has a lot of interesting insights on "zero" here: http://en.wikipedia.org/wiki/Zero Among them: Zero is an even number; Zero is the smallest non negative integer;there is no calendar year zero but there is a cosmological year zero. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Gold Member You may wish to read this post from the FAQ. Blog Entries: 4 The nature of zero...great argument You can have "1 of nothing" and "nothing of 1" both terms have a value of "1" this being a quantitive value. 1 x 0 = 1x [where x=0] ## What is the nature of zero? Quote by Godswitch The nature of zero...great argument You can have "1 of nothing" and "nothing of 1" both terms have a value of "1" this being a quantitive value. 1 x 0 = 1x [where x=0] Well ... you can have one zero. Or you can have two zeros. Or a billion zeros. But in each case you have zero. Recognitions: Gold Member Well ... you can have one zero. Or you can have two zeros. Or a billion zeros. But in each case you have zero. really? Are they ALL equal?? All infinities are not equal, so I'm still wondering if there are other subtle aspects of "zero". Apparently I did not make my question clear: I posted..... Any difference between approaching from plus, or minus, or "i" ?? Is a 1 x 1 matrix with zero the same as, say, an n x n matrix with all zeros? It seems to me a 1 x 1 matrix of zeros, say describing some some apect of spacetime in a region, has a different connotation that an n x n matrix of all zeros describing a region of spacetime. And zero in quantum mechanics just can't be as simple as in classical physics....or maybe it is?? Quote by Naty1 It seems to me a 1 x 1 matrix of zeros, say describing some some apect of spacetime in a region, has a different connotation that an n x n matrix of all zeros describing a region of spacetime. Spacetime? What does that have to do with matrices? There is precisely one 1 by 1 matrix of zero, namely the zero matrix. Similarly for any n, there is precisely one n x n matrix with all elements zero, namely the zero matrix. Heck you wrote "1 x 1 matrix of zeros". How can you have more than one zero in a matrix with one element? You seem to have muddled up something very fundamental. And zero in quantum mechanics just can't be as simple as in classical physics....or maybe it is?? Why would it be different? Why should QM affect arithmetic? Recognitions: Gold Member Spacetime? What does that have to do with matrices? You should read up on, say, tensors: http://en.wikipedia.org/wiki/Tensor there is a pretty technicolor picture for you to see. anyway, the zero issue is no big deal.... Recognitions: Gold Member Science Advisor Staff Emeritus Quote by Naty1 With the last year or so I came across a great discussion in these forums about the different types of infinity. It had never occurred to me there would necessarily be different types with different "values". And to briefly state the point -- there are lots of different kinds of things one would do that need a notion of infinity. How about zero? Are there different mathematical concepts regarding it or are all zeros "equal" ? Zero is somewhat more uniform. It is virtually always used in the following two situations: • There's an abelian group with operation '$+$', and 0 is its additive identity. ($0+x = x$) • There's a monoid with operation written multiplicatively (e.g. '$\times$', '$\cdot$', or just by juxtaposition), and 0 is a nullary element. ($0x = 0$) There are lots of different kinds of systems that have a zero -- the zero integer, the zero rational number, the zero real number, the zero complex number, the zero matrix of each size, the zero vector of a vector space, the zero polynomial, the zero rational function, the zero scalar field, the zero vector field, and so forth. While each of these is technically a different zero, they really do all appear very similar. The only oddity I can think of is in some situations, you might have xy=0, despite neither x=0 nor y=0. The only substantially different examples I can think of is if you move up a level in algebra, and start talking about things like "the zero group" or "the zero ring" or the "zero vector space". But even those aren't too dissimilar from the usual examples from the perspective of 2-algebra or higher. Thread Tools \| \| \| \| \|--------------------------------------------------\|-----------------\|---------\| \| Similar Threads for: What is the nature of zero? \| \| \| \| Thread \| Forum \| Replies \| \| \| Quantum Physics \| 18 \| \| \| General Physics \| 0 \| \| \| General Physics \| 0 \| \| \| General Physics \| 5 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9235508441925049, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=2681337	Physics Forums ## What is transversity and transverse spin? I am reading through this article for a paper and ran across the term transversity. I did some google searching but didn't come up with anything I could understand. I think it might be linked with transverse spin but I am also having trouble getting clarification on what that is. I know what spin is, and I'm wondering if it's just spin along a latitude instead of a longitude. My ultimate goal is to be able to explain what these guys are doing on the COMPASS project. I understand their aim is to try to solve the "proton spin crisis" and I have a pretty good understanding of what that is from some other articles I've read. So if you happen to have knowledge of that experiment as well as transversity and transverse spin, I'd really appreciate it. PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus You are probing a hadron with a lepton, and you can work in the approximation where a single virtual photon is exchanged. Say the virtual photon is along the horizontal direction. Basically, transversity is the following quantity : So if your hadron is polarized upwards, it's the difference between the probability of finding a quark polarized upwards minus the probability of finding a quark polarized downwards. Transversity can be measured as a function of $x_B$ (x-Bjorken), the fraction of longitudinal momentum carried by the active quark. It can also be measured as a function of the resolution of your probe, or "wavelength" of the virtual photon, usually written $Q^2$, the opposite of the invariant mass of the virtual photon (it is positive in this regime). The point is that it will not be the same as if the green and red arrows above both point horizontally (the direction of the probe), because it is a relativistic effect and there is essentially an infinite boost along the horizontal direction (the Lorentz contraction make your hadron look like a pancake). Tags cern, compass, proton spin, transverse spin, transversity Thread Tools \| \| \| \| \|----------------------------------------------------------------\|-------------------------------------\|---------\| \| Similar Threads for: What is transversity and transverse spin? \| \| \| \| Thread \| Forum \| Replies \| \| \| Quantum Physics \| 5 \| \| \| Atomic, Solid State, Comp. Physics \| 0 \| \| \| Quantum Physics \| 0 \| \| \| Biology, Chemistry & Other Homework \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9399421215057373, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/78433-radian-measure.html	Thread: 1. Radian Measure? This question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question: Two points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$. Find the values of r and $\theta$. [5] Any help towards answering this is greatly appreciated. 2. Originally Posted by db5vry This question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question: Two points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$. Find the values of r and $\theta$. [5] Any help towards answering this is greatly appreciated. arc length, $s = r\theta$ , $\theta$ is in radians sector area, $A = \frac{r^2 \theta}{2}$ , $\theta$ in radians $A = \frac{r^2 \theta}{2} = \frac{r \cdot r\theta}{2} = \frac{r \cdot s}{2}$ $r = \frac{2A}{s}$ evaluate $r$, then go back to either of the original equations and solve for $\theta$. 3. Originally Posted by db5vry This question is something to do with radian measure. I've never studied this before (I understand a few of the formulas generally used with no idea how they are applied) and want to solve this question: Two points P and Q lie on a circle with centre O. The radius of the circle is r cm and angle POQ = $\theta$ radians. The length of the arc PQ is 6 cm and the area of the sector PQ is $22.5cm^2$. Find the values of r and $\theta$. [5] Any help towards answering this is greatly appreciated. A radian is a length of the radius on the circumference. You're told the radius is $r$ and that the length of the arc $PQ$ is $6\textrm{bf}$. Therefore the angle $\theta$, in radians, is $\frac{6}{r}$. The area of a sector, if the angle is given in radians, is $\frac{\theta}{2\pi}\pi r^2 = \frac{\theta}{2}r^2$. We know that $\theta = \frac{6}{r}$ and that the area is $22.5\textrm{cm}^2$. So $\frac{\frac{6}{r}}{2}r^2 = 22.5$ $\frac{3}{r}r^2 = 22.5$ $3r = 22.5$ $r = 67.5\textrm{cm}$. From this, we can see $\theta = \frac{6}{67.5}^C$. 4. Originally Posted by Prove It A radian is a length of the radius on the circumference. You're told the radius is $r$ and that the length of the arc $PQ$ is $6\textrm{bf}$. Therefore the angle $\theta$, in radians, is $\frac{6}{r}$. The area of a sector, if the angle is given in radians, is $\frac{\theta}{2\pi}\pi r^2 = \frac{\theta}{2}r^2$. We know that $\theta = \frac{6}{r}$ and that the area is $22.5\textrm{cm}^2$. So $\frac{\frac{6}{r}}{2}r^2 = 22.5$ $\frac{3}{r}r^2 = 22.5$ $3r = 22.5$ $r = 67.5\textrm{cm}$. correction, r = 7.5 cm From this, we can see $\theta = \frac{6}{67.5}^C$. . 5. Originally Posted by skeeter . Oops... Duh... I was just testing you :P	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 43, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9516586065292358, "perplexity_flag": "head"}
http://physics.aps.org/articles/large_image/f1/10.1103/Physics.3.14	Figure 1: Global fit to all published data sensitive to the Higgs boson mass. This comprises direct searches as well as electroweak precision data. The left vertical axis shows goodness-of-fit relative to the minimum $χ2$ value, while the right vertical axis shows $σ$ values as a function of Higgs mass. The dashed line shows the situation before the Tevatron publications, while they are included in the solid line. The left shaded portion is the mass range excluded by LEP, and the central shaded region is the range excluded by the Tevatron. High masses are excluded by precision measurements of the weak mixing angle and the $W$ mass, leaving only the range $115$–$150GeV/c2$ for future searches if the standard model is the correct theory.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 5, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9147240519523621, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/59457?sort=newest	## Examples of “Unusual” Classifications ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) When one says "classification" in math, usually one of a handful of examples springs to mind: -Classification of Finite Simple Groups with 18 infinite families and 26 sporadic examples (assuming one believes the Classification is indeed complete) -Classification of finte-dimensional semisimple Lie algebras with 4 infinite families and 5 exceptional examples -Classification of (Simple, Formally Real) Jordan Algebras with 4 infinite families and 1 exceptional example I'm sure there are other examples that non-algebraists would think of before these. All the examples I cited take the basic form of having several infinite families and some number of exceptional examples which do not fall into any of these families. Thus I am wondering: Question: Does anyone know of examples of classifications of some mathematical objects such that the classification consists (A) only of infinite families or (B) only of a finite number of examples/ an infinite number of examples which do not seem to be closely related to one another (i.e. they do not "appear" to form any infinite families). One example of case (B) that comes to mind would be finite dimensional division algebras over $\mathbb{R}$ of which there are 4. On the other hand, for case (B) I would like to rule out way too specific "classifications" such as "finite simple groups with an involution centralizer of such and such a form" since this is really a subclassification within the classification of FSG's. Although I am an algebraist, I would like to hear about examples from any branch of math, for comparison's sake. (If anyone thinks of better tags for this, feel free to add or suggest them). - ## 6 Answers String theories. For example, if you confine yourself to bosonic string theory you find it only works in dimension 26 despite the definition being completely independent of the number of dimensions. Similarly there are just 5 superstring theories. All of these theories are closely related to other interesting classifications in mathematics. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There are a number of interesting classifications associated with tilings, e.g., the 17 wallpaper groups, the 230 space groups, or the 14 convex pentagons that tile the plane. (In the latter case I'm not sure if the classification has been rigorously proven to be complete.) - There are 5 platonic polyhedra... - 6 But these is a piece of a larger classification, which has a few exceptions and three families. – Mariano Suárez-Alvarez Mar 24 2011 at 19:34 Mariano: What's the third family? I can only think of the $A$ and $D$ families and the three platonic exceptions. – José Figueroa-O'Farrill Mar 25 2011 at 1:21 1 Simplices, n-cubes and generalized octahedra. – Jim Conant Mar 25 2011 at 1:53 A1 Classification of linear spaces over a field. A2 Classification of indecomposable linear actions of Z (over any field) = Jordan blocks. B1 Classification of geometries (à la Thurston) in dimention 3 (?) - Related to your B1, B2: Classification of two-dimensional geometries (Spherical, Euclidean and Hyperbolic). – Simon Rose Mar 24 2011 at 21:09 As an example of (B), I'd mention Connes's classification of injective (type $II_{1}$) factors from the theory of von Neumann algebras. In this case, many apparently disparate constructions turn out to give a single object. Classification of injective factors. Cases II1,II_\infty,III_\lambda,\lambda \not=1\$. Ann. of Math. (2) 104 (1976), no. 1, 73-115. - This feels like a somewhat silly example, but what about the classification of two (real) dimensional manifolds? They are two families of these, the orientable and non-orientable families classified by their genus/Euler characteristic. - I should add that this is an example of (A). – Simon Rose Mar 24 2011 at 18:53 Wow, I guess I totally overlooked such an obvious example from undergrad/grad topology classes. – ARupinski Mar 24 2011 at 18:53 @Simon: I'm not really sure that this is an example of (A). I think one could argue that that orientable surfaces with genera $g>1$ are the ones making up the generic infinite family, whereas $g=0$ and $g=1$ are the two "sporadic" examples. It is just that in this case, the sporadic examples came first. – José Figueroa-O'Farrill Mar 25 2011 at 1:18 How about classification of irreducible closed 2-manifolds: just $S^2, T^2, and RP^2$. All others are (connected) sums of these. So maybe this is an example of (B). @Jose why do you view g=1 as "sporadic"? (hyperbolic metric, presumably?) topologically they aren't much more special than their higher genus siblings. – Paul Mar 25 2011 at 2:48 @José - I think it depends on how you view the classification. Purely topologically, I would argue that the two families are: $A_g$ = the connected sum of $g$ copies of $T^2$, and $B_k$ = the connected sum of $k$ copies of $RP^2$. Ignoring the geometry, this is perfectly valid. The only dodgy part of it, perhaps, is that $B_0 = A_0$, but this sort of thing also occurs in the classification of Lie Algebras as well, so... – Simon Rose Mar 25 2011 at 14:51	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9185419678688049, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/98474/what-is-the-limit-of-lim-limits-n-rightarrow-infty-frac1n4-left-sum-k	# What is the limit of $\lim\limits_{n\rightarrow\infty}\frac{1}{n^4}\left(\sum_{k=1}^{n}\ k^2\int_{k}^{k+1}x\ln\big((x-k)(k+1-x)\big)dx\right)$ As the topic how to find the limit of $$\lim_{n\rightarrow\infty}\frac{1}{n^4}\left(\sum_{k=1}^{n}\ k^2\int_{k}^{k+1}x\ln\big((x-k)(k+1-x)\big)dx\right)\;.$$ - 4 The title and the body ask two different questions, let me answer the one in the body: decompose the log into the sum of two logs, transform every integral into an integral on $(0,1)$, perform the cancellations which appeared, use the fact that the sum of the $n$ first cubes is approximately $\frac14n^4$ and compute the integral of the log on $(0,1)$. The answer should be $-\frac12$. – Did Jan 12 '12 at 14:58 i dont quite get how to manipulate it. – Mathematics Jan 12 '12 at 15:05 @Mathematics The integral is $$\begin{eqnarray} &&\int_{k}^{k+1}x\ln \left( \left( x-k\right) \left( k+1-x\right) \right) dx \\ &=&\int_{k}^{k+1}x\ln \left( x-k\right) dx+\int_{k}^{k+1}x\ln \left( k+1-x\right) dx \\ &=&\int_{0}^{1}\left( u+k\right) \ln u\,du+\int_{0}^{1}\left( k+1-u\right) \ln u\,du \\ &=&\ldots \\ &=&-k-\frac{1}{4}. \end{eqnarray}$$ – Américo Tavares Jan 12 '12 at 15:42 @Américo: Not $=-(2k+1)$? Maybe I am wrong, I did this too quickly and in my head, so... – Did Jan 12 '12 at 16:07 @DidierPiau: Yes, you are right. It is $(-k-1/4)+(-k-3/4)=-(2k+1)$ – Américo Tavares Jan 12 '12 at 16:39 show 3 more comments ## 2 Answers Put $I_k:=\int_k^{k+1}x\ln ((x-k)(k+1-x))dx$; making the substitution $t=x-k$ we have, following @Didier Piau-'s idea $$I_k=\int_0^1(t+k)\ln(t(1-t))dt=\int_0^1t\ln (t(1-t))dt+k\int_0^1\ln (t(1-t))dt.$$ Since $0\leq \frac 1{n^4}\sum_{k=1}^nk^2\leq \frac{n\cdot n^2}{n^4}$, the limit we are looking for is $$l:=\lim_{k\to\infty}\frac 1{n^4}\sum_{k=1}^nk^3\int_0^1(\ln t+\ln(1-t))dt.$$ Since $\sum_{k=1}^nk^3=\frac{n^2(n+1)^2}4$, we have $$l=\frac 14\cdot 2\int_0^1\ln tdt=\frac 12\left([t\ln t]_0^1-\int_0^1t\frac 1tdt\right)=-\frac 12.$$ - 1 If you apply Stolz Cezaro to the original sequence you get exactly $\lim_n \frac{I_n}{n}$. – N. S. Jan 24 '12 at 16:40 Two substitutions, first $x\mapsto x+k$ and then $x\mapsto1-x$ yields $$\begin{align} &\int_k^{k+1}x\log\left((x-k)(k+1-x)\right)\,\mathrm{d}x\\ &=\int_0^1(x+k)\log\left(x(1-x)\right)\,\mathrm{d}x\\ &=\int_0^1(x+k)\left[\log\left(x\right)+\log\left(1-x\right)\right]\,\mathrm{d}x\\ &=\int_0^1(2k+1)\log(x)\,\mathrm{d}x\\ &=-(2k+1)\tag{1} \end{align}$$ This gives $$\begin{align} &\lim_{n\to\infty}\frac1{n^4}\sum_{k=1}^nk^2\int_k^{k+1}x\log\left((x-k)(k+1-x)\right)\,\mathrm{d}x\\ &=-\lim_{n\to\infty}\frac1{n^4}\sum_{k=1}^nk^2(2k+1)\\ &=-\lim_{n\to\infty}\sum_{k=1}^n\left(\frac kn\right)^2\left(2\frac kn+\frac1n\right)\frac1n\tag{2} \end{align}$$ and $(2)$ is a Riemann sum for $$-\int_0^12x^3\,\mathrm{d}x=-\frac12\tag{3}$$ -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8943420052528381, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/38639?sort=newest	## Thinking and Explaining ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) How big a gap is there between how you think about mathematics and what you say to others? Do you say what you're thinking? Please give either personal examples of how your thoughts and words differ, or describe how they are connected for you. I've been fascinated by the phenomenon the question addresses for a long time. We have complex minds evolved over many millions of years, with many modules always at work. A lot we don't habitually verbalize, and some of it is very challenging to verbalize or to communicate in any medium. Whether for this or other reasons, I'm under the impression that mathematicians often have unspoken thought processes guiding their work which may be difficult to explain, or they feel too inhibited to try. One prototypical situation is this: there's a mathematical object that's obviously (to you) invariant under a certain transformation. For instant, a linear map might conserve volume for an 'obvious' reason. But you don't have good language to explain your reason---so instead of explaining, or perhaps after trying to explain and failing, you fall back on computation. You turn the crank and without undue effort, demonstrate that the object is indeed invariant. Here's a specific example. Once I mentioned this phenomenon to Andy Gleason; he immediately responded that when he taught algebra courses, if he was discussing cyclic subgroups of a group, he had a mental image of group elements breaking into a formation organized into circular groups. He said that 'we' never would say anything like that to the students. His words made a vivid picture in my head, because it fit with how I thought about groups. I was reminded of my long struggle as a student, trying to attach meaning to 'group', rather than just a collection of symbols, words, definitions, theorems and proofs that I read in a textbook. Please note: I'm not advocating that we turn mathematics into a touchy-feely subject. I'm not claiming that the phenomenon I've observed is universal. I do think that paying more attention than current custom to how you and others are really thinking, to the intuitions, is helpful both in proving theorems and in explaining mathematics. I'm very curious about the varied ways that people think, and I would like to hear. What am I really thinking? I'm anxious about offending the guardians of the forum and being scolded (as they have every right to do) for going against clearly stated advice with a newbie mistake. But I can't help myself because I'm very curious how you will answer, and I can endure being scolded. - 99 What I am really thinking is that, if you had posted this anonymously, this question would have been closed in five minutes. Back to the question: I'd not have trouble saying what I think about some math, except I often have a complicated mental picture which is impossible to convey. – Felipe Voloch Sep 14 2010 at 2:22 23 Just for the record, many people do use the vivid internal analogies they have when conversing with students. One of my favorites from undergrad was describing a linear transformation as a commander-in-chief, who told the generals (a basis) where to go, who in turn tells all the soldiers (the rest of the vectors) where to go. The chain of command in action in a linear algebra class. – Cam McLeman Sep 14 2010 at 2:24 19 (By the way, I really think this is a great question!) – François G. Dorais♦ Sep 14 2010 at 2:32 21 @Felipe Voloch. Yes, I too realized that I could get away with more than most people. But that also means that mathematicians sometimes say less of what's on their mind to me than they might to someone less intimidating to them. The phenomenon of the complicated mental picture is exactly what I'm asking about. I can't expect you to convey the picture except in general terms, but just to talk more of its role in your mind, so why don't you write an actual "answer". @Cam McLeman: that's exactly what I'm asking for. Why not turn it into a full answer? – Bill Thurston Sep 14 2010 at 2:49 40 Felipe is absolutely right: posed by somebody else, this question would have been closed within a picosecond. The rationale, ironically, being that professional mathematicians of quality would flee this site in droves because the question is "discussiony", "vague", "has no right answer", etc. This is the Matthew effect in all its glory: "For to all those who have, more will be given, and they will have an abundance; but from those who have nothing, even what they have will be taken away" (Matthew 25:29) Needless to say, I'm euphoric at the thought of Bill's participation in MathOverflow. – Georges Elencwajg Sep 14 2010 at 7:46 show 13 more comments ## 21 Answers Somehow my experience, at least these days, is quite different. I do not see much difference between the way I think about things and the way I talk or write about them. Part of the reasons is being involved in semi-formal Internet acticities like blogs, polymath projects, MO, etc., another reason is that I think more about conceptual and meta questions (like this one), yet another reason is that I am quite interested in the process (for me and even more for others) leading to ideas and answers. Also, getting older may have something to do with it: On the one hand, you are less surprised from your own partial-thoughts and unexplained intuition, and, on the other hand, you are less surprised from your half-thoughts and unexplained intuitions. - I like (and immediately feel familiar) with the idea of change via lifetime. In fact, that may be a result of the practise of teaching, at least in my case. When I think back, I remember to have written about some intuitive "pattern-identification" to solve a problem, which was hardly communicable. After 10 years teaching (and 30 years lifetime after my introspective text) your input makes me aware, that that process has become more rare and the focus in my thinking has strongly changed to the communicative/understandable mode... – Gottfried Helms Jun 8 at 0:00 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. People have mentioned examples which are hard to share due to some kind of prerequisites. Here's one: I learned PDE from a professor who, in his mind, was always thinking about distribution theory, but officially could not talk about it until after he covered the material relevant to the exams. In distribution theory, whenever you see an integral over a domain $\int_\Omega u(x) dx$ you actually picture the characteristic function $\int \chi_\Omega(x) u(x) dx$ or $\int H(f(x)) u(x) dx$ if $f$ is a defining function for $\Omega$ and $H$ is a heaviside function. From this point of view, you imagine that all functions are smooth and compactly supported (or you can imagine their approximations), so that if you integrate by parts on $\int \chi_\Omega \nabla u(x) dx = - \int \nabla \chi_\Omega u(x) dx = \int \delta(f(x)) \nabla f(x) u(x) dx$. The boundary terms come when the derivative hits the characteristic function. Same thing for Stokes' theorem, Gauss's divergence theorem. It's pretty handy to compute this way. For a little while this was all I understood until I later found out what was going on. The limit of difference quotients of $\chi_\Omega$ is clearly supported on the boundary of $\Omega$ and it's clear, especially if you picture an approximation, that $\nabla \chi_\Omega$ points in the direction of increase of $\chi_\Omega$ -- i.e. the inward normal. More simply: there are two points of view -- if you were to take difference quotients of $u$, you use a Lagrangian point of view in which the point at position $x$ moves in the direction $i$, and you observe a change in $u$ between those points; instead, you can take an Eulerian point of view, (where the adjoint difference quotients go on the characteristic function) and you can instead look at movement of the region with $u$ fixed. Until I understood this point of view in a simpler way, it would not really be sensible to explain it to others. But now I know that giving a watered down version of the same proof when "proving" the fundamental theorem of calculus / Gauss's divergence for a calculus class in fact does not lose any key ideas (except for technicalities like how you need the mean value theorem to ensure the difference quotients are bounded). Of course, I would also talk about characteristic functions to any math student, since it is a nice point of view. By the way, in the calculus of variations, when your $u(x) = L(x, \phi(x) )$ is a Lagrangian and $\phi(x)$ is a solution to the Euler-Lagrange equations, and you take difference quotients using the flow of a vector field whose flow preserves the Lagrangian (a "symmetry"), you end up with Noether's theorem through only this one variation (there are only boundary terms in what I called "the Lagrangian point of view" because you vary through a family of solutions except at the boundary). So it's also a nice way to prove conservation laws in one swoop. My point: for a little while, distribution theory seemed like a magical theory with prerequisites that made it unexplainable in everyday talk, but once I really understood the ideas I could usually discard the vocabulary (actually, the whole theory can often be replaced by cutoffs, partitions of unity, Taylor expansion, and changes of variable -- although I still think it's great to learn). I suspect that this phenomenon is not uncommon for elementary applications of "fancy" mathematical theories. I believe that often once one has a more basic understanding, one can throw away the new words but still fully reveal the ideas (but maybe that's completely due to my own background). People here have talked about Feynman -- he was good at doing this in the context of physics. If you watch his (outstanding) lectures on Project Tuva you will see more or less the proof of Noether's theorem about which I just wrote. A second point: Another thing I think happens to me is that I feel some pressure not to convey just how often I rely on geometric modes of thought, especially when they go against the usual way of explaining things, or the background of a typical student, and are not completely necessary. Example 1: When you row-reduce a matrix, you make a bunch of changes (most importantly some "transvections") in the basis of the image space until a few of your basis vectors (say $v_1 = T e_1, v_2 = T e_2$) span the image of the matrix $T$. When you picture the domain of $T$ foliated by level sets (which are parallel to the null space of $T$), you know that the remaining basis vectors $e_3, e_4, ...$ can be translated by some element in the span of $e_1, e_2$ (i.e. whichever one lies on the same level set) in order to obtain a basis for the null space. Now, this is how we visualize the situation, but is it how we compute and explain? Or do we just do the algebra, which at this point is quite easy? If the algebra is easy and the geometry takes a while to explain and is not "necessary" for the computation, why explain it? This is a dilemma because once algebra is sufficiently well-developed it's possible that the necessity of (completely equivalent) geometric thinking may become more and more rare; and algebra seems to be more "robust" in that you can explore things you can't see very well. But then, when students learn the implicit function theorem, somehow I feel like having relied on that kind of foliation much more often would help understand its geometric content. Still, even if it's in your head and very important, are you going to draw a foliation every time you do row operations? We know the geometry, know the algebra, but it would take a while to repeatedly explain how to rely on the geometry while executing computations. Example 2: (Things that aren't graphs) Another problem geometric thinking faces is that modern math often seems to regard pictures as not being proofs, even if they are more convincing, so there is a bias regarding how to choose to spend class time. Let's say you want to differentiate $x^3$. You can draw a cube, and a slightly larger cube, and then look at the difference of the cubes and subdivide it into a bunch of small regions, three larger slabs taking up most of the volume. Algebraically, this subdivision corresponds to multiplying out $(x+h)^3$; collecting the terms uses the commutativity, which corresponds to rotating the various identical pieces. It is no different to write this proof out algebraically, the difference is that the algebraic one is a "proof" but the geometric one is.. not? Even if it's more convincing. So it's like the picture is only there for culture. Maybe I have the lecture time to teach both, I will. But I would like to go farther than that. When I differentiate the cube root function, the same cube appears and I go through it again if I feel like it just to convince myself of the truth. Actually, every time I ever use the product rule I always picture the same rectangle with a slightly larger rectangle. My point of view is that one important "definition" of multiplication is in terms of areas, and that a linear function is not necessarily a graph. When you think of a linear function, you should also picture things like rectangles, sectors, similar triangles like the kind that come up when "proving" basic differentiation formulas. Differentiating the integral may seem like a magical trick, but it's really just a continuation of the point of view that multiplication can look like an area/volume and differentiation means taking a small change in the input. Now, I'd like that point of view to be absorbed, but it's not exactly in the textbook, or completely consistent with what students' other teachers taught them. It's hard to go against the idea that "you should think graphically" -- if I ever think about the sine or tangent function now, it might be the area of a triangle, it might be the length of some vertical line segment, but it's basically never using the graph, which contains basically no additional information. If I have more than one shot at it, I'll try to explain both, but is it really of service to go around saying all the time why graphs aren't the end-all-be-all? Also, while I can express the pictures in my head one at a time, the fact that I repeatedly, repeatedly see this pictures is something that I feel is harder to express. After all, can't you just do algebra and get through this stuff more quickly? The algebra is "easier" too; it takes up less space. - 1 Really nice examples. – Michael Oct 13 2011 at 7:19 Thank you for starting this discussion. I think any kind of pedagogical tool should be shared with students and collegues, especially in writing form. When I read "the classics", i.e. works of famous mathematicians I always wondered the process they went through to reach those conclusions, what imagery went through their heads while they proved a theorem, maybe that would be useful to me, or not, but I always wanted to know. I think during a lecture saying something like "here is how I do it, imagine group elements breaking into a formation organized into circular groups" is no discomfort to anyone. Maybe this explanation can help one student, or two (or all) students to grok the topic just a tad more, and that's still important. Richard Feynman used to say (paraphrasing) that he never really knew in advance how his students would understant quantum mechanics, he did not have any single method, he'd only try to explain the topic from many different angles hoping that one of those angles provide an entry point for a student, into the subject. - The ring ${\mathbb Z}/N{\mathbb Z}$ is usually defined in a rather cumbersome way, and it takes some time (infinite in most cases) before students realize that you can think of it as ${\bf Z}$ with just one added relation $N=0$ to do the computations (and the problem that $xy=0$ does not necessarily imply that $x=0$ or $y=0$), so that it is indeed a very simple object and not some horribly abstract invention. - There is a huge gap. I've always been interested in this question from the point of not just the "hermeneutics" of mathematics, but also from the standpoint of motivation for mathematicians. I wonder to what degree doing mathematics is constructing a mental model for a mathematical object, comparing the properties of that model to the facts associated to the object and then trying to reconcile the model with the facts? Some personal examples: I often read a definition, and then try to write the definition in an equivalent form in my own words...based on whatever vague impression/model the definition inspired. After this, I compare my definition with the original and then try to see what in my conception/mental picture needs altering. Another thing I commonly do is to read the statement of a theorem and try to prove it for myself...however I am not simply trying to prove the theorem formally, but am trying to build a conceptual construct or point of view that will make the proof of the theorem evident in light of that construct. Personally, this (often failing) attempt to build a sharp intuitive mental model of a mathematical object is the primary motivation for doing mathematics at all. This all reminds me of what was (purportedly) written on Richard Feynman's blackboard at the time of his death: If you cannot create it, you don't understand it. One is forced to wonder what constitutes "creation" here. EDIT/ADDENDUM: In the interview with Alain Connes here there is a great description of internal mental process. This has inspired me to distill that, for me, mathematics is the triumph of concept over brute computation. The uncommunicated mental models that allow us to organize and complete computations and proofs that seem impossible is, the central source of joy and surprise in mathematics. Paraphrasing Gowers, philosophy of mathematics is useful in that it affects the practice of mathematics. The above viewpoint is liberating as it points at what to spend time trying to do. It is frustrating to watch introductory analysis students stumble around with formalism with no apparent "picture" (not necessarily geometric) of what is going on in a proof. The typical undergraduate seems to spend very little time trying to find a mental model that generates a vivid proof. This may be due to the fact that a fig leaf covers the essential part of what we do! - Looking back a few answers here, I see that mental models (This language I appropriated from something Bill Thurston wrote, BTW) and Feynman have already been mentioned...I guess what I'd like to know is whether the mathematical work of others is motivated by the process I mentioned. – Jon Bannon Oct 21 2010 at 18:29 This phenomenon occurs not just in advanced mathematics but also right at the very bottom in simple mental arithmetic. If I have to do a moderately complicated calculation such as adding two three-digit numbers, there's often a part of my brain that jumps ahead to the answer before another more cautious part has got there with carefully checked calculations. The first part just sort of feels the answer and then says "I told you so" to the second part, except occasionally when the first part gets it wrong and the second part says "Now you know why I bother to be careful" to the first part. And there are also aspects of how I carry out integer addition and subtraction that I would normally be a bit embarrassed to verbalize, such as that if I subtract 48 from 135 then there's a preliminary answer, 97, that I know from experience is wrong and has to be corrected by subtracting 10. (The justification for the preliminary answer is that 13-4=9 and that the answer must end in a 7.) It's not quite what's going on in my head, but it's almost as though I say, "OK I'll subtract 58 instead so as to get the right answer." But if I were teaching this to a child then I'd tell a slightly different story, such as borrowing 1, or first subtracting 50 and then adding 2. - 2 I have similar problems when explaining graphing to calculus students. When I see a functional expression like $x\sin(x)$ my mind jumps to decomposing it as a composite $x \to (x,\sin(x))$ with with $(x,y)\longmapsto xy$, and since I know the graphs of these two functions, assembling the graph of the composite is quick, like pasting two images together. But when teaching calculus we're deliberately not thinking in a multi-variable context, and in particular trying to deduce everything via a linear outlook: differentiate, find zeros, check 2nd derivative there, find where increasing... etc – Ryan Budney Oct 3 2010 at 21:13 7 Actually, I like explaining ways of solving problems that involve making mistakes, intentional or not, and then figuring out how to correct them. I like this better than trying to teach error-free algorithms, because it incorporates the error-checking as a natural part of the process. I believe we should be teaching more systematic methods for finding and correcting errors. Students should learn when guessing, checking, and correcting is faster and easier than a more direct algorithm. Integrals that require more than one integration by parts is an obvious example of this. – Deane Yang Oct 3 2010 at 22:08 2 I also have the same divided brain when doing arithmetic, especially when adding up 8 or so 1-digit numbers, when there are often several good orders to add them. (This mostly comes up around exam time.) The attitudes of the two halves to each other are also the same. Amazing. – James Borger Oct 4 2010 at 0:18 7 I agree, beginning math is a very rich and intriguing area. I've discussed arithmetic questions with many young children, and they are often very creative in strategies to think their way to an answer. It really requires being on your toes to discern their thought processes, because the words do not match adult expectations; they often take phrases with logical meanings that I've suppressed because of convention. To teach math to kids, I think it's paramount to encourage them to think, rather than teach conventional "borrowing 1" type stories. Early math teaching usually suppresses thinking. – Bill Thurston Oct 6 2010 at 12:05 8 @Deane Yang: Have you read the classic Proofs and Refutations: the Logic of Mathematical Discovery, by Imre Lakatos? It makes a compelling case for the dialectical method, the value of mistakes and corrections. @Ryan Budney: It's very tricky to figure out what will convey better to students. Often students have complex thought processes we're unaware of, struggling to fit into simplified explanations we give. It might well work better if you explained your way of thinking. I'm not yet convinced, though, that going to 2 dimensions is a clearer mental image than scaling sin(x) by factor x. – Bill Thurston Oct 6 2010 at 12:26 The well known situation of language translation is I believe akin to the tension between thinking and explaining. I am French, I can understand, write and explained myself in English yet I am a bit at a loss when required to translate some piece of English into French so I call this translating ability a third language. My guess or feeling or personal view is that the thinking is more semantic and - for weird (and sad) cultural reasons- mathematical explaining is too often required to be on a syntactic level. - 2 Language manifests itself in syntax, you can't escape it. But while your statement makes no sense to me if taken literally, it gets across a point - it is this courage to make superficially meaningless or wrong statements, but which nevertheless kick the brain into the right direction, which would do good sometimes in explaining mathematics. – Peter Arndt Sep 22 2010 at 12:46 @Jérôme: Is the French-English barrier you mention a matter of compartmentalization, so it feels disruptive for French to intrude if you're speaking or writing English? I think people vary a lot in the degree of immersion in different modes. I'm one who tends to get buried in a mode and feel a strong instinct to deflect all inputs from other modes. In mathematics, equations often feel intrusive in this way, unless I'm in a more symbolic mode. Syntax: sometimes I think mathematics has its own syntax, a kind of flow of thoughts and images. To think well, I need to get outside usual language. – Bill Thurston Sep 23 2010 at 3:11 A lot of the discussion going on above is about the fact that we do not understand our mind's working. I do not even know if mathematical thinking is language-bound. I have the impression that it is not: mathematical discussions are more tedious in foreign languages but not more difficult. An analogy is perhaps a walk on a barely visible path. Sometimes you lose it and have to search for it. Glimpsing it again you tell yourself: "Ah, here it is again" but language is completely irrelevant even if it can be used to describe it to a friend. How do other people feel about this? Perhaps we are all different and do not see not the "same" reality. (Is your color "red" the same as my color "red"). Perhaps we are making a huge mistake assuming that different people think very similarly: For a dog, reality is made of a lot of scents, for a horse it is perhaps prairies and vast avenues where it can gallop. Some of us are perhaps "dogs" and other "horses" of mathematics and since most scents and prairies have not been named, we have to communicate through a limited common denominator, even among people of the same "mathematical species". - 7 I'm convinced, through probing, that the role of language for different mathematicians varies considerably. I include symbolic manipulation (algebraic formula, symbolic integration, etc.) with language, because I think it makes strong use of our brains' linguistic modules. But, I believe that for many, mathematical thoughts are often rather detached from language---thus, formal discussions of mathematics are often a pale shadow of the actual thoughts. People with similar backgrounds can reconstruct ideas from their written shadows; however, when backgrounds differ, this needlessly fails. – Bill Thurston Sep 16 2010 at 11:56 I find there is a world of difference between explaining things to a colleague, and explaining things to a close collaborator. With the latter, one really can communicate at the intuitive level, because one already has a reasonable idea of what the other person's mental model of the problem is. In some ways, I find that throwing out things to a collaborator is closer to the mathematical thought process than just thinking about maths on one's own, if that makes any sense. One specific mental image that I can communicate easily with collaborators, but not always to more general audiences, is to think of quantifiers in game theoretic terms. Do we need to show that for every epsilon there exists a delta? Then imagine that you have a bag of deltas in your hand, but you can wait until your opponent (or some malicious force of nature) produces an epsilon to bother you, at which point you can reach into your bag and find the right delta to deal with the problem. Somehow, anthropomorphising the "enemy" (as well as one's "allies") can focus one's thoughts quite well. This intuition also combines well with probabilistic methods, in which case in addition to you and the adversary, there is also a Random player who spits out mathematical quantities in a way that is neither maximally helpful nor maximally adverse to your cause, but just some randomly chosen quantity in between. The trick is then to harness this randomness to let you evade and confuse your adversary. Is there a quantity in one's PDE or dynamical system that one can bound, but not otherwise estimate very well? Then imagine that it is controlled by an adversary or by Murphy's law, and will always push things in the most unfavorable direction for whatever you are trying to accomplish. Sometimes this will make that term "win" the game, in which case one either gives up (or starts hunting for negative results), or looks for additional ways to "tame" or "constrain" that troublesome term, for instance by exploiting some conservation law structure of the PDE. For evolutionary PDEs in particular, I find there is a rich zoo of colourful physical analogies that one can use to get a grip on a problem. I've used the metaphor of an egg yolk frying in a pool of oil, or a jetski riding ocean waves, to understand the behaviour of a fine-scaled or high-frequency component of a wave when under the influence of a lower frequency field, and how it exchanges mass, energy, or momentum with its environment. In one extreme case, I ended up rolling around on the floor with my eyes closed in order to understand the effect of a gauge transformation that was based on this type of interaction between different frequencies. (Incidentally, that particular gauge transformation won me a Bocher prize, once I understood how it worked.) I guess this last example is one that I would have difficulty communicating to even my closest collaborators. Needless to say, none of these analogies show up in my published papers, although I did try to convey some of them in my PDE book eventually. ADDED LATER: I think one reason why one cannot communicate most of one's internal mathematical thoughts is that one's internal mathematical model is very much a function of one's mathematical upbringing. For instance, my background is in harmonic analysis, and so I try to visualise as much as possible in terms of things like interactions between frequencies, or contests between different quantitative bounds. This is probably quite a different perspective from someone brought up from, say, an algebraic, geometric, or logical background. I can appreciate these other perspectives, but still tend to revert to the ones I am most personally comfortable with when I am thinking about these things on my own. ADDED (MUCH) LATER: Another mode of thought that I and many others use routinely, but which I realised only recently was not as ubiquitious as I believed, is to use an "economic" mindset to prove inequalities such as $X \leq Y$ or $X \leq CY$ for various positive quantities $X, Y$, interpreting them in the form "If I can afford $Y$, can I therefore afford $X$?" or "If I can afford lots of $Y$, can I therefore afford $X$?" respectively. This frame of reference starts one thinking about what types of quantities are "cheap" and what are "expensive", and whether the use of various standard inequalities constitutes a "good deal" or not. It also helps one understand the role of weights, which make things more expensive when the weight is large, and cheaper when the weight is small. - 10 Wonderful answer. I've noticed that my thought processes go into a higher gear, with heightened mathematical intuition, when I'm explaining something to a collaborator or colleague. I'm sure I sometimes abuse this by bending people's ear too long. I too sometimes think in terms of games and adversaries, in the situations as you describe involving quantifiers. I imagine it harnesses a different module of our brains that we share with wily foxes and others. Barriers: I think similar structures are often unknowingly studied in different fields. We'd do better if we could share more. END. – Bill Thurston Sep 15 2010 at 22:43 The issue seems, to me, that a lot of these mental pictures are very personal. I am reminded of an anecdote by Richard Feynman, from "The Pleasure of Finding Things Out". He explains how counting, for him, is a verbal process (he speaks the numbers to himself as he goes along), but that a friend of his would manage visually. (Text here) He finishes by saying: I often think about that, especially when I'm teaching some esoteric technique such as integrating Bessel functions. When I see equations, I see the letters in colors — I don't know why. As I'm talking, I see vague pictures of Bessel functions from Jahnke and Emde's book, with light-tan j's, slightly violet-bluish n's, and dark brown x's flying around. And I wonder what the hell it must look like to the students. Because of this, I think there might not always be a significant value in trying to pass those mental pictures over - the real aim is to provoke the student into developing his own mental pictures, that he can strongly relate to. Some words such as "homological" or "homotopical" spark up very distinctive feelings in me, in a similar way as hearing "mountain" would make me visualise various mountains, hills, cliffs, etc. But whereas the meaning of "mountain" came to me through vision (mainly, but also other senses), the origin of my mental images of mathematical ideas comes through the practice of mathematics. As such, it seems harder to convey these mathematical pictures: they must be backed up by precise mathematical understanding, which at any rate should end up conjuring these mental pictures. Of course, many mental pictures are simple enough or "canonical" enough that one might imagine everyone would come to develop very similar ones upon understanding of one particular concept; the previously mentioned example of cyclic groups comes to mind. So there might be value in passing that on, but in the end I would think that understanding accompanied by the attention to what meaning it provides already goes a long way towards developing personal mental images. - 1 @Sam: I agree, we often have very different and personal mental models. Random models are confusing. I often have several different models at once. I often work on them in my head, trying to get them to fit better to the context. I find that they can change, and improve, dramatically through this process, and when they are happy enough, they're worth communicating. But, for me: reliable precision comes only after finding a good overall mental model. I think many students are often blocked by false or inadequate mental models --- they're hard to excavate and hard to convey, but important. – Bill Thurston Sep 14 2010 at 19:27 @Bill: Indeed, there is a lot of value in finding methods to present ideas in a fashion that really enables the formation of good mental models. Indeed, I often have the impression of understanding what is presented before me, but feeling I am missing the "big picture", of having a convenient state of mind within which to understand the whole theory. But I think it might be more about finding this fertile environment than trying to convey our own mental models (even though the latter certainly contributes to bettering the environment). Great question. Wish I had good examples to contribute! – Sam Derbyshire Sep 14 2010 at 20:05 4 I find that you expressed this perfectly above: "What's important is not the process by which you arrived at an idea, but a story that gives the idea context and meaning. It's a story you make: a setting of meaning and reason for the idea, rather than the history of how you stumbled on the idea." – Sam Derbyshire Sep 14 2010 at 20:07 1 Sam: There's another Feynman passage you may be interested in, where he talks about trying to count at a steady pace and talk at the same time and finds he's unable to do it. However, a friend of his does so easily. There's another task (I forget the exact details, and I don't want to go grab the book and look for it!) that Feynman can do, and his friend can't, however. The two of them try and figure out what the exact barriers are, when they realize that Feynman is counting verbally in his head while his friend is watching numbers go by on a tape! – drvitek Sep 15 2010 at 22:01 1 @drvitek. Thanks for the reference. When I was a teenager, I realized that the standard way people count with fingers is inefficient, so I developed skill at counting in binary. It works without much conscious attention up to 1024. With that method, I can count while carrying on a conversation, or thinking of something else. In the days of typewriters when I had to manually center the title, I was unable to do it reliably in my head using word counting, but finger counting makes it trivial. ALSO: I think most people count small numbers subconsciously by rhythm, as when climbing stairs. – Bill Thurston Sep 15 2010 at 22:48 show 1 more comment I'd like to take the occasion and sketch my view on reconstruction problems in graph theory: I see a graph as a set of subjects with a relation between (some of) them. Each node (= subject) has a limited knowledge of the whole graph. The question is how many of those subjects have to put their knowledge together to know the whole graph. Of course, this depends on the kind of limited knowledge each subject (= node) has. In case it knows everything but its own relations to the rest of the graph, we have Ulam's reconstruction problem. Another natural kind of limited knowledge would be: all relations except those between the most distant nodes. Or: all relations within a neighbourhood of fixed size and nothing else. I find it enlightening to compare this situation to the case of reconstructing a 3D object from its 2D projections (another kind of limited knowledge). - @Hans: Thanks, that's an interesting mental metaphor --- I like it. – Bill Thurston Sep 14 2010 at 19:20 1 This very much reminds me of "testable" properties of graphs: arxiv.org/abs/0907.5302 The idea is that you want to compute something (say normalized dimension of the kernel of an operator) up to epsilon. Then it turns out that for many properties the following is true: no matter how big your graph is its enough to choose k vertices (k is independent of how big the graph is) and look at their r-neighbourhoods (r is also independent of how big the graph is). – Łukasz Grabowski Sep 16 2010 at 12:07 Final addition: Since I've produced many rambles, I thought I'd close my (anti-)contribution with a distilled version of the example I've attempted below. It's still something very standard, but, I hope, in the spirit of the original question. I'll describe it as if it were a personal thing. Almost always, I think of an integer as a function of the primes. So for 20, say, 20(2)= 0 20(3)=2 20(5)=0 20(7)=6 . . 20(19)=1 20(23)=20 20(29)=20 20(31)=20 20(37)=20 . . . It's quite a compelling image, I think, an integer as a function that varies in this way for a while before eventually leveling off. But, for a number of reasons, I rarely mention it to students or even to colleagues. Maybe I should. Original answer: It's unclear if this is an appropriate kind of answer, in that I'm not putting forward anything very specific. But I'll take the paragraph in highlight at face value. I find it quite hard to express publicly my vision of mathematics, and I think this is a pretty common plight. Part of the reason is the difficulty of putting into words a sense of things that ultimately stems from a view of the landscape, as may be suggested by the metaphor. But another important reason is the disapprobation of peers. To appeal to hackneyed stereotypes, each of us has in him/her a bit of Erdos, a bit of Thurston, and perhaps a bit of Grothendieck, of course in varying proportions depending on education and temperament. I think I saw somewhere on this site the sentiment that 'a bad Erdos still might be an OK mathematician, but a bad Grothendieck is really terrible,' or something to that effect. This opinion is surrounded by a pretty broad consensus, I think. If I may be allowed some cliches now from the world of finance, it's almost as though definite mathematical results are money in the bank. After you've built up some savings, you can afford to spend a bit by philosophizing. But then, you can't let the balance get too low because people will start looking at you in funny, suspicious ways. I know that on the infrequent occasions* that I get carried away and convey at any length my vision of how a certain area of mathematics should work, what should be true and why, compelling analogies, and so on, I feel rather embarrassed for a little while. It feels like I am indeed running out of money and will need to back up the highfalutin words with some theorems (or at least lemmas) relatively soon. (And then, so many basically sound ideas are initially mistaken for trivial reasons.) Now, I wish to make it clear that unlike Grothendieck (see the beginning paragraphs of this letter to Faltings) I find this quite sensible a state of affairs. For myself, it seems to be pretty healthy that my tendency to philosophize is held in check by the demand of the community that I have something to show for it. I grant that this may well be because my own visions are so meagre in comparison to Grothendieck's. In any case, the general phenomenon itself is interesting to observe, in myself and in others. Incidentally, I find the peer pressure in question remarkably democratic. Obviously, a well-established mathematician typically has more money than average in the bank, so to speak. But it's not a few times I've observed eminent people during periods of slowdown, being gradually ignored or just tolerated in their musings by many young people, even students. Meanwhile, if you're an energetic youngster with some compelling vision of an area of mathematics, it may not be so bad to let loose. If you have a really good business idea, it may even make sense to take out a large loan. And provided you have the right sort of personality, the pressure to back up your philosophical bravado with results may spur you on to great things. This isn't to say you won't have to put up with perfectly reasonable looks of incredulity, even from me, possibly for years. Maybe it seems frequent to my friends. Added: Since I commented above on something quite general, here is an attempt at a specific contribution. It's not at all personal in that I'm referring to a well-known point of view in Diophantine geometry, whereby solutions to equations are sections of fiber bundles. Some kind of a picture of the fiber bundle in question was popularized by Mumford in his Red Book. I've discovered a reproduction on this page. The picture there is of $Spec(\mathbb{Z}[x])$, but interesting equations even in two variables will conjure up a more complicated image of an arithmetic surface fibered over the 'arithmetic curve' $Spec(\mathbb{Z})$. A solution to the equation will then be a section of the bundle cutting across the fibers, also in a complicated manner. Much interesting work in number theory is concerned with how the sections meet the singular fibers. Over the years, I've had many different thoughts about this perspective. For me personally, it was truly decisive, in that I hadn't been very interested in number theory until I realized, almost with a shock, that the study of solutions to equations had been 'reduced' to the study of maps between spaces of a quite rigid sort. In recent years, I think I've also reconciled myself with the more classical view, whereby numbers are some kinds of algebraic gadgets. That is, thinking about matters purely algebraically does seem to provide certain flexible modes that can be obscured by the insistence on geometry. I've also discovered that there is indeed a good deal of variation in how compelling the inner picture of a fiber bundle can be, even among seasoned experts in arithmetic geometry. Nevertheless, it's clear that the geometric approach is important, and informs a good deal of important mathematics. For example, there is an elementary but key step in Faltings' proof of the Mordell conjecture referred to as the 'Kodaira-Parshin trick,' whereby you (essentially) get a compact curve $X$ of genus at least two to parametrize a smooth family of curves $$Y\rightarrow X.$$ Then, whenever you have a rational point $$P:Spec(\mathbb{Q})\rightarrow X$$ of $X$, you can look at the fiber $Y_P$ of $Y$ above $P$, which is itself a curve. The argument is that if you have too many points $P$, you get too many good curves over $\mathbb{Q}$. What is good about them? Well, they all spread out to arithmetic surfaces over the spectrum of $\mathbb{Z}$ that are singular only over a fixed set of places. This part can be made obvious by spreading out both $Y$, $X$, and the map between them over the integers as well, right at the outset. If you don't have that picture in mind, the goodness of the $Y_P$ is not at all easy to explain. Anyways, what I wanted to say is that the picture of solutions as sections to fiber bundles is really difficult to explain to people without a certain facility in scheme theory. Because it seems so important, and because it is a crucial ingredient in my own thinking, I make an attempt every now and then in an exposition at the colloquium level, and fail miserably. I notice almost none of my colleagues even try to explain it in a general talk. Now, I've mentioned already that this is far from a personal image of a mathematical object. But it still seems to be a good example of a very basic picture that you refrain from putting into words most of the time. If it really had been only a personal vision, it may even have been all but maddening, the schism between the clarity of the mental image and what you're able to say about it. Note that the process of putting the whole thing into words in a convincing manner in fact took thousands of pages of foundational work. Added again: Professor Thurston: To be honest, I'm not sure about the significance of competing mental images in this context. If I may, I would like to suggest another possibility. It isn't too well thought out, but I don't believe it to be entirely random either. Many people from outside the area seem to have difficulty understanding the picture I mentioned because they are intuitively suspicious of its usefulness. Consider a simpler picture of the real algebraic curve that comes up when one studies cubic equations like $$E: y^2=x^3-2.$$ There, people are easily convinced that geometry is helpful, especially when I draw the tangent line at the point $P=(3,5)$ to produce another rational point. What is the key difference from the other picture of an arithmetic surface and sections? My feeling is it has mainly to do with the suggestion that the point itself has a complicated geometry encapsulated by the arrow $$P:Spec(\Bbb{Z})\rightarrow E.$$ That is, spaces like $Spec(\Bbb{Q})$ and $Spec(\Bbb{Z})$ are problematic and, after all, are quite radical. In $Spec(\Bbb{Q})$, one encounters the absurdity that the space $Spec(\Bbb{Q})$ itself is just a point. So one has to go into the whole issue that the point is equipped with a ring of functions, which happens to be $\Bbb{Q}$, and so on. At this point, people's eyes frequently glaze over, but not, I think, because this concept is too difficult or because it competes with some other view. Rather, the typical mathematician will be unable to see the point of looking at these commonplace things in this way. The temptation arises to resort to persuasion by authority then (such and such great theorem uses this language and viewpoint, etc.), but it's obviously better if the audience can really appreciate the ideas through some first-hand experience, even of a simple sort. I do have an array of examples that might help in this regard, provided someone is kind enough to be still interested. But how helpful they really are, I'm quite unsure. At the University of Arizona, we once had a study seminar on random matrices and number theory, to which I was called upon to contribute a brief summary of the analogous theory over finite fields. Unfortunately, this does involve some mention of sheaves, arithmetic fundamental groups, and some other strange things. Afterwards, my colleague Hermann Flaschka, an excellent mathematician with whom I felt I could speak easily about almost anything, commented that he couldn't tell if the whole language just consisted of word associations or if some actual geometry was going on. Now, I'm sure this was due in part to my poor powers of exposition. But further conversation gave me the strong impression that the question that really went through his mind was: 'How could it possibly be useful to think about these objects in this way?' To restate my point, I think a good deal of conceptual inhibition comes from a kind of intuitive utilitarian concern. Matters are further complicated by the important fact that this kind of conceptual conservatism is perfectly sensible much of the time. By the way, my choice of example was somewhat motivated by the fact that it is quite likely to be difficult for people outside of arithmetic geometry, including many readers of this forum. This gives it a different flavor from the situations where we all understand each other more or less well, and focus therefore on pedagogical issues referring to classroom practice. Yet again: Forgive me for being a bore with these repeated additions. The description of your approach to lectures seems to confirm the point I made, or at least had somewhat in mind: When someone can't understand what we try to explain, it's maybe in his or her best interest (real or perceived) not to. It's hard not to feel that this happens in the classroom as well oftentimes. This then brings up the obvious point that what we try to say is best informed by some understanding of who we're speaking to as well as some humility. As a corollary, what we avoid saying might equally well be thus informed. My own approach, by the way, is almost opposite to yours. Of course I can't absorb technical details just sitting there, but I try my best to concentrate for the whole hour or so, almost regardless of the topic. (Here in Korea, it's not uncommon for standard seminar lectures to be two hours.) If I may be forgiven a simplistic generalization, your approach strikes me as common among deeply creative people, while perennial students like me tend to follow colloquia more closely. I intend neither flattery nor modesty with this remark, but only observation. Also, I am trying to create a complex picture (there's that word again) of the problem of communication. As to $Spec(\Bbb{Z})$, perhaps there will be occasion to bore you with that some other time. Why don't you post a question (assuming you are interested)? Then you are likely to get a great many perspectives more competent than mine. It might be an interesting experiment relevant to your original question. *I realize it's hardly my place to tell anyone else to be humble. - 1 @Minhyong Kim: Thanks for your thoughtful and appropriate response to the question. I've been aware that when I drift too long into philosophizing or meta-discussions, people's eyes glaze over. My eyes often glaze over at that kind of stuff, as well. I don't advocate that people spend a lot of time philosophizing. I do advocate that we try to make their mathematical discussions informed by awareness of how they are thinking, and an attempt to see how it might appear to the other person. I think it can help us get to the point more quickly and effectively. – Bill Thurston Sep 14 2010 at 19:15 3 The example you added to your answer is very telling. As an outsdier to the field, I have only a vague sense of the mathematical meaning of Mumford's picture, but I think I understand its role and its effect that you described. It's important to ask yourself "Why?" when others don't comprehend something of this nature. Usually it's because they have competing mental models where your words are nonsense. It requires some excavation of the old before you can embrace the new. – Bill Thurston Sep 15 2010 at 4:38 6 If only I could give a +1 to each of the three thirds of this post. The "money in the bank" analogy is particularly fantastic. – Cam McLeman Sep 15 2010 at 13:47 6 After "Added again": When listening to a lecture, I can't possibly attend to every word: so many words blank out my thoughts. My attention repeatedly dives inward to my own thoughts and my own mental models, asking 'what are they really saying?' or 'where is this going?'. I try to shortcut through my own understanding, then emerge to see if I'm still with the lecture. It's the only way for me, and it often works. For something like Spec(\mathbb Z) (for me, now), I need extra guidance to avoid diving into unhelpful places. Perhaps explicit instructions "don't think __1, think __2", and why. – Bill Thurston Sep 15 2010 at 14:00 2 After final addition: I think you should mention your way of thinking to students and colleagues. It may seem like a small thing, but mall things are important, and I suspect it would help set the right frame of mind. Isn't this method used in computer algorithm for computations with large integers? That could help motivate/justify it, not that there aren't more mathematical justifications. I'm enjoying this conversation, which obviously could continue for a long time following threads that have started, but this being MO, we should probably interrupt it and wait for other chances. – Bill Thurston Sep 16 2010 at 3:22 show 2 more comments When I talk about determinants, I generally talk about something on the spectrum between "it measures how much volume scales" and "it's the induced action on the top exterior power." But the way I think about determinants (especially in combinatorics) is the picture coming from the Lindstrom-Gessel-Viennot lemma: I imagine that the entries of the matrix describe transition amplitudes and that the determinant is an alternating sum over transition amplitudes in which "histories" of $n$ particles can constructively or destructively interfere. I have a hard time making this picture precise so I rarely talk about it, but for me it gives some intuition for why determinants should be useful in combinatorics (which the elegant basis-free definition, at least for me, does not). Edit: Let me also mention that something I really like about this perspective is that it makes intuitive not only the multiplicativity of the determinant but even the Cauchy-Binet formula. - @Qiaochu Yuan: Thanks very much. Your remark of how you think of determinants is intriguing to me --- I haven't really gotten why determinant formulas work in combinatorics, even though I've encountered them in situations I wanted to understand, and you've suggested a mental model to try to make better sense of it. – Bill Thurston Sep 14 2010 at 19:18 Don't ask me why but I stumbled over en.wikipedia.org/wiki/Cayley-Hamilton_theorem today, and for the outsider it is really, really astonishing at very first sight. (Is every mathematician aware of it, by the way? And/or is it in some way trivial, eventually?) So I want to ask you (honestly!): Did your picture of determinants give you a hint to the CH-theorem? – Hans Stricker Sep 16 2010 at 21:59 2 @Hans : The Cayley-Hamilton theorem is (or, at least, should be) taught in every rigorous undergraduate course in linear algebra, and hence essentially all working mathematicians know about it. It's easier to prove than you might think. Here's one possible set of hints, at least for matrices over C. First, prove it for diagonal matrices (it's easy!). Next, prove it for diagonalizable matrices (almost as easy!). Finally, use the fact that every matrix can be approximated arbitrarly well by a diagonalizable matrix to deduce the general case. – Nikita Sep 17 2010 at 19:37 One facet of the problem is when you replace the word others by yourself. Did it ever happen to you that you find a nice construction, a nice concept, a nice property, giving you a research paper of which you are proud ? Once the exciting period of intensive research ends, you come back and ask yourselves How did I come to find this trick ? Why did I look in this direction ? And sometimes, you just cannot remember How and When. But you did it !! If you can't explain yourself this process, do you expect to explain others we way you think ? - 7 @Denis: I've learned that when I go back to look at something, my thinking has usually rounded off too many corners, so my understanding is much fuzzier. I sometimes find things I have written to be very obtuse. I was too wrapped up in my then state of mind to express ideas clearly even to myself, reading it much later. What's important is not the process by which you arrived at an idea, but a story that gives the idea context and meaning. It's a story you make: a setting of meaning and reason for the idea, rather than the history of how you stumbled on the idea. – Bill Thurston Sep 14 2010 at 18:25 I am immediately reminded of the following phenomenon: "Let's consider an elliptic curve $E$ over $\mathbf{Q}_p$." ~Speaker begins to draw a donut at the board~ Clearly this is wrong. The speaker may even mention how wrong it is. But there's enough of a kernel of truth to the picture where it may be helpful for the audience. Justifying why the picture is valid in the situation being considered would involve some hard and not undue model theory, but doing anything more than waving your hands and saying the magic words ("a-la-ca-Lefschetz!") is likely to derail your talk so badly as to effectively destroy it. Even if the picture is not completely valid for what you're talking about, it's a visual aid which is at least an easily described first approximation to the truth. - 3 I remember listening to someone talk about singular complex spaces: he started the talk with «Let us consider a complex space with isolated singularities...» and made a picture on the blackbloard of what I looked like clouds as a kid would draw. Watching him as he continued, it was clear that the picture actually depicted for him the richness of the phenomenon he was talking about. I envied him a little :) – Mariano Suárez-Alvarez Sep 14 2010 at 13:28 3 @stankewicz: Thanks, that's an interesting example. "A-la-ca-Lefschetz" --- that's a great phrase. Should it be delivered with a little tune? I agree that it's a mistake to dwell too long on the meta level, excavating correctness of mental props--that's like giving a long explanation of a joke that someone doesn't "get". The trick is in setting it up so the mental prop's seem natural, but without having to say very much. – Bill Thurston Sep 14 2010 at 18:39 I think the root of the phenomenon is that we can only communicate to others what we know, not what we understand. Also, it is not unreasonable to think that one's mental images are not going to be of any help to others (In fact, they may well make things more complicated, or confusing for others: I have been told mental images by others---sometimes indirectly, by the choice of the word introduced in a definition---and been thereby misled; here «misled» means «led in a direction different to the one I personally would follow in order to form my own mental image of the concept».) For example, for me resolving the singularities of algebraic varieties makes a clicking (or clacking) sound: this is quite significant for me in a way, but when talking to others I doubt I'd make any mention of this, for seriously doubt it would help :) - 1 Isn't this a rather pessimistic point of view: that we can only communicate to others what we know, not what we understand. Don't we wish that it is understanding that is taught/trained at school, not only knowledge? – Hans Stricker Sep 14 2010 at 15:41 3 @Hans, You/we may wish all we want: I don't think it is possible! This does not mean---at all---that understanding is doomed to disappear: as I see it, understanding is the result of knowledge, when it is properly nurtured. In fact, I am far from a pessimist in this respect. But I have watched too many a seminar where people try to convey what they understand... – Mariano Suárez-Alvarez Sep 14 2010 at 16:21 2 @Mariano: I like the image of a little mental clicking sound when a singularity is resolved. It's perhaps undignified, but I expect other people would register the ideas better if you actually verbalized the sound --- I suspect the main barrier is embarassment about loss of dignity. However: I do think we can and do communicate understanding, but it's hard challenge to do it well; it requires developing a sense of how the other person is thinking, otherwise as you say it's just confusing. – Bill Thurston Sep 14 2010 at 18:11 1 Do small resolutions make softer sounds? – Vivek Shende Sep 14 2010 at 23:24 1 Vivak: the sounds are of a higher pitch :) – Mariano Suárez-Alvarez Sep 14 2010 at 23:59 show 6 more comments I have a worse problem than having unspoken thought processes: some of my best thought processes are simply beneath the level of consciousness and I don't notice them at all until they're finished. Even then, I often get only an answer and not an explanation out of them. Surely this happens to everyone: the problem solved during sleep, the idea on a walk in the woods, the conviction that a conjecture is true on utterly minimal evidence, the argument that pops up full formed in the middle of a conversation. My mathematical process is roughly this: consciously, I try a lot of stupid things which essentially have no chance of working but do have the benefit of exposing me to lots of examples; these examples pile up and are subconsciously masticated for days, weeks, months -- I'm not old enough mathematically to put "years" here yet -- and eventually by some inner and unobservable process I just have a feeling about what to do. Perhaps that's an exaggeration. But I certainly do feel that way sometimes, and to the extent that it's true, it means that the whole project of trying to communicate how I thought of something is just telling stories, at least if I say anything other than "well, I just knew one day." - 2 @Vivek: I think everyone has thought processes below the level of awareness. This is the subject of some of Poincaré's famous essays. It's very hard to be aware of your thought processes. I think of this as related to the difficulty in remembering dreams: ordinarily people quickly forget dreams, but when people pay attention, they start to remember more. So, you can become more aware, but never very fully aware. Whether it does you good, who is to say --- possibly it's like studying stars at night, where it's a mistake to look directly at them. But, I find it interesting to try to understand. – Bill Thurston Sep 14 2010 at 18:17 1 @Bill: do you, or Poincaré, (or anyone else) have suggestions as to how to reach this awareness? With dreams, people say that writing things down when you wake up helps -- not that I've ever tried it. Is it the same with this? And have you in fact managed to become more aware of how your thoughts work over time? – Vivek Shende Sep 14 2010 at 23:22 2 @Vivek: I've developed better sensitivity to how my thoughts are working, enough to know my awareness is very weak. More important is to attend to the two-way process. I work on my intuition and mental imagery to make them true, and when I'm thinking about mathematics, I've become more sensitive to intuitive dissonance and resonance. When I write, I try to get logic and intuition to work together. This increases my awareness. I'm happy when logic and intuition start to converge, as in my answer to mathoverflow.net/questions/38105, but the process never ends. – Bill Thurston Sep 15 2010 at 3:16 I am a visual thinker and I often try to describe what I see to my students. I've been known to say things like "everyone knows that HF looks like a rectangle" as I proceed to draw a rectangle on the board. (By the way, HF is the set of all hereditarily finite sets.) I find that I naturally associate different shapes with different properties of objects. Angular shapes correspond to well-defined objects whereas rounded shapes correspond to variable objects. The number of angles or curves is a measure of how complex an object is. I don't explain my scheme to my students, but I suspect the consistency of the presentation becomes transparent over time. I recall one instance where I deliberately concealed the true nature of my illustration to my students. I was describing a complex construction on infinite trees. I began the description by drawing five vertical lines that I promptly explained were "infinite trees viewed sideways." It so happens that the simplest case of the construction was when the trees consisted of single branches in which case the picture was completely accurate. This is the case I secretly had in mind for the entire description but I never said that since the result was utterly trivial in that case. This was a subtle way to reduce the complex construction to the trivial case. - 1 +1 for the infinite trees viewed sideways! – Mariano Suárez-Alvarez Sep 14 2010 at 11:37 10 HF looks like a triangle, just like V only smaller. – Andreas Blass Sep 14 2010 at 16:37 1 @Akhil: As always, you just do it!? – François G. Dorais♦ Sep 14 2010 at 16:55 10 @François: I love your example, partly because it sounds so implausible on its face. I think pictures can have a role in thinking that is parallel to pronouns in speech --- they gives a hook to refer to a particular things, even in situations when you can't explain exactly why the picture is what it is. If you're thinking with the geometric parts of your brain (regardless of whether your'e thinking about geometry), it's disruptive to use words as hooks, and often the words or symbols are too long and complicated to be effective hooks. – Bill Thurston Sep 14 2010 at 18:06 1 I would also have pictured HF as a rectangle...but for me "HF" is the set of Hilbert-Frege logical axioms for deduction rules. A mathematical formal theory to which HF is "paired" to carry on mathematics, such as ZFC, would be included in ellipse contour... :) – Qfwfq Sep 15 2010 at 14:19 show 4 more comments One observation: People understandably hesitate telling a half-truth. When you teach a heuristic picture to someone, you also need to teach them about how fuzzy it is and when it starts to break down. A more calculational proof has the virtue of being self-contained and robustly transmissible. This is even more important when writing a textbook. Then there are other cases where I'm puzzled why certain heuristic means of understanding and organizing knowledge don't seem to be usually taught. Take the concept of normal subgroups. In one of his books, V.I. Arnold says that a subgroup is normal when it is relativistically invariant, and he doesn't develop that line of thought any deeper. That statement is a good example of a heuristic analogy that is specific in its detail but general in its spirit. However you phrase it, certainly you should give your students the idea that a normal subgroup is something whose structure is invariant with respect to the parent group's symmetries. As a litmus test, your students should be able to tell whether these subgroups are normal at a glance, without calculation: Let $E^2$ be the Euclidean group of the plane and let $O^2$ be the subgroup fixing some point. Subgroups of $E^2$: • Translations along some particular direction. • Translations along every direction. • Translations and glides along every direction. • Reflections in every line. • Rotations around some particular point. • Symmetries of a tessellation. Subgroups of $O^2$: • Symmetries of a regular polygon. • Reflections in a line. The way I think about the non-normal cases is that there is something something non-isotropic about them, some structure that the subgroup preserves that is not preserved by the parent group. For example: • Translations along some particular direction: Rotations don't preserve the direction. • Translations along every direction: No special directions, so it's normal. • Translations and glides along every direction: Ditto. • Reflections in every line: Ditto. (This combines the two previous cases.) • Rotations around some particular point: Translations don't preserve the point. - 12 AS: Actually the principal definition from the Concise Oxford English dictionary is (adjective) "enabling a person to discover or learn something for themselves". Plenty much broad scope there to zero your aagghhument. – Q.Q.J. Sep 14 2010 at 8:41 3 I was using it in the sense of something like a rule of thumb that may not always work but is nevertheless useful and suggestive. – Per Vognsen Sep 14 2010 at 9:52 4 I was thinking how to frame my response to QQJ (varying from "Well, if you will use the concise OED" to more stringent) and then I remembered that MO is completely unsuited to discussion and I like it like that. I would consider deleting my original remark, but then QQJ's wouldn't make sense. All I will say is that anyone who doesn't know that there is an issue with this word should find out what it means. In its strictest form, it embodies a concept that we really ought to teach our students consciously. So I'm not bothered really about a word, but about losing an idea. – Andrew Stacey Sep 14 2010 at 13:58 7 @Per Bognsen, thanks for you interesting thoughts. "People understandably hesitate telling a half-truth. When you teach a heuristic picture to someone, you also need to teach them about how fuzzy it is and when it starts to break down" It's neither possible nor desirable to try to pour out all your thoughts. It's a challenge to present a mental images in a way that is clarifying, rather than confusing. But: I think people's failure to understand often comes from misleading and unstated mental models they have, so airing these things can be important. – Bill Thurston Sep 14 2010 at 18:00 6 This view of normal subgroups is very close to my mental picture about conjugate subgroups: "a conjugate subgroup $gHg^{-1}$ does the same thing as $H$, but somewhere else". – Qfwfq Sep 15 2010 at 14:26 show 4 more comments Professor Thurston, In my own study, I've struggled with giving meaning to mathematical objects for many years. It's likely I've learned more slowly than most, because I find it difficult to move on unless I've found something I can rely on. Usually, when I try to explain mathematics, I will give the person I'm speaking to a simple problem to give him or her the experience of doing mathematics. Almost always, I will ask to show that $\pi > 3.$ The experience is usually familiar. One person has asked me, "how is this different from finding the answer to any question?" Another example I like to give people is one of the several graphical proofs of the Pythagorean theorem. The one I present is #9 from this site, since it seems to be the simplest: http://www.cut-the-knot.org/pythagoras/index.shtml I've always had trouble explaining things like group without giving simple geometric examples, like the dihedral groups. That the turns of the Rubik's cube is offered as an example of a group is well known. Conjugation, for instance, immediately makes sense. Elohemahab Solomon offered this video http://mathoverflow.net/questions/24794/does-a-connections-blog-podcast-exist-for-math/27919#27919 where Mr. du Sautoy, (at about 12:00) in the context of groups of symmetry, compares the founding of the concept of group with that of number. To summarize Mr. du Sautoy, the symmetry a group measures is analogous to the quantity number measures. For example, we have chairs and tables, which are different, but if we have 3 of each, the quantity is the same. If we have two walls of the Alhambra with different looking designs, but have the same symmetry, then one group acts on the designs. He discusses $S_3$ and $\mathbb Z_6.$ I accepted complex numbers for a long time, but after experience with math past calculus, and the requirement for rigor slowly pervaded my thinking, I couldn't accept them as more than ad hoc constructions. Finally, I was satisfied with their construction as elements of the field $\mathbb R[x]/(x^2+1).$ I talked about this in my answer to the question: http://mathoverflow.net/questions/30156/demystifying-complex-numbers/37817#37817 I now think of complex numbers as an abstract construction from things I have intuition about, that is, real numbers and polynomials. The usual interpretations, as vectors in the complex plane, vectors in $\mathbb R^2$ that one can multiply, are still of course, extremely useful in understanding them. It's likely you know this already: it looks like Hilbert coined the term "ring" from Mr. Gleason's interpretation of cyclic groups. http://en.wikipedia.org/wiki/Ring_(mathematics)#History A distribution is something, with whose definition I've worked successfully, but still have little sense of. I know they are generalized functions, but my lack of understanding might just be lack of experience. - 2 @Anthony: Different people have different mental styles, and different standards of when they accept something enough to move on. I've watched people who are very fast on the uptake sometimes flame out, and people who absorb slowly stick to their instincts and over time do really fabulous work, picking up on things that the fast people zoomed past without noticing. I think we need the variety of styles --- they contribute to progress in ways that build on each other. Thanks for your interesting examples. – Bill Thurston Sep 14 2010 at 18:33 @Bill: Thank you very much for your clear-headed and encouraging words. I agree that we need the variety of styles for the reason you mentioned. I'll certainly remember it. – Anthony Pulido Sep 15 2010 at 0:03 Just to talk about something fresh for which I still have a good memory of what I actually thought and what I wrote, let's take this example Thoughts: 1) Ivan Fesenko teased me with the puzzle without the gradient condition 10 years ago. I solved it with the standard $(1-xy)^2+x^2$, but it would be nice to tease him with the upgraded puzzle when I talk to him next. Also, it is high time to finish it off. 2) The standard polynomial is even and the only critical point is a saddle. If you think of it, this saddle is inevitable: we have low points on the landscape on both sides of the $x$-axis and high values on the $x$-axis going to $+\infty$ both ways. Thus the mountain pass lemma will ensure a saddle somewhere. 3) This works every time when the sequence of points where the polynomial goes to $0$ has two different limiting directions. Then we can separate them by a line and run the same argument. So, the limiting direction must be unique. 4) This seems impossible because the highest (even) degree homogeneous part should vanish in this direction but then it also vanishes in the opposite direction. This makes hte second highest (odd) degree homogeneous part to vanish on the entire line too. The next degree is unclear though... 5) We certainly need something to break the symmetry here. A polynomial family $P_y(x)$ of polynomials in $x$ that have roots close to $0$ when $y\to+\infty$ and no roots when $y\to-\infty$ would be nice. 6) Hey, I know this one: $yx^2-1$. Let's try $x^2+(x^2y-1)^2$. 7) Damn, it doesn't work. The origin is still a critical point. 8) Yeah, what else would you expect: the low points on the landscape are accumulating to one direction but still are separated by the line, so the mountain pass lemma is as powerful as before. To kill it, we need to shift both descends to one side. 9) Add $x$ to $P_y(x)$. That won't change the limiting direction but will shift the zeroes a bit. So, let's try $f(x,y)=x^2+(x^2y+x+1)^2$. 10) The origin is good now: $f_x=2$ there. Actually, it is $2$ everywhere where $x=0$. 11) If $x\ne 0$, then $f_y=0$ only if $x^2y+x+1=0$ but then $f_x=2x\ne 0$ by the chain rule. 12) OK, let's post the example and keep it in memory for teasing people... That's what I actually thought for the last hour or so (interlaced with some personal thoughts that are of no interest for this thread). What I wrote can be seen easily if you follow the link. Why such discrepancy? a) Some steps in the chain like 1) and 10) are too personal to be of interest to anybody. You need them to "start the engine running" and to "vent the steam", but they aren't, strictly speaking, mathematics and do not make me look any better, so why to publish them? b) Some steps like 8) and the heuristics in 9) are actually false. To publish them would be ridiculous. c) 4) and 7) are "failures" on the way. There is no point in telling anyone where and how I failed. I could fill volumes with my failed attempts if I started doing it. d) 10) and 11) are trivial computations. Everyone can do those himself. e) 2), 3), 5) are left. 9) is the counterexample. 2), 3) are steps in the direction of the affirmative answer. Once the final answer is negative, there is no point in talking of the steps in the opposite direction. f) 5) is a nice idea but everybody knows that $y$ is not even. One can see the whole mechanics in the answer itself, so there is no need to explain it separately. I don't know if this account of one personal affair with one relatively simple problem can really shed much light on why we do not tell/write exactly what we see/think of, but you asked and I answered. - 5 @fedja, thank you for a clear account of your approach to an interesting problem. Earlier, I thought about this problem a bit, and got trapped in a false direction. When I read phrases like "One can see the whole mechanics in the answer itself, so there is no need to explain it separately." --- I usually ask myself, "Which one?" If you don't know how to think about it, the formula on its own is very disorienting. – Bill Thurston Sep 14 2010 at 17:53 23 Well I'm glad your answer came with a bit more thought than Ramanujan: "The minute I heard the question, I knew the answer was a continued fraction. 'Which continued fraction?' I asked myself, and the answer came to my mind." Gives the rest of us hope... – Dylan Wilson Sep 14 2010 at 20:58	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 99, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9652079939842224, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/10709/deriving-or-justifying-fundamental-constants?answertab=votes	# Deriving or justifying fundamental constants Is there a fundamental way to look at the universal constants ? can their orders of magnitude be explained from a general points of view like stability, causality, information theory, uncertainty? for example, what sets the relative magnitudes of Planck's constant compared to say charge of electron or is it just a matter of choice of units. Does the physics become more or less cumbersome, insightful if we set all fundamental constants to 1 in appropriate units. - 1 If a constant can be derived, it isn't fundamental any more. If one of the former fundamentals will be derived, You will know about! Even ordinary daily papers would have that as a page one headline. – Georg Jun 2 '11 at 18:32 Physics become no more or less insightful for setting important constants to one, but the formulas become rather less cluttered and it takes less time to talk about them. – dmckee♦ Jun 2 '11 at 19:56 ## 1 Answer Your last couple of comments, about units, are incredibly important. It only makes sense to compare two things if they have compatible units. So, to use your example, it doesn't really make sense to talk about the size of Planck's constant relative to the electron charge, but it does make sense to talk about the mass of the muon relative to the mass of the electron. There are a number of fundamental dimensionless constants in physics. Some are ratios of similar things, such as $m_\mu/m_e$. Others are not so obviously expressed in that form, such as the "fine-structure constant" $\alpha=e^2/(4\pi\epsilon_0\hbar c)$. It makes sense to wonder why the dimensionless constants have the values they do -- one could imagine that a future theory would explain these values. (As Georg says, there is no such theory right now -- by definition the "fundamental" constants are those that don't have a deeper explanation.) On the other hand, one would not expect any fundamental theory to explain why a dimensionful constant, such as the speed of light, had the value that it did: the fact that $c=2.9979\times 10^8$ m/s is an artifact of our definitions of the meter and second. (Actually just the meter in that particular case.) To answer your last question, most people who work in fundamental physics do choose their units so as to set as many fundamental constants equal to 1 as possible. This makes the equations look cleaner, and also makes it easier to see what the interesting dimensionless constants are. Sometimes people try to "explain" the values of various physical constants via anthropic reasoning, pointing out that if the values were quite different from what they are the Universe would be inhospitable to life. Other people find those arguments to be logically suspect or even morally reprehensible. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9473366141319275, "perplexity_flag": "head"}
http://mathoverflow.net/questions/41262/maximal-ideals-of-kx-1-x-2/41275	maximal ideals of $k[x_1,x_2,…]$ Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) What can be said about the structure of maximal ideals of $R=k[\{x_i\}_{i \in I}]$, or geometric properties of $\text{Spm } k[\{x_i\}_{i \in I}]$? Here $k$ is an arbitrary field and $I$ is an infinite set. Kernels of evaluation homomorphisms yield an injective map $\overline{k}^I / Aut(\overline{k}/k) \to \text{Spm } k[\{x_i\}_{i \in I}]$. The image consists of those maximal ideals whose residue field is algebraic over $k$. If $I$ is finite, every residue field is algebraic (Noether Normalization). However, if $I$ is infinite and $\|I\| \geq \|k\|$, for example $k(t)$ is a residue field which is not algebraic. What happens if $\|k\| > \|I\|$? Is there a description in the general case? - Is this the one where $\operatorname{Spec}(R)\overset{homeo}{\cong} \omega + 1$? – Harry Gindi Oct 6 2010 at 12:56 1 @Harry: This would mean that $R$ has many many idempotents. – Martin Brandenburg Oct 6 2010 at 13:49 Yeah, I didn't really look at the problem for anything more than a second. Just notation-wise, it looks like a problem I did where you adjoin a whole bunch of idempotents. – Harry Gindi Oct 6 2010 at 16:14 1 Answer If $\|k\| > \|I\|$ then the usual cheap proof of Nullstellensatz still works: let $K$ be a residue field. Then $\dim_k K \le \dim_kR = \|I\|$, but if $t\in K$ is transcendental over $k$, the elements $1/(t-a)$ for $a\in k$ are $k$-linearly independent. So $K/k$ is algebraic. - Nice! I accept your answer, since I think it's unlikely that there is a description in the general case. However, special cases such as $\text{Spm} \mathbb{Q}[x_1,x_2,...]$ would be interesting. – Martin Brandenburg Oct 6 2010 at 14:46 @Martin: thanks! Your example for $\|I\|\ge\|k\|$, strikingly illustrates another reason to be very careful working with non-noetherian schemes. – Tony Scholl Oct 6 2010 at 18:33	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9209638237953186, "perplexity_flag": "head"}
http://nrich.maths.org/2382&part=Ben's%20Game	### Pebbles Place four pebbles on the sand in the form of a square. Keep adding as few pebbles as necessary to double the area. How many extra pebbles are added each time? ### Adding All Nine Make a set of numbers that use all the digits from 1 to 9, once and once only. Add them up. The result is divisible by 9. Add each of the digits in the new number. What is their sum? Now try some other possibilities for yourself! ### GOT IT Now For this challenge, you'll need to play Got It! Can you explain the strategy for winning this game with any target? # Ben's Game ##### Stage: 3 Challenge Level: Ben, Jack and Emma were playing a game with a box of $40$ counters - they were not using all of them. They each had a small pile of counters in front of them. All at the same time Ben passed a third of his counters to Jack, Jack passed a quarter of his counters to Emma and Emma passed a fifth of her counters to Ben. They all passed on more than one counter. After this they all had the same number of counters. How many could each of them have started with? The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice. More information on many of our other activities can be found here.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9709700345993042, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php?title=Projection_of_a_Torus&diff=28277&oldid=3405	# Projection of a Torus ### From Math Images (Difference between revisions) \| \| \| \| \| \|----------------------------------------\|----------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\|----------------------------------------------------------\|-------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------\| \| \| \| Current revision (13:10, 29 November 2011) (edit) (undo) \| \| \| (19 intermediate revisions not shown.) \| \| \| \| \| Line 1: \| \| Line 1: \| \| \| - \| {{Image Description \| + \| {{Image Description Ready \| \| - \| \|ImageName=Projection of a Torus \| + \| \|ImageName=4-Dimensional Torus \| \| \| \|Image=4dtorus.jpg \| \| \|Image=4dtorus.jpg \| \| - \| \|ImageIntro=A four-dimensional torus projected into three-dimensional space. \| + \| \|ImageIntro=A torus in four dimensions projected into three-dimensional space. \| \| - \| \|ImageDescElem=It is impossible to visualize a complete four-dimensional object, since we have only ever lived in three-dimensional space. However, there are ways to capture parts of the four-dimensional object in three-dimensional space. A useful analogy is a world map. We can capture the essence of the three-dimensional globe on a two-dimensional map, but only by using a projection, which distorts the three-dimensional object in some way to fit on a two-dimensional surface. \| + \| \|ImageDescElem=It is impossible to visualize an object in four-dimensions, since we have only ever lived in three-dimensional space. However, there are ways to capture features of the four-dimensional object in three-dimensional space. \| \| \| \| \| \| \| - \| A similar process is carried out to create this page's main image. A four-dimensional object, described further below, is projected into three-dimensions using two different projections. \| + \| A useful analogy is a world map. We can capture the essence of the three-dimensional globe on a two-dimensional map, but only by using a projection, which translates a three-dimensional object onto a two-dimensional surface at the expense of distorting the object in some way. \| \| - \| \|ImageDesc=The four-dimensional object is defined parametrically by <math> (x_1,x_2,x_3,x_4)=(cos(u),sin(u),cos(v),sin(v)) </math> \| + \| \| \| \| \| + \| A similar process is carried out to create this page's main image. An object in four-dimensional space, described further below, is projected into three-dimensions using two different projections. \| \| \| \| + \| \| \| \| \| + \| \|ImageDesc= A [[torus]] is commonly known as the surface of a doughnut shape. It can be described using [[Parametric Equations\|parametric equations]]. While it is a <balloon title="Other examples of two dimensional surfaces are planes and the surface of a sphere. They are surfaces in the plain English sense of the word"> two dimensional surface </balloon>, it lives in three dimensional space. \| \| \| \| + \| \| \| \| \| + \| A four-dimensional torus is an analogous object that lives in four dimensional space. The main image contains two images which ways of visualizing a four dimensional torus in three dimensions. \| \| \| \| + \| \| \| \| \| + \| The four-dimensional torus is defined [[Parametric Equations\|parametrically]] by <math> (x_1,\,x_2,\,x_3,\,x_4)=(cos(u),\,sin(u),\,cos(v),\,sin(v)) </math>. The first two coordinates of the parametrization give a circle in u-space, and the second two coordinates give a circle in v-space. The torus is thus the [[Cartesian Product]] of two circles. \| \| \| \| + \| \| \| \| \| + \| A [[Stereographic Projection\| stereographic projection]] is used to map this object, which lives in four-dimensional space, into three-dimensional space, using a projection point of <math> (0,0,0,\sqrt{2})</math> for the first object in this page's main image. This projection is centered above the object, projecting the symmetric torus into three-dimensional space. For the second object, the projection point is shifted to be closer to one part of the four-dimensional object than the other, creating an uneven object in three dimensions. This projection's unevenness is similar to the shadow of a symmetric object becoming asymmetric because of the light source's positioning. \| \| \| \|AuthorName=Thomas F. Banchoff \| \| \|AuthorName=Thomas F. Banchoff \| \| - \| \|AuthorDesc=Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967. \| + \| \|AuthorDesc=Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967. \| \| \| \|SiteName=The Mathematics of In- and Outside the Torus \| \| \|SiteName=The Mathematics of In- and Outside the Torus \| \| \| \|SiteURL=http://www.math.brown.edu/~banchoff/art/PAC-9603/tour/torus/torus-math.html \| \| \|SiteURL=http://www.math.brown.edu/~banchoff/art/PAC-9603/tour/torus/torus-math.html \| \| \| \|Field=Algebra \| \| \|Field=Algebra \| \| - \| \|InProgress=Yes \| + \| \|References=http://www.math.brown.edu/~banchoff/art/PAC-9603/tour/torus/torus-math.html \| \| \| \| + \| \|InProgress=No \| \| \| }} \| \| }} \| ## Current revision 4-Dimensional Torus Field: Algebra Image Created By: Thomas F. Banchoff Website: The Mathematics of In- and Outside the Torus 4-Dimensional Torus A torus in four dimensions projected into three-dimensional space. # Basic Description It is impossible to visualize an object in four-dimensions, since we have only ever lived in three-dimensional space. However, there are ways to capture features of the four-dimensional object in three-dimensional space. A useful analogy is a world map. We can capture the essence of the three-dimensional globe on a two-dimensional map, but only by using a projection, which translates a three-dimensional object onto a two-dimensional surface at the expense of distorting the object in some way. A similar process is carried out to create this page's main image. An object in four-dimensional space, described further below, is projected into three-dimensions using two different projections. # A More Mathematical Explanation [Click to view A More Mathematical Explanation] A torus is commonly known as the surface of a doughnut shape. It can be described using [[Paramet [...] [Click to hide A More Mathematical Explanation] A torus is commonly known as the surface of a doughnut shape. It can be described using parametric equations. While it is a two dimensional surface , it lives in three dimensional space. A four-dimensional torus is an analogous object that lives in four dimensional space. The main image contains two images which ways of visualizing a four dimensional torus in three dimensions. The four-dimensional torus is defined parametrically by $(x_1,\,x_2,\,x_3,\,x_4)=(cos(u),\,sin(u),\,cos(v),\,sin(v))$. The first two coordinates of the parametrization give a circle in u-space, and the second two coordinates give a circle in v-space. The torus is thus the Cartesian Product of two circles. A stereographic projection is used to map this object, which lives in four-dimensional space, into three-dimensional space, using a projection point of $(0,0,0,\sqrt{2})$ for the first object in this page's main image. This projection is centered above the object, projecting the symmetric torus into three-dimensional space. For the second object, the projection point is shifted to be closer to one part of the four-dimensional object than the other, creating an uneven object in three dimensions. This projection's unevenness is similar to the shadow of a symmetric object becoming asymmetric because of the light source's positioning. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. # About the Creator of this Image Thomas F. Banchoff is a geometer, and a professor at Brown University since 1967. # References http://www.math.brown.edu/~banchoff/art/PAC-9603/tour/torus/torus-math.html Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 2, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.831369161605835, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/203367-find-limit-if-exists-print.html	# Find the limit if it exists Printable View • September 12th 2012, 06:48 PM calculus123 Find the limit if it exists I'm having a really difficult time understanding limits from the graph. I don't completely understand how to find if the limit exists or not. I tried to do these for my homework but I don't know if they are right. Could someone please help me understand limits? http://i45.tinypic.com/152hn36.jpg • September 12th 2012, 06:54 PM SworD Re: Find the limit if it exists B and M are incorrect. Can you recheck those and see what the problem is? Also, the graph is a bit unclear as x -> 4... is that supposed to be an asymptote, or do the curves actually stop at $\pm5$? If theres an asymptote, the one sided limits will be $\pm\infty$, the regular two-sided limit won't exist. But I think that's an asymptote, so F and G are wrong too. To answer your question, a one sided limit will exist if the function approaches a particular value (or to infinity), and gets arbitrarily close as you get closer and closer, FROM that direction. The usual two-sided limit will exist ONLY if the left-hand and right-hand limits exist and are equal. The function will be continuous if the limit exists, AND the limit actually equals the function value. Notice that this does actually happen at x=6. Edit: A looks incorrect as well. • September 12th 2012, 06:56 PM MarkFL Re: Find the limit if it exists By my count, you have 4 incorrect. • September 12th 2012, 07:14 PM Soroban Re: Find the limit if it exists Hello, calculus123! Most of your answers are correct . . . good work! Quote: I'm having a really difficult time understanding limits from the graph. I don't completely understand how to find if the limit exists or not. I tried to do these for my homework but I don't know if they are right. http://i45.tinypic.com/152hn36.jpg Your answer for (a) is incorrect. Trace the graph approaching -3 from the left. . . The graph approaches $f(x) =1.$ Trace the graph approaching -3 from the right. . . The graph approaches $f(x) = 1.$ Therefore: . $\lim_{x\to\text{-}3}f(x) \;=\;1$ Your answer for (g) is off. It looks like it goes to 5, . . but I believe there is a vertical asymptote at $x = 4.$ Therefore: . $\lim_{x\to4^+}}f(x) \:=\:+\infty \quad\text{ and }\,\lim_{x\to4^-}f(x) \:=\:-\infty$ Your answer to (m) is incorrect. $f(x)$ is continuous at $x = 6.$ (The graph doesn't "break" there, does it?) • September 12th 2012, 08:01 PM calculus123 Re: Find the limit if it exists Thank you Soroban and SworD! I seem to understand them a little bit better now. I made all the changes that you guys suggested, are they correct now? I tried to redo them and they seem to make more sense now. Thanks! http://i50.tinypic.com/33k4ya1.jpg • September 12th 2012, 08:08 PM SworD Re: Find the limit if it exists Only B is still incorrect. Notice that they want the limit from the right. • September 12th 2012, 08:32 PM calculus123 Re: Find the limit if it exists So the answer to b would be 2 since it's coming from the right? • September 12th 2012, 08:36 PM SworD Re: Find the limit if it exists Yes. Limits don't care at all about the value at that point. • September 13th 2012, 02:56 AM Plato Re: Find the limit if it exists Quote: Originally Posted by SworD Only B is still incorrect. Notice that they want the limit from the right. That in not correct. Quote: Originally Posted by calculus123 So the answer to b would be 2 since it's coming from the right? That in not correct. b) ${\lim _{x \to {2^ + }}}f = 1$. • September 13th 2012, 12:37 PM calculus123 Re: Find the limit if it exists For b), It's ${\lim _{x \to {-2^ + }}}f = 1$, not ${\lim _{x \to {2^ + }}}f = 1$. Are you sure the answer is 1 and not 2? I'm looking at the graph again and it seems like the answer is 2. • September 13th 2012, 02:09 PM Plato Re: Find the limit if it exists Quote: Originally Posted by calculus123 For b), It's ${\lim _{x \to {-2^ + }}}f = 1$, not ${\lim _{x \to {2^ + }}}f = 1$. Are you sure the answer is 1 and not 2? I'm looking at the graph again and it seems like the answer is 2. The image you posted is very hard to read. You should try to make questions clear to the reader. • September 13th 2012, 02:32 PM SworD Re: Find the limit if it exists It's pretty clear that the number being approached in (b) is -2... unless they decided to make the -> arrow twice as big and have a wedge drawn on it. • September 13th 2012, 02:42 PM Plato Re: Find the limit if it exists Quote: Originally Posted by SworD It's pretty clear that the number being approached in (b) is -2... unless they decided to make the -> arrow twice as big and have a wedge drawn on it. It may be different on different computer screens. On mine it appears $\to 2$ as opposed to $\to -2$. I cannot understand why any serious user of this site does not learn to use LaTeX. • September 13th 2012, 07:16 PM calculus123 Re: Find the limit if it exists Quote: Originally Posted by Plato It may be different on different computer screens. On mine it appears $\to 2$ as opposed to $\to -2$. I cannot understand why any serious user of this site does not learn to use LaTeX. I don't blame you for not seeing the negative sign because I have the sheet in front of me printed and I can't see it either. Thanks for your help! All times are GMT -8. The time now is 01:08 PM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 18, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9434118270874023, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=215287	Physics Forums ## Finding the power supplied by a battery I have a multiple loop circuit, which I was able to apply Kirchhoff's rules to in order to find the current of each resister. However, got stuck trying to determine the power supplied by the batteries (one in each loop). I tried using the equation p= I $$\epsilon$$, where $$\epsilon$$ is the volts of the battery and I is volts of the battery divided by equivalent resistance of the loop that I am solving. Something seems to be wrong in my calculations, and I am not quite sure what I am doing wrong. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 3 Quote by snoweangel27 I have a multiple loop circuit, which I was able to apply Kirchhoff's rules to in order to find the current of each resister. However, got stuck trying to determine the power supplied by the batteries (one in each loop). I tried using the equation p= I $$\epsilon$$, where $$\epsilon$$ is the volts of the battery and I is volts of the battery divided by equivalent resistance of the loop that I am solving. Something seems to be wrong in my calculations, and I am not quite sure what I am doing wrong. You probably did your nodal analysis wrong. If you calculated the currents correctly then you can use $$P=IV=I(IR)=I^2R$$ . Note that V is the power across the resister. Recognitions: Homework Help Note that battery voltage and total capacity will drop as the load increases. Recognitions: Science Advisor ## Finding the power supplied by a battery Note that battery voltage and total capacity will drop as the load increases I suspect that he's using the term "battery" where he really means ideal DC voltage source. I is volts of the battery divided by equivalent resistance of the loop that I am solving Well that sounds like the problem right there, why the heck would you do that? Why did you even bother calculating all the currents (presumably) correctly if you were then going to just ignore them and do something silly like what you stated above? The correct value of I to use here is the actual value of I flowing through the voltage source (or battery) in question. You can use Kirchovs current law to find the current in each voltage source from the other currents you have calculated. Thread Tools Similar Threads for: Finding the power supplied by a battery Thread Forum Replies Electrical Engineering 8 Introductory Physics Homework 5 Classical Physics 21 General Physics 12 Introductory Physics Homework 3	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9534509181976318, "perplexity_flag": "middle"}
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http://mathhelpforum.com/advanced-statistics/157168-probability-random-k-digit-number-contains-least-one-0-1-2-a.html	# Thread: 1. ## Probability that a random k-digit number contains at least one 0, 1, and 2 Let $k\geq 3$ be any given integer. What is the probability that a random k-digit number will have at least one 0, at least one 1, and at least one 2? I'm thinking that this could be answered via the principle of inclusion-exclusion, but I could use some clarification (or possible correction). Here's what I've got so far. Assuming that leading zeroes are not allowed, # of k-digits containing no 0’s: $9^k$ # of k-digits containing no 1’s: $8×9^{k-1}$ # of k-digits containing no 2’s: $8×9^{k-1}$ # of k-digits containing no 0’s and no 1’s: $8^k$ # of k-digits containing no 0’s and no 2’s: $8^k$ # of k-digits containing no 1’s and no 2’s: $7×8^{k-1}$ # of k-digits containing no 0’s, 1’s or 2’s: $7^k$ I then applied the following: $\|A\cup B\cup C\|=\|A\|+\|B\|+\|C\|-\|A\cap B\|-\|A\cap C\|-\|B\cap C\|+\|A\cap B\cap C\|$ This would obtain the complement set of what I'm looking for, so this would be subtracted from $9×10^{k-1}$. Am I on the right track? 2. You are. 3. Thanks for confirming that. I currently have the answer written out as follows: $P=(9×10^{k-1} )-[9^k+2(8×9^{k-1})-2(8^k)-(7×8^{k-1})+7^k]$ This probably isn't the cleanest way to show the answer. Is there a more condensed version I could use? 4. There is probably no shorter formula for this case.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8994777798652649, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?t=580561	Physics Forums Page 1 of 2 1 2 > ## how to count elements say you are multiplying integers from 1 to 1000 and you want to count the number of "unique" elements. How would you do that? Is there a closed form expression for that? counting the unique or non-unique elements is the same since we know the total number of elements and can get the unique or non-unique by simple substraction. an example of unique element is 23 because there is only one way to generate it by multiplication. But 24 is a non-unique element since there are few ways to generate it (38, 212, 64...) and you don't want to count it in as many ways as you can generate it or you simply want to count it once. In other words, a 1010 matrix will have 100 elements but we want to know how many of those 100 are unique and how many are non unique. PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Hey there! It seems to me that your definition of unique is lacking something. why would 6 be unique? 23 and 16 are two ways to generate it. if "unique" means "there is only one combination of numbers that multiplied generate this number" then you'd be looking for prime numbers (every non-prime number has at least two ways to generate it) Quote by bluekuma Hey there! It seems to me that your definition of unique is lacking something. why would 6 be unique? 23 and 16 are two ways to generate it. if "unique" means "there is only one combination of numbers that multiplied generate this number" then you'd be looking for prime numbers (every non-prime number has at least two ways to generate it) You are right of course about 6. 16 is discounted because it applies to every number. I am interested in counting non unique number of the form, with p primes p1p2p3p4...pn=p1(p2p3...pn)=(p1p2)(p2p3...pn)=.... my question is how many ways ( exactly not just order of magnitude ) are there to generate such a number ( as a function of n ) Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus ## how to count elements Are you asking for the prime factorization of 1000! ??? See homepage.smc.edu/kennedy_john/NFACT.PDF Quote by micromass Are you asking for the prime factorization of 1000! ??? See homepage.smc.edu/kennedy_john/NFACT.PDF nope. take this number as an example: 375=105. This number can also be written as: 215=105 and 335=105, and 715=105. So in a matrix of say 40x40, the number 105 will appear 3 times ( forget the 1105). I do not want to count it 3 times because the value is the same, 105, but there are three ways to produce this number. The total number of matrix elements is 4040=1600 but the total number of "unique numbers" is less because it makes no sense counting an element with the same value 3 times ( or as many times as it appears in the matrix). My question is how do we count the number of non unique numbers like 105. Is there a closed form expression for those elements? Blog Entries: 2 Quote by epsi00 nope. take this number as an example: 375=105. This number can also be written as: 215=105 and 335=105, and 715=105. So in a matrix of say 40x40, the number 105 will appear 3 times ( forget the 1105). I do not want to count it 3 times because the value is the same, 105, but there are three ways to produce this number. The total number of matrix elements is 4040=1600 but the total number of "unique numbers" is less because it makes no sense counting an element with the same value 3 times ( or as many times as it appears in the matrix). My question is how do we count the number of non unique numbers like 105. Is there a closed form expression for those elements? If i understand you correctly, the numbers ab are unique if and only if b is always a prime greater than or equal to than a, and a is either prime or 1. In other words 1,2,3,4,5,6,7,9,10,11,13... are unique. So just count the number of primes less than 1000 then multiply that number by that number plus 1. Edit If you want to denote the cube of a prime as unique , say 8 = 24, then add to the previous number the number of primes less than 100. 2nd edit: If one can say 37 is not unique since it also equals 73 since you can't order the pair by size then the number of unique elements is the number of primes under 1000. That is by counting all numbers pp as the only unique elements. We can perhaps interpret the question as follows: Let u(n) be the cardinality of the set {ab: a,b in {1,2,...,n}}. For example, u(3)=6 and u(7)=25. Is there some kind of general formula, or a clever way to determine u(n)? In particular, what is u(1000)? Quote by Norwegian We can perhaps interpret the question as follows: Let u(n) be the cardinality of the set {ab: a,b in {1,2,...,n}}. For example, u(3)=6 and u(7)=25. Is there some kind of general formula, or a clever way to determine u(n)? In particular, what is u(1000)? I think the word "unique" is being interpreted two different ways. I agree with you, the number of unique products is what the word "unique" should mean here. So if 30 = 2x15 = 3x10 we only count it once. In this case it's easy to see that these are just all numbers between 1 and n, plus the number of composites between n and n^2. (Because you can't generate any of the primes between n and n^2 as a product of numbers less than n, but you can so generate ALL of the composites). But I believe the OP means something else. He's calling a product "unique" if it only has one expression. He already said he means 6 = 2x3 is a "unique" because it can only be expressed one way. (OP rejects 1x6) So the question is now reduced to how many numbers less than n^2 have at least three distinct prime factors. Also, OP has not said what to do about exact powers such as 2^4 = 28 = 44. I think there's enough ambiguity in what the OP is asking to create some confusion in this thread. Quote by SteveL27 In this case it's easy to see that these are just all numbers between 1 and n, plus the number of composites between n and n^2. (Because you can't generate any of the primes between n and n^2 as a product of numbers less than n, but you can so generate ALL of the composites). I just want to remark that the statement in the above paragraph is wrong. All composites up to n2 can certainly not be written as a product of two numbers n or less. (n=3, 8= ?) Quote by Norwegian I just want to remark that the statement in the above paragraph is wrong. All composites up to n2 can certainly not be written as a product of two numbers n or less. (n=3, 8= ?) Yes thanks for the correction. Blog Entries: 2 Quote by epsi00 nope. take this number as an example: 375=105. This number can also be written as: 215=105 and 335=105, and 715=105. So in a matrix of say 40x40, the number 105 will appear 3 times ( forget the 1105). I do not want to count it 3 times because the value is the same, 105, but there are three ways to produce this number. The total number of matrix elements is 4040=1600 but the total number of "unique numbers" is less because it makes no sense counting an element with the same value 3 times ( or as many times as it appears in the matrix). My question is how do we count the number of non unique numbers like 105. Is there a closed form expression for those elements? To get an understanding of what you want, let me set forth my understanding. You want to count all the numbers that can be obtained by multiplying two integers each of which are less than 1001, and you don't want to count a given number more than once. When you keep talking of a 10 by 10 matrix, 40 by 40 matrix, I envision a multiplication table. I think you are in fact referring to an a 1000 by 1000 matrix, which is a table for multiplication by numbers from 1 to 1000. In such a table primes are not unique because they appear twice. On column 1 as P1 and on row 1 as 1P. Composites of two or more primes also appear twice, since AB appears in row A, column B and in row B, column A. The only unique elements are 1 and the squares of prime numbers greater than 31 and less than 1000 and certain other squares e.g. 1000000. So 3737 lies on the diagonal and is unique because it is greater than 1000. 3131 is not unique since it appears in the multiplication table three times: 1961, 961* 1 and 3131. I don't understand how to count the non-unique numbers only once though. The numbers range up to 1000000 but certainly do not include every number up to 1000000 such as primes greater than 1000. Blog Entries: 2 Quote by SteveL27 I think the word "unique" is being interpreted two different ways. I agree with you, the number of unique products is what the word "unique" should mean here. So if 30 = 2x15 = 3x10 we only count it once. In this case it's easy to see that these are just all numbers between 1 and n, plus the number of composites between n and n^2. (Because you can't generate any of the primes between n and n^2 as a product of numbers less than n, but you can so generate ALL of the composites). Of course you dont mean to include composites of primes between n and (n^2)/2 as those also can not be counted since no product of two numbers less than a prime P can equal 2P, 3P etc. Quote by Norwegian We can perhaps interpret the question as follows: Let u(n) be the cardinality of the set {ab: a,b in {1,2,...,n}}. For example, u(3)=6 and u(7)=25. I believe the OP's intention was to exclude products by 1; maybe a,b should be taken from {2,3,...,n}. Other than that, this was also how I interpreted the OP's request. An alternative formulation of the same thing could be this (though I don't know if it helps towards a solution): How many numbers 'k' between 4 and n2 have a divisor d<=n, such that k/d is also <= n? I will try to rephrase my question in the hope that we get an answer. Take an nn matrix (products of integers not necessarily primes ), how many of those numbers are distinct. We know the total number of matrix elements is n^2 but we also know that the number of distinct numbers ( out of the n^2) is m<n. The important word here is Distinct. We are not allowed to count any number more than once even it occurs more than once. Is there a way to calculate m or ( n-m) whichever is easier to calculate? OK, so you take an nn-matrix, the entries of which are products of integers, and you ask about how many distinct numbers are to be found in the matrix. Now, you have posted four times in this thread, and you have yet to inform us what those entries are. In particular, what is the entry in row i, colomn j? Blog Entries: 8 Recognitions: Gold Member Science Advisor Staff Emeritus Maybe you need to give some examples of the matrix you're looking at... Blog Entries: 1 Recognitions: Homework Help Quote by Norwegian Now, you have posted four times in this thread, and you have yet to inform us what those entries are. In particular, what is the entry in row i, colomn j? It's ij Here's a small matrix example of what he's talking about [tex] \left( \begin{array}{ccc} 11 & 12 & 13 \\ 21 & 22 & 23 \\ 31 & 32 & 33 \end{array} \right) [/tex] The question can be rephrased as: $$u(n) = \| \{ a*b \\|\ a\leq n,\ b\leq n \}\|$$ calculate u(n). Page 1 of 2 1 2 > Thread Tools \| \| \| \| \|--------------------------------------------\|------------------------------------\|---------\| \| Similar Threads for: how to count elements \| \| \| \| Thread \| Forum \| Replies \| \| \| Precalculus Mathematics Homework \| 2 \| \| \| Linear & Abstract Algebra \| 2 \| \| \| Atomic, Solid State, Comp. Physics \| 3 \| \| \| Chemistry \| 8 \| \| \| Linear & Abstract Algebra \| 0 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9402077794075012, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/10474?sort=newest	## Killing Chern classes ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $G$ be a compact connected Lie group and let $E\to B$ be a principal $G$-bundle. Suppose $a$ is a rational cohomology class of $E$ such that its pullback $b$ under an orbit inclusion map $G\to E$ is one of the standard multiplicative generators of $H^{{\bullet}}(G,\mathbf{Q})$ . Let $E'=E\times EG/G$ be the Borel construction (corresponding to the action of $G$ on $E$) and let $(E^{pq}_r,d_r)$ be the Leray spectral sequence corresponding to the fiber bundle $E'\to BG$. The class $a$ gives an element $a'\in E^{0,2k-1}_2$ for some $k$. Assume that $d_i(a')=0,i< 2k$. Is it true that $d_{2k}(a')$ is what has remained in $E_{2k}$ of the multiplicative generator of $H^{{\bullet}}(BG,\mathbf{Q})$ corresponding to $b$? For simplicity one can assume $G=U(n)$, in which case what remains in $E_\infty$ of the generator of $H^{{{\bullet}}}(BG,\mathbf{Q})\cong E^{{\bullet},0}_2$ corresponding to $b$ is precisely the $k$-th Chern class of $E$, under the natural isomorphism $H^{{\bullet}}(E',\mathbf{Q})\cong H^{{\bullet}}(B,\mathbf{Q})$. This is probably standard, but for some reason I don't see how to prove it nor can construct a counter-example off hand. upd: here is a weaker version, which would be easier to (dis)prove: take $G=U(n)\times H$ where $H$ is another Lie group and suppose that the pullback of $a$ to $G$ is the canonical generator of $H^{\bullet}(U(n),\mathbf{Q})\subset H^{\bullet}(G,\mathbf{Q})$ in degree $2k-1$. Is it true that $d_{2k}(a')$ is mapped to zero under the mapping of the spectral sequences induced by the pullback of $E'$ to $BH$ i.e. by the map $$(E\times EG)/H\to (E\times EG)/G$$ To prove this it would suffice, of course, to show that $d_{2k}(a')$ is represented in $E_2$ by a class in $$H^{\bullet}(BG,\mathbf{Q})\cong H^{\bullet}(BU(n),\mathbf{Q})\otimes H^{\bullet}(H,\mathbf{Q})$$ that is mapped to zero under $H^{\bullet}(BG,\mathbf{Q})\to H^{\bullet}(BH,\mathbf{Q})$. - ## 1 Answer The inclusion $G \to E$ induces a map $EG \to E'$ of spaces over $BG$, where the map of fibers is $G \to E$, and there is a map backwards of Serre spectral sequences. Because $a$ lifts a standard generator of the cohomology of $G$, and $a'$ is a cycle up to the `$E_{2k}$`-page, the differential of this element maps to the differential of the standard generator in the spectral sequence $H^p(BG;H^q(G)) \Rightarrow H^{p+q}()$. The `$d_{2k}$`-differential on this class in $H^{2k-1}(G)$ is the "corresponding generator" in $H^{2k}(BG)$. - Thanks, Tyler! The map $EG\to E'$ induces an isomorphism of the $q=0$ rows of thet $E_2$ pages of the spectral sequences: both are just the $H^(BG,\mathbf{Q})$. Why is the map of the $E_{2k}$ pages injective on the zero row? (If it isn't, then we can't deduce the image of $a'$ in $E^{2k,0}_{2k}$ from its image in the spectral sequence of $EG\to BG$.) Or did you mean something else? – algori Jan 2 2010 at 18:22 1 Right, but "the" corresponding element, as I usually understand it, in the cohomology of BG is not well-defined - it's only unique up to the indeterminacy from the images of differentials in the spectral sequence associated to G -> EG -> BG. Did you have some specific corresponding element in mind? – Tyler Lawson Jan 2 2010 at 19:57 For $G$ a product of $U(n)$'s (the case I'm mainly interested in) it is possible to fix the generators of $H^*(BG,\mathbf{Q})$ as follows: for a single $U(n)$ the generator in degree $2l$ is the $l$-th Chern class of the tautological bundle. These are well defined if the Chern classes are required to satisfy the usual axioms (the only ambiguity is the sign of the $c_1$ of the tautological bundle on $\mathbf{P}^1(\mathbf{C})$). Then, if we have a product of several $U(n)$'s, we take the tensor products of the generators constructed above. There should be something similar should for general $G$ – algori Jan 2 2010 at 21:55 Actually, I don't have to be that rigid about the generators. I've updated the posting including a weaker version of the problem. Very roughly speaking, it says that if $G$ is a product and $a$ comes from one factor, then its differential is zero restricted to $B$ of the other factor. – algori Jan 3 2010 at 0:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 71, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9361343383789062, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/35541/in-copenhagen-can-this-idea-preserve-locality-for-bell-inequalities	# In Copenhagen, can this idea preserve locality for Bell inequalities? Generate an entangled pair of qubits. Send to Alice and Bob far away from each other. Both measure along basis in one of two possible orientations. The result is sent to Charlie at some later time, who compares the corellations and concludes the Bell inequalities have been violated. But in Copenhagen, Charlie can say, the pair of results didn't become real until I collected both observations from Alice and Bob. They weren't real when either of them measured them. They only became real once I observed them. Then, both results, when they materialized, materialized at the same place. So, no nonlocality? - Maybe Charlie can say the results aren't real yet if he believes that Alice and Bob were still in superposition states until they report their findings to him (like Wigner's friend). But I don't think that Alice and Bob, who have seen the outcome of the measurement apparatus, would be of the opinion that the measurements aren't real yet. But I've never been completely clear on the exact beliefs of the Copenhagenists. – Dan Stahlke Sep 4 '12 at 1:41 ## 3 Answers Yes, this restores locality, and is sometimes called "Solipsistic Copenhagen". It is described on the first page of Everett's thesis and paper on many worlds. The solipsistic Copenhagen interpretation is the only you can collapse the wavefunction, everyone else is in superposition. From this interpretation to many worlds is only the step of moving the collapsing entity away from youself, and noting there is no contradiction with experiment, so the whole collapse thing is not really required, and one can logical positivistically remove it from the theory, at the cost of saying that the notion of reality and definite events is not physics, but how the branches get selected by the memories of observers. It moves the question away from physics to philosophy of mind. While I think this is a neat trick, it might not be nature's trick. One has to be sure quantum mechanics is exact before declaring it so. - It depends how they communicated the results to Charlie. Compenhagen interpretation makes a distinction between the classical and quantum world. So if they communicated the results to Charlie using a classical device then a Copenhagenist Charlie would have to conclude that the results must have been real at the time Alice and Bob sent them. On the other hand, if they sent the results using quantum communication, then they only became real when he measured the received system. My conclusion from this is that a Copenhagenist can still demonstrate nonlocality by communicating results classically. - If it's really all just a manifestation in Charlie's mind (What!? Is He God, or what?), then why can Alice and Bob only score at most ${1\over 2} + {1\over 2\sqrt 2}$ in the CHSH game? Why would Charlie impose such a limitation on himself? OK, let's just suppose for the moment that Charlie has self-imposed limitations on his mind. Wouldn't that limitation be none other than "quantum locality" by another name? -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9643940925598145, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/179240-hahn-banach-theorem-normed-spaces.html	# Thread: 1. ## Hahn-Banach Theorem (Normed Spaces) Dear Colleagues, I have a problem in the attachment and could you please help me in solving it. Best Regards. Raed. Attached Files • Hahn-Banach Theorem _Normed Spaces_.pdf (38.2 KB, 21 views) 2. Since $\mathbb{R}^n$ is an inner-product space, every linear functional is given by an inner product. In fact, if $\alpha = (\alpha_1,\alpha_2)$ then the functional $f(x) = \alpha_1\xi_1+\alpha_2\xi_2$ is given by $f(x) = \langle x,\alpha\rangle$. Use that fact to show that $\\|f\\|_{\mathbb{R}^2} = \sqrt{\alpha_1^2+\alpha_2^2}$. Now that you know $\\|f\\|$, you have to figure out how to extend f to a linear functional on $\mathbb{R}^3$ without increasing the norm. 3. Thank you very much for your reply I have already proved the norm of f. But the problem is in finding a linear extension of f that preserves the norm. Best Regards. 4. Originally Posted by raed I have already proved the norm of f. But the problem is in finding a linear extension of f that preserves the norm. So the extension to $\mathbb{R}^3$ must be of the form $\tilde{f}(x) = \alpha_1\xi_1+\alpha_2\xi_2+\alpha_3\xi_3$, and its norm will be $\\|\tilde{f}\\| = \sqrt{\alpha_1^2+\alpha_2^2+\alpha_3^2}.$ That leaves only one choice for $\alpha_3.$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9283086061477661, "perplexity_flag": "head"}
http://mathoverflow.net/questions/60667/presenting-the-hyperfinite-ii-1-factor	## Presenting the Hyperfinite II_1 Factor ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It's well known that all hyperfinite $\mathrm{II}_1$ factors are isomorphic. I risk the wrath of MathOverflow elders to ask if a particular isomorph is easier than others to handle. In particular, is there a presentation for which it's obvious that the commutant is also the hyperfinite $\mathrm{II}_1$ factor (presented in the same way)? - ## 2 Answers Another classical construction is to see the hyperfinite $\mathrm{II}_1$ factor as the infinite tensor product of the two by two matrices $$\mathcal{R}=\otimes_{n=1}^{\infty}{M_{2}(\mathbb{C})}$$ acting on its $L^{2}$ closure $L^{2}\Big(\otimes_{n=1}^{\infty}M_{2}(\mathbb{C})\Big)$ by right multiplication. Then its commutant is given by the left multiplication. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I suppose what you mean is that you want the commutant of the hyperfinite $II_{1}$ coming from a certain construction to have arisen from essentially the same construction. The closest I can think of to this is the classic one: if you take the left group von Neumann algebra of the group of those permutations of $\mathbb{Z}$ having finite support, then the commutant will have arisen as the right group von Neumann algebra of this group. - Or any discrete ICC amenable group, but I agree that your example is probably the easiest and most natural of such - as far as I know – Yemon Choi Apr 5 2011 at 19:13	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9492089152336121, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/29546/p-value-vs-confidence-interval-obtained-in-bootstrapping	p-value vs. confidence interval obtained in Bootstrapping I ran a simple randomized experiment with 1 control and 1 experimental condition (final N = 80). The dependent variable (frequency of a shown behavior) is clearly not normally distributed so I thought about bootstrapping my analysis (an independent t-test). The t-test without bootstrapping resulted in a significant effect between the conditions (p < .05). Using SPSS, the p-value based on bootstrapping (5000 resamples) is only marginally significant (p < .10). However, the 95% confidence interval does not include zero. I just apply statistical methods and sometimes I don't know what is actually right (unfortunately!). But that's why I ask this question. When I tried to learn how Bootstrapping works, I thought that one has to look at the confidence intervals to detect whether an effect is not zero. In my example above, the 95% CI does not accompany with the bootstrapped p-value. So I don't know whether I should report the bootstrapped CI and/or the bootstrapped p-value and/or the typical unbootstrapped p-value. What would you say? - I am not an SPSS expert, but to me it looks like it will be difficult to help you if people do not know what is your data, how you bootstrapped it and what score you measured in the end. – gui11aume May 31 '12 at 9:46 2 Answers I am not a real Bootstrap expert, but I can tell you about the two main things: 1. Bootstrap confidence intervals are usually more robust and accurate then the ones estimated without bootstrap. 2. If you estimate the parameter with bootstrap, your confidence interval (CI) usually evaluated in a different way then in a regular t-test. For example, in a regular case CI is $[ \hat{\theta} - \hat{q}_{1-\alpha/2}, \hat{\theta} + \hat{q}_{\alpha/2} ]$ (here $\hat{\theta}$ is an estimate of the parameter, $\hat{q}_{\alpha}$ is an $\alpha$-quantile). But for bootstrap it is $[ \hat{\theta} - \hat{q}_{1-\alpha/2}, \hat{\theta} - \hat{q}_{\alpha/2} ]$ (minus sign in both cases). From that all I would suggest you to recheck whether with this formulas bootstrap CI accompany with the p-value. And if you will find that it's ok now, report bootstrapped results. If no, it is better to ask SPSS experts about how bootstrap works there. - You have many choices for bootstrap confidence intervals. All bootstrap confidence intervals are approximate and do not always do well in small samples (usually 80 is not considered small). also if you read Hall and Wilson's paper you will find that testing hypotheses assuming the bootstrap distribution under the null hypothesis works better than inverting confidence intervals. It is an issue about how to center the pivotal quantity in the test statistics. Schenker in 1985 showed that bootstrap methods such as Efron's percentile method and even the BC method severely under cover the true parameter for certain chi square populations when the sample size is not very large. Chernick and LaBudde in 2010 American Journal of Mathematical and Management Science showed that in small samples there can even be problems with BCa and bootstrap t for highly skewed distributions such as the lognormal. So based on the literature including my own research I suggest doing the hypothesis test with the centering approach recommended by Hall and Wilson and base your conclusions on that p-value. You can find detailed coverage of this in my recent book "An Introduction to the Bootstrap with Applications to R" published by Wiley in 2011. - Since this isn't an academic paper, a link to referenced articles is better than a citation IMO. – Michael McGowan May 31 '12 at 15:29 2 Yes but it also takes very little effort to use the citation on Google to find any link I might give and others. Some of the links might have the paper available to read either for free or for some small fee. – Michael Chernick May 31 '12 at 15:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9199661612510681, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4044527	Physics Forums ## Longest streaks of successes Let's say you have 5 successes in 7 trials. There are a total of 21 different ways to arrange them. How many times will the longest streak of successes be 5, 4, 3, and 2? I figured this out by brute force, but wondered if there was a formula to calculate it, one that could be used for any x successes in y trials. The answers for the example are: 5 in a row: 3 4 in a row: 6 3 in a row: 9 2 in a row: 3 Thanks for any help. Ken PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug It's fairly complicated. See http://mathworld.wolfram.com/Run.html Thank you for the link! I am trying to understand the formulas and adapt them for my purposes. ## Longest streaks of successes For small numbers (say less than 10), it is probably easiest to do it recursively. Even by hand. For somewhat larger numbers, you would need a computer. Take a look at this thread: http://www.physicsforums.com/showthread.php?t=405379 In it the original poster discovers several recursive solutions to a problem that is similar to yours, but slightly different. That might give you some ideas. Thank you, I appreciate the helpful reply. Looking at that older thread, I believe this is going to be too involved for me to tackle. I want to apply this to a baseball season where a team goes 100-62 (but the W-L sequence is unknown), then calculate the probabilities of maximum winning streaks of various lengths. If we change the problem to independent trials with the probability of winning each game at 100/162, there is a great site that calculates this for you: http://www.pulcinientertainment.com/...tor-enter.html There are even Excel sheets to download in the site that show in detail how the results are determined. That is a handy website. If you are still interested in doing the problem as you originally stated it (where the number of successes and the number of trials are given), it is doable. If you are comfortable with counting techniques and binomial coefficients, I believe it is even possible to obtain a formula (kinda nasty looking...but still manageable) in terms of those two given quantities. For fairly large values like 162, it may be easier to have a computer evaluate the formula than to have it solve the problem recursively. If you want to pursue this, I'll try to help. EDIT: I'll go ahead and post the formula I have in mind, and you can decide if it's something you want to look into. Here n is the number of trials, s the number of successes, and R the longest run or streak of successes. Then the number of ways of getting a run of at least r is $$N(R \geq r) = \sum_{b=1}^{B} \binom{n-s+1}{b}\sum_{j=1}^{J}(-1)^{j+1}\binom{b}{j}\binom{s-j(r-1)-1}{b-1}$$ here J = min{b, integer part of s/r} and B = min{s, n-s-1} To find the number of ways N(R=r) where the longest streak is exactly r, you can subtract N(R >= r+1) from N(R >= r). Thanks for the formula! I minored in statistics in college many many years ago, so should be able to make sense of it. I can get estimates of some of these values via simulation, so that will help in checking against what I get from the formula. You're welcome. I hope it's correct. I posted it soon after I was convinced that the argument leading to it made sense, but I didn't try it out on any numbers. I ran 1000 seasons of a baseball team going exactly 100-62, recording the longest winning streak in each. This would be a with replacement experiment. In the table below, the first column lists the maximum streak lengths. The second column lists the actual probabilities of at least that long of a streak for without replacement (probability of winning each game exactly 100/162), which I got from the site linked in post 5. The third column lists the percentages from the 1000 trials. As you would expect, without replacement has a better chance for longer streaks, but for the shorter ones, the results are almost identical. Overall, these numbers are a lot closer than I thought they would be. Code: ``` Random Specific Streak Pct Pct 4 100.0 100.0 5 99.9 99.9 6 98.0 98.0 7 89.9 89.3 8 74.3 73.1 9 55.7 52.5 10 38.8 37.2 11 25.8 24.8 12 16.6 14.4 13 10.5 8.7 14 6.5 5.8 15 4.1 4.0 16 2.5 2.0 17 1.5 1.0 18 0.9 0.6 19 0.6 0.5``` Interesting. For long streaks, which are infrequent, it looks like you can treat it as a Poisson process. If p is the probability of success and q = 1-p, then the expected number of streaks of length at least r in a sequence of n trials is given by $$\lambda = p^r + (n-r)qp^r \approx (n-r)qp^r$$ Then the probability of at least one streak of length r or more is given by $$P(R \geq r) \approx 1-e^{-\lambda}$$ Using p = 100/162, I tried it for r=10 and r = 15 and got answers that agree well with what you got by simulation. here are the values I get with the Poisson approximation (which asssumes independent trials with probability of success p = 100/162 each time): Code: ```Streak length 5 .996 6 .965 7 .873 8 .717 9 .539 10 .378 11 .253 12 .164 13 .104 14 .065 15 .040 16 .025 17 .015 18 .009 19 .006``` I mistakenly thought the Poisson formula was for dependent trials; for independent ones, the results do agree quite closely for streaks of 11 or more. See second column of the chart in post 9 . It appears there is no easy way to calculate or estimate streak probabilities for dependent trials such as the 100-62 baseball season. Yeah, the Poisson approximation works quite well as long as trials are only weakly dependent. For your problem, I think this condition holds for streaks of moderate length, say 10 to 15 or 20. Shorter streaks happen too frequently, and so the occurrence of a streak of length 5 starting on the 10th trial eliminates the possibility of another streak of length 5 starting on the 14th trial, for example (it can't be the start of one because you are already in the middle of one). On the other hand, since we are given the total number of successes, 100, a very long streak like 30 or more means that successes are rare from there on out. So again, the trials are too strongly dependent for the Poisson approximation to apply. Still, I am surprised that the values you gave for dependent and independent trials are as close as they are over such a large range of streak lengths. Are you primarily interested in very long streaks? EDIT: Also, wouldn't independent trials be a better assumption for a season of baseball games, anyway? Assuming that the team was drawing from a bag with a predetermined number of "win" and "lose" marbles in it isn't all that realistic. I'm just curious to know if you are primarily interested in the mathematics, which is pretty interesting itself, or if you are more interested in an application (like wagering?). Thread Tools \| \| \| \| \|---------------------------------------------------\|--------------------------------------------\|---------\| \| Similar Threads for: Longest streaks of successes \| \| \| \| Thread \| Forum \| Replies \| \| \| General Physics \| 2 \| \| \| General Physics \| 3 \| \| \| General Discussion \| 15 \| \| \| Set Theory, Logic, Probability, Statistics \| 18 \| \| \| Current Events \| 65 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9529234170913696, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Matrix_calculus	# Matrix calculus Calculus Definitions Concepts Rules and identities Integral calculus Definitions Integration by Formalisms • Matrix • Tensor • Exterior • Geometric Definitions Specialized calculi In mathematics, matrix calculus is a specialized notation for doing multivariable calculus, especially over spaces of matrices. It collects the various partial derivatives of a single function with respect to many variables, and/or of a multivariate function with respect to a single variable, into vectors and matrices that can be treated as single entities. This greatly simplifies operations such as finding the maximum or minimum of a multivariate function and solving systems of differential equations. The notation used here is commonly used in statistics and engineering, while the tensor index notation is preferred in physics. Two competing notational conventions split the field of matrix calculus into two separate groups. The two groups can be distinguished by whether they write the derivative of a scalar with respect to a vector as a column vector or a row vector. Both of these conventions are possible even when the common assumption is made that vectors should be treated as column vectors when combined with matrices (rather than row vectors). A single convention can be somewhat standard throughout a single field that commonly use matrix calculus (e.g. econometrics, statistics, estimation theory and machine learning). However, even within a given field different authors can be found using competing conventions. Authors of both groups often write as though their specific convention is standard. Serious mistakes can result when combining results from different authors without carefully verifying that compatible notations are used. Therefore great care should be taken to ensure notational consistency. Definitions of these two conventions and comparisons between them are collected in the layout conventions section. ## Scope Matrix calculus refers to a number of different notations that use matrices and vectors to collect the derivative of each component of the dependent variable with respect to each component of the independent variable. In general, the independent variable can be a scalar, a vector, or a matrix while the dependent variable can be any of these as well. Each different situation will lead to a different set of rules, or a separate calculus, using the broader sense of the term. Matrix notation serves as a convenient way to collect the many derivatives in an organized way. As a first example, consider the gradient from vector calculus. For a scalar function of three independent variables, $f(x_1, x_2, x_3)$, the gradient is given by the vector equation $\nabla f = \frac{\partial f}{\partial x_1} \hat{x_1} + \frac{\partial f}{\partial x_2} \hat{x_2} + \frac{\partial f}{\partial x_3} \hat{x_3}$, where $\hat{x_i}$ represents a unit vector in the $x_i$ direction for $1\le i \le 3$. This type of generalized derivative can be seen as the derivative of a scalar, f, with respect to a vector, $\mathbf{x}$ and its result can be easily collected in vector form. $\nabla f = \frac{\partial f}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial f}{\partial x_1} & \frac{\partial f}{\partial x_2} & \frac{\partial f}{\partial x_3} \\ \end{bmatrix}.$[citation needed] More complicated examples include the derivative of a scalar function with respect to a matrix, known as the gradient matrix, which collects the derivative with respect to each matrix element in the corresponding position in the resulting matrix. In that case the scalar must be a function of each of the independent variables in the matrix. As another example, if we have an n-vector of dependent variables, or functions, of m independent variables we might consider the derivative of the dependent vector with respect to the independent vector. The result could be collected in an m×n matrix consisting of all of the possible derivative combinations. There are, of course, a total of nine possibilities using scalars, vectors, and matrices. Notice that as we consider higher numbers of components in each of the independent and dependent variables we can be left with a very large number of possibilities. The six kinds of derivatives that can be most neatly organized in matrix form are collected in the following table.[1] Types of Matrix Derivatives Types Scalar Vector Matrix Scalar $\frac{\partial y}{\partial x}$ $\frac{\partial \mathbf{y}}{\partial x}$ $\frac{\partial \mathbf{Y}}{\partial x}$ Vector $\frac{\partial y}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{y}}{\partial \mathbf{x}}$ Matrix $\frac{\partial y}{\partial \mathbf{X}}$ Here, we have used the term "matrix" in its most general sense, recognizing that vectors and scalars are simply matrices with one column and then one row respectively. Moreover, we have used bold letters to indicated vectors and bold capital letters for matrices. This notation is used throughout. Notice that we could also talk about the derivative of a vector with respect to a matrix, or any of the other unfilled cells in our table. However, these derivatives are most naturally organized in a tensor of rank higher than 2, so that they do not fit neatly into a matrix. In the following three sections we will define each one of these derivatives and relate them to other branches of mathematics. See the layout conventions section for a more detailed table. ### Relation to other derivatives The matrix derivative is a convenient notation for keeping track of partial derivatives for doing calculations. The Fréchet derivative is the standard way in the setting of functional analysis to take derivatives with respect to vectors. In the case that a matrix function of a matrix is Fréchet differentiable, the two derivatives will agree up to translation of notations. As is the case in general for partial derivatives, some formulae may extend under weaker analytic conditions than the existence of the derivative as approximating linear mapping. ### Usages Matrix calculus is used for deriving optimal stochastic estimators, often involving the use of Lagrange multipliers. This includes the derivation of: ## Notation The vector and matrix derivatives presented in the sections to follow take full advantage of matrix notation, using a single variable to represent a large number of variables. In what follows we will distinguish scalars, vectors and matrices by their typeface. We will let M(n,m) denote the space of real n×m matrices with n rows and m columns. Such matrices will be denoted using bold capital letters: A, X, Y, etc. An element of M(n,1), that is, a column vector, is denoted with a boldface lowercase letter: a, x, y, etc. An element of M(1,1) is a scalar, denoted with lowercase italic typeface: a, t, x, etc. XT denotes matrix transpose, tr(X) is the trace, and det(X) is the determinant. All functions are assumed to be of differentiability class C1 unless otherwise noted. Generally letters from first half of the alphabet (a, b, c, …) will be used to denote constants, and from the second half (t, x, y, …) to denote variables. NOTE: As mentioned above, there are competing notations for laying out systems of partial derivatives in vectors and matrices, and no standard appears to be emerging yet. The next two introductory sections use the numerator layout convention simply for the purposes of convenience, to avoid overly complicating the discussion. The section after them discusses layout conventions in more detail. It is important to realize the following: 1. Despite the use of the terms "numerator layout" and "denominator layout", there are actually more than two possible notational choices involved. The reason is that the choice of numerator vs. denominator (or in some situations, numerator vs. mixed) can be made independently for scalar-by-vector, vector-by-scalar, vector-by-vector, and scalar-by-matrix derivatives, and a number of authors mix and match their layout choices in various ways. 2. The choice of numerator layout in the introductory sections below does not imply that this is the "correct" or "superior" choice. There are advantages and disadvantages to the various layout types. Serious mistakes can result from carelessly combining formulas written in different layouts, and converting from one layout to another requires care to avoid errors. As a result, when working with existing formulas the best policy is probably to identify whichever layout is used and maintain consistency with it, rather than attempting to use the same layout in all situations. ### Alternatives The tensor index notation with its Einstein summation convention is very similar to the matrix calculus, except one writes only a single component at a time. It has the advantage that one can easily manipulate arbitrarily high rank tensors, whereas tensors of rank higher than two are quite unwieldy with matrix notation. All of the work here can be done in this notation without use of the single-variable matrix notation. However, many problems in estimation theory and other areas of applied mathematics would result in too many indices to properly keep track of, pointing in favor of matrix calculus in those areas. Also, Einstein notation can be very useful in proving the identities presented here, as an alternative to typical element notation, which can become cumbersome when the explicit sums are carried around. Note that a matrix can be considered a tensor of rank two. ## Derivatives with vectors Main article: Vector calculus Because vectors are matrices with only one column, the simplest matrix derivatives are vector derivatives. The notations developed here can accommodate the usual operations of vector calculus by identifying the space M(n,1) of n-vectors with the Euclidean space Rn, and the scalar M(1,1) is identified with R. The corresponding concept from vector calculus is indicated at the end of each subsection. NOTE: The discussion in this section assumes the numerator layout convention for pedagogical purposes. Some authors use different conventions. The section on layout conventions discusses this issue in greater detail. The identities given further down are presented in forms that can be used in conjunction with all common layout conventions. ### Vector-by-scalar The derivative of a vector $\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \\ \end{bmatrix}$, by a scalar x is written (in numerator layout notation) as $\frac{\partial \mathbf{y}}{\partial x} = \begin{bmatrix} \frac{\partial y_1}{\partial x}\\ \frac{\partial y_2}{\partial x}\\ \vdots\\ \frac{\partial y_m}{\partial x}\\ \end{bmatrix}.$ In vector calculus the derivative of a vector y with respect to a scalar x is known as the tangent vector of the vector y, $\frac{\partial \mathbf{y}}{\partial x}$. Notice here that y:R $\rightarrow$ Rm. Example Simple examples of this include the velocity vector in Euclidean space, which is the tangent vector of the position vector (considered as a function of time). Also, the acceleration is the tangent vector of the velocity. ### Scalar-by-vector The derivative of a scalar y by a vector $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{bmatrix}$, is written (in numerator layout notation) as $\frac{\partial y}{\partial \mathbf{x}} = \left[ \frac{\partial y}{\partial x_1} \ \ \frac{\partial y}{\partial x_2} \ \ \cdots \ \ \frac{\partial y}{\partial x_n} \right].$ In vector calculus the gradient of a scalar field y, in the space Rn whose independent coordinates are the components of x is the derivative of a scalar by a vector. In physics, the electric field is the vector gradient of the electric potential. The directional derivative of a scalar function f(x) of the space vector x in the direction of the unit vector u is defined using the gradient as follows. $\nabla_{\bold{u}}{f}(\bold{x}) = \nabla f(\bold{x}) \cdot \bold{u}$ Using the notation just defined for the derivative of a scalar with respect to a vector we can re-write the directional derivative as $\nabla_\mathbf{u} f = \frac{\partial f}{\partial \mathbf{x}}\mathbf{u}.$ This type of notation will be nice when proving product rules and chain rules that come out looking similar to what we are familiar with for the scalar derivative. ### Vector-by-vector Each of the previous two cases can be considered as an application of the derivative of a vector with respect to a vector, using a vector of size one appropriately. Similarly we will find that the derivatives involving matrices will reduce to derivatives involving vectors in a corresponding way. The derivative of a vector function (a vector whose components are functions) $\mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \\ \end{bmatrix}$, of an independent vector $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \\ \end{bmatrix}$, is written (in numerator layout notation) as $\frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} & \cdots & \frac{\partial y_1}{\partial x_n}\\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} & \cdots & \frac{\partial y_2}{\partial x_n}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y_m}{\partial x_1} & \frac{\partial y_m}{\partial x_2} & \cdots & \frac{\partial y_m}{\partial x_n}\\ \end{bmatrix}.$ In vector calculus, the derivative of a vector function y with respect to a vector x whose components represent a space is known as the pushforward or differential, or the Jacobian matrix. The pushforward along a vector function f with respect to vector v in Rm is given by $d\,\mathbf{f}(\mathbf{v}) = \frac{\partial \mathbf{f}}{\partial \mathbf{x}} \mathbf{v}.$ ## Derivatives with matrices There are two types of derivatives with matrices that can be organized into a matrix of the same size. These are the derivative of a matrix by a scalar and the derivative of a scalar by a matrix respectively. These can be useful in minimization problems found many areas of applied mathematics and have adopted the names tangent matrix and gradient matrix respectively after their analogs for vectors. NOTE: The discussion in this section assumes the numerator layout convention for pedagogical purposes. Some authors use different conventions. The section on layout conventions discusses this issue in greater detail. The identities given further down are presented in forms that can be used in conjunction with all common layout conventions. ### Matrix-by-scalar The derivative of a matrix function Y by a scalar x is known as the tangent matrix and is given (in numerator layout notation) by $\frac{\partial \mathbf{Y}}{\partial x} = \begin{bmatrix} \frac{\partial y_{11}}{\partial x} & \frac{\partial y_{12}}{\partial x} & \cdots & \frac{\partial y_{1n}}{\partial x}\\ \frac{\partial y_{21}}{\partial x} & \frac{\partial y_{22}}{\partial x} & \cdots & \frac{\partial y_{2n}}{\partial x}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y_{m1}}{\partial x} & \frac{\partial y_{m2}}{\partial x} & \cdots & \frac{\partial y_{mn}}{\partial x}\\ \end{bmatrix}.$ ### Scalar-by-matrix The derivative of a scalar y function of a matrix X of independent variables, with respect to the matrix X, is given (in numerator layout notation) by $\frac{\partial y}{\partial \mathbf{X}} = \begin{bmatrix} \frac{\partial y}{\partial x_{11}} & \frac{\partial y}{\partial x_{21}} & \cdots & \frac{\partial y}{\partial x_{p1}}\\ \frac{\partial y}{\partial x_{12}} & \frac{\partial y}{\partial x_{22}} & \cdots & \frac{\partial y}{\partial x_{p2}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y}{\partial x_{1q}} & \frac{\partial y}{\partial x_{2q}} & \cdots & \frac{\partial y}{\partial x_{pq}}\\ \end{bmatrix}.$ Notice that the indexing of the gradient with respect to X is transposed as compared with the indexing of X. Important examples of scalar functions of matrices include the trace of a matrix and the determinant. In analog with vector calculus this derivative is often written as the following. $\nabla_\mathbf{X} y(\mathbf{X}) = \frac{\partial y(\mathbf{X})}{\partial \mathbf{X}}$ Also in analog with vector calculus, the directional derivative of a scalar f(X) of a matrix X in the direction of matrix Y is given by $\nabla_\mathbf{Y} f = \operatorname{tr} \left(\frac{\partial f}{\partial \mathbf{X}} \mathbf{Y}\right).$ It is the gradient matrix, in particular, that finds many uses in minimization problems in estimation theory, particularly in the derivation of the Kalman filter algorithm, which is of great importance in the field. ### Other matrix derivatives This section's . (July 2009) The three types of derivatives that have not been considered are those involving vectors-by-matrices, matrices-by-vectors, and matrices-by-matrices. These are not as widely considered and a notation is not widely agreed upon. As for vectors, the other two types of higher matrix derivatives can be seen as applications of the derivative of a matrix by a matrix by using a matrix with one column in the correct place. For this reason, in this subsection we consider only how one can write the derivative of a matrix by another matrix. The differential or the matrix derivative of a matrix function F(X) that maps from n×m matrices to p×q matrices, F : M(n,m) $\to$ M(p,q), is an element of M(p,q) ? M(m,n), a fourth-rank tensor (the reversal of m and n here indicates the dual space of M(n,m)). In short it is an m×n matrix each of whose entries is a p×q matrix.[citation needed] $\frac{\partial\mathbf{F}} {\partial\mathbf{X}}= \begin{bmatrix} \frac{\partial\mathbf{F}}{\partial X_{1,1}} & \cdots & \frac{\partial \mathbf{F}}{\partial X_{n,1}}\\ \vdots & \ddots & \vdots\\ \frac{\partial\mathbf{F}}{\partial X_{1,m}} & \cdots & \frac{\partial \mathbf{F}}{\partial X_{n,m}}\\ \end{bmatrix},$ and note that each $\frac{\partial \mathbf{F}}{\partial \mathbf{X}_{ij}}$ is a p×q matrix defined as above. Note also that this matrix has its indexing transposed; m rows and n columns. The pushforward along F of an n×m matrix Y in M(n,m) is then $d\mathbf{F}(\mathbf{Y}) = \operatorname{tr}\left(\frac{\partial\mathbf{F}} {\partial\mathbf{X}}\mathbf{Y}\right),$ as formal block matrices. Note that this definition encompasses all of the preceding definitions as special cases. According to Jan R. Magnus and Heinz Neudecker, the following notations are both unsuitable, as the determinant of the second resulting matrix would have "no interpretation" and "a useful chain rule does not exist" if these notations are being used:[2] Given $\phi$, a differentiable function of an $n \times m$ matrix $\mathbf X = (x_{i, j})$, $\frac{\partial \mathbf\phi(\mathbf X)} {\partial \mathbf{X}}= \begin{bmatrix} \frac{\partial \mathbf\phi}{\partial x_{1,1}} & \cdots & \frac{\partial \mathbf\phi}{\partial x_{1,q}}\\ \vdots & \ddots & \vdots\\ \frac{\partial \mathbf\phi}{\partial x_{n,1}} & \cdots & \frac{\partial \mathbf\phi}{\partial x_{n,q}}\\ \end{bmatrix}$ Given $\mathbf F = (f_{s,t})$, a differentiable $m \times n$ function of an $n \times m$ matrix $\mathbf X$, $\frac{\partial \mathbf F (\mathbf X)} {\partial \mathbf{X}}= \begin{bmatrix} \frac{\partial f_{1,1}}{\partial \mathbf X} & \cdots & \frac{\partial f_{1,p}}{\partial \mathbf X}\\ \vdots & \ddots & \vdots\\ \frac{\partial f_{m,1}}{\partial \mathbf X} & \cdots & \frac{\partial f_{m,p}}{\partial \mathbf X}\\ \end{bmatrix}$ The Jacobian matrix, according to Magnus and Neudecker,[2] is $\mathrm D\, \mathbf F\left(\mathbf X\right) = \frac{\partial\, \mathrm{vec}\ \mathbf F\left(\mathbf X\right)}{\partial\left(\mathrm{vec}\ \mathbf X\right)^{\prime}}.$ ## Layout conventions This section discusses the similarities and differences between notational conventions that are used in the various fields that take advantage of matrix calculus. Although there are largely two consistent conventions, some authors find it convenient to mix the two conventions in forms that are discussed below. After this section equations will be listed in both competing forms separately. The fundamental issue is that the derivative of a vector with respect to a vector, i.e. $\frac{\partial \mathbf{y}}{\partial \mathbf{x}}$, is often written in two competing ways. If the numerator y is of size m and the denominator x of size n, then the result can be laid out as either an m×n matrix or n×m matrix, i.e. the elements of y laid out in columns and the elements of x laid out in rows, or vice-versa. This leads to the following possibilities: 1. Numerator layout, i.e. lay out according to y and xT (i.e. contrarily to x). This is sometimes known as the Jacobian formulation. 2. Denominator layout, i.e. lay out according to yT and x (i.e. contrarily to y). This is sometimes known as the Hessian formulation. Some authors term this layout the gradient, in distinction to the Jacobian (numerator layout), which is its transpose. (However, "gradient" more commonly means the derivative $\frac{\partial y}{\partial \mathbf{x}},$ regardless of layout.) 3. A third possibility sometimes seen is to insist on writing the derivative as $\frac{\partial \mathbf{y}}{\partial \mathbf{x}'},$ (i.e. the derivative is taken with respect to the transpose of x) and follow the numerator layout. This makes it possible to claim that the matrix is laid out according to both numerator and denominator. In practice this produces results the same as the numerator layout. When handling the gradient $\frac{\partial y}{\partial \mathbf{x}}$ and the opposite case $\frac{\partial \mathbf{y}}{\partial x},$ we have the same issues. To be consistent, we should do one of the following: 1. If we choose numerator layout for $\frac{\partial \mathbf{y}}{\partial \mathbf{x}},$ we should lay out the gradient $\frac{\partial y}{\partial \mathbf{x}}$ as a row vector, and $\frac{\partial \mathbf{y}}{\partial x}$ as a column vector. 2. If we choose denominator layout for $\frac{\partial \mathbf{y}}{\partial \mathbf{x}},$ we should lay out the gradient $\frac{\partial y}{\partial \mathbf{x}}$ as a column vector, and $\frac{\partial \mathbf{y}}{\partial x}$ as a row vector. 3. In the third possibility above, we write $\frac{\partial y}{\partial \mathbf{x}'}$ and $\frac{\partial \mathbf{y}}{\partial x},$ and use numerator layout. Not all math textbooks and papers are consistent in this respect throughout the entire paper. That is, sometimes different conventions are used in different contexts within the same paper. For example, some choose denominator layout for gradients (laying them out as column vectors), but numerator layout for the vector-by-vector derivative $\frac{\partial \mathbf{y}}{\partial \mathbf{x}}.$ Similarly, when it comes to scalar-by-matrix derivatives $\frac{\partial y}{\partial \mathbf{X}}$ and matrix-by-scalar derivatives $\frac{\partial \mathbf{Y}}{\partial x},$ then consistent numerator layout lays out according to Y and XT, while consistent denominator layout lays out according to YT and X. In practice, however, following a denominator layout for $\frac{\partial \mathbf{Y}}{\partial x},$ and laying the result out according to YT, is rarely seen because it makes for ugly formulas that do not correspond to the scalar formulas. As a result, the following layouts can often be found: 1. Consistent numerator layout, which lays out $\frac{\partial \mathbf{Y}}{\partial x}$ according to Y and $\frac{\partial y}{\partial \mathbf{X}}$ according to XT. 2. Mixed layout, which lays out $\frac{\partial \mathbf{Y}}{\partial x}$ according to Y and $\frac{\partial y}{\partial \mathbf{X}}$ according to X. 3. Use the notation $\frac{\partial y}{\partial \mathbf{X}^{\rm T}},$ with results the same as consistent numerator layout. In the following formulas, we handle the five possible combinations $\frac{\partial y}{\partial \mathbf{x}}, \frac{\partial \mathbf{y}}{\partial x}, \frac{\partial \mathbf{y}}{\partial \mathbf{x}}, \frac{\partial y}{\partial \mathbf{X}}$ and $\frac{\partial \mathbf{Y}}{\partial x}$ separately. We also handle cases of scalar-by-scalar derivatives that involve an intermediate vector or matrix. (This can arise, for example, if a multi-dimensional parametric curve is defined in terms of a scalar variable, and then a derivative of a scalar function of the curve is taken with respect to the scalar that parameterizes the curve.) For each of the various combinations, we give numerator-layout and denominator-layout results, except in the cases above where denominator layout rarely occurs. In cases involving matrices where it makes sense, we give numerator-layout and mixed-layout results. As noted above, cases where vector and matrix denominators are written in transpose notation are equivalent to numerator layout with the denominators written without the transpose. Keep in mind that various authors use different combinations of numerator and denominator layouts for different types of derivatives, and there is no guarantee that an author will consistently use either numerator or denominator layout for all types. Match up the formulas below with those quoted in the source to determine the layout used for that particular type of derivative, but be careful not to assume that derivatives of other types necessarily follow the same kind of layout. When taking derivatives with an aggregate (vector or matrix) denominator in order to find a maximum or minimum of the aggregate, it should be kept in mind that using numerator layout will produce results that are transposed with respect to the aggregate. For example, in attempting to find the maximum likelihood estimate of a multivariate normal distribution using matrix calculus, if the domain is a kx1 column vector, then the result using the numerator layout will be in the form of a 1xk row vector. Thus, either the results should be transposed at the end or the denominator layout (or mixed layout) should be used. Result of differentiating various kinds of aggregates with other kinds of aggregates Scalar y Vector y (size m) Matrix Y (size m×n) Notation Type Notation Type Notation Type Scalar x $\frac{\partial y}{\partial x}$ scalar $\frac{\partial \mathbf{y}}{\partial x}$ (numerator layout) size-m column vector (denominator layout) size-m row vector $\frac{\partial \mathbf{Y}}{\partial x}$ (numerator layout) m×n matrix Vector x (size n) $\frac{\partial y}{\partial \mathbf{x}}$ (numerator layout) size-n row vector (denominator layout) size-n column vector $\frac{\partial \mathbf{y}}{\partial \mathbf{x}}$ (numerator layout) m×n matrix (denominator layout) n×m matrix $\frac{\partial \mathbf{Y}}{\partial \mathbf{x}}$ ? Matrix X (size p×q) $\frac{\partial y}{\partial \mathbf{X}}$ (numerator layout) q×p matrix (denominator layout) p×q matrix $\frac{\partial \mathbf{y}}{\partial \mathbf{X}}$ ? $\frac{\partial \mathbf{Y}}{\partial \mathbf{X}}$ ? The results of operations will be transposed when switching between numerator-layout and denominator-layout notation. ### Numerator-layout notation Using numerator-layout notation, we have:[1] $\frac{\partial y}{\partial \mathbf{x}} = \left[ \frac{\partial y}{\partial x_1} \frac{\partial y}{\partial x_2} \cdots \frac{\partial y}{\partial x_n} \right].$ $\frac{\partial \mathbf{y}}{\partial x} = \begin{bmatrix} \frac{\partial y_1}{\partial x}\\ \frac{\partial y_2}{\partial x}\\ \vdots\\ \frac{\partial y_m}{\partial x}\\ \end{bmatrix}.$ $\frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} & \cdots & \frac{\partial y_1}{\partial x_n}\\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} & \cdots & \frac{\partial y_2}{\partial x_n}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y_m}{\partial x_1} & \frac{\partial y_m}{\partial x_2} & \cdots & \frac{\partial y_m}{\partial x_n}\\ \end{bmatrix}.$ $\frac{\partial y}{\partial \mathbf{X}} = \begin{bmatrix} \frac{\partial y}{\partial x_{11}} & \frac{\partial y}{\partial x_{21}} & \cdots & \frac{\partial y}{\partial x_{p1}}\\ \frac{\partial y}{\partial x_{12}} & \frac{\partial y}{\partial x_{22}} & \cdots & \frac{\partial y}{\partial x_{p2}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y}{\partial x_{1q}} & \frac{\partial y}{\partial x_{2q}} & \cdots & \frac{\partial y}{\partial x_{pq}}\\ \end{bmatrix}.$ The following definitions are only provided in numerator-layout notation: $\frac{\partial \mathbf{Y}}{\partial x} = \begin{bmatrix} \frac{\partial y_{11}}{\partial x} & \frac{\partial y_{12}}{\partial x} & \cdots & \frac{\partial y_{1n}}{\partial x}\\ \frac{\partial y_{21}}{\partial x} & \frac{\partial y_{22}}{\partial x} & \cdots & \frac{\partial y_{2n}}{\partial x}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y_{m1}}{\partial x} & \frac{\partial y_{m2}}{\partial x} & \cdots & \frac{\partial y_{mn}}{\partial x}\\ \end{bmatrix}.$ $d\mathbf{X} = \begin{bmatrix} dx_{11} & dx_{12} & \cdots & dx_{1n}\\ dx_{21} & dx_{22} & \cdots & dx_{2n}\\ \vdots & \vdots & \ddots & \vdots\\ dx_{m1} & dx_{m2} & \cdots & dx_{mn}\\ \end{bmatrix}.$ ### Denominator-layout notation Using denominator-layout notation, we have:[3] $\frac{\partial y}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial y}{\partial x_1}\\ \frac{\partial y}{\partial x_2}\\ \vdots\\ \frac{\partial y}{\partial x_n}\\ \end{bmatrix}.$ $\frac{\partial \mathbf{y}}{\partial x} = \left[ \frac{\partial y_1}{\partial x} \frac{\partial y_2}{\partial x} \cdots \frac{\partial y_m}{\partial x} \right].$ $\frac{\partial \mathbf{y}}{\partial \mathbf{x}} = \begin{bmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_2}{\partial x_1} & \cdots & \frac{\partial y_m}{\partial x_1}\\ \frac{\partial y_1}{\partial x_2} & \frac{\partial y_2}{\partial x_2} & \cdots & \frac{\partial y_m}{\partial x_2}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y_1}{\partial x_n} & \frac{\partial y_2}{\partial x_n} & \cdots & \frac{\partial y_m}{\partial x_n}\\ \end{bmatrix}.$ $\frac{\partial y}{\partial \mathbf{X}} = \begin{bmatrix} \frac{\partial y}{\partial x_{11}} & \frac{\partial y}{\partial x_{12}} & \cdots & \frac{\partial y}{\partial x_{1q}}\\ \frac{\partial y}{\partial x_{21}} & \frac{\partial y}{\partial x_{22}} & \cdots & \frac{\partial y}{\partial x_{2q}}\\ \vdots & \vdots & \ddots & \vdots\\ \frac{\partial y}{\partial x_{p1}} & \frac{\partial y}{\partial x_{p2}} & \cdots & \frac{\partial y}{\partial x_{pq}}\\ \end{bmatrix}.$ ## Identities As noted above, in general, the results of operations will be transposed when switching between numerator-layout and denominator-layout notation. To help make sense of all the identities below, keep in mind the most important rules: the chain rule, product rule and sum rule. The sum rule applies universally, and the product rule applies in most of the cases below, provided that the order of matrix products is maintained, since matrix products are not commutative. The chain rule applies in some of the cases, but unfortunately does not apply in matrix-by-scalar derivatives or scalar-by-matrix derivatives (in the latter case, mostly involving the trace operator applied to matrices). In the latter case, the product rule can't quite be applied directly, either, but the equivalent can be done with a bit more work using the differential identities. ### Vector-by-vector identities This is presented first because all of the operations that apply to vector-by-vector differentiation apply directly to vector-by-scalar or scalar-by-vector differentiation simply by reducing the appropriate vector in the numerator or denominator to a scalar. Identities: vector-by-vector $\frac{\partial \mathbf{y}}{\partial \mathbf{x}}$ Condition Expression Numerator layout, i.e. by y and xT Denominator layout, i.e. by yT and x a is not a function of x $\frac{\partial \mathbf{a}}{\partial \mathbf{x}} =$ $\mathbf{0}$ $\frac{\partial \mathbf{x}}{\partial \mathbf{x}} =$ $\mathbf{I}$ A is not a function of x $\frac{\partial \mathbf{A}\mathbf{x}}{\partial \mathbf{x}} =$ $\mathbf{A}$ $\mathbf{A}^{\rm T}$ A is not a function of x $\frac{\partial \mathbf{x}^{\rm T}\mathbf{A}}{\partial \mathbf{x}} =$ $\mathbf{A}^{\rm T}$ $\mathbf{A}$ a is not a function of x, u = u(x) $\frac{\partial a\mathbf{u}}{\partial\, \mathbf{x}} =$ $a\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ A is not a function of x, u = u(x) $\frac{\partial \mathbf{A}\mathbf{u}}{\partial \mathbf{x}} =$ $\mathbf{A}\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}\mathbf{A}^{\rm T}$ u = u(x), v = v(x) $\frac{\partial (\mathbf{u} + \mathbf{v})}{\partial \mathbf{x}} =$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}} + \frac{\partial \mathbf{v}}{\partial \mathbf{x}}$ u = u(x) $\frac{\partial \mathbf{g(u)}}{\partial \mathbf{x}} =$ $\frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}} \frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}} \frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}}$ u = u(x) $\frac{\partial \mathbf{f(g(u))}}{\partial \mathbf{x}} =$ $\frac{\partial \mathbf{f(g)}}{\partial \mathbf{g}} \frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}} \frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}} \frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}} \frac{\partial \mathbf{f(g)}}{\partial \mathbf{g}}$ ### Scalar-by-vector identities The fundamental identities are placed above the thick black line. Identities: scalar-by-vector $\frac{\partial y}{\partial \mathbf{x}} = \nabla_\mathbf{x} y$ Condition Expression Numerator layout, i.e. by xT; result is row vector Denominator layout, i.e. by x; result is column vector a is not a function of x $\frac{\partial a}{\partial \mathbf{x}} =$ $\mathbf{0}^{\rm T}$ [4] $\mathbf{0}$ [4] a is not a function of x, u = u(x) $\frac{\partial au}{\partial \mathbf{x}} =$ $a\frac{\partial u}{\partial \mathbf{x}}$ u = u(x), v = v(x) $\frac{\partial (u+v)}{\partial \mathbf{x}} =$ $\frac{\partial u}{\partial \mathbf{x}} + \frac{\partial v}{\partial \mathbf{x}}$ u = u(x), v = v(x) $\frac{\partial uv}{\partial \mathbf{x}} =$ $u\frac{\partial v}{\partial \mathbf{x}} + v\frac{\partial u}{\partial \mathbf{x}}$ u = u(x) $\frac{\partial g(u)}{\partial \mathbf{x}} =$ $\frac{\partial g(u)}{\partial u} \frac{\partial u}{\partial \mathbf{x}}$ u = u(x) $\frac{\partial f(g(u))}{\partial \mathbf{x}} =$ $\frac{\partial f(g)}{\partial g} \frac{\partial g(u)}{\partial u} \frac{\partial u}{\partial \mathbf{x}}$ u = u(x), v = v(x) $\frac{\partial (\mathbf{u} \cdot \mathbf{v})}{\partial \mathbf{x}} = \frac{\partial \mathbf{u}^{\rm T}\mathbf{v}}{\partial \mathbf{x}} =$ $\mathbf{u}^{\rm T}\frac{\partial \mathbf{v}}{\partial \mathbf{x}} + \mathbf{v}^{\rm T}\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ • assumes numerator layout of $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}, \frac{\partial \mathbf{v}}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}\mathbf{v} + \frac{\partial \mathbf{v}}{\partial \mathbf{x}}\mathbf{u}$ • assumes denominator layout of $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}, \frac{\partial \mathbf{v}}{\partial \mathbf{x}}$ u = u(x), v = v(x), A is not a function of x $\frac{\partial (\mathbf{u} \cdot \mathbf{A}\mathbf{v})}{\partial \mathbf{x}} = \frac{\partial \mathbf{u}^{\rm T}\mathbf{A}\mathbf{v}}{\partial \mathbf{x}} =$ $\mathbf{u}^{\rm T}\mathbf{A}\frac{\partial \mathbf{v}}{\partial \mathbf{x}} + \mathbf{v}^{\rm T}\mathbf{A}^{\rm T}\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ • assumes numerator layout of $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}, \frac{\partial \mathbf{v}}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}\mathbf{A}\mathbf{v} + \frac{\partial \mathbf{v}}{\partial \mathbf{x}}\mathbf{A}^{\rm T}\mathbf{u}$ • assumes denominator layout of $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}, \frac{\partial \mathbf{v}}{\partial \mathbf{x}}$ a is not a function of x $\frac{\partial (\mathbf{a}\cdot\mathbf{x})}{\partial \mathbf{x}} = \frac{\partial (\mathbf{x}\cdot\mathbf{a})}{\partial \mathbf{x}} =$ $\frac{\partial \mathbf{a}^{\rm T}\mathbf{x}}{\partial \mathbf{x}} = \frac{\partial \mathbf{x}^{\rm T}\mathbf{a}}{\partial \mathbf{x}} =$ $\mathbf{a}^{\rm T}$ $\mathbf{a}$ A is not a function of x b is not a function of x $\frac{\partial \mathbf{b}^{\rm T}\mathbf{A}\mathbf{x}}{\partial \mathbf{x}} =$ $\mathbf{b}^{\rm T}\mathbf{A}$ $\mathbf{A}^{\rm T}\mathbf{b}$ A is not a function of x $\frac{\partial \mathbf{x}^{\rm T}\mathbf{A}\mathbf{x}}{\partial \mathbf{x}} =$ $\mathbf{x}^{\rm T}(\mathbf{A} + \mathbf{A}^{\rm T})$ $(\mathbf{A} + \mathbf{A}^{\rm T})\mathbf{x}$ A is not a function of x A is symmetric $\frac{\partial \mathbf{x}^{\rm T}\mathbf{A}\mathbf{x}}{\partial \mathbf{x}} =$ $2\mathbf{x}^{\rm T}\mathbf{A}$ $2\mathbf{A}\mathbf{x}$ A is not a function of x $\frac{\partial^2 \mathbf{x}^{\rm T}\mathbf{A}\mathbf{x}}{\partial \mathbf{x}^2} =$ $\mathbf{A} + \mathbf{A}^{\rm T}$ A is not a function of x A is symmetric $\frac{\partial^2 \mathbf{x}^{\rm T}\mathbf{A}\mathbf{x}}{\partial \mathbf{x}^2} =$ $2\mathbf{A}$ $\frac{\partial (\mathbf{x} \cdot \mathbf{x})}{\partial \mathbf{x}} = \frac{\partial \mathbf{x}^{\rm T}\mathbf{x}}{\partial \mathbf{x}} =$ $2\mathbf{x}^{\rm T}$ $2\mathbf{x}$ a is not a function of x, u = u(x) $\frac{\partial (\mathbf{a} \cdot \mathbf{u})}{\partial \mathbf{x}} = \frac{\partial \mathbf{a}^{\rm T}\mathbf{u}}{\partial \mathbf{x}} =$ $\mathbf{a}^{\rm T}\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ • assumes numerator layout of $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}\mathbf{a}$ • assumes denominator layout of $\frac{\partial \mathbf{u}}{\partial \mathbf{x}}$ a, b are not functions of x $\frac{\partial \; \textbf{a}^{\rm T}\textbf{x}\textbf{x}^{\rm T}\textbf{b}}{\partial \; \textbf{x}} =$ $\textbf{x}^{\rm T}(\textbf{a}\textbf{b}^{\rm T} + \textbf{b}\textbf{a}^{\rm T})$ $(\textbf{a}\textbf{b}^{\rm T} + \textbf{b}\textbf{a}^{\rm T})\textbf{x}$ A, b, C, D, e are not functions of x $\frac{\partial \; (\textbf{A}\textbf{x} + \textbf{b})^{\rm T} \textbf{C} (\textbf{D}\textbf{x} + \textbf{e}) }{\partial \; \textbf{x}} =$ $(\textbf{D}\textbf{x} + \textbf{e})^{\rm T} \textbf{C}^{\rm T} \textbf{A} + (\textbf{A}\textbf{x} + \textbf{b})^{\rm T} \textbf{C} \textbf{D}$ $\textbf{D}^{\rm T} \textbf{C}^{\rm T} (\textbf{A}\textbf{x} + \textbf{b}) + \textbf{A}^{\rm T}\textbf{C}(\textbf{D}\textbf{x} + \textbf{e})$ a is not a function of x $\frac{\partial \; \\|\mathbf{x}-\mathbf{a}\\|}{\partial \; \mathbf{x}} =$ $\frac{(\mathbf{x}-\mathbf{a})^{\rm T}}{\\|\mathbf{x}-\mathbf{a}\\|}$ $\frac{\mathbf{x}-\mathbf{a}}{\\|\mathbf{x}-\mathbf{a}\\|}$ ### Vector-by-scalar identities Identities: vector-by-scalar $\frac{\partial \mathbf{y}}{\partial x}$ Condition Expression Numerator layout, i.e. by y, result is column vector Denominator layout, i.e. by yT, result is row vector a is not a function of x $\frac{\partial \mathbf{a}}{\partial x} =$ $\mathbf{0}$[4] a is not a function of x, u = u(x) $\frac{\partial a\mathbf{u}}{\partial x} =$ $a\frac{\partial \mathbf{u}}{\partial x}$ A is not a function of x, u = u(x) $\frac{\partial \mathbf{A}\mathbf{u}}{\partial x} =$ $\mathbf{A}\frac{\partial \mathbf{u}}{\partial x}$ $\frac{\partial \mathbf{u}}{\partial x}\mathbf{A}^{\rm T}$ u = u(x) $\frac{\partial \mathbf{u}^{\rm T}}{\partial x} =$ $\left(\frac{\partial \mathbf{u}}{\partial x}\right)^{\rm T}$ u = u(x), v = v(x) $\frac{\partial (\mathbf{u} + \mathbf{v})}{\partial x} =$ $\frac{\partial \mathbf{u}}{\partial x} + \frac{\partial \mathbf{v}}{\partial x}$ u = u(x), v = v(x) $\frac{\partial (\mathbf{u} \times \mathbf{v})}{\partial x} =$ $\mathbf{u} \times \frac{\partial \mathbf{v}}{\partial x} + \frac{\partial \mathbf{u}}{\partial x} \times \mathbf{v}$ u = u(x) $\frac{\partial \mathbf{g(u)}}{\partial x} =$ $\frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}} \frac{\partial \mathbf{u}}{\partial x}$ $\frac{\partial \mathbf{u}}{\partial x} \frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}}$ Assumes consistent matrix layout; see below. u = u(x) $\frac{\partial \mathbf{f(g(u))}}{\partial x} =$ $\frac{\partial \mathbf{f(g)}}{\partial \mathbf{g}} \frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}} \frac{\partial \mathbf{u}}{\partial x}$ $\frac{\partial \mathbf{u}}{\partial x} \frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}} \frac{\partial \mathbf{f(g)}}{\partial \mathbf{g}}$ Assumes consistent matrix layout; see below. NOTE: The formulas involving the vector-by-vector derivatives $\frac{\partial \mathbf{g(u)}}{\partial \mathbf{u}}$ and $\frac{\partial \mathbf{f(g)}}{\partial \mathbf{g}}$ (whose outputs are matrices) assume the matrices are laid out consistent with the vector layout, i.e. numerator-layout matrix when numerator-layout vector and vice-versa; otherwise, transpose the vector-by-vector derivatives. ### Scalar-by-matrix identities Note that exact equivalents of the scalar product rule and chain rule do not exist when applied to matrix-valued functions of matrices. However, the product rule of this sort does apply to the differential form (see below), and this is the way to derive many of the identities below involving the trace function, combined with the fact that the trace function allows transposing and cyclic permutation, i.e.: ${\rm tr}(\mathbf{A}) = {\rm tr}(\mathbf{A^{\rm T}})$ ${\rm tr}(\mathbf{ABCD}) = {\rm tr}(\mathbf{BCDA}) = {\rm tr}(\mathbf{CDAB}) = {\rm tr}(\mathbf{DABC})$ For example, to compute $\frac{\partial {\rm tr}(\mathbf{AXBX^{\rm T}C})}{\partial \mathbf{X}}:$ $\begin{align} d\, {\rm tr}(\mathbf{AXBX^{\rm T}C}) &= d\, {\rm tr}(\mathbf{CAXBX^{\rm T}}) = {\rm tr}(d(\mathbf{CAXBX^{\rm T}})) \\ &= {\rm tr}(\mathbf{CAX} d(\mathbf{BX^{\rm T}}) + d(\mathbf{CAX})\mathbf{BX^{\rm T}}) \\ &= {\rm tr}(\mathbf{CAX} d(\mathbf{BX^{\rm T}})) + {\rm tr}(d(\mathbf{CAX})\mathbf{BX^{\rm T}}) \\ &= {\rm tr}(\mathbf{CAXB} d(\mathbf{X^{\rm T}})) + {\rm tr}(\mathbf{CA}(d\mathbf{X})\mathbf{BX^{\rm T}}) \\ &= {\rm tr}(\mathbf{CAXB} (d\mathbf{X})^{\rm T}) + {\rm tr}(\mathbf{CA}(d\mathbf{X})\mathbf{BX^{\rm T}}) \\ &= {\rm tr}\left((\mathbf{CAXB} (d\mathbf{X})^{\rm T})^{\rm T}\right) + {\rm tr}(\mathbf{CA}(d\mathbf{X})\mathbf{BX^{\rm T}}) \\ &= {\rm tr}((d\mathbf{X})\mathbf{B^{\rm T}X^{\rm T}A^{\rm T}C^{\rm T}}) + {\rm tr}(\mathbf{CA}(d\mathbf{X})\mathbf{BX^{\rm T}}) \\ &= {\rm tr}(\mathbf{B^{\rm T}X^{\rm T}A^{\rm T}C^{\rm T}}(d\mathbf{X})) + {\rm tr}(\mathbf{BX^{\rm T}}\mathbf{CA}(d\mathbf{X})) \\ &= {\rm tr}\left((\mathbf{B^{\rm T}X^{\rm T}A^{\rm T}C^{\rm T}} + \mathbf{BX^{\rm T}}\mathbf{CA})d\mathbf{X}\right) \end{align}$ Therefore, $\frac{\partial {\rm tr}(\mathbf{AXBX^{\rm T}C})}{\partial \mathbf{X}} = \mathbf{B^{\rm T}X^{\rm T}A^{\rm T}C^{\rm T}} + \mathbf{BX^{\rm T}}\mathbf{CA} .$ Identities: scalar-by-matrix $\frac{\partial y}{\partial \mathbf{X}}$ Condition Expression Numerator layout, i.e. by XT Denominator layout, i.e. by X a is not a function of X $\frac{\partial a}{\partial \mathbf{X}} =$ $\mathbf{0}^{\rm T}$ [5] $\mathbf{0}$ [5] a is not a function of X, u = u(X) $\frac{\partial au}{\partial \mathbf{X}} =$ $a\frac{\partial u}{\partial \mathbf{X}}$ u = u(X), v = v(X) $\frac{\partial (u+v)}{\partial \mathbf{X}} =$ $\frac{\partial u}{\partial \mathbf{X}} + \frac{\partial v}{\partial \mathbf{X}}$ u = u(X), v = v(X) $\frac{\partial uv}{\partial \mathbf{X}} =$ $u\frac{\partial v}{\partial \mathbf{X}} + v\frac{\partial u}{\partial \mathbf{X}}$ u = u(X) $\frac{\partial g(u)}{\partial \mathbf{X}} =$ $\frac{\partial g(u)}{\partial u} \frac{\partial u}{\partial \mathbf{X}}$ u = u(X) $\frac{\partial f(g(u))}{\partial \mathbf{X}} =$ $\frac{\partial f(g)}{\partial g} \frac{\partial g(u)}{\partial u} \frac{\partial u}{\partial \mathbf{X}}$ U = U(X) [6] $\frac{\partial g(\mathbf{U})}{\partial X_{ij}} =$ ${\rm tr}\left( \frac{\partial g(\mathbf{U})}{\partial \mathbf{U}} \frac{\partial \mathbf{U}}{\partial X_{ij}}\right)$ ${\rm tr}\left( \left(\frac{\partial g(\mathbf{U})}{\partial \mathbf{U}}\right)^{\rm T} \frac{\partial \mathbf{U}}{\partial X_{ij}}\right)$ Both forms assume numerator layout for $\frac{\partial \mathbf{U}}{\partial X_{ij}},$ i.e. mixed layout if denominator layout for X is being used. $\frac{\partial {\rm tr}(\mathbf{X})}{\partial \mathbf{X}} =$ $\mathbf{I}$ U = U(X), V = V(X) $\frac{\partial {\rm tr}(\mathbf{U}+\mathbf{V})}{\partial \mathbf{X}} =$ $\frac{\partial {\rm tr}(\mathbf{U})}{\partial \mathbf{X}} + \frac{\partial {\rm tr}(\mathbf{V})}{\partial \mathbf{X}}$ a is not a function of X, U = U(X) $\frac{\partial {\rm tr}(a\mathbf{U})}{\partial \mathbf{X}} =$ $a\frac{\partial {\rm tr}(\mathbf{U})}{\partial \mathbf{X}}$ g(X) is any polynomial with scalar coefficients, or any matrix function defined by an infinite polynomial series (e.g. eX, sin(X), cos(X), ln(X), etc. using a Taylor series); g(x) is the equivalent scalar function, g′(x) is its derivative, and g′(X) is the corresponding matrix function $\frac{\partial {\rm tr}(\mathbf{g(X)})}{\partial \mathbf{X}} =$ $\mathbf{g}'(\mathbf{X})$ $(\mathbf{g}'(\mathbf{X}))^{\rm T}$ A is not a function of X [7] $\frac{\partial {\rm tr}(\mathbf{AX})}{\partial \mathbf{X}} = \frac{\partial {\rm tr}(\mathbf{XA})}{\partial \mathbf{X}} =$ $\mathbf{A}$ $\mathbf{A}^{\rm T}$ A is not a function of X [6] $\frac{\partial {\rm tr}(\mathbf{AX^{\rm T}})}{\partial \mathbf{X}} = \frac{\partial {\rm tr}(\mathbf{X^{\rm T}A})}{\partial \mathbf{X}} =$ $\mathbf{A}^{\rm T}$ $\mathbf{A}$ A is not a function of X [6] $\frac{\partial {\rm tr}(\mathbf{X^{\rm T}AX})}{\partial \mathbf{X}} =$ $\mathbf{X}^{\rm T}(\mathbf{A}+\mathbf{A}^{\rm T})$ $(\mathbf{A}+\mathbf{A}^{\rm T})\mathbf{X}$ A is not a function of X [6] $\frac{\partial {\rm tr}(\mathbf{X^{-1}A})}{\partial \mathbf{X}} =$ $-(\mathbf{X}^{-1})^{\rm T}\mathbf{A}(\mathbf{X}^{-1})^{\rm T}$ $-\mathbf{X}^{-1}\mathbf{A}^{\rm T}\mathbf{X}^{-1}$ A, B are not functions of X $\frac{\partial {\rm tr}(\mathbf{AXB})}{\partial \mathbf{X}} = \frac{\partial {\rm tr}(\mathbf{BAX})}{\partial \mathbf{X}} =$ $\mathbf{BA}$ $\mathbf{A^{\rm T}B^{\rm T}}$ A, B, C are not functions of X $\frac{\partial {\rm tr}(\mathbf{AXBX^{\rm T}C})}{\partial \mathbf{X}} =$ $\mathbf{BX^{\rm T}CA} + \mathbf{B^{\rm T}X^{\rm T}A^{\rm T}C^{\rm T}}$ $\mathbf{A^{\rm T}C^{\rm T}XB^{\rm T}} + \mathbf{CAXB}$ n is a positive integer [6] $\frac{\partial {\rm tr}(\mathbf{X}^n)}{\partial \mathbf{X}} =$ $n\mathbf{X}^{n-1}$ $n(\mathbf{X}^{n-1})^{\rm T}$ A is not a function of X, n is a positive integer [6] $\frac{\partial {\rm tr}(\mathbf{A}\mathbf{X}^n)}{\partial \mathbf{X}} =$ $\sum_{i=0}^{n-1} \mathbf{X}^i\mathbf{A}\mathbf{X}^{n-i-1}$ $\sum_{i=0}^{n-1} (\mathbf{X}^i\mathbf{A}\mathbf{X}^{n-i-1})^{\rm T}$ [6] $\frac{\partial {\rm tr}(e^\mathbf{X})}{\partial \mathbf{X}} =$ $e^\mathbf{X}$ $(e^\mathbf{X})^{\rm T}$ [6] $\frac{\partial {\rm tr}(\sin(\mathbf{X}))}{\partial \mathbf{X}} =$ $\cos(\mathbf{X})$ $(\cos(\mathbf{X}))^{\rm T}$ [8] $\frac{\partial \|\mathbf{X}\|}{\partial \mathbf{X}} =$ $\operatorname{cofactor}(X)^{\rm T} = \|\mathbf{X}\|\mathbf{X}^{-1}$ $\operatorname{cofactor}(X) = \|\mathbf{X}\|(\mathbf{X}^{-1})^{\rm T}$ a is not a function of X [6]$\frac{\partial \ln \|a\mathbf{X}\|}{\partial \mathbf{X}} =$[9] $\mathbf{X}^{-1}$ $(\mathbf{X}^{-1})^{\rm T}$ A, B are not functions of X [6] $\frac{\partial \|\mathbf{AXB}\|}{\partial \mathbf{X}} =$ $\|\mathbf{AXB}\|\mathbf{X}^{-1}$ $\|\mathbf{AXB}\|(\mathbf{X}^{-1})^{\rm T}$ n is a positive integer [6] $\frac{\partial \|\mathbf{X}^n\|}{\partial \mathbf{X}} =$ $n\|\mathbf{X}^n\|\mathbf{X}^{-1}$ $n\|\mathbf{X}^n\|(\mathbf{X}^{-1})^{\rm T}$ (see pseudo-inverse) [6] $\frac{\partial \ln \|\mathbf{X}^{\rm T}\mathbf{X}\|}{\partial \mathbf{X}} =$ $2\mathbf{X}^{+}$ $2(\mathbf{X}^{+})^{\rm T}$ (see pseudo-inverse) [6] $\frac{\partial \ln \|\mathbf{X}^{\rm T}\mathbf{X}\|}{\partial \mathbf{X}^{+}} =$ $-2\mathbf{X}$ $-2\mathbf{X}^{\rm T}$ A is not a function of X, X is square and invertible $\frac{\partial \|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|}{\partial \mathbf{X}} =$ $2\|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|\mathbf{X}^{-1}$ $2\|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|(\mathbf{X}^{-1})^{\rm T}$ A is not a function of X, X is non-square, A is symmetric $\frac{\partial \|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|}{\partial \mathbf{X}} =$ $2\|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|(\mathbf{X^{\rm T}A^{\rm T}X})^{-1}\mathbf{X^{\rm T}A^{\rm T}}$ $2\|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|\mathbf{AX}(\mathbf{X^{\rm T}AX})^{-1}$ A is not a function of X, X is non-square, A is non-symmetric $\frac{\partial \|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|}{\partial \mathbf{X}} =$ $\|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|((\mathbf{X^{\rm T}AX})^{-1}\mathbf{X^{\rm T}A}$ $+ (\mathbf{X^{\rm T}A^{\rm T}X})^{-1}\mathbf{X^{\rm T}A^{\rm T}})$ $\|\mathbf{X^{\rm T}}\mathbf{A}\mathbf{X}\|(\mathbf{AX}(\mathbf{X^{\rm T}AX})^{-1}$ $+ \mathbf{A^{\rm T}X}(\mathbf{X^{\rm T}A^{\rm T}X})^{-1})$ ### Matrix-by-scalar identities Identities: matrix-by-scalar $\frac{\partial \mathbf{Y}}{\partial x}$ Condition Expression Numerator layout, i.e. by Y U = U(x) $\frac{\partial a\mathbf{U}}{\partial x} =$ $a\frac{\partial \mathbf{U}}{\partial x}$ A, B are not functions of x, U = U(x) $\frac{\partial \mathbf{AUB}}{\partial x} =$ $\mathbf{A}\frac{\partial \mathbf{U}}{\partial x}\mathbf{B}$ U = U(x), V = V(x) $\frac{\partial (\mathbf{U}+\mathbf{V})}{\partial x} =$ $\frac{\partial \mathbf{U}}{\partial x} + \frac{\partial \mathbf{V}}{\partial x}$ U = U(x), V = V(x) $\frac{\partial (\mathbf{U}\mathbf{V})}{\partial x} =$ $\mathbf{U}\frac{\partial \mathbf{V}}{\partial x} + \frac{\partial \mathbf{U}}{\partial x}\mathbf{V}$ U = U(x), V = V(x) $\frac{\partial (\mathbf{U} \otimes \mathbf{V})}{\partial x} =$ $\mathbf{U} \otimes \frac{\partial \mathbf{V}}{\partial x} + \frac{\partial \mathbf{U}}{\partial x} \otimes \mathbf{V}$ U = U(x), V = V(x) $\frac{\partial (\mathbf{U} \circ \mathbf{V})}{\partial x} =$ $\mathbf{U} \circ \frac{\partial \mathbf{V}}{\partial x} + \frac{\partial \mathbf{U}}{\partial x} \circ \mathbf{V}$ U = U(x) $\frac{\partial \mathbf{U}^{-1}}{\partial x} =$ $-\mathbf{U}^{-1} \frac{\partial \mathbf{U}}{\partial x}\mathbf{U}^{-1}$ U = U(x,y) $\frac{\partial^2 \mathbf{U}^{-1}}{\partial x \partial y} =$ $\mathbf{U}^{-1}\left(\frac{\partial \mathbf{U}}{\partial x}\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial y} - \frac{\partial^2 \mathbf{U}}{\partial x \partial y} + \frac{\partial \mathbf{U}}{\partial y}\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial x}\right)\mathbf{U}^{-1}$ A is not a function of x, g(X) is any polynomial with scalar coefficients, or any matrix function defined by an infinite polynomial series (e.g. eX, sin(X), cos(X), ln(X), etc.); g(x) is the equivalent scalar function, g′(x) is its derivative, and g′(X) is the corresponding matrix function $\frac{\partial \, \mathbf{g}(x\mathbf{A})}{\partial x} =$ $\mathbf{A}\mathbf{g}'(x\mathbf{A}) = \mathbf{g}'(x\mathbf{A})\mathbf{A}$ A is not a function of x $\frac{\partial e^{x\mathbf{A}}}{\partial x} =$ $\mathbf{A}e^{x\mathbf{A}} = e^{x\mathbf{A}}\mathbf{A}$ ### Scalar-by-scalar identities #### With vectors involved Identities: scalar-by-scalar, with vectors involved Condition Expression Any layout (assumes dot product ignores row vs. column layout) u = u(x) $\frac{\partial g(\mathbf{u})}{\partial x} =$ $\frac{\partial g(\mathbf{u})}{\partial \mathbf{u}} \cdot \frac{\partial \mathbf{u}}{\partial x}$ u = u(x), v = v(x) $\frac{\partial (\mathbf{u} \cdot \mathbf{v})}{\partial x} =$ $\mathbf{u} \cdot \frac{\partial \mathbf{v}}{\partial x} + \frac{\partial \mathbf{u}}{\partial x} \cdot \mathbf{v}$ #### With matrices involved Identities: scalar-by-scalar, with matrices involved[6] Condition Expression Consistent numerator layout, i.e. by Y and XT Mixed layout, i.e. by Y and X U = U(x) $\frac{\partial \|\mathbf{U}\|}{\partial x} =$ $\|\mathbf{U}\|{\rm tr}\left(\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial x}\right)$ U = U(x) $\frac{\partial \ln\|\mathbf{U}\|}{\partial x} =$ ${\rm tr}\left(\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial x}\right)$ U = U(x) $\frac{\partial^2 \|\mathbf{U}\|}{\partial x^2} =$ $\|\mathbf{U}\|\left[{\rm tr}\left(\mathbf{U}^{-1}\frac{\partial^2 \mathbf{U}}{\partial x^2}\right) + \left({\rm tr}\left(\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial x}\right)\right)^2-{\rm tr}\left(\left(\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial x}\right)\left(\mathbf{U}^{-1}\frac{\partial \mathbf{U}}{\partial x}\right)\right)\right]$ U = U(x) $\frac{\partial g(\mathbf{U})}{\partial x} =$ ${\rm tr}\left( \frac{\partial g(\mathbf{U})}{\partial \mathbf{U}} \frac{\partial \mathbf{U}}{\partial x}\right)$ ${\rm tr}\left( \left(\frac{\partial g(\mathbf{U})}{\partial \mathbf{U}}\right)^{\rm T} \frac{\partial \mathbf{U}}{\partial x}\right)$ A is not a function of x, g(X) is any polynomial with scalar coefficients, or any matrix function defined by an infinite polynomial series (e.g. eX, sin(X), cos(X), ln(X), etc.); g(x) is the equivalent scalar function, g′(x) is its derivative, and g′(X) is the corresponding matrix function. $\frac{\partial \, {\rm tr}(\mathbf{g}(x\mathbf{A}))}{\partial x} =$ ${\rm tr}(\mathbf{A}\mathbf{g}'(x\mathbf{A}))$ A is not a function of x $\frac{\partial \, {\rm tr}(e^{x\mathbf{A}})}{\partial x} =$ ${\rm tr}(\mathbf{A}e^{x\mathbf{A}})$ ### Identities in differential form It is often easier to work in differential form and then convert back to normal derivatives. This only works well using the numerator layout. Differential identities: scalar involving matrix [1][6] Condition Expression Result (numerator layout) $d({\rm tr}(\mathbf{X})) =$ ${\rm tr}(d\mathbf{X})$ $d(\|\mathbf{X}\|) =$ $\|\mathbf{X}\|{\rm tr}(\mathbf{X}^{-1}d\mathbf{X})$ $d(\ln\|\mathbf{X}\|) =$ ${\rm tr}(\mathbf{X}^{-1}d\mathbf{X})$ Differential identities: matrix [1][6] Condition Expression Result (numerator layout) A is not a function of X $d(\mathbf{A}) =$ $0$ a is not a function of X $d(a\mathbf{X}) =$ $a\,d\mathbf{X}$ $d(\mathbf{X}+\mathbf{Y}) =$ $d\mathbf{X}+d\mathbf{Y}$ $d(\mathbf{X}\mathbf{Y}) =$ $(d\mathbf{X})\mathbf{Y}+\mathbf{X}(d\mathbf{Y})$ (Kronecker product) $d(\mathbf{X} \otimes \mathbf{Y}) =$ $(d\mathbf{X})\otimes\mathbf{Y}+\mathbf{X}\otimes(d\mathbf{Y})$ (Hadamard product) $d(\mathbf{X} \circ \mathbf{Y}) =$ $(d\mathbf{X})\circ\mathbf{Y}+\mathbf{X}\circ(d\mathbf{Y})$ $d(\mathbf{X}^{\rm T}) =$ $(d\mathbf{X})^{\rm T}$ (conjugate transpose) $d(\mathbf{X}^{\rm H}) =$ $(d\mathbf{X})^{\rm H}$ To convert to normal derivative form, first convert it to one of the following canonical forms, and then use these identities: Conversion from differential to derivative form [1] Canonical differential form Equivalent derivative form $dy = a\,dx$ $\frac{dy}{dx} = a$ $dy = \mathbf{a}\,d\mathbf{x}$ $\frac{dy}{d\mathbf{x}} = \mathbf{a}$ $dy = {\rm tr}(\mathbf{A}\,d\mathbf{X})$ $\frac{dy}{d\mathbf{X}} = \mathbf{A}$ $d\mathbf{y} = \mathbf{a}\,dx$ $\frac{d\mathbf{y}}{dx} = \mathbf{a}$ $d\mathbf{y} = \mathbf{A}\,d\mathbf{x}$ $\frac{d\mathbf{y}}{d\mathbf{x}} = \mathbf{A}$ $d\mathbf{Y} = \mathbf{A}\,dx$ $\frac{d\mathbf{Y}}{dx} = \mathbf{A}$ ## Notes 1. ^ a b Magnus, Jan R.; Neudecker, Heinz (1999). Matrix Differential Calculus with Applications in Statistics and Econometrics. Wiley Series in Probability and Statistics (2nd ed.). Wiley. pp. 171–173. 2. ^ a b c Here, $\mathbf{0}$ refers to a column vector of all 0's, of size n, where n is the length of x. 3. ^ a b Here, $\mathbf{0}$ refers to a matrix of all 0's, of the same shape as X. 4. Petersen, Kaare Brandt and Michael Syskind Pedersen. The Matrix Cookbook. November 14, 2008. http://matrixcookbook.com. [3] This book uses a mixed layout, i.e. by Y in $\frac{\partial \mathbf{Y}}{\partial x},$ by X in $\frac{\partial y}{\partial \mathbf{X}}.$ 5. Duchi, John C. "Properties of the Trace and Matrix Derivatives". University of California at Berkeley. Retrieved 19 July 2011. 6. The constant a disappears in the result. This is intentional. In general, $\frac{d\,\ln au}{dx} = \frac{1}{au}\frac{d(au)}{dx} = \frac{1}{au}a\frac{du}{dx} = \frac{1}{u}\frac{du}{dx} = \frac{d\,\ln u}{dx}.$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 358, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8953205943107605, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/11829/gravititonal-fields-compared-to-electromagnetic-fields-are-they-infinite-in-ra	# Gravititonal fields compared to electromagnetic fields - are they infinite in range? me and my friend has a discussion last night, and he argued that both an electromagnetic field and gravititonal field are infinite in their area of effect, but with diminishing effects as you get farther away from the object. I argued that only gravity field is infinite in range. So which one of us is correct, and please explain in laymen's terms, as we're not physicists :) - ## 2 Answers Actually, your friend is right! The electromagnetic field has an infinite range, for example for a charged point particle the electric field is proportional to $\frac 1{r^2}$, where $r$ is the distance to the particle. However in very special circumstances, like inside a superconductor, the electromagnetic field will be short-range because the photons will effectively be massive. See this table, where you can see both the electromagnetic and gravitational fields have infinite range of interaction, but the Weak doesn't. It's the famous Higgs boson which renders the weak field short range (by making the corresponding gauge bosons massive), very similar to what happens inside a superconductor. - 2 You've got a typo here: the electric field is proportional to $1/r^2$, not to $1/r$. Alternatively, the electric potential is proportional to $1/r$. – Ted Bunn Jul 3 '11 at 15:45 Also, it's kind of misleading to say that the Higgs boson makes the weak force short-range. It's really due to the mass of the W and Z bosons. Plus, the strong force is kind of strange since it actually increases with distance; you could say that means its range is infinite, but not in the same way as gravity and the EM force. – David Zaslavsky♦ Jul 3 '11 at 17:29 @Ted, thanks for the correction. @David Yes, but the masses of the W and Z bosons are generated through interaction with the Higgs field. Regarding the strong force, I agree. It's more subtle than I made it appear. – Heidar Jul 4 '11 at 11:53 The key difference is that there are positive and negative electric charges and the universe is basically neutral. The opposite effects of the charges cancel out so on large scales, ie astronomical, the net electrical forces are much less than the gravitational forces, which are all one sign and hence never "cancel out." -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9501971006393433, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/32069/list	3 edited tags 2 title change 1 # Nakajima quiver varieties Are deformations of Nakajima quiver varieties also Nakajima quiver varieties ? In case the answer to this is (don't k)no(w), here are some simpler things to ask for. 1. (If you're a differential geometer) Is any hyperkahler rotation / twistor deformation of a Nakajima quiver variety also a Nakajima quiver variety ? 2. (An example, for algebraic geometers) Consider the Hilbert scheme $H$ of $k$ points on the minimal resolution of $\mathbb C^2/(\mathbb Z/n)$. Or restrict to $n=2=k$ and consider $Hilb^2T^\mathbb P^1$. Its exceptional divisor over the symmetric product defines a class in $H^1(\Omega_H)$ (despite the noncompactness). Using the holomorphic symplectic form, we get a deformation class in $H^1(T_H)$. (The corresponding deformation is not so far from the twistor deformation, and can be realised as a composition of a deformation of the ALE space followed by a twistor deformation.) Is there a Nakajima quiver variety in the direction of this deformation ? For instance if I take the quiver $\bullet^{\ \rightrightarrows}_{\ \leftleftarrows}\bullet$, dimension vector (1,1), and an appropriate stability condition (or value of the real moment map) then I get $T^\mathbb P^1$ as the moduli space over the value $0$ of the complex moment map, and the smoothing of the surface ordinary double point over nonzero values. Now if I take dimension vector (2,2) I can presumably get $Hilb^2$ of these surfaces, for an appropriate stability condition. However, as I vary the value of the complex moment map I simply vary the surface that I take $Hilb^2$ of, rather than getting the deformation I'm after. (The hyperkahler rotation is not a $Hilb^2$, since the exceptional divisor disappears in this deformation.) But is there another quivery way of producing this deformation ?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8571521043777466, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/116420/list	## Return to Answer 2 added 1789 characters in body N0. It is NOT the case that it is true in general. It is true for all cases with both $m \le 5$ and all prime $p n \lt 100$but may always be false . It is also true for $m=6$ and $n = 31$. But actually it is false ( which requires for $p\ge 31.$) At m=6$) over$\mathbb{Z}$and also over$\mathbb{Z}/{n}\mathbb{Z}$for any rate$n \ge 36\$, a specific example is the prime or composite. The sets $$A=\{0, 1, 4, 10, 12, 17 \} \text{ and } B=\{ 0, 1, 8, 11, 13, 17\}$$ for $n= 37.$ Here are Sidon sets with $A-A=B-B$ is everything in $A \subset \mathbb{Z}/{37}\mathbb{Z}$ except \mathbb{Z}$. So the only restriction on$\{14, 15, 18, 19n$is that they are Sidon sets. Since the elements of$A-A=B-B$range from$-17$to$17$, they are Sidon sets with$A-A=B-B$in$\mathbb{Z}/{n}\mathbb{Z}$for any modulus, 22prime or not, 23\}.$ starting with $n=35.$ It can does not work for $n=32,33,34$. It does also work for $n=31.$ However for $n=31$ we have $B=11A.$ Also, for $n=35$ we have $B=17A-11.$ These are the only cases where an affine map takes $A$ to $B$. I found this example looking in $\mathbb{Z}/{n}\mathbb{Z}$ but it turns out to be checked that known. I think it could be decided if there are no any exceptions for $m,b$ with m=4,5$(other then the special case$B=\{ma+b \mid m=4$and$n=13$) however the method seems tedious. Here is a \in A \}.$ There sketch: We may be assume that $A=\{0,1,a,b\}.$ If $B-B=A-A$ then $s-r=1$ for some construction here but I did not find one yet$r,s\in B.$ We can translate to have $r=0,s=1$ so that $B=\{0,1,w,x\}.$ Further more, $w,x$ are elements of $A-A.$ Three of the 13 elements ($-1,0,1$) are ruled out, leaving $\binom{10}{2}=55$ possible cases. Many of these (like $w=-x$, $w=1-x$ and $w=x-1$) are immediately seen to be impossible. The ratios between elements rest come in pairs $A$ \{0,1,w,x\}$and$\{0,1,1-w,1-x\}\$ which are all squares reflections of each other. If we try $B=\{0,1,1-a,a-b\}$ then $A-A$ and $B-B$ overlap in $9$ elements leaving $\{b,-b,1-b,b-1\}$ unmatched in $A-A$ and $\{1-2a+b,-1+2a-b,a-b-1,1-a+b\}$ in $B-B$ If $b=1-2a+b$ then $n=2a-1$ and $B$ is seen to be a translated reflection of A. We can't have $b=1-a+b$ because $a \subset ne 1.$ The other two possibilities for $b$ also fail. But if we try $B=\{0,1,-a,b-1\}$ there are $6$ unmatched expressions in each of $A-A$ and $B-B$ and running through the possible matchups of pairs one case is $b=-a-1,a-1=1-b-a$ which comes out to $a=3,b=-4$ which leaves the remaining elements as $\pm2,\pm 4,\pm 7$ for $A-A$ and $\pm2, \mathbb{Z}/{37}\mathbb{Z}$ but pm4, \pm6$for$B-B$. So this sub-sub-case is possible for, but only partially the case for,$B.$n=13$ where, it can be checked, $B$ is a translated diolation of $A$. 1 N0. It is NOT the case. It is true for $m \le 5$ and all $p \lt 100$ but may always be false for $m=6$ (which requires $p\ge 31.$) At any rate, a specific example is the sets $$A=\{0, 1, 4, 10, 12, 17 \} \text{ and } B=\{ 0, 1, 8, 11, 13, 17\}$$ for $n= 37.$ Here $A-A=B-B$ is everything in $A \subset \mathbb{Z}/{37}\mathbb{Z}$ except $\{14, 15, 18, 19, 22, 23\}.$ It can be checked that there are no $m,b$ with $B=\{ma+b \mid a \in A \}.$ There may be some construction here but I did not find one yet. The ratios between elements in $A$ are all squares in $A \subset \mathbb{Z}/{37}\mathbb{Z}$ but this is only partially the case for $B.$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 88, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9485884308815002, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/220186/ideas-of-linear-independence-in-the-context-of-modules-and-vector-spaces	# Ideas of Linear Independence in the context of modules and vector spaces Let $k$ be a ring and suppose $M$ is a module over $k$. A set $X \subseteq M$ is a minimal generating set if it generates $M$ and no proper subset of $X$ generates $M$. It is easy to see this means that no element of $X$ can be written as a finite $k$-linear combination of the other elements in $X$. However this does NOT correspond to "linear independence" as is the case for vector spaces. For example if you consider $\mathbb Z_3$ as a $\mathbb Z$-module then $\{ 1 \}$ is a minimal generating set but $$6 \cdot 1 = 3 \cdot 1 = 0$$ but $3 \not= 6$. However don't these notions coincide when we look at modules over a field, i.e. vector spaces? Why does $\sum \alpha_i x_i = 0 \implies$ every $\alpha_i = 0$ if $\{x_i \}$ is a basis for a space $V$, but it doesn't hold for arbitrary modules? - Turns out: $$\sum \alpha_i x_i = 0 \implies \text{ each } \alpha_i \text{ is a non-unit}$$ is equivalent to $X$ being a minimal generating set – Paul Slevin Oct 25 '12 at 17:44 ## 2 Answers If $X$ is a minimal generating set in a vector space or more generally in a module $M$ over a skew-field, then $M$ is linearly independent and therefore a basis of $M$: The reason is that if $\sum_{x} \lambda_x x = 0$ is a non-trivial linear combination, then at least one $\lambda_x$ is non-zero and therefore invertible. But then $x$ is a linear combination of the other elements, a contradiction. As you have already seen, this does not work over general rings (since non-zero does not imply invertible there). Basically you have answered the question yourself. By the way, the correct definition of a basis $X$ of a module $M$ is that it is a linearly independent generating system. Equivalently, the canonical map $k^{\oplus X} \to M$ is an isomorphism. - but why does ax = bx imply a = b if x is an element from a min generator set of a vector space? – Paul Slevin Oct 24 '12 at 16:58 If $a \neq b$ then $a - b$ is invertible (if we work over a skew-field), so $(a-b)x=0$ implies $x=0$, impossible. Perhaps you should Linear Algebra before Module Theory. – Martin Brandenburg Oct 24 '12 at 19:42 For that set of generators $x_i$, $1\leq i\leq n$, let $N$ be the submodule of $M$ that it generates. The linear independence condition basically says that the kernel of the obvious map from a free module rank $n$ onto $N$ is zero, and makes $N$ isomorphic to the free module of rank $N$. Of course, for nonfree modules, it will then be impossible to find a generating set that is also linearly independent. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 41, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9037718176841736, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/7070/r-dimensional-mesh-question?answertab=active	# r-dimensional mesh question I am trying to calculate the average distance between any two nodes in an $r$-dimensional mesh. This is a $3$-dimensional mesh with $n=3$ To choose any two points in the mesh there are $\left(n^r \left(n^r - 1\right)\right)/2$ ways of doing this. So if we choose 2 points in the mesh $p\lbrace 1, 2, ..., r \rbrace$ and $q\lbrace 1, 2, ..., r \rbrace$ The distance is calculated as the Manhattan Distance $\sum_{i=1}^{r} \left\| {q_i - p_i} \right\|$ Now i have to find the total distance between all nodes and divide that by the the number of ways of choosing two nodes (given above). I know how to do this by looking at this 3-dimensional case which should work on the r-dimensional case also but i am having trouble expressing it in some sort of sum notation. Here is how i think i can get the total distance between all nodes, i hope this is clear. In this 3-d mesh let's assume the nodes are labelled $\left( 1, 1, 1 \right)$ to $\left( 3, 3, 3\right)$. Lets start at $\left( 1, 1, 1 \right)$ and sum up the distance between this node and all other nodes. Then we move to $\left( 1, 1, 2 \right)$ and sum up the distances between this node and all other nodes except for $\left( 1, 1, 1 \right)$ because we have already counted the distance between $\left( 1, 1, 1 \right)$ and $\left( 1, 1, 2 \right)$. Then we move onto $\left( 1, 1, 3 \right)$ and sum up all except for $\left( 1, 1, 1 \right)$ and $\left( 1, 1, 2 \right)$. We continue this until we permute through in order $\left( 1, 1, 1 \right)$, $\left( 1, 1, 2 \right)$, $\left( 1, 1, 3 \right)$, $\left( 1, 2, 1 \right)$, $\left( 1, 2, 2 \right)$, ect... Does that make sense? That should give me the total distance without counting anything twice. Then i divide that by the number of ways of choosing 2 nodes and i will have the average distance. Does this sound correct to you? Any help would be appreciated. - ## 1 Answer This approach is correct, but more work than required. Using the taxicab metric each dimension is independent. Taking a cube of side n you can count the vertical segments used as follows: from the bottom layer to the next layer up, we count $n^2$ (points on the bottom layer) times $n^2(n-1)$ (points above the bottom layer) = $n^4(n-1)$. For the next layer you have $2n^2n^2(n-2)$ and so on. The $n^4$s distribute out and you can just think about the total distance along a line of length n. This is $\sum_{i=1}^{i=n}i(n-i)$. Then multiply by $n^4$ and you have the total usage in the vertical direction. Then multiply by 3 for the total usage in each direction. And divide by the number of pairs of points, which you have calculated already. The extension to different number of dimensions or non-cubical meshes should be clear. - does this approach work for all values on n, not just n=3 in the example above? – gprime Oct 18 '10 at 11:44 Yes, you just have to change some of the constants. The $n^4$ becomes $n^{(2*(r-1))}$ in r dimensions. The 3 to multiply by becomes r. And the number of points to divide by becomes the value you already calculated. – Ross Millikan Oct 18 '10 at 15:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9415808916091919, "perplexity_flag": "head"}
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http://samjshah.com/2013/03/12/polar/	# polar! Posted on March 12, 2013 by After my last unit, which didn’t go as well as I’d have liked, I have been teaching a unit on polar. Where one space (rectangular) transmogrifies into another space (polar). I’ve been content with what the kids have done this unit. Again, they did all the discovery. • How to change a point in rectangular coordinates to polar • How to change a point in polar to rectangular • How to graph a polar point — and find a bunch of different ways to represent that point in polar coordinates (showing that polar coordinates are not unique, unlike rectangular coordinates) • How to convert from a rectangular equation to the polar equation • How to convert from a polar equation to a rectangular equation • How to graph simple polar equations by hand • How to conceptually understand what the polar graphs will look like, and why • How to graph complex numbers on the complex plane • How to represent complex numbers in polar form (“cis”) • How to multiply and divide complex numbers using polar form • How to take find roots of unity (haven’t taught yet… throwing it in tomorrow…) This unit, I’ve been more conscientious about collecting their work almost each night, so I could see their problems. One big thing was that I identified that rectangular points in the 2nd and 3rd quadrants provided difficulty when converting to polar (because of the inverse tangent function, which only yields angles in the 1st and 4th quadrant). Because of that, I was able to target that and bring that up in class, and incorporate more points there. I think I made some solid — but very basic — materials for polar. I’m going to share them below, and explain each one of them… [In Word format, in case you want to use/modify them, they are here: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 1. Introduction to a new space: This is my introduction to polar. I introduce the idea of isotropy (without using the word) and say that we need things to anchor us so we can refer to them. One way — the way we’ve always done it — is using how for north/south we’ve gone from the origin, and how far east/west we’ve gone from the origin. Our regular rectangular coordinates. But… BUT there is another way. I have them play a few rounds of polar battleship on Geogebra to get a sense of it, without telling them anything. Then, without any direct instruction, students move on to actually convert from rectangular coordinates to polar. I do a little direct instruction about how to plot polar coordinates, when $r$ is negative and when $\theta$ is negative. And then they do some plotting, and then convert from polar coordinates to rectangular coordinates. Most importantly, they figure out how to do these conversions on their own. 2. Graphing in Polar: The warm up refreshes kids on the equation for a circle, but it also importantly has them graph a horizontal and a vertical line. When we talk about the warm up, I really emphasize that for the equation $x=-2$, what does that mean about the $y$ coordinate? (“It can be ANYTHING! There is no restriction placed on $y$!”) The reason this is so important is that when they graph $r=2$, I hope they say “well $r$ is stuck at $2$, but the angle can be anything” (and similarly, for $\theta=\pi/4$, “I suppose $\theta$ is stuck at $\pi/4$, but $r$ can be anything.”). The rest is pretty much saying: okay, we’ve got our conversions for individual points… can we use them to convert whole equations? (Yes.) 3. Understanding Polar Graphs: Now, after all of this, we’re finally going to graph our first not-so-simple polar equation by hand. We do the very basic first one together, and then they’re off to the races! I wasn’t sure how much I truly cared about them graphing by hand, but it’s clear to me now that doing a few of these by hand really brings to life what graphing in polar truly means. Especially when you get negative r values. 4. Polar Graphing Exploration: This was a day and a half of conceptual lifting, done by the students. There is the most amazing polar applet created by David Little. More than anything else, this one applet has truly let me understand how polar graphs are created. It’s simple, and amazing. I didn’t want to teach my kids about the different names for the different graphs, and to identify what the graph will look like based on the equation. Instead, I wanted them to be able to understand why some graphs have a dent, why some look like a loop with a loop-de-loop inside of it, why some are spirals, and why some are flowers. So I had them use the applet — and the packet — to explore. And I have to say: they were really making solid connections, and having good conversations, as they were going through it. If you use it, the one thing you want to tell the kids is to “not touch a” and “after you finish analyzing one graph, change b back to b=0.1\$”. Honestly, this was the most “risky” of my classes because instead of staying traditional, I said “here, learn it.” 5. Shape of a Polar Graph: After they finished their exploration, I wanted to see what they took away from it, so I gave this out. I collected it and read through them. Some kids fared better on it than others. I didn’t have time to mark them up and give lots of individual feedback, so instead I created a solution packet and I gave it to them — letting them compare, ask questions, etc. 6. Polar Practice: After the last assessment, because I wasn’t totally confident on my kids’s ability to do the basic types of problems, I whipped this set of problems up, which I gave to them, collected, and harshly marked up. This was the best feedback for me, because I got to see what kids were confident about, and what they struggled with. 7. Complex Numbers, the Complex Plane, and Polar Coordinates: I start bringing complex numbers into the mix. I first have them spend a few minutes trying to multiply and divide complex numbers, but it starts getting tedious and annoying (it’s supposed to). We talk about what a complex plane looks like, how I can graph complex numbers on them, and how complex numbers not only have a rectangular (a+bi) form, but also a polar form (involving $r$ and $\theta$). They don’t seem to have any trouble getting that. Then: BAM! I immediately have them do a problem, look for a pattern, and make a conjecture. The theorem we’re going for says that if you have two complex numbers written in polar form, if you multiply these numbers, you simply have to multiply the $r$ values and add the $\theta$ values. BAM. Awesome. So they see this. We talk about whether it’s a proof or not (it’s not). We talk about another example to evidence it, and then I let them loose on complex multiplication and division problems. 8. Basic Practice Problems for Complex Numbers: These are the practice problems I gave for students to work on. I also taught them how to use their calculators to input, store, and multiply/divide complex numbers. 9. Advanced Practice with Complex Numbers: This is a short sheet that has kids prove DeMoivre’s theorem. 10. Complex Roots: I wanted to teach roots of unity, but I don’t have a lot of time, so I made this sheet up — and we’re going to walk through it together. (Tomorrow.) We almost never do that. But it’s our last class before the test, and I think it’s just so cool that I have to show it to them. Reflecting back, I feel kinda bad that I didn’t design this sheet backwards. Start by having students draw a perfect, regular pentagon on the coordinate plane (letting one point be (1,0) and the center at (0.0)). Then have them find the coordinate of the vertices of the pentagon. Then talking through the vertices to conclude they all are roots of unity (if we consider the plane a complex plane). But eh, I didn’t. ### Like this: This entry was tagged Pre-Calculus. Bookmark the permalink. ## 7 thoughts on “polar!” 1. Pingback: Blogging as Therapy \| Reason and Wonder 2. Eric How long did you take to go through the 10 sheets? 3. I counted 9 instructional days. I go at the pace of my kids. I didn’t quite do much with the last sheet, and we spent a solid 2 days on the applet activity. In reflections I had my kids write about what they felt stood out to them this quarter, a number of them talked about the connections and the “a ha” moment they were able to draw with that activity. 4. Eric Thanks for the reply. I’m in my 1st year teaching math/physics (taught 3rd/5th grade prior to this year for 5 years) so I’m trying to figure out how to not just lecture/practice/rinse/repeat every day. I’m only planning on 4-5 days of polar and don’t have access to class computers, but your stuff at least opened some conceptual doors that I’ll try with my students. On another note, could you approve me follow request on Twitter, I’m @swedenese 5. Good luck with it! Also, followed you back. In general I don’t approve anyone who doesn’t have “math teacher” written in their description… because there are a lot of spam bots out there… So that’s why I hadn’t clicked approve. But now I know you’re a real math teacher so yay! 6. Pingback: Introducing Conic Sections \| Continuous Everywhere but Differentiable Nowhere 7. Thanks for all of this, Sam! I’ve been stealing a packet here and a packet there for our current work on Polar coordinates because I didn’t like the way I did it last year. Of course, we’re also 1 month behind where I was this time last year–good thing I don’t have an end-of-year exam to answer to (Instead, I’d like to believe that my students get to learn more problem-solving abilities than others who have an end-of-year exam…) Unfortunately my class isn’t set up the same way as yours, so using these are a little like trying to fit a square peg in a round hole, but I do appreciate the discussions that are arising from my students, and the many different ways you build intuition of polar coordinates. # Tags This is a work in progress (not all posts are tagged yet). But it's all due to the efforts of @crstn85. Thank you! Algebra II Pre-Calculus Calculus Multivariable Calculus Standards Based Grading General Ideas for the Classroom Big Teaching Questions Good Math Problems Mathematical Communication Other # Email Subscription Join 250 other followers # Blogroll (0.6014,-0.1169) (-0.4777,0.1747) Blog at WordPress.com. \| Theme: Confit by Automattic. Follow ### Follow “Continuous Everywhere but Differentiable Nowhere” Get every new post delivered to your Inbox. Join 250 other followers %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 16, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9392416477203369, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/22030/can-someone-explain-the-science-behind-mits-230-efficient-leds	# Can someone explain the science behind MIT's 230% efficient LEDs? I was reading Gizmodo the other day and I didn't quite understand the Physics behind this. Could anybody shed some light on how this effect actually works? - Good question. Though I don't know enough to make an answer, it seems to me that what is going on is that the electricity is essentially catalyzing light production in the LED. The light production itself is merely the result of an endoenergetic event and that energy can come from many sources. – AdamRedwine Mar 7 '12 at 20:01 2 This is really interesting. It sounds crazy at first but they basically invented a phonon to photon heat pump and I am curious what the limits of these processes are. – Alexander Mar 7 '12 at 20:41 The paper is here (behind paywall) link.aps.org/doi/10.1103 and the synopsis at Physics is here /PhysRevLett.108.097403 physics.aps.org/synopsis-for/10.1103/PhysRevLett.108.097403 . The synopsis confirms its a heat pump. – Frédéric Grosshans Mar 7 '12 at 21:38 2 The only real answer here is more incompetent science journalism. – Colin K Mar 8 '12 at 4:03 @ColinK : It is not incompetent science journalism. The abstract of the Physical Review Letter itself and the answer below show that the efficiency is indeed above 1. – Frédéric Grosshans Mar 13 '12 at 14:26 show 2 more comments ## 5 Answers As others have said, a heat pump with an efficiency greater than unity doesn't have any problem with the laws of thermodynamics. There are still a few problems with this which mean that it isn't as good as it seems. The efficiency of a heat pump depends on the temperature difference - you can get higher efficiencies when you're just increasing the temperature a bit. Now the effective temperature of white light is the temperature of the sun's surface, so there's no way that this can be a light emitting diode. In fact it emits at a wavelength of about 2.5 microns, which is in the infrared. However, that still corresponds to a temperature of about 1200 K. The device is heated to 135 degrees C, i.e about 400 K. Thats still too much of a temperature difference for a heat pump to have 230% efficiency, so there's clearly something else going on. Were they adding in the normal black body radiation from the device? No, they were subtracting it, but the interesting thing is that it is much larger than the emission from the device. In the paper they say that the black body emission in the appropriate wavelength range is about 40 nW, while the emission from the device is 69pW. This explains how they can have a heat pump with such high efficiency. If you think of the effective temperature for the emission, well 40 nW is 135 Celsius, so 40.069 nW will be a bit more than 135 Celsius. So there's no problem in having a heat pump with 230% efficiency. - I think that this is the most coherent and easy to understand answer – Kian Mar 13 '12 at 11:57 I can only hypothesize that the extra energy is due to cooling of environment. As there is some input of energy, that would not contradict the second law of thermodynamics. - That seems right -- quoting from the abstract, "The extra energy comes from lattice vibrations, so the device should be cooled slightly, as occurs in thermoelectric coolers." What they actually claim is that their device outputs more light than the electrical energy it consumes. Why this is not in violation of the second law of thermodynamics -- I still don't understand. – Anonymous Mar 7 '12 at 21:17 2 – akhmeteli Mar 7 '12 at 21:51 I skimmed their paper, one thing that I noticed is that they heat their LED well above ambient temperature. I'm curious if it could still work if it was in equilibrium or if blackbody radiation would disrupt this effect. – user2963 Mar 7 '12 at 22:28 @Anonymous : The first line of the paper is clear about the second law : "The presence of entropy in incoherent electromagnetic radiation permits semiconductor light-emitting diodes (LEDs) to emit more optical power than they consume in electrical power, with the remainder drawn from lattice heat." – Frédéric Grosshans Mar 9 '12 at 11:53 I just read the paper today and here's my take, with several points still not really worked out, any help would be appreciated. The last paragraph of the paper is extremely illuminating, I'm quoting it here (it should be fine right?): In conclusion, a new regime for electroluminescent cooling of a semiconductor diode was experimentally demonstrated. In this regime application of a forward bias voltage V less than the thermal voltage $k_BT/q$ im- poses a small deviation from thermodynamic equilibrium on the device. In response, the rates of both radiative and nonradiative recombination in the device’s active region have contributions at linear order in V, so that the external quantum efficiency EQE is voltage independent. As a result, the LED’s optical output power scales linearly with voltage while the input power scales quadratically, resulting in arbitrarily efficient photon generation accom- panied by net electroluminescent cooling of the solid at low bias. Experimental evidence was presented for small band gap emitters at high temperature. So, the clever part of this approach is operating at the low-bias region (where the thermal voltage is a lot greater than the forward-bias). In this regime then, the steady-state carrier concentration is raised and this quantity ($\delta n$ and $\delta p$) are then proportional to the forward-bias (from Taylor series of the $e^{qV/k_BT}$ relationship). Both radiative (producing light) and non-radiative (producing heat) recombination processes are proportional to $$np=n_0p_0e^{qV/k_BT}$$ This means the rate of recombination is also proportional to the voltage applied. I'm not exactly convinced that $\eta_{EQE}$, the external quantum efficiency is subsequently voltage independent. But mathematically it probably comes from: $$\eta_{EQE}=\frac{recombination_{radiative}}{recombination_{total}}$$, voltage drops out and the result is indepedent from V. Finally, the above unity efficiency comes from how optical power and input power scales differently according to the equation they used: $$P_{cool}=I(\hbar\omega/q)\eta_{EQE}-I^2R$$, where the first term is the total optical power, the second term is the total electrical power from the plug. I think the equation makes more sense if we arrange the first term into: $$\frac{I}{q}(\hbar\omega)\eta_{EQE}$$, which says the total optical power is the emitted energy from $\eta_{EQE}$ of the total electron-hole recombinations. So following this train of thought, the paper makes sense mathematically. A few questions remain for me though: 1) This relationship reveals that there is an optimal forward-bias at which the heat-pumping is optimal. Yet in the abstract, it states "the device's wall-plug efficiency diverges as V approaches 0". Am I missing something here? 2) The whole train of math makes sense. The physical picture I'm getting is that, in low-bias, excess carriers are created and a fixed fraction of them will combine radiatively. And the total light emitted somehow exceed the total input electrical power, thus leaving the $(1-\eta_{EQE})$ carriers...have no idea... Glancing through Ram's other papers (http://www.rle.mit.edu/sclaser/Publications/Bias-dependent.pdf), seems like this would be explained pretty easily with a knowledge of optical cooling. Can anyone explain physically how low forward-bias FORCES extracting lattice energy? - Again, another hypothesis: LED's work on semiconductors. If the outside environment is hotter, we get excitation of majority carriers, which will supplement the light. It may be possible that the majority carriers are not excited enough when there is no electricity. Could be a laser-like stable-metastable effect. - 2012-03-08.17:37 EST: Kian Mayne: Good question! Since everyone knows energy is conserved, the title of the Gizmodo article summary sounds instantly self-contradictory. It certainly makes you want to figure out what the authors really meant, since unless they completely misunderstood the original paper, they clearly did not mean that you get 2.3 times as much energy out as light as you put in as electrical energy. A careful reading of the summary articles points out near the end that what they are really talking about is higher overall efficiency achieved by converting not just electricity into photons, but also "waste heat" into light also. The article defines "waste heat" as "... vibrations in the bulb's atomic lattice ...", which is certainly true since that pretty much is the definition of heat. They also mention ambient temperature affecting the process, but the summary itself doesn't give enough information to assess that part of the claim. So, what's almost certainly going on is this: Sending electricity through their device is generating both light and heat, with more of the energy going into heat than into light. That's not surprising, since all light systems generate heat also. The clever part of what this group has done -- and it really is clever, with some nice potential if it can be scaled up to regular scale LEDs -- is that they've found a way to recapture a large chunk of that heat energy and convert that into light also. So, if you do the percentages in the more obvious fashion of looking at joules of energy in the form of light over joules of electrical power sent into the device while it is at equilibrium operating temperature (to avoid that ambient heat confusion), you will get a percentage that is less than 100%, as you would expect given conservation of energy. So, the title of the Gizmodo summary is technically accurate, but it is also confusing because it refers to relationship that exists within the overall energy budget of the electrical current driving it. For example -- and I'm making these figures up! -- if the device converted 30% of its input electrical power directly into photons via traditional hole-electron recombination, and 39% of input power to photons via the new heat-capture, the overall budget would still lose 31% of the original input power, but would nonetheless radiate 2.3 times as many photons (230%) as would a 30% traditional LED for the same electrical power input. So why might the be important if it is scaled up? Two reasons, really. The first of course is flat-out efficiency. A 130% additional output for the same energy inputs is a very, very impressive gain if it can be scaled up! But the second is every bit as important, and possibly even more important in the broader scheme of things: Such a bulb would run far, far cooler than a regular LED. Most people don't realize that at this point, the main reason why LED lights have not yet displaced other forms of lighting is that despite how cool they are in comparison both to incandescent and florescent lights, heat is still a huge factor in creating large, bright LED lights, since the LEDs are solid state devices with some definite limits on how much heat they can handle before the lose efficiency or begin to incur physical damage. So, this is good stuff. Scale up is always the issue in such cases, but simply showing that something like this is even possible is the real start. This group seems to have done just that. - If an article receives an 'Editor's suggestion' from PRL it is seems that this is not only due to some 'superb press corp' from MIT. Harshly critizing someone else's publication without even reading it is not constructive in any way. – Alexander Mar 8 '12 at 19:31 Alexander, you are correct, and I apologize. Please see my complete rewrite above. – Terry Bollinger Mar 8 '12 at 23:31 Reading the PRL paper, it's not what you describe : the 230% efficiency is from the plug power: So sending electricity in the device does not generate heat, ti transfers it from the device to the light. – Frédéric Grosshans Mar 9 '12 at 11:53 Sort of a heat pump then, hmm? Interesting. Heat pumps of course still require electrical energy to run, but are more efficient. – Terry Bollinger Mar 10 '12 at 0:51 1 Terry: It really seems to be an LED heat pump with a 230% efficiency from the input electric power. Unfortunately their effect only showed these properties at an 'ambient' temperature level above 100°C, but it is just a prove of concept so far. Thank you for your rewrite, removing the -1. – Alexander Mar 11 '12 at 14:52 show 1 more comment ## protected by Qmechanic♦Dec 19 '12 at 12:53 This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9499267339706421, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/09/15/power-series-expansions/?like=1&source=post_flair&_wpnonce=7b62a7b6bf	The Unapologetic Mathematician Power Series Expansions Up to this point we’ve been talking about power series like $\sum\limits_{n=0}^\infty c_nz^n$, where “power” refers to powers of $z$. This led to us to show that when we evaluate a power series, the result converges in a disk centered at ${0}$. But what’s so special about zero? Indeed, we could just as well write a series like $\sum\limits_{n=0}^\infty c_n(z-z_0)^n$ for any point $z_0$. The result is just like picking up our original power series and carrying it over a bit. In particular, it still converges — and within the same radius — but now in a disk centered at $z_0$. So when we have an equation like $f=\sum\limits_{n=0}^\infty c_n(z-z_0)^n$, where the given series converges within the radius $R$, we say that the series “represents” $f$ in the disk of convergence. Alternately, we call the series itself a “power series expansion” of $f$ about $z_0$. For example, consider the series $\sum\limits_{n=0}^\infty\left(\frac{2}{3}\right)^{n+1}\left(z+\frac{1}{2}\right)^n$. A simple application of the root test tells us that this series converges in the disk $\left\|z+\frac{1}{2}\right\|<\frac{3}{2}$, of radius $\frac{3}{2}$ about the point $z_0=-\frac{1}{2}$. Some algebra shows us that if we multiply this series by $1-z=\frac{3}{2}-\left(z+\frac{1}{2}\right)$ we get ${1}$. Thus the series is a power series expansion of $\frac{1}{1-z}$ about $z_0=-\frac{1}{2}$. This new power series expansion actually subsumes the old one, since every point within ${1}$ of ${0}$ is also within $\frac{3}{2}$ of $-\frac{1}{2}$. But sometimes disks overlap only partly. Then each expansion describes the behavior of the function at values of $z$ that the other one cannot. And of course no power series expansion can describe what happens at a discontinuity. Like this: Posted by John Armstrong \| Analysis, Calculus, Power Series 6 Comments » 1. [...] Power Series So we know that we can have two power series expansions of the same function about different points. How are they related? An important step in this [...] Pingback by \| September 16, 2008 \| Reply 2. [...] Now, to be specific: if the power series converges for to a function , then has a derivative , which itself has a power series expansion [...] Pingback by \| September 17, 2008 \| Reply 3. [...] we have power series expansions of functions around various points, and within various radii of convergence. We even have formulas [...] Pingback by \| September 18, 2008 \| Reply 4. [...] Okay, we know that power series define functions, and that the functions so defined have derivatives, which have power series expansions. And thus [...] Pingback by \| September 27, 2008 \| Reply 5. [...] what functions might we try finding a power series expansion for? Polynomials would be boring, because they already are power series that cut off after a finite [...] Pingback by \| October 7, 2008 \| Reply 6. [...] to stop what I was working on about linear algebra. Instead, I set off on power series and how power series expansions can be used to express analytic functions. Then I showed how power series can be used to solve [...] Pingback by \| October 16, 2008 \| Reply « Previous \| Next » About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 24, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9397340416908264, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/247082/what-can-be-said-about-the-galois-group-of-fgx	# What can be said about the Galois group of $f(g(x))$? Supposing we know the Galois groups of $f(x)$ and $g(x)$ over $K$, what can be said about the Galois group of $f(g(x))$? I suppose we can restrict the question to normal polynomials over $\mathbb{Q}$, though the general case would be interesting also. - 1 Not much, I suppose. With fixed $g$ cubic and varying $f$ even just linear, you can obtain $S_3$, $C_3$ and even $S_2$ (i.e. produce a reducible cubic). – Hagen von Eitzen Nov 29 '12 at 7:29 ## 1 Answer If $f(x)$ is a separable polynomial of degree $n$ then its Galois group is a subgroup of $S_n$ (via the action on the roots). And in fact any subgroup $H$ of $S_n$ occurs as a Galois group of a separable polynomial of degree $n$ over some field. From this we get that if $f,g$ are polynomials of respective degrees $n,m$ then the Galois group $G$ of $f(g(x)$ is a subgroup of $S_{nm}$. However here not every subgroup can be achieved, because the action of $G$ on the roots of $f(g(x))$ is imprimitive (provided $n,m>1$). The group that should replace $S_{nm}$ is the permutational wreath product $S_m\wr_n S_n$. Then one can show that $G$ always embeds into $S_m\wr_n S_n$, via the action on the roots, and that for any subgroup $H$ of $S_m\wr_n S_n$, there are a field $K$ and $f,g\in K[X]$ of respective degrees $n,m$ such that $Gal(f(g(x)),K)\cong H$. I will end with the definition of the permutational wreath product: it is the semi-direct product $S_m^n \rtimes S_n$ where the action of $S_n$ on $S_m^{n}$ is the translation action, that is $$(\tau_1, \ldots, \tau_n)^\sigma := (\tau_{\sigma(1)}, \ldots, \tau_{\sigma(n)})$$ for all $\tau_i\in S_m$ and $\sigma\in S_n$. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 40, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.92661452293396, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/23151/winds-sources-and-drains-see-live-map	# Wind's Sources and Drains (see live map!!) I was pointed out by a friend to this website that shows live map of wind in US. It sometimes show interesting places where all the wind seems to converge and vanish. What's the origin of such "wind drains"? Do pure geological phenomena lead to simple cancellation of wind vectors or more complex effects? Is that (can that be) somehow explored in terms of wind energy? - – Manishearth♦ Apr 2 '12 at 4:34 I would probably say that these are low pressure regions, so the wind converges there. Or it may have to do with temperature. The wind isn't converging there, the wind is caused by the pressure/temperature difference. (I think) So it's already being harnessed in the form of wind energy. – Manishearth♦ Apr 2 '12 at 5:13 1 @Manishearth: can't be--- the pressure wouldn't stay low. This must be a surface measurement, the air heats and flows the other way on top of the atmosphere. It wouldn't be possible any other way. – Ron Maimon Apr 2 '12 at 5:33 @RonMaimon: Good point... It would be pressure if the spot was a one-time phenomenon, but its still there so yeah, its probably different directions of flow on the atmosphere. – Manishearth♦ Apr 2 '12 at 5:34 ## 3 Answers Air is more or less incompressible for the purposes of this discussion, and there is no way that the average wind speed can converge onto one spot, or else air would pile up there. But this suggests the resolution, since the piled up air will simply move the other way at the top of the atmosphere. There is a third dimension suppressed, and the opposing divergence at higher altitude makes the total divergence zero. I cannot think of another reasonable reason why this structure would persist. But this would mean that at some places near the convergence line in the central USA, you will observe winds flowing in opposite directions at the bottom and top of the atmosphere. - Wind is, in general, powered by convection. The ground heats up because the sun shines on it. The top of the atmosphere is generally cooler because heat escapes from it into space in the form of infra-red radiation. (The ground can't radiate into space in this way because of greenhouse gases in the atmosphere, which provide it with a kind of thermal blanket.) Since hot air is more buoyant than cold air, it tends to want to rise, and this results in the formation of convection cells, where hot air rises in some places and cool air comes down in other places. As the other answers have pointed out, the site you linked to shows only the wind speeds at the surface. There are plumes of hot air rising to the top of the atmosphere, and what you see on the map is the air being sucked into the bottom of the plume. If you could see the whole thing in 3D you would see that air rising and then spreading out again at the top of the atmosphere, and then eventually coming back down again in the high-pressure regions where the surface-level winds are diverging. Because the Earth rotates, there is also a Coriolis force which makes the air tend to rotate as it gets drawn inward toward the plume. You can see this rotation on the map as well. When the same thing happens with a much greater intensity over the ocean, we call it a hurricane. - To be clear, Ron's answer is quite correct already, I just wanted to go into a bit more detail about why the wind moves in a three-dimensional way. – Nathaniel Apr 2 '12 at 19:56 That map is showing surface winds (typically measured 10 metres above the ground). It is quite possible for the surface wind that's been graphed here to be two-dimensional: without poking around in the original metadata, and in the algorithm used to graph it, it's rather hard to tell. When many winds seem to converge at a point, and there looks to be a wide-area rotation going on, then either surface winds seems to converge and vanish at a point, because the energy in surface winds is getting diverted from lateral to vertical movement; or surface winds appear to emerge from a point, because the reverse is happening. More broadly, when winds vanish, any combination of several things may be happening: • when fronts collide, they can push the moving air upwards, thus increasing the kinetic energy of the air higher up, but making the wind at the surface still • all winds blow themselves out: surface friction turns kinetic wind energy into low-grade heat. The higher the roughness of the surface, the faster this happens. • particular configurations of landscape can cause wind to change direction and height, meaning it might vanish from the surface-wind charts • it's not unusual for the source data to come in the form of zonal and meridional components ($u$ and $v$ vectors), averaged over an hour. To take an extreme example of how that can mislead: if you have a wind that's 10 m/s south for 30 minutes, that then changes direction to be 10 m/s north for 30 minutes, then the average wind in that hour would be shown as zero, in $u$-$v$ wind vectors. - This is complicating things for no reason---- it is not a combination of anything, all your reasons are pointless and don't change anything except for the one I pointed out. For example, averaging over a 30 minute window does not change the fact that the divergence has to be zero (because averaging is linear), "making heat" doesn't change anything, and points 1 and 3 are the same and the same as mine. The notable thing is that the 3rd dimension matters, One would think that the average flows are 2d on this scale, and they aren't. – Ron Maimon Apr 2 '12 at 12:47 Thanks for expanding, +1. – Ron Maimon Apr 3 '12 at 7:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9490766525268555, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/queueing-theory+probability	# Tagged Questions 2answers 130 views ### When does the next bus come? People arrive at a bus stop according to a Poisson process at rate $\lambda$ per minute. The bus leaves every $n$ minutes, but you have no idea when the last bus left. You observe that there are $k$ ... 0answers 119 views ### Boundedness of expected reward Markov chain (may be related to discret $M/M/\infty$ queue) [EDIT]: I read a bit on $M/M/\infty$ queue and it may not be the right comparison and my notation may be confusing (I'm in discrete time and $\lambda,\mu$ look likes rates when they are probability). ... 0answers 87 views ### Is Queueing Theory dead? [closed] I was studying queueing theory for my class and noticed that we are now able to either solve with certainity most queiening problems or simulate them. is queueing a dead research area? I read this ... 0answers 43 views ### Proof of $\text{Poisson}(\lambda p)$-arrivals. We have a queue where people pass out of it with $\text{Poisson}(\lambda)$ and they come in with probability $p$. I understand that the arrivals follow $\text{Poisson}(\lambda p)$, but how can ... 1answer 113 views ### Markov Chain Transition Intensity Conversion I have a question about converting a 3-state discrete state, continuous-time, markov chain to a 2-state. My 3-state model has states: Well (state 1), Ill (state 2) and Dead (state 3). ... 1answer 145 views ### Find the probability that the second customer to arrive has to wait to be served if arrival time is exponential and serving time is uniform Customers line up to be serviced according to a Poisson process at an average rate of five per hour. If the time it takes to serve one customer is a continuous uniform random variable on $[0,4]$, ... 0answers 46 views ### Probability that a presentation will be on time I'm visiting a grand conference soon which has 30 lectures in a single day. Unfortunately they don't say when each of the lectures will be held. Only that each will be held in the order written ... 2answers 512 views ### One vs multiple servers - problem Consider the following problem: We have a simple queueing system with $\lambda%$ - probabilistic intensity of queries per some predefined time interval. Now, we can arrange the system as a single ... 1answer 100 views ### Queueing Theory - Probability that all jobs have been served? Suppose I have M/M/1 system with $\lambda = 4$ per hour and $\mu = 5$ per hour. How can I find out if all jobs have been served after, say, 8 hours? At first I thought about doing $P(n > 40)$ since ... 2answers 254 views ### The processing time of an M/M/1 queue Suppose I have a queue with $\lambda$ and $\mu$. I can calculate the probability that there are 2 objects in the queue trivially, but how can I compute, for example, the probability that it takes an ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9455891251564026, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/feynman-diagram+wick-rotation	# Tagged Questions 1answer 209 views ### Wick Rotation, interpretation of $\bar{p}^2$ vs the usual $p^2=m^2$ Suppose we use the metric $(+,-,-,-)$ thus the momentum squared is $p^2 = p_0^2-\vec{p}^2 = m^2>0$ Defining $p_E:=\mathrm{i}\cdot p_0$ and $\bar{p}:=(\,p_E,\vec{p})$ with Euclidean norm ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8315695524215698, "perplexity_flag": "head"}
http://mathoverflow.net/questions/72173?sort=votes	## Coderivations of S(V) correspond to linear maps S(V) -> V. Only over characteristic 0? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Definition. Let $k$ be a commutative ring. Let $V$ be a $k$-module. We turn the symmetric algebra $\mathrm{S}\left(V\right)$ of $V$ into a graded Hopf algebra by defining the comultiplication $\Delta : \mathrm{S}\left(V\right) \to \mathrm{S}\left(V\right) \otimes \mathrm{S}\left(V\right)$ by $\Delta\left(v_1v_2...v_n\right) = \sum\limits_{i=0}^n \sum\limits_{\sigma\in\mathrm{Sh}\left(i,n-i\right)} \left(v_{\sigma\left(1\right)}v_{\sigma\left(2\right)}...v_{\sigma\left(i\right)}\right) \otimes \left(v_{\sigma\left(i+1\right)}v_{\sigma\left(i+2\right)}...v_{\sigma\left(n\right)}\right)$, where $\mathrm{Sh}\left(i,n-i\right)$ denotes the set $\left\lbrace \sigma\in S_n \mid \sigma\left(1\right) < \sigma\left(2\right) < ... < \sigma\left(i\right) \text{ and }\sigma\left(i+1\right) < \sigma\left(i+2\right) < ... < \sigma\left(n\right) \right\rbrace$ of all $\left(i,n-i\right)$-shuffles. The counit of this Hopf algebra is simply the projection from $S\left(V\right)$ onto $k$. Definition. Let $k$ be a commutative ring, and $C$ be a $k$-coalgebra. A coderivation of $C$ means a $k$-linear map $c:C\to C$ such that $\Delta \circ c = \left(c\otimes\mathrm{id} + \mathrm{id}\otimes c\right)\circ \Delta$. Remark. Coderivations behave, in some sense, dually to derivations (which is not surprising since the condition $\Delta \circ c = \left(c\otimes\mathrm{id} + \mathrm{id}\otimes c\right)\circ \Delta$ is a kind of dual to the Leibniz identity, when the latter is written in pointfree notation): First of all, if $c:C\to C$ is a coderivation, then $c^{\ast} : C^{\ast}\to C^{\ast}$ is a derivation. The converse holds at least when $C$ is finite-dimensional. As an exercise in reversing arrows, the reader can prove that $\varepsilon\circ c=0$ for every coderivation $c$ of a coalgebra (in analogy to the equality $d\left(1\right)=0$ which holds for every derivation $d$ of an algebra). Theorem. Let $k$ be a field of characteristic $0$. Let $V$ be a $k$-vector space. Then, the maps $\mathrm{Hom}\left(S\left(V\right),V\right) \to \mathrm{Coder}\left(S\left(V\right)\right)$, $X\mapsto \mu \circ \left(\mathrm{id}\otimes X\right) \circ \Delta$ (where $\mu$ is the multiplication morphism $S\left(V\right)\otimes S\left(V\right)\to S\left(V\right)$) and $\mathrm{Coder}\left(S\left(V\right)\right) \to \mathrm{Hom}\left(S\left(V\right),V\right)$, $c\mapsto \pi_1\circ c$ (where $\pi_1$ is the projection from $\mathrm{S}\left(V\right)=\bigoplus\limits_{i\in\mathbb N}\mathrm{S}^i\left(V\right)$ onto the addend $\mathrm{S}^1\left(V\right)=V$) are mutually inverse isomorphisms. This is how I understand Chapter 5, Theorem 4.19 in Eckhard Meinrenken, Clifford algebras and Lie groups. (Unfortunately, the statement of the theorem in Meinrenken's text is obscured by the fact that one direction of the isomorphism - the easy one - is not written out explicitly.) The proof given in this text uses the characteristic-$0$ assumption: first, by assuming "WLOG" that a generic element of $\mathrm{S}\left(V\right)$ has the form $e^v$ for some $v\in V$ (this is made formal by going over to formal power series, but stripped of this formality, this is exactly what is known as umbral calculus for over a century), and second, by using the fact that the primitives of $\mathrm{S}\left(V\right)$ all come from $V$. Question. Does the above theorem hold in arbitrary characteristic? I am sure this is intimately related to the question whether $\mathrm{S}\left(V\right)$ is the cofree graded cocommutative coalgebra over $V$ (or at least whether it is "cogenerated" in degree $1$, whatever this means precisely!). Unfortunately, the only case when I am sure of this is the characteristic-$0$ case, so this is of no help to me. Loday-Valette does not seem to care for positive characteristic too much, either, and it is difficult for me to find any other source. - ## 1 Answer Let us assume $k$ has characteristic $p$. The problem is (to me at least) easier to understand by dualising (assume that we are only looking at homgeneous derivations so that we can take the graded dual and have no problems at least if $V$ is finite-dimensional, things will go wrong even here). Then the statement would say that any linear map $V\to V\subset S(V^\ast)^\ast$ has a unique extension to a derivation of $S(V^\ast)^\ast$. However, $S(V^\ast)^\ast$ is the divided power algebra on $V$ which is not generated as an algebra by its degree $1$ elements so the uniqueness doesn't follow as for the symmetric algebra. In fact for $V=kx$ we have that the divided power algebra is generated as a commutative algebra by $x_n:=\gamma_{p^n}(x)$ and relations $x_n^p=0$. Hence one can arbitrarily choose the value of a derivation on the $x_n$ which makes it very clear that the derivation is not determined by its value on $x_1$. - How do you know the generators and relations for the divided power algebra? – darij grinberg Aug 5 2011 at 13:36 Ah, I see. Lucas' theorem for binomial coefficients modulo $p$ is more useful than I thought... – darij grinberg Aug 5 2011 at 13:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 64, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.94102543592453, "perplexity_flag": "head"}
http://mathoverflow.net/questions/71143/when-a-sequence-of-coefficients-converges-to-the-coefficients-of-a-rational-funct	## When a sequence of coefficients converges to the coefficients of a rational function $R$, when does the sequence $R_n$ converge uniformly to $R$? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $R$ be a rational function of degree $d$ mapping the Riemann sphere to itself:$$R(z) = \frac{a_d z^d + a_{d-1} z^{d-1} + \cdots + a_0}{b_d z^d + b_{d-1} z^{d-1} + \cdots + b_0}$$ where $a_d$ and $b_d$ are not both zero. And suppose that a sequence of coefficients ${(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n}$ converges to the coefficients of $R$ in $\mathbb{C}^{2d+2}$. Let $R_n$ be the rational function with coefficients $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n$. Is it a theorem that: The sequence ${R_n}$ converges uniformly to $R$ on the Riemann sphere if and only if for all $n$ sufficiently large the degree of $R_n$ matches the degree of $R$? I ask because I am reading Beardon's Iteration of Rational Functions and in section 2.8 he introduces a mapping $\Psi:\mathcal{R}\to\mathbb{C}\mathbb{P}^{2d+1}$, which takes a rational function to its coefficients. Beardon omits the details, but claims that $\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$, where $\mathcal{R}_d$ is the space of rational functions having degree exactly $d$. My thought is, if for $n$ sufficiently large $R_n$ has degree $d$, then $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n \in \Psi(\mathcal{R}_d)$. Now $\Psi^{-1}$ is defined and continuous at $(a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n$, because $\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$. And (up to multiplication by a constant) $R_n = \Psi^{-1}((a_d, a_{d-1}, \ldots, a_0; b_d, b_{d-1}, \ldots, b_0)_n)$. Since $\Psi^{-1}$ is continuous on the sequence of coefficients and at the limit of the sequence of coefficients, $R_n$ converges to $R$. But I would like to know: Is the convergence always uniform under these circumstances? Conversely, if for all $n$ there exists $R_{m>n}$ in the sequence with degree not equal to $d$, then ${R_n}$ cannot converge to $R$ because (Beardon also states that) the spaces ${\mathcal{R}_0, \mathcal{R}_1, \ldots, \mathcal{R}_d}$ are the connected components of the space of rational functions with degree at most $d$. The reason that I think this might be a theorem is that Beardon introduces a metric $\rho(R,S) = \sup_{z\in\mathbb{C}_{\infty}} \sigma_0 (R(z),S(z))$, $\sigma_0$ the chordal metric on $\mathbb{C}_{\infty}$, just a couple of pages earlier than the point where he introduces the mapping $\Psi$. It is tempting to read "$\Psi$ is a homeomorphism of $\mathcal{R}_d$ onto $\Psi(\mathcal{R}_d)$" as "$\Psi$ is a homeomorphism of the metric space $(\mathcal{R}_d,\rho)$ onto $\Psi(\mathcal{R}_d)$." If that is the correct reading then $R_n$ converges to $R$ under the metric $\rho$ and the convergence is clearly uniform. But I am not sure that this is the correct reading since Beardon is only explicitly treating $\mathcal{R}_d$ and $\mathbb{C}\mathbb{P}^{2d+1}$ as topological spaces at the point where he introduces $\Psi$. Note: I have edited the question in response to Margaret Friedland's answer, in an attempt to clarify my assumptions (and hopefully my source of confusion). I am a little worried that the reason my point of confusion was unclear is that it is actually trivial. I apologize in advance if that is the case. - Except in trivial cases, your rational function $\bf R$ has a pole somewhere, and generally the poles of $\bf R_n$ will be different. How can it converge uniformly in a neighbourhood of the pole? Or are you thinking of rational functions as mappings of the Riemann sphere to itself? – Robert Israel Jul 24 2011 at 20:36 I am thinking of rational functions as mapping the Riemann sphere to itself. I forgot to state that in the question. I will edit the question to clarify the setting. – Aaron Golden Jul 25 2011 at 3:51 I think you need to clean up your assumptions a little bit, especially at the beginning of the question, and things will become clearer. E.g. Are you considering the coefficients in projective space or not? Do you assume that the limit has degree exactly d or at most d? Margaret Friedland's answer suggests that the answer may change depending on the specific assumptions. – Thierry Zell Jul 26 2011 at 14:36 Thierry, do my assumptions seem more clear now? I am tempted to completely rewrite the question to try to simplify/clarify it, but I don't know if that is appropriate since there are already comments and an answer that depend on the original version of the question for context. – Aaron Golden Jul 28 2011 at 1:55 ## 3 Answers Since you reformulated your question, it deserves a new answer. Now, the theorem in the gray box in the new version is true, but what does it really say? It says that the map $\Psi^{-1}: \Psi(\mathcal{R}_d) \mapsto \mathcal{R}_d$ is continuous! (The topology on $\mathcal{R}_d$ is inherited from $\mathbb{P}^{2d+2}$; on $\Psi(\mathcal{R}_d)$ we have, as you state, the topology of uniform convergence with respect to e.g. chordal metric). So what you are really asking is: how to prove that $\Psi^{-1}$ is continuous in these topologies? This is not so hard if you are used to working with homogeneous coordinates on projective spaces, but there is a subtle point involved. Here is the argument: Let $[r:Z]=[a_0:...:a_d:b_0:...:b_d:Z_0:Z_1] \in \mathbb{P}^{2d+2}\times \mathbb{P}^1$ and define $p([r:Z] \in \mathbb{P}^1$ as $p([r:Z]=[a_0{Z_0}^d+a_1{Z_0}^{d-1}Z_1+...+a_d{Z_1}^d:b_0{Z_0}^d+b_1{Z_0}^{d-1}Z_1+...:b_d{Z_1}^d]$. Then the map $p$ is continuous (being given by homogeneous polynomials in homogeneous coordinates), so as $[r_n:Z^{(n)}] \to [r,Z]$, we have $p[r_n:Z^{(n)}] \to p[r:Z]$. Note that $p[r:Z]$ is the value of the rational function $R$ with the coefficients $r$ at the point $[Z_0:Z_1]\in \mathbb{P}^1$. Note also that the continuity of $p$ means that $R_n$ converge continuously to $R$ on $\mathbb{P}^1$. On a compact space , continuous convergence of a sequence of continuous functions is equivalent to its uniform convergence. The subtlety is the following: the domain of the map $p$ is $\mathcal{R}_d \times \mathbb{P}^1$, not the whole product $\mathbb{P}^{2d+2}\times \mathbb{P}^1$. (The ratio of two polynomials is not defined at their common zeros. We saw this in Beardon's counterexample: the limit expression $z/z$ (or $[Z_1:Z_1]$) in the limit is not defined at the point $0=[1:0]$.) So the assumption that all $R_n$ and $R$ have the same degree $d$ guarantees that there are no factors in common for each numerator and denominator and we do not consider fractional expressions that are not rational functions on $\mathbb{P}^1$. - Thank you very much for this new answer. The relationship between the sequence of coefficients and the sequence of rational functions is clear when everything is spelled out explicitly with homogeneous coordinates, as you've done here. – Aaron Golden Aug 11 2011 at 9:45 I am glad it helps. Homogeneous coordinates are very useful indeed. And even if your interest is in one-dimensional complex dynamics, it pays to have some insight into higher-dimensional setting. – Margaret Friedland Aug 11 2011 at 13:44 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The equivalence you state in your first question cannot be a theorem, because the "if" part is false (take $R_n(z):=(n+1)z, R_n(\infty)=\infty), n \geq 1$ and $R(z)=z$; the degrees certainly match, but $R_n$ does not converge to $R$, even pointwise). The "only if" part is true and can be proved using Rouche's theorem (Beardon does include some detail on this). The convergence of rational functions in the sense of coefficients does not imply uniform convergence: let $R_n(z)=(z+1/n)/z, R(z)=1$; the degree drops for the limit function (this example is right there in Beardon).\ Edit: And of course all $R_n$ have a pole at $0$, while $R$ does not (as Robert Israel suggested) so this is the true reason why uniform convergence does not hold. Still, $(1/n,1,0,1)$ converge to $(0,1,0,1)$. The map between rational functions and their coefficients that Beardon introduces is not the same as the map you wrote above, if by $\mathcal{R}$ you denote the space of all rational functions of degree at most $d$. Your map cannot have a continuous inverse, since the projective space is connected and the space $\mathcal{R}$ is a union of non-trivial connected components $\mathcal{R}_j$ , of functions of degree exactly $j$, $j=0,...,d$. - Thank you for the answer. I see now that my question was unclear. I mean to require that: (1) $R$ has degree exactly $d$. (2) $R_n$ is the sequence of functions associated with a sequence of coefficients converging to the coefficients of $R$ in $\mathbb{C}^{2d+2}$. And (3) For sufficiently large $n$ the degree of $R_n$ is exactly $d$. Then your first sequence is not a counterexample because the coefficients of $R_n$ do not converge to the coefficients of $R$, and in the second example, as you and Beardon both point out, the degree drops in the limit, violating (3). – Aaron Golden Jul 26 2011 at 16:40 After rereading section 2.8 in Iteration of Rational Functions I am convinced that when Beardon writes "$\Psi$ is a homeomorphism of $\mathcal R_d$ onto $\Psi(\mathcal R_d)$," he means $\mathcal R_d$ to be the space of rational functions with degree exactly $d$, with the topology induced by the metric: $$\rho(R,S) = \sup_{z\in\mathbb C_\infty} \sigma(R(z), S(z))$$ where $\sigma$ is the spherical metric on $\mathbb{C}_\infty$ (but Beardon also says that he could equally well have used the chordal instead of the spherical metric because the two differ only by a bounded factor, and I'll make $\sigma$ the chordal metric in what follows). Beardon explicitly treats $\mathcal{R}$ as a metric space (with metric $\rho$) in his proof that the degree function, $\deg:\mathcal{R}\to\mathbb{Z}$, is continuous, and to show that $\mathcal R_d$ is an open subset of $\mathcal R$. He couldn't possibly mean any other topology by the time he introduces $\Psi$. So If $R$ is a rational function of degree exactly $d$ mapping $\mathbb{C}_\infty$ to itself and $R_n$ converges to $R$ in the sense of coefficients, and for all $n$ sufficiently large $\deg{R_n} = d$, then $R_n$ converges to $R$ uniformly on $\mathbb{C}_\infty$ under the spherical (or equivalently the chordal) metric. Proof: Let $\Theta = \Psi(R)$ be the coefficients of the limit function $R$, and let $\Theta_n = \Psi(R_n)$. Since $R_n$ converges to $R$ in the sense of coefficients $\Theta_n$ converges to $\Theta$. Since both $R$ and $R_n$ (for sufficiently large $n$) have degree exactly $d$, $\Theta\in\Psi(\mathcal{R}_d)$ and $\Theta_n\in\Psi(\mathcal{R}_d)$ for sufficiently large $n$. Now, $\Psi$ is a homeomorphism of the topological space $\mathcal{R}_d$ with the topology induced by the metric $\rho$ onto $\Psi(\mathcal{R}_d)$. So $\Psi^{-1}$ is defined and continuous at $\Theta$ and at $\Theta_n$ for large $n$, and $R = \Psi^{-1}(\Theta)$ and $R_n = \Psi^{-1}(\Theta_n)$, so $R_n$ converges to $R$ with respect to the metric $\rho$. Since $\rho$ is a metric of uniform convergence on $\mathbb{C}_\infty$, the convergence of $R_n$ is uniform on $\mathbb{C}_\infty$. QED. ## Issues with poles of $R$ and $R_n$ Some of the responses to this question have indicated that the convergence cannot be uniform in the neighborhood of a pole of $R$ when the $R_n$ do not share that pole. It is true that if $z$ is a pole of $R$ and no pole of $R_n$ approaches $z$, then $R_n$ does not converge to $R$ at $z$. For if $R(z) = \infty$ and $\left\| R_n(z)\right\| < M < \infty$ for all $n$, then as $n\to\infty$, using the chordal metric: $$\sigma(R_n(z), R(z)) = \sigma(R_n(z), \infty) = \frac{2}{(1 + \left\| R_n(z) \right\|^2)^{1/2}} > \frac{2}{(1 + M^2)^{1/2}} \not\to 0.$$ And the same calculation works if all the $R_n$ share a pole not shared by $R$. Margaret Friedland's answer gives an explicit example of this. If $R_n(z) = (z + 1/n)/z$, $R(z) = 1$, then $\sigma(R_n(0), R(0)) = \sigma(\infty, 1) = 2 / \sqrt{2}$. But what if the poles of $R_n$ do not match those of $R$, but approach them? Then $R_n$ may converge uniformly to $R$ on the sphere. We can show this directly with an example. Let $R(z) = 1/z$ and $R_n(z) = 1/(z - 1/n)$. Then, under the chordal metric, for $z\neq 0$: $$\sigma(R_n(z),R(z)) = \frac{2 \left\| 1/(z - 1/n) - 1/z \right\|}{(1 + \left\| 1/(z-1/n)\right\|^2)^{1/2} (1 + \left\| 1/z \right\|^2)^{1/2}}$$ simplifies (try running it through WolframAlpha) to: $$2 / ((\|z\|^2 + 1)^{1/2} (\|n z - 1\|^2 + n^2)^{1/2})$$ which is less than or equal to $2/n$ for $n>0$. And when $z=0$ we have $$\sigma(R_n(0),\infty) = \sigma(-n,\infty) = 2/\sqrt{1 + \|n\|^2} \leq 2/n.$$ So for any given $\epsilon>0$, choose $n>2/\epsilon$, then $\rho(R_n, R) = \sup_{z\in\mathbb{C}_\infty} \sigma(R_n(z), R(z)) \leq 2/n < \epsilon$. The discussion of above indicates that the requirement that $\deg R_n = \deg R$ for all $n$ sufficiently large is actually a requirement that all the poles of $R_n$ converge to the poles of $R$. (Beardon even hints at this, it turns out. He mentions "sliding" the roots and poles of one function to another, when the degrees match, as a way to show that each $\mathcal{R}_d$ is connected.) ## Summary 1. Convergence of the coefficients, on its own, is not even a sufficient condition for point-wise convergence of $R_n$ to $R$ on the sphere. In Margaret Friedland's example, $R_n(z) = (z + 1/n)/z$, $R(z) = 1$ and the coefficients $(1/n, 1; 0, 1)$ converge to $(0, 1; 0, 1)$, but $R_n(0)$ is bound away from $R(0)$ for all $n$. 2. In general if the poles of $R_n$ fail to converge to the poles of $R$ then $R_n$ does not converge to $R$ at some pole of $R$. 3. If the poles of $R_n$ do converge to the poles of $R$ then it is possible that $R_n$ converges uniformly to $R$ on the sphere (for example when $R(z) = 1/z$ and $R_n(z) = 1/(z - 1/n)$). 4. If the coefficients converge and we require that $\deg R_n$ matches $\deg R$ for all $n$ sufficiently large, then $R_n$ converges uniformly to $R$ on the sphere (based on the discussion of Beardon's mapping, $\Psi$). 5. (2) and (4) together yield: If the coefficients of $R_n$ converge to the coefficients of $R$ and $\deg R_n$ matches $\deg R$ for all $n$ sufficiently large, then all the poles of $R_n$ converge to poles of $R$. I don't think that any of these points contradict Margaret Friedland's answer and I hope that this answer makes my original question and thinking a little more clear. I would still like to have a direct proof for item (4). It is unsatisfying to rely on Beardon's $\Psi$ mapping, even assuming that I have interpreted the text correctly, since Beardon's discussion of $\Psi$ is so light on details. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 244, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9460051655769348, "perplexity_flag": "head"}
http://dorigo.wordpress.com/2009/01/12/1939/	# A Quantum Diaries Survivor private thoughts of a physicist and chessplayer ## No CHAMPS in CDF data January 12, 2009 Posted by dorigo in news, physics, science. Tags: CDF, Champ, dark matter, new physics, SUSY trackback A recent search for long-lived charged massive particles in CDF data has found no signal in 1.0 inverse femtobarns of proton-antiproton collisions produced by the Tevatron collider at Fermilab. Most subnuclear particles we know have very short lifetimes: they disintegrate into lighter bodies by the action of strong, or electromagnetic, or weak interactions. In the first case the particle is by necessity a hadron- one composed of quarks and gluons-, and the strength of the interaction that disintegrates it is evident by the fact that the life of the particle is extremely short: we are talking about a billionth of a trillionth of a second, or even less time. In the second case, the electromagnetic decay takes longer, but still in most instances a ridiculously small time; the neutral pion, for instance, decays to two photons ($\pi^\circ \to \gamma \gamma$) in about $8 \times 10^{-17}$ seconds: eighty billionths of a billionth of a second. In the third case, however, the weakness of the interaction manifests itself in decay times that are typically long enough that the particle is indeed capable of traveling for a while. Currently, the longest-living subnuclear particle is the neutron, which lives about 15 minutes before undergoing the weak decay $n \to p e \nu$, the well-studied beta-decay process which is at the basis of a host of radioactive phenomena. The neutron is very lucky, however, because its long life is not only due to the weakness of virtual W-boson exchange, but also by the fact that this particle happens to have a mass just a tiny bit larger than the sum of the bodies it must decay into: this translates in a very, very small “phase space” for the decay products, and a small phase space means a small decay rate. Of course, we have only discussed unstable particles so far: but the landscape of particle physics includes also stable particles, i.e. the proton, the electron, the photon, and (as far as we know) the neutrinos. We would be very surprised if this latter set included particles we have not discovered yet, but we should be more possibilistic. A stable, electrically-neutral massive particle would be less easy to detect than we could naively think. In fact, most dark-matter searches aimed at detecting a signal of a stable massive particle are tuned to be sensitive to very small signals: if a gas of neutralinos pervaded the universe, we might be unaware of their presence until we looked at rotation curves of galaxies and other non-trivial data, and even then, a direct signal in a detector would require extremely good sensitivity, since a stable neutral particle would be typically very weakly interacting, which means that swarms of such bodies could easily fly through whatever detector we cook up unscathed. Despite that, we of course are looking for such things, with CDMS, DAMA, and other dark-matter-dedicated experiments. The existence of a charged massive stable particle (CHAMP for friends), however, is harder to buy. An electrically-charged particle does not go unseen for long: its electromagnetic interaction is liable to betray it easily. However, there is no need to require that a CHAMP is by force THE reason of missing mass in the universe. These particles could be rare, or even non-existent in the Universe today, and in that case our only chance to see them would be in hadron-collision experiments, where we could produce them if the energy and collision rate are sufficient. What would happen in the event of a creation of a CHAMP in a hadron collision is that the particle would slowly traverse the detector, leaving a ionization trail. A weak-interacting CHAMP (and to some extent even a strongly-interacting one) would not interact strongly with the heavy layers of iron and lead making up the calorimeter systems of which collider experiments are equipped, and so it would be able to punch through, leaving a signal in the muon chambers before drifting away. What we could see, if we looked carefully, would be a muon track which ionizes the gas much more than muons usually do -because massive CHAMPS are heavy, and so they kick atoms around as they traverse the gas. Also, the low velocity of the particle (be it clear, here “low” means “only few tenths of the speed of light”!) would manifest itself in a delay in the detector signals as the particle traverses them in succession. CDF has searched for such evidence in its data, by selecting muon candidates and determining whether their crossing time and ionization is compatible with muon tracks or not. More specifically, by directly measuring the time needed for the track to cross the 1.5 meter-radius of the inner tracker, and the particle momentum, the mass of the particle can be inferred. That is easier said than done, however: a muon takes about 5 nanoseconds to traverse the 1.5 meters of the tracker, and to discern a particle moving half that fast, one is requred to measure this time interval with a resolution better than a couple of nanoseconds. The CDF Time-Of-Flight system (TOF) is capable of doing that. One just needs to determine the production time with enough precision, and then the scintillation bars which are mounted just outside of the tracking chamber (the COT, for central outer tracker) will measure the time delay. The problem with this technique, however, is that the time resolution has a distinctly non-Gaussian behaviour, which may introduce large backgrounds when one selects tracks compatible with a long travel time. The redundancy of CDF comes to the rescue: one can measure the travel time of the particles through the tracker by looking at the residuals of the track fit. Let me explain. A charged particle crossing the COT leaves a ionization trail. These ions are detected by 96 planes of sense wires along the path, and from the pattern of hit wires the trajectory can be reconstructed. However, each wire will have recorded the released charge at a different time, because they are located at different distances from the track, and the ions take some time to drift in the electric field before their signal is collected. The hit time is used in the fit that determines the particle trajectory: residuals of these time measurements after the track is fit provide a measurement of the particle velocity. In fact, a particle moving slowly creates ionization signals that are progressively delayed as a function of radius; these residuals can be used to determine the travel time. The resulting measurement has a three-times-worse precision than that coming from the dedicated TOF system (fortunately, I would say, otherwise the TOF itself would be a rather useless tool); however, the uncertainty on this latter measurement has a much more Gaussian behaviour! This is an important asset, since by requiring that the two time measurements are consistent with one another, one can effectively remove the non-Gaussian behavior of the TOF measurement. By combining crossing time -i.e. velocity- and track momentum measurement, one may then derive a mass estimate for the particle. The distribution of reconstructed masses for CHAMP candidates is shown in the graph below. Overimposed to the data, the distribution expected for a 220-GeV CHAMP signal has been overlaid. It is clear to see that the mass resolution provided by the method is rather poor: despite of that, a high-mass charged particle would be easy to spot if it were there. One note of warning about this graph: the distribution above shows masses ranging all the way from 0 to 100 GeV, but that does not mean that these tracks have similar masses: the vast majority of tracks are real muons, for which the velocity is underestimated due to instrumental effects: in a sense, the very shape of the curve describes the resolution of the time measurement provided by the analysis. The absence of tracks compatible with a mass larger than 120 GeV in the data allows to place model-independent limits on the CHAMP mass. Weak-interacting CHAMPS are excluded, in the kinematic region $\|\eta\|<0.7$ covered by the muon chambers, and with $P_T>40 GeV$, if they are produced with a cross section larger than 10 fb. For strongly-interacting CHAMPS the search considers the case of a scalar top R-hadron, a particle which is predicted by Supersymmetric theories when the stable stop quark binds together with an ordinary quark. In that case, the 95% CL limit can be set at a mass of 249 GeV. It is interesting to note that this analysis, while not using the magnitude of the ionization left by the track in the gas chamber (the so-called $dE/dx$ on which most past searches of CHAMPS have been based, e.g. in CDF (Run I) and ALEPH) to identify the CHAMP signal candidates, still does use the $dE/dx$ to infer the (background) particle species when determining the resolution of the time measurement from COT residuals. So the measurement shows once more how collider detectors really benefit from the high redundancy of their design! [Post scriptum: I discuss in simple terms the ionization energy loss in the second half of this recent post.] It only remains to congratulate with the main authors of this search, Thomas Phillips (from Duke University) and Rick Snider (Fermilab), for their nice result, which is being sent for publication as we speak. The public web page of the analysis, which contains more plots and an abstract, can be browsed here. ## Comments 1. onymous - January 12, 2009 Any idea why a limit for stop is quoted but not for stau? Staus are likely CHAMPs in a wide variety of models…. 2. onymous - January 12, 2009 (Rough estimate: 10 fb xsec for staus translates to a limit of around 100 GeV on the mass, which is a bit better than LEP did.) 3. dorigo - January 12, 2009 Yep ominous. No explicit limits on staus, but my guess is it does not improve the 99.5 GeV of LEP combined. The 10 fb is a limit for weak CHAMPs with some acceptance cuts applied, so the limit is a bit weaker than that overall. Cheers, T. 4. ervin - January 12, 2009 Tommaso, Can you please elaborate on the implications (if any) of this finding for current theories regarding the CDF anomaly? Specifically, how does the search for CHAMPS affect models that explain multi-muon events starting from various Dark Matter candidates? Thank you. Ervin 5. Andrea Giammanco - January 12, 2009 > It is interesting to note that this analysis, while not using the magnitude of the ionization left by the track in the gas chamber (the so-called dE/dx on which most past searches of CHAMPS have been based, e.g. in CDF (Run I) and ALEPH), does use the dE/dx to infer the particle species when determining the resolution of the time measurement from COT residuals. Do you mean that the dE/dx in the CDF tracking system is able to discriminate SM particles between them in some kinematical regime, but this ability is of no use in the search for CHAMPs? Probably I misunderstood, can you clarify better how dE/dx is used in this analysis? (I’ve already had a quick look at the webpage and at the note linked there.) 6. dorigo - January 13, 2009 Hi Ervin, I have no idea on the implications for the multi-muon interpretations… As you know, there are a few, but they are not very clear nor do they make very specific predictions. The authors have their own model, which implies the cascade of several bodies which are long-lived in the sense that they have a lifetime of about 20 picoseconds. These particles are however neutral. It seems to me that these limits can well coexist with a neutral higgs boson which produces some cascade to hidden valley particles. Andrea, yes, the CDF COT does measure dE/dx rather well and it discriminates protons from deuterons, for instance. With the latter, the method of determining residuals from the track fit has been tuned and verified. That is, they select deuterons with the dE/dx, and then, by knowing the mass of the particle, they can verify that the residuals are consistent with the velocity they infer from the momentum measurement for deuterons. The search does not make use of dE/dx other than that (contrarily to what was done in Run I, see Phys. Rev. Lett. 90, 131801 (2003)), although I think it could well do it. I am not sure whether avoiding dE/dx info one retains more model-independence. Cheers, T. 7. Andrea Giammanco - January 13, 2009 Thanks a lot for the clarification. I should probably explain why I am interested in that: I’m involved in the set-up of such a search in CMS, specifically working on the dE/dx side, and as far as I know the dE/dx by ionization has no significant model dependence (while, of course, the energy deposit in the calorimeters is totally model dependent – indeed it would provide a wonderful way to discriminate between lepton-like and hadron-like CHAMPs in case of discovery). My guess: maybe the TOF subgroup in CDF finished its analysis before the dE/dx subgroup, and when both will have completely validated their respective methods they will come up with a combined note? (A bidimensional plot of TOF versus dE/dx would be the ultimate smoking gun for a discovery – or for killing a claim of discovery coming from only one of the two methods – since the betas inferred from the two variables are uncorrelated or even have opposite behaviours for the main backgrounds.) 8. dorigo - January 13, 2009 Hi Andrea, well, the dE/dx by ionization of a stau or a stop R-hadron does depend only on the charge, velocity, and the mass of the particle, so in principle there is model independence even if one uses that. For a fractionary charge things would be different… But I doubt anybody is interested in such a scenario. I believe that the reason dE/dx was not used is really due to the fact that the combination of timing measurements has smaller backgrounds -the one in the paper is indeed a novel technique for CDF. As for “TOF subgroup” and “dE/dx subgroup”, CDF does not work that way. There is shortage of manpower these days, and I doubt there will be another search for CHAMPS in the same data. Maybe a future one with 3,4, 5 /fb will revert to using the dE/dx, but I doubt that too -much more likely that the same authors will repeat the present analysis with the larger dataset. Cheers, T. 9. Daniel de França MTd2 - January 14, 2009 Hi Tommaso, any gossip about the anomalous multi muon on D0? Cheers, Daniel dorigo - January 14, 2009 Not anything that I know of, but somebody said they’re working at it, and I also know that D0 might (just might) have a slot to talk about their own results on this matter at La Thuile (or was it Moriond ?). Cheers, T. 10. Daniel de França MTd2 - January 15, 2009 La Thuile, Aosta Valley, Italy March 1-7, 2009: http://www.pi.infn.it/lathuile/lathuile_2009.html 44th Rencontres de Moriond La Thuile (Val d’Aosta, Italy) February 1 – 8, 2009 http://moriond.in2p3.fr/J09/ It seems that Moriond is more oriented to sky surveys… 11. Philipum - January 15, 2009 Hi Dorigo, Do you know when we expect these results to appear as a journal publication ? I am working on a stable stop analysis in ATLAS and often quote the limits from the Tevatron. best, Philippe 12. dorigo - January 15, 2009 Yes Daniel, prolly it was La Thuile after all. Philipum, yes, we are done with the paper review, and I think it will be sent to publication very soon. I will add a pointer here when it comes out (if I don’t forget) Cheers, T. 13. Philipum - January 15, 2009 Thanks ! 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http://mathoverflow.net/questions/34882/why-are-divisible-abelian-groups-important/34963	## Why are divisible abelian groups important? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I just quote wikipedia: "Divisible groups are important in understanding the structure of abelian groups, especially because they are the injective abelian groups." I am asking for detail explanation of this application (i.e. concrete theorems, methods and so forth), or/and sources on that issue. Are there some other applications of DAG besides abelian groups theory or even besides algebra? Thanks in advance. - 5 Over any PID, divisible and injective are the same thing. I think over $\mathbb{Z}$, things are nice because all divisible groups (and therefore all injective $\mathbb{Z}$-modules) are known up to isomorphism. There are lots of books which talk about this. Personally, I would recommend (if you can get your hands on it) Kaplansky's "Infinite Abelian Groups". Besides having a ton of info in about 100 pgs, I think the way he presents divisible groups is very slick and easy-to-follow. – Steve D Aug 8 2010 at 4:04 3 An example where divisibility is used crucially that came to mind first for me is in constructing the invariant map $Br(K)\simeq\mathbb{Q}/\mathbb{Z}$, where $Br(K)=H^2(K,\mathbb{G}_m)$ is the Brauer group of the non-Archimedean local field $K$. One uses the short exact sequence $0\rightarrow\mathbb{Z}\rightarrow\mathbb{Q}\rightarrow\mathbb{Q}/\mathbb{Z}$ of trivial $G_K$-modules, and the fact that $\mathbb{Q}$ is uniquely divisible implies that its $\geq 1$-dimensional cohomology vanishes. – Keenan Kidwell Aug 8 2010 at 12:43 3 <<Continued>> The unique divisibility of actually implies that $H^n(G,\mathbb{Q})=0$ for all finite groups $G$ and $n\geq 1$, since the latter group is killed by $\vert G\vert$, but the unique divisibility of $\mathbb{Q}$ implies that multiplication by $\vert G\vert$ is an isomorphism on cohomology. Taking direct limits one gets the same result for profinite $G$. This is crucial because upon taking the long exact cohomology sequence one gets $\mathbb{Q}/\mathbb{Z}\simeq Hom(\hat{\mathbb{Z}},\mathbb{Q}/\mathbb{Z})\simeq H^2(\hat{\mathbb{Z}},\mathbb{Z})$, which is a part of the invariant map. – Keenan Kidwell Aug 8 2010 at 12:51 As Steve has mentioned, divisible modules and injective modules are the same over a PID. More generally, this is true over any Dedekind domain. – Beren Sanders Aug 9 2010 at 4:17 ## 3 Answers In my work I come upon divisible abelian groups all the time, especially those of the form $\mathbb Q_p/\mathbb Z_p$, or direct sums of such groups. One frequently uses the fact that they are injective (to conclude that certain sequences of Hom groups, which a priori would only be left exact, are in fact exact on the right as well), and other properties as well; for example, groups of type mentioned above, related to $\mathbb Q_p/\mathbb Z_p$, have non-zero $p$-adic Tate modules, which can often be useful. This last property is related to the role that divisible groups play in duality theory (in the sense of Pontrjagin duality), and this is another reason that divisible groups are important. I think I probably speak for a lot of algebraic number theorists when I say that a lot of what we do is a kind of applied algebra. We don't care about algebraic properties of structures for their own sake, but in our work we encounter lots of different groups, rings, and modules, and we like to understand their properties as well as we can, so that we can use them to gain control of the number theoretic computations that we are trying to make. Divisibility of an abelian group is one such useful property, which we are trained to recognize and exploit. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The meaning of the WP quote is that a divisible subgroup of an abelian group is always a direct summand. On the other hand, any divisible abelian group is a direct sum of copies of $\mathbb{Q}$ and $\mathbb{Q}_p/\mathbb{Z}_p$, thus unlike general infinite abelian groups, they can be completely classified. See the books of Kurosh and Kargapolov–Merzlyakov on group theory for the proofs. As Steve D mentioned in the comments, there is a straightforward generalization to modules over PIDs. Here are two related "concrete applications outside of algebra". 1 Pontryagin duality for locally compact abelian groups, $G\mapsto\text{Hom}(G,T),$ where $T=\mathbb{R}/\mathbb{Z}$ is the circle group and Hom denotes continuous homomorphisms. 2 Tate duality for Galois modules in algebraic number theory, $A\mapsto\text{Hom}(A,\mu)$, where $\mu$ is the group of roots of unity ($\mu\simeq\mathbb{Q}/\mathbb{Z}$ as abelian groups, but the Galois module structure is different). In both cases, divisibility of the target is needed to assure that the natural map into the double dual is injective. - Per FAQ, this question is too general for MO. – Victor Protsak Aug 8 2010 at 4:38 Victor: Quote from FAQ: <<The site works best for well-defined questions: math questions that actually have a specific answer.>> This question is about concrete and, I think, well known for algebraists (I am not the one) relationship between two concrete algebraic structures expressible in precise mathematical theorems. On the other hand I agree that it is not a technical question, but it is community wiki. – Sergei Tropanets Aug 8 2010 at 5:08 Victor: <<A question should be made community wiki if you don't think that people should gain reputation for their answers. A typical case is requests for references where it is the reference that is being judged by the voting system rather than the person who supplied it.>> I think the question satisfy this. Note that according to FAQ the final characteristics of CW questions aren't defined yet. – Sergei Tropanets Aug 8 2010 at 5:11 3 Sergei, I am not that worried about CW and all that. But you are, in essence, asking people to teach you a subject, without even providing a compelling motivation ("I just quote from Wikipedia"). While I am generally "on the left" in the matters which questions are suitable, I disagree with such an open-ended use of MO. Other people might even characterize it as a "fishing expedition". In any case, this is going to be my last answer to your questions: I've put some effort in answering your question, however imperfect it was, but you haven't reciprocated by reading and upvoting it. – Victor Protsak Aug 8 2010 at 5:40 I've read it and now I'm studying the details you provided. Thanks for the answer. – Sergei Tropanets Aug 8 2010 at 5:45 show 1 more comment Suppose we have a category (that is, a collection of objects and arrows ["morphisms"] between those objects; for example, the collection of abelian groups and all homomorphisms between those abelian groups). An object $I$ is injective if whenever we have an injection $i:A \to B$ and a homomorphism $f:A \to I$, then there is a morphism $g:B \to I$ such that $f = g \circ i$. In the category of abelian groups, the divisible groups are precisely the injective objects in that category. To see a sketch of why, suppose $B$ is an infinite cyclic groups, and $A$ is the set of powers of $n$ resting within that cyclic group (hence also isomorphic to an infinite cyclic groups). Suppose $B$ has generator $b$, and $A$ has generated $a:=b^n$. Then a homomorphism $\varphi:A \to I$ amounts to choosing an element of $I$ where $a$ will be sent. Extending this to a homomorphism from $B$ to $I$ amounts to choosing an element of $I$ whose $n$th power is $\varphi(a)=\varphi(b)^n$; that is, dividing by $n$. This is why they are related. Now, you might ask, why are injective objects important? Injective objects are some of the most important objects in a category which underlie many applications of homological algebra. Suppose we have an exact sequence $0 \to A \to B \to C \to 0$. It is well known that the sequence $0 \to Hom(C,I) \to Hom(B,I) \to Hom(A,I)$ is exact no matter what. However, we know that that last map is surjective if $I$ is injective. Thus, $Hom(-,I)$ is an exact (contravariant) functor iff $I$ is injective. It follows from this that $Ext(-,I)$ is trivial iff $I$ is injective, and in general, right derived functors are trivial on injective objects. This latter fact is interesting because derived functors appear all over higher mathematics. To give an example, the group cohomology is always trivial on injective objects in the category of $G$-modules. However, all we talked about was injective objects in the category of $\mathbb{Z}$-modules (i.e. abelian groups). So why is this interesting? Well, the functor which sends an abelian group $M$ to the induced $G$-module $\mathrm{Ind}^G(M)$ (essentially $\mathbb{Z}[G] \otimes_{\mathbb{Z}} M$ by extension of scalars) is exact and sends injectives to injectives. This means that we get a whole host of injective $G$-modules by considering injective abelian groups, and what's more, it allows us to show that the category of $G$-modules has enough injectives (i.e. that every $G$-module can be embedded in an injective $G$-module). Derived functors show up in many other areas, including sheaf cohomology (where injective objects are also very important), and divisible groups are the most basic example of the injective objects which underlie right derived functors. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 65, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.945942759513855, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/trees+homework	# Tagged Questions 2answers 44 views ### Prove there is a tree with $n$ vertices having degrees $d_1, d_2…d_n$ For $n ≥ 2$ suppose $d_1, d_2,....d_n$ are positive integers with sum $2n - 2$. Prove there is a tree with n vertices having degrees $d_1, d_2....d_n$. I'm at a loss on this one. I'm sure it's pretty ... 2answers 141 views ### Suppose there are two different spanning trees for a simple graph. Must they have an edge in common? My instinct is yes, but I don't know how to formalize it into a proof. I still haven't wrapped my head around spanning trees yet. Any thoughts are appreciated! 1answer 101 views ### M-ary tree problem A full $m$-ary tree $T$ has 81 leaves and height 4 1) Give the upper and lower bounds for $m$ 2) What is $m$ if T is also balanced? [with $m^h=l$ for maximum leaf in a m-ary tree $m^4=81$ then m=3 ... 0answers 47 views ### How can I prove this property of a $d$-ary tree? I have the following homework (algorithms lecture): Every $d$-ary tree $G=(V,E)$ contains a vertex $v$ such that the size of the subtree with root $v$ is at least $\frac{1}{d+1} \vert V \vert$ and at ... 1answer 35 views ### If $G$ is a tree and $\forall v_1, v_2 : dist(v_1, v_2)$ is even, where $v_1, v_2-$ are leaves $\Rightarrow \exists!$ a maximal independent set. If $G$ is a tree and $\forall v_1, v_2 : dist(v_1, v_2)$ is even, where $v_1, v_2-$ are leaves of the tree $\Rightarrow \exists!$ a maximal independent set. Give some clue please! Thanks anyway! 1answer 75 views ### Prove equivalence of conditions for a tree Let $G=(V,E)$ denote a nonempty graph. Show that the following conditions are all equivalent. $G$ is a tree. Any two vertices in $G$ can be connected by a unique simple path. $G$ is ... 0answers 64 views ### Recurrence relation induction [duplicate] Possible Duplicate: Solving the recurrence $t(n)=(t(n-1))^2 + 1$ Show that the number of binary trees of height less than or equal to $n$ is given by the recurrence \begin{align*} ... 1answer 720 views ### Number of Trees with n Nodes I am struggling with a question that asks the number of trees that exist with x nodes and max level z. During my research I found that the number of binary trees with x nodes can be obtained by ... 1answer 229 views ### Graph Theory - Spanning Trees Consider a graph $G$ composed of two cycles which share an edge. $C_x$ is the cycle of length $x$ and $C_y$ is the cycle length $y$, for $x,y \ge 3$. (for example, if $x = 6$ and $y = 5$, then $C_x$ ... 3answers 1k views ### How to show that every connected graph has a spanning tree, working from the graph “down” I am confused about how to approach this. It says: Show that every connected graph has a spanning tree. It's possible to find a proof that starts with the graph and works "down" towards the ... 1answer 175 views ### Proof involving a minimum weight spanning tree. Please help with the following homework problem: Let G be an undirected graph, $v: E\to R$ and $w: E\to R$ be two weight functions on the edges of $G$. Let $z: E\to R$ be defined as the sum of ... 3answers 85 views ### Comparison trees You have 60 coins. You know that 1 coin is either lighter or heavier than the other coins. How many comparisons are needed in a worst case scenario to discover which coin is the false one and ... 2answers 357 views ### Graph - Minimum spanning tree I have a graph with a cycle ($v_1,\ldots,v_k, v_1=v_k$). Claim: If there is a cycle with 2 edges of the same weight, and they are the heaviest edges in this cycle, then there is more than one Minimum ... 3answers 480 views ### How do you calculate the average length of a random binary tree? Assuming that you start out with a root node, and decide with 50% probability whether or not to add two children nodes. If they do, repeat this process for them. How can you find the average length of ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 46, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9493270516395569, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/38630/list	## Return to Question 2 added 214 characters in body; added 14 characters in body; edited body; added 4 characters in body; edited title # homotopy invariant andcoinvariant Let $V$ be a chain complex, which is either $Z$ or $Z/2$ graded. A circle action on $V$ is by definition an action of the dga $H_\ast(S^1)$. This consists of a map $D : V → V$ , which is of square zero, commutes with $d$, and in the $Z$ graded case is of degree 1. the homotopy invariant is $V^{hS^1}= V [[t]]$ with differential $d+tD$. That is if $vf(t)$ is an element $V [[t]]$, then $d(vf(t)) = (dv)f(t) + (Dv)tf(t)$ The Tate complex is $V_{Tate} = V ((t))$ again with differential $d+tD.$ The homotopy coinvariants is the space $V_{hS^1} = t^−1V [t^−1] = V_{Tate}/V^{hS^1}$ with differential induced from that on $V_{Tate}$ Could anyone explain to me how this works? why homotopy invariant (co)invariant is that? I am not quite familiar with homotopy invariant.(but (co)invariant.(but I know what it is) 1 # homotopy invariant Let $V$ be a chain complex, which is either $Z$ or $Z/2$ graded. A circle action on $V$ is by definition an action of the dga $H_\ast(S^1)$. This consists of a map $D : V → V$ , which is of square zero, commutes with $d$, and in the $Z$ graded case is of degree 1. the homotopy invariant is $V^{hS^1}= V [[t]]$ with differential $d+tD$. That is if $vf(t)$ is an element $V [[t]]$, then $d(vf(t)) = (dv)f(t) + (Dv)tf(t)$ Could anyone explain to me how this works? why homotopy invariant is that? I am not quite familiar with homotopy invariant.(but I know what it is)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9351692199707031, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2010/02/22/proving-the-classification-theorem-i/?like=1&source=post_flair&_wpnonce=c1c9c39881	# The Unapologetic Mathematician ## Proving the Classification Theorem I This week, we will prove the classification theorem for root systems. The proof consist of a long series of steps, and we’ll split it up over a number of posts. Our strategy is to determine which Coxeter graphs can arise from actual root systems, and then see what Dynkin diagrams we can get. Since Coxeter graphs ignore relative lengths of roots, we will start by just working with unit vectors whose pairwise angles are described by the Coxeter graph. So, for the time being we make the following assumptions: $V$ is a real inner-product space (of arbitrary dimension), and $\mathfrak{D}=\{\epsilon_1,\dots,\epsilon_n\}$ is a set of $n$ linearly independent unit vectors satisfying $\langle\epsilon_i,\epsilon_j\rangle\leq0$ and $4\langle\epsilon_i,\epsilon_j\rangle^2$ is one of $0$, $1$, $2$, or $3$ for $i\neq j$. For the duration of this proof, we will call such a set of vectors “admissible”. The elements of a base $\Delta$ for a root system $\Phi$, divided by their lengths, are an admissible set. We construct a Coxeter graph $\Gamma$, just as before, for every admissible set, and these clearly include the Coxeter graphs that come from root systems. Our task is now to classify all the connected Coxeter graphs that come from admissible sets of vectors. 1. If we remove some of the vectors from an admissible set, then the remaining vectors still form an admissible set. The Coxeter graph of the new set is obtained from that of the old one by removing the vertices corresponding to the removed vectors and all their incident edges. 2. This should be straightforward. The remaining vectors are still linearly independent, and they still have unit length. The angles between each pair are also unchanged. All we do is remove some vectors (vertices in the graph). 3. The number of pairs of vertices in $\Gamma$ connected by at least one edge is strictly less than $n$ — the number of overall vertices. 4. Let us define $\displaystyle\epsilon=\sum\limits_{i=1}^n\epsilon_i$ Since the $\epsilon_i$ are linearly independent, $\epsilon$ cannot be zero. Thus we find that $\displaystyle\begin{aligned}0<\langle\epsilon,\epsilon\rangle&=\left\langle\sum\limits_{i=1}^n\epsilon_i,\sum\limits_{j=1}^n\epsilon_j\right\rangle\\&=\sum\limits_{i=1}^n\sum\limits_{j=1}^n\langle\epsilon_i,\epsilon_j\rangle\\&=\sum\limits_{i<j}\langle\epsilon_i,\epsilon_j\rangle+\sum\limits_{i=j}\langle\epsilon_i,\epsilon_j\rangle+\sum\limits_{i>j}\langle\epsilon_i,\epsilon_j\rangle\\&=n+2\sum\limits_{i<j}\langle\epsilon_i,\epsilon_j\rangle\end{aligned}$ Now, for any pair $\epsilon_i$ and $\epsilon_j$ joined by an edge, we must find that $4\langle\epsilon_i,\epsilon_j\rangle^2$ is $1$, $2$, or $3$, and thus that $2\langle\epsilon_i,\epsilon_j\rangle\leq-1$. Clearly, we can have no more than $n-1$ of these edges if we are to satisfy the above inequality. 5. The graph $\Gamma$ contains no cycles 6. Indeed, if we have a cycle of length $n$ in the graph, we could (using our step 1) discard all the vertices not involved in the cycle and still have the graph of an admissible set. But this graph would have $n$ vertices and at least $n$ edges — the ones involved in the cycle. But this is is impossible by step 2! About these ads ### Like this: Like Loading... Posted by John Armstrong \| Geometry, Root Systems ## 8 Comments » 1. Hi – I think you missed out a minus sign in step two, near the end: 2\langle\epsilon_i,\epsilon_j\rangle\leq1. Paul Comment by Paul Scriven \| February 23, 2010 \| Reply 2. Ah yes, thanks. Comment by \| February 23, 2010 \| Reply 3. [...] Proving the Classification Theorem II We continue with the proof of the classification theorem that we started yesterday. [...] Pingback by \| February 23, 2010 \| Reply 4. Students love to have a Big Theorem at the end of the semester, from which mountaintop to survey the landscape that they have mastered. This classification result is such a theorem. Comment by \| February 23, 2010 \| Reply 5. Well it’s not done yet. After proving what was stated on Friday, we have to actually show whether or not any of these possible diagrams actually do occur. Comment by \| February 23, 2010 \| Reply 6. [...] continue with the proof of the classification theorem. The first two parts are here and [...] Pingback by \| February 25, 2010 \| Reply 7. [...] Theorem IV We continue proving the classification theorem. The first three parts are here, here, and [...] Pingback by \| February 25, 2010 \| Reply 8. [...] Today we conclude the proof of the classification theorem. The first four parts of the proof are here, here, here, and [...] Pingback by \| February 26, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now. • ## RSS Feeds RSS - Posts RSS - Comments • ## Feedback Got something to say? Anonymous questions, comments, and suggestions at Formspring.me! %d bloggers like this:	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 31, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9420426487922668, "perplexity_flag": "head"}
http://en.m.wiktionary.org/wiki/Hermitian_matrix	# Hermitian matrix ## English Wikipedia has an article on: ### Etymology Named after Charles Hermite (1822–1901), French mathematician. ### Pronunciation • "her mission matrix" • () IPA: /hɝ.ˈmɪ.ʃən ˈmeɪ.tɹɪks/ ### Noun Hermitian matrix (plural ) 1. (linear algebra) a square matrix with complex entries that is equal to its own conjugate transpose, i.e., a matrix such that $A = A^\dagger\,,$ where $A^\dagger$ denotes the conjugate transpose of a matrix A Hermitian matrices have real diagonal elements as well as real eigenvalues.[1] If a Hermitian matrix has a simple spectrum (of eigenvalues) then its eigenvectors are orthogonal.[2]On the other hand, a set of two or more eigenvectors with the same eigenvalue can be orthogonalized (e.g., through the Gram–Schmidt process, since any linear combination of equal-eigenvalue eigenvectors will also be an eigenvector) and will already be orthogonal to other eigenvectors which have different eigenvalues. If an observable can be described by a Hermitian matrix $H$, then for a given state $\|A\rangle$, the expectation value of the observable for that state is $\langle A\|H\|A\rangle$. #### Synonyms • (hermitian matrix): self-adjoint matrix #### Translations mathematics: square matrix that is equal to its own conjugate transpose Icelandic: sjálfoka fylki n, hermískt fylki n ### References ↑Jump back a section ## Read in another language This page is available in 1 language Last modified on 18 January 2013, at 14:37	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7891261577606201, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/tagged/srp-6?sort=unanswered&pagesize=15	# Tagged Questions The srp-6 tag has no wiki summary. In SRP6 $B$ is calculated as $B=kv+ g^b, k=3$. What is the purpose of $k$, and why was it fixed as $3$? (In SRP 6b, this value $3$ is replaced by $k = H(N,g)$, but this question is about SRP 6.)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9897021055221558, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/111583/generalised-formula-for-sum-of-first-n-tetranacci-numbers	generalised formula for sum of first $n$ tetranacci numbers In the case of Fibonacci numbers, the formula for the sum of first $n$ numbers of the series is $f(n+2)-1$, but in the case of tetranacci numbers I am unable to arrive at such formula. Thanks. - One can express the $n$-th whatever-nicci number as a linear combination of the $n$-th powers of the roots of a certain polynomial. Then the sum is just that linear combination of the sum of finite geometric series. So in a sense we get a closed-form formula. Perhaps that can be expressed as a linear combination of explicit whatever-nicci numbers. – André Nicolas Feb 21 '12 at 6:52 In fact, we can always express the sum as a linear combination of $n$ terms in the series and a constant. For the Fibonacci numbers, something nice happens and we only need one. – Yuval Filmus Feb 21 '12 at 7:05 – user25383 Feb 21 '12 at 9:51 Upps, I hope there was nothing wrong with my answer... Why didn't the OP add a remark? (P.s. my firefox-browser does not "trust" that link. Is it ok to proceed to there? (added: I can open that site if I use "http" instead of "https")) – Gottfried Helms Feb 21 '12 at 9:57 2 I see no evidence that the OP got the problem from the SPOJ site; it’s a perfectly reasonable mathematical question independent of any programming aspects. – Brian M. Scott Feb 22 '12 at 6:45 show 1 more comment 3 Answers Marcellus E. Waddill, The Tetranacci Sequence and Generalizations, gives the following identity: $$\sum_{i=0}^n\mu_i=\frac13\Big(\mu_{n+2}+2\mu_n+\mu_{n-1}+2\mu_0+\mu_1-\mu_3\Big)\;,\tag{1}$$ where $\mu_0,\mu_1,\mu_2,\mu_3$ are arbitrary initial values and $\mu_n=\mu_{n-1}+\mu_{n-2}+\mu_{n-3}+\mu_{n-4}$ for $n\ge 4$; it is formula $(39)$ in the paper. It can be proved by induction, but Waddill gives a nicer proof by summing the identities $$\mu_k+\mu_{k+1}+\mu_{k+2}=\mu_{k+2}-\mu_{k+1}$$ for $k=0,\dots,n$ to obtain $$\sum_{k=0}^n\mu_k+\left(\sum_{k=0}^n\mu_k+\mu_{n+1}-\mu_0\right)+\left(\sum_{k=0}^n\mu_i+\mu_{n+1}+\mu_{n+2}-\mu_0-\mu_1\right)=\mu_{n+4}-\mu_3$$ and then $$3\sum_{k=0}^n\mu_k+2\mu_{n+1}+\mu_{n+2}-2\mu_0-\mu_1=\mu_{n+4}-\mu_3\;,$$ which can be rearranged to yield $$\begin{align} 3\sum_{k=0}^n\mu_k&=\mu_{n+4}-2\mu_{n+1}-\mu_{n+2}-\mu_3+2\mu_0+\mu_1\\ &=(\mu_{n+3}+\mu_{n+2}+\mu_{n+1}+\mu_n)-2\mu_{n+1}-\mu_{n+2}+2\mu_0+\mu_1-\mu_3\\ &=\mu_{n+3}-\mu_{n+1}+\mu_n+2\mu_0+\mu_1-\mu_3\\ &=(\mu_{n+2}+\mu_{n+1}+\mu_n+\mu_{n-1})-\mu_{n+1}+\mu_n+2\mu_0+\mu_1-\mu_3\\ &=\mu_{n+2}+2\mu_n+\mu_{n-1}+2\mu_0+\mu_1-\mu_3\;, \end{align}$$ as desired. If you set $\mu_0=\mu_1=\mu_2=0$ and $\mu_3=1$, $(1)$ becomes $$\sum_{i=0}^n\mu_i=\frac13\Big(\mu_{n+2}+2\mu_n+\mu_{n-1}-1\Big)\;.$$ - if we take sum of first 9 terms of the series: then sum = 0+0+0+1+1+2+4+8+15=31 ; but by the formula it comes as : 221/3 – pranay Feb 21 '12 at 7:28 @pranay: The formula yields $\frac13(56+2\cdot 15+8-1)=\frac13(93)=31$. – Brian M. Scott Feb 21 '12 at 7:34 thanks a lot :) – pranay Feb 21 '12 at 8:56 This can also be solved by a matrix-ansatz (and then generalized in a completely obvious way). Example: assume a vector A containing your first four values, say $\qquad \small A=[1,3,4,5]$ . Next consider the transfermatrix, say T which defines the composition of the next entry by $\small 11 + 31 + 41 + 51 =13$, and simply shifts the old entries $\small [1,3,4,5] \to [3,4,5,13]$ . The required matrix T looks like $\qquad \small T=\begin{bmatrix} 0&0&0&1\\1&0&0&1 \\0&1&0&1 \\0&0&1&1\\ \end{bmatrix}$ Then we have the iteration for the computation of consecutive elements of the tetranacci-sequence simply by $\qquad \small A_{k+1} = A_k \cdot T$ or, even better: $\qquad \small A_k = A \cdot T^k$ To sum the consecutive entries we can simply use the sum of the powers of T: $\qquad \small S_k = A \cdot ( T^0 + T^1 + T^2 + ... + T^{k-1})$ and the leading part of the geometric-series for the matrix T (also known as Neumann-series) is then $\qquad \small U_k = (T^k - I) \cdot (T-I)^{-1}$ . Thus the sum of the first k elements of the tetranacci-sequence can be found by $\qquad \small S_k = A \cdot U_k$ and for k=5 I get $\qquad \small S_k = [26, 50, 94, 180]$ where 26 (=1+3+4+5+13) is the sum of the first 5 elements of the sequence. It is obvious, how this can be generalized in two ways: • different starting values: either modify the (example) init-values in A (or keep them symbolic) • different type of m-nacci-sequences: just increase the size of T in the obvious way - Denote the tetranacci numbers by $t(n)$. Then $$\sum_{k=0}^n t(k) = \frac{t(n) - t(n+1) + t(n+3) - 1}{3}.$$ If you prefer your identity to involve $t(n+a),t(n+b),t(n+c),t(n+d)$ instead, just solve linear equations, not forgetting the constant term. As commented above, since $t(n)$ is a linear combination of four powers $$t(n) = c_1 \lambda_1^n + c_2 \lambda_2^n + c_3 \lambda_3^n + c_4 \lambda_4^n,$$ if we take sums then we get $$\sum_{k=0}^n t(k) = \frac{c_1}{\lambda_1 - 1} \lambda_1^{n+1} + \cdots + \frac{c_4}{\lambda_4 - 1} \lambda_4^{n+1} - \left( \frac{c_1}{\lambda_1 - 1} + \cdots + \frac{c_4}{\lambda_4 - 1} \right).$$ If you put in $$\sum_{k=0}^n = A t(n) + B t(n+1) + C t(n+2) + D t(n+2) + E$$ then you can solve a linear system to get the values of $A,B,C,D$: $$\begin{align} \frac{c_1}{\lambda_1 - 1} \lambda_1 &= A + B\lambda_1 + C\lambda_1^2 + D\lambda_1^3 \\ \cdots \\ \frac{c_4}{\lambda_4 - 1} \lambda_4 &= A + B\lambda_4 + C\lambda_4^2 + D\lambda_4^3 \end{align}$$ The system must have a solution since the coefficients on the right-hand side form a Vandermonde matrix. The remaining coefficient $E$ can be read off directly: $$E = - \left( \frac{c_1}{\lambda_1 - 1} + \cdots + \frac{c_4}{\lambda_4 - 1} \right).$$ Given that we know that such a representation exists, we can forget about $c_1,\ldots,c_4,\lambda_1,\ldots,\lambda_4$ and calculate the coefficients $A,B,C,D,E$ directly by solving a different system of equations: $$\begin{align} At(0) + Bt(1) + Ct(2) + Dt(3) + E &= t(0), \\ At(1) + Bt(2) + Ct(3) + Dt(4) + E &= t(0) + t(1), \\ \cdots \\ At(4) + Bt(5) + Ct(6) + Dt(7) + E &= t(0) + t(1) + t(2) + t(3) + t(4). \end{align}$$ In order to find five unknowns we need five equations. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 32, "mathjax_display_tex": 13, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9067938327789307, "perplexity_flag": "middle"}
http://en.wikibooks.org/wiki/Operations_Research/Graphical_LP_solution	# Operations Research/Graphical LP solution We will now attempt to find an optimal solution to the linear programming model we introduced in the previous section. The method we will employ is known as the graphical method and can be applied to any problem with two decision variables. It basically consists of two steps: Finding the feasible region or the feasible space (which is the region in the plane where all the feasible solutions to the problems lie) and then identifying the optimal solution among all the feasible ones. To begin the procedure we first graph the lines $6x_1+4x_2= 24$, $x_1+2x_2= 6$, $x_2-x_1= 1$, $x_2= 2$, in the first quadrant. Note that for our purpose $x_1=x$ and $x_2=y$ on the graph. We now will shade the feasible region. To do that consider the constraints one by one. The first one is $6x_1+4x_2\le 24$. To determine the region it represents, choose any point which does not pass through the line $6x_1+4x_2= 24$ say (0,0). Substitute it in the constraint $6x_1+4x_2\le 24$ to get $0\le 24$. Since this is true we conclude that (0,0) lies in the region represented by $6x_1+4x_2\le 24$. We conclude that all the points on the side of $6x_1+4x_2= 24$ containing (0,0) actually represent $6x_1+4x_2\le 24$. This is suggested by the fact that the line $6x_1+4x_2 = 24$ divides the plane in 2 distinct halfs: One of points satisfying the inequality and one of those which don't. In this way all the inequalities can be shaded. The region which is shaded under all inequalities is the feasible region of the whole problem. Clearly in this region all the constraints of the problem hold. (The non negativity restrictions hold since we are working in the first quadrant.) We are now ready to find out the optimal solutions. To do this graph the line $5x_1+4x_2=10$. Since $z=5x_1+4x_2$ represents the objective function so $5x_1+4x_2=10$ represents the points where the objective function has value 10 (i.e. the total profit is 10). Now plot the line $5x_1+4x_2=15$ which represents the points where the objective function has value 15. This gives us an idea of the direction of increase in z. The optimal solution occurs at the point X which is the point beyond which any further increase will put z outside the boundaries of the feasible region. The coordiantes of X can be found by solving $6x_1+4x_2= 24$ and $x_1+2x_2= 6$ so that $x_1=3$ and $x_2=1.5$. This is the optimal solution to the problem and indicates that the amounts of salts X and Y should be 3 and 1.5 respectively. This will give the maximum profit of 21 which is the optimal value. A point to note is that the optimal solution in a LP model always occurs at a corner point of a feasible region. This is true even if the line z=c comes out to be parallel to one of the constraints. Although a mathematical proof of this fact would involve considerable linear algebra we will satisfy ourselves of it by noting that that any objective function in the feasible region would have glided out of the region just after touching one of the corner points. ## A minimization example Let us look at a minimization problem. It can occur in actual practice when instead of the profits associated with the salts X and Y we are given the costs of their production. All we have to do is now move the line z=c in the direction of its decrease and we have the optimal solution at the point ((0,0) in our example) where any further decrease will take z outside the feasible region. Another way to solve the problem is to convert the min problem into a max one. To do that simple consider the negative of the objective function.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 22, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9324200749397278, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/tagged/neural-networks+pattern-recognition	# Tagged Questions 2answers 132 views ### The most common extracted features for image recognition I want to build a neural network, but because I have high resolution pictures, I rejected the idea of passing the entire image to the NN. I was wondering what are the most common extracted features ... 2answers 339 views ### What sort of problems is backpropagation best suited to solving, and what are the best alternatives to backprop for solving those problems? I am developing a neuroscience-inspired training algorithm for feed-forward neural networks. The natural benchmark for comparison is backpropagation. So I need to know to know what sort of ... 1answer 150 views ### What is the best way to use a 2-class classifier for a multi-category case? I have a 3 class sample labeled data set, which I have divided into 2 parts. I am using the first part to train the two-class perceptron classifier. One approach is to train $\binom{3}{2}$ two-class ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9393537044525146, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/46324/to-which-extent-is-general-relativity-a-gauge-theory?answertab=active	# To which extent is general relativity a gauge theory? In quantum mechanics, we know that a change of frame -- a gauge transform -- leaves the probability of an outcome measurement invariant (well, the square modulus of the wave-function, i.e. the probability), because it is just a multiplication by a phase term. I was wondering about general relativity. Is there something left invariant by a change of frame? (of course, energy, momentum, ... are invariants of Lorentz transform, but these are special relativity examples. I guess there is something else, more intrinsic and related to the mathematical structure of the theory, like a space-time interval or something). I've tried looking at the Landau book on field theory, but it is too dense for me to have a quick answer to this question. I have bad understanding about GR -- I apologize for that. I'm trying to understand to which respect one calls the GR theory a gauge theory: for me a gauge transform leaves something invariant. Best regards. EDIT: Thanks to the first answers, I think I should refine my question and first ask this: To which extent is general relativity a gauge theory? If you have good references to this topic, that would be great(The Wikipedia pages are obscure for me for the moment). Thanks in advance. Best regards. - 2 Parallel transport involves moving around in a manifold, while a change of frame is, well just that. The title of the question is a bit misleading. – Chris White Dec 9 '12 at 2:13 1 Thanks Chris White, that's partly my problem: I see the parallel transport as a change of frame with one extra condition (the moving frame remains as much as possible parallel to the original one). My second problem was that in QM, the invariant quantity is an extra structure (the wave function modulus, whereas the wave function is displaced / covariant). This you partly answered with Namehere: only scalar are invariant, tensor are covariant. I was asking if it exists an extra structure kept invariant when changing frame of reference. Thanks to both of you. – Oaoa Dec 9 '12 at 9:40 2 @Oaoa Tensors are not covariant, they are invariant. It is their components that are covariant. And by the way there are many scalars that can be constructed from tensors, such as norms of vectors, traces of matrices and others. Oh, and was there anything else you think I missed or could improve in my answer(I didn't edit my question to include Chris White's comment since I felt it would be like stealing)? – namehere Dec 9 '12 at 9:59 2 I faintly remember that there is a nice way to think about GR as a gauge theory (or gauge theories as geometry), and it had to do with viewing the Levi-Civita connection as a gauge field. Unfortunately I don't know enough GR to write down the argument. – jdm Dec 9 '12 at 12:10 1 – twistor59 Dec 9 '12 at 15:30 show 6 more comments ## 4 Answers Consider a gauge theory with gauge group $GL(n,R)$. First of all, let me remind you the basics of gauge transformations: Let $G$ be a gauge group, $g(x)∈ G$ be an element of $G$. Then: $\psi (x)→g(x)\psi (x)$ $A_\alpha→g(x)A_\alpha g^{-1}(x)-\frac{∂g(x)}{∂x^{\alpha}}g^{-1}(x)$ is a gauge transformation, and covariant derivative defined as ${\nabla}_{\alpha}\psi = \frac{∂\psi}{∂x^{\alpha}}+A_\alpha \psi$ Now consider coordinates $(x^1, ..., x^n)$ in region $U$. They define basis of vectors space $\frac{∂}{∂x^1},...,\frac{∂}{∂x^n}$. So, tangent vector fields in region $U$ can be considered as vector-valued functions: $\xi = (\xi^1,...,\xi^n)$. Change of coordinates in $U$: $x^{\nu}→x^{{\nu^{\prime}}}=x^{{\nu^{\prime}}}(x)$ defines local transformation: $\xi^\nu→\xi^{\nu^\prime} = \frac{∂x^{\nu^\prime}}{∂x^\nu}\xi^\nu = g(x)\xi$. Here matrix $g(x) = (\frac{∂x^{\nu^\prime}}{∂x^\nu})$ belongs to $GL(n,R)$, and inverse matrix has the form $g^{-1}(x)=(\frac{∂x^\nu}{∂x^{\nu^\prime}})$. Lie algebra of $GL(n,R)$ is formed by all matrices of degree $n$, so the "gauge field" $A_\mu (x)$ is also matrix of degree $n$. Let us denote it's elements as follows: $(A_\mu)^{\nu}_{\lambda}=\Gamma^{\nu}_{\lambda \mu}$. Covariant derivative of the vector $\xi$ reads as follows: $(\nabla_{\mu}\xi)^\nu=\frac{∂\xi^\nu}{∂x^\mu}+\Gamma^{\nu}_{\lambda \mu}\xi^\lambda ↔ \nabla_\mu \xi=\frac{∂\xi}{∂x^\mu}+A_\mu \xi$ (right side is in matrix form!) There is only one thing left to check, namely the form of gauge field transformation. Using general rule of transforming gauge field we obtain: $\Gamma^{\nu}_{\lambda \mu}→\Gamma^{\nu^\prime}_{\lambda^\prime \mu}=\frac{∂x^{\nu^\prime}}{∂x^\nu}\Gamma^{\nu}_{\lambda \mu}\frac{∂x^\lambda}{∂x^{\lambda^\prime}}+\frac{∂x^{\nu^\prime}}{∂x^\nu}\frac{∂}{∂x^\mu}(\frac{∂x^\nu}{∂x^{\lambda^\prime}})$. Since $A_\mu$ is a covariant vector, then $A_{\mu^\prime}=\frac{∂x^\mu}{∂x^{\mu^\prime}}A_\mu$. Hence we obtain: $\Gamma^{\nu^\prime}_{\lambda^\prime \mu^\prime}=\frac{∂x^\mu}{∂x^{\mu^\prime}}\frac{∂x^{\nu^\prime}}{∂x^\nu}\Gamma^{\nu}_{\lambda \mu}\frac{∂x^\lambda}{∂x^{\lambda^\prime}}+\frac{∂x^{\nu^\prime}}{∂x^\nu}\frac{∂^2 x^\nu}{∂x^{\lambda^\prime}∂x^{\mu^\prime}}$. Q.E.D. And final remark: the commutator of two covariant derivatives leads to expression of the Riemann tensor: $(F_{\mu\nu})^\rho_\lambda = R^\rho_{\lambda ,\mu\nu}$ EDIT: Dear Oaoa, I’m not a GR specialist, so what I have written below might be wrong. My first advise is as follows: do not read Landau who is mixing up two fundamental concepts: connection and metric. Instead I encourage you to read "Space-time structure" by Erwin Schrödinger. In order to answer your question let us first separate the roles of connection and metrics. 1. Connection is used for parallel transport and enables to compare two vectors in different points. Important consequence is that using connection one can introduce the curvature tensor (that can further be contracted to curvature scalar). Curvature appears when you transport vector along the closed curve and then compare with the initial vector. Curvature scalar is then used to construct “field action” just like in all gauge theories. As shown in Schrödinger’s book, connection can also be used to measure distance along geodesic line (it worth noting that the expression for such “measure” is so much similar to the expression of Feynman’s path integral action!). But in general, connection cannot be used for measuring distances between arbitrary points. 1. Metric is introduced for measuring distances between arbitrary points and defining vector products. 2. Connection and Metric are independent concepts. Only additional condition of their consistency (i.e. when you require that vector product is invariant when both vectors are parallel transported) allows to express connection via metric tensor. Let’s get back to your question now. All that is written about $GL(n,R)$ above is related to connection only. In the first place, it allows expressing “field action” in terms of a scalar curvature. But what you are most interested in is probably not this, but conservation laws related to matter fields. In the theory with point particles functions $\xi$ (or $\psi$) can be associated with vectors $\frac{dx^\nu}{ds}$. I’m not sure, but consequent conservation law is probably energy-momentum conservation. I think it is the same in Special relativity where space is flat and all connections are zero, but indirectly the conservation of energy-momentum in SR might be a consequence of “conservation” of null curvature by Lorentz transformations (please note that homogeneity of space-time means zero curvature). I know you expect to see some other conserved quantities similar to “electric charge” conservation in Dirac theory of electron. But please note that in Dirac theory the “global” conservation of “charge” is practically indistinguishable from conservation of energy-momentum. As for local theories – I do not know, concrete model need to be considered. - thanks. So you've shown that the Christofell symbols behave mathematically like a connection under a general GL gauge group transform. What is the field $\psi(x)$ in your notations ? Is it by some extent conserved ? Thanks in advance. – Oaoa Dec 9 '12 at 17:21 1 But $GL(n)$ doesn't admit finite dimensional unitary representations, which cause problems when we want to write down components of fields...because you end up with an infinite number of components...which is why people usually use $SO(4)$ (or $SO(3,1)$ depending on your religion) instead. – Alex Nelson Dec 9 '12 at 21:28 @AlexNelson Thank you for this comment. Indeed, that's part of the problem when trying to use gauge theory. – Oaoa Dec 10 '12 at 9:31 @MurodAbdukhakimov Thank again and so much for your new update. Thanks also for the two references (I actually tried to read Utiyama paper last year, without big success). I will try to read the book by Schrödinger for the holidays. Thanks a lot again. – Oaoa Dec 10 '12 at 9:33 @Oaoa you are welcome! – Murod Abdukhakimov Dec 10 '12 at 12:12 show 3 more comments In arXiv:physics/9801019, gravity is classified as a parameterized classical field theory with dynamic metric. The following quote is the best one I could find on short notice: Consider a classical field theory with gauge group $\mathcal G$. Suppose that $\mathcal G ⊂ \mathrm{Aut}(Y)$, the automorphism group of the covariant configuration bundle $Y$. We may distinguish two basic types of field theories based upon the relationship between the gauge group $\mathcal G$ and the (spacetime) diffeomorphism group $\mathrm{Diff}(X)$. The first consists of those which are parametrized in the sense that the natural homomorphism $\mathrm{Aut}(Y) \to \mathrm{Diff}(X)$ given by $η_Y \mapsto η_X$ maps $\mathcal G$ onto $\mathrm{Diff}(X)$ (or at least onto a “sufficiently large” subgroup thereof, such as the compactly supported diffeomorphisms). This terminology reflects the fact that such a theory is invariant under (essentially) arbitrary relabeling of the points of the parameter “spacetime” $X$. (In relativity theory, cf. Anderson [1967], one would say that $X$ is a relative object in the theory.) The parametrized theory par excellence is of course general relativity, in which case $\mathcal G$ equals the spacetime diffeomorphism group. - In general relativity, tensor fields(e.g. the metric tensor, the Riemann curvature tensor, the energy momentum tensor) are left unchanged by changes of frames. Tensors are naturally unchanged under changes of frames. They form the basis of general relativity and is the key to how general relativity is generally covariant. Update: A note on the invariance of tensors: while tensors are invariant their components certainly do vary under different frames. There are also scalars which are 'single numbers' that don't change at all. I do not think general relativity can be considered a gauge theory at all. In general, there are no local or global symmetries in the Langrangian of general relativity. Any spacetime can occur in general relativity; just define the energy momentum tensor according to Einstein's field equations. - 2 Also, while tensors as a whole are "invariant" in their own geometric sense, their components certainly change. If you want a single number to be unchanged, then you have to restrict yourself to scalars. – Chris White Dec 9 '12 at 2:16 1 @ChrisWhite Yes, I should have included that too. Thank you for reminding me. – namehere Dec 9 '12 at 2:54 @namehere: Thanks for your update ! – Oaoa Dec 9 '12 at 13:19 @Oaoa No problem :) – namehere Dec 9 '12 at 13:25 @namehere Feel free to incorporate comments (mine at least) into your answers at your discretion. Single coherent answers are usually better than scattered comments, and besides the comment record is preserved :) – Chris White Dec 9 '12 at 20:50 show 2 more comments GR has some formal resemblance to Yang-Mills gauge theory. But it isn't quite the same thing. We formulate YM in terms of gauge fields, AKA, connections on G-bundles on our manifold. We also make use of a connection when we formulate GR, the Levi-Civita connection on the tangent bundle of our spacetime, which is determined by the metric and some assumptions (metric is covariantly constant, no torsion). But the metric is the more fundamental degree of freedom, and there's nothing like this in YM theory. (You can do functional integration on the space of YM fields with enough mathematical rigor to satisfy most physicists, but in 4d, it's not possible to do this with metrics.) Another similarity: The observables in YM theory are invariants of the group of gauge transformations. Similarly, in GR, the true observables are generally supposed to be invariant under the group of spacetime diffeomorphisms (n.b., not the same thing as the group of gauge transformations for the tangent bundle). These observables aren't generally local observables, like the curvature at a point, but instead more complicated expression constructed from local observables, like the average of the curvature over the spacetime. This is also in contrast to Yang-Mills theory, where there are plenty of local observables, like the energy density of the YM field at a point. The common theme is that we have to introduce weird auxilliary unphysical variables into both theories to make locality and Lorentz invariance manifest; physical observables are then obtained by forgetting the redundant information. - Thanks a lot for the addenda – Oaoa Dec 10 '12 at 9:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9303795695304871, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/156921-f-z-derivable-then-s-continuous-prove.html	# Thread: 1. ## f(z) is derivable - then it's continuous - Prove it! Prove that if $f$ is derivable in $z_0$ then $f$ is continuous in $z_0$. $\Omega \in \mathbb{C}, ~f:\Omega \rightarrow \mathbb{C}, ~ z_0 \in \Omega$ I started with the definition. $\lim_{z \to 0} \frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}$ Using Cauchys integral formula. $\frac {f(z_0+\Delta z) - f(z_0)}{\Delta z}$ = $\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-(z_0+\Delta z)} \, dz -\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)}{z-z_0} \, dz$ = $\frac{1}{2\pi i \Delta z} \oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)} \, dz$ = $\frac{1}{2\pi i } \oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \, dz$ Now comparing the result with Cauchy's Integral formula, first order. $\oint_C \frac{f(z)}{(z-z_0-\Delta z)(z-z_0)} \, dz - \oint_C \frac{f(z)}{(z-z_0)^2} \, dz$ = $\oint_C \frac{f(z)\Delta z}{(z-z_0-\Delta z)(z-z_0)^2} \, dz$ What happens with the right side if $\Delta z \rightarrow 0$? I'm sort of in the dark here... I thought maybe I proved something by showing how Cauchy's and the definition of derivation dance so nicely together. Help or any pointers would be appreciated! 2. You're overcomplicating it! Note that $lim_{z \to z_0} f(z)-f(z_0) = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}\cdot (z-z_0)$ Also, I think that the proper term is 'differentiable', not 'derivable'. 3. Well as far as languages, I like Math and my native tongue. Bare with me... Tomorrow I'll dig into the stuff you suggested. 4. Another person who wants us to "bare" with him! This website is getting too raw for me! liquidFuzz, in order that a function be differentiable at z= a, you must have $\lim_{\Delta z\to 0}\frac{f(a+ \Delta z)- f(a)}{\Delta z}$ exist. Since that denominator obviously goes to 0, in order that the limit exist, the numerator must also exist (a necessary though not sufficient condition). That is, we must have $\lim_{\Delta z\to 0} f(a+ \Delta z}- f(a)= 0$, whence $\lim_{\Delta z\to 0} f(a+ \Delta z)= f(a)$. Letting $z= a+ \Delta z$ that is the same as $\lim_{z\to a} f(z)= f(a)$. 5. Mhmm, but by showing that the limit exists have I proved that a differentiable function is continuous?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 19, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9524263143539429, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2010/05/03/more-measurable-real-valued-functions/?like=1&source=post_flair&_wpnonce=9916d21231	# The Unapologetic Mathematician ## More Measurable Real-Valued Functions We want a few more convenient definitions of a measurable real-valued function. To begin with: a real-valued function $f$ on a measurable space $(X,\mathcal{S})$ is measurable if and only if for every real number $c$ the set $N(f)\cap\{x\in X\vert f(x)<c\}$ is measurable. Indeed, say $f$ is measurable. If we take $M=\left(-\infty,c\right)$ then $M$ is a Borel set and $f^{-1}(M)=\{x\in X\vert f(x)<c\}$. The measurability of $f$ tells us that $N(f)\cap f^{-1}(M)$ is measurable as a subset of $X$. Conversely, suppose that the given sets are all measurable. If $c_1\leq c_2$ are real numbers, then we can write $\left[c_1,c_2\right)=\left(-\infty,c_2\right)\setminus\left(-\infty,c_1\right)$, and thus $\displaystyle\{x\in X\vert c_1\leq f(x)<c_2\}=\{x\in X\vert f(x)<c_2\}\setminus\{x\in X\vert f(x)<c_1\}$ That is, if $M$ is any semiclosed interval then $N(f)\cap f^{-1}(M)$ is the difference of two measurable sets, and is thus measurable itself. If $\mathcal{M}$ is the collection of all the subsets $M\subseteq\mathbb{R}$ for which $N(f)\cap f^{-1}(M)$ is measurable, then $\mathcal{M}$ is a $\sigma$-ring containing all semiclosed intervals. It must then contain all Borel sets, and so $f$ is measurable. The same statement will hold true if we replace $<$ by $\leq$, or by $>$, or by $\geq$. We walk through the exact same method as before, constructing left- or right-semiclosed intervals — and thus all Borel sets — from open or closed rays as needed. In fact, we can even restrict $c$ to lie in some everywhere-dense subset $C\subseteq\mathbb{R}$. For example, we might only check this condition for rational $c$. Indeed, say we want to construct the closed interval $\left[a,b\right]$. By density, we can find sequences $\{a_i\}$ (increasing) and $\{b_i\}$ (decreasing) of points in $C$ converging to $a$ and $b$, respectively. Then we can construct the intervals $\left[a_i,b_i\right)$ or $\left(a_i,b_i\right]$, and their intersection is the closed interval $\left[a,b\right]$ we want. Then the closed intervals generate the Borel sets, and we’re done. All of these proofs, by the way, hinge on the fact that taking preimages and intersections commute with all of our set-theoretic constructions. Now, if $f$ is a nonzero constant function $f(x)=c$, then $f$ is measurable if and only if $X$ is a measurable subset of itself. Indeed, $N(f)=X$, and $f^{-1}(M)$ is either $X$ or $\emptyset$, according as $M$ does or does not contain $c$. And since every $c$ must be contained in some measurable set, $X$ must be measurable for $f$ to be measurable. More generally, the characteristic function $\xi_E$ of a set $E\subseteq X$ is measurable if and only if $E$ is a measurable subset of $X$. This time, $N(f)=E$, and $f^{-1}(M)$ is either $E$ or $\emptyset$, according as $M$ contains $1$ or not. If $f$ is a measurable function and $k$ is a nonzero real number, then the function $kf$ is also measurable. Indeed, it’s clear that $N(kf)=N(f)$. We must check that $N(kf)\cap\{x\in X\vert kf(x)<c\}$ is measurable, but this set is equal to $N(f)\cap\{x\in X\vert f(x)<\frac{c}{k}\}$, which is measurable. Finally, every continuous function $f:\mathbb{R}\to\mathbb{R}$ is Borel measurable. Indeed, we can write any Borel set $B$ as a limit of open sets. The preimage of each open set is open, and thus Borel, and the preimage of the limit is the limit of the preimages, which is again Borel. ### Like this: Posted by John Armstrong \| Analysis, Measure Theory ## 6 Comments » 1. [...] the function . If is a positive integer, we have the function . These are all continuous, which implies that they’re Borel measurable, and they send back to itself. We conclude that any positive [...] Pingback by \| May 4, 2010 \| Reply 2. [...] continuous function such that if and only if . And since it’s continuous, it’s Borel measurable, and any Borel measurable function is Lebesgue [...] Pingback by \| May 5, 2010 \| Reply 3. [...] now we turn to our convenient condition for measurability. Since we’ve handled the sets where and are infinite, we can assume that they’re [...] Pingback by \| May 7, 2010 \| Reply 4. [...] now let’s use our convenient condition. Given a real number we know that if and only if for some . That is, we can [...] Pingback by \| May 10, 2010 \| Reply 5. [...] is positive, is negative, and is a Hahn decomposition. Indeed, we know that and are measurable. Thus if is measurable, then is measurable, and we [...] Pingback by \| June 29, 2010 \| Reply 6. [...] so that (or ) is a measurable set, then is a measurable function. We will show this using an equivalent definition of measurability — that will be measurable if we can show that for every real number the [...] Pingback by \| July 21, 2010 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). 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http://mathoverflow.net/questions/75689?sort=oldest	## Uniform distribution with respect to Hausdorff measure ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Suppose I have some nicely defined "fractal" subset of (to make life simpler) Euclidean space $\mathbb{E}^n,$ of some arbitrary Hausdorff dimension $s,$ such that the corresponding Hausdorff measure $H_s$ has positive mass. The first question is whether it makes sense to talk of a "uniform sample" (or Poisson point process) with respect to $H_s$ (since it is only an outer measure), and second question is: if it makes sense, have people figured out how to generate uniform $H_s$ variates (in simple cases, like the von Koch snowflake, or the more interesting cases, like limit sets of Kleinian groups)? - What do you mean "only an outer measure"? Isn't it a measure on the borel sets? Also, does "variate" here just mean an algorithm to produce uniform random variables? If so, it is easy in the case of the snowflake. – Ori Gurel-Gurevich Sep 17 2011 at 19:12 What Ori said. Use a Borel set. All the usual "nicely fractal" sets in Euclidean space are Borel sets. Indeed, they are compact or at least sigma-compact. – Gerald Edgar Sep 17 2011 at 19:16 I am a babe in the woods of measure theory... As for variates, yes, an algorithm. In particular, what is the "easy method" for the snowflake? And how do you go about proving that some algorithm gives uniform points (for snowflake or dynamically defined sets)? – Igor Rivin Sep 17 2011 at 19:20 Other people have answered, but still: To pick a uniform point on one side (out of three) of the snowflake, you just have to pick 1 of the 4 parts of the curve, then 1 of 4 parts of that part, etc. It is really no different than picking a uniform point on $[0,1]$, by iteratively choosing halves of the interval. – Ori Gurel-Gurevich Sep 19 2011 at 18:44 @Ori: I sort of guessed that, but why is that true? What is the general version of the construction? Presumably, it's some iterated function system thing, but if I knew all there was to know about such, I would not be asking the question... – Igor Rivin Sep 19 2011 at 19:25 ## 3 Answers There is a standard map from the interval $[0,1]$ onto the snowflake curve, that maps Lebesgue measure into (constant multiple of) Hausdorff measure. So take a sample in $[0,1]$ and map it to the snowflake. This works for all of the standard "self-similar" IFS constructions of fractals, when the IFS satisfies the open set condition. There is a standard "parameter space" or "code space" with a well-understood measure on it that maps onto the fractal, and the image measure is (constant multiple of) Hausdorff measure, and also (constant multiple of) packing measure. - @Gerald: thanks, but saying that the construction is "standard" does not tell me what it is. Would you happen to have any references you would particularly recommend? – Igor Rivin Sep 17 2011 at 20:20 Ken Falconer's books are a great place to start. – BSteinhurst Sep 19 2011 at 14:21 @BSteinhurst: thanks, I have started looking at one of them already... – Igor Rivin Sep 19 2011 at 15:21 I could also plug my own book, Measure, Topology, and Fractal Geometry – Gerald Edgar Sep 19 2011 at 18:31 @Gerald: I don't think there is anything morally wrong about suggesting your own book (especially if it is the best book on the subject :)) – Igor Rivin Sep 19 2011 at 19:26 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. Given $A$ a loc. compact space and $\mu$ a loc. finite measure on $A$, there exists a unique Poisson point process $\eta$ with control measure $\mu$. (see for instance th.3.2.1 in "stochastic geometry", by Schneider & Weil, if there are no atoms, but it holds in full generality) . If $A$ is your fractal and $\mu$ Hausdorff measure, then $\eta$ is what you are looking for, isn't it? For the construction, everything depends on how you define your fractal set. The difficulty is exactly the same as drawing a uniform random point in a subset that has finite Hausdorff measure. - Firstly, thanks for the reference! As for "nicely defined", the canonical example is a limit set of a kleinian group, but again, the snowflake is a nice simple example... – Igor Rivin Sep 19 2011 at 15:19 If one interprets "nicely defined" correctly, the answer is "yes, duh!". One possible interpretation is to be the attractor of a Horseshoe map. Then the dynamics on this attractor can be conjugated to symbolic dynamics and it is clear what "drawing a random point" means, and how to generate them numerically. Of course, one also needs to check that the invariant measure has maximal Hausdorff dimension. In order to illustrate this, let me give the simplest example. Consider the map $f(x) = x^2 - c$ for $c > 4$. It is well known, that there exists an invariant set $\Sigma = f^{-1}(\Sigma)$ with Hausdorff dimension $\in (0,1)$. $\Sigma$ can be described as $$\Sigma = \{x:\quad \exists C>0 \forall n\geq 1, \|f^n(x)\|\leq C\}$$ so the set of points with bounded orbit. If you define for $x\in \Sigma$ the sequence $$x_n = \begin{cases} 1 & f^n(x) > 0, \\ - 1,& f^n(x) < 0 \end{cases}$$ Then standard results imply that the Bernouilli measure on ${-1,1}^{\mathbb{N}}$ is conjugate to the measure of maximal Hausdorff dimension, which is ergodic with respect to $(\Sigma, f)$. Hence, picking a random point just amounts to choosing a random string of $\pm 1$. - Thanks! I will try to digest... – Igor Rivin Sep 19 2011 at 15:20	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9174684286117554, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/3347?sort=oldest	## Is the set of primes “translation-finite”? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The definition in the title probably needs explaining. I should say that the question itself was an idea I had for someone else's undergraduate research project, but we decided early on it would be better for him to try adjacent and less technical questions. So it's not of importance for my own work per se, but I'd be interested to know if it easily reduces to a known conjecture/fact/counterexample in number theory. Apologies if the question is too technical/localized/unappealing/bereft of schemes. Given a subset $X$ of the natural numbers $N$, and given $n \in N$, we write $X-n$ for the backward translate of $X$, i.e. the set {${x-n : x\in X}$}. We say that $X$ is translation-finite if it has the following property: for every strictly increasing sequence n1 < n2 < in $N$, there exists k (possibly depending on the sequence) such that $(X-n_1) \cap (X-n_2) \cap \dots\cap (X-n_k)$ is finite or empty. Thus every finite set is trivially translation-finite: and if the elements of $X$ form a sequence in which the difference between successive terms tends to infinity, then $X$ is translation-finite and we can always take k=2. Moreover: • if $X$ contains an infinite arithmetic progression, or if it has positive (upper) Banach density, then it is NOT translation finite; • there exist translation-finite sets which, when enumerated as strictly increasing sequences, grow more slowly than any faster-than-liner function. • there exist translation-finite sets containing arbitrarily long arithmetic progressions. These resultlets suggest the question in the title, but I don't know enough about number theory to know if it's a reasonable question. Note that if, in the definition, we were to fix k first (i.e. there exists k such that for any sequence (nj)...) then we would get something related to Hardy-Littlewood conjectures; but I was hoping that this might not be necessary to resolve the present question. EDIT (2nd Nov) It's been pointed out below that the question reduces in some sense to a pair of known, hard, open problems. More precisely: if the answer to the question is yes, then we disprove the Hardy-Littlewood k-tuples conjecture; if the answer is no, then there are infinitely many prime gaps bounded by some absolute constant, and this is thought to be beyond current techniques unless one assumes the Eliott-Halberstam conjecture. - 1 What do you mean by "the elements of X form a sequence tending to infinity"? Does that imply something other than X being an infinite set? – Darsh Ranjan Oct 30 2009 at 0:07 aargh, typo/brain spasm - I missed out some important words. I meant that the gap size has to tend to infinity. I've edited the original post to correct this. – Yemon Choi Oct 30 2009 at 2:00 5 Yay for non-scheme questions! – Theo Johnson-Freyd Oct 30 2009 at 2:12 I think "policy" says that upon questions being found to be equivalent to a known open problem, they should be wiki'd. This doesn't seem appropriate in this case. Feel free to disagree over at meta: meta.mathoverflow.net/discussion/8/… – Scott Morrison♦ Nov 3 2009 at 0:55 @Scott: makes sense to me. @Yemon: thanks for nicely summarizing th results in the question body! – Ilya Nikokoshev Nov 27 2009 at 23:52 ## 1 Answer As you mention, this is related to the Hardy-Littlewood k-tuple conjecture. In particular, if their conjecture is true, then the primes are not translation-finite. Indeed, it is possible to find an increasing sequence n1 < n2 < n3 < ⋯ so that for every k, the first k nis form an admissible k-tuple. (For example, I think ni = (i+1)! works.) Then, by the k-tuple conjecture, infinitely many such prime constellations exist and so for all k, (X-n1) ∩ (X-n2) ∩ ⋯ ∩ (X-nk) is infinite. (Here and below, X is the set of primes.) However, maybe we can prove that the primes are not translation finite by some other means. Unfortunately, the technology is not quite good enough to do that. Proving that the primes are not translation finite would, in particular, prove that there exist n1 < n2 such that (X-n1) ∩ (X-n2) is infinite. In particular, this implies that the gap n2-n1 occurs infinitely often in primes, and so pn+1-pn is constant infinitely often. (The standard notation pn indicates the nth prime.) The best known upper bound for the size of small gaps in primes is that lim infn→∞ (pn+1-pn)/log pn = 0. This was established by Goldston and Yildirim around 2003 and the proof was later simplified. To the best of my knowledge, the best conditional result is by the same authors; they show that given the Elliott-Halberstam conjecture, the prime gap is infinitely often at most 20 or so. - Thanks! That's very clear. So, if I understand you correctly: a proof of HL's k-tuple conjecture would imply the primes are not translation finite; while a proof that the primes are not translation finite would imply a bound on prime gaps that is beyond current knowledge. So my question essentially reduces to known open problems I guess. – Yemon Choi Nov 1 2009 at 23:52 Which probably means it deserves a CW status + open-problems tag. – Ilya Nikokoshev Nov 2 2009 at 0:04 As well as rewriting the question so the two open problems become clear. – Ilya Nikokoshev Nov 2 2009 at 0:04 1 @Yemon Choi: That's correct. Proving the primes are translation finite would disprove the k-tuples conjecture, while proving the primes aren't translation finite would prove bounds on small prime gaps. – aorq Nov 2 2009 at 3:27 Why does the infinitude of $(X-n_1)\cap(X-n_2)$ imply that the gap $n_2-n_1$ occurs infinitely often? Couldn't some smaller gaps occur infinitely often instead? – François G. Dorais♦ Dec 27 at 4:59	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9433725476264954, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Canonical_class	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Canonical bundle (Redirected from Canonical class) In mathematics, the canonical bundle of a non-singular algebraic variety V of dimension n is the line bundle $\,\!\Omega^n = \omega$ which is the nth exterior power of the cotangent bundle Ω on V. That is, it is the bundle of holomorphic n-forms on V, if V is defined over the complex number field. This is the dualising object for Serre duality on V. It may equally be considered an invertible sheaf. The canonical class is the divisor class of a Cartier divisor K on V giving rise to the canonical bundle — it is an equivalence class for linear equivalence on V, and any divisor in it may be called a canonical divisor. An anticanonical divisor is any divisor $-\!K$ with $\!\,K$ canonical. The anticanonical bundle is the corresponding inverse bundle $\,\!\omega^{-1}$. 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 4, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7320778369903564, "perplexity_flag": "middle"}
http://outofthenormmaths.wordpress.com/tag/taxicab-distance/	# Tag Archives: taxicab distance November 29, 2011 · 10:09 pm ## Getting into norms This blog is named “Out of the Norm”, and the logo is the list of axioms that a norm satisfies (with some extra graffiti), but there’s a fair chance that it’s just mathematical gobbledegook to you. So what is a norm? At its heart, a norm is essentially a measure of distance, though not necessarily a spatial distance. Here are a couple of examples to give you some informal sense of what a norm is, and how they might be used. Distance in space If I asked you to go find some buried treasure 3 units of distance East, and 4 units North, how far would you have to travel? That is, from the origin (0, 0) to the point (3, 4) on the graph or ’map’ below? It’s not a trick question: the distance as the crow flies, by Pythagoras’ theorem, is simply $\sqrt{3^2 + 4^2}=5$. But what if it were a trick question? Let’s say you were actually a taxi driver in some Manhattan-like city with square blocks (each a single unit in size), and were taking a passenger three blocks East and four blocks North (perhaps to the buried treasure, if you insist). Then the shortest route would be seven units long (7=3+4; there are many different shortest routes). Two of the shortest routes So now you’ve seen two different notions of distance in 2-dimensional space: the first very familiar; the second probably less so. Just to reinforce the point, you can do the same thing in 3-dimensional space. The distance as the crow flies directly to a point 5 units above the unnecessary treasure, that is, the point (3, 4, 5), is approximately seven units: $\sqrt{3^2+4^2+5^2}=\sqrt{50}$. However, our taxi cab driver has to drive around the buildings and then take some sort of car-lift, and travels $3+4+5=12$ units. Diagram not at all to scale. You can extend this to as many spatial dimensions as you want, though as far as I know it’s impossible for humans to imagine a diagram with the point (3, 4, 5, 2) hovering somewhere. But that doesn’t mean the notion is useless. Guess the age of the lecturers The next example has been used by one of my former colleagues to introduce norms to his students. Firstly, he gets the students to each write down a guess of the ages of all the lecturers in the maths department, and writes the guesses up on the board. Then he reveals the true ages of the lecturers, and asks the students how they should decide who has won. Here’s some completely made-up numbers: Let’s say there are four lecturers: a PhD student, a lecturer, a senior lecturer, and a professor, who are 25, 35, 45 and 60 years old respectively. We’ll write this as the vector (25, 35, 45, 60). Now the three students make their guesses: Adam guesses in order 27, 36, 55 and 66: again, write this as a vector (27, 36, 55, 66). Beth guesses (26, 35, 53, 51), completely misjudging the age order of the older pair. And, because there’s always one in every class, Charlie guesses (27.2, 34, 56.4, 62.4), which reminds us that we needn’t work only with whole numbers. The students decide to judge the results Continue reading → 3 Comments Filed under Accessible Tagged as distance, manhattan distance, maths, metric, norms, taxicab distance, \ell_1, \ell_2, \ell_{infty}	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 3, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9458191394805908, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/192674-linear-regression-residual-sum-squares.html	# Thread: 1. ## Linear Regression, Residual Sum of Squares Suppose the columns of a rank 4 design matrix $X = [X_{1}, X_{2}, X_{3}, X_{4}]$ come in three groups Group 1: $\{X_{1}\}$, Group 2: $\{X_{2}, X_{3}\}$ and Group 3: $\{X_{4}\}$ so that $X^T_{i}X_{j} = 0$ if i and j are indicies from different groups. Next consider 4 models, with design matricies: Model 1: $X_{(1)} = [X_{1}]$ Model 2: $X_{(2)} = [X_{1}, X_{2}, X_{3}]$ Model 3: $X_{(3)} = [X_{1}, X_{2}, X_{3}, X_{4}]$ Model 4: $X_{(4)} = [X_{1}, X_{4}]$ Now then the question asks to show that $SS_{2} - SS_{3} = SS_{1} - SS_{4}$ where $SS_{i}$ is the residual sum of squares (SS error) for the ith model. $SS_{error} = Y^T Y - \hat\theta X^T Y$ is the only formula for SS error that I know, which I believe does not apply in this situation. So it means that I'm absolutely stuck! I think I'm looking at this the wrong way to be honest. Where should I look into? Any hints?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 12, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.919986367225647, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Dirac_equation	# All Science Fair Projects ## Science Fair Project Encyclopedia for Schools! Search Browse Forum Coach Links Editor Help Tell-a-Friend Encyclopedia Dictionary # Science Fair Project Encyclopedia For information on any area of science that interests you, enter a keyword (eg. scientific method, molecule, cloud, carbohydrate etc.). Or else, you can start by choosing any of the categories below. # Dirac equation The Dirac equation is a relativistic quantum mechanical wave equation invented by Paul Dirac in 1928. It provides a description of elementary spin-½ particles, such as electrons, that is fully consistent with the principles of quantum mechanics and largely consistent with the theory of special relativity. It also accounts in a natural way for the nature of particle spin and the existence of antiparticles. Contents ## Introduction Since the Dirac equation was originally invented to describe the electron, we will generally speak of "electrons" in this article. Actually, the equation applies to other types of elementary spin-½ particles, such as quarks (but not neutrinos because the left-handed and the hypothetical right-handed neutrinos have different Majorana masses). A modified Dirac equation can be used to approximately describe protons and neutrons, which are made of smaller particles called quarks and are therefore not elementary particles. The Dirac equation is $\left(\alpha_0 mc^2 + \sum_{j = 1}^3 \alpha_j p_j \, c\right) \psi (\mathbf{x},t) = i \hbar \frac{\partial\psi}{\partial t} (\mathbf{x},t)$ where m is the rest mass of the electron, c is the speed of light, p is the momentum operator, $\hbar$ is Planck's constant, x and t are the space and time coordinates respectively, and ψ(x, t) is a four-component wavefunction. (The wavefunction has to be formulated as a four-component spinor, rather than a simple scalar, due to the demands of special relativity. The physical meanings of the components are discussed below.) The α's are linear operators that act on the wavefunction, written as a column matrix, as 4×4 matrices known as Dirac matrices. There is more than one way to choose a set of Dirac matrices, a convenient choice being $\alpha_0 = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} \quad \alpha_1 = \begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{bmatrix}$ $\alpha_2 = \begin{bmatrix} 0 & 0 & 0 & -i \\ 0 & 0 & i & 0 \\ 0 & -i& 0 & 0 \\ i & 0 & 0 & 0 \end{bmatrix} \quad \alpha_3 = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \end{bmatrix}$ All possible choices are related by similarity transformations because Dirac spinors are unique representation theoretically. The Dirac equation describes the probability amplitudes for a single electron. This single-particle theory gives a fairly good prediction of the spin and magnetic moment of the electron and explains much of the fine structure observed in atomic spectral lines. It also makes the peculiar prediction that there exists an infinite set of quantum states in which the electron possesses negative energy. This strange result led Dirac to predict, via a remarkable hypothesis known as "hole theory", the existence of particles behaving like positively-charged electrons. This prediction was verified by the discovery of the positron in 1932. Despite these successes, the theory is flawed by its neglect of the possibility of creating and destroying particles, one of the basic consequences of relativity. This difficulty is resolved by reformulating it as a quantum field theory. Adding a quantized electromagnetic field to this theory leads to the modern theory of quantum electrodynamics (QED). For a more detailed discussion of the field formulation, refer to the article on Dirac field theory . Similar equation for spin 3/2 particles is called Rarita-Schwinger equation. ## A handwaving derivation of the Dirac equation The Dirac equation is a special case of the Schrödinger equation, which describes the time-evolution of a quantum mechanical system: $H \left\| \psi (t) \right\rangle = i \hbar {d\over d t} \left\| \psi (t) \right\rangle$ For convenience, we will work in the position basis, in which the state of the system is represented by a wavefunction, ψ(x,t). In this basis, the Schrödinger equation becomes $H \psi (\mathbf{x},t) = i \hbar \frac{\partial\psi}{\partial t} (\mathbf{x},t)$ where the Hamiltonian H now denotes an operator acting on wavefunctions rather than state vectors. We have to specify the Hamiltonian so that it appropriately describes the total energy of the system in question. Let us consider a "free" electron isolated from all external force fields. For a non-relativistic model, we adopt a Hamiltonian analogous to the kinetic energy of classical mechanics (ignoring spin for the moment): $H = \sum_{j=1}^3 \frac{p_j^2}{2m}$ where the p's are the momentum operators in each of the three spatial directions j=1,2,3. Each momentum operator acts on the wavefunction as a spatial derivative: $p_j \psi(\mathbf{x},t) \equiv - i \hbar \, \frac{\partial\psi}{\partial x_j} (\mathbf{x},t)$ To describe a relativistic system, we have to find a different Hamiltonian. Assume that the momentum operators retain the above definition. According to Albert Einstein's famous mass-momentum-energy relationship, the total energy of a system is given by $E = \sqrt{(mc^2)^2 + \sum_{j=1}^3 (p_jc)^2}$ This prescribes something like $\sqrt{(mc^2)^2 + \sum_{j=1}^3 (p_jc)^2} \; \psi = i \hbar \frac{d\psi}{d t}$ This is not a satisfactory equation, for it does not treat time and space on an equal footing, one of the basic tenets of special relativity. Dirac reasoned that, since the right side of the equation contains a first-order derivative in time, the left side should contain equally simple first-order derivatives in space (i.e., in the momentum operators). One way for this to happen is if the quantity in the square root is a perfect square. Suppose that $(mc^2)^2 + \sum_{j=1}^3 (p_jc)^2 = \left( \alpha_0 mc^2 + \sum_{j=1}^3 \alpha_j p_j \, c \right)^2$ where the α's are constants to be determined. Expanding the square and comparing coefficients on each side, we obtain the following conditions for the α's: $\alpha_\mu^2 = I \,,\qquad\qquad\quad\;\; \mu = 0,1,2,3$ $\alpha_\mu \alpha_\nu + \alpha_\nu \alpha_\mu = 0 \,,\quad \mu \ne \nu$ Here, I stands for the identity element. These conditions may be written more concisely as $\left\{\alpha_\mu , \alpha_\nu\right\} = 2\delta_{\mu\nu} \cdot I$ where {...} is the anticommutator, defined as {A,B}≡AB+BA, and δ is the Kronecker delta, which has the value 1 if its two subscripts are equal and 0 otherwise. See Clifford algebra. These conditions cannot be satisfied if the α's are ordinary numbers, but they can be satisfied if the α's are matrices. The matrices must be Hermitian, so that the Hamiltonian is Hermitian. The smallest matrices that work are 4×4 matrices, but there is more than one possible choice, or representation, of matrices. Although the choice of representation does not affect the properties of the Dirac equation, it does affect the physical meaning of the individual components of the wavefunction. In the introduction, we presented the representation used by Dirac. This representation can be more compactly written as $\alpha_0 = \begin{bmatrix} I & 0 \\ 0 & -I \end{bmatrix} \quad \alpha_j = \begin{bmatrix} 0 & \sigma_j \\ \sigma_j & 0 \end{bmatrix}$ where 0 and I are the 2×2 zero and identity matrices, respectively, and the σj's (j=1,2,3) are the Pauli matrices. It is now straightforward to carry out the square root, which gives the Dirac equation. The Hamiltonian in this equation, $H = \,\alpha_0 mc^2 + \sum_{j = 1}^3 \alpha_j p_j \, c$ is called the Dirac Hamiltonian. ### Nature of the wavefunction Since the wavefunction ψ is acted on by the 4×4 Dirac matrices, it must be a four-component object. We will see, in the next section, that the wavefunction contains two sets of degrees of freedom, one associated with positive energies and the other with negative energies, with each set containing two degrees of freedom that describe the probability amplitudes for the spin to be pointing "up" or "down" along a specified direction. We may explicitly write the wavefunction as a column matrix: $\psi(\mathbf{x},t) \equiv \begin{bmatrix}\psi_1(\mathbf{x},t) \\ \psi_2(\mathbf{x},t) \\ \psi_3(\mathbf{x},t) \\ \psi_4(\mathbf{x},t) \end{bmatrix}$ The dual wavefunction can be written as a row matrix: $\psi^\dagger(\mathbf{x},t) \equiv \begin{bmatrix}\psi_1^(\mathbf{x},t) & \psi_2^(\mathbf{x},t) & \psi_3^(\mathbf{x},t) & \psi_4^(\mathbf{x},t) \end{bmatrix}$ where the * superscript denotes complex conjugation. By comparison, the dual of a scalar (one-component) wavefunction is just its complex conjugate. As in ordinary single-particle quantum mechanics, the "absolute square" of the wavefunction gives the probability density of the particle at each position x and time t. In this case, the "absolute square" is obtained by matrix multiplication: $\psi^\dagger \psi \, (\mathbf{x},t) = \sum_{j = 1}^4 \psi_j^*(\mathbf{x},t) \psi_j(\mathbf{x},t)$ The conservation of probability gives the normalization condition $\int \psi^\dagger \psi \, (\mathbf{x},t) \; d^3x = 1$ By applying Dirac's equation, we can examine the local flow of probability: $\frac{\partial}{\partial t} \psi^\dagger \psi \, (\mathbf{x},t) = - \nabla \cdot \mathbf{J}$ The probability current J is given by $J_j = c \psi^\dagger \alpha_j \psi$ Multiplying J by the electron charge e yields the electric current density j carried by the electron. The values of the wavefunction components depend on the coordinate system. Dirac showed how ψ transforms under general changes of coordinate system, including rotations in three-dimensional space as well as Lorentz transformations between relativistic frames of reference. It turns out that ψ does not transform like a vector under rotations and is in fact a type of object known as a spinor. ### Energy spectrum It is instructive to find the energy eigenstates of the Dirac Hamiltonian. To do this, we must solve the time-independent Schrödinger equation, $H \psi_0 (\mathbf{x}) = E \psi_0(\mathbf{x})$ where ψ0 is the time-independent part of the energy eigenfunction: $\psi (\mathbf{x}, t) = \psi_0 (\mathbf{x}) e^{- i E t / \hbar}$ Let us look for a plane-wave solution. For convenience, we align the z axis with the direction in which the particle is moving, so that $\psi_0 = w e^{\frac{ipz}{\hbar}}$ where w is a constant four-component spinor and p is the momentum of the particle, as we can verify by applying the momentum operator to this wavefunction. In the Dirac representation, the equation for ψ0 reduces to the eigenvalue equation: $\begin{bmatrix} mc^2 & 0 & pc & 0 \\ 0 & mc^2 & 0 & -pc \\ pc & 0 & -mc^2 & 0 \\ 0 & -pc & 0 & -mc^2 \end{bmatrix} w = E w$ For each value of p, there are two eigenspaces, both two-dimensional. One eigenspace contains positive eigenvalues, and the other negative eigenvalues, of the form: $E_\pm (p) = \pm \sqrt{(mc^2)^2 + (pc)^2}$ The positive eigenspace is spanned by the eigenstates: $\left\{ \begin{bmatrix}pc \\ 0 \\ \epsilon \\ 0 \end{bmatrix} \,,\, \begin{bmatrix}0 \\ pc \\ 0 \\ - \epsilon \end{bmatrix} \right\} \times \frac{1}{\sqrt{\epsilon^2+(pc)^2}}$ and the negative eigenspace by the eigenstates: $\left\{ \begin{bmatrix}-\epsilon \\ 0 \\ pc \\ 0 \end{bmatrix} \,,\, \begin{bmatrix}0 \\ \epsilon \\ 0 \\ pc \end{bmatrix} \right\} \times \frac{1}{\sqrt{\epsilon^2+(pc)^2}}$ where $\epsilon \equiv \|E\| - mc^2$ The first spanning eigenstate in each eigenspace has spin pointing in the +z direction ("spin up"), and the second eigenstate has spin pointing in the −z direction ("spin down"). In the non-relativistic limit, the ε spinor component reduces to the kinetic energy of the particle, which is negligible compared to pc: $\epsilon \sim \frac{p^2}{2m} <\!\!< pc$ In this limit, therefore, we can interpret the four wavefunction components as the respective amplitudes of (i) spin-up with positive energy, (ii) spin-down with positive energy, (iii) spin-up with negative energy, and (iv) spin-down with negative energy. This description is not accurate in the relativistic regime, where the non-zero spinor components have similar sizes. ### Hole theory The negative E solutions found in the preceding section are problematic, for relativistic mechanics tells us that the energy of a particle at rest (p = 0) should be E = mc² rather than E = −mc². Mathematically speaking, however, there seems to be no reason for us to reject the negative-energy solutions. Since they exist, we cannot simply ignore them, for once we include the interaction between the electron and the electromagnetic field, any electron placed in a positive-energy eigenstate would decay into negative-energy eigenstates of successively lower energy by emitting excess energy in the form of photons. Real electrons obviously do not behave in this way. To cope with this problem, Dirac introduced the hypothesis, known as hole theory, that the vacuum is the many-body quantum state in which all the negative-energy electron eigenstates are occupied. This description of the vacuum as a "sea" of electrons is called the Dirac sea. Since the Pauli exclusion principle forbids electrons from occupying the same state, any additional electron would be forced to occupy a positive-energy eigenstate, and positive-energy electrons would be forbidden from decaying into negative-energy eigenstates. Dirac further reasoned that if the negative-energy eigenstates are incompletely filled, each unoccupied eigenstate – called a hole – would behave like a positively charged particle. The hole possesses a positive energy, since energy is required to create a particle–hole pair from the vacuum. Dirac initially thought that the hole was a proton, but Hermann Weyl pointed out that the hole should behave as if it had the same mass as an electron, whereas the proton is over a thousand times heavier. The hole was eventually identified as the positron, experimentally discovered by Carl Anderson in 1932. However, it is not entirely satisfactory to describe the "vacuum" using an infinite sea of negative-energy electrons. We must postulate that the negative-energy electrons do not contribute to the total energy and momentum of the vacuum, which would otherwise be infinite, and that the negative-energy electrons do not produce an electric field, although they can be affected by an external field. These difficulties led physicists to abandon hole theory in favour of Dirac field theory , which bypasses the problem of negative energy states by treating positrons as true particles. (Caveat: in certain applications of condensed matter physics, the underlying concepts of "hole theory" are certainly valid. The sea of conduction electrons in an electrical conductor, called a Fermi sea , contains electrons with energies up to the chemical potential of the system. An unfilled state in the Fermi sea behaves like a positively-charged electron, though it is referred to as a "hole" rather than a "positron". The negative charge of the Fermi sea is balanced by the positively-charged ionic lattice of the material.) ## Electromagnetic interaction So far, we have considered an electron that is not in contact with any external fields. Proceeding by analogy with the Hamiltonian of a charged particle in classical electrodynamics, we can modify the Dirac Hamiltonian to include the effect of an electromagnetic field. The revised Hamiltonian is (in SI units): $H = \alpha_0 mc^2 + \sum_{j=1}^3 \alpha_j \left[p_j - e A_j(\mathbf{x}, t) \right] c + e \phi(\mathbf{x}, t)$ where e is the electric charge of the electron (in this convention, e is negative), and A and φ are the electromagnetic vector and scalar potentials, respectively. By setting φ = 0 and working in the non-relativistic limit, Dirac solved for the top two components in the positive-energy wavefunctions (which, as discussed earlier, are the dominant components in the non-relativistic limit), obtaining $\left( \frac{1}{2m} \sum_j \|p_j - e A_j(\mathbf{x}, t)\|^2 - \frac{\hbar e}{2mc} \sum_j \sigma_j B_j(\mathbf{x}) \right) \begin{bmatrix}\psi_1 \\ \psi_2 \end{bmatrix}$ $= (E - mc^2) \begin{bmatrix}\psi_1 \\ \psi_2 \end{bmatrix}$ where B = $\nabla$ ×A is the magnetic field acting on the particle. This is precisely the Pauli equation for a non-relativistic spin-½ particle, with magnetic moment $\hbar e/2mc$ (i.e., a spin g-factor of 2). The actual magnetic moment of the electron is larger than this, though only by about 0.12%. The shortfall is due to quantum fluctuations in the electromagnetic field, which have been neglected. See vertex function. For several years after the discovery of the Dirac equation, most physicists believed that it also described the proton and the neutron, which are both spin-½ particles. However, beginning with the experiments of Stern and Frisch in 1933, the magnetic moments of these particles were found to disagree significantly with the predictions of the Dirac equation. The proton has a magnetic moment 2.79 times larger than predicted (with the proton mass inserted for m in the above formulas), i.e., a g-factor of 5.58. The neutron, which is electrically neutral, has a g-factor of −3.83. These "anomalous magnetic moments" were the first experimental indication that the proton and neutron are not elementary particles. They are in fact composed of smaller particles called quarks. Incidentally, quarks are spin-½ particles, which are exactly described by the Dirac equation ! ### Interaction Hamiltonian It is noteworthy that the Hamiltonian can be written as the sum of two terms: $H = H_{\mathrm{free}} + H_{\mathrm{int}} \,$ where Hfree is the Dirac Hamiltonian for a free electron and Hint is the Hamiltonian of the electromagnetic interaction. The latter may be written as $H_{\mathrm{int}} = e \phi(\mathbf{x}, t) - ec \sum_{j=1}^3 \alpha_j A_j(\mathbf{x}, t)$ It has the expected value $\langle H \rangle = \int \, \psi^\dagger H_{\mathrm{int}} \psi \, d^3x = \int \, \left(\rho \phi - \sum_{i=1}^3 j_i A_i \right) \, d^3x$ where ρ is the electric charge density and j is the electric current density defined earlier. The integrand in the final expression is the interaction energy density. It is a relativistically covariant scalar quantity, as we can see by writing it in terms of the current-charge four-vector j = (ρc,j) and the potential four-vector A = (φ/c,A): $\langle H \rangle = \int \, \left( \sum_{\mu,\nu = 0}^3 \eta^{\mu\nu} j_\mu A_\nu \right) \; d^3r$ where η is the metric of flat spacetime: η00 = 1 $\eta^{ii} \;= -1 \quad\, \forall \, i=1,2,3$ $\eta^{\mu\nu} = 0 \qquad \forall \, \mu \ne \nu$ ## Relativistically covariant notation Let us return to the Dirac equation for the free electron. It is often useful to write the equation in a relativistically covariant form, in which the derivatives with time and space are treated on the same footing. To do this, first recall that the momentum operator p acts like a spatial derivative: $\mathbf{p} \psi(\mathbf{x},t) = - i \hbar \nabla \psi(\mathbf{x},t)$ Multiplying each side of the Dirac equation by α0 (recalling that α0²=I) and plugging in the above definition of p, we obtain $\left[ i\hbar c \left(\alpha_0 \frac{\partial}{c \partial t} + \sum_{j=1}^3 \alpha_0 \alpha_j \frac{\partial}{\partial x_j} \right) - mc^2 \right] \psi = 0$ Now, define four gamma matrices: $\gamma^0 \equiv \alpha_0 \,,\quad \gamma^j \equiv \alpha_0 \alpha_j$ These matrices possess the property that $\left\{\gamma^\mu , \gamma^\nu \right\} = 2\eta^{\mu\nu} \cdot I\,,\quad \mu,\nu = 0, 1, 2, 3$ where η once again stands for the metric of flat spacetime. These relations define a Clifford algebra called the Dirac algebra. The Dirac equation may now be written, using the position-time four-vector x = (ct,x), as $\left(i\hbar c \, \sum_{\mu=0}^3 \; \gamma^\mu \, \partial_\mu - mc^2 \right) \psi = 0$ With this notation, the Dirac equation can be generated by extremising the action $\mathcal{S} = \int \bar\psi(i \hbar c \, \sum_\mu \gamma^\mu \partial_\mu - mc^2)\psi \, d^4 x$ where $\bar\psi \equiv \psi^\dagger \gamma_0$ is called the Dirac adjoint of ψ. This is the basis for the use of the Dirac equation in quantum field theory. A notation called the "Feynman slash" is sometimes used. Writing $a\!\!\!/ \leftrightarrow \sum_\mu \gamma^\mu a_\mu$ the Dirac equation becomes $(i \hbar c \, \partial\!\!\!/ - mc^2) \psi = 0$ and the expression for the action becomes $\mathcal{S} = \int \bar\psi(i \hbar c \, \partial \!\!\!/ - mc^2)\psi \, d^4 x$ ## References ### Selected Papers • P.A.M. Dirac, Proc. R. Soc. A117 610 (1928) • P.A.M. Dirac, Proc. R. Soc. A126 360 (1930) • C.D. Anderson, Phys. Rev. 43, 491 (1933) • R. Frisch and O. Stern, Z. Phys. 85 4 (1933) ### Textbooks • Dirac, P.A.M., Principles of Quantum Mechanics, 4th edition (Clarendon, 1982) • Shankar, R., Principles of Quantum Mechanics, 2nd edition (Plenum, 1994) • Bjorken, J D & Drell, S, Relativistic Quantum mechanics • Thaller, B., The Dirac Equation, Texts and Monographs in Physics (Springer, 1992) 03-10-2013 05:06:04	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 52, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9075637459754944, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/35013/exponentiated-logistic-regression-coefficient-different-than-odds-ratio	# Exponentiated logistic regression coefficient different than odds ratio As I understand it, the exponentiated beta value from a logistic regression is the odds ratio of that variable for the dependent variable of interest. However, the value does not match the manually calculated odds ratio. My model is predicting stunting (a measure of malnutrition) using, amongst other indicators, insurance. ````// Odds ratio from LR, being done in stata logit stunting insurance age ... etc. or_insurance = exp(beta_value_insurance) // Odds ratio, manually calculated odds_stunted_insured = num_stunted_ins/num_not_stunted_ins odds_stunted_unins = num_stunted_unins/num_not_stunted_unins odds_ratio = odds_stunted_ins/odds_stunted_unins ```` What is the conceptual reason for these values being different? Controlling for other factors in the regression? Just want to be able to explain the discrepancy. - 1 Are you putting additional predictors into the logistic regression model? The manually calculated odds ratio will only match the odds ratio you get out of logistic regression if you include no other predictors. – Macro Aug 24 '12 at 14:54 That's what I figured, but wanted confirmation. That is because the result from the regression is accounting for variation in other predictors? – mike Aug 24 '12 at 15:11 Yes, @mike. Assuming the model is correctly specified, you can interpret it as the odds ratio when the other predictors are all fixed. – Macro Aug 24 '12 at 15:13 Great, thanks!! – mike Aug 24 '12 at 15:16 @Macro: would you mind restating your comment as an answer? – jrennie Aug 24 '12 at 15:39 ## 2 Answers If you're only putting that lone predictor into the model, then the odds ratio between the predictor and the response will be exactly equal to the exponentiated regression coefficient. I don't think a derivation of this result in present on the site, so I will take this opportunity to provide it. Consider a binary outcome $Y$ and single binary predictor $X$: $$\begin{array}{c\|cc} \phantom{} & Y = 1 & Y = 0 \\ \hline X=1 & p_{11} & p_{10} \\ X=0 & p_{01} & p_{00} \\ \end{array}$$ Then, one way to calculate the odds ratio between $X_i$ and $Y_i$ is $${\rm OR} = \frac{ p_{11} p_{00} }{p_{01} p_{10}}$$ By definition of conditional probability, $p_{ij} = P(Y = i \| X = j) \cdot P(X = j)$. In the ratio, he marginal probabilities involving the $X$ cancel out and you can rewrite the odds ratio in terms of the conditional probabilities of $Y\|X$: $${\rm OR} = \frac{ P(Y = 1\| X = 1) }{P(Y = 0 \| X = 1)} \cdot \frac{ P(Y = 0 \| X = 0) }{ P(Y = 1 \| X = 0)}$$ In logistic regression, you model these probabilities directly: $$\log \left( \frac{ P(Y_i = 1\|X_i) }{ P(Y_i = 0\|X_i) } \right) = \beta_0 + \beta_1 X_i$$ So we can calculate these conditional probabilities directly from the model. The first ratio in the expression for ${\rm OR}$ above is: $$\frac{ P(Y_i = 1\| X_i = 1) }{P(Y_i = 0 \| X_i = 1)} = \frac{ \left( \frac{1}{1 + e^{-(\beta_0+\beta_1)}} \right) } {\left( \frac{e^{-(\beta_0+\beta_1)}}{1 + e^{-(\beta_0+\beta_1)}}\right)} = \frac{1}{e^{-(\beta_0+\beta_1)}} = e^{(\beta_0+\beta_1)}$$ and the second is: $$\frac{ P(Y_i = 0\| X_i = 0) }{P(Y_i = 1 \| X_i = 0)} = \frac{ \left( \frac{e^{-\beta_0}}{1 + e^{-\beta_0}} \right) } { \left( \frac{1}{1 + e^{-\beta_0}} \right) } = e^{-\beta_0}$$ plugging this back into the formula, we have ${\rm OR} = e^{(\beta_0+\beta_1)} \cdot e^{-\beta_0} = e^{\beta_1}$, which is the result. Note: When you have other predictors, call them $Z_1, ..., Z_p$, in the model, the exponentiated regression coefficient (using a similar derivation) is actually $$\frac{ P(Y = 1\| X = 1, Z_1, ..., Z_p) }{P(Y = 0 \| X = 1, Z_1, ..., Z_p)} \cdot \frac{ P(Y = 0 \| X = 0, Z_1, ..., Z_p) }{ P(Y = 1 \| X = 0, Z_1, ..., Z_p)}$$ so it is the odds ratio conditional on the values of the other predictors in the model and, in general, in not equal to $$\frac{ P(Y = 1\| X = 1) }{P(Y = 0 \| X = 1)} \cdot \frac{ P(Y = 0 \| X = 0) }{ P(Y = 1 \| X = 0)}$$ So, it is no surprise that you're observing a discrepancy between the exponentiated coefficient and the observed odds ratio. Note 2: I derived a relationship between the true $\beta$ and the true odds ratio but note that the same relationship holds for the sample quantities since the fitted logistic regression with a single binary predictor will exactly reproduce the entries of a two-by-two table. That is, the fitted means exactly match the sample means, as with any GLM. So, all of the logic used above applies with the true values replaced by sample quantities. - 2 Wow, thanks for taking the time to write out such a complete explanation. – mike Aug 24 '12 at 16:35 You have a really nice answer from @Macro (+1), who has pointed out that the simple (marginal) odds ratio calculated without reference to a model and the odds ratio taken from a multiple logistic regression model ($\exp(\beta)$) are in general not equal. I wonder if I can still contribute a little bit of related information here, in particular explaining when they will and will not be equal. Beta values in logistic regression, like in OLS regression, specify the ceteris paribus change in the parameter governing the response distribution associated with a 1-unit change in the covariate. (For logistic regression, this is a change in the logit of the probability of 'success', whereas for OLS regression it is the mean, $\mu$.) That is, it is the change all else being equal. Exponentiated betas are similarly ceteris paribus odds ratios. Thus, the first issue is to be sure that it is possible for this to be meaningful. Specifically, the covariate in question should not exist in other terms (e.g., in an interaction, or a polynomial term) elsewhere in the model. (Note that here I am referring to terms that are included in your model, but there are also problems if the true relationship varies across levels of another covariate but an interaction term was not included, for example.) Once we've established that it's meaningful to calculate an odds ratio by exponentiating a beta from a logistic regression model, we can ask the questions of when will the model-based and marginal odds ratios differ, and which should you prefer when they do? The reason that these ORs will differ is because the other covariates included in your model are not orthogonal to the one in question. For example, you can check by running a simple correlation between your covariates (it doesn't matter what the p-values are, or if your covariates are $0/1$ instead of continuous, the point is simply that $r\ne0$). On the other hand, when all of your other covariates are orthogonal to the one in question, $\exp(\beta)$ will equal the marginal OR. If the marginal OR and the model-based OR differ, you should use / interpret the model-based version. The reason is that the marginal OR does not account for the confounding amongst your covariates, whereas the model does. This phenomenon is related to Simpson's Paradox, which you may want to read about (SEP also has a good entry, there is a discussion on CV here: Basic-simpson's-paradox, and you can search on CV's simpsons-paradox tag). For the sake of simplicity and practicality, you may want to just only use the model based OR, since it will be either clearly preferable or the same. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 8, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9184677004814148, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/109688/examples-of-non-constructive-results/109700	# Examples of non-constructive results I'm giving a talk on constructive mathematics, and I'd like some snappy examples of weird things that happen in non-constructive math. For example, it would be great if there were some theorem claiming $\neg \forall x. \neg P(x)$, where no $x$ satisfying $P(x)$ were known or knowable. But other examples are welcome as well. - A standard example is the Intermediate Value Theorem, with $P(x)$ being "$f(x)=0$". But that one is not "weird"... Banach-Tarski is a pretty weird example. – Arturo Magidin Feb 15 '12 at 18:21 @ArturoMagidin, the IVT has a constructive version based on nested intervals. – lhf Feb 15 '12 at 18:22 – lhf Feb 15 '12 at 18:23 @lhf: The IVT ultimately relies on the supremum property; the proofs of which that I can think of right now are all by contradiction and non-constructive – Arturo Magidin Feb 15 '12 at 18:24 1 – lhf Feb 15 '12 at 18:25 show 1 more comment ## 6 Answers At a fun level, there is the two-player game of Chomp. Briefly, you have an $m\times n$ chocolate bar, divided into squares as usual. The lower left-hand little square is poisoned. The two players, A and B, play alternately. At any move, a player picks the lower left-hand corner of a square, and eats all squares above and/or to the right of that corner. The objective is not to eat the poisoned square. One can prove quite simply that A has a winning strategy for any chocolate bar except the $1\times1$. But the proof is indirect. It is clear that for any specific bar, one of the two players has a winning strategy. One then shows that if B had a winning strategy, then A could adapt that strategy and win, by taking the square in the upper right-hand corner. However, for even modest-sized chocolate bars, say $19\times 19$, no winning strategy for A is known. I may be out of date on the $19$, but know that computer searches for strategies have not had great success. - 2 More generally, every two-player game of perfect information in which every play is finite and there are no ties has a winning strategy for one player or the other, by the Gale/Stewart theorem. E.g. if we declare that white wins any draw, then chess has this property and we have no idea which player has the winning strategy. – Carl Mummert Feb 16 '12 at 0:46 I think you mean that no efficient winning strategy for A is known? – Dan Brumleve Feb 16 '12 at 4:57 @Dan Brumleve: Yes, I was using the "no known" perhaps too casually, for any board one can try everything. – André Nicolas Feb 16 '12 at 5:04 1 Actually today one student told me a strategy that works for square bars (like $19\times 19$): Let's say the squares are numbered, the poisoned being (1,1). A picks $(2,2)$ as his first move. Now at each move B can only pick $(1,k)$ or $(k,1)$, and A just responds by reflecting the move, picking $(k,1)$ or $(1,k)$. It's easy to see that B is forced to take the last square. On a non-square board this strategy would fail - if A picked $(2,2)$, B would just pick the necessary squares to make the board diagonally symmetric and subsequently win the game. – Petr Pudlák Oct 29 '12 at 18:21 The existence of a Hamel Basis, that is, a basis for $\mathbb R$ as a vector space over $\mathbb Q$. No one knows a Hamel basis; it's probably unknowable in some sense. The existence of a basis for every vector space is equivalent to the axiom of choice, which is the non-constructive piece of math by excellence. - 2 There are plenty of constructive systems, such as Martin-Lof type theory, which include the axiom of choice. It is only nonconstructive when combined with various other principles of classical mathematics, such as the law of the excluded middle. – Carl Mummert Feb 15 '12 at 22:43 If $\Phi$ is any statement, the following is a consequence of the law of the excluded middie: $$(\exists n \in \mathbb{N})[(n = 0 \land \Phi) \lor (n \not = 0 \land \lnot \Phi)]$$ It will only be provable constructively if either $\Phi$ or $\lnot \Phi$ is provable constructively, because to prove it constructively you would have to produce an actual value of $n$, which means you would have to decide $\Phi$. - At least some time ago (I'm not sure if this has been cleared up recently), it was not known which of the quantities $\sqrt 2^\sqrt 2$ and $(\sqrt 2^\sqrt 2)^\sqrt 2$ furnishes an example of an irrational number raised to an irrational power that is rational. - 3 – Steven Stadnicki Feb 15 '12 at 19:43 Brouwer's fixed point theorem in 2 dimensions is equivalent the fact that the game of Hex has a winning strategy but no one knows what that strategy is. - It is equivalent to at least one player winning eventually, it has nothing to do with strategy. But we do know that the first player can always win in Hex. – Michael Greinecker Feb 16 '12 at 1:31 @MichaelGreinecker Perhaps Wikipedia is wrong on this one then, but it says "John Nash proved in 1952 that a game of Hex cannot end in a tie, and that for a symmetric board there exists a winning strategy for the player who makes the first move (by the strategy-stealing argument). However, the argument is non-constructive : it only shows the existence of a winning strategy, without describing it explicitly. Finding an explicit strategy has been the main subject of research since then." – Ragib Zaman Feb 16 '12 at 3:45 That is true, but the relation between Hex and the (two-dimensional) Brouwer fixed point theorem is due to David Gale. Nash's "stealing strategy"-argument is however a nice example of a nonconstructive proof. – Michael Greinecker Feb 16 '12 at 9:22 @MichaelGreinecker I understand now, I was grouping two distinct results together. Thanks. – Ragib Zaman Feb 16 '12 at 11:55 It has been shown that almost all real numbers are normal in all bases (ref?), but I don't think that anyone has ever exhibited such a number. - 1 @CarlMummert: You may be thinking of a different result; this one is pretty easy to prove. My favorite proof is probabilistic. Fix a base $b$. Since the base-$b$ digits of a $U(0,1)$ random variable $U$ are iid uniform on $\{0, 1, \dots, b-1\}$, the strong law of large numbers gives that $U$ is normal in base $b$, almost surely. Taking a countable intersection, $U$ is normal in every base, almost surely. Or in other words, Lebesgue-almost-every $x \in [0,1]$ is normal in every base. – Nate Eldredge Feb 15 '12 at 23:55 1 @Nate Eldredge: the question I had asked about in a deleted comment was about the Champernowne number .123456789101112... which is known to be normal in base 10, but AFAIK not known to be normal in other bases. – Carl Mummert Feb 16 '12 at 0:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9597520232200623, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/15183/where-did-the-idea-of-a-neutrino-come-from	# Where did the idea of a neutrino come from? Exactly what paper by W. Pauli introduced the idea for the existence of a neutrino and how was its existence confirmed experimentally (who did that and in what paper)? - ## 2 Answers Pauli didn't publish the idea--- it seems to be in a private correspondence with Meitner, according to Wikipedia. Enrico Fermi named the particle and published the four-fermion interaction theory of beta decay in 1934, in this article: E. Fermi "Versuch einer theorie der $\beta$-strahlen" I.Z. Physik 88,161 (1934) I found a quick review of the history here:http://cdsweb.cern.ch/record/779702/files/CERN-ARCH-PBC-328-5.pdf (there are some confusing editing bloopers in the layout). The detection of the neutrino is well documented on Wikipedia, because it won the detecting team the 1995 Nobel, and I don't have anything to add to that. The documentation on the earlier history is more scant. The prehistory of this affair is Bohr's reluctance to accept energy conservation as fundamental. He thought of energy conservation as a thermodynamic statistical law in the 1920s, and was willing to throw it away in order to get quantum theory to make sense. One of the paradoxes of early quantum theory is that you can have a spread out quantum of light, which passes over many atoms. Then the photon must be absorbed by one and only one atom, by conservation of energy. But locality seems to imply that if an atom over here can absorb the photon, and an atom over there can absorb the photon, then if they cannot signal, they can both absorb the same photon. I am not sure that Bohr would have said it this way in 1924, I don't think he believed in photons yet. He would have talked about a classical electromagnetic field washing over many atoms, and the response going at the speed of light, but the absorption being quantum, so the energy cannot be conserved. In modern quantum mechanics, this is not true, because you have a global wavefunction which means that the different atoms absorb the photon in different Everett branches, or, if you prefer, the absorption of the photon collapses the wavefunction nonlocally to prevent the absorption somewhere else. But in 1924, all this was two years in the future, and Kramers Bohr and Slater proposed that energy is just not conserved, and that atomic transitions are governed by something similar to a semiclassical electromagnetic field interacting with a quantum atom. This description only gives energy conservation on average, because the same electromagnetic field can dump energy into far away atoms at once, and in the semi-classical electromagnetic field approximation, you ignore the quantum structure of the field, so if the field represents a single photon, that single photon can be absorbed by different atoms. This is now interpreted as saying that single photons are not well described by a semiclassical field, but that wasn't BKS's interpretation. Heisenberg's 1925 quantum theory put an end to BKS, and restored energy conservation. This was a major motivation for Heisenberg--- he knew he was on the right track when he realized that the interpreting energy as the Hamiltonian matrix makes energy conserved. Bohr was very unhappy with the fate of his theory, which really was the first coherent attempt at a description of photons and atoms, and deserved more recognition. It was attacked by Einstein, and it died even before Matrix mechanics. In Feynman's 1950's book on quantum electrodynamics, he shows how far you can take the semi-classical field idea, that you can reproduce all of quantum electrodynamics just from semi-classical fields interacting with particles, so long as at the end you understand that the semiclassical calculations are only guidelines for getting the proper Feynman rules for single photons. I think this type of presentation is descended from BKS (although it is quantum mechanically correct of course, and it is ab-initio, as was Feynman's way). The detection and confirmation that the electrons in nuclear beta decay can come out with any energy less than the maximum possible, by Meitner and collborators, led Bohr to revive the earlier discredited statistical conservation of energy. This idea was now much less motivated than before, and Pauli decided that there is just something carrying away the extra energy. Pauli had a publication aversion, for some reason, and tended to write things in letters. This is a great pity, because his published papers are so good. I couldn't find the neutrino letter online, and I don't read German anyway. The Fermi paper is probably the best source for the early ideas, but this theory is replaced by the V-A theory in the 1950s. The paper I linked to suggests that Pauli didn't want to publish a four-Fermi theory because he knew that this theory would have terrible perturbation theory infinities. He does write that Fermi's theory leads to terrible infinities at higher order of perturbations, but this is a property shared by all quantum field theories in a non-covariant formalism. Now we know that Fermi type theories are nonrenormalizable in 4d, but if you attribute this view to the 1930's Pauli, I think it is an anachronism. Even quantum electrodynamics is infinite at one loop, after all, and the order of the divergence was only established to be logarithmic by Weisskopf in the 1940s. In the 1930s, people weren't sure exactly how it blew up, but probably they expected a power law, like the self-energy in classical electrodynamics. The divergence in quantum electrodynamics is only softened to a log because of the contributions of positrons, and it is important to have a relativistically invariant formulation to know the exact divergence order, because you don't get the right answer if you separate positron and electron contributions. - According to the German Wikipedia, the neutrino was first suggested by Pauli in a private letter on December 4th 1930. The Letter is available in the CERN archives, although, unfortunately, it is in German. A link to the PDF is here. The first paragraph basically mentions that in order to explain the continuous spectrum of beta decay, Pauli suggests the existence of a neutral spin 1/2 particle obeying the exclusion principle inside the nucleus. He calls the particle "neutron" and according to him, it should roughly have the same mass as the electron, but certainly not higher than 1% of the proton mass. Later he mentions that this may seem like a cheap trick to explain the spectrum, but in such desperate times one should examine all possible solutions to a problem. Wikipedia does not give any source as to wether this really is the first suggestion of the existence of the neutrino. Again, according to Wikipedia, in 1956 Cowan and Reines confirmed the existence of a particle with the right properties. They published their result in 'Science' and the paper is available for free here. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9589895009994507, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/81653/simple-but-potentially-fishy-ode	# Simple but potentially fishy ODE? Is the solution of ${1\over x}(y''-y)=0$ simply $y(x)=a\cosh{x}+b\sinh{x}$? Or is there something fishy to do with the $1\over x$? What are the solutions s.t. ${1\over x}y(x)$ is bounded as $x\to 0$? I am guessing $y(x)=b\sinh{x}$? And the ones bounded as $x\to \infty$? Thanks. - "Oops, error" is not very clear. If there's an error, please remove it or mark it clearly. – joriki Nov 13 '11 at 17:36 @joriki: the "oops error" bit is just to bring attention to that (following) bit where I have made changes :-). This version of the question should be error-free. – magna Nov 13 '11 at 17:49 @joriki: I have deleted the misleading "oops, error" :-) – magna Nov 13 '11 at 18:06 ## 1 Answer Strictly speaking, the differential equation isn't defined at $x=0$. Thus the maximal domain on which you can consider this is $\mathbb R\setminus\{0\}$, and there's nothing to constrain the solutions on both sides of $0$ to have anything to do with each other; thus you can choose their parameters $a$ and $b$ independently of each other. You can also use $c_+\mathrm e^x+c_-\mathrm e^{-x}$ if you like, by the way. Your guess $y(x)=b\sinh x$ is correct, except again you can choose different parameters $b$ to the left and right of $0$. For $x\to\infty$, the exponential representation above is more conducive for seeing the answer immediately. - Thanks, @joriki ! – magna Nov 13 '11 at 20:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9298714399337769, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/196494-prove-space-geometry-question.html	# Thread: 1. ## Proof for a space geometry question Show that there exist one point equidistant from four non complainer distinct points (maybe it can be solved using axis of a circle ) 2. ## Re: Proof for a space geometry question Originally Posted by lebanon Show that there exist one point equidistant from four non complainer distinct points (maybe it can be solved using axis of a circle ) 1. Four non-complanar points are located on a sphere. Thus you are looking for it's center. 2. If the center of the sphere is $C(c_1, c_2, c_3)$ then the equation of the sphere is: $(x-c_1)^2+(y-c_2)^2+(z-c_3)^2=r^2$ Plug in the coordinates of the four points. You'll get a system of 4 equations which can be easily reduced to a system of 3 linear equations. Solve for $(c_1, c_2, c_3)$.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9125584363937378, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/110942-cantor-set-question.html	Thread: 1. Cantor set question Let $C$ represent the Cantor set. Q: Show that there exists $x_{1},y_{1}\in{C}$ satisfying $x_{1}+y_{1}=s\in[0,2]$. I found some work that goes like this... $C_{1}=[0,1/3]\cup[2/3,1]$. Then we have, $[0,1/3]+[0,1/3]=[0,1/3]\cup[0,2/3]+[2/3,1]=[2/3,4/3]\cup[2/3,1]<br /> +[2/3,1]$. So, $C_{1}+C_{1}=[0, 2/3]\cup[2/3, 4/3]\cup[4/3, 2]=[0, 2]$. Thus, for all $s\in[0, 2]$, we can find an $x_{1} , y_{1}$ $C_{1}$ satisfying $x_{1} + y_{1} = s$ I am not sure that I follow exactly what is being done. Can someone help me to understand the work above. Do I think of these intervals as only lengths? thanks 2. Originally Posted by Danneedshelp Let $C$ represent the Cantor set. Q: Show that there exists $x_{1},y_{1}\in{C}$ satisfying $x_{1}+y_{1}=s\in[0,2]$. I found some work that goes like this... $C_{1}=[0,1/3]\cup[2/3,1]$. Then we have, $\color{blue}[0,1/3]+[0,1/3]=[0,1/3]\cup[0,2/3]+[2/3,1]=[2/3,4/3]\cup[2/3,1]<br /> +[2/3,1]$. So, $C_{1}+C_{1}=[0, 2/3]\cup[2/3, 4/3]\cup[4/3, 2]=[0, 2]$. Thus, for all $s\in[0, 2]$, we can find an $x_{1} , y_{1}$ $C_{1}$ satisfying $x_{1} + y_{1} = s$ I am not sure that I follow exactly what is being done. Can someone help me to understand the work above. Do I think of these intervals as only lengths? $C_1+C_1$ means the set of all numbers that are equal to an element of $C_1$ plus an element of $C_1$. But $C_1 = [0,1/3]\cup[2/3,1]$. By adding two numbers from the interval [0,1/3] you can get anything in the interval [0,2/3]. By adding a number from the interval [0,1/3] and a number from the interval [2/3,1] you can get anything in the interval [2/3,4/3]. By adding two numbers from the interval [2/3,1] you can get anything in the interval [4/3,2]. Putting those facts together, you see that by adding two numbers from the set $C_1$ you can get anything in the interval [0,2]. That is what is meant by the line $C_{1}+C_{1}=[0, 2/3]\cup[2/3, 4/3]\cup[4/3, 2]=[0, 2]$. (The previous line, which I have highlighted in blue, seems totally confused.) Therefore, for any $s\in[0,2]$, we can find $x_1,\ y_1 \in C_1$ with $x_1+y_1=s$. That is just the first stage of this proof. You need to go on to show that $C_n+C_n = [0,2]$ for each of the sets $C_n$ used in the construction of the Cantor set. So there exist $x_n,\ y_n \in C_n$ with $x_n+y_n=s$. Finally, you need to show that a subsequence of the sequence $(x_n)$ converges to a point x in the Cantor set. The corresponding subsequence of the $y_n$s will then converge to a point y in the Cantor set, with x+y=s.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 35, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": false, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9466156959533691, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/117453/list	## Return to Answer 2 deleted 14 characters in body If your filter is generated by $\kappa$ many sets, then indeed the conclusion you seek can be made, by a direct argument that does not go through strong compactness. Theorem. The following are equivalent, for any uncountable regular cardinal $\kappa$. 1. $\kappa$ is a measurable cardinal. 2. Every $\kappa$ complete filter $F$, generated by at most $\kappa$-many sets, extends to a $\kappa$-complete ultrafilteron that set. Proof: It is easy to see that $2$ implies $1$, since the filter of co-bounded sets in $\kappa$ is $\kappa$-complete and generated by the tails, so there is a $\kappa$-complete non-principal ultrafilter on $\kappa$. For the main direction, assume $\kappa$ is measurable and $F$ is a $\kappa$-complete filter on a set $D$ with $F$ generated by at most $\kappa$ many sets $X_\alpha$, for $\alpha\lt\kappa$. Let $j:V\to M$ be an elementary embedding with critical point $\kappa$. By applying $j$ to $\vec X=\langle X_\alpha\lt\kappa\rangle$ and restricting to $\kappa$, we see that $\langle j(X_\alpha)\mid\alpha\lt\kappa\rangle$ is in $M$. And since this is fewer than $j(\kappa)$ many elements of $j(F)$, which is $j(\kappa)$-complete in $M$, it follows that $\bigcap_{\alpha\lt\kappa}j(X_\alpha)\in j(F)$, and in particular, there is some $a\in \bigcap_\alpha j(X_\alpha)$. Define $U=\{X\subset D\mid a\in j(X)\}$. It is easy to verify that $U$ is a $\kappa$-complete ultrafilter on $D$ and $F\subset U$, as desired. QED For $\theta$-generated filters, one generally needs $\theta$-strong compactness, as mentioned in the comments, and this is in fact equivalent to $\theta$-strong compactness. The essence of the argument above, then, is that a cardinal $\kappa$ is measurable if and only if it is $\kappa$-strongly compact. That said, if you want this filter extension property, then I encourage you to go ahead and make the strong compactness assumption. There are many beautiful theorems using strongly compact cardinals. 1 If your filter is generated by $\kappa$ many sets, then indeed the conclusion you seek can be made, by a direct argument that does not go through strong compactness. Theorem. The following are equivalent, for any uncountable regular cardinal $\kappa$. 1. $\kappa$ is a measurable cardinal. 2. Every $\kappa$ complete filter $F$, generated by at most $\kappa$-many sets, extends to a $\kappa$-complete ultrafilter on that set. Proof: It is easy to see that $2$ implies $1$, since the filter of co-bounded sets in $\kappa$ is $\kappa$-complete and generated by the tails, so there is a $\kappa$-complete non-principal ultrafilter on $\kappa$. For the main direction, assume $\kappa$ is measurable and $F$ is a $\kappa$-complete filter on a set $D$ with $F$ generated by at most $\kappa$ many sets $X_\alpha$, for $\alpha\lt\kappa$. Let $j:V\to M$ be an elementary embedding with critical point $\kappa$. By applying $j$ to $\vec X=\langle X_\alpha\lt\kappa\rangle$ and restricting to $\kappa$, we see that $\langle j(X_\alpha)\mid\alpha\lt\kappa\rangle$ is in $M$. And since this is fewer than $j(\kappa)$ many elements of $j(F)$, which is $j(\kappa)$-complete in $M$, it follows that $\bigcap_{\alpha\lt\kappa}j(X_\alpha)\in j(F)$, and in particular, there is some $a\in \bigcap_\alpha j(X_\alpha)$. Define $U=\{X\subset D\mid a\in j(X)\}$. It is easy to verify that $U$ is a $\kappa$-complete ultrafilter on $D$ and $F\subset U$, as desired. QED For $\theta$-generated filters, one generally needs $\theta$-strong compactness, as mentioned in the comments, and this is in fact equivalent to $\theta$-strong compactness. The essence of the argument above, then, is that a cardinal $\kappa$ is measurable if and only if it is $\kappa$-strongly compact. That said, if you want this filter extension property, then I encourage you to go ahead and make the strong compactness assumption. There are many beautiful theorems using strongly compact cardinals.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 88, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9540143609046936, "perplexity_flag": "head"}
http://mathoverflow.net/questions/117759/streamlined-probability-measure-for-tossing-infinitely-many-coins	## Streamlined probability measure for tossing infinitely many coins ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The standard probability measure over countably many independent coin tosses (i.e., the probability that you get a prescribed prefix of length $v$ is $2^{-v}$) is usually obtained via results in measure theory (at least, that's what I have seen). Is there a streamlined presentation out there that uses the least possible amount of results from measure theory (ideally, none) to show that this is indeed a valid probability measure? - What does it mean to prove a statement whose very formulation requires measure theory without using any measure theory? – Michael Greinecker Jan 5 at 23:53 ## 1 Answer This version is due to Emile Borel ... Sequence of heads & tails encoded as 0s and 1s, then sequence is taken to represent a number in $[0,1]$ in its binary expansion. The measure is Lebesgue measure. So you still need to know that Lebesgue measure exists. - More in detail: let $D\subset 2^{\mathbb {N} _ +}$ be the countable set of eventually zero sequences $\{x_k \}_ {k\ge 1}$, which correspond to dyadic rationals of $[0,1]$, i.e. numbers with double binary representation. Then $2^{\mathbb {N} _ +}\setminus D\ni x\mapsto \sum_k\ 2^{-k}x_k \in (0,1]$ is a bi-measurable bijection. – Pietro Majer Jan 1 at 12:29 Thanks, but I was hoping for something more basic. Real numbers are even more difficult than infinite coin tosses, in my opinion. How would you explain this probability measure to a high-school student? – Emanuele Viola Jan 2 at 14:01	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9500386714935303, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/3009/how-to-fit-probability-density-function-from-sample-moments	# How to fit probability density function from sample moments? If I have calculated the sample mean, variance, skew and kurtosis of a set of data, how would I go about fitting a probability distribution to match these moments (i.e. choosing a probability distribution and optimizing its parameters to fit the sample moments). Are there any packages in R/MATLAB/etc. that are capable of this? For context, I believe I can calculate these moments for a portfolio's return distribution, but I actually need a whole probability density function for the portfolio's returns in order to perform additional analysis. - ## 5 Answers I asked an almost identical question on the Cross Validated site here. I think my accepted answer, given by Whuber, might be what you are looking for. - You should be able to do this with the `fitdistr` function in the MASS package. You will certainly be able to hold the mean and variance constant, but I'm less sure about skewness and kurtosis (they would need to be arguments to the density function). The actuar package may also be useful, as it contains additional density functions. - If you have a formula giving you the moments as a function of the parameters of the distribution, you can use `gmm`, in the `gmm` package: there is a detailed example for the Gaussian distribution in the documentation. (Of course, in this case, you are only solving a system of equations, and could probably do it by hand: the generalized method of moments (GMM) is typically used when there are more moments than parameters.) - Often, we are interested to check if our data is close to normal, then you can use the Jarque-Bera test, where skewness and kurthosis are directly deployed. Look up Matlab implementation. If you are not constrained to use the moments, you can calculate histogram and use Komogorov-Smirnov test, which can test similarity to any distribution, not necessarily normal. Look up Matlab implementation. - It may never have been done, but it seems to me that the Cornish-Fisher expansion for moments up to 4 could be used to trace a density function from a tabulation of the percentiles of the normal distribution, for instance: $$w_{\alpha}\cong z_{\alpha}+\frac{1}{6}\left(z_{\alpha}^{2}-1\right)AS+\frac{1}{24}\left(z_{\alpha}^{3}-3z_{\alpha}\right)EKUR-\frac{1}{36}\left(2z_{\alpha}^{3}-5z_{\alpha}\right)AS^{2}$$ where $w_{\alpha}$ is the revised percentile, $z_{\alpha}$ is the percentile for the standard normal distribution, AS is the skewness coefficient and EKUR is the excess kurtosis coefficient (from a kurtosis coefficient of 3 for a normal distribution). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9190283417701721, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/128046-prove-every-element-g-has-finite-order-print.html	# Prove that every element of G has finite order Printable View • February 9th 2010, 02:49 PM crushingyen Prove that every element of G has finite order If $N$ is a normal subgroup of $G$ and if every element of $N$ and of $G/N$ has finite order, prove that every element of $G$ has finite order. I started off doing this: Let $g \in G$. By showing that $g^n = e$, $n \in \mathbb{Z}$, $\|G\| = n <\infty$. $(Ng)^k = Ne$ for some $k \in \mathbb{Z}$ $\Rightarrow (eg)^k \in (Ng)^k = Ne = N$ $\Rightarrow (eg)^k \in N$ $\Rightarrow (e^k)(g^k) \in N$ $\Rightarrow eg^k \in N$ $\Rightarrow g^k \in N$ $\Rightarrow \exists$ $n \in \mathbb{Z}$ such that $(g^k)^n = e$ I don't know how to finish off this problem after point. Any help would be greatly appreciated. • February 9th 2010, 03:10 PM Drexel28 Quote: Originally Posted by crushingyen If $N$ is a normal subgroup of $G$ and if every element of $N$ and of $G/N$ has finite order, prove that every element of $G$ has finite order. I started off doing this: Let $g \in G$. By showing that $g^n = e$, $n \in \mathbb{Z}$ $(Ng)^k = Ne$ for some $k \in \mathbb{Z}$ $\Rightarrow (eg)^k \in (Ng)^k = Ne = N$ $\Rightarrow (eg)^k \in N$ $\Rightarrow (e^k)(g^k) \in N$ $\Rightarrow eg^k \in N$ $\Rightarrow g^k \in N$ $\Rightarrow \exists$ $n \in \mathbb{Z}$ such that $(g^k)^n = e$ I don't know how to finish off this problem after point. Any help would be greatly appreciated. I think this is what you did... Lemma: Let $N\lhd G$ and $m=\left[G:N\right]$, then $g^m\in N$ for every $g\in G$ Proof: Since $N\lhd G$ we know that $G/N$ is a group of order $m$ and so given any element $K$ in $G/N$ we have that, by basic group theory, that $K^m=e=N$. In particular, $\left(Ng\right)^m=Ng^m=N$ from where it follows that $g^m\in N$. $\blacksquare$. Is that what you wre getting at?? • February 9th 2010, 04:25 PM crushingyen Sorry, I edited the my post up above. Pretty much, I'm trying to show that the group G is finite, not the lemma. • February 9th 2010, 07:21 PM crushingyen Never mind, I found out how to do it. Since $g^k$ has finite order, $\|g^k\| = n$ for some $n \in \mathbb{Z}$ $\Rightarrow (g^k)^n = e$ $\Rightarrow g^{kn} = e$ $\Rightarrow \|g\| = kn < \infty$ All times are GMT -8. The time now is 05:27 AM.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 56, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9602195620536804, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/180559/why-gradient-descent-works?answertab=votes	# Why gradient descent works? On Wikipedia, this is the following description of gradient descent: Gradient descent is based on the observation that if the multivariable function $F(\mathbf{x})$ is defined and differentiable in a neighborhood of a point $\mathbf{a}$, then $F(\mathbf{x})$ decreases fastest if one goes from $\mathbf{a}$ in the direction of the negative gradient of $F$ at $\mathbf{a}$. Now I have several doubts in this description. First of all I have an example of $f(x)=x^2$ in my mind and my starting point is, say, $x=5$. 1. What is the meaning of "decreases fastest"? I mean, I can go straight from $x=5$ to $x=0$ (which is minimum point), then what's the point of fastest decrease? What is the notion of fast here? 2. Where did this observation come from? I didn't see the proof of this observation. - 1 – J. M. Aug 9 '12 at 8:16 ## 3 Answers In two dimensions, like in the example of $f(x)=x^2$,at the point $(5,25)$ on the curve, there are only two ways to "move" and still be on the curve: to the left and to the right, and in our case, the value of the function decreases as we move to the left, and increases as we go to the right. No problems. But, when we are on, for example the following function $z=f(x,y)$ which is a two variable function, which looks like the inverted dome below. Imagine that you are on top of the dome. Then, There are an infinite number of curves(see figure) you can move on and still stay on the surface. And, since our location is so symmetric, it does not matter which direction we move in- it will always be the same rate of change in all directions. But, suppose we were not on such a symmetric location, say a little lower in any direction. Then, with a little more thought we can see that there is exactly one direction in which you must move in order to achieve maximum descent. That is downward on the curve connecting the point you are on to the top of the dome. As for rigorous proof of this phenomenon,J.M's link in the comments is an excellent resource. - Suggestion: Assuming that the context of the OP is minimization, it might add clarity if the illustration was inverted so that it had a definite minimum. – Tpofofn Aug 25 '12 at 12:41 The concept of a direction of fastest descent only makes sense in more than one dimension. In your one-dimensional example, you can only go left or right, and the function increases in one direction and decreases in the other; there's nothing else to compare. In several dimensions, however, you have a continuum of directions to choose from, and the directional derivative is maximal in the direction of the gradient. - In your example, the gradient at $x= 5$ is $10$, so the negative gradient is $-10$. Since the negative gradient is negative (not neccesarily the case, thinkk about $x= -5$), it decreases fastest in the negative direction. The gradient can be interpreted as the slope of a tangent. In the one-variable case, the tangent is a line and the slope can be represented as a single number. In the multivariable case, let's say with a function like $f(x, y, z) = x^2 + zy^3 - \frac{x}{y^2}$, the gradient will be a vector with corresponding to the direction of fastest ascent. Which means the fastest descent will be the opposite direction: the direction of the negative gradient. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9436973333358765, "perplexity_flag": "head"}
http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.jdg/1180135680	### Higher Noether-Lefschetz loci of elliptic surfaces Remke Kloosterman Source: J. Differential Geom. Volume 76, Number 2 (2007), 293-316. #### Abstract We calculate the dimension of the locus of Jacobian elliptic surfaces over $P\sb 1$} with a given Picard number, in the corresponding moduli space. First Page: We're sorry, but we are unable to provide you with the full text of this article because we are not able to identify you as a subscriber. If you have a personal subscription to this journal, then please login. If you are already logged in, then you may need to update your profile to register your subscription. Read more about accessing full-text	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8397240042686462, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/84397/list	Return to Answer 3 fixed typos If you mean a (real) analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin. All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}$$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin. Added note about complex solutions: I neglected to mention that, over the complex numbers, there are, of course, two solutions satisfying $y(0)=0$. I doubt that there is a closed form expression for them in elementary terms, but one easily obtains that they have convergent power series expansions of the form $$y_{\pm}(x) = {} \pm i\sqrt{2}\ x f_0(x^4) + x^3 f_1(x^4),$$ where $$f_0(t) = 1 - \frac{t}{3\cdot2^6} + \frac{13\ t^2}{5\cdot3^2\cdot2^{13}} - \cdots$$ and $$f_1(t) = \frac{1}{2^4} frac{1}{2^2} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots.$$ 2 Added information about complex solutions If you mean an a (real) analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin. All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}$$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin. Added note about complex solutions: I neglected to mention that, over the complex numbers, there are, of course, two solutions satisfying $y(0)=0$. I doubt that there is a closed form expression for them in elementary terms, but one easily obtains that they have convergent power series expansions of the form $$y_{\pm}(x) = {} \pm i\sqrt{2}\ x f_0(x^4) + x^3 f_1(x^4),$$ where $$f_0(t) = 1 - \frac{t}{3\cdot2^6} + \frac{13\ t^2}{5\cdot3^2\cdot2^{13}} - \cdots$$ and $$f_1(t) = \frac{1}{2^4} + \frac{t}{3\cdot2^9} - \frac{11\ t^2}{5\cdot3^3\cdot2^{15}} + \cdots.$$ 1 If you mean an analytical solution with $y(0)=0$, then the answer is 'no'. If you write this as the problem of looking for integral curves of $\omega = (2x-x^2y)\ dx + y\ dy$ in the $xy$-plane, you'll immediately see that the origin is an isolated singular point of elliptic type (i.e., the eigenvalues of the linearization at $(x,y)=(0,0)$ are pure imaginary). In particular, there is no nontrivial integral curve of $\omega$ that passes through the origin. All of the integral curves of $\omega$ near the origin spiral around it and converge inward as they go counterclockwise around the origin. To see this, just consider the vector field $$Z = (2x - x^2y)\frac{\partial\ }{\partial y} - y\frac{\partial\ }{\partial x}$$ which is tangent to the integral curves of $\omega$, and consider the convex function $H = 2x^2 + y^2$. The derivative of $H$ with respect to $Z$ is $-2x^2y^2$, which is non-positive and vanishes only on the axes, which are not tangent to $Z$. Since $H$ is strictly decreasing along each integral curve of $Z$, the integral curves of $Z$ spiral in to the origin.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 47, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9580082893371582, "perplexity_flag": "head"}
http://mathoverflow.net/questions/73984/classical-invariants-involving-exterior-powers-of-standard-representation/74024	## Classical invariants involving exterior powers of standard representation ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) While investigating certain conformal blocks line bundles on $\overline{M}_{0,n}$, I was led to what seems to be an identification between two spaces of invariants, and I am curious if there is a direct way to see this identification. Statement: for any integers $n\ge 4$ and $r\ge 2$, and any integers `$i_1,\ldots,i_n$` such that `$1 \le i_j \le r-1$` and `$r=\frac{1}{2}\sum_{j=1}^n i_j$`, I believe there is a vector space isomorphism `\[(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes \wedge^{i_n}\mathbb{C}^r)^{SL(r)} \cong (S^{i_1}\mathbb{C}^2\otimes\cdots\otimes S^{i_n}\mathbb{C}^2)^{SL(2)},\]` where $\mathbb{C}^m$ denotes the standard representation of $SL(m)$. The invariants on the RHS are classical and well-known: a basis is given by all $2\times r$ semi-standard tableaux with entries in ${1,\ldots,n}$ such that $j$ occurs exactly `$i_j$` times. I wonder if the invariants on the LHS are also known, and if there's a conceptual reason why they might be in bijection with those on the RHS. Background: This is not relevant for the question itself, but I am including it in case you are curious how this purported identity arose. It seems likely that for conformal blocks bundles on `$\overline{M}_{0,n}$` of level 1 and Lie algebra $\mathfrak{sl}(r)$, the global sections are naturally identified with a space of covariants. Specifically, the conformal blocks line bundle with weights `$(\omega_{i_1},\ldots,\omega_{i_n})$`, where `$\omega_i$` are fundamental weights, should have global sections `$(\wedge^{i_1}\mathbb{C}^r\otimes\cdots\otimes\wedge^{i_n}\mathbb{C}^r)_{\mathfrak{sl}(r)}$`, since the irreducible representation associated to $\omega_i$ is $\wedge^i\mathbb{C}^r$. This space of $\mathfrak{sl}(r)$-covariants is isomorphic to the corresponding space of $\mathfrak{sl}(r)$-invariants, which in turn is the same as the space of $SL(r)$-invariants for this representation. On the other hand, it is known (by a result of Fakhruddin) that when `$\sum_{j=1}^n i_j = 2r$` then this conformal blocks line bundle induces the GIT morphism `$\overline{M}_{0,n} \rightarrow (\mathbb{P}^1)^n//_{(i_1,\ldots,i_n)}SL(2)$`, so we know that its space of global sections is $H^0((\mathbb{P}^1)^n,\mathcal{O}(i_1,\ldots,i_n))^{SL(2)}$. - ## 2 Answers To follow up on Sasha's answer, yes there is a natural isomorphism of vector spaces which lifts the combinatorial equality. All isomorphisms in this answer will be natural. Schur-Weyl duality: Let $\lambda$ be a partion; set $d = \|\lambda\|$ and let $m$ be greater than or equal to the number of parts of $\lambda$. Let $V$ be a vector space of dimension $m$, let $V_{\lambda}$ be the irrep of $GL(V)$ with highest weight $\lambda$ an let $M_{\lambda}$ be the $S_d$-irrep (aka Specht module) indexed by $\lambda$. Schur-Weyl duality is the isomorphism of $GL(V)$ representations: `$$\mathrm{Hom}_{S_d}(M_{\lambda}, V^{\otimes d}) \cong V_{\lambda}.$$` I'll take this to be the definition of $V_{\lambda}$. Tensor product and restriction: Let $\mu$, $\lambda_1$, ..., $\lambda_k$ be partitions, with $\|\mu\| = \sum \|\lambda_i\|$. Let $m$ be greater than or equal to the number of parts of $\mu$. Set $d_i = \|\lambda_i\|$. Then the above shows that ```$$\begin{array}{rl} \mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) &\cong \mathrm{Hom}_{GL(V)} \left( \mathrm{Hom}_{S_{\sum d_i}} \left( M_{\mu}, V^{\otimes \sum d_i} \right), \mathrm{Hom}_{S_{d_1} \times \cdots S_{d_k}}\left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, V^{\otimes \sum d_i} \right) \right) \\ & \cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)\|_{S_{d_1} \times \cdots \times S_{d_k}} \right) \end{array}.$$``` In other words, every Hom in the second expression is induced from composition with one of the Hom's in the third expression. Here, if $U$ and $V$ are $G$ and $H$-representations, then $U \boxtimes V$ denotes $U \otimes V$ with $G$ and $H$ acting on the first and second factors respectively. Relation to transpose: Now let $\lambda_i$ and $\mu$ be as above. Let $\epsilon(d)$ be the sign representation of $S_d$, and let $\lambda^T$ be the transpose of $\lambda$. Then $M_{\lambda} \cong M_{\lambda^T} \otimes \epsilon(\|\lambda\|)$. How to make this natural depends on exactly how you define $M_{\lambda}$. For example, if you use the Vershik-Okounkov approach, the bases they construct for $M_{\lambda}$ and $M_{\lambda^T}$ correspond to each other. Now, $\epsilon(d_1) \boxtimes \cdots \boxtimes \epsilon(d_k) \cong \epsilon(\sum d_i)\|_{S_{d_1} \times \cdots \times S_{d_k}}$, and is irreducible. So we deduce that $$\mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1} \boxtimes \cdots \boxtimes M_{\lambda_k}, \left( M_{\mu} \right)\|_{S_{d_1} \times \cdots \times S_{d_k}} \right) \cong \mathrm{Hom}_{S_{d_1} \times \cdots \times S_{d_k}} \left( M_{\lambda_1}^T \boxtimes \cdots \boxtimes M_{\lambda_k}^T, \left( M_{\mu}^T \right)\|_{S_{d_1} \times \cdots \times S_{d_k}} \right)$$ and this isomorphism is natural if we define $M_{\lambda} \cong M_{\lambda^T} \otimes \epsilon(\|\lambda\|)$ in a natural way. Putting it all together, if $\dim V$ is greater than or equal to the number of parts in $\mu$, and $\dim W$ is greater than or equal to the number of parts in $\mu^T$, then we have a natural isomorphism: `$$\mathrm{Hom}_{GL(V)}\left( V_{\mu}, V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k} \right) \cong \mathrm{Hom}_{GL(W)}\left( W_{\mu^T}, W_{\lambda^T_1} \otimes \cdots \otimes W_{\lambda^T_k} \right).$$` Take $\lambda_j = 1^{i_j}$ and $\mu = 2^r$. Let $\dim V = r$ and $\dim W=2$. Since $M_{\lambda_j}$ is the sign rep, we have $V_{\lambda_j} \cong \bigwedge^{i_j} V$ naturally. Similarly, since $M_{\lambda_j^T}$ is the trivial rep, we have $W_{\lambda_j^T} \cong \mathrm{Sym}^{i_j}(W)$. Finally, one needs to use the fact that $V_{\mu}$ and $W_{\mu^T}$ are trivial one-dimensional $SL(V)$ and $SL(W)$ reps. (More precisely, they are $\det^2$ and $\det^r$, as $GL$-reps.) I'm not sure what the easiest proof of this is, but it can only introduce a scalar factor to your isomorphisms. - @David: great post. Do you see a way of doing this by hand? One can use the FFT of invariant theory to describe a linearly generating set for the SL(V) side as follows: take 2r column vectors v_1,...,v_2r in C^r. For a partition of 1,...,2r in two sets of r indices take the product of the two r by r determinants corresponding to this partition. Then antisymmetrize the blocks {1,...,i1} etc. A subset corresponds to the 2 by r SSYT. The rows give the selection of indices for each of the two determinants. Can one prove by hand using Grassmann-Plucker that the subset is a basis. I wonder also... – Abdelmalek Abdesselam Sep 1 2011 at 20:00 ...if one can make a direct relation to the binary domain (the SL2 side) by specializing to v's of the form 1,x,...,x^{r-1}. – Abdelmalek Abdesselam Sep 1 2011 at 20:01 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The dimension of the spaces of invariants are given by the number of semistandard Young tableaux of a specific form. If one applies the Littlewood-Richardson rule for tensoring wedge powers, one sees that for the first space one needs to count the tableaux of the rectangle form with $r$ rows and $2$ columns which a filled with $i_1$ of $1$'s, $i_2$ of $2$'s, ..., $i_n$ of $n$'s such that the numbers increase strictly in rows and non-strictly in columns. And if one applies the Littlewood-Richardson rule for tensoring symmetric products, one sees that For the second space one has to count the tableaux of the rectangle form with $2$ rows and $r$ columns which a filled with $i_1$ of $1$'s, $i_2$ of $2$'s, ..., $i_n$ of $n$'s such that the numbers increase strictly in columns and non-strictly in rows. Now you can see that the transposition turns the tableaux of the first type into the tableaux of the second type. Which give the equality you want. - Quite clean and simple, excellent! – Noah Giansiracusa Aug 30 2011 at 8:02	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 91, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9179966449737549, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/04/14/bilinear-forms/?like=1&source=post_flair&_wpnonce=8cb43e0c56	The Unapologetic Mathematician Bilinear Forms Now that we’ve said a lot about individual operators on vector spaces, I want to go back and consider some other sorts of structures we can put on the space itself. Foremost among these is the idea of a bilinear form. This is really nothing but a bilinear function to the base field: $B:V\times V\rightarrow\mathbb{F}$. Of course, this means that it’s equivalent to a linear function from the tensor square: $B:V\otimes V\rightarrow\mathbb{F}$. Instead of writing this as a function, we will often use a slightly different notation. We write a bracket $B(v,w)=\langle v,w\rangle$, or sometimes $\langle v,w\rangle_B$, if we need to specify which of multiple different inner products under consideration. Another viewpoint comes from recognizing that we’ve got a duality for vector spaces. This lets us rewrite our bilinear form $B:V\otimes V\rightarrow\mathbb{F}$ as a linear transformation $B_1:V\rightarrow V^$. We can view this as saying that once we pick one of the vectors $x\in V$, the bilinear form reduces to a linear functional $\langle v,\underbar{\hphantom{X}}\rangle:V\rightarrow\mathbb{F}$, which is a vector in the dual space $V^$. Or we could focus on the other slot and define $B_2(v)=\langle\underbar{\hphantom{X}},v\rangle\in V^$. We know that the dual space of a finite-dimensional vector space has the same dimension as the space itself, which raises the possibility that $B_1$ or $B_2$ is an isomorphism from $V$ to $V^$. If either one is, then both are, and we say that the bilinear form $B$ is nondegenerate. We can also note that there is a symmetry on the category of vector spaces. That is, we have a linear transformation $\tau_{V,V}:V\otimes V\rightarrow V\otimes V$ defined by $\tau_{V,V}(v\otimes w)=w\otimes v$. This makes it natural to ask what effect this has on our form. Two obvious possibilities are that $\tau_{V,V}\circ B=B$ and that $\tau_{V,V}\circ B=-B$. In the first case we’ll call the bilinear form “symmetric”, and in the second we’ll call it “antisymmetric”. In terms of the maps $B_1$ and $B_2$, we see that composing $B$ with the symmetry swaps the roles of these two functions. For symmetric bilinear forms, $B_1=B_2$, while for antisymmetric bilinear forms we have $B_1=-B_2$. This leads us to consider nondegenerate bilinear forms a little more. If $B_2$ is an isomorphism it has an inverse $B_2^{-1}$. Then we can form the composite $B_2^{-1}\circ B_1:V\rightarrow V$. If $B$ is symmetric then this composition is the identity transformation on $V$. On the other hand, if $B$ is antisymmetric then this composition is the negative of the identity transformation. Thus, the composite transformation measures how much the bilinear transformation diverges from symmetry. Accordingly, we call it the asymmetry of the form $B$. Finally, if we’re working over a finite-dimensional vector space we can pick a basis $\left\{e_i\right\}$ for $V$, and get a matrix for $B$. We define the matrix entry $B_{ij}=\langle e_i,e_j\rangle_B$. Then if we have vectors $v=v^ie_i$ and $w=w^je_j$ we can calculate $\displaystyle\langle v,w\rangle=\langle v^ie_iw^je_j\rangle=v^iw^j\langle e_i,e_j\rangle=v^iw^jB_{ij}$ In terms of this basis and its dual basis $\left\{\epsilon^j\right\}$, we find the image of the linear transformation $B_1(v)=\langle v,\underbar{\hphantom{X}}\rangle=v^iB_{ij}\epsilon^j$. That is, the matrix also can be used to represent the partial maps $B_1$ and $B_2$. If $B$ is symmetric, then the matrix is symmetric $B_{ij}=B_{ji}$, while if it’s antisymmetric then $B_{ij}=-B_{ji}$. Like this: Posted by John Armstrong \| Algebra, Linear Algebra 9 Comments » 1. [...] Inner Products Now that we’ve got bilinear forms, let’s focus in on when the base field is . We’ll also add the requirement that our [...] Pingback by \| April 15, 2009 \| Reply 2. [...] Inner Products Now consider a complex vector space. We can define bilinear forms, and even ask that they be symmetric and nondegenerate. But there’s no way for such a form to [...] Pingback by \| April 22, 2009 \| Reply 3. [...] start with either a bilinear or a sesquilinear form on the vector space . Let’s also pick an arbitrary basis of . I want [...] Pingback by \| June 24, 2009 \| Reply 4. [...] of matrices and forms and try to tie both views of our matrix together. We’re considering a bilinear form \$ on a vector space over the real or complex numbers, which we can also think of as a linear [...] Pingback by \| June 25, 2009 \| Reply 5. [...] back in on a real, finite-dimensional vector space and give it an inner product. As a symmetric bilinear form, the inner product provides us with an isomorphism . Now we can use functoriality to see what this [...] Pingback by \| October 29, 2009 \| Reply 6. [...] I said above, this is a bilinear form. Further, Clairaut’s theorem tells us that it’s a symmetric form. Then the spectral [...] Pingback by \| November 24, 2009 \| Reply 7. [...] this as a bilinear function which takes in two vectors and spits out a number . That is, is a bilinear form on the space of tangent vectors at [...] Pingback by \| September 20, 2011 \| Reply 8. [...] the next three families of linear Lie algebras we equip our vector space with a bilinear form . We’re going to consider the endomorphisms such [...] Pingback by \| August 9, 2012 \| Reply 9. [...] can now define a symmetric bilinear form on our Lie algebra by the [...] Pingback by \| September 3, 2012 \| Reply « Previous \| Next » About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 45, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.91927170753479, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/217160/how-can-a-markov-chain-be-written-as-a-measure-preserving-dynamic-system	How can a Markov chain be written as a measure-preserving dynamic system From http://masi.cscs.lsa.umich.edu/~crshalizi/notabene/ergodic-theory.html irreducible Markov chains with finite state spaces are ergodic processes, since they have a unique invariant distribution over the states. (In the Markov chain case, each of the ergodic components corresponds to an irreducible sub-space.) By "ergodic processes", I understand it to be the same as "ergodic measure-preserving dynamic system", if I am correct. As far as I know an ergodic measure-preserving dynamic system is a mapping $\Phi: T \times S \to S$ that satisfies a couple of properties, where $S$ is the state space, and $T$ is the time space. Sometimes there is a measure preserving mapping on $S$ that can generate the system by repeating itself. So I wonder how a Markov chain can be written as a mapping $\Phi: T \times S \to S$, and what the measure preserving mapping that generates the Markov chain is? Thanks! - 2 Answers The article talks about a (stationary) Markov chain ${(X_n)}_{n \in \mathbb{Z}}$ in discrete time with each $X_n$ taking its values in a finite set $E$. The canonical space of the Markov chain is the product set $E^{\mathbb{Z}}$. The trajectory $X=(\ldots, X_{-1}, X_{0}, X_1, \ldots)$ of the Markov chain is a random variable taking its values in $E^{\mathbb{Z}}$. Denoting by $\mu$ its distribution (which could be termed as the law of the Markov process) then $\mu$ is invariant under the classical shift operator $T \colon E^{\mathbb{Z}} \to E^{\mathbb{Z}}$. Then the Markov chain can be considered as the dynamical system $(T,\mu)$. In fact here we only use the fact that ${(X_n)}_{n \in \mathbb{Z}}$ is a stationary process. In the Markov case we can say in addition that the ergodicity of $T$ is equivalent to the irreducibility of ${(X_n)}_{n \in \mathbb{Z}}$. - – Tim Nov 18 '12 at 19:06 Did you find this question interesting? Try our newsletter Sign up for our newsletter and get our top new questions delivered to your inbox (see an example). email address The invariant probability measure is the one that assigns to state $i$ the probability $p_i$ defined as the limit, as time goes to infinity, of the fraction of time steps that the chain has been in state $i$. Strictly speaking $p_i$ is the "almost sure" limit. With probability 1 (in the probability measure on infinitely long walks of the Markov chain) the fraction will converge to $p_i$, and statistically unusual walks where something else happens have probability 0. - Thanks! But what is the $\Phi$ for a Makrov chain? – Tim Oct 20 '12 at 1:42 – Tim Oct 20 '12 at 1:58 Wikipedia is talking about the ergodic theory of a non-random map T that preserves a probability measure. The ergodic theorem is that this invariant measure can be defined in the same way as for the Markov chain, as occupation frequency of regions in the state space. Here, you either consider T as stochastic (advance one step in the random walk) or as a fixed, non-random map on the set PM of probability distributions on the Markov chain state space M. – zyx Oct 20 '12 at 2:08 Why isn't T (i.e. my $\Phi$) a mapping from state space of the Markov chain to itself? – Tim Oct 20 '12 at 2:11 It can be that, but then it has to be stochastic (with the state to state transition probabilities defined by the chain), otherwise you are taking a deterministic walk on the state space. Markov chains model a type of random walk. – zyx Oct 20 '12 at 2:15 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9263113141059875, "perplexity_flag": "head"}
http://math-a-magic.tumblr.com/	I feel stupid. I just realized it says you like guys in your bio. I apologize. Can we be tumblr boyfriend and girlfriend? Anonymous Yes we can <3 :D If you're Gay I'm upset. All of the guys I find attractive are Gay. I love Gay people but I wish some were straight. I'm not calling you Gay. I'm just wondering. I hope this doesn't sound offensive. I don't mean for it to be offensive. Oh boy I am awkward. But if your straight...I'm your girl. Anonymous I’m sorry, I happen to be gay <3 It’s nice of you to ask though :) Hey, while I’m on this subject (sorry to hijack this ask), I thought I would take a moment to talk about this, as I don’t often speak openly about my sexuality on here other than what I identify as. Quick thing here, I am a romantically bi, homosexual, meaning I love the idea of relationships with anyone, but kind of reserve the whole sexual arousal thing for guys. Now that that’s out of the way, I’m going to clear up another thing. I’ve gotten a lot of requests to go shopping from girls in my school, as well as people who openly state that I should get into fashions (simply for my sexuality), or wonder why I don’t “act gay”. For one, you (not you anon, I’m not directing this at you in any way <3 ) need to know that there are varying degrees of identification. They do not come with the ratio of attraction to males (e.g. the idea that bisexual males act more like males than gay males do), but are a function of the person’s personality. I identify as fully homosexual, and you know what; I have seen people genuinely surprised when I come out to them, because I don’t act or look in any way different from straight males. I do this not because I want to look and act “normal” but because I am just that, an average person; so you can see why it is frustrating to have to go through the explanation that I have no intentions of going shopping with anyone, I don’t know what colors match (I’ve worn red brown and black out of the house all at once), and that I can be gay without the “stereotypical gay voice” (I actually find it kind of off putting when people have a “stereotypical gay voice”). questions like “I thought gay people like fashions how can you like math if you are gay?” are actually getting on my nerves; it’s not necessarily that I find them offensive, as I know what is causing someone to ask them is just ignorance, but it’s starting to make me wonder if many people outside of the gay community actually know that there are people like me out there. P.S. I can’t even spell fashions without auto correct. I spell it fassions. Also, I kind of relate to this guy http://adventuresingay.tumblr.com/ <3 Do you live in the states or Europe? Anonymous Canada actually :) Ontario So is that your real hair? Anonymous Real hair, no dye :) You're really attractive. Just so you know. ;) Anonymous Aw shucks :3 you are so sweet :) GUYS! I just had an idea of how elliptic curves could be related to ring theory questions. I might take the second half of my math class to work it out, and I’ll try and get it on here by the end of the long weekend :) Fanmail ;-) Hey man! I just wanted to tell you that I’ve discovered your blog today (since you reblogged a post of mine), and I really love it! :) Although I had lot of things to do, I couldn’t stop exploring your blog. I admire how exceptionally much (for a 16 year old guy) you already know about mathematics, and the passionate way you speak about it. There’s still lots of things to learn of course, but I think you’re a very promising maths-talent :). Personally, I’m a maths student. I’m in my second year of university now. Of course you don’t need to answer if you don’t want to, but I think it would be nice to talk to you, so feel free to answer :). I hope I’ll hear from you! Christophe (PS: Actually I wanted to send you this by ‘Fanmail’, but I couldn’t do that because I only started following you a few hours ago. So you don’t have to submit this on your blog :p) ABC conjecture If you would like to read the papers on the abc conjecture, there are four of them, each with different meanings. The first two set up the definitions for a new kind of math called inter-universal geometry, the third constructs objects in an inter-universal geometry setting, and the forth actually solves the problem. I urge you to check it out and see if you can read them yourselves. So far, I’m able to read the first three sentences XD http://www.kurims.kyoto-u.ac.jp/~motizuki/Inter-universal%20Teichmuller%20Theory%20I.pdf http://www.kurims.kyoto-u.ac.jp/~motizuki/Inter-universal%20Teichmuller%20Theory%20II.pdf http://www.kurims.kyoto-u.ac.jp/~motizuki/Inter-universal%20Teichmuller%20Theory%20III.pdf http://www.kurims.kyoto-u.ac.jp/~motizuki/Inter-universal%20Teichmuller%20Theory%20IV.pdf EDIT! Tumblr user twocubes (cool blog btw) has added to the pile of papers by providing me with the overview for the papers! http://www.kurims.kyoto-u.ac.jp/~motizuki/Panoramic%20Overview%20of%20Inter-universal%20Teichmuller%20Theory.pdf velushomaz: A magic spell. #mathematics #number #theory #uni idobelieveininfinity: New stuff to read! Totally bought this book when I was in Washington :D On August 31, 2012, Japanese mathematician Shinichi Mochizuki posted four papers on the Internet. The titles were inscrutable. The volume was daunting: 512 pages in total. The claim was audacious: … GUYS! Okay, the math departments at Universities across the world are having a bit of an “wat are dis??” (Jenna Marbles quote) moment because of the newly given proof for the ABC conjecture. Quick history of the problem; it was proposed by Joseph Oesterle and David Masser,and it holds so much weight, that 13 of the hardest mathematics questions to date rely on this theorem for their solution to exist. NOW FOR THE COOL PART The ABC conjecture begins with the diophantine equation A+B=C (pretty easy right?) Now, the terms don’t share any common factors, meaning that if I can divide a by 2, I can’t divide b by 2. With that in mind, I’d like to throw out some really easy to follow examples of this: $1024+81=1105$ $125+3=128$ Let’s factor them into their prime factors to make sure they don’t have any common factors. $2%5E%7B10%7D+3%5E4=5%5Ctimes%2013%5Ctimes%2017$ $5%5E3+3=2%5E7$ Notice the differences here, there are 14 primes on the left in the first example and 3 on the right, there are 4 primes on the left in the second example and 7 on the right. It’s been observed that the condition in the second example occurs less frequently than that of the condition in the first example; so the second example is considered the “exception” and the first one the “rule”. (there isn’t really a way to call one the rule and the other the exception as there are infinitely many of each, but it is similar to how there are infinitely many primes, but they seem to occur “less frequently” than the composite numbers.) From here, we define something called a radical.This isn’t the radical from math class, which means that there is a root function involved, instead, it’s defined this way: $rad(ABC)=p(A)%5Ctimes%20p(B)%5Ctimes%20p(C),%20p(x)=p_0%5Ctimes%20p_1...p_n$ Or incase that was too vague, the radical basically takes all of the prime factors of the numbers involved, and multiplies the distinct ones. For instance, rad(ABC) in the first example is: 2x3x5x13x17=6630 and, rad(ABC) in the second example is: 2x3x5=30 Looking at differences again, in the first example, rad(ABC)>C, and in the second example, rad(ABC)<C. Knowing this, we can form the statement: $rad(ABC)%5Ek%3EC,%20k%5Cgeq%201$ The ABC conjecture builds off of this by saying that there are an infintly many number of exceptions when k=1, and a countable number of solutions when k>1! All that being said, the ABC conjecture could shed some new light on a deep connection between the fundamental operations of addition and multiplication, and provide us with some new mathematics, if the paper ever gets read that is.The proof provided by , Shinichi Mochizuki was so complicated, that whole teams of people have given up trying to decipher it, and what’s worse, Mochizuki refuses to give a lecture on the new subject he created, or even run through the proof. It could be years before the proof is checked, and many more years until the subject that was introduced in the proof; inter-universal geometry, is ever fully explored. science-junkie: Oscillate by Daniel Sierra “Oscillate” is the title of my thesis animation done at the MFA Computer Art program in the School of Visual Arts located in New York City. My goal with “Oscillate” was to visualize waveform patterns that evolve from the fundamental sine wave to more complex patterns, creating a mesmerizing audio-visual experience in which sight and sound work in unison to capture the viewer’s attention… —Daniel Sierra. This is the most beautiful video I have watched in most likely my whole life. For some reason I feel so calmed now after watching this; and whether it is from the music or from the visuals, this video has really given me the spiritual experience I’ve been looking for. I love the way the dust seems to flake off the strings as they vibrate with the noise; how complex musical aspect is without straining your brain with indecipherable melodies and rhythms; the sudden changes between movements that come with a new visual beauty to behold; and what I found the most profound, the fact that I can completely relate to the changes in music and visuals, as I go through the same abrupt mood swings every day of my life. To give you an idea of how profound this was for me, I have a bit of auditory-visual synesthesia, where sounds and colors are constantly interchanging in my head; and as I followed through, I slowly discovered that the creator of this video must have seen this music through the same eyes I did. The strings changed color in the same way that my brain would think to seem them turn; producing a lovely euphoria for me, where my brain no longer had to fight to see the music the way I thought it should look, as it was already displayed for me. I’m so calm I could almost cry again, I love this so much. Basic summary of my last two weeks in case you guys were wondering where I have been! :D Washington was great, and no, I am not as fat as I look, I was just standing with my hips out in the second one XD You guys rock! :’) Leave an honest opinion of me in my ask :) Math will come soon, just need to get caught up on school work :)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9338805675506592, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/29262/lepton-number-conservation/29265	# Lepton Number Conservation What is the global symmetry of the electroweak Lagrangian that gives rise to lepton number conservation? As I understand it, electric charge is some linear combination of the conserved quantities corresponding to the $SU(2)$ symmetry and the $U(1)$ symmetry of the electroweak theory. If something similar is true for lepton number, then perhaps it would not be easiest to give the global symmetry directly. This is perfectly fine of course. - ## 1 Answer The symmetry is not a mix-up of the gauge symmetries, it's a global symmetry. The symmetry is a phase rotation of the lepton fields in one generation, so lepton number, muon number, and tau number are each conserved, ignoring neutrino oscillations. You multiply the electron-neutrino doublet and electron-singlet field by an opposite phase. Since you choose the fields to diagonalize the Higgs couplings, this doesn't affect either their electroweak or (standard dimension 4) Higgs interaction, because each spinor action term multiplies two pieces $\bar{\psi}$ and $\psi$ that rotate in opposite sense. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 4, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8843615651130676, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/190134-closed-subsets-connected-space-x.html	Thread: 1. closed subsets of a connected space X Let $A\& B$ be closed subsets of a connected space $X$ such that $X = A \cup B$. How to show $A\& B$ are connected, if $A \cap B$ is connected. I traied this: Suppose $B$ is not connected and let $B = S \cup T$ be sepration of $B$, then, $A \cap B$ must lie in $S$ (say) So, $X = \left( {A \cup S} \right) \cup T$ form separation of x $\Rightarrow \Leftarrow$ is this right 2. Re: closed subsets of a connected space X Originally Posted by rqeeb Let $A\& B$ be closed subsets of a connected space $X$ such that $X = A \cup B$. How to show $A\& B$ are connected, if $A \cap B$ is connected. I traied this: Suppose $B$ is not connected and let $B = S \cup T$ be sepration of $B$, then, $A \cap B$ must lie in $S$ (say) So, $X = \left( {A \cup S} \right) \cup T$ form separation of x $\Rightarrow \Leftarrow$ Now say that likewise $A$ must be connected. 3. Re: closed subsets of a connected space X Originally Posted by Plato Now say that likewise $A$ must be connected. but now the problem is: To show AUS and S are spen in X we know that S & T are open in B, and B is closed in X. Bur how to show they r open in X?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 26, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9617899060249329, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/96558/does-the-measure-preserving-property-commute-with-ultralimits	Does the “measure-preserving property” commute with ultralimits ? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $(X, \mathcal{B}, \mu, T)$ be a measure-preserving system, with $T$ invertible, where the $\sigma$-algebra $\mathcal{B}$ is a Borel algebra arising from a topology which makes $T$ continuous, and such that $X$ is compact Hausdorff. Then there's a well-defined action of the semigroup $\beta \mathbb{Z}$ on $X$ : $\forall p \in \beta \mathbb{Z}, \forall x \in X,\quad T^p x := p\lim\limits_{n \in \mathbb{Z}} T^n x$ If $p$ is a principal ultrafilter, it is clear that $T^p$ is measure-preserving. Is it still true for any ultrafilter on $\mathbb{Z}$ ? The measure preserving property of a $T^p$ would be a consequence of the corresponding dominated convergence theorem along the ultrafilter $p$ ; as far as I can see, this is not a direct consequence of the (usual) dominated convergence theorem, and I didn't succeed in proving or disproving it. Any ideas ? - 1 There should exist a non-normal pure state on l^{\infty}(\mathbb{Z}), and that should provide you with a counterexample. I hope this helps. – Jon Bannon May 10 2012 at 13:35 1 A non-normal pure state on $l^\infty({\bf Z})$ is the same as a non-principal ultrafilter on ${\bf Z}$. – Nik Weaver May 10 2012 at 15:25 1 @Nik: I suspected so. How does one get that every nonprincipal ultrafilter isn't normal (a reference will do)? – Jon Bannon May 10 2012 at 15:44 3 Let $f_n \in l^\infty({\bf Z})$ be the bi-infinite sequence which is zero between $-n$ and $n$, and one outside that interval. This sequence converges weak* to the zero sequence, but each $f_n$ evaluates to 1 along any nonprincipal ultrafilter. So evaluating on any nonprincipal ultrafilter cannot be a normal linear functional on $l^\infty({\bf Z})$. – Nik Weaver May 10 2012 at 15:51 Thank you Nik!! – Jon Bannon May 10 2012 at 16:26 show 1 more comment 1 Answer Looks false to me. Let $X$ be the product of countably many copies of `$\{0,1\}$` with the standard Bernoulli measure (each point has measure $1/2$). Index these copies with ${\bf Z}$, and let $T$ be the shift operator. I claim that evaluating along any free ultrafilter won't be measure-preserving. To see this, let $A_0 \subset X$ be the set of sequences $x = (\ldots, x_{-1}, x_0, x_1, \ldots)$ with $x_0 = 0$ and let $A_1$ be its complement, the set of sequences $x$ with $x_0 = 1$. Both of these sets have measure $1/2$. However, it is easy to check that $T^p(A_0) = T^p(A_1) = T^p(X)$, using the fact that the limit of a sequence along a free ultrafilter doesn't change if you modify the sequence in finitely many coordinates. So there's no way $T^p$ can be measure preserving. (Also, it isn't a homeomorphism.) - 1 Thanks for the neat counterexample. It also disproves the "dominated convergence theorem" for convergence along a free ultrafilter. – js May 10 2012 at 19:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9249894618988037, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/03/04/jordan-normal-form/?like=1&source=post_flair&_wpnonce=7b62a7b6bf	# The Unapologetic Mathematician ## Jordan Normal Form Okay, let’s put everything together now. Start with a linear endomorphism $T:V\rightarrow V$ on a vector space $V$ of finite dimension $d$ over an algebraically closed field $\mathbb{F}$. If you want to be specific, use the complex numbers $\mathbb{C}$. Now we can calculate the characteristic polynomial of $T$, whose roots are the eigenvalues of $T$. For each eigenvalue $\lambda$, we can define the generalized eigenspace as the kernel $U_\lambda=\mathrm{Ker}\left((T-\lambda1_V)^d\right)$, since if some power of $T-\lambda1_V$ kills a vector then the $d$th power will. These generalized eigenspaces do not overlap, and each one is invariant under $T$. The dimension of the generalized eigenspace $U_\lambda$ associated to $\lambda$ is the multiplicity of $\lambda$, which is the number of times $\lambda$ shows up on the diagonal of an upper-triangular matrix for $T$. Since the total number of diagonal entries is $d$, we see that the dimensions of all the generalized eigenspaces add up to the dimension of the entire space $V$. Thus, we have a decomposition $\displaystyle V=\bigoplus\limits_{\lambda}U_\lambda$ of $V$ as the direct sum of these generalized eigenspaces, where $\lambda$ runs over the roots of the characteristic polynomial. If we restrict $T$ to the generalized eigenspace $U_\lambda$ with eigenvalue $\lambda$, the transformation $T\vert_{U_\lambda}-\lambda1_{U_\lambda}$ is nilpotent. Thus we can find a Jordan basis for $U_\lambda$, which puts $T\vert_{U_\lambda}-\lambda1_{U_\lambda}$ into the block-diagonal form $\displaystyle\begin{pmatrix}A_1&&{0}\\&\ddots&\\{0}&&A_n\end{pmatrix}$ where each block has the form $\displaystyle\begin{pmatrix}{0}&1&&&{0}\\&{0}&1&&\\&&\ddots&\ddots&\\&&&{0}&1\\{0}&&&&{0}\end{pmatrix}$ We can now add back in the eigenvalue $\lambda$ times the identity transformation to the restriction of $T$. Now we still have the block-diagonal form, but the blocks themselves now have the form $\displaystyle\begin{pmatrix}\lambda&1&&&{0}\\&\lambda&1&&\\&&\ddots&\ddots&\\&&&\lambda&1\\{0}&&&&\lambda\end{pmatrix}$ where, of course, a block could be a single $1\times1$ matrix whose only entry is $\lambda$. Putting these together for all the different eigenvalues, we have a Jordan basis for $V$. This puts the matrix $T$ into “Jordan normal form”. That is, the matrix of $T$ with respect to a Jordan basis is block-diagonal, with each block consisting of one eigenvalue $\lambda$ down its diagonal, and ${1}$s just above the diagonal. Unfortunately, if the base field $\mathbb{F}$ is not algebraically closed, we may not have any upper-triangular matrix for $T$, and so we can only put the portion of $T$ captured by generalized eigenspaces into Jordan normal form. There may still be another direct summand which contains no generalized eigenvectors at all. Over an arbitrary field, this sort of thing gets complicated quickly, but it will be useful for us to consider what happens over the real numbers $\mathbb{R}$. We’ll come back to this. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 8 Comments » 1. [...] of Jordan Normal Forms So we’ve got a Jordan normal form for every linear endomorphism on a vector space of finite dimension over an algebraically closed [...] Pingback by \| March 5, 2009 \| Reply 2. [...] we know how to do this! Use a matrix in Jordan normal form! We know that within a given conjugacy class, the Jordan normal form is unique — up to [...] Pingback by \| March 6, 2009 \| Reply 3. [...] seen a lot about Jordan normal forms, which can pretty much capture the behavior of any single linear transformation over an [...] Pingback by \| March 13, 2009 \| Reply 4. Hello! Very Interesting post! Thank you for such interesting resource! PS: Sorry for my bad english, I’v just started to learn this language See you! Your, Raiul Baztepo Comment by RaiulBaztepo \| March 29, 2009 \| Reply 5. [...] Real Linear Transformations Finally, we come to the analogue of Jordan normal form over the real [...] Pingback by \| April 9, 2009 \| Reply 6. [...] looking at an endomorphism , we have found that we can pick a basis so that the matrix of is in Jordan normal form, which is almost diagonal. That is, we can find an invertible transformation and a “Jordan [...] Pingback by \| August 24, 2009 \| Reply 7. [...] of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of blocks that look [...] Pingback by \| August 28, 2012 \| Reply 8. [...] of a finite-dimensional vector space over an algebraically closed field can be put into Jordan normal form: we can find a basis such that its matrix is the sum of blocks that look [...] Pingback by \| August 28, 2012 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 44, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9083238840103149, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/commutator+symmetry	# Tagged Questions 1answer 127 views ### Why Must Conserved Currents of Lorentz Symmetry Satisfy the Lorentz Algebra I've seen it written many times that the commutation relation $[M^{I-},M^{J-}]=0$ is required for Lorentz invariance in the light cone gauge quantisation of the bosonic string. This follows ... 2answers 225 views ### Symmetries, Generators, Commutators and Observables I'm learning about generators and conservation laws and have derived the equation (1) $$[Q,A]=-i\hbar f(A)$$ which is satisfied by the observable generator $Q$ for a transformation group with ... 1answer 101 views ### transformations with commutators and anticommutators that generate displacements is well known that composition of point reflections generate pure displacements. This implies that the commutator of two point reflections will be a pure displacement. Are there similar elemental ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8733559846878052, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/250686/probability-of-lambda-given-k-in-poisson-distribution	# Probability of $\lambda$ given k, in Poisson distribution. We know the PMF of poisson distribution is $P_K(k) = e^{-\lambda} \frac{\lambda ^k}{k!}$, now, given $k$ arrivals in a unit time, what is the PDF of the arriving rate being $\lambda$? - I don't know this notation: What does $K$ stands for? – math Dec 4 '12 at 13:48 It means $P(K=k)$, $K$ stands for the random variable, $k$ stands for the value of $K$. – CravingSpirit Dec 4 '12 at 14:07 ## 1 Answer There's no such thing unless you specify a prior. Wikipedia has a section that gives a conjugate prior for this problem. - Thanks. This is kind of deep for me at the moment. – CravingSpirit Dec 4 '12 at 14:08 @CravingSpirit: OK; to put it in more mundane terms: What you think about $\lambda$ after you learn that there were $k$ arrivals in a unit time depends on what you that about $\lambda$ before you learned that. For instance, if you think a priori that very low values of $\lambda$ are very unlikely, then even after you observe $0$ arrivals the probability you assign a posteriori to low values of $\lambda$ will be less than the probability you would have assigned a posteriori after observing $0$ arrivals if you had a priori believed these low values to be quite likely. – joriki Dec 4 '12 at 14:30 Ok, you mean the probability of $\lambda$ depends both on our a priori judgment and the value of $k$. – CravingSpirit Dec 4 '12 at 14:35 @CravingSpirit: Yes, that just about summarizes it :-) – joriki Dec 4 '12 at 14:35	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 18, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9161763191223145, "perplexity_flag": "middle"}
http://mathhelpforum.com/pre-calculus/88942-needing-help-vector-word-problem.html	# Thread: 1. ## Needing help with a vector word problem I'm taking my word problem final in two days and I really need to understand this ASAP. Here' s the problem: A place is flying on a bearing of South 10 degrees East at 460 mph. A tail wind is blowing in the direction of South 20 degrees West at 80 mph. Find the resultant velocity of the plane and give a graphic representation of the resultant vector. I'm just having problems finding angles and understanding so that I can use the Law of Cos. Thank you. 2. Originally Posted by Shanizzle I'm taking my word problem final in two days and I really need to understand this ASAP. Here' s the problem: A place is flying on a bearing of South 10 degrees East at 460 mph. A tail wind is blowing in the direction of South 20 degrees West at 80 mph. Find the resultant velocity of the plane and give a graphic representation of the resultant vector. I'm just having problems finding angles and understanding so that I can use the Law of Cos. Thank you. Here is a diagram $V_1=(460\cos(350),460\sin(350))$ $V_2=(80\cos(200),80\sin(200))$ $V_1+V_2 \approx (377.8,-107.2)$ So the resultant velocity is the magnitute of the sum of the vectors or $\|\|V\|\| \approx \sqrt{(377.8)^2+(-107.2)^2}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 4, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9257943630218506, "perplexity_flag": "middle"}
http://[email protected]/news/math-in-the-media/06-2011-media	\| Tony Phillips' Take on Math in the Media A monthly survey of math news # This month's topics: ## Taming Complexity "The mathematics of network control--from cell biology to cellphones" is the cover story for the May 12, 2011 Nature. The authors, Yang-Yu Liu, Jean-Jacques Slotine and Albert-László Barabási, present new theoretical results with surprising practical consequences. The image on the left shows a small but typical network. There are nodes numbered $i = 1, 2, 3, 4$, each in a state measured by the function $x_i(t)$; so the state of the entire network is represented by the time-dependent vector ${\bf x}(t) = (x_1(t), x_2(t), x_3(t), x_4(t))$. The nodes interact: the coefficient $a_{ij}$ measures the strength of the effect of node $j$ on node $i$; $a_{ij}=0$ if the nodes do not interact. Additionally the node $i$ may have input ($b_i = 0,1$) of strength $u_i(t)$ from outside the network. In this model, the interactions are linear: the state $x_1$ obeys the differential equation $dx_1/dt = a_{12}x_2 + a_{13}x_3 + a_{14}x_4 + u_1$, and similarly for the other nodes. In vector form, $$\frac{\displaystyle d{\bf x}}{\displaystyle dt} = A{\bf x} + B{\bf u}$$ where $A$ is the matrix of the $a_{ij}$, and $B$ is the matrix specifying which nodes have inputs. For our network, $$A = \left ( \begin{array}{cccc} 0&0&0&0\\ a_{21}&0&0&0\\ a_{31}&0&0&a_{34}\\ a_{41}&0&0&0 \end{array}\right ); ~~~~~B = \left ( \begin{array}{cc} b_1&0\\ 0&b_2\\ 0&0\\ 0&0 \end{array}\right );$$ The network is said to be controllable if the functions $u_i(t)$ can drive the network from any one state ${\bf x} = (x_1, x_2, x_3, x_4)$ to any other. The authors begin their exposition with the statement of Kalman's controllability rank condition: a linear system like ours with $N$ nodes is controllable if the rank of the matrix $C = (B, AB, A^2B, \dots, A^{N-1}B)$ is maximum, i.e. $N$. In our case $$C = \left ( \begin{array}{cccccccc} b_1&0 & 0 & 0 & 0 &0 &0 &0\\ 0 & b_2 & a_{21}b_1 & 0 & 0 &0 &0 &0\\ 0&0& a_{31}b_1 & 0 & a_{34}a_{41}b_1 &0&0&0\\ 0&0 & a_{41}b_1 &0 & 0 &0 &0 &0 \end{array}\right ).$$ clearly has rank 4, so the system is controllable. Note that without the nonzero $a_{34}$ connection this would not be the case. The article first describes an efficient algorithm for determining, in an arbitrary network, the minimum number $N_D$ of driver nodes: a set of nodes whose inputs control the entire network. Then they show (they consider this their most important finding) that $N_D$ is determined mainly by the degree distribution, a feature of the topology of the network: the probabilistic analysis of the number of ingoing and outgoing links per node. A salient feature of the article (picked up in a "News & Views" piece by Maurice Egerstedt in the same issue of Nature), is its analysis of a large collection of real-world networks, including the counter-intuitive finding that "Social networks are much easier to control than biological regulatory networks, in the sense that fewer driver nodes are necessary to fully control them." Furthermore, "engineered networks such as power grids and electronic circuits are overall much easier to control than social networks and those involving gene regulation." Which, as Egerstedt remarks, "may or may not be a good thing, depending on who is trying to control the network." ## "Capturing Rogue Waves" "Rogue Wave Observation in a Water Wave Tank" by A. Chabchoub, N. P. Hoffmann (Hamburg University of Technology) and N. Akhmediev (Australian National University, Canberra) appeared in Physical Review Letters for May 20, 2011. Rogue waves are anomalous, giant waves (typically three times the height of the ambient wave pattern) that seem to appear out of nowhere and to vanish equally mysteriously. The authors give evidence that a rogue wave is in fact an example of Peregrine's soliton, a solution to the nonlinear Schrödinger equation discovered by the mathematician D. Howell Peregrine in 1983; it is localized both in space and in time. Peregrine's solitons are also called "breather waves" because they temporarily suck up energy from their neighbors. The authors managed to program their wave-maker to reproduce the initial conditions for a breather wave in a water tank. Breather wave (detail). Read from the bottom: as the wave packet progresses down the tank (each record of height/time is taken 1 m. beyond the one below it) the soliton develops, up to its maximum. The almost symmetrical decay of the anomaly would have required a tank more than twice as long to work itself out. Tick-marks are 10 sec. apart. Click for larger and more complete image. Images courtesy of Amin Chabchoub. The waveform at its maximum almost exactly matches the shape predicted by Peregrine: the measured surface height at the position of maximum amplitude (solid blue line) compared with the theoretical Peregrine solution (dashed red line). This work was picked up by Devin Powell for Science News, under the title "Rogue Waves Captured: Re-creating monster swells in a tank helps explain their origin." He quotes Chabchoub: "It's possible that the wind could generate a similar ... perturbation in the open sea." Tony Phillips Stony Brook University tony at math.sunysb.edu	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9278585314750671, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2007/09/18/more-on-kernels-and-cokernels/?like=1&source=post_flair&_wpnonce=18b5244a97	# The Unapologetic Mathematician ## More on Kernels and Cokernels The best-known abelian categories are categories of modules over various rings. And as modules, these objects are structured sets. Now, even though we’re willing to elide the difference between a hom-group and a hom-set, we would at least like to avoid talking about the objects as sets, and the morphisms as functions. So let’s try to focus on studying the morphisms and use them to understand the properties of the category. Luckily, in an abelian category we know a lot about morphisms. For one thing, let’s consider an object $C\in\mathcal{C}$ and the set $P_C=\bigcup\limits_{C'\in\mathcal{C}}\hom_\mathcal{C}(C',C)$ of all the arrows coming into $C$. We can put the structure of a preorder $\preceq$ on this set by saying $(f:A\rightarrow C)\preceq(g:B\rightarrow C)$ if there is $h:A\rightarrow B$ with $f=g\circ h$. It’s straightforward to show that this relation is reflexive and transitive. Then, as usual we can symmetrize this relation to get an equivalence relation: $f\equiv g$ if and only if $f\preceq g$ and $g\preceq f$. When we pass to the equivalence classes, the preorder becomes a partial order, which we’ll also call $P_C$. In particular, this includes all monomorphisms, which we know as subobjects. Similarly, we can construct another preorder $Q^C=\bigcup\limits_{C'\in\mathcal{C}}\hom_\mathcal{C}(C,C')$ of all the arrows leading out of $C$ with $f\succeq g$ if there is an $h$ with $g=h\circ f$. We can also turn this into a partial order, which includes the partial order of all quotient objects of $C$. Now for every arrow $g$ from $C$ pick a kernel and for every $f$ to $C$ pick a cokernel. Then $f$ and $\mathrm{Ker}(g)$ live in $P_C$, while $\mathrm{Cok}(f)$ and $g$ live in $Q^C$. And we have three equivalent statements: • $f\preceq\mathrm{Ker}(g)$ • $g\circ f=0$ • $\mathrm{Cok}(f)\succeq g$ Now the partial orders $Q^C$ and $P_C$ can be considered as categories, as any partial order can be. And the maps $\mathrm{Ker}:Q^C\rightarrow P_C$ and $\mathrm{Cok}:P_C\rightarrow Q^C$ reverse these orders, so they are contravariant functors. So let’s flip the order on $P_C$ ($x\geq y$ instead of $x\preceq y$) and instead consider them as covariant functors $\mathrm{Ker}:Q^C\rightarrow P_C^\mathrm{op}$ and $\mathrm{Cok}:P_C^\mathrm{op}\rightarrow Q^C$. The above equivalence now reads $f\geq\mathrm{Ker}(g)\Leftrightarrow\mathrm{Cok}(f)\succeq g$. That is, we have an adjunction. The identities satisfied by the unit and counit read, in this case: • $\mathrm{Ker}(\mathrm{Cok}(\mathrm{Ker}(g)))=\mathrm{Ker}(g)$ • $\mathrm{Cok}(\mathrm{Ker}(\mathrm{Cok}(f)))=\mathrm{Cok}(f)$ ### Like this: Posted by John Armstrong \| Category theory No comments yet. « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 42, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9277997016906738, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/29791/how-to-measure-usage-data-across-cohorts-segments	# How to measure usage data across cohorts/segments? Netflix and Google among others use customer usage (number of videos streamed or unique searches conducted) as a proxy for value, or for determining which version of a test is best. I am curious how they would measure this but I'm not great at statistics. If they split their customer base into even groups, the usage within the group will still follow a distribution. So just comparing the mean usage in each group might be misleading if one group has a Bill Gates in it (or someone whose usage dominates the distribution). What are some proxies I can use to help measure "this segment's usage is higher than the other segment's"? The distribution I am trying to measure is much more likely to have extreme usage numbers than Netflix or Google whose usage numbers are limited by the number of hours in a day. Would a Q-Q plot be appropriate? - ## 3 Answers Statistically, this question seems reducible to: how can you compare two distributions to see which one is bigger? You are right to question the naive use of the mean as a blind summary of the distribution. However, there are still very many ways you could approach this. Here are a couple of options: 1) The two-sample Kolgomorov-Smirnov test. You can think of this as a way to plot the two empirical cumulative distributions and then search for the point with the largest maximum vertical deviation between the distributions. This has the advantage of being agnostic about precisely where in the distribution you look for the differences. I actually think this page is better than the Wikipedia page at explaining it, because it has better graphics. 2) A Bayesian comparison between the case that both samples are drawn from the same distribution and the case that both are drawn from different distributions. Here you will likely want to bin the distributions into histograms. This paper discusses and demonstrates a nice way of performing this. It models each histogram as coming from a Poisson distribution, and then infers whether they share the same rate parameter or whether the rate parameters differ. This is captured by a parameter $\pi$, whose probability mass is near one in the case that the histograms are the same, and is near zero in the case that the histograms are different, (The code is available online. You'll need ROOT installed.) This method is likely to give better answers, but will take more time to implement. Whether you should use it or something like it depends on how crucial it is to you to get an extremely accurate answer rather than a good enough one. Q-Q plots This might be a nice way to visualize the differences between the two usage distributions. From what I can tell, Q-Q plots are more often used to compare one empirical distribution to a theoretical one, e.g. to a normal distribution, to justify its use. If you want to do a Q-Q plot with two empirical distributions you will first need to estimate the quantiles. Q-Q plots may be worth investing some time into, but as a descriptive plot (i.e., you can't easily get a summary statistic out of them), you will either have to convert what you learn from them into a form that is easier to interpret, or make sure that whoever you are presenting the data to can interpret them. And they are not that easy to interpret. So, bottom line: yes, they could work, but unless you are already good at them I think it'd be easier to use another option. - ## Quick answer Compare medians instead of means, which will be more robust to the effects of extreme values. ## More details The best way to compare the distributions of usage between your customer groups depends on what you are trying to achieve by comparing them. For example, say you have the following groups: ````A: 4, 5, 6 B: 0, 0, 0, 0, 50 ```` There is no one answer to which group is has higher usage, but here are a few scenarios: ### Mean Comparing the mean of B (10) to the mean of A (5) would be appropriate if you want to know, for example, whether the increase in total usage would be greater if you were to add 5 new customers that would fall into group A (expected increase of 25) versus 5 new customers from group B (expected increase of 50). The extreme values will only contribute to larger variance on these expectations. ### Median Comparing the median of A (5) to the median of B (0) would be appropriate if you just want a quick way of saying whether a "typical" customer from group A would have higher usage than a "typical" customer from group B. Calculating the median is easy and gives good results, and in most cases will provide nearly the same results as the following more rigorous approach. ### Mann-Whitney U / Wilcoxon rank-sum / AUC... a.k.a "test statistic of many names" Calculating the U statistic is my personal favorite (although the proliferation of names is confusing). The U statistic is the probability that the usage of a uniform randomly chosen customer from group A is greater than the usage of a randomly chosen customer from group B. It's useful in a wide variety of situations. In addition to having an intuitive meaning, it's easy to test if the value of this U statistic is significant. It generalizes well to more than two groups, under the name Multiclass AUC. ## Significance You will probably also want to check that any differences between groups are significant. The Mann-Whitney U works for this, as does the Kolgomorov-Smirnov (KS) test. The way-too-common T test makes assumptions about the distributions that don't sound justified with your data. If, like me, you can't stand the proliferation of tests named "{Dead statistician} {Letter of alphabet} Test", (and you are interested in really understanding the concept of statistical significance), you can choose your software and simulate the process many times and see how unlikely the observed value of the test statistic is (bootstrapping). For example, if you want to see if the measured value of your U statistic is unlikely to occur if the groups are not different at all, try this: 1. Combine the data from group A and group B into one group, `4, 5, 6, 0, 0, 0, 0, 50` 2. Randomly pick values from the combined list, with replacement, to form two groups with the same sizes as A and B, for example `A': 0, 4, 4` and `B': 50, 0, 0, 6, 5`. Since you know that these `A'` and `B'` were just chosen randomly from the same population, this is an instance of a "null model" that the groups are not different. 3. Measure the "test statistic" for these two groups (this could be the difference in means, the difference in medians, the AUC, the KS value, or anything else you dream up) and write it down. 4. Repeat 2 and 3 many times. 5. Measure the "test statistic" for your actual groups, and compare it to the distribution of test statistics that you just generated. If it it falls outside the range, then the actual value of your test statistic was unlikely to occur in your null model, and the value is statistically significant. NOTE: I couldn't put more than two links in here, because I don't have enough street cred on Cross Validated yet. I'll edit when I do. - It sounds like you are looking for a robust test of the difference between two measures of location. You are right that using the mean poses all sorts of problems, not least of them extreme outliers. Research in recent decades has shown that comparison of means, relying on large sample properties, is much more problematic in many real-life situations than had been appreciated. A good alternative is a comparison of 20% trimmed means. A percentile bootstrap around a mean with at least 0.2 trimming is "one of the most effective methods for obtaining accurate probability coverage and achieving relatively high power" (p.336 of Wilcox 2012, Modern Statistics for the Social and Behavioral Sciences, CRC Press, thoroughly recommended). There are straightforward implementations available in R. For example, consider the mixed log-normal distributions in the R code below. x1 is generally much bigger than x2, but x2 is contaminated with 20 very large values that blow the mean out of the water. Either the median or the trimemd mean give a much better sense of the general centre of location. ````> x1 <- exp(rnorm(1000,10,1)) > x2 <- exp(rnorm(1000,8,2)) > x2[sample(1:1000,20)] <- exp(rnorm(20,12,5)) > combined <- melt(data.frame(x1,x2)) Using as id variables > tmp <- round(rbind(with(combined, tapply(value, variable, mean)), + with(combined, tapply(value, variable, mean, tr=.2)), + with(combined, tapply(value, variable, median)))) > row.names(tmp) <- c("Mean", "TrimmedMean", "Median") > tmp x1 x2 Mean 35229 1282017 TrimmedMean 25077 5992 Median 22454 3896 > qplot(value, data=combined, log="x") + facet_wrap(~variable) ```` -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9405401349067688, "perplexity_flag": "middle"}

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