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http://www.physicsforums.com/showthread.php?t=12999	Physics Forums ## The distributive law Is there any way to prove the distributive law for integers? I heard that there is yet I don't understand how being that the distributive law is an axiom and therefore what the understanding our number system is based on. 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 Originally posted by Ed Quanta Is there any way to prove the distributive law for integers? I heard that there is yet I don't understand how being that the distributive law is an axiom and therefore what the understanding our number system is based on. When working with the integers, you usually just consider the distributive law to be an axiom. So you don't prove it, since you've assumed it to be true. However, in some cases you wish to prove that using axioms for some other objects you can construct the integers. For example, you might want to prove that you can construct the integers using just sets and natural numbers. In that case you would have to prove that your construction obeys the distributive law. This approach has some advantages. Instead of having to assume that sets exist and that natural numbers exist, that integers exist, that rationals exist, etc. we can just instead assume that one exists (such as sets) and use them to construct all the other objects. But, as I said, there really isn't much point to trying to prove the axioms of the integers when all you want to work with are the integers. So to just prove a(b+c)= ab + ac in general would not be possible? ## The distributive law Originally posted by Ed Quanta So to just prove a(b+c)= ab + ac in general would not be possible? Right. The best you can do is to assume something else which is equivalent to the distributive law, and prove it from that. But if you aren't satisified with assuming the distributive law is true, you probably won't be satisified with assuming something else equivalent. Recognitions: Gold Member Science Advisor Staff Emeritus On the contrary, it is possible to prove the distributive law starting from "Peano's axioms" for the natural numbers. That is basically equivalent to "induction" and all proofs of properties of natural numbers are inductive. Here's a link to a PDF file that contains such proofs: http://academic.gallaudet.edu/course...5688700561c74/\$file/NUMBERS.pdf Click on "Numbers". Originally posted by HallsofIvy On the contrary, it is possible to prove the distributive law starting from "Peano's axioms" for the natural numbers. That is basically equivalent to "induction" and all proofs of properties of natural numbers are inductive. Here's a link to a PDF file that contains such proofs: http://academic.gallaudet.edu/course...5688700561c74/\$file/NUMBERS.pdf Click on "Numbers". Well, we were talking about the integers, not the natural numbers. But my point was that we choose these axioms because they produce distributivity, so we aren't really "proving" the distributive law in a meaningful sense. It's sort of like proving 1+1=2. Of course we can do it, but it isn't really that satisifying. The constructions and definitions we use to prove it were specifically chosen because they give that result. Recognitions: Gold Member Science Advisor Staff Emeritus I see your point but I disagree. The axioms for the counting numbers were not specifically chosen so that they give the distributive law but so that they give "induction" (i.e. "counting"). It happens that they also give the distributive law but that has to be proven by someone! I wouldn't expect every algebra student to do it but it's a nice "five finger exercise" for mathematicians. The properties of the integers are derived from those of the counting numbers specifically to give the group properties (existance of an additive identity, additive inverses) and, again, it happens that the distributive law is true and that has to be proven. By the way, one does not have to prove that "1+ 1= 2" because that is basically how "2" is define. That "2+ 2= 4" is a theorem and has to be proven (by someone). It a simple two or three line proof, of course. Originally posted by HallsofIvy By the way, one does not have to prove that "1+ 1= 2" because that is basically how "2" is define. That "2+ 2= 4" is a theorem and has to be proven (by someone). It a simple two or three line proof, of course. Most (minimal) constructions of N that I have ever seen define 2 as the successor of 1, not as 1+1. Of course, the successor of any number n is n+1, but strictly speaking, 1+1=2 is a theorem. But I guess that depends on the details of how you decide to construct N and define +. I guess you are right in that any construction that provides induction should provide distribution. It is a derivation from a simpler principle and not just a restatement. I would have realized that if I didn't have a temporary rectal-cranial inversion. Recognitions: Gold Member Science Advisor Staff Emeritus Hmm, Yes, 2 is defined as the sucessor of 1 and n+1 is defined as "the sucessor of n" therefore-- Gosh, I just might be forced to agree with you! Originally posted by HallsofIvy Hmm, Yes, 2 is defined as the sucessor of 1 and n+1 is defined as "the sucessor of n" therefore-- Gosh, I just might be forced to agree with you! Actually, I think it's even worse. Addition is usually defined with something like $n+0=n$ and $n+s(m)=s(n+m)$. So you have to use that definition along with $1=s(0)$ to prove that $n+1=s(n)$ first. It is kind of neat sometimes to see some horribly counterintuitive construction turn into the natural numbers we know and love. Thread Tools \| \| \| \| \|-------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: The distributive law \| \| \| \| Thread \| Forum \| Replies \| \| \| Linear & Abstract Algebra \| 6 \| \| \| Introductory Physics Homework \| 0 \| \| \| Current Events \| 25 \| \| \| Linear & Abstract Algebra \| 3 \| \| \| General Math \| 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": 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": 0.963284432888031, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=4254610	Physics Forums Reactive power with electromagnetic sources in free space Good morning, in circuit theory I know that reacting power arise from phasors and represents a power which can't be used, because not delivered to any load, but continuously flows back and forth between the load and the generator with a zero mean during one period. I can't understand very well, anyway, the meaning of this power in a field context, with electromagnetic sources in free space. Let's consider a hertzian dipole, which has several fields component with several dipendence from the distance $r$ (in spherical coordinates). [itex]H_{\varphi} = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{jk}{r} + \frac{1}{r^2} \right) \sin \theta \\ E_r = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{2 \eta}{r^2} + \frac{2}{j \omega \epsilon r^3} \right) \cos \theta \\ E_{\theta} = \displaystyle \frac{I_0 h}{4 \pi} e^{-jkr} \left( \displaystyle \frac{j \omega \mu}{r} + \frac{\eta}{r^2} + \frac{1}{j \omega \epsilon r^3} \right) \sin \theta[/itex] Components proportional to $1/r^2$ and $1/r^3$ are called near field components; components proportional to 1/r are called far-field components. $I_0$ is the phasor of the current in the dipole and $h$ its length. I calculate the power as flux of the Poynting vector through a surface $S$: [itex] P = \displaystyle \oint_S \mathbf{E} \times \mathbf{H}^* \cdot d\mathbf{S} [/itex] where the dipole is in the origin and $S$ is a sphere with radius $r$ centered in the origin too. I find for the power $P$ a real part, which is the power that goes away from the dipole, and an imaginary part, which is the reactive power and is only stored near the dipole. In the waveguide theory, reactive power is carried by modes that cannot propagate, that is, modes that have a purely imaginary propagation constant: they attenuate exponentially along the waveguide. But now the reactive power is carried by a field that propagate, because it has the $e^{-jkr}$ term like the "far field" components: how can it is possible? Moreover, what about the meaning of this reactive power? Books say that it is due to the $1/r^2$ and $1/r^3$ components, which are related to static fields. So, should I state that reactive power is that carried from static fields? A static field stores an energy and I can use this energy if I place a charge in the field, because the field will move the charge, making a work: but this is a sort of active power, a suitable power, isn't it? Thank you for having read and sorry for my confusion. Bye! Emily 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 Recognitions: Gold Member Science Advisor I feel that the term 'reactive power' is not strictly appropriate because no energy is being transferred. I think that the term 'reactive energy' is a better one because it is an energy build up between the inductive and capacitive elements in the structure. It takes a finite time, after switch-on for the waves / oscillating fields to build up - this can be either an antenna or a transmission line feeding into a load, where there are mismatches along the way. When the dipole (or any other antenna structure) is at resonance, the reactive energy is maximum and the power source 'sees' just a resistance (the radiation resistance). Quote by sophiecentaur it is an energy build up between the inductive and capacitive elements in the structure. Thank you! What "elements" can you consider as inductive/capacitive in the free space? Quote by sophiecentaur It takes a finite time, after switch-on for the waves / oscillating fields to build up And I suppose this time is necessary to "charge" the inductors and/or the capacitors. Emily Recognitions: Gold Member Science Advisor Reactive power with electromagnetic sources in free space Quote by EmilyRuck Thank you! What "elements" can you consider as inductive/capacitive in the free space? And I suppose this time is necessary to "charge" the inductors and/or the capacitors. Emily For a line, Z0 =√(L/C) The time I was referring to is the time for the waves to reach a steady state of energy flow in a transmission line - taking into account any reflections there might be at interfaces, due to mis-matches. The 'standing waves' need to establish themselves and this will take a number of journeys along the section of line. The 'charge / discharge' times of reactive elements is included in the transmission equations, I think. I don't know how relevant this is, actually, to your post, now I think about it. Quote by sophiecentaur The are no 'elements' in free space but there is a characteristic impedance for free space of 377Ω. For a line, Z0 =√(L/C) Ok, that's right, we can establish an analogy between transmission lines and free space. Quote by sophiecentaur I don't know how relevant this is, actually, to your post, now I think about it. This is not really about my topic, but is anyway a useful in-depth analysis. Now I would like to know more about the question: reactive power is that carried from static fields? I observed that a static field stores an energy and I can use this energy if I place a charge in the field, because the field will move the charge, making a work. Which is the difference between the energy provided by a static field and the energy provided by an electro-magnetic field in dynamic conditions? Emily Recognitions: Gold Member Science Advisor I don't think the term "reactive" applies to static fields, does it? For a static situation, you will either have a reactance of zero or infinity. It's more 'potential energy' that is involved in that case, I think. Also, does a 'static field' actually carry energy (implies from place to place?) But in your OP, you are referring to changing fields - EM waves associated with a Hertzian dipole. I think my post about transmission lines is, in fact, relevant in as far as the quadrature E and H fields in the near field region must take time to build up to the steady state. These fields store Energy and not Power (you don't store power, by definition because power is rate of energy transfer). Because they exist at a distance from the dipole, they take time to establish themselves. The in-phase components of the field are the ones involved with radiating power. (I don't think this is being too pedantic, is it?) Which is the difference between the energy provided by a static field and the energy provided by an electro-magnetic field in dynamic conditions? I don't think there needs to be any difference - after all, a very slowly varying field is indistinguishable from a static field so where would you draw the line, apart from where QM becomes relevant? Quote by sophiecentaur I don't think the term "reactive" applies to static fields, does it? For a static situation, you will either have a reactance of zero or infinity. You're right. But if we substitute "reactive" with "storing and giving back energy", we can talk also about static fields. Quote by sophiecentaur Also, does a 'static field' actually carry energy (implies from place to place?) A charge in the space is subjected to the field and the field can make (theoretically) everywhere a work on this charge: this is a sort of transfer of energy from the source of the fields to the place where the work is made, so, yes, I think from place to place. But the difference with the dynamic case is that this energy, stored by the field, is used only in the volume where the field lies, whereas the active energy "moves away" from the hertzian dipole. Is it right? Quote by sophiecentaur Because they exist at a distance from the dipole, they take time to establish themselves. This is more and more similar to the transmission line theory with a reactive load: here the reactive load is space and there is a sort of standing wave with one end in the source and the other end in the surrounding space, until the near-field components are significant. Quote by sophiecentaur The in-phase components of the field are the ones involved with radiating power. (I don't think this is being too pedantic, is it?) No, I completely agree! Quote by sophiecentaur I don't think there needs to be any difference - after all, a very slowly varying field is indistinguishable from a static field so where would you draw the line, apart from where QM becomes relevant? My line was $\omega = 0$ for static fields and simply $\omega \neq 0$ for dynamic ones. But now it is not important, because you gave me an excellent point of view: "a very slowly varying field is indistinguishable from a static field". Thank you! Recognitions: Gold Member Science Advisor So we're all happy - excellent! Tags active, antenna, power, poynting, reactive Thread Tools \| \| \| \| \|--------------------------------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Reactive power with electromagnetic sources in free space \| \| \| \| Thread \| Forum \| Replies \| \| \| Electrical Engineering \| 7 \| \| \| Advanced Physics Homework \| 1 \| \| \| Calculus & Beyond Homework \| 1 \| \| \| Electrical Engineering \| 8 \| \| \| Electrical Engineering \| 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": 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": 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": 0.9408690929412842, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/74545?sort=newest	## Approximation to the ratio of a Gaussian CDF to PDF ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Johnstone and Silverman (2005) claimed that for large x $\frac{1-\Phi(x)}{\phi(x)} \approx \frac{1}{x}$ where $\Phi(x)$ and $\phi(x)$ are the CDF and PDF for a normal random variable. I was able to verify the claim numerically. Q: But how would I show this analytically? This seems like it should be easy, but I can't figure it out. Also, Q: Is there a symbolic logic system (e.g., Mathematica) that can generate these sort of approximations? - 1 Hint: integration by parts in the integral for the CDF. This is a very standard trick for finding asymptotic approximations for many kinds of integrals. – Zen Harper Sep 5 2011 at 1:30 @Zen: Ahh, nice trick. Then you generate a power series in $1/x$---the one given by @Robert below. – brianjd Sep 5 2011 at 18:40 ## 4 Answers If you interpret this as the existence of the limit $$\lim_{x \rightarrow \infty} \frac{x(1-\Phi(x))}{\phi(x)}$$ then it is easy to verify using l'Hopital's rule. - @Deane: Ahh yes. Easy. I suspected. :( Second derivatives of numerator and denominator do the trick. – brianjd Sep 5 2011 at 18:19 ### 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. Reproducing a lemma from the classic Feller book, first we can write $$(1-3x^{-4})\phi(x)<\phi(x)<(1+x^{-2})\phi(x).$$ Integrating this from $x$ to $+\infty$, we obtain $$(x^{-1}-x^{-3})\phi(x)<1-\Phi(x)< x^{-1}\phi(x),$$ so you easily get an approximation rate $x^{-3}\phi(x)$, too. - If $Y(x)=(1-\Phi(x))/\phi(x)$, it is easy to check that $Y'(x)=xY(x)-1$ and from this anything you like follows by standard methods. - @Brendan. Then by L'hospital's rule on $Y'(x)$ we get that $\lim_{\infty}Y'(x)=0$ so that $\lim_{\infty}xY(x)=1$ or $Y(x) \approx \frac{1}{x}$ as x becomes large. Thx! – brianjd Sep 5 2011 at 19:01 In Maple: with(Statistics): Phi:= CDF(Normal(0,1),x): phi:= PDF(Normal(0,1),x): asympt((1-Phi)/phi,x,10); $\frac{1}{x} - \frac{1}{x^3} + \frac{3}{x^5} - \frac{15}{x^7} + \frac{105}{x^9} + O\left(\frac{1}{x^{11}}\right)$ See also http://oeis.org/A001147 for the sequence of coefficients - @Robert. VERY useful. I must get Maple (or Mathematica). Thx! – brianjd Sep 5 2011 at 19:03	{"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": 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": 0.8976638317108154, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/71188?sort=newest	Probabilistic (and other mathematical) methods of physics without the physics? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Many of the methods of physics are vastly more general than their use in that discipline. For example, information theory overlaps with a lot of statistical mechanics, and the latter actually developed first. ET Jaynes wrote a famous paper illustrating the connections. However, each is comprehensible without the language and intuition of the other (though I do not deny that a richer understanding comes from knowing both). What other methods of physics (particularly those with a statistical or computational bent) have interpretations (Please mention useful introductory texts!) that are completely physics free? I understand that various field theories meet this criterion; any good non-physics introductions? - 2 I seem to recall that quaternions were originally intended for use in physics. Today they're used for computer graphics. Does that count as physics-free? – Michael Hardy Jul 25 2011 at 6:10 2 You might enjoy Michel Talagrand's introductory article on spin glasses: people.math.jussieu.fr/~talagran/spinglasses/… – Simon Lyons Aug 22 2011 at 18:52 7 Answers I found Complexity and Criticality by Christensen and Moloney to by quite excellent. It gives a much more computational approach to the percolation phase transition, the ising phase change and issues of self organized criticality (via the sand pile and rice pile model). As a computer science student, I found this book to be invaluable for my work on phase transitions of NP-Complete problems. - 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. Statisticians use Tracy-Widom laws in the new developments in random matrix theory. The asymptotic behavior of spectra of random matrices was understood (by statisticians) back in the 1960s when the rows of the matrix are independent and identically distributed row-vectors from a fixed distribution (most importantly, with a fixed dimension), and produced asymptotically normal/Gaussian laws typical for the Central Limit Theorem and its generalizations. Tracy-Widom laws apply to matrices in which both row and column dimensions are allowed to grow to infinity (proportionally to one another). See e.g. doi:10.1214/aos/1009210544. - This was a "comment", but I may as well make it an "answer": Quaternions were intended for use in physics. Today they are used in computer graphics. - This is interesting because I originally learned about Quaternions solely in their context as an interesting mathematical construct, a non-commutative algebra of historical interest. – DoubleJay Jul 29 2011 at 16:47 Hamiltonian Monte Carlo/Hybrid Monte Carlo (HMC) uses Hamiltonian dynamics to construct MCMC algorithms for sampling from complicated probability distributions. It is quite useful in Bayesian statistics. - Percolation (several kinds), Ising model and other probabilistic models on arbitrary (transitive) graphs came from physics and are purely mathematical theories now. - Distilled to its linear algebraic core, Quantum Computing can be presented and understood in a completely physics-free way easily accessible to Computer Scientists. Two papers taking this point of view are Fortnow and Fenner. Driving this linear algebraic point of view even further, one can see multilinear algebra, which deals with the contraction of tensor networks as a core concept, as fundamental. It suffices to model quantum computing, simulation of quantum systems (Projected Entangled Pair States, or PEPS), statistical mechanical models (partition functions), etc. A good text to get started is probably this. Tensor networks have also an intuitive yet precise graphical calculus as explored by Bob Coecke, et al. which abstracts manipulations of tensor networks to operations in compact closed monoidal categories. - Euclidean geometry was originally supposed to model physical space. Now suppose $\overline{X} = (1/n)\sum_{i=1}^n X_i$ and $S^2 = (1/(n-1))\sum_{i=1}^n(X_i - \overline{X})^2$ are the sample mean and sample variance of an i.i.d. sample from a normally distributed (or "Gaussian") population. How do we know that $\overline{X}$ is probabilistically independent of $S^2$? Maybe the quickest way is via two facts from Euclidean geometry: (1) the two mappings $(X_1,\ldots,X_n) \mapsto (\overline{X},\ldots,\overline{X})$ and $(X_1,\ldots,X_n) \mapsto (X_1-\overline{X},\ldots,X_n-\overline{X})$ are complementary orthogonal projections, and (2) the probability distribution of $(X_1,\ldots,X_n)$ is spherically symmetric, i.e the density depends on the coordinates only through the sum of their squares. -	{"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": 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": 0.9341549277305603, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/105012/if-one-sequence-is-convergent-so-is-the-other-one	# if one sequence is convergent, so is the other one [duplicate] Possible Duplicate: If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$ I have a question, can anyone help me? Let $\{ a_i\} _{i=1} ^ { + \infty }$ be a sequence of positive real numbers such that for every sequence $\{ b_i\} _{i=1} ^ { + \infty }$ of positive real numbers satisfying the condition $\sum {b_n ^2} <+ \infty$ we have $\sum {a_n b_n} < + \infty$ . Prove that $\sum {a_n^2} < + \infty$. It appeared in Iran's 3rd round Olympiad exam 2009, but I think it's a well-known result. - 1 – Nate Eldredge Feb 2 '12 at 18:19 ## marked as duplicate by Davide Giraudo, Aryabhata, JavaMan, Nate Eldredge, DidFeb 2 '12 at 19:07 This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question. ## 1 Answer For any $x \in \ell^2$, $\sum_n a_n x_n$ converges absolutely, and therefore converges. Now apply the Uniform Boundedness Principle to the sequence of "partial sum" functionals $x \to \sum_{n \le N} a_n x_n$, together with the Riesz-Fischer theorem that identifies bounded linear functionals on $\ell^2$ as members of $\ell^2$. If you consider all this too functional-analytic for an Olympiad problem, a more elementary way would be this. If $\sum_n a_n^2 = \infty$, consider an increasing sequence $N_j$ such that $N_1 = 1$ and $c_j = \sum_{n=N_j}^{N_{j+1}-1} a_n^2 > 1$. Define a sequence $b_n$ by $b_n = a_n/(j \sqrt{c_j})$ for $N_j \le n < N_{j+1}$. Note that $\sum_{n=N_j}^{N_{j+1}-1} b_n^2 = 1/j^2$ and $\sum_{n=N_j}^{N_{j+1}-1} a_n b_n = \sqrt{c_j}/j > 1/j$, so $\sum_n b_n$ converges but $\sum_n a_n b)n$ diverges. -	{"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": 24, "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": 0.9301471710205078, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/48779/computing-an-integral-where-the-poles-of-the-integrand-are-the-roots-of-unity	# Computing an integral where the poles of the integrand are the roots of unity I am trying to compute the following integral: $$\int_0^{2\pi} \frac{y}{y^n-1} dy$$ I've tried to decompose $y^n-1$ into $(y-1)(y-e^{i\theta})(y-e^{i2\theta})...(y-e^{i(n-1)\theta})$ but I don't know what to do with this factorization. I've read some others similar questions with the answers but I don't know if the same methods apply. - You are integrating on an interval and I am not sure what the roots of unity have do with anything. Also, the integral doesn't exist. – SteveH Jun 30 '11 at 22:24 That is weird. I was originally trying to integrate $\int_{S^1} dz/(z^n-a)$ and that's how I got this integral (I didn't include the factor $a^{2/(n-1)}$ that I got from the substitution I did for simplicity). $S^1$ is the unit circle. – user786 Jun 30 '11 at 22:36 OK. Assume $\|a\|\not=1$. Treat the cases $\|a\|>1$ and $\|a\|<1$ separately. One case is easy. Try the other case by thinking Cauchy's Integral formula, small circles around the poles and a suitable contour that incorporates these small circles and the unit circle. – SteveH Jun 30 '11 at 22:44 Thank you! I'll come back if I still can't make substantial progress. – user786 Jun 30 '11 at 22:47 2 The integral in your question isn't the integral you want. You want to replace $y$ with $z = e^{iy}$ in the integrand or something similar. – Qiaochu Yuan Jun 30 '11 at 23:23 show 2 more comments ## 2 Answers Let $f(z) = \frac{z}{z^n - 1}$. You're looking for $\oint_{\gamma} f(z) dz$ taken over a contour that is outside of the unit circle, loosely speaking. We will use the Residue Theorem to compute the integral. Define $\omega_k = e^{2 \pi i k / n}$ for $k = 0, \dots, n-1$ to be the roots of unity, which are also the poles of $f(z)$. We calculate the residues using the factorization $z^n - 1 = \prod_{k=0}^{n-1} (z-\omega_k)$ as follows: $$R_k = \lim_{z\to \omega_k} (z-\omega_k)f(z) = \omega_k \prod_{j \ne k} (\omega_k - \omega_j)^{-1} = \omega_k \prod_{l = 1}^{n-1} \left[ \omega_k^{-1} ( 1 - \omega_l)^{-1} \right] = \omega_k^2 R_0 .$$ Note that the above uses $\omega_k^{-n} = 1$ and the index substitution $l = j-k$. We could use the geometric formula to calculate $R_0 = 1/(n-1)$, but it's actually unnecessary because we know that $\sum \omega_k^2 = 0$ (for $n > 2$), since it is the $z^2$ coefficient of the polynomial $z^n-1$. Hence the integral vanishes. - Thanks for your answer. It may be trivial but why is the sum of the squares of the nth roots equal to 0? – user786 Jul 1 '11 at 20:10 To that end: recall the connection between the zeros of a polynomial and its coefficients. If $$p(z)=(z-z_1)(z-z_2)\cdots(z-z_n)=z^n+a_1z^{n-1}+a_2z^{n-2}+\cdots+a_n,$$ then the sum of the zeros is the negative of $a_1$ et cetera. The sum of squares is a combination of the two highest coefficients. Not sure that you need to look at that, though? – Jyrki Lahtonen Jul 1 '11 at 21:56 Hint: Let $\xi_k, k=1,2,\ldots,n,$ be any one of the zeros of $z^n-a$, and $f(z)=1/(z^n-a)$, so $\xi_k$ is a simple pole of $f(z)$ (assuming that $a\neq0$). Then by l'Hospital $$Res(f(z),\xi_k)=\lim_{z\to\xi_k}\frac{z-\xi_k}{z^n-a}= \lim_{z\to\xi_k}\frac{1}{nz^{n-1}}=\frac{1}{n\xi_k^{n-1}}=\frac{\xi_k}{n\xi_k^n}=\frac{\xi_k}{na}.$$ - Thanks for the hint. That's what I wanted to do but which poles are located in the upper half plane? – user786 Jul 1 '11 at 20:09 Why worry about the upper half plane? Aren't you integrating along the unit circle? Read the comment by SteveH. – Jyrki Lahtonen Jul 1 '11 at 20:53	{"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": 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": 0.3, "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": 0.9449338912963867, "perplexity_flag": "head"}
http://mathoverflow.net/questions/65068?sort=votes	## Why is the base manifold of a Lie groupoid required to be second-countable? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I wonder why one requires that the base manifold of a Lie groupoid is second-countable? - 1 Isn't second-countability in the definition of manifold? en.wikipedia.org/wiki/… – Zev Chonoles May 15 2011 at 20:41 2 Usually manifolds tout court are required to be second countable. – Mariano Suárez-Alvarez May 15 2011 at 20:42 2 Sorry, I should state this question more carefully. Of course, Zev Chonoles and Mariano Suarez-Alvarez are right: the usual definition of a manifold requires second-countability and Hausdorff and locally euclidean. My question should merely be: At which point in the theory of Lie groupoids does one really need that the base manifold is second-countable? When constructing a Lie groupoid from a foliation one actually has to be a bit careful at this point. If one takes uncountably many charts the base manifold of the Lie groupoid won't be second-countable. – Dave Lewis May 15 2011 at 20:57 2 @Dave Lewis: Can I request that you edit your question to include your comments above? (Mark the edit as an edit, so that @Zev and @Mariano 's comments still make sense.) It sounds like you have a more specific direction that you're thinking about, and in any case clearly recognize that "When constructing a Lie groupoid from a foliation one actually has to be a bit careful at this point", for example. I do not know of a good reason to have questions on MO that are only one sentence long, and there are many good reasons for including a few paragraphs. – Theo Johnson-Freyd May 15 2011 at 22:50 2 @Zev, Mariano, and Dave: If you require manifolds to be second countable, then a disjoint union of manifolds is not always a manifold. Replacing second countability by paracompactness allows you to keep all good properties of second countable manifolds and makes the category of manifolds closed under coproducts, which seems like a good property to have. – Dmitri Pavlov May 16 2011 at 4:19 show 2 more comments ## 1 Answer Answer #1: There is no real reason for imposing that the base manifold of a groupoid be second countable. Answer #2: You lose some desirable properties if you don't impose second countability: For example, without it, the homotopy type of the geometric realisation of the nerve will no longer be an invariant of the Morita equivalence class of the groupoid. - Re Answer #2: Weird! I would have expected that the homotopy type of the nerve was a well-defined invariant for any topological groupoid, and that the construction should factor through forgetting from Manifolds to Homotopy Types. Could you either explain more, or include a reference? – Theo Johnson-Freyd May 15 2011 at 22:52 @Theo: I take my favourite non-second countable manifold: the long line $L$, and I look at the cover consisting of all of its bounded connected open subsets. The corresponding Cech groupoid is Morita equivalent to $L$. There is an obvious projection from the geometric realization of the Cech groupoid back to $L$. But there is no section of that map: that's because the cover does not admit partitions of unity. More generally, you can show that the projection does not admit a homotopy inverse. – André Henriques May 15 2011 at 23:12 Ah, Andre - that is why you redefine Morita equivalence not to use 'local sections', but 'local sections wrt a numerable cover'. This class of weak equivalences of topological/Lie groupoids (take your pick) is closer to what people think of when they restrict to paracompact spaces. – David Roberts May 16 2011 at 0:35	{"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": 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": 0.9230931401252747, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/182198/given-a-banach-algebra-a-and-elements-x-y-in-a-if-x-and-xy-are-inverti	# Given a Banach Algebra $A$ and elements $x,y \in A$. If $x$ and $xy$ are invertible, then so is $y$. A silly question, but I don't see the answer. This question is from Rudin's Functional Analysis, Chapter 10, Exercise 1a). It's obvious that $y$ has a left inverse as $((xy)^{-1}x)y=e$, where $e$ is the unit of the algebra. However, starting with $y((xy)^{-1}x)$, I cannot proceed further. Does this suggest that I need to 'venture beyond' the multiplicative group structure, maybe introduce ideas from the ring structure on the Banach Algebra? ( I'm thinking of something analogous to the proof where $e-xy$ is invertible if and only if $e-yx$ is invertible. There, if $z$ is the inverse of $e-xy$, then, $e+yzx$ is the inverse of $e-yx$.) - 1 Hint: a product of two invertible elements is certainly invertible. Given the information you have, can you write y as the product of two invertible elements? – Manny Reyes Aug 13 '12 at 20:28 ## 2 Answers I am guessing you meant Exercise 2(a)? Let $p = y((xy)^{-1}x)$, then $xp = xy((xy)^{-1}x) = x$, since $x$ is invertible, it follows that $p=e$. - The inverse of $x^{-1}xy$ is $(xy)^{-1}x$, since $x^{-1}xy\cdot (xy)^{-1}x=e$ and $(xy)^{-1}xx^{-1}xy=e$. Hence $y$ is invertible. -	{"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": 24, "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": 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": 0.9353256821632385, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/118136/serge-langs-remarks-on-the-superiority-of-algebra-what-it-actually-means?answertab=oldest	Serge Lang´s remarks on the superiority of algebra. What it actually means? [closed] I read two comments of Lang that basically places algebra over other math subjects. One of this comments is on his calculus book preface (see Remark 1 below); I am not finding his other comment, but it was an interaction he had with someone at Yale´s math department coffee break and is written somewhere. It basically says that algebra is superior to any other math subject, if I recall it correctly. My problem with his comments is that I have no idea what he is talking about. They baffle me. I suppose I stand in the exact opposite from his viewpoints. For me, you can´t compare the applicability and importance of analysis and differential equations to that of modern algebra. Hence: 1) Is there an article of Lang explaining in detail his viewpoints? 2) or, do you know what is his point? Remark 1) On the preface of his Calculus book, Serge Lang basically says that he thinks bright students may benefit more in studying abstract algebra before or at the same time they learn calculus. My book is in portuguese, so I give a rough translation: " when I was a student I didn´t like calculus nor analysis. I probably woundn´t like this book either... [today I think] that calculus and analysis are overestimated, with a loss to algebra, mainly because of historical accidents." He goes on to say that a beginner course in algebra should consist in a study of vector spaces and groups, that this is independent of calculus and has important applications to other fields, and that some people prefer this material over calculus. He also says that there is no reason someone should be forced to study calculus before algebra. This being true specially for the most talented students. Remrak 2) "I remember one time when I was a grad student, I was standing next to him at tea while he was explaining to a first-year that analysis is just “number theory at infinity”. I said Come on, that’s not true. He immediately turned up the volume, challenging me to stop bullshitting and give an example. I said OK, p-adic analysis, and then walked away. But I’ve always wished I had stayed to see what his reaction would have been. We need more trouble makers like him. ", from the site Not Even Wrong. - 8 It's an interesting question. But nothing you have quoted by Lang gets anywhere near claiming the superiority of algebra. Rather, he is questioning the primacy of analysis in the mathematical curriculum and the fact that one must wade through it to get to other branches of mathematics. From my pure mathematician's perspective, I agree that this seems to be largely due to historical inertia: one could equally well learn algebra first, or some topology, or number theory, or... (Like many other people, I did learn some number theory before I studied university-level mathematics.) – Pete L. Clark Mar 9 '12 at 6:37 What I wrote is perhaps very long and you missed that the "superiority" part was expressed in his Yale common room exchange. Unfortunately, I did not find it. – Espinho da Flor Mar 9 '12 at 6:42 3 $@$Espinho: I am responding to what you directly quoted. If you are going to ask a question about what someone (especially no someone no longer living) said, it seems only reasonable to supply a direct reference. Otherwise we're just trafficking in rumors and hearsay. – Pete L. Clark Mar 9 '12 at 6:44 8 I would advise extreme caution about "philosophical" comments by celebrated mathematicians. Some of them, like Lang and Arnold, delight in controversies and provocative statements. I've heard both make outrageous claims that didn't strike me as particularly convincing, to put it mildly. Fortunately other mathematicians show remarkable restraint and totally abstain from pontificating, Deligne being the ultimate example of such wise behaviour. – Georges Elencwajg Mar 9 '12 at 8:31 1 @Espinho: Your title is Serge Lang's remarks on the superiority of algebra. But you haven't given any documentation of this whatsoever. You've only alluded to a coffee break conversation at which you were not present: that's pretty much the definition of hearsay. Further, it is not really clear to me that insight into a deceased mathematician's mental math map is on topic for this site: how are the rest of us to judge the correctness of such an answer? Finally, as to what Lang thought...he had (and documented) a lot stranger thoughts than topology before analysis, to be sure. – Pete L. Clark Mar 13 '12 at 5:22 show 4 more comments closed as not constructive by Pete L. Clark, lhf, Benjamin Lim, Asaf Karagila, Willie Wong♦Mar 14 '12 at 13:39 As it currently stands, this question is not a good fit for our Q&A format. We expect answers to be supported by facts, references, or specific expertise, but this question will likely solicit debate, arguments, polling, or extended discussion. If you feel that this question can be improved and possibly reopened, see the FAQ for guidance. 3 Answers I don't know about the remark at Yale, but as for his comment in the preface I agree 100%. I think you are misunderstanding his remark though, as his point seem to be mostly pedagogical and not asserting the superiority of algebra over analysis. It is true that in many departments, the first year of an undergraduate program is largely calculus, and serious algebra is left until later. For example at the University of Sydney, in first year the focus is on Differential and Integral Calculus being done with more rigor and topics than in high school, and other than that there is Statistics, Discrete Math and "Linear Algebra" which is actually a course in basic matrix algebra. The axioms of a vector space never appear. This is not a dig at the department here, it is simply a statement of the reality of things - universities often leave serious algebra until after calculus because the majority of first year students could not handle the new level of abstraction so quickly, and also many of them never intend to - most people in the class intend to be majors in other scientific fields. This comes at a disadvantage to the students that could handle it, and indeed would thrive. Algebra has a distinctly different flavor to analysis (not always true, but certainly calculus vs group theory). Some students are more inclined to thinking algebraically, and some students are adept at thinking in both mindsets. For these students, a beginners course in algebra would go a long way in expanding their mathematical horizons and may increase their interest in mathematics. Lang's remark is simply noting that it is a pity that these students don't receive this. - Serge Lang is simply expressing his disappointment with universities pushing for calculus courses as the first mathematics courses available for first year students. In fact at my university the first serious proof based algebra course is not till the second semester of second year. Linear algebra is taught in first year but alas - computing eigenvalues, taking determinants and row reducing (though important techniques to know) I believe do not show students the power of linear algebra. It was not till I did a reading course in algebra (of which required a lot of linear algebra) that I absolutely loved it. I realised that with an operator on say a complex vector space $V$, one can decompose $V$ as a direct sum of invariant subspaces. If these are cyclic, you have Jordan Canonical Form. This for me was true power. However I still had to go through classes where we learned 10 different methods to evaluate integrals and another 15 methods on solving differential equations. I think Lang's remark can equally be applied to the case where students are exposed to cookbook clases( matrix algebra or calculus). As Ragib has already mentioned, Serge Lang is just simply saddened that the more talented students have to endure these classes (calculus or matrix algebra) as their first ones at university. Their is not mention of the superiority of Algebra over Analysis. I don't think from what he provided did he ever claim that say Commutative Algebra was more powerful than Functional Analysis. - Serge Lang had a number of unusual beliefs that he would promote... he even believed that the HIV virus doesn't cause AIDS and was very vocal about it. Thinking that abstract algebra should be covered at the same time as calculus is relatively minor in comparison, assuming that's what he believed (it's not 100% clear to me from your comments.) Here we have someone whose research and way of thinking was highly algebraic, and in addition like many mathematicians he was, shall we say, somewhat out of touch with reality. So it's not too surprising he might think abstract algebra was an appropriate subject to teach college freshmen along with calculus. He may not have realized how much easier calculus is to grasp for normal people than abstract algebra is. I don't know if this means he thought algebra is "superior". -	{"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": false, "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": 0.9735528230667114, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/3618/can-maxwells-equations-be-derived-from-coulombs-law-and-special-relativity/3674	# Can Maxwell's equations be derived from Coulomb's Law and Special Relativity? As an exercise I sat down and derived the magnetic field produced by moving charges for a few contrived situations. I started out with Coulomb's Law and Special Relativity. For example, I derived the magnetic field produced by a current I in an infinite wire. It's a relativistic effect; in the frame of a test charge, the electron density increases or decreases relative to the proton density in the wire due to relativistic length contraction, depending on the test charge's movement. The net effect is a frame-dependent Coulomb field whose effect on a test charge is exactly equivalent to that of a magnetic field according to the Biot–Savart Law. My question is: Can Maxwell's equations be derived using only Coulomb's Law and Special Relativity? If so, and the B-field is in all cases a purely relativistic effect, then Maxwell's equations can be re-written without reference to a B-field. Does this still leave room for magnetic monopoles? - I have a vague recollection of, when I was in high school, finding a book that did undergrad E&M by assuming SR is correct from the beginning and doing something like this. I don't recall the title, though (or if it was any good), but if you want to see this worked out in detail you might try to look for it? – Mr X Jan 22 '11 at 18:05 6 @Jeremy -- The book you're thinking of is probably Electricity and Magnetism by E. Purcell (part of the Berkeley Physics series). A very good book, by the way. – Ted Bunn Jan 22 '11 at 18:32 Yes! I believe it is the one I was thinking of. – Mr X Jan 22 '11 at 18:33 Yes, I used this book in College. Very good book. But it didn't go "all the way" and derive Maxwell's equations (If I recall correctly). – user1247 Jan 22 '11 at 18:58 2 You also need the assumption that charge is a scalar and same-charges repel. Then the derivation is contained in Purcell's EM book. – Ron Maimon Jun 15 '12 at 2:06 ## 9 Answers Maxwell's equations do follow from the laws of electricity combined with the principles of special relativity. But this fact does not imply that the magnetic field at a given point is less real than the electric field. Quite on the contrary, relativity implies that these two fields have to be equally real. When the principles of special relativity are imposed, the electric field $\vec{E}$ has to be incorporated into an object that transforms in a well-defined way under the Lorentz transformations - i.e. when the velocity of the observer is changed. Because there exists no "scalar electric force", and for other technical reasons I don't want to explain, $\vec{E}$ can't be a part of a 4-vector in the spacetime, $V_{\mu}$. Instead, it must be the components $F_{0i}$ of an antisymmetric tensor with two indices, $$F_{\mu\nu}=-F_{\nu\mu}$$ Such objects, generally known as tensors, know how to behave under the Lorentz transformations - when the space and time are rotated into each other as relativity makes mandatory. The indices $\mu,\nu$ take values $0,1,2,3$ i.e. $t,x,y,z$. Because of the antisymmetry above, there are 6 inequivalent components of the tensor - the values of $\mu\nu$ can be $$01,02,03;23,31,12.$$ The first three combinations correspond to the three components of the electric field $\vec{E}$ while the last three combinations carry the information about the magnetic field $\vec{B}$. When I was 10, I also thought that the magnetic field could have been just some artifact of the electric field but it can't be so. Instead, the electric and magnetic fields at each point are completely independent of each other. Nevertheless, the Lorentz symmetry can transform them into each other and both of them are needed for their friend to be able to transform into something in a different inertial system, so that the symmetry under the change of the inertial system isn't lost. If you only start with the $E_z$ electric field, the component $F_{03}$ is nonzero. However, when you boost the system in the $x$-direction, you mix the time coordinate $0$ with the spatial $x$-coordinate $1$. Consequently, a part of the $F_{03}$ field is transformed into the component $F_{13}$ which is interpreted as the magnetic field $B_y$, up to a sign. Alternatively, one may describe the electricity by the electric potential $\phi$. However, the energy density from the charge density $\rho=j_0$ has to be a tensor with two time-like indices, $T_{00}$, so $\phi$ itself must carry a time-like index, too. It must be that $\phi=A_0$ for some 4-vector $A$. This whole 4-vector must exist by relativity, including the spatial components $\vec{A}$, and a new field $\vec{B}$ may be calculated as the curl of $\vec{A}$ while $\vec{E}=-\nabla\phi-\partial \vec{A}/\partial t$. You apparently wanted to prove the absence of the magnetic monopoles by proving the absence of the magnetic field itself. Well, apologies for having interrupted your research plan: it can't work. Magnets are damn real. And if you're interested, the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. In particular, two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will inevitably possess the (opposite) magnetic monopole charges. The lightest possible (Planck mass) black holes with magnetic monopole charges will be "proofs of concept" heavy elementary particles with magnetic charges - however, lighter particles with the same charges may sometimes exist, too. - 1 Your previous answer was only unclear in that it does not address why I am able (in a few contrived examples, admittedly) to solve for the equations of motion without reference to a B-field. All I need to do is show how the E-field transforms under Lorentz boosts, and I can do that without introducing a B-field. Did I not do those examples correctly, or are they lucky exceptions because they are contrived? – user1247 Jan 22 '11 at 18:20 4 I read your answer thoroughly, and you are still not answering my questions. When I do my examples, nowhere must I postulate a "new field". I merely start with Coulomb's law and SR, and do the math, and the math shows that a particle experiences forces that can be effectively described by a "new field". This is analogous to the Coriolis force. Does gravity plus a rotating reference frame imply a new "Coriolis field"? Of course not, but it can be effectively described by one. – user1247 Jan 22 '11 at 18:44 6 Lubos answer is very good and very precise. I fully subscribe. I am only a bit puzzled by the last paragraph of the answer, where he says that the existence of magnetic monopoles is inevitable in any consistent theory of quantum gravity. The argument given is that two poles of a dumbbell-shaped magnet may collapse into a pair of black holes which will "inevitably possess the (opposite) magnetic monopole charges". If I break a magnet, I do not get two monopoles, of course. I get two magnets. What would prevent the two black holes to do exactly the same, and behave like two magnets, without – Carlo Rovelli Jan 27 '11 at 8:07 3 ...any monopole around? (that's the rest of Dr. Rovelli's comment, which was cut off by the system) – David Zaslavsky♦ Jan 27 '11 at 8:09 2 – Luboš Motl Jan 27 '11 at 10:50 show 9 more comments Lubos Motl's answer is very good, but I think it's worth saying one or two additional things. You can regard magnetism as simply a byproduct of electricity, in the following sense: if you assume that Coulomb's Law is correct, and that special relativity is correct, and that charge is a Lorentz scalar (so that charge and current density form a 4-vector), then you can derive all of Maxwell's equations. (Actually, you probably also need to assume the theory is linear as well, now that I think about it.) The undergraduate-level textbook by Purcell works this out very explicitly in a nice, pleasing way, and it's also in more advanced textbooks. Some books gloss over the need to postulate that charge is a scalar. At least one textbook -- I don't remember which -- does emphasize it, and makes a convincing case that it's worth paying attention to. One way to see that it's not a trivial condition to impose is to consider the analogy with gravity -- that is, substitute mass for charge and gravity for electric field, and try to run the same argument. (Assume weak fields so that everything can be treated as linear if you like.) There are "gravitomagnetic" effects, but they're not related to regular gravity in the same way as the magnetic field is related to the electric field -- i.e., the gravitational analogues of Maxwell's equations look different from the regular Maxwell equations). One reason is the sign differences, of course -- like charges repel in one case and attract in the other. But a bigger reason is that the source of gravity is not a scalar: its density doesn't form part of a 4-vector, but rather of a rank-2 tensor. But on a more philosophical (or perhaps semantic) level, I wouldn't jump from this fact to the conclusion that magnetism is "merely" a byproduct of electricity. At the very least, such language doesn't appear to be useful in understanding the theory or in using it! For instance, understanding how an electromagnetic wave can propagate from a distant galaxy to your eye is much easier and more natural if you look at it from the "usual" point of view. - Thanks Ted. So, if, as you say, you can derive all of Maxwell's equations as a byproduct of electricity, it seems to follow trivially that one can write Maxwell's equations without reference to a B-field. (Just as I can write, as in the exercise I described, the forces due to a current I without reference to a B-field). This is my question that Lubos seems to refuse to want to address. I understand that this doesn't change the physics, and may not be the most parsimonious way of expressing electromagnetism -- I'm just interested if it can and has been done. – user1247 Jan 23 '11 at 1:39 5 Yes, it can be done. With sufficient effort, you can go further and express all of electricity and magnetism without reference to either an E or B field -- just as a very strange and complicated force law between charges, in which the force on each charge depends on the properties of the other charge at the retarded time. Griffiths's textbook writes the force law out explicitly in one of the later chapters. You give up a lot by doing this -- the biggest thing that comes to mind is that I have no idea how you'd even try to talk about energy-momentum conservation in this language. – Ted Bunn Jan 23 '11 at 16:03 by Hans de Vries (): The simplest, and the full derivation of Magnetism as a Relativistic side efect of ElectroStatics He uses only Electrostactic field and the non-simultaneity to obtain the Magnetic Field. He does explain it better than Purcell. Magnetic field is a side effect of movement in the electric field. () Hans de Vries has a very interesting online book (not yet finished) in his site, and he offers another pearl, not related to this post, but I feel compelled to share: The Lorentz contraction is a real effect and not only 'a referential effect' as we are tempted to believe. - @JxB I can not comment your answer and quoting "To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone." – Helder Velez Feb 8 '11 at 15:50 @JxB continuing previous comment (trouble with Enter key versus ShiftEnter, sorry) electric field = light One can not dissociate electric field from light. "at c speed " here en.wikipedia.org/wiki/Electric_field electr 28 times here: en.wikipedia.org/wiki/Photon_polarization explore the Radiation2D.exe from here www-xfel.spring8.or.jp I do not have any doubt that electric field,and gravity,propagates at c speed. – Helder Velez Feb 8 '11 at 16:15 @JxB quoting "arising from time-varying electric field, and not a result of the motion of electric charge." Motion is a relative concept and a 'time-varying' electric field is always resultant of charges in motion. – Helder Velez Feb 8 '11 at 16:35 – Helder Velez Feb 8 '11 at 18:10 With Coulomb's law and special relativity you can derive Ampere's law, which gives you magnetostatics. What's missing for electrodynamics is the displacement current ($\frac{1}{c^2} \frac{\partial E}{\partial t}$), which is a source of magnetic field arising from time-varying electric field, and not a result of the motion of electric charge. Relativity has only two postulates: 1. The laws of physics are the same in all inertial reference frames 2. All inertial observers measure the same speed for light in vacuum. Relativity, by itself, does not mandate that electric fields (or electric potential for that matter) must travel at the speed of light. To derive the Maxwell equations, you need an additional postulate, and that is provided by the wave equation (for electric potential) in Section 4 of the reference in Helder's answer. Without this additional postulate (that changes in electric potential propagate at the speed of light), you cannot derive the displacement current from Coulomb's law and relativity alone. - I know that Purcell and others have used Lorentz symmetry as a pedagogical device to motivate the introduction of magnetic fields, but I do not recall ever having seen an axiomatic derivation of Maxwell's equations. It might be an interesting exercise to see precisely what assumptions beyond Lorentz symmetry and Coulomb's Law are necessary to reconstruct Maxwell's equations. B fields are not fictitious fields If you know the electric and magnetic fields in one inertial frame, you can determine the electric and magnetic fields in any other frame via Lorentz transformation. If the magnetic field happens to vanish in a given inertial frame, you could think of magnetic effects in other frames as fictitious. However, it is not always possible to find a frame in which the magnetic fields vanish. The fastest way to see this is to note that E^2 - B^2 c^2 is a Lorentz invariant quantity (see Wikipedia). If we find that B^2 > E^2/c^2 at a given spacetime point in a given inertial frame, it follows that B^2 > 0 at that point in all inertial frames. In fact, you could begin in a frame where the electric field vanishes but the magnetic field does not; the electric fields observed in other frames could then be considered fictitious. In general, neither the electric field nor the magnetic field can be made to vanish under a Lorentz boost. To see this quickly, note that the dot product of the E field vector with the B field vector at a given spacetime point is a Lorentz invariant quantity (see Wikipedia). If this dot product is nonzero at a given spacetime point in a given inertial frame, the electric and magnetic field vectors will both be nonzero at that spacetime point in all inertial frames. As Einstein pointed out, you can understand the motion of a charged particle by referring to the electric field in the rest frame of that particle. However, if you have multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle. Since Lorentz boosts mix the E field with the B field, the only way to keep track of the E field in the rest frame of each of your particles in terms of local quantities in one inertial frame is by reference to the E field and the B field. Locality Even if it is possible, it is not clear to me that it would be desirable to use Coulomb's law as an axiom in electromagnetic theory. Maxwell's equations explain the motion of particles by referring to local degrees of freedom, the fields. Coulomb's law, on the other hand, is a form of action-at-a-distance, and is manifestly non-local. It is certainly possible to rewrite both the E and B fields in terms of integrals over charge density and current density (I can't post another link, so google "Jefimenko's equations"), and then to use these expressions to interpret electromagnetic forces as a form of retarded action-at-a-distance. However, to obtain these expressions requires assumptions about the boundary conditions on the E and B fields. We can always obtain another valid solution of Maxwell's equations by simply changing the boundary conditions on the fields, which demonstrates that the fields have independent existence, and are not mere book-keeping variables to simplify a more fundamental non-local interaction. Monopoles As usually written, Maxwell's equations do not contain terms corresponding to magnetic charge, but it would be consistent to add such terms. In fact, Dirac showed that the quantization of electric charge could be due to the existence of magnetic monopoles (I can't post another link, so google "magnetic monopole dirac quantization condition"). Maxwell's equations do not tell us whether magnetic monopoles exist or could exist, but the quantization of electric charge could be evidence that magnetic monopoles exist somewhere in the universe. - In my answer I linked a paper by Hans de Vries where he did as user1247 is saying, and you can check his validity. In motionmoutain ch 18 - Motion in GR, we find also that GravitoElectric is a fundamental field and the GravitoMagnetic is a relativistic effect by the same reason. The motion induces it. To me is a force and not a 'field' (there is no Coriolis field, but force). How the particle can do this? "multiple particles with different velocities, you need to keep track of the electric field in the instantaneous rest frame of each particle" and 'in advance'? – Helder Velez Mar 8 '11 at 3:39 Not a direct answer to your question but still a surprising derivation of Maxwells equations: Feynman's proof of the Maxwell equations (FJ Dyson - Phys. Rev. A, 1989) shows, that it is possible to derive Maxwells equations from Newtons second law of motion and the uncertainty principle. - No, you can't. For several reasons. First, if you have E, to get the B field, you need additional assumptions about the structure of the theory, ie in more detail the field strength tensor, see above reply by Lubos. But in addition to this, even if you had the solution for a point charge, to get Maxwell's equations you need to know more than just having one solution. For example that they're linear, second order, and what the symmetry group is. And if you've added that, you can derive the Maxwell equations from these assumptions anyway without even starting with the Coulomb field. - 1 I have to agree that some extra assumptions are needed. – Philip Gibbs Jan 23 '11 at 14:45 You cannot. B is not just a relativistic side-effect of E. Jackson, Electrodynamics, Section 12.2 has a nice discussion, in which he refutes the "proofs" given in some undergraduate texts. "The confusion arises chiefly because the Lorentz transformation properties of the force are such that a magnetic-like force term appears when the force in one inertial frame is expressed in terms of the force in another frame. It is tempting to give this extra force term an independent existence and so identify the magnetic field as a separate entity. But such a step is unwarranted without additional assumptions." Jackson goes on to exhibit an explicit counterexample, based on a Lorentz scalar potential. This field looks like electrostatics (or even Newtonian gravitation!) in the non-relativistic limit. It also has "an apparent magnetic-like force. But there is no independent entity B." So in this "theory" B is indeed only a relativistic effect, but this theory does not apply to Nature. - Yes, you can make it, but you also need to use a superposition principle. 1. You determine that Couloms's law, $$\mathbf F = \frac{qQ\mathbf r}{\|\mathbf r \|^{3}},$$ is a boundary case of the relativistic force, which acts on the charge q by the field of a Q-charge. 2. Using Lorentz transformation for the force and for the radius-vector, $$\mathbf F = \mathbf F' + \gamma \mathbf u \frac{(\mathbf F' \cdot \mathbf v')}{c^{2}} + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf F')}{c^{2}},$$ $$\mathbf r' = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} - \gamma \mathbf u t = \mathbf r + \Gamma \mathbf u \frac{(\mathbf u \cdot \mathbf r)}{c^{2}} (t = 0),$$ where u is the speed of inertial system, v is the charge speed, you can assume, that relative to the other inertial system with relative speed u the force looks as $$\mathbf F = q\mathbf E + \frac{q}{c}[\mathbf v \times \mathbf B],$$ where $$\mathbf E = \frac{\gamma Q \mathbf r}{(r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf r \cdot \mathbf u)^{2})^{\frac{3}{2}}}, \quad \mathbf B = \frac{1}{c}[\mathbf u \times \mathbf E].$$ Of course, magnetic field is a relativistiс kinematic effect, but a procedure described above are the relativistic kinematiс transformation of Coulomb's law. So some people made a mistake by giving negative answer. 3. After that, using primary theoremes of vector analisys and regularization procedure, you can "take" rot and div of the E and B expressions above. After that you can earn Maxwell's equations. You must use superposition principle, when you move from a field of one charge to multi-charge continuously 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": 30, "mathjax_display_tex": 7, "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": 0.9409857988357544, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/7329/fair-game-and-dice?answertab=active	# Fair Game and Dice Two players put a dollar in a pot. They decide to throw a pair of dice alternatively. The first one who throws a total of $5$ on both dice wins the pot. How much should the player who starts add to the pot to make this a fair game? So my interpretation of this problem is that I first throw a dice and then the other person throws a dice afterwards. E.g. if I throw a 1, and if the other person throws a 4 wins the pot. So we are trying to find the expected payout? - "Throws a total of \$5 on both dice"? Please clarify. Also 0% accept rate? – Sev Oct 20 '10 at 18:32 Should just be 5. – PEV Oct 20 '10 at 18:40 Trevor, you should tell us how the rules of the game are. The answer to the question of course depends on that. – Rasmus Oct 20 '10 at 18:42 This is all you are given. – PEV Oct 20 '10 at 18:46 1 -1 for not defining the problem, even when asked – Ross Millikan Oct 21 '10 at 14:13 show 2 more comments ## 3 Answers Player 1 throws on odd throws and player 2 throws on even throws. This is a geometric distribution. So his distribution is $\left(\frac{8}{9} \right)^{2k} \cdot \frac{1}{9}$ and player 2's distribution is $\left(\frac{8}{9} \right)^{2k-1} \cdot \frac{1}{9}$. So player 1 must give $1/8$. - Building on your interpretation, let us define the rules as follows: A and B alternately throw a single die. If the sum of the last two throws is 5, the one who just threw wins. Note that as stated, an initial 5 could win, or a series of throws of a different length than 2. The analysis would change, but follow the same route. The below requires precisely two throws to add to 5. Let p be the probability that the first player wins. Let q be the probability that the next player to throw wins given that he has received a chance of winning, that is that the last throw is less than 5. Then if you receive a throw of 5 or 6 your chance of winning is p. So q=1/6 (that you win on this throw) + (1-p)/3 (that you throw 5 or 6 and then win) +(1-q)/2 (that you throw <5, don't win this throw, but finally win. p=(1-p)/3 (that you throw 5 or 6 and win)+2(1-q)/3 (that you throw <5 and win). $q=\frac{1}{6}+\frac{1-p}{3}+\frac{1-q}{2}$ $p=\frac{1-p}{3}+\frac{2(1-q)}{3}$ If I have the algebra right, p=15/32 and q=9/16, so the first player should withdraw \$2/17. - First, find the probability p that a 5 is thrown and define q = 1 - p. The chance That the first player wins is C=p+pq^2+pq^4+\cdots; calculate this value. Now set C=(1-C)(1+x) and solve for the extra contribution x. - So player $1$ throws on odd throws? On the first throw, the probability that he wins is $p$. On the third throw, the probability that he wins is $pq^2$ (e.g 2 failures and 1 success)? But he has only thrown the dice once right? So how can he have 2 failures? – PEV Oct 20 '10 at 19:22 1 For player 1 to win on turn 3, player 1 must have failed on turn 1, player 2 must have failed on turn 2, and player 1 must have won on turn 3. – Charles Oct 20 '10 at 19:34	{"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": 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": 0.9623588919639587, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/150420/a-mouse-leaping-along-the-square-tile	# A mouse leaping along the square tile A $n \times n$ square is made of square tiles of dimensions $1\times1$. A mouse can leap along the diagonal or along the side of square tiles. In how many ways can the mouse reach the right lower corner vertex of the square from the lower left corner vertex of the square leaping exactly $n$ times? In one of my exam, I encountered a particular version of this problem with $n=5$. With a semi-brute force (case counting) kind of approach I derived the answer as $21$. How to derive the general solution for any $n \in \mathbb{N}$ ? - Is there any particular reason that it is a square? if I get the problem correctly the mouse will never get higher than $\frac n2$ for any suitable path. – Simon Markett May 27 '12 at 16:52 What do you mean by "leap"? By diagonal leap, do you mean just single steps, or any move to any other point on the same diagonal? Does the mouse move like a chess king, or a chess queen? – Thomas Andrews May 27 '12 at 17:23 Thomas: more like a king – Henry May 27 '12 at 17:27 ## 2 Answers The general solution is the $n$th Motzkin number $-$ the number of ways of drawing non-intersecting chords between $n$ points on a circle. There is a Wikipedia article on Motzkin numbers, and an entry (A001006) in the OEIS database. The OEIS entry gives several recurrence relations and generating functions, but they are all very messy. - From the Wiki article, that seems like it might be a little off - the "path" version is the number of paths of length $n$ from $(0,0)$ to $(0,n)$. But the OP's question is asking for paths of length $n$ from $(0,0)$. to $(0,n-1)$, or else you'd be dealing with an $n+1\times n+1$ grid. In particular, the Motzkin example never requires all steps to be at least to the right. – Thomas Andrews May 27 '12 at 17:18 1 @Thomas: I think you have misunderstood the OP's question. Also, it seems that you didn't read the Wikipedia article I linked to (or even look at the pictures). – TonyK May 27 '12 at 17:21 You will find a mass of formulae at OEIS A001006 If I were working this out from scratch I would use $$\sum_{k=0}^{\lfloor n/2 \rfloor} {n \choose 2k} \frac{1}{k+1} {2k \choose k} =\sum_{k=0}^{\lfloor n/2 \rfloor} \frac{n!}{k! (k+1)!(n-2k)!}$$ as a combination of $k$ diagonal up steps, $k$ diagonal down [but not going below the starting point so involving Catalan numbers] and $2n-k$ horizontal. So for $n=5$ this gives $\dfrac{5!}{0!1!5!} + \dfrac{5!}{1!2!3!} + \dfrac{5!}{2!3!1!} = 1 + 10 + 10 =21.$ -	{"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": 22, "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": 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": 0.9279356598854065, "perplexity_flag": "head"}
http://mathoverflow.net/questions/81040/do-etale-neighhbourhoods-of-a-subvariety-descend-along-base-field-extensions-doe/81089	## Do etale neighhbourhoods of a subvariety descend along base field extensions; does normalization commute with etale base change? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have some questions such that the corresponding statements are well-known for affine varieties, and I wonder whether they hold for projective ones. 1. Let $Z\subset X$ be a closed subvariety of a (projective) variety over a field $K$. Let $L/K$ be a finite field extension, and let $Y/X_L$ be an etale neighbourhood of $Z_L$ in $X_L$ i.e. $Y/X_L$ is etale and $Y\times_{X_L}Z_L=Z_L$. Is it true that $Y$ descends to an etale neighbourhood $U$ of $Z$ in $X$ i.e. that there exists an (etale neighbourhood) $U$ such that the morphism $U_L\to X_L$ factorizes through $Y$? Would it help if I will demand that $L/K$ is separable or Galois? 2. A reference question. For a domain $R$ and an extension $L$ of the fraction field of $R$ one can consider the integral closure (or the normalization) of $R$ in $L$. Now, Theorem 5.1 here http://mathsci.kaist.ac.kr/~jinhyun/note/normalization/normalization.pdf yields that a similar fact holds for any (irreducible) variety $V$ (instead of the spectrum of $R$) and a finite extension of the function field of $V$. Is there a 'canonical' reference for this fact? How would you call the variety obtained? 3. The integral closure opeation for (commutative) rings commutes with etale base change by [EGAIV, Prop. 18.12.15]. Does this statement generalize to varieties? - ## 3 Answers For question 1, you can use Weil restriction. More generally, let $X'\to X$ be a finite locally free morphism of schemes (in the present case it will be `$X_L\to X$`), and let $Y$ be an $X'$-scheme. Recall that the Weil restriction functor $U$ of $Y\to X'$ (relative to $X'\to X$) associates to an $X$-scheme $T$ the set `$$U(T):=\mathrm{Hom}_{X'}(T\times_X X',Y);$$` in other words, it is right adjoint to the base change functor. In particular we have a canonical $X'$-morphism `$U\times_X X'\to Y$`. Now (see Bosch-Lütkebohmert-Raynaud Néron Models, 7.6): (a) If $Y\to X'$ is quasiprojective, then $U$ is (representable by) a quasiprojective $X$-scheme. (b) If $Y\to X'$ is smooth (resp. étale), so is $U\to X$. (c) This construction clearly commutes with every base change $Z\to X$. In particular, in your situation, `$Y\times_X Z\to Z':=X'\times_X Z$` is an isomorphism, so `$U\times_X Z$` is the Weil restriction of $Z'$ relative to $Z'\to Z$, which is $Z$. - ### 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. 1. I don't know. Do you have a reference for the affine case ? 2. The variety obtained is called "integral closure" (or "normalization") of $V$ in $L$. ;-) A reference is EGA II 6.3, which seems pretty canonical. EGA II 6.3.4 tells you that for any scheme $X$, for any quasi-coherent $\mathcal{O}_X$-algebra $\mathcal{A}$, you can construct the integral closure of $X$ in $\mathcal{A}$, which is an affine scheme $X'$ over $X$. 3. It seems that it should generalize trivially, maybe I am missing something ? If I understand your question correctly, you have a scheme $X$, a quasi-coherent `$\mathcal{O}_X$`-algebra $\mathcal{A}$, the integral closure $X'\rightarrow X$ of $X$ in $\mathcal{A}$ and an étale map $f:Y\rightarrow X$. You define $\mathcal{B}$ to be the pull-back of $\mathcal{A}$ (i.e. `$f^{-1}\mathcal{A}\otimes_{f^{-1}\mathcal{O}_X}\mathcal{O}_Y$`), and you want to show that $Y':=X'\times_X Y$ is the integral closure of $Y$ in $\mathcal{B}$. You can assume that $X$ is affine. Then, for every affine open $U$ of $Y$, the proposition you quote tells you that $Y'_U\rightarrow U$ is the integral closure of $U$ in $\mathcal{B}_U$, which, if I understand EGA II 6.3.4, is just saying that $Y'$ is the integral closure of $Y$ in $\mathcal{B}$. - 1. The statement in the affine case is that finite base change of a Henselian pair yields a Henselian pair. So, base change of the 'smallest pro-etale neighbourhood' of $Z$ in $X$ yields the smallest pro-etale neighbourhood of $Z_L$ in $X_L$. – Mikhail Bondarko Nov 16 2011 at 17:23 See the beginning of Gabber's springerlink.com/content/c3x37474162663hp for this statement – Mikhail Bondarko Nov 16 2011 at 17:39 The answers above completely answer your questions. Let me just point out the following useful reference (to me at least). For question 2, you could look at Chapter 4.1.2. of Liu's Algebraic geometry and arithmetic curves. See Definition 4.1.24. Using a standard argument with the trace form one can show that, for any integral normal noetherian scheme $X$ with function field $K(X)$, the normalization $X^\prime \to X$ of $X$ in a finite separable extension of $K(X)$ is a finite morphism (Proposition 4.1.25). -	{"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": 80, "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": 0.8955138921737671, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/63382/list	Return to Answer 2 I added material in response to the questioner's comments. A Killing field is preserved by the Ricci flow. By a theorem of Daskalopoulos, Hamilton and Sesum (arXiv:0902.1158), on a compact surface an ancient (defined for all negative time) solution to the Ricci flow which is not a shrinking soliton is diffeomorphism equivalent to the Fateev-Onofri-Zamolodchikov-King-Rosenau one-paremeter one-parameter family of metrics. These sausage metrics have the form (The long list of names attached to these metrics is explained as follows. For a rotationally symmetric metric on the sphere the Ricci flow can be rewritten as the logarithmic diffusion equation $u_{t} = (\log u)_{zz}$ (not the same $z$ as above). P. Rosenau and J.R. King independently found the above solution in the context of such diffusion equations. In the Ricci flow literature they are usually called the Rosenau metrics or the King-Rosenau metrics. However, Fateev-Onofri-Zamolodchikov found this metric, which they called the sausage metric, earlier, in the context of studying the renormalization group flow for a two-dimensional sigma model, in their paper Integrable deformations of the O(3) sigma model. The sausage model, Nucl. Phys. B 406 (3), 521-565 (1993). (the Ricci flow is the one-loop renormalization group flow).flow).) On a compact oriented surface of genus at least 2 there is no non-zero Killing field, as follows, for instance, from the classical Bochner argument. Similarly, on a torus, a non-zero Killing field must be parallel. On $S^{2}$ a Killing field generates an isometric $S^1$ action fixing two distinct points; that is, the metric is rotationally symmetric (a proof can be found in the paper of Chen, Lu, and Tian). What this shows is that in the compact case an answer to your question has to be on the sphere or torus. Since the Ricci flow preserves isometries, it is natural to ask what the are interesting rotationally symmetric Ricci flows (A recent survey is arXiv:1103.4669); the theorem above about ancient solutions is one sort of answer. The following is added to my original answer in response to comments by the original questioner asking for metrics related to the hyperbolic metric on the disk. Rescaling the sausage metrics $g(t)$ to have constant volume (in this case this means multiplying by a constant multiple of $t^{-1}$), they solve the volume normalized Ricci flow, and by a theorem of Chow-Hamilton as $t\to 0$ these rescaled metrics converge (as here can be checked directly) to a round metric on the sphere. Consider the metrics$$\tilde{g}(t) = \frac{-4\sin(2t)(dx^{2} + dy^{2})}{1 + 2\cos(2t)r^{2} + r^{4}}$$which are rotationally symmetric and solve the Ricci flow for $t \in (-\pi/2, 0)$. (Formally they are related to the sausage metrics by a complex rotation $t \to it$). Their scalar curvatures $R_{\tilde{g}(t)}$ are bounded as follows$$-2\cot(2t) = \min_{S^{2}}R(t) \leq R(t) \leq \max_{S^{2}}R(t) \leq -2\csc(t2).$$$R(t)$ is everywhere positive for $t \in (-\pi/4, 0)$, but for $t \in (-\pi/2, -\pi/4)$ it is both positive and negative. More precisely, it is positive in an equatorial band, and negative on the complementary disks. As $t \to -\pi/2$ these disks expand to fill the complement of the equator. The homothetic metrics $$h(t) = -\frac{1}{\sin(2t)}\tilde{g}(t) = \frac{4(dx^{2} + dy^{2})}{1 + 2\cos(4t)r^{2} + r^{4}}$$are determined by the normalization $\max_{S^{2}}R_{h(t)} = 2$, and satisfy the lower bound $-2 \leq 2\cos(2t)\leq \min_{S^{2}}R_{h(t)}$. As $t \to 0$ these metrics converge pointwise to the round metric on the sphere of volume $4\pi$, while as $t \to -\pi/2$ they converge pointwise on either of the disks complementary to the equator to the hyperbolic metric $4(1 - r^{2})^{-1}(dx^{2} + dy^{2})$ of scalar curvature $-2$. Thus the family $h(t)$ interpolates between the hyperbolic metric and the round metric. There is a similar rescaling of the sausage metrics which interpolates in a similar way between the flat metric on the punctured plane and the round metric on the sphere. There are similar families of metrics on the torus, though I am not going to write them down here. I came across the metrics $h(t)$ thinking about Einstein-Weyl structures on surfaces (this is explained in the arXiv:1011.5723, although there the relations with the Ricci flow and sausage metrics are not mentioned, as I was not then aware of them). I don't know what the characterization of the $\tilde{g}(t)$ in Ricci flow terms is, though I suspect they've been described somewhere in the literature on 2-d sigma models and RG flows. 1 A Killing field is preserved by the Ricci flow. By a theorem of Daskalopoulos, Hamilton and Sesum (arXiv:0902.1158), on a compact surface an ancient (defined for all negative time) solution to the Ricci flow which is not a shrinking soliton is diffeomorphism equivalent to the Fateev-Onofri-Zamolodchikov-King-Rosenau one-paremeter of metrics. These metrics have the form $$g(t) = \frac{-4\sinh(2t)(dx^{2} + dy^{2})}{1 + 2\cosh(2t) r^{2} + r^{4}}$$ in which $z = x + iy$ is a standard coordinate on the complement of a point in the complex projective line, and $r = \sqrt{x^{2} + y^{2}}$. There is an obvious rotational symmetry. The scalar curvature $R(t)$ of $g(t)$ satisfies the bounds $$\frac{-2}{\sinh(2t)} = \min_{S^{2}}R(t) \leq R(t) \leq \max_{S^{2}}R(t) = -2\coth(2t).$$ For a rotationally symmetric metric on the sphere the Ricci flow can be rewritten as the logarithmic diffusion equation $u_{t} = (\log u)_{zz}$ (not the same $z$ as above). P. Rosenau and J.R. King independently found the above solution in the context of such diffusion equations. However, Fateev-Onofri-Zamolodchikov found this metric, which they called the sausage metric, earlier, in the context of studying the renormalization group flow for a two-dimensional sigma model, in their paper Integrable deformations of the O(3) sigma model. The sausage model, Nucl. Phys. B 406 (3), 521-565 (1993). (the Ricci flow is the one-loop renormalization group flow). On a compact oriented surface of genus at least 2 there is no non-zero Killing field, as follows, for instance, from the classical Bochner argument. Similarly, on a torus, a non-zero Killing field must be parallel. On $S^{2}$ a Killing field generates an isometric $S^1$ action fixing two distinct points; that is, the metric is rotationally symmetric (a proof can be found in the paper of Chen, Lu, and Tian). What this shows is that in the compact case an answer to your question has to be on the sphere or torus. Since the Ricci flow preserves isometries, it is natural to ask what the are interesting rotationally symmetric Ricci flows (A recent survey is arXiv:1103.4669); the theorem above about ancient solutions is one sort of answer.	{"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": 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": 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": 0.9219263792037964, "perplexity_flag": "head"}
http://cms.math.ca/10.4153/CMB-2004-021-1	Canadian Mathematical Society www.cms.math.ca \| \| \| \| \| \|----------\|----\|-----------\|----\| \| \| \| \| \| \| \| \| Site map \| \| \| CMS store \| \| location: Publications → journals → CMB Abstract view # Countable Amenable Identity Excluding Groups Read article [PDF: 193KB] http://dx.doi.org/10.4153/CMB-2004-021-1 Canad. Math. Bull. 47(2004), 215-228 Published:2004-06-01 Printed: Jun 2004 • Wojciech Jaworski Features coming soon: Citations (via CrossRef) Tools: Search Google Scholar: Format: HTML LaTeX MathJax PDF PostScript ## Abstract A discrete group $G$ is called \emph{identity excluding\/} if the only irreducible unitary representation of $G$ which weakly contains the $1$-dimensional identity representation is the $1$-dimensional identity representation itself. Given a unitary representation $\pi$ of $G$ and a probability measure $\mu$ on $G$, let $P_\mu$ denote the $\mu$-average $\int\pi(g) \mu(dg)$. The goal of this article is twofold: (1)~to study the asymptotic behaviour of the powers $P_\mu^n$, and (2)~to provide a characterization of countable amenable identity excluding groups. We prove that for every adapted probability measure $\mu$ on an identity excluding group and every unitary representation $\pi$ there exists and orthogonal projection $E_\mu$ onto a $\pi$-invariant subspace such that $s$-$\lim_{n\to\infty}\bigl(P_\mu^n- \pi(a)^nE_\mu\bigr)=0$ for every $a\in\supp\mu$. This also remains true for suitably defined identity excluding locally compact groups. We show that the class of countable amenable identity excluding groups coincides with the class of $\FC$-hypercentral groups; in the finitely generated case this is precisely the class of groups of polynomial growth. We also establish that every adapted random walk on a countable amenable identity excluding group is ergodic. MSC Classifications: 22D10 - Unitary representations of locally compact groups 22D40 - Ergodic theory on groups [See also 28Dxx] 43A05 - Measures on groups and semigroups, etc. 47A35 - Ergodic theory [See also 28Dxx, 37Axx] 60B15 - Probability measures on groups or semigroups, Fourier transforms, factorization 60J50 - Boundary theory	{"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": 20, "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": 0.8291443586349487, "perplexity_flag": "middle"}
http://mathhelpforum.com/trigonometry/184125-help-points-circle.html	Thread: 1. help with points on a circle If I know a coordinate of a point (red dot on image) of a circle with radius x and i want to rotate it y degrees to the right, how do I get the coordinate of the new point (blue dot on image) ? thanks Attached Thumbnails 2. Re: help with points on a circle As you are seeking a pair of co-ordinates $\left(x_2,\;y_2\right)$ you could set up a pair of equations in 2 unknowns. If the circle centre is the origin, then $\left(x_2\right)^2+\left(y_2 \right)^2=\left(x_1 \right)^2+\left(y_1 \right)^2=r^2$ where the other co-ordinates are known. Then the area of the triangle between the origin and the 2 points is $0.5r^2sin\theta$ and so, your 2nd equation is, using the area of a triangle for which one vertex is the origin $0.5r^2sin\theta=0.5\|x_1y_2-x_2y_1\|$ However, a trigonometric solution will be much simpler. If you label the acute angle between the negative part of the x-axis, the origin and the red dot "A", and the obtuse angle between the positive part of the x-axis and the blue dot "C", then use the given co-ordinates to calculate the circle radius and the angle "A", then calculate "C" and use $x=rcosC,\;\;y=rsinC$ 3. Re: help with points on a circle Originally Posted by Sneaky If I know a coordinate of a point (red dot on image) of a circle with radius x and i want to rotate it y degrees to the right, how do I get the coordinate of the new point (blue dot on image) ? thanks p-radius(not x, for this matter) x=pcos (a) y=psin(a) you have red point, (x1,y1) put it above, find angle a. blue point is: x=pcos (a-b) y=psin(a-b) I think that would work... 4. Re: help with points on a circle Originally Posted by Sneaky If I know a coordinate of a point (red dot on image) of a circle with radius x and i want to rotate it y degrees to the right, how do I get the coordinate of the new point (blue dot on image) ? Let's change some of the notation here. Say that the the coordinates of the red point are $(x,y)$ then the coordinates of the blue point $\left( {x',y'} \right)$ can be found using: $\begin{array}{*{20}c}{x' = x\cos ( - \phi ) - y\sin ( - \phi )} \\ {y' = x\sin ( - \phi ) + y\cos ( - \phi )} \\ \end{array}$ NOTE. I have changed the notation. $\phi$ is the angle you have called $y$. Also note that you said that you know $\color{red}(x,y)$	{"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": 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": 0.8888199329376221, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/34749/is-causality-the-equivalent-of-a-claim-that-the-future-is-predictable-based-on/34767	# Is “Causality” the equivalent of a claim that the future is predictable based on the present and the past? In classical (Newtonian) mechanics, every observer had the same past and the same future and if you had perfect knowledge about the current state of all particles in the universe, you could (theoretically) compute the future state of all particles in the universe. With special (and general) relativity, we have the relativity of simultaneity. Therefore the best we can do is to say that for an event happening right now for any particular observer, we can theoretically predict the event if we know everything about the past light cone of the observer. However, it tachyons (that always travel faster than the speed of light) are allowed, then we cannot predict the future since a tachyon can come in from the space-like region for the observer and can cause an event that cannot be predicted by the past light cone. That is, I believe, why tachyons are incompatible with causality in relativity. Basically, the future cannot be predicted for any given observer so the universe is in general unpredictable - i.e. physics is impossible. Now in quantum mechanics, perfect predictability is impossible in principle. Instead all we can predict is the probability of events happening. However, Schrodinger's equation allows the future wavefunction to be calculated given the current wavefunction. However, the wavefunction only allows for the predictions of probabilities of events happening. Quantum mechanics claims that this is the calculations of probabilities is the best that can be done by any physical theory. So the question is: "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" Since prediction is the goal of physics and science in general, causality is necessary for physics and science to be possible. I am really not asking for a philosophical discussion, I want to know if there are any practical results of the principle of causality other than this predictability of the future of the universe. Please don't immediately close this as being a subjective question, let's see if anyone can come up with additional implications for causality besides future predictability. - It seems to be the time between starting to accumulate this information and having all of it something will change preventing you from reaching your goal in a finite time. – Argus Aug 23 '12 at 0:21 To elaborate if you can ably predict the future accurately if you have "perfect" knowledge thus makeing perfect knowledge impossible because the knowledge you already have is not enough to perfectly predict the future of your current knowledge. – Argus Aug 23 '12 at 0:25 The intent is not to actually and accurately measure the entire universe to enough accuracy to predict the entire universe. Rather it is that in principle it could be done. What can actually be accomplished is to do small scale experiments and to measure the starting state well enough to predict the result of the experiment. – FrankH Aug 23 '12 at 4:39 The definition of causality in the 1960s was analyticity of the S-matrix. This is not exactly the same as "future and past", because it only relates the infinite future to the infinite past. One part of this idea is that asymptotically successive scattering events can be thought of as one scattering event followed by another. Another part is that the scattering is never superluminal information transfer asymptotically. There's more, and it's hard to review properly, but this is what "causality" means to an S-matrixer. In field theory, it just meant spacelike boson fields commute. – Ron Maimon Aug 23 '12 at 4:50 ## 8 Answers Your question "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" has the trivial answer ''no'' as the qualification ''to whatever extent is possible'' turns your assumption into a tautology. The tautology makes your statement false, as your question asks whether the universally true statement is equivalent to causality. An answer "true" would make any theory causal, thus making the concept meaningless. Why is your assumption a tautology? No matter which theory one considers, the future is always predictable to precisely the extent this is possible (based on whatever knowledge one has). In particular, this is the case even in a classical relativistic theory with tachyons or in theories where antimatter moves from the future to the past. However, in orthodox quantum mechanics and quantum field theory, causality is related to prepareability, not to predictability. On the quantum field theory level (from which all higher levels derive), causality means that arbitrary observable operators $A$ and $B$ constructed from the fields of the QFT at points in supports $X_A$ and $X_B$ in space-time commute whenever $X_A$ and $X_B$ are causally independent, i.e., if (x_A-x_B is spacelike for arbitrary $x_A\in X_A$ and . $x_B\in X_B$. Loosely speaking, this is equivalent to the requirement that that, at least in principle, arbitrary observables can be independently prepared in causally independent regions. Arguments from representation theory (almost completely presented in Volume 1 of the QFT books by Weinberg) then imply that all observable fields must realize causal unitary representations of the Poincare group, i.e., representations in which the spectrum of the momentum 4-vector is timelike or lightlike. This excludes tachyon states. While the latter may occur as unobservable unrenormalized fields in QFTs with broken symmetry, the observable fields are causal even in this case. - Thank you for your answer. I need to think about what you say some more but I already up voted since yours is the first answer to actually address the question I asked, in my opinion. BTW, if my statement is a tautology, that would make it trivially TRUE, not false. The qualification of whenever possible was to allow for QM where only probabilities are predicted, not actual exact future results. – FrankH Sep 14 '12 at 11:33 @FrankH: The tautology makes your statement false, not true, as your question asks whether the universally true statement is equivalent to causality. An answer "true" would make any theroy causal, thus making the concept meaningless. – Arnold Neumaier Sep 14 '12 at 12:54 How can this be generalized to quantum gravity, where the metric itself fluctuates and where we therefore not know a priori, whether two spacetime points are spacelike seperated or not ? – jjcale Sep 15 '12 at 14:32 @jjcale: Far too little is known about quantum gravity to give an answer at that level. – Arnold Neumaier Sep 15 '12 at 18:44 Causality (the effect occurs after the cause) is often used with the Kramers Koenig relations: http://en.wikipedia.org/wiki/Kramers%E2%80%93Kronig_relations Which in turns are often used in electromagnetism. As you can see, the structure of many usual functions is directly linked to the universe being causal. From a purely sci-fi point of view, if at present, we only know the probability of a particle to be here or there, we would be foolish to ask where is the particle in ten years. The wiser question would be what is the probability of the particle to be here or there in ten years. By the way, QM is causal, once you know the initial conditions you can compute whatever you want. - So, I cannot figure out if your answer to the question is yes or no. The question is: "Is causality the equivalent of the claim that the future is predictable based on the present and the past?" Please answer with Yes or No... – FrankH Aug 23 '12 at 11:05 A causal theory states that knowing the initial conditions (initial state), you can compute the state at a latter time, and reciprocally you can also determine the past. However, QM is causal but if you only know the probability of the particle to be at a point, all you can do is compute the probability at a latter time. – Shaktyai Aug 23 '12 at 15:22 So @shaktyai - what you say is correct, you can only compute probabilities in QM. So is the answer Yes or No? Is QM causal? – FrankH Aug 23 '12 at 16:33 QM is causal in a sense that: if you know a probability at t0, then the probability at any subsequent time is known. In classical mechanics if you know the position and momemtum at t0 you know them at t>t0. Now you need to define what you mean by predict. You can not overcome the limitation of knowledge in QM.: Probability at t0 and t>t0. But the path from t0 to t is unique and well determined. – Shaktyai Aug 23 '12 at 22:51 Unlike quantum mechanics and relativity, causality is something we have a strong intuition for but somewhat weak theoretical basis, so I doubt you will get an answer that isn't a bit vague and philosophical. The problem is that the laws of physics are (mostly) time-symmetric, meaning that both past and future should be equally "predictable". That includes quantum mechanics if you don't believe in wave function collapse, and even general relativity (a time-reversed black hole is a white hole, which is not forbidden by GR). So what distinguishes cause and effect? The apparent distinction between past and future is called the arrow of time, which seems to be related to the second law of thermodynamics - the statistical tendency of systems to evolve from a more ordered state to a more disordered one. Why was the universe more ordered in the direction of time that we call the past? I don't think anybody knows. - +1 Thanks for a thoughtful answer that actually address the question! So is causality somehow related to how the initial low entropy of the universe is increasing with time (and the 2nd law of thermodynamics)? – FrankH Sep 15 '12 at 5:43 You have to distinguish two different meanings of "causality". First, a theory is (Einstein-) causal if there is a light cone structure that connects events and all the possible influence of an event comes from the past light cone. Secondly, a theory can be called causal if for every single event we can identify a cause, or causal relationship. That means such events cannot happen spontaneously without any prior state leading to them. And then there's a third but wrong meaning of causality, that says cause comes before effect. This is meaningless because cause and effect are just defined by their very time order.(There's also a correct statement that goes in the same direction, namely saying that all observers agree on the order of cause and effect. But that's not important for the discussion at hands) Another useful and related term is "determinism", which means that the future states of the system are a consequence of the current state only, and nothing more. A deterministic system does not have to allow the reconstruction of past states from the current state however, as it may lose information during the time evolution. To guarantee that the past is also 'predictable', you have to have a theory that is deterministic and its time reversal is deterministic too. Now for general relativity you have both meanings of causality and determinism. That also implies that you can use a space-like time slice to describe the state of GR and predict all events from it. A past light-cone only works for the events contained in it. Quantum theory is causal in the first case, meaning that events are only influenced by their past light cone. This is even true in the presence of nonlocal entanglement, because the actual interactions and therefore the state evolution are Einstein-local in relativistic quantum theory. For the second meaning of causality quantum theory does not seem to fit. Decay events seem to be uncaused and happen spontaneously. Quantum theory also appears to describe an indeterministic world, because we cannot predict the outcome of measurements and many events seem to be fundamentally random. The interesting question is, if these properties are just results of the incomplete knowledge of the observer about the state of the universe. Could it be possible that decays are really caused by some trigger that is too faint to be observed? Or would knowing the state of the universe allow to predict the outcome of a measurement as seen by an observer? Many physicists argue that a fundamental scientific theory should be causal (in the 2nd sense) and deterministic, so there's a lot of research going on to answer these questions. The most important approaches are Bohmian mechanics which adds additional structure in order to get determinism, and the relative state or many worlds interpretation, which assumes a deterministic evolution of the universe and attempts to derive the observed indeterminism from it. You can see that it is important to be very precise with the meanings of causality and determinism here. And in general one can say that the causality first meaning and second meaning together imply something even stronger than determinism, because they add the light cone structure. Determinism on the other hand does imply the second meaning of causality, but not the first, because a light cone structure is not necessary for determinism. If you are interested in seeing how quantum theory can be thought of as a causal and deterministic theory, I would be pleased to welcome you to my blog at http://aquantumoftheory.wordpress.com - An idealized computer is the prototypical example of a causal system. If you analyze the flow of information in a (potentially parallelized) computer program, it should basically be a directed acyclic graph. One way to destroy this property (and hence causality) for a computer program would be to run it in an endless loop and look for fixed points and other properties of this "iteration" instead of properties which depend on the actual number of the current time-step. I expect from causality that (a generalization of) a directed acyclic graph captures the essence of the flow of information. A directed graph can be topologically sorted if and only if it is a directed acyclic graph. The ordering provided by a topological ordering can take over the role of time. A topological ordering is not unique, but this is no problem. Already special relativity tells us that time is not absolute. A slightly bigger issue is that there is a physical time, and it would be nice if it would actually provide a topological ordering, but in actual reality it probably just provides an approximate topological ordering. I also want to clarify that determinism is no requirement for causality to me. It is fine if there is randomness, as long as the flow of information itself is "acyclic". - Predictability can have some randomness in it as QM requires. So, is your answer to the question yes or no. The question is: "Is causality the equivalent of the claim that the future is predictable based on the present and the past?" Please answer with Yes or No... – FrankH Aug 23 '12 at 11:11 @FrankH I think the answer is "No". If I translate my picture into the language of past, present and future, the essential point would be that the future doesn't change (or influence) the past. No information can flow from the future to the past. – Thomas Klimpel Aug 23 '12 at 11:45 Now you really confuse me. I certainly agree the future doesn't change the past, but to me that's what causality means. Tachyons violate causality because they can come back from the far future and modify what is going to happen in the next second so it is no longer determined by the past. – FrankH Aug 23 '12 at 14:59 @FrankH I wasn't talking about Tachyons. I have no opinion with respect to Tachyons. For me, a directed acyclic graph captures the essence of causality. I know how to interpret such a graph in terms of conditional independence, but it's unclear to me how to interpret it in terms of predictability. So when I try to see what this essence means in terms of past, present and future, I end up with "no information can flow from the future to the past". This looks different than "the future is predictable based on the present and past", that's why I think the answer to your question is "No". – Thomas Klimpel Aug 23 '12 at 17:07 One of the lessons from SR is that if a causal influence propagated faster than $c$, observers would disagree on what occurred first, "cause" or "effect", i.e. cause and effect would be relative. So, it might be argued that the principle of causality is perhaps equivalent to the statement that cause and effect are not relative. I think it might be the case that the principle of causality is necessary but not sufficient to predict the future. I, for one, believe that all actions are caused by entities but that the future is not determined. - If cause and effect are two events that are at a timelike separation with cause before effect then ALL observers will agree that cause came before effect. So as you point out, if faster than light propagation is allowed then cause and event could have a spacelike separation and some observers will say that effect came before the cause. That is why tachyons are assumed to not exist. But to the question: To whatever extent the future is determineable is that equivalent to the principle of causality. Thus whenever the future can be determined, causality is involved. Then they are equivalent. – FrankH Sep 14 '12 at 1:09 This is a bit subtle but your statement that some "will say effect came before cause" presumes the absoluteness of what event is the "cause" and what event is the "effect". But consider the notion that one observer's "cause" is another observer's "effect". – Alfred Centauri Sep 14 '12 at 1:16 If I shoot a tachyon gun at someone on Alpha Centauri and it kills them (assuming tachyons kill), then my shooting the gun is certainly the cause of the person on Alpha Centauri dying. However an observers moving at a high enough velocity relative to me would say that the person at Alpha Centauri died as the tachyon happened to leap out of his body and travel 4 light years to earth and end up going down the barrel of my gun just as I pulled the trigger. Nobody could possibly think that the person on Alpha Centauri dying is the cause of the tachyon being absorbed in my gun barrel, would they? – FrankH Sep 14 '12 at 2:28 If it were the case that observations were that Alpha Centaurians died first and that taychons were absorbed by your gun barrel later in all cases, it would only be scientific to conjecture the causal connection no matter how odd that would seem to you and me. If you only observed cracked eggs rising from the floor and assembling themselves into perfect sealed ovoids, you would consider it natural. Don't forget that our intuition is formed from repeated experience. – Alfred Centauri Sep 14 '12 at 2:49 Cum hoc ergo propter hoc, or in English with it therefore because of it. Future might be correlated with the present and the past but not caused by it. So one could imagine a universe in which both the future and the past are caused for example by the state of the universe at some fixed point in time $t_0$. This could cause the universe to behave in a way such that for any two times $t_2>t_1$ the state of the universe at $t_2$ is in some way correlated with the state of the universe at $t_1$. Correlation is enough for prediction. In probably the most extreme case (and assuming an infinitely lived universe), we might even have $t_0 = \infty$, meaning the point in time that causes the correlation is infinitely far away. In other words, future can always cause the past, but still be inferrable from the past. So the answer to your question is no, predictability does not imply temporally ordered causality. The converse is true though, causality implies some degree of predictability. - I am not talking about imaginary universe, how about in OUR universe? – FrankH Sep 14 '12 at 2:32 First, remember that the way science works is by constructing a theory that makes predictions, which are tested by experiments. If the experiments fail to square with predictions, the theory is thrown away. Now suppose you have two theories with equivalent predictions, but one says everything is caused by some point in the future and the other says it is caused by the past. It is impossible to distinguish the two theories by experiment because they make the same predictions. Therefore, the only thing I can tell you is if the concepts of causality and predictability are necessarily related. – SMeznaric Sep 14 '12 at 12:39 Whether the universe really acts this way or not is unfortunately outside the scope of science. – SMeznaric Sep 14 '12 at 12:39 I'm only interested in science, not philosophy. – FrankH Sep 14 '12 at 13:09 You asked and I am quoting "Is the predictability of the future to whatever extent is possible (based on the present and the past) equivalent to the principle of causality?" And my answer is no. I suspect you just don't like the answer, because you want it to be yes since you thought the idea up. Sorry for disappointing you. – SMeznaric Sep 14 '12 at 13:51 you are confusing determinisn with predictability, there are things that are deterministic (i.e. causal) but are unpredictable. unpredictability does not imply the lack of a cause. - -1: I am not confusing determinism with predictability. I am fully aware of chaos and the nondeterministic nature of QM. But what QM can predict is probabilities and in fact QM claims that the best you can do is predict probabilities. So that is why I say "to whatever extent is possible". – FrankH Sep 19 '12 at 21:25 The question is: "Is causality the equivalent of the claim that the future is predictable based on the present and the past?" Please answer with Yes or No... – FrankH Aug 23 at 11:05-----No Equivalent. – user12103 Sep 20 '12 at 4:08	{"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": 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": 0.9541829824447632, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/74724/unique-limits-of-sequences-plus-what-implies-hausdorff	## Unique limits of sequences plus what implies Hausdorff? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is known that there are non-Hausdorff spaces which admit unique limits for all convergent sequence (see here) and it is also known that unique limits for nets implies Hausdorff. What I am wondering is, if there is a (somehow weak) condition which one should add to "unique limits of sequences" to obtain a Hausdorff space. Would, for example, some countability help? Somehow in the same direction: What is the central property which is needed for a space such that it can be non-Hausdorff but has unique sequence limits? Is there a whole class of non-Hausdorff spaces which admit unique limits for convergent sequence? - The link in your question doesn't work for me. Anyway, it's easy to produce non-Hausdorff spaces with unique limits. The Frechet Topology has unique limits and is anti-Hausdorff (all open sets intersect). Another example is the co-countable topology on $\mathbb{R}$. See en.wikipedia.org/wiki/Fr%C3%A9chet_space and en.wikipedia.org/wiki/Cocountable_topology – David White Sep 7 2011 at 12:45 I believe having unique limits implies the space is $T_1$, so perhaps the question boils down to $T_1$ plus what implies $T_2$. – David White Sep 7 2011 at 12:50 I'm a bit confused. The Wikipedia page says that Frechet spaces are indeed Hausdorff. Also: Why do you have unique limits of sequences in the cocountable topology? – Dirk Sep 7 2011 at 13:46 It seems you're right about Frechet spaces. I admit that I was just quoting a line from a topology course I took some time ago, and I didn't stop to question whether or not it's true. However, the cocountable one is right and I'll post the details as an answer – David White Sep 7 2011 at 15:23 Looking back at my notes, it wasn't Frechet spaces, but rather the Frechet topology on any space $X$, where you define $A$ to be closed iff $A$ is the set of limits of sequences in $A$. This topology can be used to construct an example of an anti-Hausdorff space with unique limits, but the answer below is just as good. – David White Sep 7 2011 at 15:31 ## 4 Answers First countable is enough. Let $x\neq y$ be two points in your space that cannot be separated by neighborhoods. Let $O_1,O_2,\ldots$ form a neighborhood base of $x$ and let $U_1,U_2,\ldots$ form a neighborhood base for $y$. Choose a sequence $(z_n)$ such that $z_n\in O_n\cap U_n$ for all $n$. Now $(z_n)$ converges to both $x$ and $y$. - 2 Thanks! To clarify: First countable + non-Hausdorff implies "non-unique limits" and hence, "unique limits" + first countable implies Hausdorff, right? – Dirk Sep 7 2011 at 10:45 1 First countable is a nice answer. I don't think you're going to do better because you kind of need at least this assumption to say anything about sequences – David White Sep 7 2011 at 12:50 ### 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. Here's an example of a space which is not Hausdorff but which has unique limits... Let $X = \mathbb{R}$ with the cocountable topology, i.e. a set is open iff its complement is countable. Clearly any two open sets intersect, because $\mathbb{R}$ is uncountable. So $X$ is non-Hausdorff. Now, suppose $(x_n)$ is a sequence which converges to $x$. Then $C =$ {$x_n\;\|\;x_n\neq x$} is closed because it's countable. So $X-C$ is a neighborhood of $x$ and this means there is some $N$ such that for all $n>N$ $x_n\in X-C$, i.e. $x_n=x$ for large $n$. This means if $x_n\rightarrow y$ then $y=x$, proving limits are unique. - This past weekend, entirely by chance, I came across a published paper that used the term "US-space" for the class of topological spaces having the property that no sequence can converge to more than one point. The google search just below seems to bring up some things that might be of use to you: - The term "US-space" appears to go back to Albert Wilansky (1967) - ams.org/mathscinet-getitem?mr=208557 – François G. Dorais♦ Sep 12 2011 at 17:49 1 Thanks! Especially the topospaces.subwiki.org is pretty cool... – Dirk Sep 12 2011 at 18:52 Here is an answer to Dirk's last question, Is there a class of non_Hausdorff spaces in which convergent sequences have unique limits''? Yes. The so called KC-spaces or maximal compact spaces. These are spaces such that every compact subspace is closed. (The 1967 Monthly article of Wilansky `Between T1 and T2' subsumes, references, or implies all of the following). In a KC-space, convergent sequences have unique limits. (Suppose xn-->x in the KC space X. The set {x,x1,x2,..} is compact and hence closed. Thus if y is not in the set {x,x1,x2,..} then the open set X minus {x,x1,x2,..} shows it is false that xn-->y. Thus if xn-->y then y=x or y=xn for some n. If y=xn for infinitely many indices n then y=x (since every KC space is T1 (since singletons are compact) and since constant sequences have unique limits in a T1 space). If y=xn for finitely many indices then (deleting y from the sequence x1,x,2...) we are left with a subsequence zn-->x, the knowledge that y is not zn, and the knowledge that y is in the set {x,z1,z2,...} and we conclude y=x). To exhibit a large class of non-Hausdorff KC spaces let X be a non-locally-compact metric space ( for example the rationals) and let Y=X U {y} denote the Alexandroff compactification of X ( i.e. V is open in Y if V is open in X or if Y\V is a compact subspace of X). The space Y is a KC space but Y is not Hausdorff. -	{"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": 35, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "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": 0.9362621903419495, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/34586/limiting-distribution-of-alternating-renewal-process	# Limiting distribution of alternating renewal process Consider an alternating renewal system that can be in one of two states: on or off. Initially it is on and it remains on for a time $Z_1$, it then goes off and remains off for a time $Y_1$, it then goes on for a time $Z_2$, then off for a time $Y_2$ ; then on, and so forth. Suppose that the random vectors $(Z_n, Y_n), n > 1$, are i.i.d.. Then $Z_n, n>1$ are i.i.d. and $Y_n, n> 1$ are also i.i.d.. Suppose that the distribution of $Z_n$ is a Geometric distribution and the distribution of $Y_n$ a Poisson distribution. My question is whether it is possible to compute $\lim_{t\rightarrow \infty} P(\text{system is on at time }t)$? I am tempted to apply Theorem 3.4.4 of Stochastic processes by Sheldon M. Ross, which states that If $E[Z_n + Y_n] < \infty$ and $Z_n + Y_n$ is nonlattice, then $$\lim_{t\rightarrow \infty} P(\text{system is on at time }t) = \frac{E(Z_n)}{E(Z_n)+E(Y_n)}$$ But $Z_n + Y_n, n\geq 1$ are nonnegative integer valued random variables, and therefore lattice, which violates the condition of the theorem. - ## 1 Answer There is a remark at the end of Theorem 8.23 in Modeling and analysis of stochastic systems By Vidyadhar G. Kulkarni: If $Y_n+Z_n$ is periodic with period $d$, then the above result is true if $t$ is an integral multiple of $d$. - What is a periodic random variable? – Did Apr 22 '11 at 19:57 A lattice process is one whose distribution is concentrated at multiples of a period $d$: $f_X(x)=\sum_n c_n \delta (x-n d)$ or $\sum_n P(X_i=nd) = 1$. – Emre Apr 22 '11 at 20:26 2 In other words periodic random variable means nothing. – Did Apr 22 '11 at 21:46 @Didier: I think it has meaning. From Ross's book, "A nonnegative random variable X is said to be lattice if there exists $d\geq 0$ such that $\sum_{n=0}^{\infty} P(X = nd) = 1$. That is, $X$ is lattice if it only takes on integral multiples of some nonnegative number $d$. The largest $d$ having this property is said to be the period of $X$." Based on this, I think a nonnegative random variable is called periodic, if it is lattice. – Ethan Apr 22 '11 at 23:18 Based on this one sees that a random variable being lattice is a standard definition and that lattice random variables are very seldom called periodic. – Did Apr 23 '11 at 21:47	{"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": 26, "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": 0.9312890768051147, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3849658	Physics Forums ## Residue Theorem integral application 1. The problem statement, all variables and given/known data Compute the integral: ∫ x2/(x4-4x2+5) 2. Relevant equations Uses Residue theorem. 3. The attempt at a solution So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane. Then I used residue theorem that said Res(P(z)/Q(z); 2+i) = P(2+i)/Q'(2+i) = (2+i)2/(4(2+i)3-8(2+i)) = (4i+3)/(36i-8) and then I multiplied by 2∏i which would leave me with a value in the complex plane. I think this is wrong because it should come out with a real valued number. Does it have something to do with the zero having a multiplicity of 2? And if so, how do I go about redoing it with that in mind, I don't remember learning how to do that... 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 Quote by jsi So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane. Fourth order equation has four zeroes :) Quote by jsi 1. The problem statement, all variables and given/known data Compute the integral: ∫ x2/(x4-4x2+5) 2. Relevant equations Uses Residue theorem. 3. The attempt at a solution So I found the zeroes of x4-4x2+5 to be 2+i and 2-i, and therefore the one that is of relevance is 2+i since it is in the upper half-plane. Then I used residue theorem that said Res(P(z)/Q(z); 2+i) = P(2+i)/Q'(2+i) = (2+i)2/(4(2+i)3-8(2+i)) = (4i+3)/(36i-8) and then I multiplied by 2∏i which would leave me with a value in the complex plane. I think this is wrong because it should come out with a real valued number. Does it have something to do with the zero having a multiplicity of 2? And if so, how do I go about redoing it with that in mind, I don't remember learning how to do that... Until you analyze the entire problem meticulously with a fine-tooth comb, it's not going to happen even when you get all four. First, I assume you want the indefinite integral: $$\int_{-\infty}^{\infty} \frac{x^2}{x^4-4x^2+5}dx$$ via the Residue Theorem. Then we could write: $$\mathop\oint\limits_{C} \frac{z^2}{z^4-4z^2+5}dz=\int_{-\infty}^{\infty} \frac{x^2}{x^4-4x^2+5}dx+\lim_{R\to\infty} \int\limits_{\gamma}\frac{z^2}{z^4-4z^2+5}dz=2\pi i \sum \text{Res}f(z)$$ Now, you understand all that? Every little bit of it? Without me having to explain what all the notation is right? When you do, then analyze every part of it in detail even that "excipient" leg of the contour that I assume goes to zero but don't know for sure cus' I haven't analyzed it meticulously, then determine which of the four roots are in the contour, compute the residue of the function there, then do the final sum. Bingo-bango and we're done. Recognitions: Homework Help Science Advisor ## Residue Theorem integral application 2+i isn't even a root of your polynomial. sqrt(2+i) (for example) is. Tags complex analysis, integration, residue theorem Thread Tools \| \| \| \| \|-----------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Residue Theorem integral application \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 3 \| \| \| Calculus & Beyond Homework \| 0 \| \| \| Advanced Physics Homework \| 5 \| \| \| Calculus & Beyond Homework \| 6 \| \| \| Calculus \| 1 \|	{"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": 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": 0.9198964238166809, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/144337/lebesgue-integrable	# Lebesgue Integrable Prove that $f(x)=x$ is not Lebesgue integrable on $[1, \infty)$. (hint: use def. of Lebesgue integrability for positive functions). hint: Use integral $= \infty$ by defining simple function and $\chi_A$ characteristic function. - 4 What have you tried? – Alex Becker May 12 '12 at 20:36 ## 1 Answer It is hard to give a rigorous proof if we don't know the theorems you already know concerning Lebesgue Integrals, nor your definitions. A very easy way to see it is if you know that the Lebesgue Integral is monotone: $f(x)=x \geq 1$ on $[1,\infty)$, therefore $$\int_{[1,\infty)} f(x) d\mathcal{L}(\mathbb{R})\geq\int_{[1,\infty)} 1 d\mathcal{L}(\mathbb{R})=\mathcal{L}([1,\infty))=\infty$$ -	{"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": 6, "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": 0.8923437595367432, "perplexity_flag": "middle"}
http://mathhelpforum.com/discrete-math/187039-show-nonempty-finite-subset-r-has-max-min.html	# Thread: 1. ## Show a nonempty finite subset of R has a max and min Show that a nonempty finite subset of the Real Numbers has both a maximum and minimum. I know I can use induction to solve this but I seem to be having some problems getting started: Let P(n) be assertion: a nonempty finite subset of R has a max and min P(1) is the assertion: a set S={1} has a max and min, both of which are 1. Suppose P(k) is true for: a set S={k} containing a max and min. Show P(k+1) true for S={k+1} This is where I seem to be getting stuck. I am just not sure what to do from here (or even if my set up is correct) Any help would be greatly appreciated! Thank you 2. ## Re: Show a nonempty finite subset of R has a max and min Originally Posted by mybrohshi5 Show that a nonempty finite subset of the Real Numbers has both a maximum and minimum. Let P(n) be assertion: a nonempty finite subset of R has a max and min P(1) is the assertion: a set S={1} has a max and min, both of which are 1. Suppose P(k) is true for: a set S={k} containing a max and min. Show P(k+1) true for S={k+1} I can see how this could be tricky depending of how picky an instructor can be. But in a set $S$ of $n+1$ real numbers remove any one, say $x$. Now $T=S\setminus\{x\}$ is set of $n$ elements. So by the inductive step let $\alpha=\max(T)~\&~\omega=\min(T)$. Now what? For example what if $x<\alpha~?$ 3. ## Re: Show a nonempty finite subset of R has a max and min Now there are a few cases to cover! Here's what I have come up with now, Let $\alpha = max(T)$ and $\beta = min(T)$ if $x < \alpha$ then $\alpha$ is a maximum for T if $x > \alpha$ then $x$ is a maximum for T if $x < \beta$ then $x$ is a minimum for T if $x > \beta$ then $\beta$ is a minimum for T 4. ## Re: Show a nonempty finite subset of R has a max and min Originally Posted by mybrohshi5 Now there are a few cases to cover! Here's what I have come up with now, Let $\alpha = max(T)$ and $\beta = min(T)$ if $x < \alpha$ then $\alpha$ is a maximum for T if $x > \alpha$ then $x$ is a maximum for T if $x < \beta$ then $x$ is a minimum for T if $x > \beta$ then $\beta$ is a minimum for T You are trying to show that $S$ has a max & min. $T$ is the set in the inductive step and $S=T\cup \{x\}$ has one more element. 5. ## Re: Show a nonempty finite subset of R has a max and min Okay I may be getting a little confused now. so i have a set S which contains {n+1} elements. Now I let $x$ be an element of S and let T = S \ {x}. So now T is of size "n" and we know that T contains a max and min, namely $\alpha = max(T)$ and $\beta = min(T)$ if $x < \alpha$ then $\alpha$ is a maximum for S if $x > \alpha$ then $x$ is a maximum for S similar reasoning for minimum Thus we have shown that S will always contain a max and min (whether that be x, $\alpha$ or $\beta$) Am I missing something? 6. ## Re: Show a nonempty finite subset of R has a max and min Originally Posted by mybrohshi5 so i have a set S which contains {n+1} elements. Now I let $x$ be an element of S and let T = S \ {x}. So now T is of size "n" and we know that T contains a max and min, namely $\alpha = max(T)$ and $\beta = min(T)$ if $x < \alpha$ then $\alpha$ is a maximum for S if $x > \alpha$ then $x$ is a maximum for S similar reasoning for minimum Thus we have shown that S will always contain a max and min (whether that be x, $\alpha$ or $\beta$) Am I missing something? No, you have it now. You may want to add that $\alpha\in T\subseteq S$ and $\beta\in T\subseteq S$ 7. ## Re: Show a nonempty finite subset of R has a max and min Great! I will do that. Thanks again Plato. Your knowledge is greatly appreciated. Have a great day!	{"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": 50, "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": 0.9367488622665405, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/1518/light-emission-spectrum-units?answertab=oldest	# Light emission spectrum units Do someone knows the units of the spectra provided here ? It seems obvious enough that it's said nowhere, but even Wikipedia and other sites are quite blurry on this point. So, is it power ($W$), radiance($\:\rm{W/m^2sr}$), or something else ? Thanks ! - The README file provided with the samples seems to give all the information you might need. What precisely do you not understand? – Marek Dec 1 '10 at 17:46 It doesn't specify in what units the numbers in the other files are. But I suppose knowing the apparatus that did the measurement can help resolve that issue. – Raskolnikov Dec 1 '10 at 17:54 Yeah, I meant READMEs included with other measurements. But then I noticed that different apparatuses were used in those cases and also different format of data so it's not really very useful. – Marek Dec 1 '10 at 20:20 ## 3 Answers Usually, the spectrum is in arbitrary unit, proportional to the power, the radiance or whatever. The information the spectrum provides is the relative radiance of various wavelengths which is contained in the spectrum. If you want the radiance of one wavelength band, you simply multiply the relevant band of a normalized spectrum (of integral 1) by the total radiance of your source. - 2 – Calvin1602 Dec 2 '10 at 8:10 Most spectrometers work by spatially separating light into its component frequencies by means of a diffraction grating, and measuring that with a CCD array. That means measuring the intensity, or power per unit area, the units of which are watts per square meter. However, there is usually an unknown scaling factor involved, so don't expect the values to actually be in W/m². I would guess that the values in these files are normalized so that a value of 1 is the intensity at which the corresponding CCD pixel saturates. Strictly speaking, you are measuring the intensity present in one range of frequencies, but it's simpler and usually justified to just treat it as one frequency. - From the author : Hi, The data is in W/sr/m2. Best, Jussi So, radiance. Thanks anyway :) - It should be a radiance/nm. So you have to take into account the 2 nm spectral window width. – Frédéric Grosshans Dec 3 '10 at 16:49 @Frédéric : Yes, sure, it's the integral of the spectrum which is in w/sr/m². – Calvin1602 Dec 4 '10 at 0:53	{"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": 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": 0.9364036917686462, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/108675-complex-numbers.html	# Thread: 1. ## Complex numbers Show that $%5Cmid%20z-2i%5Cmid%20=%204-z$ can be represented by the cartesian equation $%5Cfrac%7Bx%5E2%7D%7B3%7D+%5Cfrac%7B%28y-1%29%5E2%7D%7B4%7D=1$ 2. Originally Posted by usagi_killer Show that $%5Cmid%20z-2i%5Cmid%20=%204-z$ can be represented by the cartesian equation $%5Cfrac%7Bx%5E2%7D%7B3%7D+%5Cfrac%7B%28y-1%29%5E2%7D%7B4%7D=1$ You're missing magnitude brackets on the right hand side of the relation: it should be \|4 - z\|. Furthermore, the locus of \|z - 2i\| = \|4 - z\| is NOT a circle so the result you're trying to prove is wrong.	{"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": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 4}, "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": 0.9273965358734131, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/22323/unknown-isotope?answertab=oldest	# Unknown isotope Problem description: The most common isotope of a single nucleus stripped of its electrons is accelerated through a potential difference of 1225V and fired horizontally into a B-field directed perpendicularly into the page (flat page lying on desk). The nucleus is observed to travel in a circular path of radius 9.973 cm. Afterward an alpha particle is accelerated through the same potential difference and fired in the same manner into the magnetic field and observed to travel in a circular path with radius 8.980 cm. What element is the unknown nucleus? (For this problem determine the masses of the nuclei--unknown and alpha--by summing the individual masses of the protons and neutrons. Also, include a diagram with your solution showing all aspects of the problem description.) This is what I have so far: . . This is the point where I am confused as to how to find the mass and charge of the unknown particle, or where exactly I'm supposed to go from this point to solve the problem. (I have heard from other students that there is some guessing involved (from the periodic table)) Help would be appreciated. Thank you. - Note that the radius of the path taken by the unknown ion is larger than that of the alpha, so it's charge/mass ratio is smaller (i.e. it is neutron rich). From there there are only a few hundred isotopes in a few score elements...worst case use an exhaustive search. Expressing charge to mass as A/Z neglects the nucleon mass difference but might make the search faster. – dmckee♦ Mar 13 '12 at 15:21 ## 1 Answer First part: From the formula for the radius, and the fact that magnetic field is the same in both cases, you get: $$B = \frac{m_1 v_1}{q_1 r_1} = \frac{m_2 v_2}{q_2 r_2}$$ Because you don't know the velocities, you want to get them from the potential difference. You also have $$v = \sqrt\frac{2q V}{m}$$ You put that back into the first equation, and isolate the two terms you don't know: $q_1$ and $m_1$: $$\frac{m_1}{q_1} = \frac{m_2}{q_2} \left(\frac{r_1}{r_2}\right)^2$$ So, you have for your unknown particle $m/q = 2.4677$. Note that you could also get there from your calculations, because you have calculated $B$. The above is simpler because by writing that $B$ and $V$ are constant, we use invariants to relate quantities relating to particles 1 and 2. Second part, the search! We don't want to check all elements one by one. To get an idea of where that element is, we take advantage of the fact that, for an isotope, mass is roughly equal to $A$ (atomic number). So, we want $A/Z \approx 2.4677$. Then, we use what we know about stable isotopes: they are situated mostly along a curve near (but not along) the $A=2Z$ axis. Check this diagram from Wikipedia: The most stable isotopes (which include the most common isotope of each element) are the marroon ones. They are not randomly distributed, but follow a rough curve. Where does this curve intersect the line for $A/Z \approx 2.4677$? Well, you can check a few values here and there. In my case, I used Mathematica so you can see better the idea (the red line is $m/Z = 2.4677$): ````Show[ ListPlot[ Table[ElementData[ElementData[i], "AtomicWeight"], {i, 1, 100}]], Plot[2.4677*x, {x, 0, 100}, PlotStyle -> Red] ] ```` The crossing is thus somewhere between 60 and 70. I have experimented myself and found that the answer is Gd (gadolinium, element 64). It is the element that best fits the bill: its most common isotope is 158Gd, has a mass of 157.9241039 and its $q/m$ ratio is thus: 2.4675. That's within the uncertainty of the measurements given. -	{"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": 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": 0.9480723738670349, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-statistics/119645-poisson-process-r-2-a-print.html	# Poisson process on R^2 Printable View • December 9th 2009, 05:05 PM kingwinner Poisson process on R^2 Consider a Poisson process of rate λ on R^2. Let N(r) be the number of points in a circle of radius r centered at the origin and Y2 be the distance from the origin to the 2nd closest point. Calculate E(Y2) and E[Y2 \|N(1)=10]. I have no problem with computing E(Y2), but I am stuck with calcuating E[Y2 \|N(1)=10]. So my first step is to try to compute the tail distribution function P(Y2>y \|N(1)=10), and then differentiate to get the density function and then take the conditional expectation using the density. By definition, P(Y2>y \|N(1)=10) = P(Y2>y and N(1)=10) / P(N(1)=10) Now I am having trouble computing P(Y2>y and N(1)=10). How can we express it in terms of N(y) (which is the number of points in a circle of radius y about the origin)? Can someone please go through the idea of how to compute this? Any help is greatly appreciated! :) [note: also under discussion in talk stats forum] • December 10th 2009, 01:06 AM Laurent Quote: Originally Posted by kingwinner Consider a Poisson process of rate λ on R^2. Let N(r) be the number of points in a circle of radius r centered at the origin and Y2 be the distance from the origin to the 2nd closest point. Calculate E(Y2) and E[Y2 \|N(1)=10]. You should probably use the fact that, conditionally to $N(1)=10$, the points of the Poisson process inside the circle of radius 1 have same joint distribution as 10 independent random variables uniformly distributed in the unit disk. (In particular, $Y_2$ depends only on these points) So you can forget about the Poisson process and only consider a situation involving uniform distribution in the unit disk. • December 10th 2009, 03:42 AM kingwinner Quote: Originally Posted by Laurent You should probably use the fact that, conditionally to $N(1)=10$, the points of the Poisson process inside the circle of radius 1 have same joint distribution as 10 independent random variables uniformly distributed in the unit disk. (In particular, $Y_2$ depends only on these points) So you can forget about the Poisson process and only consider a situation involving uniform distribution in the unit disk. hmm...I haven't encountered this theorem in my course, so I think I would have to do the computations by direct method. I know that N(y)~Poisson(λ pi y^2), and I know that for Poisson process, things happening in non-overlapping sets are independent. I think the use of these would allow me to compute directly. But I am not sure how to express the event {Y2>y and N(1)=10} entirely in terms of N(y). Could you please give me some help on this? [I don't really bother with all the computations, but I know, at least in theory, that if we can express the event {Y2>y and N(1)=10} entirely in terms of N(y), then the problem is basically solved.] Thank you very much! • December 10th 2009, 09:23 AM Laurent Quote: Originally Posted by kingwinner hmm...I haven't encountered this theorem in my course, so I think I would have to do the computations by direct method. (in fact, this theorem is the exact analog to the theorem about ordered uniform random variables in the 1-dimensional case) You could find the distribution of $Y_2$ given $N(1)=10$. To that aim, for any $0<r<1$, you have: $P(Y_2> r, N(1)=10)= P(N(r)\leq 1, N(1)=10)=$ $P(N(r)=0, N(1)=10) + P(N(r)=1, N(1)=10)$ and all these probabilities can be computed easily using annuli. For instance, $P(N(r)=1,N(1)=10)$ is the probability that there is 1 point in $D(0,r)$ and 9 points in $D(0,1)\setminus D(0,r)$, and since these sets are disjoint, the numbers of points therein are independent (and Poisson distributed with parameter proportional to the area...). Finally, you can use $E[Y_2\|N(1)=10]=\int_0^1 P(Y_2>t\|N(1)=10)dt$ to compute the expectation. I let you try that. • December 10th 2009, 02:19 PM kingwinner Quote: Originally Posted by Laurent (in fact, this theorem is the exact analog to the theorem about ordered uniform random variables in the 1-dimensional case) You could find the distribution of $Y_2$ given $N(1)=10$. To that aim, for any $0<r<1$, you have: $P(Y_2> r, N(1)=10)= P(N(r)\leq 1, N(1)=10)=$ $P(N(r)=0, N(1)=10) + P(N(r)=1, N(1)=10)$ and all these probabilities can be computed easily using annuli. For instance, $P(N(r)=1,N(1)=10)$ is the probability that there is 1 point in $D(0,r)$ and 9 points in $D(0,1)\setminus D(0,r)$, and since these sets are disjoint, the numbers of points therein are independent (and Poisson distributed with parameter proportional to the area...). Finally, you can use $E[Y_2\|N(1)=10]=\int_0^1 P(Y_2>t\|N(1)=10)dt$ to compute the expectation. I let you try that. Thanks! :) If I use E(Y2\|N(1)=10) = 1 ∫ y f(y) dy 0 where f(y) is the conditional density function of Y2 given that N(1)=10, will I get the same answer? • December 10th 2009, 02:27 PM Laurent Quote: Originally Posted by kingwinner If I use E(Y2\|N(1)=10) = 1 ∫ y f(y) dy 0 where f(y) is the conditional density function of Y2 given that N(1)=10, will I get the same answer? Sure, even if it's less direct. It's like applying an integration by part from the formula I gave. All times are GMT -8. The time now is 12:14 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": 22, "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": 0.9100563526153564, "perplexity_flag": "head"}
http://mathoverflow.net/questions/13240?sort=votes	Do finite places of a number field also correspond to embeddings? Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Something that seems to be pretty standard in every introductory treatment is that the infinite places correspond to embeddings into $\mathbb{C}$. Do the finite places correspond to embeddings as well? I can envision two possibilities. My first guess is that the primes sitting above $p \in \mathbb{Q}$ correspond to embeddings into $\overline{\mathbb{Q}_p}$, and thus also to embeddings into $\mathbb{C}$ by some messy non-canonical field isomorphism. My second guess, which I think would imply the first, is that the places of $\mathbb{Q}[\alpha]$ above $p \in \mathbb{Q}$ correspond to embeddings into $\mathbb{Q}_p[\alpha]$. I've never been able to find a precise statement about this in any of the texts I've been studying (mostly Milne's notes and Frohlich & Taylor) and would appreciate if anyone could let me know where to learn more about this -- or if I'm just plain wrong. One other thing is that the embeddings into $\mathbb{C}$ play a central role in analyzing the basic structure of a number field by way of Minkowski theory. Is there some analog for the finite places, or does that even make any sense? - 1 One way to make your second question precise is Leopoldt's conjecture (whose status is currently up in the air, at least last I heard; maybe someone here has an update); see e.g. its Wikipedia page. – D. Savitt Jan 28 2010 at 11:13 Thanks everybody for all the great responses. Every one of them is helpful and I wish I could accept more than one. However, I am going to wait a bit before accepting any, so as not to discourage any other good answers. That is, if there's even any more to be said at this point anyway.. – Jon Yard Jan 29 2010 at 6:50 6 Answers The Archimedean places of a number field K do not quite correspond to the embeddings of K into $\mathbb{C}$: there are exactly $d = [K:\mathbb{Q}]$ of the latter, whereas there are $r_1 + r_2$ Archimedean places, where: if $K = \mathbb{Q}[t]/(P(t))$, then $r_1$ is the number of real roots of $P$ and $r_2$ is the number of complex-conjugate pairs of complex roots of $P$. In other words, $r_1$ is the number of degree $1$ irreducible factors and $r_2$ is the number of degree $2$ irreducible factors of $P(t) \in \mathbb{R}[t]$. There is a perfect analogue of this description for the non-Archimedean places. Namely, the places of $K$ lying over the $p$-adic place on $\mathbb{Q}$ correspond to the irreducible factors of `$\mathbb{Q}_p[t]/(P(t))$`; or equivalently, to the prime ideals in the finite-dimensional `$\mathbb{Q}_p$`-algebra $K \otimes_{\mathbb{Q}} \mathbb{Q}_p$. More generally: if $L = K[t]/P(t)/K$ is a finite degree field extension and $v$ is a place of $K$ (possibly Archimedean), then the places of $L$ extending $v$ correspond to the prime ideals in $L \otimes_K K_v$ or, if you like, to the distinct irreducible factors of $P(t)$ in $K_v$, where $K_v$ is the completion of $K$ with respect to $v$. See e.g. Section 9.9 of Jacobson's Basic Algebra II. By coincidence, this is exactly the result I'm currently working towards in a course I'm teaching at UGA. I'll post my lecture notes when they are finished. (But I predict they will bear a strong resemblance to the treatment in Jacobson's book.) - Thanks Pete. I didn't know the formulation using prime ideals in the tensor product algebra. Looking forward to seeing your lecture notes when they're done. – Jon Yard Feb 1 2010 at 5:08 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'm not sure why Pete stopped short of coming back to the original poster's description in terms of embeddings. The point is that the infinite places of a number field $K$ correspond to embeddings of $K$ into $\mathbb{C}$ up to equivalence, where two embeddings are considered equivalent if one can be obtained from the other by post-composing with an element of `$\textrm{Aut}(\mathbb{C}/\mathbb{R})$`. In exactly the same way the places of $K$ lying over $p$ correspond to embeddings of $K$ into $\overline{\mathbb{Q}}_p$ counted up to equivalence, where two embeddings are considered equivalent if one can be obtained from the other by post-composing with an element of $\textrm{Aut}(\overline{\mathbb{Q}}_p/\mathbb{Q}_p)$. This is not hard to derive this from what Pete wrote. (Consider `$K = \mathbb{Q}[t]/P(t)$` and the description of the places above $p$ as irreducible factors of $P(t)$ over $\mathbb{Q}_p$. An embedding of $K$ into $\overline{\mathbb{Q}}_p$ amounts to choosing a root of $P(t)$ in $\overline{\mathbb{Q}}_p$, and two embeddings will be equivalent if and only if the corresponding roots are roots of the same irreducible factor of $P(t)$.) And of course by the same argument there's a similar description of the places of L lying over a place of K. - 4 "I'm not sure why..." Because I was laboring under the delusion that if I stopped typing up answers on Math Overflow, I might get some sleep? Also, because I saw something in the question that reminded me so strongly of what's already on my mind, I sort of ignored the rest. And finally, I'm not completely sure I knew explicitly what you said [although if you told me that I did, I would believe you]. I think you have given me an exercise for my course: +1. – Pete L. Clark Jan 28 2010 at 11:26 OK, yes, upon further reflection, I did know this. But thanks for reminding me! – Pete L. Clark Jan 28 2010 at 16:48 Thanks David. I really like the notion of embeddings up to equivalence. It makes complete sense. – Jon Yard Feb 1 2010 at 4:27 Another way to think of a place of a number field is in terms of equivalence classes of absolute values. You should work out how this definition is the same as the one given by David Savitt. Let $k$ be any field. An absolute value on $k$ is a homomorphism $\|\ \|:k^\times\to\mathbb{R}^{\times\circ}$ from the multiplicative group of $k$ into the ordered group of stritctly positive reals which is not trivial and such that the triangular inequality $$\|x+y\|\le \|x\|+\|y\|$$ is satisfied for all $x,y\in k$, with the convention that $\|0\|=0$. An absolute value $\|\ \|$ gives rise to a distance $d(x,y)=\|x-y\|$ on $k$, making it into a metric space. It can thus be completed to a field $k_{\|\ \|}$ in which $k$ is a dense subfield. Two absolute values on $k$ are equivalent if they induce the same topology on $k$, and thus give rise to the same completion. Ostrowski (1918) determined all absolute values on a number field, and Artin (1932) gave a beautifully simple proof of his theorem. You can read all about it in many places, including my online notes arXiv:0903.2615 - Thanks for the link, Chandan. Your notes will be of use to me for the class I'm teaching. – Pete L. Clark Jan 28 2010 at 16:47 Regarding DLS: he hath revealed unto us his true name. – Pete L. Clark Jan 29 2010 at 8:38 Ah ! He turns out to be a friend ! Welcome, David. – Chandan Singh Dalawat Jan 29 2010 at 9:40 Hi! Likewise, good to see you here. – D. Savitt Jan 29 2010 at 11:51 Thanks to the pointer to your notes Chandan. – Jon Yard Feb 1 2010 at 4:29 A more precise way of expressing your second guess is as follows. Given a number field $K$ (a finite extension of $\mathbb{Q}$), and a place $v$ of $\mathbb{Q}$ (so that $v$ is either a prime number of the sybol $\infty$, corresponding to the completion $\mathbb{R}$ of $\mathbb{Q}$), we may ask what $K\otimes\mathbb{Q}_v$ is, where $\mathbb{Q}_v$ is the completion of $\mathbb{Q}$ at $v$. Pete beat me to the rest. - Just to tack something on about your final question: There exists a rather concrete "yes" answer to your last question in the form of Arakelov theory. One defines an Arakelov divisor as a finite formal linear combination of places, with integral coefficients on the finite places and real coefficients on the infinite places. You map $K^\times$ into the divisor group via the natural definition of principal Arakelov divisors, and if you compose this map with the projection on to the infinite components of the divisor group, you recover the logarithm map from Minkowski theory. So at the very least, ignoring the extra stuff you just chucked in doesn't prevent you from recovering Minkoswki theory. But there's much more to be gained, namely, by considering the quotient of the Arakelov divors by the principal ones and dubbing this the Arakelov class group. The power of this construction, at least as I see it, is to simultaneously deal with two of the most powerful ideas in basic algebraic number theory -- the class group and the Minkowski embedding -- within the context of a single object, the Arakelov class group. This line of thought is pursued in depth beautifully in Chapter 3 of Neukirch's Algebraic Number Theory. Roughly, the Arakelov class group looks a lot more like the divisor class groups arising in function fields, and motivated by this analogy, you can define line bundles, Euler characteristics, genus, canonical bundles, Chern classes, etc., for number fields, culminating with a pretty compelling analogue for Riemann-Roch for number fields. (Again, see Neukirch). - Thanks to pointing me to Neukirch for the Arakelov theory. I knew (loosely) about the analog between number and function fields, but didn't know where to look to find out more. This was actually part of the motivation for my question; since places are the points of a topological space, it seems they should be treated on the same footing as much as possible. – Jon Yard Feb 1 2010 at 4:33 I would say that the finite primes of $K$ play a very strong role in analyzing the structure of a number field! So-called local methods are one of the most powerful tools in developing basic algebraic number theory! It may be that they are not 'essential' in developing the theory, but I think they bring a great deal of clarity to the picture. There are a bunch of good answers already, but I want to throw a couple things on the pile. To begin with, one can think about the primes of $K$ as being given by equivalence classes of metrics on $K$. Equivalently, you can think of them as being given by all the equivalence classes of embeddings of $K$ into complete fields. We know that studying the embeddings of $K$ into $\mathbb{C}$ leads to Minkowkski theory and, among other things, a proof of Dirichlet's Unit Theorem, finiteness of the classgroup, etc. What can we get out of studying the embeddings of $K$ into $K_v$, the completion of $K$ at a finite prime $v$? One nice thing is that when you complete $K$, you also complete $\mathcal{O}_K$, the ring of integers in $K$. If $\mathcal{O}_v$ is this completion than it turns out that $\mathcal{O}_v$ is always a complete discrete valuation ring and as such, questions about ideals in this ring are simpler than they are in $\mathcal{O}_K$. The theory that one develops allows you to easily pass back and forth between questions about ideals in $\mathcal{O}_K$ and ideals in $\mathcal{O}_v$, at the cot of having to consider all of the finite primes. Off the top of my head, this technique makes it very easy to prove functoriality properties of the different associated to an extension of number fields. Also, when working with complete objects, the techniques of analysis become available! An important step in at least one proof of class field theory involves using $p$-adic analogues of the logarithm and exponential functions to relate the structure of the multiplicative group of units in $\mathcal{O}_v$ with the structure of the additive group of $\mathcal{O}_v$. - "using $p$-adic analogues of the logarithm and exponential functions to relate the structure of the multiplicative group of units in $\mathfrak{o}_v$ with the structure of the additive group of $\mathfrak{o}_v$". You can do without. See Part III of arXiv:0711.3878. – Chandan Singh Dalawat Jan 29 2010 at 10:30 Thanks. I hadn't realized that completing the field completes its ring of integers as well, but I guess this makes complete sense in hindsight. – Jon Yard Feb 1 2010 at 4:35	{"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": 102, "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": 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": 0.9368751645088196, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/151785-stationary-points-question.html	# Thread: 1. ## stationary points question "find the stationary points of the following curves and determine whether the points are maxima, minima, or points of inflexion" y= (x + 1)/ (x - 1) I know how to these sorts of questions but for this example i find it tricky, I set it as: y= (x + 1) (x - 1)^-1 but then differentiating this is tricky - not sure how to use the chain rule on the latter, or the product rule overall. thanks for any help. 2. Originally Posted by porge111 "find the stationary points of the following curves and determine whether the points are maxima, minima, or points of inflexion" y= (x + 1)/ (x - 1) I know how to these sorts of questions but for this example i find it tricky, I set it as: y= (x + 1) (x - 1)^-1 but then differentiating this is tricky - not sure how to use the chain rule on the latter, or the product rule overall. thanks for any help. In this case I would use the quotient rule: $f(x) = \dfrac{u(x)}{v(x)}~\implies~f'(x)=\dfrac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}$ You should come out with: $f'(x)=-\dfrac2{(x-1)^2}$	{"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": 2, "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": 0.8792294263839722, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/240582/bernoulli-trials-with-two-random-variables-run-of-successes-and-faliures?answertab=votes	# Bernoulli Trials with two random variables - Run of successes and faliures Consider a sequence of Bernoulli trials with probability of success $p$. Suppose you started the game with a run of succsses followed by the run of faliures (note that you can learn an unlucky run is over if and only if it is followed by a success). Let the random variable $X$ be the number of successful trials any $Y$ be the number of unsuccessful ones (where a run is a sequence of one or more identcal outcome). Find: i) Joint probability $P(X=n, Y=m)$? ii) $E(X)$ & $E(Y)$? iii) Correlation function of $E(XY)$? iv) Covariance $Cov(XY)$? Now after doing some research i believe i have got myself in the right direction for solving these. Here is what i think so far. i) I thik this would be..... $p^n$$^+$$^1$ $q^m$$^+$$^1$?? usinf the fact it will take one extra trial to realise the end to a run ii) not sure iii) have to use this formula prehaps, but not quite sure.... Σ $f(n,m) p (X=n, Y=m)$ iv) I take it this can only be solved if i know the earlier using...$Cov(xy) = E(xy) -E(x) E(y)$ i think these are the right formulas etc i have to use but am struggeling to get past forming problem s which i can then solve. -	{"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": 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": 0.8936731815338135, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/8132/phase-shifts-in-scattering-theory?answertab=active	# Phase shifts in scattering theory I have been studying scattering theory in Sakurai's quantum mechanics. The phase shift in scattering theory has been a major conceptual and computational stumbling block for me. How (if at all) does the phase shift relate to the scattering amplitude? Also, any literature or book references that might be more accessible than Sakurai would be greatly appreciated. - 3 What do you need to know? Its used in partial wave analysis, a common orthogonal expansion . Any function can be decomposed into infinitely many partial waves, the different partial waves correspond to different angular momenta physically. The phase shifts come up as one of the constants that need to determined from the boundary conditions for each partial wave. The scattering amplitude can be expanded in terms of the phase shifts of the waves and spehrical harmonics. I am not writing this as an answer and cluttering it with equations because its there in all standard texts. Eg.-Griffiths etc – yayu Apr 6 '11 at 4:00 1 Also, I don't think that Sakurai is a good way to learn these topics if you are learning about them for the first time. Try the more accessible texts first. I would recommend Shankar\Griffiths. – yayu Apr 6 '11 at 4:09 2 I think, in retrospect my real problem is not really understanding the partial wave expansion. – Cogitator Apr 6 '11 at 14:23 2 upvoting your question as I think a good explanation for partial waves will be good for the site.. you may wish to change your question slightly perhaps to get new answers though – yayu Apr 6 '11 at 17:42 1 And I'm adding a +100 bounty. – Carl Brannen Apr 8 '11 at 21:13 ## 1 Answer Suppose you treat scattering of a particle in a central potential. This means that the Hamiltonian $H$ commutes with the angular momentum operators $L^2$ and $L_z$. Hence, you can find simultaneous eigenfunctions $\psi_{k,l,m}$. You might know, for example from the solution of the hydrogen atom, that these functions can be expressed in terms of the spherical harmonics: $$\psi_{k,l,m}(x) = R_{k,l}(r) \Psi_m^l(\theta, \varphi)$$ where the radial part satisfies $$\frac{1}{r^2} \frac{d}{dr} \left( r^2 \frac{dR_{k,l}}{dr}\right) +\left(n^2 - U(r) - \frac{l(l+1)}{r^2}\right) R_{k,l} = 0$$ with $U(r) = 2m/\hbar^2 V(r)$, your central potential, and $k$ is the particle's wavenumber, i.e., $E = \frac{\hbar^2 k^2}{2m}.$ The first step is to look for a special case with simple solutions. This would be the free particle, with $U(r) = 0$. Then, the radial equation is a special case of Bessel's equation. The solutions are the spherical Bessel functions $j_l(kr)$ and $n_l(kr)$, where the $j_l$ are regular at the origin whereas the $n_l$ are singular at the origin. Hence, for a free particle, the solutions are superpositions of the $j_l$: $$\psi(x) = \sum_{l,m} a_{l,m} j_l(kr) Y^l_m(\theta, \varphi)$$ If we also have axial symmetry, only $m = 0$ is relevant. Then we can rewrite the spherical harmonics using Legendre polynomials. This will lead to $$\psi(x) = \sum_{l,m} A_{l} j_l(kr) P_l(\cos \theta)$$ One important special case of such an expansion is the Rayleigh plane wave expansion $$e^{ikz} = \sum_l (2l+1) i^l j_l(kr) P_l(\cos\theta)$$ which we will need in the next step. We move away from free particles and consider scattering from a potential with a finite range (this excludes Coulomb scattering!). So, $U(r) = 0$ for $r > a$ where $a$ is the range of the potential. For simplicity, we assume axial symmetry. Then, outside the range, the solution must be again that of a free particle. But this time, the origin is not included in the range, so we can (and, in fact, must) include the $n_l(kr)$ solutions to the Bessel equations: $$\psi(r) = \sum_l (a_l j_l(kr) + b_l n_l(kr)) P_l(\cos \theta)$$ Note how the solution for a given $l$ has two parameters $a_l$ and $b_l$. We can think of another parametrization: $a_l = A_l \cos\delta_l$ and $b_l = -A_l \sin \delta_l$. The reason for doing this becomes apparent in the next step: The spherical Bessel functions have long range approximations: $$j_l(kr) \sim \frac{\sin(kr - l\pi/2)}{kr}$$ $$n_l(kr) \sim \frac{\cos(kr - l\pi/2)}{kr}$$ which we can insert into the wavefunction to get a long range approximation. After some trigonometry, we get $$\psi(r) \sim \sum_l \frac{A_l}{kr} \sin(kr - l\pi/2 + \delta_l) P_l(\cos \theta)'$$ So, this is how our wavefunction looks like for large $r$. But we already know how it should look like: If the incoming scattered particle is described as a plane wave in $z$-direction, it is related to the scattering amplitude $f$ via $$\psi(\vec{x}) \sim e^{ikz} + f(\theta) \frac{e^{ikr}}{r}.$$ Obviously, both forms for writing down a long-range approximation for $\psi$ should give the same, so we use the Rayleigh plane wave expansion to rewrite the latter form. We also rewrite the $\sin$ function using complex exponentials. The ensuing calculations are a bit tedious, but not complicated in itself. You just insert the expansions. What we can do afterwards is comparing the coefficients in both expressions for the same terms, e.g. equation the coefficients for $e^{-ikr}P_l(\cos\theta)$ will give you $$A_l = (2l+1)i^l e^{i\delta_l}$$ whereas equating coefficients for $e^{ikr}$ gives you $$f(\theta) = \frac{1}{2ik} \sum_l (2l+1) \left( e^{2i\delta_l} - 1 \right) P_l(\cos \theta).$$ Interpretation of the Phase Shift: Remember the long range limit of the wavefunction. It lead to an expression for the $l$-th radial wavefunction in the long-range of $$u_l(r) = kr\psi_l(r) \sim A_l \sin(kr - l\pi/2 +\delta_l).$$ For a free particle, the phase shift $\delta_l$ would be $0$. One could therefore say that the phase shift measures how far the asymptotic solution of your scattering problem is displaced at the origin from the asymptotic free solution. Interpretation of the Partial Wave Expansion: In the literature, you will often come across terms such as $s$-wave scattering. The partial wave expansion decomposes the scattering process into the scattering of incoming waves with definite angular momentum quantum number. It explains in which way $s$-, $p$-, $d$-waves etc. are affected by the potential. For low energy scattering, only the first few $l$-quantum numbers are affected. If all but the first term are discarded, only the $s$-waves take part in the scattering process. This is an approximation that is, for example, made in the scattering of the atoms in a Bose-Einstein condensate. - "For a typical potential" is a very vague statement. It would be better perhaps to say "in the limit of low energy scattering". – Raphael R. Apr 13 '11 at 21:40	{"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": 39, "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": 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": 0.9234552979469299, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/302132/thoughts-about-sectional-curvature	# Thoughts about sectional curvature I'm currently trying to understand the sectional curvature of riemannian manifolds and I don't know if I'm thinking correctly. So, say we have a riemannian manifold $(M,g)$ with constant sectional curvature $k$. As far as i know the curvature at any given point is completely determined by the metric $g$. After doing some calculations it seems true to me, that if we equip $M$ with a new metric $$\tilde{g} = \lambda \cdot g,$$ where $\lambda$ is just some strictly positive scalar, the curvature of the manifold $(M,\tilde{g})$ is scaled in the same way, i.e. $(M,\tilde{g})$ has constant sectional curvature $\lambda \cdot k$. Now, I know that the sectional curvature of the sphere with radius $1$ is equal to $1$ and the curvature of the sphere with radius $2$ is equal to $\frac{1}{4}$. I'm wondering if, from a riemannian geometry viewpoint, it's somehow the same to look at the radius $2$ sphere, embedded in euclidean space OR the radius $1$ sphere, equipped with the "compressed" metric $\frac{1}{4}g$, where $g$ is the usual scalar product of $\mathbb{R}^n$. This is probably a rather soft question, but maybe someone has something to add or can correct my reasoning :) - ## 1 Answer You probably made a mistake when you computed the curvature of the scaled manifold. In local coordinates $\mathrm{Riem} \approx \partial \Gamma + \Gamma^2$ and $\Gamma \approx g^{-1} \partial g$. So the Riemann curvature as a $(1,3)$ tensor is scale invariant. This means that the sectional curvature $$K(u,v) = \frac{\langle \mathrm{Riem}(u,v)v,u\rangle}{\langle u,u\rangle\langle v,v\rangle - \langle u,v\rangle^2}$$ scales like $g * g^{-2}$ or that $\tilde{K} = \lambda^{-1} K$ if $\tilde{g} = \lambda g$. -	{"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": 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": 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": 0.9399664998054504, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/65185/are-textbfset-and-textbfab-distributive-categories/65186	# Are $\textbf{Set}$ and $\textbf{Ab}$ distributive categories? A quick one-line comment on wikipedia says that the category $\textbf{Set}$ is distributive, but does not offer proof. I think I intuitively see this, as the product of sets is just the cartesian product, and the coproduct is the disjoint union, and I believe that the cartestian product distributes over the disjoint union. However, is there a more rigorous proof to show this? Perhaps that for the canonical morphism $(A\times B)\oplus(A\times C)\to A\times (B\oplus C)$ there is a morphism going in the reverse direction which gives the identity on both objects when composed? It's also mentioned that $\textbf{Grp}$ is not distributive. I don't think I'm quite ready to tackle the concept of the free product, so I want to just look at $\textbf{Ab}$, the category of abelian groups. I know that here the product is just the direct product of groups, and the coproduct is the direct sum of groups. These products look similar in nature, so my hunch is that $\textbf{Ab}$ is also distributive, but I'm not sure either way. Can someone provide a proof or reference that $\textbf{Set}$ is distributive, and whether $\textbf{Ab}$ is or not? - ## 1 Answer You're overthinking the case of $\text{Set}$. You probably know explicit descriptions of the product and coproduct, and it's not hard to show that the canonical morphism is an isomorphism by inspection (show by hand that it's injective, show by hand that it's surjective). $\text{Ab}$ is not distributive. There is a counterexample using finite cyclic groups. - Thanks, do you have a reference for the counterexample with finite cyclic groups? – groops Sep 18 '11 at 0:00 Nope. It's not particularly hard to come up with; I imagine it's an exercise somewhere. – Qiaochu Yuan Sep 18 '11 at 0:22 2 @groops: Let $A=\mathbb{Z}_2$, and let $B$ and $C$ be the trivial group. $(A\times B)\oplus(A\times C)$ is the Klein $4$-group, and $A\times(B\oplus C)$ is $\mathbb{Z}_2$, so they aren’t even the same size. – Brian M. Scott Sep 18 '11 at 23:15 Thanks for the example! – groops Sep 22 '11 at 23:45	{"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": 0.9603349566459656, "perplexity_flag": "head"}
http://mathoverflow.net/questions/3700/explanation-for-e-8s-torsion/5306	## Explanation for E_8’s torsion ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) To study the topology of Lie groups, you can decompose them into the simple compact ones, plus some additional steps, such as taking the cover if necessary. After that, the structure of $SO(n)$'s is rather straightforward, but the exceptional groups are more interesting. Any simple compact Lie group, by means of Hopf algebra theory, has the rational homology of a product $$S^a \times S^b \times \dots \times S^z$$ where the numbers are called exponents. Other than that, their cohomology could also have torsion. Now the torsion for all groups is known: • Among classical groups, only 2-torsion is possible and only for $Spin(n)$ • Exceptional groups can only have 2 and 3-torsion (most do), with the exception of: • $E_8$ which has 2-, 3-, and 5- torsion. Well, this is bound to be related to $E_8$'s Coxeter number, which is 30, but are there any hints as to why? My reference would be math-ph/0212067 but it can't relate this to Coxeter number either. For the reference, exponents are known to be related to Coxeter number, see Kostant, The Principal Three-Dimensional Subgroup and the Betti Numbers of a Complex Simple Lie Group (google search). Is this an open problem? Maybe yes, but maybe it's been explained, so I'm posting it as it is for now. - On the related problem of torsion in the cohomology of BG, there has been recent results by Totaro for spin groups and E_8, see dpmms.cam.ac.uk/~bt219/spin.pdf dpmms.cam.ac.uk/~bt219/torsion.pdf – Simon Pepin Lehalleur Aug 7 2010 at 22:18 ## 6 Answers This doesn't directly address your question, but it does give you a way of thinking about torsion in the cohomology of Lie groups in general. (This is all coming from Borel and Serre's Sur certains sous-groupes des groupes de Lie, which can be found in Commentarii mathematici Helvetici Volume 27, 1953) As you mentioned above, every compact lie group is rationally a product of odd spheres. But how many odd spheres? Turns out, if G is compact and rank k, then it is rationally a product of k spheres (of various dimensions). There is an analogous result for torsion. That is, one can define the 2-group of G to be any subgroup which is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$ or some n. One defines the 2-rank of a group as the maximal $n$ of any 2-group in G. (On can show that for connected $G$, the 2-rank is bounded by twice the rank, and is thus finite). Just to point out something that really threw me when I first learned of these - while the rank is an invariant of the algebra (i.e., all Lie groups with the same algebra have the same rank), the 2-rank of a Group is NOT an invariant of the algebra. For example, the 2-rank of SU(2) is 1 (in fact, -Id is the UNIQUE element of SU(2) of order 2), while the 2-rank of SO(3) is 2 (generated by diag(-1,-1,1) and diag(-1,1,-1) ). The 2-rank of O(3) is 3 (generated by diag(-1,1,1), diag(1,-1,1), and diag(1,1,-1) ). Now, given $T\subseteq G$, the maximal torus, it's clear that simply by taking the maximal 2-group in T, that the 2-rank of G is AT LEAST the rank of G. When is it strictly bigger? Precisely when the group G contains 2-torsion. The analogous result for p-groups and p-torsion (p any prime) also holds. In short, to understand the existence of the 5-torsion in $E_{8}$, one need only understand why there is a subgroup isomorphic to $(\mathbb{Z}/5\mathbb{Z})^n\subseteq E_8$ for some $n\geq 9$. - ### 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. JP Serre, in his june 1999 Bourbaki talk "Sous-groupes finis des groupes de Lie", gives the following two references for torsion in Lie groups R. STEINBERG - Torsion in reductive groups, Adv. in Math. 15 (1975), 63-92 and A. BOREL - Sous-groupes commutatifs et torsion des groupes de Lie compacts connexes, T^ohoku Math. J. 13 (1961), 216-240. Btw, the Bourbaki talk is on Serre's College de France page http://www.college-de-france.fr/media/ins_pro/UPL61366_Serre_Bourbaki_864.pdf Hope this helps. - The last one, at least, seems to be about `G(F_q)` where `G` is a simple algebraic group and `F_q` a field of `q` elements, which is a very different story, as long as I know! I haven't seen the first two though. – Ilya Nikokoshev Nov 2 2009 at 19:34 I don't know the answer to your question, so the following may simply be a repackaging of the mystery, or may be wholly related, and at any rate, is probably already well-known to you. The numbers 2, 3, 5 remind one of the symmetries of the icosahedron, which is related to E8 by the McKay correspondence (see http://math.ucr.edu/home/baez//ADE.html). The ADE Dynkin diagrams are related to finite subgroups of SU(2) which include cyclic groups, dihedral groups, and the three exceptional symmetries: tetrahedral, octahedral, and icosahedral. - I've heard about McKay, but indeed how to apply it here would be a mystery! – Ilya Nikokoshev Nov 2 2009 at 18:48 Take a look at "Finite H-spaces and Lie Groups" by Frank Adams, particularly the letter from E8 and the appendix which follows it. - To throw in a bit more numerology from another Serre's Bourbaki seminar: Cohomologie galoisienne : progrès et problèmes. Séminaire Bourbaki, 36 (1993-1994), Exposé No. 783, 29 p. [available at numdam]. In \S 2.2 he refers to torsions related to Lie groups in two senses, as far as I understand, the second one is related to group of automorphisms of the completed Dynkin diagram. - Sorry, I forgot to mention in my answer that only \S 1.3 in Serre 99 is about torsion in Lie or algebraic groups in zero characteristic (which doesn't prevent him to try embed some finite groups of Lie Type into them !). PS: didn't found how to append this to your answer to my answer. - You'll be able to comment once you get more reputation (50, I think) – Ilya Nikokoshev Nov 3 2009 at 15:13 Also, I think there's some strange bug or perhaps you created two logins by mistake. This one has 1 rep and the other one has 21 rep. – Ilya Nikokoshev Nov 3 2009 at 15:14	{"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": 11, "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": 0.9268255233764648, "perplexity_flag": "middle"}
http://mathforum.org/mathimages/index.php/Russell's_Antinomy	Russell's Antinomy From Math Images The Set of All Sets Which Do Not Contain Themselves Fields: Algebra, Number Theory, and Algebra Created By: Peter Weck Website: [ ] The Set of All Sets Which Do Not Contain Themselves The blob on the right represents the set of all sets which are not elements of themselves. At first such a set might seem logically acceptable, but it leads straight to a famous contradiction known as Russell’s Antinomy or Russell’s Paradox. Basic Description In the most basic sense of the word, a set is just a collection of objects, or elements. When we say a set contains or includes something, we mean the thing is an element of the set. These elements could be teacups, ideas, numbers, or other sets. Sets can even contain themselves as elements. For example, the set of all ideas is itself an idea. Of course, not all sets contain themselves. The set of all teacups is most definitely not a teacup. This allows us to divide all imaginable sets into two distinct categories, two enormous sets: the set of all sets which contain themselves (e.g., the set of all ideas) and the set of all sets which do not contain themselves (e.g., the set of all teacups). The blob in the main image represents the set of all sets which do not contain themselves. We can see a few examples of elements of this set floating around the blob's interior. The set of all numbers is an element because it is not a number, and the set of all people is an element because it is obviously not a person. This is all well and good, until we consider whether the set of all sets which do not contain themselves contains itself as an element. The resulting antinomy is best illustrated with an analogy. The Barber Analogy Who shaves the barber? Let me tell you a story... Once upon a time there was a town with strict laws on shaving. Everyone was required by law to shave daily. Those who didn't feel like shaving themselves went to the town barbershop. The barber who owned the shop had been legally appointed “the man who shaves all and only those who don’t shave themselves”. This was all well and good, until the barber was arrested for being unshaven. You might think the barber could have just shaven himself, but the law said that the barber could only shave those who don’t shave themselves. And if he didn't shave himself, he was supposed to be shaved by the barber. But he is the barber, and once again, the barber was not supposed to shave men who shave themselves! When the judge realized that it was impossible for the barber to follow the law, he ruled that the law be revised to resolve the contradiction and that the barber be released from jail. The set of all sets which do not contain themselves is analogous to the barber who only shaves men who do not shave themselves. If the barber shaves himself, he is no longer a person who is allowed to be shaved by the barber; if the set of all sets which do not contain themselves contains itself, it is no longer allowed to contain itself. If the barber does not shave himself, he should be shaved by the barber; if the set does not contain itself, it should be contained in itself. This is Russell’s Antimony. Just as the law is impossible for the barber to follow, the set of all sets which do not contain themselves is a logical impossibility. See the image below for a visual presentation of the contradiction. It must either contain itself or not contain itself, and since both possibilities yield contradictions, this set is a logical impossibility. A little History The year was 1901. Just when the world of mathematics had dealt with one set of paradoxes and foundational crises regarding limits in calculus, a new batch had emerged from set theory.[1] Like many of his peers at the time, young philosopher and mathematician Bertrand Russell was hard at work trying to construct a firm logical foundation for all of mathematics. While working on the first of his books in the area, the Principles of Mathematics, Russell came upon the idea of a set of all sets which do not include themselves.[2] He soon realized that when you consider whether this, in his words, "very peculiar class" includes itself, "each alternative leads to its opposite and there is a contradiction".[2] Russell's Antinomy came to be the most famous paradox in set theory.[1] The concept of a set, or class as Russell called it, was crucial for the program of deriving the foundation of mathematics from logic.[2] Russell's discovery of a paradox stemming from the accepted conception of a set was like a crack in this foundation. It prompted a refinement of the concept of a set and much later work in logic, set theory, and the philosophy of the foundations of mathematics. [3] For more on the implications of Russell's Antinomy, see this section. A More Mathematical Explanation [Click to view A More Mathematical Explanation] Some Basic Notation Before describing Russell’s Antinomy mathematically, we need to get acqua [...] [Click to hide A More Mathematical Explanation] Some Basic Notation Before describing Russell’s Antinomy mathematically, we need to get acquainted with some basic set theory notation. Membership Relation The fundamental concept of set theory is that of inclusion, or membership, of an object in a set. The symbol ∈ denotes the membership relation between an object and a set. To say that the set A contains the element a, or equivalently that a is in A, we write $a \in A$ To say that Bertrand Russell was a person using set theory notation, where Russell is represented r and the set of all people P, we write $r \in P$ We can use a similar notation to indicate that something is not included in a set. For example, to say that Bertrand Russell was not a fish, where the set of all fish is F, we write $r \notin F$ List Notation Sets are often represented as lists of elements separated by commas and contained in brackets. For example, a set A containing only the elements a and b is written $A = \{ a,b \}$ A set Y consisting of the numbers 1, 2, 3, and 4 looks like $Y = \{ 1,2,3,4 \}$ Representing a set with a list like this can be tedious if the set is especially large. Some sets have infinitely many elements, and can't be written this way at all. In cases like these, set-builder notation is usually used. Set-builder Notation Set-builder notation describes a set by stating the properties that its elements must satisfy. This allows us to represent sets of any size without trouble. For example, while the set of all people P would take quite a long time to write out as a list, we can represent it in set-builder notation with $P = \{ x:x \text{ is a person} \}$ This formula would be read "P is the set of all x such that x is a person". The set Y we defined earlier as a list of the numbers 1, 2, 3, and 4 could also be written in set-builder notation. Where the set of all integers is $\mathbb{Z}$, $Y = \{ n \in \mathbb{Z}:0<n<5 \}$ Which would be read "Y is the set of all integers n such that n is greater than 0 and less than 5". Russell's Antinomy If S is some set, then Russell's set of all sets which do not contain themselves, which we will call R, is represented $R = \{ S:S \notin S \}$ Which means that saying $S \in R$ is equivalent to saying $S \notin S$ or formally $S \in R \Leftrightarrow S \notin S$ Where the symbol $\Leftrightarrow$ is read "if and only if". As discussed earlier, the paradox arises when we consider whether R includes itself. We can represent this question of self-inclusion formally by substituting the set R for the variable set S in the expression above. This gives us $R \in R \Leftrightarrow R \notin R$ which is a contradiction. Why It's Interesting hello The Implications of the Paradox Russell's Antinomy was like a crack in the logical foundation of mathematics. In so-called naïve set theory prior to the discovery of Russell’s Antinomy, a set was simply a collection of objects with some common property. It didn’t matter what the property was or what the objects were. Mathematically, this definition was embodied by the Comprehension Axiom of set theory. This axiom essentially says that any propositional function P(x) can be used to determine a set. So for any statement of a property P, there exists a set whose elements are all the things x which satisfy the property.[3] We will return to this axiom of naïve set theory later when discussing the resolution of the paradox. For now, the takeaway is that the generality of the classical defintion of a set is what makes a set of all sets which do not contain themselves an acceptable notion, and thus makes Russell’s Antinomy possible. When Russell discovered the paradox, Gottlob Frege was working on his Grundgesetze der Arithmetik, which attempted to establish a foundation for mathematics using symbolic logic. Unfortunately, much of Frege’s work relied on the conception of a set which leads to Russell’s Antinomy.[2] Russell wrote to Frege after discovering his paradox, just when the second volume of Frege’s Grundgesetze was going to press. Realizing the significance of the paradox, Frege added a note to the end of his book saying “A scientist can hardly meet with anything more undesirable than to have the foundations give way just as the work is finished. I was put in this position by a letter from Mr. Bertrand Russell when the work was nearly through the press.”[2] It wasn't just Frege's work that was jeopardized by Russell's Antinomy. As mentioned earlier, the entire enterprise of shoring up mathematics on a logical foundation, including Russell's own work, needed set theory to have any hope of succeeding. And set theory wasn't going to be up to the task until Russell's paradox had been resolved. The Resolution of the Paradox This was no easy task. Various resolutions have been developed over the years, most focused on restricting the principles governing what qualifies as a set, like the Comprehension Axiom, so that a set of all sets which do not contain themselves is not an acceptable notion in the first place.[3] The Theory of Types Russell's own response to the paradox was his theory of types, which can be illustrated with a continuation of the barber story told earlier... After the trial of the barber, the town leaders met to resolve the confusion created by their shaving law. After much debate, they agreed on a new shaving law establishing a caste system in the town. Each and every townsperson thenceforth belonged to a caste, and was only allowed to be shaven by someone belonging to a lower caste. The barber, for example, ended up in caste 3, and while his business was restricted to only shaving people in higher castes like 4, 5, and 6, he no longer had to worry about whether or not he could shave himself since he was shaven by members of castes 1 and 2. The law was quite controversial, since it meant that nobody was allowed to shave themselves anymore. They instead had to pay to be shaven by members of a lower caste. The members of caste 1 were especially unhappy. There was no lower caste to shave them, and as a result they became quite scruffy. As before, the people in the story are like sets, and shaving someone is like being contained in a set. Russell's theory of types does away with the Comprehension Axiom and imposes a hierarchy on sets, such that sets can only contain objects of lower types, just like the caste system for shaving imposed on the townspeople. The equivalent of caste 1 in the story are objects called individuals in the theory of types. Sets containing only individuals are like members of caste 2 who can only be shaven by members of caste 1. Sets of sets, or families of sets, are analogous to members of caste 3, and so on. Just as the caste system abolishes self-shaving and with it the barber's paradox, Russell's theory of types makes it so that self-inclusion and Russell's antimony can't occur in the first place. Connection to Principia Mathematica Russell and Whitehead's Principia was so exact that it took them until page 362 to prove 1 + 1 = 2. As you can see, the language used is more symbolic logic than English. Russell's theory of types served as the vehicle for his groundbreaking work formalizing the foundations of mathematics.[4] The theory was first discussed in Russell's original Principles of Mathematics published in 1903, but wasn't fully formed for about five years after that. It was eventually laid out by the monumental Principia Mathematica he co-authored with Alfred North Whitehead. The Principia was the product of years of work seeking to succeed where Frege had failed and establish a foundation of symbolic logic for all of mathematics. It presented a system of axioms and rules of reasoning from which all of mathematics was to be formulated and proved.[5] While the Principia was certainly a milestone in mathematics, it is essentially an unfinished work and many would even say a lost cause.[1] The hierarchy that the theory of types imposes has been criticized for being too ad hoc to eliminate Russell's paradox successfully. And even if the Principia does do away with the paradox, it has other problems which make it fall short of the immense ambitions of its authors. Kurt Gödel In 1931, the 25-year-old mathematician Kurt Gödel proved two theorems, known as Gödel's Incompleteness Theorems, that dealt a crushing blow to the mission of the Principia Mathematica. The first of these theorems states that if the system laid out by the Principia is consistent, it must necessarily be incomplete. Incomplete means that there is a true statement in the language of the system (symbolic logic) that can neither be proven nor disproven using the system. Any such statement is said to be Undecidable within the system. The second theorem states that the consistency of the system used by the Principia cannot be established within the system. If you could prove the system was consistent using the rules of the system, it would in fact be an indication that the system is inconsistent. So even if Russell's Antinomy and all other paradoxes could be eradicated from Principia, it would still not describe all the truths in number theory. Furthermore, the Principia could never prove itself to be free of paradoxes. The foundations of mathematics could never be completed. Other Resolutions of Russell's Antinomy The theory of types is not the only possible resolution of the paradox. Various alternative set theories, most modifying or excluding the Comprehension Axiom, have been developed to deal with Russell's Antinomy. Among them are Gödel-Bernays set theory, Zermelo-Fraenkel set theory, and Quine's New Foundations.[6] Although Russell's theory of types is still used in areas of computer science and some philosophical investigations, it is no longer considered to be the best mathematical formulation of set theory. Today a version of Zermelo-Fraenkel theory is generally used.[3] Zermelo-Fraenkel Theory Zermelo-Fraenkel theory, or ZF, resolves the paradox in a way not dissimilar to that used by Russell. Instead of the “top down” approach to sets embodied by the Comprehension Axiom, where any property is sufficient to define a set, ZF adopts a “bottom up” approach, where only those sets exist that can be explicitly constructed from already-constructed sets, starting with individual elements and following carefully constrained operations. This ZF conception of a set is said to be iterative, since new sets are built up from old sets. In ZF, the Comprehension Axiom is restricted so that instead of being able to define a set of all things satisfying some property, we can only define a set of all things from an existing set which satisfy the property. This new axiom is known as the Restricted Comprehension Axiom or the Separation Axiom.[7] With the Separation Axiom, we can’t form a set of all sets which do not contain themselves, only a set of all sets from some preexisting set which do not contain themselves. Without a set of all sets which do not contain themselves, the contradiction discovered by Russell cannot be proven and Russell’s Antinomy doesn’t exist. Although one axiom of ZF called the Axiom of Choice was initially highly controversial, today ZF is the generally accepted form of set theory.[7] It too is subject to the constraints proven by Gödel's Incompleteness Theorems, and thus cannot provide the complete and consistent foundation for all of mathematics that Russell and his peers sought, but set theory remains a vital underlying branch of mathematics. Self-Reference Self-reference can get pretty loopy. Self-reference lies at the heart of many paradoxes, Russell's Antinomy among them. Another famous example is the liar paradox, arising from the self-referential statement "This sentence is not true", or "I am lying". In the context of set theory, self-reference isn't about truth of statements but membership of sets. From this perspective, the set of all sets which do not contain themselves is like the set theory version of the liar sentence. Russell recognized the role of self-reference in generating his paradox, concluding that "whatever involves all members of a collection must not itself be a member of the collection".[2] In other words, self-reference had to be banished from the realm of set theory, a goal which Russell accomplished with his theory of types in the Principia. As Douglas Hofstadter puts it, "Principia Mathematica was a mammoth exercise in exorcising strange loops from logic, set theory, and number theory".[1] An example of the "infinite hallway" effect. We can visualize self-reference, or "strange loopiness" in Hofstadter's lingo, using pictorial self-containment. A picture is said to be self-referential if it contains a copy of itself in miniature, which then must contain another even smaller copy, which itself contains another, ad infinitum. We can see this "infinite hallway" effect in the animation on the left of a self-referential picture of a room being constructed step by step. We also see this visual consequence of self-reference in this page's main image illustrating Russell's Antinomy. When we visualize a self-containing set with a blob containing a smaller copy of itself, that copy must include a smaller copy, and that copy a smaller copy, and so on. Unfortunately this is hard to draw beyond two or three copies. As a side note, not all self-referential things are paradoxical and not all paradoxes are self-referential. For example, this page is self-referential, because of sentences like this, and is about a paradox, but I wouldn't go so far as to say it is paradoxical. Teaching Materials There are currently no teaching materials for this page. Add teaching materials. Related Links Additional Resources • For a good story about the life of Bertrand Russell and the search for the foundations of mathematics, I'd suggest reading the graphic novel Logicomix. • If you're interested in self-reference, you should definitely look into Douglas Hofstadter's book I am a Strange Loop. • For more about paradoxes exhibiting self-reference, check out the Stanford Encyclopedia of Philosophy's page about self-reference. • For more on resolutions for Russell's paradox besides the theory of types, look at the third section of this page. References 1. ↑ 1.0 1.1 1.2 1.3 Hofstadter, D. (1979). Gödel, Escher, Bach: an eternal golden braid. New York: Basic Books, Inc. 2. ↑ 2.0 2.1 2.2 2.3 2.4 2.5 Dunham, W. (1994). The Mathematical Universe: an alphabetical journey through the great proofs, problems, and personalities. New York: Jon Wiley & Sons, Inc. 3. ↑ 3.0 3.1 3.2 3.3 Irvine, A. D. "Russell's Paradox". The Stanford Encyclopedia of Philosophy (Summer 2009 Edition). Edward N. Zalta (ed.). Web. 10 Jul. 2012. http://plato.stanford.edu/archives/sum2009/entries/russell-paradox/. 4. ↑ Baldwin, J. and Lessmann, O. "What is Russell's paradox?". (1998) Scientific American. Web. 10 Jul. 2012. http://www.scientificamerican.com/article.cfm?id=what-is-russells-paradox. 5. ↑ Franzen, T. (2005). Gödel's Theorem: an incomplete guide to its use and abuse. Wellesley, MA: A K Peters, Ltd. 6. ↑ Bolander, T. "Self-Reference". The Stanford Encyclopedia of Philosophy (Summer 2009 Edition). Edward N. Zalta (ed.). Web. 10 Jul. 2012. http://plato.stanford.edu/archives/win2009/entries/self-reference/. 7. ↑ 7.0 7.1 "Zermelo-Fraenkel set theory". Encyclopedia Brittanica (Encyclopedia Brittanica Online Academic 2012 Edition). Web. 10 Jul. 2012. http://www.britannica.com/EBchecked/topic/656629/Zermelo-Fraenkel-set-theory. 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": 14, "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": 0.9478551149368286, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/248693/for-a-finite-group-g-and-field-k-of-char-p-if-p-p-are-projective-kg	# For a finite group $G$ and field $k$ of char$=p$, if $P,P'$ are projective $k[G]$-modules with $[P]=[P']$, is it true that $P=P'$? That is -- is it true that if projective $k[G]$-modules have same composition factors then they are isomorphic? This is easy to see for $\text{char}(k)=0$, or if $G$ is a composition of a $p$-group and a $p'$-group. Serre in "Linear Representations of Finite Groups" (a remark in 16.2 after Corr.2) states this as a well-known fact: "Indeed we know that the equality $[P] = [ P']$ (...) is equivalent to $P = P'$)". But unfortunately no references. -	{"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": 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": 0.9281941056251526, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/7780-z60.html	# Thread: 1. ## Z60 The ring Z60 is a field right? So a list of all its ideals would be 0 and all the elements in Z60. Is this correct? And could someone just explain quickly what a homomorphic image is? Thank you. 2. Originally Posted by Smitey42 The ring Z60 is a field right? No, is it to a power of a prime? No it is not. So a list of all its ideals would be 0 and all the elements in Z60. A field has no ideal, since this is it not a field it might. (Note: I worked on this problem before when you posted it and it got a little messy by looking at all the possible ideal. What I was doing was finding all the cyclic subgorups of this ring because they need to be cyclic for the additive group is cylcic) Is this correct? No And could someone just explain quickly what a homomorphic image is? I think you mean when you have a ring homomorphism the image of $\phi[R]$ is a homomorphic image.	{"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": 1, "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": 0.9675725698471069, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/7713/predicting-decay-rates-via-the-standard-model	# Predicting Decay Rates via the Standard Model Question 7584 illustrated a procedure to forecast the decay rates of isotopes with known long average lifetimes. Lifetimes of the many U isotopes vary from micoseconds to gigayears. F has only one stable isotope while Sn has 10. Can Standard Model principles be used to predict the stability of isotopes and the average lifetimes for unstable isotopes, or can this only be done by measurement? - ## 2 Answers The nuclear forces are a complex amalgam of primarily Quantum Chromodynamics forces and electromagnetic ones, but to deal with the diagramatic way of calculating in Quantum Field Theory, is not possible. The weak force responsible for beta decays should also be in the calculations.Too many diagrams and too convoluted. Quantum mechanical models with a potential well to estimate the collective forces are used for this. Nuclear physics has been using various models successfully, like the shell model. to predict energy levels in nuclei. Ways of estimating lifetimes are taught in nuclear engineering, for example: Course Outcomes: Students must be able to... calculate the consequences of radioactive growth and decay and nuclear reactions. calculate estimates of nuclear masses and energetics based on empirical data and nuclear models. calculate estimates of the lifetimes of nuclear states that are unstable to alpha-,beta- and gamma decay and internal conversion based on the theory of simple nuclear models. use nuclear models to predict low-energy level structure and level energies. - In principle, the Standard Model contains the answer to all these questions about half-lives. In practice, it is impossible to calculate at this point - as far as I know. However, it is very easy to explain why the lifetimes span such an exponentially huge interval of possible time scales. Here's why. An alpha-decay - and only alpha-decaying nuclei with "really long-lived" half-lives will be explained by the simple argument below - may be approximated as the confinement and quantum tunneling of an alpha particle. Imagine that a big nucleus $M$ decays to a smaller one $N$ and an alpha-particle: $$M \to N + \alpha$$ In this context, $M$ may be viewed as a bound state of $N$ and $\alpha$. The alpha-particle is confined by the attractive potential of the nucleus $N$. However, we know that this bound state isn't quite stable - it decays. It follows that a free $N$ plus a free $\alpha$ must have a lower total mass/energy. Consequently, the bound state of $N$ and $\alpha$, and that's what we called $M$, must be metastable and the alpha-particle is confined by a potential wall. Classically, it would stay there forever. Quantum mechanically, there is quantum tunneling. There's a nonzero probability that the alpha-particle tunnels through the potential wall and gets out. However, the rate of this event is exponentially small because the wave function, having an imaginary momentum inside the potential wall (a negative kinetic energy needed to go through the wall), exponentially decreases in the wall. The probability or decay rate $\Gamma$ is then comparable to $$\Gamma = dP/dt \sim \exp (-V/V_0) \times (10^{-24}{\rm seconds})^{-1}$$ where $V$ is the total "thickness" of the potential created by the nucleus $N$ inside the bound state $M$, and $V_0$ is some typical QCD-scale unit for such thickness. To get the right dimensions, I also included $10^{-24}$ seconds as the typical "QCD time scale". However, as you can see, the half-lives of the isotopes will be exponentially larger. Relatively moderate changes of $V/V_0$ - from 20 to 100 - may change the lifetime from a fraction of a second to billions of years. This qualitative pattern is surely well understood. In fact, approximate estimates of the lifetimes can be done by models in nuclear physics, and those models can be at least approximately be showed compatible with the Standard Model. But no one has calculated the half-life of uranium-235 via a bound state of many quarks and gluons - as far as I know. Also, however, there is not a single glimpse of a demonstrable discrepancy. Well, I was cheating a bit: even beta-decay may lead to incredibly long lifetimes. An example is potassium-40 which can decay by any three types of the beta-decay - and its lifetime is still over a billion of years. Microscopically, the beta-decay is a bit different than the alpha-decay because it requires a real transformation of the types of particles - not just a different state of protons' and neutrons' motion and position. At the same moment, beta-decay is somewhat simpler because its "microscopic process" is just a single weak interaction vertex with a W-boson - and the rest of the nucleus is less changed (only 1 nucleon is added/killed, instead of 4). In some sense, the beta-decays combine the quantum tunneling above with the elementary W-boson-induced weak interaction and they may have very diverse lifetimes, too. - This is a rather small gain in reputation compared to Your standards, Lubos, I wonder why? (tongue in cheek) :=( – Georg Apr 1 '11 at 11:27 You know, people suck compared to what they normally do. ;-) – Luboš Motl May 20 '11 at 15:08	{"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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9406489729881287, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/tagged/descriptive-set-theory	## Tagged Questions 0answers 109 views ### From universal measurability to measurability Let $(\Omega,\Sigma)$ be a measurable space and $K$ be a compact metrizable space endowed with its Borel $\sigma$-algebra $\mathcal{B}(K)$. Let $A\subseteq\Omega\times K$ be … 1answer 325 views ### How long can it take to generate a $\sigma$-algebra? I want to know if there is a $\sigma$-algebra such that for every countable ordinal $\alpha$ the $\sigma$-algebra can be generated in more than $\alpha$ steps but less than \$\omega … 0answers 100 views ### Restrictions of null/meager ideal Let I denote the null (resp. meager) ideal on reals. Is it consistent that for any pair of non null (resp. meager) sets A and B, there is a null (resp. meager) preserving bijection … 1answer 75 views ### Borel ideals on $\omega$ are meager? Let $\mathcal{I}$ be a proper ideal on $\omega$. If $\mathcal{I}$ is Borel as a subset of $2^\omega$, does it follow that $\mathcal{I}$ is meager? Edit: What if $\mathcal{I}$ cont … 2answers 218 views ### Cohen algebra (generalization) Let Bor($X$) = class of all borel subsets of $X$. Cohen algebra is defined as Bor(X) modulo the ideal of meager sets. The Cohen algebra has a combinatorial : it is the uniqu … 1answer 101 views ### Basis theorem (due to Solovay?) I'm finishingg up my bibliography and I'm looking for a reference for the statement that, working in $L(\R)$, the $\Delta^2_1$ sets form a basis for the $\Sigma^2_1$ predicates. I … 2answers 408 views ### Image of L^1 under the Fourier Transform The Fourier Transform $\mathcal{F}:L^1(\mathbb{R})\to C_0(\mathbb{R})$ is an injective, bounded linear map that isn't onto. It is known (if I remember correctly) that the range is … 2answers 124 views ### Obtaining conditional probabilities as pushforwards of [0,1] It is standard that every Borel probability measure on a polish space $X$ can be obtained as pushforward of the uniform measure $\lambda$ on $[0,1]$ along an almost-everywhere-defi … 0answers 208 views ### Godel on recursion-theoretic hierarchies At the end of his excellent article, "The Emergence of Descriptive Set Theory" (http://math.bu.edu/people/aki/2.pdf), Kanamori writes: "Another mathematical eternal return: To … 1answer 184 views ### Models of Determinacy Today we have that $L(\mathbb{R}) \models AD$ (assuming there are $\omega$ many Woodin cardinals and a measurable above them all). I was wondering what other models of $AD$ might l … 1answer 81 views ### Product of Baire sigma-algebras Suppose that $X$ is a Polish space and $\mathcal{E}$ is the $\sigma$-algebra of subsets of $X$ with the property of Baire. Consider the product $\sigma$-algebra \$\mathcal{E}\otim … 1answer 126 views ### Complexity of winning strategies for open games (for open player) If $G\subseteq\omega^{<\omega}$ is a computable clopen game, then $G$ has a winning strategy which is hyperarithmetic $(\Delta^1_1)$, by an inductive ranking process. The key ob … 1answer 136 views ### When do substructures have computable copies? Say that a class $\mathcal{C}$ of countable first-order structures in some finite signature has the effective substructure property if $\mathcal{C}$ is closed under isomorphism and … 1answer 146 views ### Ensuring nonempty lightface Borel sets have elements via theories of second-order arithmetic This question is an outgrowth of this MathSE question: http://math.stackexchange.com/questions/276068/members-of-lightface-borel-sets. A Borel set $X\subseteq 2^\omega$ is a memb … 0answers 87 views ### Open games formed by pasting together infinitely many clopen games Throughout, I think of games and their underlying trees as the same: so a "clopen game" and a "well-founded tree" mean the same thing. Fix a sequence of clopen games \$\lbrace T_i: … 15 30 50 per 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": 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": 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": 0.9058524966239929, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/27489/do-any-entanglement-measures-for-mixed-states-exist-that-use-only-single-site-co/27491	Do any entanglement measures for mixed states exist that use only single site correlation functions? For a pure state $\rho_{AB}$, the entropy of entanglement of subsystem $A$ is \begin{equation} S( \rho_A) = -tr (\rho_A \log \rho_A) \end{equation} where $\rho_A$ is the reduced density matrix of A. For a single site of a spin chain, $\rho_A$ can be written in terms of single site correlation functions $\langle \sigma_l^\alpha \rangle$ where $\alpha = x,y,z$. Are there any entanglement measures for mixed states that use only the same correlation functions, $\langle \sigma_l^\alpha \rangle$ where $\alpha = x,y,z$? - @PiotrMigdal Thanks, but no, that's not what I'm asking. I know that the density matrix can be written in terms of the Pauli matrices. I want to know whether there are any entanglement measures for mixed states which only use the 'single site' correlation functions (i.e. the expectation values of the Pauli matrices). I hope that's clearer. – Calvin Jan 25 '12 at 21:45 2 I guess one would not call that "correlation functions", so maybe that's what caused the confusion. With this clarification, the answer to your question is no: The maximally entangled and the (unentangled) maximally mixed state have the same reduced single site density matrices (and thus the same expectation values for any local operator), but completely different entanglement. – Norbert Schuch Jan 26 '12 at 4:42 @NorbertSchuch Slightly off topic, but why wouldn't you call them correlation functions? – Calvin Jan 26 '12 at 16:52 @Calvin: Wikipedia says "A correlation function is the correlation between random variables at two different points in space or time". – Norbert Schuch Jan 26 '12 at 20:00 2 Answers It seems that such a measure for mixed states is fundamentally impossible, since you can have both entangled and separable states which have exactly the local expectation values. For pure states, monogamy of entanglement ensures that the impurity of a reduced density matrix (which can be infered from the expectation values of local Pauli operators) is directly related to entanglement. However for mixed states, this is not the case, as the following example will homefully make clear: Consider a two qubit system, in which the two reduced density matrices are maximally mixed. In this case, it is possible the system is separable, composed of two copies of the maximally mixed state, or it is maximally entangled, composed of a single EPR pair, or anything in between. Thus no function of local expectation values can distinguish separable from entangled states in general. However, purity (which is a function of single site correlation functions), can indeed be used as a bound on the entanglement of a system, again due to monogomy of entanglement. If the local system is not maximally mixed, then it is not maximally entangled, and hence maximum amount of entanglement possible for a system is a monotonic function of its (im)purity. - Thank you. Just to check, by local expectation value, you mean one made up of just one Pauli matrix, while something like $\langle \sigma_l^\alpha \sigma_{l+m}^\beta \rangle$ with $\alpha,\beta = x,y,z$ is not. – Calvin Jan 26 '12 at 16:50 @Calvin: Exactly, although there is no need to restrict to just Pauli operators. – Joe Fitzsimons Jan 26 '12 at 20:10 As Joe says, the local spectrum will still tell you something about entanglement if you have some kind of promise/information about the global purity. For example, if $S(\rho_{AB})<S(\rho_A)$, then the state is $A:B$-entangled – Marco Jan 26 '12 at 21:06 Joe: Sorry to reply here, but the website won't let me edit your post (not enough characters); you've made a typo with 'homefully' (hopefully) and also 'monogomy' (monogamy). Please delete this comment, if you can, as it's otherwise useless. – Noon Silk Jan 26 '12 at 22:45 1 @calvin: As long as the expectation values are all within the same part of the system, they all won't tell you anything about the entanglement (unless you are given additional information). – Norbert Schuch Jan 27 '12 at 5:42 For any multipartite state $\rho_{ABC...}$ the product state $\rho_A\otimes\rho_B\otimes\rho_C\otimes\ldots$, with $\rho_A$, $\rho_B$, $\rho_C$, ... the local reduced density matrices of $\rho_{ABC...}$, has exactly the same single-site expectation values. So, with only single-site expectation values one cannot say anything about any kind of correlation, not only about entanglement. On the other hand, thanks to the Schmidt decomposition and the invariance under local unitaries of entanglement measures, it is obvious that all bipartite entanglement measures on pure states can only depend on the spectrum of the reduced density matrix. -	{"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": 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": 0.8930370211601257, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/48024/likelihood-ratio-test-for-mle-markov-chain	# Likelihood ratio test for MLE (Markov Chain) As suggested in Calculating log-likelihood for given MLE (Markov Chains) I want to perform a likelihood ratio test for two fitted models (i.e., first and second order markov chains). Simply comparing the resulting log-likelihood values is as suggested in the other thread not enough. I know how to calculate the likelihood ratio, but I am unsure about how to determine the statistical significane. I need the degree of freedoms for both models. What exactly is this in the case of my Markov Chain MLEs. The number of non-zero probabilities in the MLE, or the number of states? - Is there any structure to the model or are you estimating the distribution of $Y_{t}\| Y_{t-1} = i, Y_{t-2} = j$ independently for every combination of $i,j$.? – Macro Jan 18 at 16:06 @Macro What do you mean exactly by structure? Could you give an example? – ph_singer Jan 19 at 15:10 I am basically asking whether you are building a model for $Y_t \| Y_{t-1} = i, Y_{t-2} = j$, e.g. $P(Y_t) = f(Y_{t-1}, Y_{t-2})$, where you estimate $f$; a simple example would be a series of ordinal (if $Y_t$ is ordinal) or multinomial logit regression models. Or, are you just estimating each of the probabilities independently? In that case, if there are $m$ states, I think you would be estimating $m^2 (m-1)$ parameters, since there are $m^2$ possible combinations of the $t-1,t-2$ states and $m-1$ cell probabilities to estimate (since the $m$'th one is determined by the others). – Macro Jan 19 at 20:56 @Macro Sorry for the late response. I am estimating each of the probabilities independently. So it should be the latter case. I am reducing a higher order MC to a k=1 MC, so this would be the case as well, right? But, $m$ states would always refer to the number of states of the input data, or would it change for the new states I create for my reduced model? Because, the reduced higher order MC has a higher states space $S'$. – ph_singer Jan 29 at 10:43	{"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": 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": 0.9221175909042358, "perplexity_flag": "head"}
http://mathoverflow.net/questions/102802/quasi-unipotent-monodromy-for-general-families	## Quasi-unipotent monodromy for general families ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) This must be a naive question, but I'm wondering about the definition of the quasi-unipotent monodromy for general (not only 1-parameter families). The problem is that usually in the books of algebraic geometry quasi-unipotent monodromy is only discussed over a disc $\Delta ^{}$ i.e. for a 1-parameter family. In this case we know that for a fibration $f: X \rightarrow \Delta$, with monodromy representation $\rho : \pi_{1}(\Delta^{}) \rightarrow Aut(H^{i}(X_{0}))$, the image $\rho (T)$ of a generator $T$ of $\pi_{1}(\Delta^{*})$ is a quasi-unipotent matrix. What is the correct generalization of this to arbitrary families? For example, in multi-parameter case, it could happen that there are several generators $T_{i}$ such that each $\rho(T_{i})$ is a quasi-unipotent matrix but for example $\rho(T_{1})\rho(T_{2})$ is not quasi-unipotent . So is the possible generalization is that all of the matrices in the monodromy group (image of the monodromy representation) should be quasi-unipotent or it suffices only that the image of the generators to be quas-unipotent? - 2 In higher dimensions, I think quasi-unipotent monodromy only makes sense in the case of normal crossings boundary. In this case the local fundamental group is abelian, hence if each the image of each generator is quasi-unipotent, so is the image of any other element. – ulrich Jul 21 at 12:58 You mean if the discriminant is not NC, the notion of quasi-unipotency is not defined? Is there a reference which discusses the quasi-unupotency in details? anyway thank you very much for your anser. – Jack Jul 21 at 14:32 You should check out the answer to this: mathoverflow.net/questions/1912/… – Igor Rivin Jul 21 at 17:55 ## 1 Answer Quasi-unipotency is a well defined notion at any point of the discriminant. If we have a proper family $f : X \to S$ of varieties with a smooth total space and a smooth base, and if $p \in D \subset S$ is a point of the discriminant, then we say that the local monodromy of the family near $p$ is quasi-unipotent if we can find a small analytic neighborhood $p \in U \subset S$ of $p$ in $S$, so that if $o \in U - D$ is a base point, then the monodromy representation $mon : \pi_{1}(U-D,o) \to GL(H^{i}(X_{o},\mathbb{C})$ has an image whose Zariski closure $G$ is a quasi-unipotent linear algebraic group (that is, the quotient of $G$ by its unipotent radical is a finite-group). In general it is rare for the local monodromy to be quasi-unipotent. If $p$ happens to be a very singular point of the discriminant, then the local monodromy tends to be big and is often as big as it can be, and not quasi-unipotent at all. However, if $p$ is at worst a normal crossing singularity of $D$, then the local monodromy is quasi-unipotent. - Thank you very much for your beautiful answer. I had never known of this general definition of quasi-unipotency. I have some more questions: First of all what is the best reference which discusses this definition and it's consequences. Secondly, if this is the definition of "quasi-unipotency", then what would be the definition of "unipotency" in general? – Jack Jul 22 at 6:46 1 Ah, I noticed I missed an adjective in the comment - I was defining what it means for $G$ to be quasi-unipotent. I edited the answer to reflect this correctly. The unipotency of $mon$ is defined in a similar manner - we say that $mon$ is unpotent, when $G$ is a connected unipotent algebraic group, i.e. when when it coincides with its unipotent radical. The references are numerous and the applications are usually Hodge theoretic. The quasi-unipotency of a local systen can is also very useful when we compute cohomology. – Tony Pantev Jul 23 at 3:00 1 Two classical references are the paper "Periods of integrals on algebraic manifolds III", Publ. Math. IHES 38 (1970) by Griffiths and the paper "Variation of Hodge structure: the singularities of the period mapping" Invent. math. 22 (1973) by Schmid. They in particular explain Borel's proof of the quasi-unipotency theorem that I mentioned above. There are many other modern references. For instance, you may want to take a look at the excellent book "Period mappings and period domains" by Carlson, Mueller-Stach, and Peters. – Tony Pantev Jul 23 at 3:07	{"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": 26, "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": 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": 0.923823893070221, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/187205-absolute-value-equations.html	# Thread: 1. ## Absolute Value Equations How, when you are given an absolute value equation, solve it graphically on a real-number line? 2. ## Re: Absolute Value Equations Originally Posted by Bashyboy How, when you are given an absolute value equation, solve it graphically on a real-number line? You should know that the absolute value corresponds to the "size" of a number... So say \|x\| = 3. That means that the size of the number is 3 units from 0. Therefore x = -3 or x = 3. Does that make sense? 3. ## Re: Absolute Value Equations Yes, it does. But I become a bit more confounded with a problem like \|x - 3\| = 5. How would I solve one like this on a real number-line? 4. ## Re: Absolute Value Equations Originally Posted by Bashyboy Yes, it does. But I become a bit more confounded with a problem like \|x - 3\| = 5. How would I solve one like this on a real number-line? Well that means that the size of x - 3 is 5 units from 0. So $\displaystyle x - 3 = -5 \implies x = -2$ or $\displaystyle x - 3 = 5 \implies x = 8$. 5. ## Re: Absolute Value Equations I have another problem. I am given a real number line with the two points -3 an 1 marked on it. How do I write an absolute value equations from this data? 6. ## Re: Absolute Value Equations Originally Posted by Bashyboy I have another problem. I am given a real number line with the two points -3 an 1 marked on it. How do I write an absolute value equations from this data? If $a<b$ then the equation $\left\| {x - \frac{{b + a}}{2}} \right\| = \frac{{b - a}}{2}$ has solutions $x=a\text{ or }x=b$. 7. ## Re: Absolute Value Equations I am truly sorry, but I don't quite follow. I have not seen anything like this before. 8. ## Re: Absolute Value Equations Originally Posted by Bashyboy I am truly sorry, but I don't quite follow. I have not seen anything like this before. Are you saying that you cannot substitute $a=-3~\&~b=1$ into that? 9. ## Re: Absolute Value Equations No, plugging and chugging is quite simple. But I have never seen that formula you have presented me with. Also, the way the book wants me to solve it is just graphically.	{"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": 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": 0.9329167604446411, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/33679/as-the-universe-expands-why-do-some-things-stretch-but-not-others/33684	# As the universe expands, why do some things stretch but not others? I got into watching a video on Olbers' Paradox a few days ago, and from there read about the origins of the universe, its expansion, and so on... it's always fascinated me, but now something about it bothers me. I've heard many analogies about it (the dot-balloon, the raisin bread loaf, and others), but none really seem to explain this question. (This comes close, but dances around the answer more than explain it.) At the beginning of the universe as we know it, the universe itself was very small, so all the stars giving off light would have made it very bright (16:29 in this video). Since that time, the wavelength of that light has been stretched (17:01, same video). I found a few explanations saying that space itself stretched (here; described as "ether" in the article), which would stretch out the wavelengths. But here's what bothers me: If space is stretching out, redshifting all the light soaring around our universe, why are we not stretching? Theoretically, the universe is expanding an incredible amount faster than the speed of light, and the edge of the universe is an unimaginably large number of megaparsecs away from us. But should we not notice some of the stretch here, too? That is to say, if the light in space (the "ether", though I'm not fond of that term) is stretching out, why is everything on Earth still the same size as it was a hundred years ago? Is it stretching uniformly, but we are just unable to notice such a small stretch? Or does mass have some property that space and light do not, that prevents it from stretching out? I've also heard about time stretching, too; does this have an impact on it? - – Qmechanic♦ Aug 8 '12 at 3:58 – Eric Aug 8 '12 at 5:25 ## 4 Answers This is not my field but the way I understand it is that the expansion involves unbound states. It does not affect bound states. For example protons, bound by the strong interaction, once generated, during the expansion, and decoupled, i.e. the quark gluon plasma has stopped existing, remain protons with the dimensions we know them. Incorporating your comment question: Is there an answer as to whether the cosmological or atomic force was larger initially? Decoupling means that as expansion progresses locally the cosmological force becomes smaller than the strong force ( in the case of protons decoupling) and therefore there is no longer a dissolution and recreation of protons from the energy soup of the Big Bang, in this case the quark gluon plasma which should exist before protons can appear. The same is true for galaxies, which are a gravitationally bound state and separate between each other due to the expansion but remain bound internally. However the only locally visible effect of the accelerating expansion is the disappearance (by runaway redshift) of distant galaxies; gravitationally bound objects like the Milky Way do not expand. Photons (and neutrinos) are not bound states, and therefore follow the expansion of space changing their wavelength due to it. Always keep in mind that this expansion happens locally at every spacetime point of what we define as space time for usual physics studies. This is a field which is researched still, but this model seems to fit observations up to now. - Excellent explanation. I'm going to keep the question open for a bit longer to see if other answers arrive, but this is definitely a more thorough explanation than I had expected. Thanks a lot! – Eric Aug 8 '12 at 5:45 The expansion of the universe is due to the expansion of spacetime. There's a good article on this here. Suppose you take two non-interacting particles, put them some distance apart and make them stationary with respect to each other. If you now watch them for a few billion years you'll see the particles start to accelerate away from each other. This happens because the spacetime between the two particles is stretching i.e. there is more "space" between them. One way of interpreting the acceleration is to say there is a force between the two particles repelling them. This is a slightly dodgy description because there isn't really a force; it's just expansion of space. Nevertheless, if you tied the two particles together with a rope and watch for a few billion years there would be a tension in the rope so the force is real in this sense. Anyhow, now we have everything we need to understand why the Earth isn't stretching. The expansion of spacetime creates a stretching force, but this will only have an effect if there is no other force to oppose it. For example you are indeed being stretched by the expansion, but the interatomic forces between the atoms in your body are vastly stronger than the stretching due to expansion, so you remain the same size. Likewise the gravitational force between the Sun and Earth is vastly greater than the stretching force so the Earth's orbit doesn't change. The stretching force is vanishingly small at small distances, but it gets greater and greater with increasing distance so at some point it wins. Galaxies and indeed galaxy clusters are still too small to be stretched, but at greater sizes than this the stretching wins. That's why galaxy clusters are the largest objects observed in the universe. At greater sizes spacetime expansion wins. A footnote: if anyone's still interested in this subject, there's a paper Local cosmological effects of order H in the orbital motion of a binary system just out claiming that the effect of the expansion on the Solar System might be measurable. - 1 It's such a darn shame I can't accept two answers... very well-written, and a great explanation of all the forces in play. Thanks so much! – Eric Aug 8 '12 at 6:22 When you model the expanding universe in cosmology, you do so with a particular solution to the Einstein field equations called the FRW metric. The defining feature of this metric is, of course, metric expansion. This means that distances will increase over time. One assumption that goes into the FRW metric is homogeneity. Since the universe is homogeneous on large scales, this works excellently for very large portions of the universe. However, galaxies are certainly not homogeneous. So, you need to use a different metric inside of galaxies - and because of this, space inside of galaxies is totally unaffected by metric expansion. It's not even that the effect is too small to be noticed, galaxies are totally unaffected by expansion. So, we can generalize this to say that expansion occurs in between bound systems. There is a good entry on this at the Usenet FAQ: http://math.ucr.edu/home/baez/physics/Relativity/GR/expanding_universe.html Dark energy, however, is a bit trickier. Since it is a negative pressure vacuum energy, it exerts an extremely small force everywhere. So, it has a small effect inside of bound systems. This is because dark energy is a cosmological constant - which is also a term in the Einstein field equations. Since these still govern gravitational interactions inside of galaxies, dark energy has an effect there. The easiest way to see by is by looking at attractive gravitational force between two objects with a cosmological constant in the Newtonian limit: $$F = {GMm \over r^2} - {\Lambda m c^2 \over 3} r$$ However, this effect is utterly negligible. - I do not think anyone at this time can answer this question with certainty because there could be a number of explanations of what others are observing. I think we need to learn from nature for some of these more difficult questions. My theory could be; that the whole universe is expanding, including you, me and everything that consists of the universe. For example instead of the big bang or the big stretch, we could have the big revolve. This option would follow the course of nature maybe a little more closely. Most things that we know of in the universe revolve and or orbit. Everything in space that we see is moving and orbiting something or another. It appears that by measuring light we can see a shift of some moving away and others moving closer. Now you read that they say that the space in between is stretching. Maybe it is expanding as we are expanding and then the observation could be similar. If it were true and we are all expanding then it can explain some parts of the laws of gravity as well but not all of them. Maybe they can be explained under a different law. In nature the earth itself is revolving and it is renewing itself by the movement if its plates. Over time all of our records of existence will be wiped off the face of the earth. Maybe our sun and our stars known as the white light is part of a dimension that we can see, and all of the particles from this source are travelling and expanding through space and then will eventually collect into the black holes to be spit out the other side ( another dimension ) to again be returned to our dimension through the sun and stars. Thus we have the big revolve. We have a related view point, it will take billions of years for this to happen. Maybe from another view it will take a split second by their time. Maybe our universe is inside and is a part of another universe, maybe someone in the other universe is cold and will throw another log on the fire, and thus our universe goes up in a puff of smoke. All I know for sure is we have a long way to go to get closer to the truth, the guys at NASA are on the right track, one step at a time and by using only proven science to build upon the next mission. - Hmm... interesting take with a lot of neat ideas. I'm anxious to get some more opinions on this, too. – Eric Aug 8 '12 at 3:13	{"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": 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": 0.18, "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": 0.9608489274978638, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/109957-polynomial-ring-print.html	# polynomial ring Printable View • October 23rd 2009, 10:19 AM knguyen2005 polynomial ring Let R be a ring, suppose that 2 non-zero functions f and g with: f = a_0 + a_1x +...+ a_mx^m belongs to R[x] , a_m not equal 0 and deg(f) = m g = b_0 + b_1x +....+ b_nx^n belongs to R[x], b_n not equal 0 and deg(g) = n (a) Let R be and Integral Domain (ID). Show that f.g is nonzero and f.g has leading term is (a_m)(b_n)x^(m+n). Deduce that R[x] is a ID and deg(f.g) = deg(f) +deg(g). If R[x] is a ID, show that R is also a ID. (b) Let R be a ID. Show that U(R[x]) = U(R) (c) Let a in R and a^n = 0 for n>1. Show that (1-a) belongs to U(R). What is (1-a)^-1? (d) Show that U(Z_4) is strictly not subset of U(Z_4[x]). Why doesn't this contradict part (b)? Note: Z_4 is set of integer mod 4 Ok, Part (a) is easy and straightforwards. f.g is nonzero sine both f and g are nonzero. Also, the highest power of f.g is (a_m)(b_n)x^(m+n) obtained when u multiplied 2 functions. R[x] is an ID because either f = 0 or g = 0 (not both equal o at the same time) deg(f.g) = m + n = deg(f) +deg(g) Part (b), (c) and (d) I dont know how to begin with Can someone show me how to do part b, c and d please? Thank you • October 23rd 2009, 02:07 PM proscientia I don’t think you’ve even done part (a) properly. Recall that the coefficient of $x^k$ in $fg$ is $\sum_{i\,=\,0}^ka_ib_{k-i}$ When $k=m+n,$ the sum is just $a_mb_n$ because $a_i=0$ for all $i>m$ and $b_{m+n-i}=0$ for all $i<m.$ Hence the coefficient of $x^{m+n}$ in $fg$ is $a_mb_n.$ And since $a_m\ne0,\ b_n\ne0$ and $R$ is an integral domain, $a_mb_n\ne0$ and so $fg$ is nonzero. Do you understand this so far? • October 23rd 2009, 08:12 PM aliceinwonderland Quote: Originally Posted by knguyen2005 (a) ..... If R[x] is a ID, show that R is also a ID. Let $f(x)=a_0 + a_1x + \cdots +a_mx^m$ and $g(x)=b_0 + b_1x + \cdots + b_nx^n$ be two nonzero polynomials in R[x]. Since R[x] is an integral domain, f(x)g(x) is nonzero and we have a non-zero leading term $a_kb_lx^{k+l}$. This follows that $a_k \neq 0$ and $b_l \neq 0$. Since R[x] is an integral domain, we obtain $a_kb_l \neq 0$. Thus R is an integral domain. Think about this argument $((a_k \neq 0 \wedge b_l \neq 0) \Rightarrow a_kb_l \neq 0) \Leftrightarrow (a_kb_l = 0 \Rightarrow (a_k = 0 \vee b_l=0))$. Quote: (b) Let R be a ID. Show that U(R[x]) = U(R) A polynomial f(x) of degree m times a polynomial g(x) of degree n is a polynomial f(x)g(x) of degree m+n in an integral domain R[x]. Suppose f(x) is a polynomial whose degree is 1 or greater. Then no polynomial g(x) exists such that f(x)g(x)=1, which is a polynomial of degree 0. This forces that units of R[x] only exists with a degree zero. Thus U(R[x])=U(R). Quote: (c) Let a in R and a^n = 0 for n>1. Show that (1-a) belongs to U(R). What is (1-a)^-1? Think about the Maclaurin series of 1/(1-a) and use a nilpotency of a. $(1-a)(1+a+ \cdots +a^{n-1}) = 1$, where $a^n=0$ Quote: (d) Show that U(Z_4) is strictly (not?) subset of U(Z_4[x]). Why doesn't this contradict part (b)? Note: Z_4 is set of integer mod 4 $U(\mathbb{Z}_4) = \{1,3\}$. They are coprime to 4. Meanwhile $U(\mathbb{Z}_4[x])$ has additional units including $U(\mathbb{Z}_4)$. For instance, 1 + [2]x is a unit in $U(\mathbb{Z}_4[x])$. We know that $\mathbb{Z}_4$ is not an integral domain. This implies that if R is not an integral domain, $U(R) \subset U(R[x])$. • October 24th 2009, 03:13 AM knguyen2005 Thank you both proscientia and aliceinwonderland I already understood and know how to do it now Thanks again for spending your time on this question All times are GMT -8. The time now is 08:40 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": 31, "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": 0.8956887125968933, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/10955/what-is-quantum-discord?answertab=votes	# What is quantum discord? What is quantum discord? I stumbled upon this term on Quantum Computing: The power of discord, but have never heard of it before. Can you give a bit more mathematical explanation of the term here? - There are some progress in calculating discord for X states in 2 qubit bipartite systems. Ali, M., Rau, a. R. P., & Alber, G. (2010). Quantum discord for two-qubit X states. Physical Review A, 81(4), 042105. doi:10.1103/PhysRevA.81.042105 Luo, S. (2008). Quantum discord for two-qubit systems. Physical Review A, 77(4), 042303. doi:10.1103/PhysRevA.77.042303 – Ars3nous Mar 15 at 20:19 ## 1 Answer It is basically a measure of the quantumness of some correlations, which is not vanishing for some separable state. It was introduced by Ollivier and Zurek (PRL/arXiv). It is the difference between two different generalizations of the classical (Shannon) conditional entropy to the quantum world, and is 0 for a pure bipartite separable state. It has been proven to be the amount of entanglement needed in the task of state-merging (PRA/arXiv and PRA/arXiv). Definition (PRL/arXiv) Classically the conditional entropy $H(A\|B)$ is a measure of the uncertainty one has on the variable $A$ once we know the variable $B$. Of course, the definition of "knowing" $B$ becomes problematic when $B$ is quantum. 1. Classically, one can define $H(A\|B)$ as the average $H(A\|B)=\sum_b {\mathcal P}(B=b)H(A\|B=b)$, each $H(A\|B=b)$ being the entropy of $A$ given that the random variable has the value $b$. If one generalizes this to the quantum world, the $B=b$ part implies a quantum measurement (a POVM) which should be specified. A natural choice is the "best" measurement, the one which minimizes the entropy. The Shannon $H$ entropy is replaced by the Von Neumann entropy, and we define $S(A\|B_c)=\min_{\text{POVM}} \sum_{b}\mathcal{P}(\text{POVM applied to B gives } b) S(A\|\text{POVM applied to B gives }b)$. 2. The previous definition leads classically to a redefinition of the conditional entropy as an entropy difference : $H(A\|B)=H(A,B)-H(B)$, which is always positive. Its quantum version, $S(A\|B)=S(AB)-S(B)$ can be negative (in contrast with $S(A\|B_c)$). Its negativity is a sufficient condition for entanglement. The discord is defined as $S(A\|B_v)-S(A\|B)$ and is always positive. You can maybe see it as the amount of correlation between $A$ and $B$ which is destroyed by a classical measurement of $B$. The state merging primitive is the following. Suppose Alice, Bob and Charly share a 3-party pure entangled state. Alice want to send her part to Bob without destroying the quantum correlations between $AB$ and $C$. Basically, she has to teleport $A$ to Bob, and the minimal amount of entanglement Alice and Bob need to perform this task is given by the quantum discord.	{"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": 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": 0.913801372051239, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/22811/charge-distribution-on-a-parallel-plate-capacitor/22828	# Charge Distribution on a Parallel Plate Capacitor If a parallel plate capacitor is formed by placing two infinite grounded conducting sheets, one at potential $V_1$ and another at $V_2$, a distance $d$ away from each other, then the charge on either plate will lie entirely on its inner surface. I'm having a little trouble showing why this is true. In the space between the two plates the field $E = ( V_1 - V_2 ) / d$ satisfies Laplace's equation and the boundary conditions, from which I can derive the surface charge density is $\pm E / 4 \pi$. But how about the space above and below the capacitor? Certainly I can't just use superposition of the inner surface charge distributions to say that the field outside the capacitor is zero, (and thus the surface area charge density is zero), for this assumes there is no charge on the outer surfaces to begin with. Any help clearing up this mental block would be greatly appreciated, thanks. - Off the bat, I would treat this problem as unsolvable because there's no such thing as an infinite capacitor, and even if one did exist, it could never be charged. Now, saying that a capacitor's radius (assume a circular plate...if it's big enough its shape doesn't really matter) compared to the plate separation is large is a different, yet much more realistic, way of characterizing the capacitor. – user11266 Jan 22 at 0:50 ## 3 Answers One could deal with the problem by being careful with how one constructs a mathematical interpretation of the physical system. I will treat the simplest case: treat the surfaces of the parallel plate capacitors as true two dimensional surfaces. In this case there is no inner or outer surface charge, just a surface charge density defined on each surface. Mathematically one could represent each conductor as an infinite plane, say $S_\pm \subset \mathbb{R}^3$, then there are two surface charge densities $\sigma_\pm$ each defined on the corresponding surface $S_\pm$. Alternatively, one may use the language of distributions and use a (volume) charge distribution defined on all of $\mathbb{R}^3$ such that $\rho(x, y, z) = \sigma_+ \delta(z - d/2) + \sigma_+ \delta(z + d/2)$ where I have put $S_\pm$ on the planes $z = \pm d/2$. More complicated models might assume each plate of the conductor has a finite thickness. One could then solve the more complicated problem and compute what happens in the limits at the thickness approaches zero. - First, a nit: if the potentials on the two plates are non-zero they are not grounded, by definition. Second, the way I think about it: in the region of interest above and below the plates, the boundary conditions are not established. To set these boundary conditions, you could imagine adding an additional two infinite conducting plates above and below the original plates, and grounding these new plates to 0 potential. • If the new plates are initially located close to the original plates, there will indeed be an electric field above and below the original plates, and a corresponding surface charge density on their outside surfaces. • Now imagine the new plates being removed to infinity. Since the potential differences are fixed, the electric field, and the outer surface charge densities, go to zero. - So the reason that setting the potential at infinity to zero would lead to zero surface charge is because the field for an infinite sheet is essentially constant? Also, if the plates were truly infinite, how would we deal with the problem? At that point talking about zero potential doesn't mean anything. – Blooper Mar 27 '12 at 17:59 Thinking about the problem, it also seems like having the same charge density on the outer surface of each plate should be a solution, because then the field inside the conductor would still be zero. – Blooper Mar 27 '12 at 18:05 To the first comment, yes, with these idealized infinitely long and wide sheets, the E-fields are just the potential difference / sheet separation (solution of Laplace's equation in a very simple geometry). I don't see any problem with grounding arbitrarily large plates. – Art Brown Mar 27 '12 at 18:17 To the second comment, such an outside surface charge density would create an electric field above and below the plates, running off to infinity, that would raise the plate potential to infinity, violating the problem conditions. (In other words, this field isn't consistent with the boundary conditions.) – Art Brown Mar 27 '12 at 19:03 Ignore inner and outer surfaces. There is just one surface. Imagine a single, infinite plane with some positive charge density. You can easily show there would be an electric field of constant strength, perpendicularly out of the plane all the way to infinity on both directions. Now imagine a single, infinite plate with the same negative charge density. There would be an electric field of constant strength perpendicularly into the plane all the way to infinity in both directions. Put these two plates on top of each other, and these fields perfectly cancel. Put these two plates in parallel, and because the field is constant strength it will perfectly cancel everywhere except between the two plates, where the electric field directions are the same and it will add to be twice as strong. [By constant strength I mean the electric field is just as strong no matter how far you are from the plate. Why is the field constant strength? Because the field lines can't ever diverge from one another. The way fields usually get weaker is the equipotential surface the field lines are normal to gets bigger as you increase the distance from the object. So the same number of field lines piercing a bigger surface means a field lines are more spread out, and thus a weaker field. In this case however, the equipotential surfaces are always a pair of infinite parallel planes, no matter what distance we are from the charged plane. No spreading means no change in field strength.] -	{"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": 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": 0.9430122375488281, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/204994/largest-number-obtained-from-products	# Largest number obtained from products? What is the largest number that can be obtained as the product of two or more positive integers that add up to 20? - 1 do they have to be distinct? – Jean-Sébastien Sep 30 '12 at 19:10 There are some IMO problems regarding things like this. – Beni Bogosel Sep 30 '12 at 19:24 The original problem doesn't state if they have to be distint, so I'm assuming it doesn't matter. – Ctrl Sep 30 '12 at 19:46 ## 1 Answer If a number $m\gt 4$ occurs in a decomposition (of $20$ in this case), the product can be increased by splitting $m$ suitably. It makes no difference whether $4$ is split as $2\cdot 2$ or not, so it might as well be. It is clear that it is no good to use any $1$'s. Thus we can assume that all the numbers in our splitting of $20$ are $2$'s and/or $3$'s. Then note that $6=2+2+2=3+3$ but $3^2\gt 2^3$. So if there are three or more $2$'s in a splitting, we can do better. Remark: If you want to prove that any number $m \gt 4$ should be split further, there are two cases to consider. (i) If $m=2k$, show that $k^2\gt 2k$ if $k\gt 2$. (ii) If $m=2k+1$, show that $k(k+1)\gt 2k+1$ if $k\ge 2$. - So 1458 seems to be optimal- 6 3s and a 2. – Geoff Robinson Sep 30 '12 at 19:39	{"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": 19, "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": 0.9428706169128418, "perplexity_flag": "head"}
http://mathoverflow.net/questions/111915/a-question-about-first-order-hyperbolic-equations	## a question about first-order hyperbolic equations ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Performing certain manipulations on pseudo-differential equations I have come across the following first order equation: $$D_{t}u-\lambda(t,x,D_{t},D_{x})u=0, \ \ ()$$ where $\lambda$ is a scalar pseudo-differential operator with the principal symbol being real-valued and independent of $D_{t}.$ But, the lower-order terms of $\lambda$ depend on $D_{t}$ (or $\tau$ at the symbol level). I was expecting a hyperbolic equation. But, I find that standard text books(like M.Taylor's 'Pseudo-differential operators') treat only equations in which $\lambda$ term is independent of $D_{t}$ (or $\tau$ at the symbol level). For the equation $()$ to be hyperbolic, is it necessary that $\lambda$ to be independent of $D_{t}$? Are there any references which discuss these issues? - ## 2 Answers Your principal symbol is $\tau-\lambda_1 (t,x,\xi)$, with a real-valued $\lambda_1$, but you have lower order terms which may depend on $\tau$. You can get rid of that dependency as follows: take for instance $\lambda_0$ of order $0$. You can find an operator $M$ of order $0$ such that $$Op\bigl(\tau-\lambda_1 (t,x,\xi)-\lambda_0(t,x,\tau,\xi)\bigr)= e^{-iM}Op\bigl(\tau-\lambda_1 (t,x,\xi)\bigr)e^{iM}+Op(S^{-1}).$$ In fact, since $e^{iM}$ is a pseudodifferential operator of order 0, the composition formula gives $$e^{-iM}Op\bigl(\tau-\lambda_1 (t,x,\xi)\bigr)e^{iM}=D_t+\frac{\partial M}{\partial t} -e^{-iM}[Op(\lambda_1),e^{iM}]-Op(\lambda_1),$$ and since $e^{-iM}[Op(\lambda_1),e^{iM}]=Op(${$\lambda_1,m$}$)+Op(S^{-1})$ where $m$ is of order 0 and {} is the Poisson bracket. To get rid of $\lambda_0$, you have only to solve $$\frac{\partial m}{\partial t}-\text{ {$\lambda_1,m$ } }=-\lambda_0,$$ which is a linear transport equation of real principal type in $m$. You can of course iterate this business to get a remainder of order as negative as you like. This is explained in the Chapter 23 on hyperbolic equations in Hörmander third volume of ALPDO. - ### 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. If I understand correctly, you have a PDE of the form $$(D_t - \lambda_1(t, x, D_x) + \lambda_0(t, x, D_t, D_x))u = f,$$ where $\lambda_1$ is a first order pseudodifferential operator and $\lambda_0$ is a zero-th order pseudodifferential operator. It seems to me that the proofs of many if not all estimates, including energy integral estimates, and theorems about regularity, uniqueness, and existence for the equation $$(D_t - a(t,x)D_x + b(t,x))u = f,$$ as presented in books like Taylor can be extended to your PDE. -	{"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": 27, "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": 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": 0.9477031826972961, "perplexity_flag": "head"}
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http://math.stackexchange.com/questions/103657/differentiability-in-mathbbrn-and-chain-rule	# Differentiability in $\mathbb{R}^n$ and chain rule I have a question: Consider a function $g:\mathbb{R}^{n}\rightarrow \mathbb{R}$ is differentiable. Find the derivative of the function: $G(x)=[ g\left ( x,x^{2},...,x^{n} \right )]^{2}$ where $x\in \mathbb{R}$. Here, G is a function of one variable, so I tried to apply the chain rule to find its derivative as follows: $G^{'}\left ( x \right )=2H^{'}\left ( x \right )H(x)$ where $H\left ( u_{1},u_{2},...,u_{n} \right )=g\left ( x,x^{2},...,x^{n} \right )$. Now to find the derivative of $H$, I did the following: $$\frac{d}{dx}H=\frac{\partial H}{\partial u_{1}}\frac{\partial u_{1}}{\partial x}+...\frac{\partial H}{\partial u_{n}}\frac{\partial u_{n}}{\partial x}.$$ Does what I did make sense? - – Zev Chonoles♦ Jan 29 '12 at 19:18 Provided one replaces $H(u_1,u_2,\ldots,u_n)$ by $H(x)$ and each $\partial H/\partial u_k$ by $\partial g/\partial u_k$, the answer is: yes. – Did Jan 29 '12 at 19:22 @ Zev Chonoles: Yes, I just mean simple brackets only. – pin296 Jan 29 '12 at 19:24 @pin296: Ah, I see that you actually did have brackets, but my browser was for some reason not displaying them correctly - apologies. – Zev Chonoles♦ Jan 29 '12 at 19:28 Here is my answer based on the above comments: $G^{'}\left ( x \right )=2.g\left ( x,x^{2},...,x^{n} \right ).\left [ \frac{\partial g}{\partial u_{1}}.1+\frac{\partial g}{\partial u_{2}}.2x +...+nx^{n-1}.\frac{\partial g}{\partial u_{n}} \right ]$. where : $u_{i}=x^{i}$. I am not convinced to the presence of the variables $u_{i}=x^{i}$ in my final answer. Does anyone have a better idea on how to solve the problem? – pin296 Jan 29 '12 at 21:01 ## 1 Answer It's almost correct. You have a function $$u(\cdot):\quad {\mathbb R}\to{\mathbb R}^n\ ,\qquad x\mapsto u(x):=(x,x^2,\ldots, x^n)$$ and a second function $$g:\quad {\mathbb R}^n\to {\mathbb R}\ ,\qquad (u_1,\ldots, u_n)\mapsto g(u_1,\ldots, u_n)\ .$$ The real-valued function $H:x\mapsto H(x)$ is defined as the composition of the two: $$H(x)\ :=\ g\bigl(u(x)\bigr)\ .$$ By the chain rule the derivative of $H$ is given by $$H'(x)={\partial g\over\partial u_1} u_1'(x) +\ldots+ {\partial g\over\partial u_n} u_n'(x) =\nabla g\bigl(u(x)\bigr)\cdot u'(x)\ ,$$ where the $\cdot$ denotes the scalar product. - I think my final answer looks like yours: $G^{'}\left ( x \right )=2.g\left ( x,x^{2},...,x^{n} \right ).\left [ \frac{\partial g}{\partial u_{1}}.1+\frac{\partial g}{\partial u_{2}}.2x +...+nx^{n-1}.\frac{\partial g}{\partial u_{n}} \right ]$. Is my expression true? – pin296 Jan 29 '12 at 21:06 Your "final answer" is correct. You can write $g_k$ instead of ${\partial g\over\partial u_k}$ if you want to avoid the appearance of the letter $u$ "out of nowhere". – Christian Blatter Jan 30 '12 at 8:52	{"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": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9281069040298462, "perplexity_flag": "head"}
http://mathoverflow.net/questions/7823/negative-gromov-witten-invariants/7873	## Negative Gromov-Witten invariants ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I understand the heuristic reason why Gromov-Witten invariants can be rational; roughly it's because we're doing curve counts in some stacky sense, so each curve $C$ contributes $1/\|\text{Aut}(C)\|$ to the count rather than $1$. However, I don't understand why or how Gromov-Witten invariants can be negative. What is the meaning of a negative GW invariant? What are some simple examples? - ## 4 Answers Gromov--Witten invariants are designed to count the "number" of curves in a space in a deformation invariant way. Since the number of curves can change under deformations, the Gromov--Witten invariants won't have a direct interpretation in terms of actual numbers of curves, even taking automorphisms into account. Here is an example of how a negative number might come up, though strictly speaking it isn't a Gromov--Witten invariant. Let M be the moduli space of maps from P^1 to a the total space of O(-4) on P^1. Call this space X. Note that I said maps from P^1, not a genus zero curve, so the source curve is rigid. That's why this isn't Gromov--Witten theory. Any such map factors through the zero section (since O(-4) has no nonzero sections), so this space is the same as the space of maps from P^1 to itself. I just want to look at degree one maps, so the moduli space is 3 dimensional. We could also compute the dimension using deformation theory: the deformations of a map f are classified by $H^0(f^\ast T)$ where T is the tangent bundle of the target. The target in this case is O(-4), not just P^1, and the tangent bundle restricts to O(2) + O(-4) on the zero section. Thus $H^0(f^\ast T)$ is indeed 3-dimensional, as we expected. However, the Euler characteristic of $f^\ast T$ is not 3 but 0, which means that the "expected dimension" is zero. The meaning of expected dimension is rather vague. Roughly speaking, it is the dimension of the moduli space for a "generic" choice of deformation. The trouble is that such a deformation might not actually exist. Nevertheless, we can still pretend that a generic deformation does exist and, if the expected dimension is zero, compute the number of curves that it "should" have. What makes this possible is the obstruction bundle E on M. Any deformation of X gives rise to a section of E and the vanishing locus of this section is the collection of curves that can be deformed to first order along with X. Even though a generic deformation might not exist, the obstruction bundle does still exist, and we can make sense of the vanishing locus of a generic section by taking the top Chern class. In our situation, the (fiber of the) obstruction bundle is $H^1(f^\ast T)$. Since O(2) does not contribute to H^1, the obstruction bundle is $R^1 p_\ast f^\ast O_{P^1}(-4) = R^1 p_\ast O_{P^3 \times P^1}(-4, -4)$ where $p : P^3 \times P^1 \rightarrow P^3$ is the projection. By the projection formula, this is $O(-4)^{\oplus 3}$ and the top Chern class is -64. This is the "Gromov--Witten invariant" of maps from P^1 to $O_{P^1}(-4)$. Unfortunately, I don't have anything to say about what this -64 means... - ### 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. Gromov Witten invariant is supposed to "count" the number of curves but it can happen that the dimension of the space of curves that you want to count is larger than you expect. For example on a 3-dimensional Calabi-Yau variety the expected dimension of any curve is 0, so in an ideal situation curves should be isolated and you should just count the number of curves of a given degree and genus. In practice this does not happen all the time and often you can deform curves (though Clemence conjectured that RATIONAL curves on generic CY 3-fold are rigid) For example if you take a quintic in $CP^4$, you can intersect it with a 2-plane and you obtain a curve that moves in the quintic. While you excpect it to be rigid. What you need to do in this case to count the GW invariant is to conisder on the moduli space of curves the obstruction bundle, which will be of the same dimesnion as the moduli space and then you need to calculate the Euler class of the bundle (this is in the best case, when the moduli space is smooth). Now, the Euler class that you calculate has no particular reason to be positive. Of course if the curve were isolated it would count with the positive weight, but it is not. I can not provide an explicite example where you get a negative answer, but there are a lot of examples when you get zero, while there are curves. This happens on algebraic surfaces. You can take a minimal surface of general type and consider a curve C of zero expected dimention, i.e. $C^2-KC=0$, where K is the canonical divisor. Then the GW invariant of this curve will be non zero only if C is a canonical divisor. This is a corrolary of Seiberg Witten theory. But there are examples when you do have curves C that satisfy $C^2-CK-0$ and $C$ is different from $K$ these curves contribute 0 to GW invariant. So, what is the meaning of negative GW invariant? If you are doing symplectic geometry, everything can be perturbed so that the final number of curves that you get is finite. It is exactly the same as to say that if you have a map of two manifolds $M^n\to N^n$, for a generic point the number of perimages is finite. Some of them count with + and some with -, add together and you get degree. Negative GW= negative degree. But if you are doing Algebraic geometry negative GW invariant has an additional consequence -- namely that the moduli space of curves (this time algebraic and not almost complex) that you consider MUST has exessive dimention. Just as in the example with CY quintic that is desribed above. Indeed, if all curves were isolated they would contribute positively and GW would be positive. - Be specific, this happens on the Fermat quintic, as I recall. However, for a generic one, the curves should have no deformations. – Charles Siegel Dec 5 2009 at 1:53 No Charles, it is clear that the example that I decirbed works for every quntic. You can always move a plane in CP^4 and its intersection with the quintic will move. I guess, you were thinking about RATIONAL curves, they indeed should be rigid on a generic quintic according to Clemmence conjecture. But the example that I give is not genus zero. This is a generic phenomena for ALL Calabi-Yaus. Take two divisors of high degree that move on CY and intersect them – Dmitri Dec 5 2009 at 2:00 Ahh, ok. Yeah, I only think about GW invariants in genus zero here, where they're generically rigid. (and it's Clemens, just for anyone who might google names, and the paper arxiv.org/abs/alg-geom/9510015 has some good info on it, if I remember right) – Charles Siegel Dec 5 2009 at 2:44 While what you say makes sense, Dmitri, it doesn't seem to answer the question as to the meaning of negative invariants. – Simon Rose Dec 5 2009 at 4:36 Simon, I added two pragraphs on the meaning of the negative tsign - as much as I can say :) Charles, really sorry for misspelling the name of Clemence. – Dmitri Dec 5 2009 at 8:57 Let me use an example. If I recall correctly (I am too lazy to look), the GW invariant for "local P^2" (i.e. the canonical bundle O(-3) over P^2) in genus zero and degree 1 is 3, and in degree 2 is -45/8. Now -45/8 breaks down, after trying to account for multiple covers, to 3/2^3 - 6, giving an effective number of degree 2 curves as -6. What up? This number is how many degree 2 curves the P^2 should "account for," when it pops into existence within a family of Calabi-Yau's. (Here of course we have a family of maps into the P^2.) Then you should ask, "Does this mean that for a compact CY to have just rigid genus zero curves and an embedded P^2 (and no other surface) and lie in a family that includes a CY with only rigid genus zero curves and no other surface, that it must have at least 6 degree 2 rational curves somewhere else?" Presumably. - The simplest example that I could find is: blow-up $P^3$ at one point and consider the following correlator $\langle pt, E^2, E^2\rangle_l=-1$ , where $l$ is the pull-back of a line from $P^3$. Proof: use the splitting axiom for the following 4 classes $(pt, E^2\|E,E)_l$. To unravel this notation take a look at page 5 in http://www.mathematik.uni-kl.de/~gathmann/pub/blowup-9804043.pdf The splitting axiom gives: $\langle pt, E^2, E^2\rangle_l=\langle pt, E, E\rangle_{l-E'}\cdot\langle E^2, E^2, E\rangle_{E'}=1\cdot(-1)=-1$ I hope this is corect. Does anybody know an explicit example in dimension 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": 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": 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": 0.9392176866531372, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-statistics/89070-poisson-process-renewal-processes.html	# Thread: 1. ## Poisson process & renewal processes Question: "Traffic on Snyder Hill Road is a Poisson process with rate 1 car per minute. That is, the times between cars are independent exponentials $t_1, t_2, . . .$ with mean 1. Let $T_k = t_i + ... + t_k$ be the time the k-th car passes. A turtle needs two minutes to cross the road. Let $N = min\{i : t_i > 2\}$ (a) Find the expected value of $T_{N-1}$, which is the time until turtle starts to cross, and the expected value of $T_N$, the time the next car passes (b) Consider a renewal reward process with $r_i = 1$ if $t_i > 2$ and $r_i = 0$ otherwise. At what rate are rewards earned?" Thanks a bunch - I really appreciate it! 2. Just to get you started, what is the distribution of N? Then use total expectation: $E[T] = E[E[T\|N]]$	{"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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "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": 0.9284691214561462, "perplexity_flag": "middle"}
http://cstheory.stackexchange.com/questions/tagged/tree	# Tagged Questions A tree is a special type of graph which only allows for a hierarchical set of edges similar to a tree . Mathematically it is actually an arborescence. Trees have a root node and children nodes. In formal terms it is described as a acyclic connected graph. 0answers 51 views ### How many Trees are Possible? [closed] We are given Two Integers P and Q. We have to find how many different T-Trees are possible. Definition of T-Tree It is a tree with Q * P Nodes. Nodes are Numbered from 0 to Q * P - 1 The Nodes a ... 2answers 192 views ### Caterpillar decomposition of trees Can any tree on $n$ nodes be decomposed into a set of $O(\log n)$ caterpillars? If not, what is the maximum number of caterpillars required? Are there efficient algorithms for finding the ... 1answer 219 views ### Multidimensional B+ tree I've got an idea for indexing multidimensional data. 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The obvious way I would go about implementing this would be to find entire subtrees whose range can fit into ... 2answers 1k views ### efficient diff algorithm for trees and Levenshtein distance I've recently read this summary of the issues involved with doing diff between trees and it got me interested in learning what is the state of the art for this problem. Also, suppose that between ... 1answer 209 views ### Remove specific edge from ST (link-cut) tree ST (or link cut) trees are a special kind of trees used for dynamic graph algorithms. They support the following operations in logarithmic time: CUT(v) Deletes the edge from v to its parent JOIN(v, ... 1answer 2k views ### unique binary tree from preorder and postorder traversals of a full binary tree [closed] If we have a preorder and postorder traversals of a full binary tree T(i.e every internal node have exactly 2 children). can we uniquely construct the corresponding full binary tree T. If so.. could ... 2answers 355 views ### P-complete problems on trees This question is related to one of my previous questions, NP-hard problems on trees. I am looking for problems that are P-complete on trees. 4answers 3k views ### Why would one ever use an Octree over a KD-tree? I have some experience in scientific computing, and have extensively used kd-trees for BSP (binary space partitioning) applications. I have recently become rather more familiar with octrees, a similar ... 1answer 266 views ### Minimum degree of the “tree graph” Given a graph $G$, define the tree graph $T(G)$ as a graph whose vertices are the spanning trees of $G$, and there is an edge between two trees if one can be obtained from the other by replacing a ... 1answer 90 views ### Is this the right strategy to convert an in-level order binary tree to a doubly linked list? [closed] So I recently came across this question - Make a function that converts a in-level-order binary tree into a doubly linked list. 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I first tried to solve it based on a greedy approach, but on a second thought, I found a counterexample to the greedy ... 1answer 230 views ### Searching nodes in semi-splay tree If you search for a node in a semi-splay tree, it's basically to push certain nodes closer to the root, to reduce future search operations. My course also says that if you search for a node and the ... 1answer 228 views ### Optimal Self Balancing Trees with Canonical Form? Are any efficient [O(log n)] self balancing trees that are canonical? By canonical I mean that for any set of data inserted into the tree, inserting it after any permutation results in the same tree. ... 0answers 287 views ### Geometric / Visual explanation that the average height of a random binary tree of given size $n$ is asymptotically $2\sqrt{\pi n}$ I just finished reading the proof that the average height of a random binary of given size $n$ is asymptotically $2\sqrt{\pi n}$. I'm now searching for an intuitive, or geometric, or visual proof of ... 2answers 620 views ### How do I choose a functional dictionary data structure? I've read a bit about the following data structures: Bagwell's Ideal Hash Tries Larson's Dynamic hash tables Red-Black trees Patricia trees ...and I'm sure there are a lot of others out there. ... 2answers 445 views ### maintaining a balanced spanning tree of a growing undirected graph I am looking for ways to maintain a relatively balanced spanning tree of a graph, as I add new nodes/edges to the graph. I have an undirected graph that starts as a single node, the "root". At each ... 1answer 269 views ### Lower bound on the number of “short” paths in a rooted tree with polynomial size Let $T$ be a rooted binary tree. Every path from the root of $T$ to a leaf has length $n$. Every node of $T$ has always a left and a right child node but it is possible that they are the same (So ... 2answers 416 views ### What is the optimal data structure for a tree of maps. I'm looking for a data structure, that is basically a tree of maps, where the map at each node contains some new elements, as well as the elements in its parent node's map. By map here I mean a ... 2answers 433 views ### What is the initialization time of a link-cut tree? Link-cut tree is a data structure invented by Sleator and Tarjan, which supports various operations and queries on a $n$-node forest in time $O(\log n)$. (For example, operation link combines two ... 0answers 256 views ### Notation for drawing rooted trees with isomorphic subtrees and recursive parts I need to specify small rooted trees with a lot of repeated parts and some recursive definitions. 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[closed] In which situations would I use a red/black tree instead of an unbalanced tree?	{"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": 42, "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": 0.9333709478378296, "perplexity_flag": "middle"}
http://en.wikibooks.org/wiki/A-level_Physics_(Advancing_Physics)/Radar_and_Triangulation	# A-level Physics (Advancing Physics)/Radar and Triangulation Radar and triangulation are two relatively easy methods of measuring the distance to some celestial objects. Radar can also be used to measure the velocity of a celestial object relative to us. ## Radar Essentially, radar is a system which uses a radio pulse to measure the distance to an object. The pulse is transmitted, reflected by the object, and then received at the site of the transmitter. The time taken for all this to happen is measured. This can be used to determine the distance to a planet or even the velocity of a spaceship. ### Distance The speed of electromagnetic waves (c) is constant in a vacuum: 3 x 108 ms-1. If we fire a pulse of radio waves to a planet within the Solar System, we know that: $d = ct$ where d is the distance to the object, and t is the time taken for the pulse to travel there and back from the object. However, the pulse has to get both there and back, so: $2d = ct$ $d = \frac{ct}{2}$ where d is the distance to the object, and t is the time taken for the pulse to return. ### Velocity The velocity of an object can be found by firing two radar pulses at an object at different times. Two distances are measured. When asked to calculate the relative velocity of an object in this way, use the following method: 1. Calculate the distance to the object at both times: $d_1 = \frac{c\Delta t_1}{2}$ $d_2 = \frac{c\Delta t_2}{2}$ 2. Calculate the distance the object has travelled between the two pulses. This is the difference between the two distances previously calculated: $\Delta d = d_2 - d_1$ 3. Calculate the time between the transmission (or reception, but not both) of the two pulses: $\Delta t = t_2 - t_1$ 4. Divide the distance calculated in step 2 by the time calculated in step 3 to find the average velocity of the object between the transmission of the two pulses: $v = \frac{\Delta d}{\Delta t}$ ## Triangulation The distance from the solar system to a relatively nearby celestial object can be found using triangulation. We know that the Earth is, on average, about 150 Gm away from the Sun. If we measure the angle between the vertical and the light from a nearby star 6 months apart (i.e. on opposite sides of the Sun), we can approximate the distance from the Solar System to the star. Let r be the radius of the Earth's orbit (assumed to be constant for simplicity's sake), a and b be the angles to the star (from the horizontal) when the Earth is on either side of the Sun, and let d be the perpendicular distance from the plane of the Earth's orbit to the star, as shown in the diagram on the right. By simple trigonometry: $2r = \frac{d}{\tan{a}} + \frac{d}{\tan{b}} = \frac{d(\tan{a} + \tan{b})}{\tan{a}\tan{b}}$ Therefore: $d = \frac{2r\tan{a}\tan{b}}{\tan{a} + \tan{b}}$ ## Questions 1. A radar pulse takes 8 minutes to travel to Venus and back. How far away is Venus at this time? 2. Why can't a radar pulse be used to measure the distance to the Sun? Regardless of (λ) wavelength, power density, or wavefront properties, the pulse would be absorbed with no reflection possible. Distances to pure energy sources are generally measured in terms of received light intensity, shifts of the light spectrum, and radio interferometry. The RF spectrum; and Laser (light) spectrum can be used to "listen" to radiation, but not bounce a pulse from an energy source having no true angle of incidence. Just as an observation, I will note that the sun can be seen on most radars either sunrise or sunset, usually when the sun is just above the horizon. But these receptions are unusable strobes (interference) and not a result of receiving a radar pulse from the sun. Radar technicians also use the sun as a "known" exact position to align the system to true north (and magnetic variations); this is called solar-boresighting and, again, only receives the radiation. 3. Radar is used to measure the velocity of a spacecraft travelling between the Earth and the Moon. Use the following data to measure this velocity: Pulse Transmission Time Reception Time 1 3:26:45.31213 3:26:45.51213 2 3:26:46.32742 3:26:46.52785 4. The angles between the horizontal and a star are measured at midnight on January 1 as 89.99980° and at midnight on July 1 as 89.99982°. How far away is the star? 5. Why can't triangulation be used to measure the distance to another galaxy? Worked Solutions	{"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": 10, "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": 0.9345549941062927, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/33911/why-linear-algebra-is-funor/80946	## Why linear algebra is fun!(or ?) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Edit: the original poster is Menny, but the question is CW; the first-person pronoun refers to Menny, not to the most recent editor. I'm doing an introductory talk on linear algebra with the following aim: I want to give the students a concrete example through which they will be able to see how many notions arise "naturally". Notions such as vector spaces, the zero vector, span, linear dependency and independency, basis, dimension, "good" bases, solving linear equations, and even linear maps and eigenvectors. A related MO question is Linear algebra proofs in combinatorics. The aim of this post is to find some more "concrete","real" and "natural" examples in this spirit that can interest everyone who loves what we do (and give them motivation to learn new definitions and formalisms). So if you have some ideas - please post them! Thanks, Menny - I think linear algebra is much more convincing when used to compute non-homogeneous inductive sequences. To compute homogeneous inductive sequences, it's probably more efficient to employ other methods (such as generating series (some MathOverflowers know much more than I about this)). [By inductive sequence I mean here a sequence $z\in\mathbb C^{\mathbb N}$ satisfying $P(S)z=a$, with $P\in\mathbb C[X]$, $P\not=0$, $S=$ shift, $a\in\mathbb C^{\mathbb N}$. And $z$ is called homogeneous iff $a=0$.] – Pierre-Yves Gaillard Jul 30 2010 at 15:45 2 I have started a discussion over on meta that mentions this question, although it is not currently focused solely on this one: meta.mathoverflow.net/discussion/566 The short version: as this question is CW, any user may edit it to improve it, and I think that the question would be improved if OP's original suggestion were moved to an answer. So I will do that. – Theo Johnson-Freyd Jul 30 2010 at 23:40 Thanks Theo. It looks much better. Menny – Menny Jul 31 2010 at 6:17 1 Per earlier discussion at meta, I state my objection to closing this question. Please, consider me as voting against closing. – Victor Protsak Jul 31 2010 at 21:38 A practial application of linear algebra: mathoverflow.net/questions/11784/… – Anweshi Aug 1 2010 at 19:20 ## 20 Answers An example that my last class loved was lossy image compression using the singular value decomposition. The SVD says that the transformation corresponding to any real matrix (not necessarily square) can be decomposed into three steps: a rotation that forgets some dimensions, a stretch along the coordinate axes, and finally a rotation. In other words, every matrix can be written the form HDA, with the rows of A being orthonormal, the columns of H being orthonormal, and D being a square diagonal matrix with nonnegative nonincreasing entries on the diagonal. Consider a photograph that is an $768\times 1024$ array of $(red,green,blue)$ triples, which we can just as well store as 3 matrices $R$, $G$, and $B$ of real numbers. Now even though the matrix $R$ has nothing to do with transforming space, we can consider it as such, and using SVD write $R=HDA$. Call the numbers on the diagonal of $D$ by $\lambda_1\geq \lambda_2 \geq \cdots \geq \lambda_s\geq 0$, and let $D_k'$ be $diag(\lambda_1,\dots,\lambda_k,0,0,\dots)$, an $s\times s$ diagonal matrix, and let $D_k$ be $diag(\lambda_1,\dots,\lambda_k)$. Let $H_k$ be the $768\times k$ matrix formed from the first $k$ columns of $H$, and similarly let $A_k$ be the $k\times 1024$ matrix formed from the first $k$ rows of $A$. Then $$R = HDA \approx HD_k' A = H_k D_k A_k,$$ where $\approx$ is because of continuity, which is appropriate if the $\lambda$'s that were replaced with zeros were small. Now for the punch-line. We need $3\cdot 768 \cdot 1024$ (about 2 megabytes) real numbers initially to store the photograph. But to store $H_k$, $D_k$, and $A_k$ for each of the three colors, we need only $3(768\cdot k+k+k\cdot 1024)=5379 k$ real numbers. With $k=25$, that gives a compression ratio of about 18. That is, file size goes from around 2 mb to around 130 kb. That is, it is faster by a factor of 20 to transmit the three matrices $H_{25}, D_{25}, A_{25}$ than it is to transmit their product. SVD is fast enough to compute that you can do this instantly (using Mathematica, say) with a picture of the audience, and they can marvel at their own blurry (but quite recognizable) faces. Also, showing the actual file sizes on disk of the original bitmap and the compressed image is quite impressive. At least, it is if your calculations come out extremely close. What's really impressive about this example (to me, at least) is that the matrix that we start with is just a table of data and not a transformation. But by considering it as if it were a transform, we gain power over it anyway. This is great motivation for linear algebra; students find it much easier to imagine encountering a table of data than a linear transformation. - 2 This of course works with any orthogonal transform. Adaptive transforms like the SVD have the disadvantage that you either need to transmit the basis separately from the coefficients or you need to multiply everything out as you did, in the process destroying much of the sparsity introduced by the zero truncations. The discrete cosine transform is surprisingly close to the KLT in practice and is based on a fixed basis, so you only need to transmit the coefficients; this is what is used in JPEG. – Per Vognsen Jul 31 2010 at 8:20 @Per: I don't understand what you mean by needing to transmit the basis. In this case, the matrix starts as a table of numbers, so the `obvious' basis is the correct one. I've rewritten some of what's above to make this more transparent. One thing to be cautious of with arbitrary transforms is that you may need to handle complex numbers, which can take up space and lecture time (and be hard to motivate, depending on the audience). On the up side, if you get to say, "and this is how JPEG works," then you have your finale written for you. – Kevin O'Bryant Jul 31 2010 at 16:52 He is referring to the basis of $\mathbb R^{n^2}$, not the basis of $\mathbb R^n$. The basis, in your case, consists of rank $1$ matrices of the form [column of first matrix]*[row of second]. – Will Sawin Nov 14 2011 at 21:44 ### 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. Menny's original version of the above question included the following example, which is better placed in an answer, so that it can be voted up and down. Like all answers to any CW question, this one is community wiki. The remainder of this post, unless someone else edit's it, consists of Menny's writing, so the first-person pronoun is Menny, not Theo. My example - the Fibonacci Sequence! I'll write it in the way I intend to present it; I hope it won't bore you and give you an idea for the type of example I'm looking for. • start by defining it. To them to wonder what is the general term. • Define $F_{a,b}$ to be the Fibonacci Sequence that starts with (a,b,a+b,...). Emphasize they know the first two, determines the other terms (but not explicitly! (yet)) • Ask them if there exists a sequence that they "really know", i.e., they can give me the general term. (someone will come up with the zero sequence (zero vector!)). • Tell them: if I give you the general term of, say $F_{2,3}$ can you use this information to find the general terms of other sequences? (hopefully, we will discover that we can multiply by scalars) • emphasize this great discovery - a scalar multiple of Fib seq is another Fib seq! • Well, assume they are given $F_{2,3}$ explicitly, can they get to any other seq. by scalar multiples? • No? OK, So I'll give you another sequence, which one do you want? (linear dependency...) • Get to the fact that you can also add them!!! • Take $F_{2,4}$ Is this enough? Yes? Well how do you get to $F_{0,1}$? and to $F_{\sqrt{2},1.5}$ (solving linear equations !!!) • Well, these $F_{2,4}$, $F_{2,3}$ must by special, if we work hard and find their general terms, we would find any general term of any given Fib. seq!!!!! • What are their main properties? you can't get to one from the other, with both you can get to everyone (this is almost the definition of a basis...!) • Can we find three seq. like the last too with similar properties? how would we phrase the property "you can't get to one from the other" for 3 seq? • Well, let them show\give as an exercise\ show it yourself that this cannot be. • Ask: any two seq with the property that you can't get to one from the other, also have the property that you can get to everything with them (using scalar multi. and addition)? • Ask: the reverse question? • Summarize: We've seen a vector space, the fact that one vector cannot span 2-dim space, the fact the 3 are linearly dependent, the fact that 2 lin.indep. span and vice-versa.... and (I didn't write) that the zero vector does not help to span and you can always get to it. • BUT.....this is becoming boring! we didn't find any general term yet and we are just assuming we did. BUT we can redefine a very glorious aim: find two linearly dependent sequences with their general term! • We only "know" two sequences from High school - let's try arithmetic progressions. Well... it doesn't work. • Let's try geometric sequence! ...work it out... It works! with q that satisfies q^2=q+1. At last, a "real" motivation for solving a quadratic equation! • find a "basis" • give the formula for $F_{0,1}$! • Summarize - this time dividing the board into "formal part" which will have words like vector space etc. and a part with seq and "bad definition" as above. If you also wish to talk about eigenvectors and give a more "natural" reason for using the geometric sequence - tell them that there is another "symmetry"/operation for the seq- The Shifting map. - Well, a sequence is geometric if and only if it is an eigenvector. Also, you can talk about larger recurrence laws, i.e. $a_n=a_{n-1}+a_{n-2}+a_{n-3}$ and get 3-dim space... I also found such examples in the first chapter of Newman's Analytic number theory book (which is amazing leisure-time reading!!) - My favorite elementary application of linear algebra is proving that the decomposition used in Calculus of rational functions into partial fractions works. Start with a polynomial $Q(x)=(x-r_1)(x-r_2)\cdots(x-r_n)$. Then the space of $P(x)/Q(x)$ with $deg P < deg Q$ is $n$-dimensional since it has a basis {$\frac{1}{Q(x)}, \frac{x}{Q(x)}, \frac{x^2}{Q(x)}, \dots, \frac{x^{n-1}}{Q(x)}$}. But {$\frac{1}{(x-r_1)},\frac{1}{(x-r_2)},\dots,\frac{1}{(x-r_n)}$} are linearly independent vectors in the space and thus a basis. Hence, $\frac{P(x)}{Q(x)}=\frac{A_1}{(1-x-r_1)}+\frac{A_2}{(x-r_2)}+\dots+\frac{A_n}{(x-r_n)}$ for some constants {$A_1,\dots,A_n$}, which then we can furthermore find by taking the limit of $(x-r_i)\frac{P(x)}{Q(x)}$ as $x$ goes to $r_i$. - Along similar lines, the Gram-Schmidt algorithm applied to the standard polynomial basis $1, x, x^2, \ldots$ with respect to the $\int_{[-1,1]} w(x) p(x) {\bar q}(x) dx$ inner product constructs the family of orthogonal polynomials defined by the weight function $w(x)$. – Per Vognsen Jul 31 2010 at 8:27 @Vladimir,Per: Hey,those are 2 great and simple ones!I got to remember those for when I teach linear algebra the first time! – Andrew L Jul 31 2010 at 20:11 1 This of course works with polynomials with repeated roots, where you just throw 1/(x-r_i)^j for 1 <= j <= k with k the degree of repetition. Asking students to extend the case for Q with distinct roots to arbitrary Q might be a fun exercise. – Vladimir Sotirov Aug 1 2010 at 20:24 Since this question was on the front page anyway, I took the liberty to texify this answer. I hope no one minds. – David White Nov 14 2011 at 14:10 Google's page rank algorithm makes use of many concepts and ideas from linear algebra. - 1 This is indeed a very good example. I taught Linear Algebra to Informatics students for a couple of years in the past and one of the tutorial questions was about the page rank algorithm and I dare say it was the most popular. – José Figueroa-O'Farrill Jul 30 2010 at 14:15 10 Well, "linear algebra makes you rich" is a great punchline for a linear algebra class :) – Mariano Suárez-Alvarez Jul 30 2010 at 14:28 I'm not sure how much that article simplifies things, but last time I checked PageRank was based off algoriths involving (spectral) graph theory. This does of course relate to linear algebra, but is a bit more involved. – Noldorin Jul 30 2010 at 15:00 @Noldorin: that's true, but fundamentally it's just an eigenvector problem. – Mark Meckes Jul 30 2010 at 15:30 My favorite application of linear algebra, as introduced to me by Fan Chung, is Oddtown (which I learned about from a manuscript of Lovasz, but may not be due to him). The $n$ residents of Oddtown love to form clubs; call the family of these $\mathcal{F}$. If $F_1$ and $F_2$ ($F_1 \neq F_2$) are in $\mathcal{F}$, then $\|F_1\|$ must be odd (this is Oddtown!) and $\|F_1 \cap F_2\|$ must be even ($\scriptsize{go\;Oddtown?}$). The question is, how many clubs may these $n$ people form? The answer (taken from Tibor Szabó's lecture notes) is this: Let $\mathcal{F} = {F_1,\ldots,F_m} \subseteq 2^{[n]}$ be a set of clubs in Oddtown. Let $\mathbf{v}_i \in \{0,1\}^n$ be the characteristic vector of $F_i$; the $j$th coordinate is 1 iff $j \in F_i$. Note that $\mathbf{v}_i^T v_j = \|F_i \cap F_j\|$. Now, $\mathbf{v}_1,\ldots,\mathbf{v}_m$ is independent over $\mathbb{F}^n_2$: if $\lambda_1\mathbf{v}_1 + \cdots + \lambda_m\mathbf{v}_m = 0$, then for each $i$ we have $$0 \;=\; (\lambda_1\mathbf{v}_1 + \cdots + \lambda_m\mathbf{v}_m)^T\mathbf{v}_i \;=\; \lambda_1\mathbf{v}_1^T\mathbf{v}_i + \cdots + \lambda_i\mathbf{v}_i^T\mathbf{v}_i + \ldots + \lambda_m\mathbf{v}_m^T\mathbf{v}_i \;=\; \lambda_i$$ Since $\mathbf{v}_1,\ldots,\mathbf{v}_m$ are linearly independent vectors over $\mathbb{F}^n_2$, $m \leq n$, and Oddtown can have at most $n$ clubs. - 2 and the upper bound is obviously achieved. – Alekk Jul 31 2010 at 2:27 1 Yes, but whether we prefer n singleton clubs or (for n even) n clubs each omitting one person, depends upon how odd the residents are. I would like to see the minutes from a meeting of a club with only one person. Particularly any objections. – Eric Tressler Jul 31 2010 at 16:49 1 You should post this over at mathoverflow.net/questions/17006/… , too. Beautiful. – Kevin O'Bryant Jul 31 2010 at 21:38 The theory of error correcting codes is a very nice and elementary context for introducing linear algebra, assuming the students know $\mathbb{F}_2$. The notion that each message of bit-length $n$ can coded as a "vector" over $\mathbb{F}_2$ of dimension $m > n$, using some linear conditions ("linear subspace") so as to provide easy error-checking conditions, should be quite motivating. Concepts such as "linear transforms" (matrices) and "null-spaces" show up naturally when considering the parity check matrix of the code. Etc., Etc. - I like using the example of magic squares when starting to go over linear algebra, usually starting with $3\times 3$ squares. They're a nice recreational maths thing that everyone has seen before, but usually not thought about. When asked for an example, most students come up with something like $\pmatrix{6&1&8\cr 7&5&3\cr 2&9&4}$, remembering a construction from before. When prodded for a second example, someone might suggest rotating or reflecting this example. Once it's suggest that we just want the rows, columns and diagonals to sum to the same thing, and that the numbers don't have to be distinct, someone usually thinks of $\pmatrix{1&1&1\cr 1&1&1\cr 1&1&1}$. It then usually becomes clear that linear combinations of what we have so far will also work, and this leads naturally into asking how many squares we need in a basis, and so on. (I then ask them to work out the dimension of the space of $n\times n$ magic squares as homework.) Another "unexpected" use of linear algebra is when they're asked to prove that things like $\sqrt2+\sqrt3$ or $\sqrt2 + \sqrt[3]2$ are algebraic. Many fiddle around until they chance upon an arrangement that works, but they all like it when we show that it's sufficient to take a few powers and say "oh, some combination of those will do". This usually goes down well, as people often like playing with numbers. - Yesterday Only I learned about Fisher's inequality and I think it is good example to show application of rank calculations. The problem is following: Fisher, a population geneticist and statistician, was concerned with the design of experiments studying the differences among several different varieties of plants, under each of a number of different growing conditions, called "blocks". Let: ```` v be the number of varieties of plants; b be the number of blocks. ```` It was required that: ````1 k different varieties are in each block, k < v; no variety occurs twice in any one block; 2 any two varieties occur together in exactly λ blocks; 3 each variety occurs in exactly r blocks. ```` Fisher's inequality states simply that $v \leq b$. And its proof (given below) involves basic linear algebra. Let the incidence matrix $M$ be a $v×b$ matrix defined so that $M_{i,j}$ is 1 if element $i$ is in block $j$ and $0$ otherwise. Then $B=MM^T$ is a $v×v$ matrix such that $B_{i,i}=r$ and $B_{i,j}=λ$ for $i \neq j$. Since $r\neq \lambda$, $det(B) \neq 0$, so $rank(B) = v$; on the other hand, $rank(B) \leq rank(M) \leq b$, so $v \leq b$. I have not been to able to link directly to Wikipedia page, so had to paste the question and answer here. Apologies for that. - Actually assumption 3 is not necessary; if we drop it we get $B_{ii}=r_i$ non-constant, as it was in the MO question in your link. (The conclusion is the same; check my answer there, and the subsequent comments). – Pietro Majer Jul 30 2010 at 20:22 I believe that determinants are worth special attention not only because they are good indicators on whether a given matrix has full rank. The famous example is the Vandermonde determinant (with application, for example, to the Lagrange interpolation) but please make a brief look at the Advanced Determinant Calculus and Advanced Determinant Calculus: A Complement by Christian Krattenthaler. The methods of computing determinants as well as their numerious applications all over maths belong to Arts.... - 1 Seconded. Determinants and minors can be treated very effectively and intuitively from the dual combinatorial and geometric perspective of exterior algebra. It's a shame this isn't done more often in introductory courses. – Per Vognsen Jul 31 2010 at 12:54 1 I think that interpolation is a great application to present! – Victor Protsak Jul 31 2010 at 21:42 1 For interpolation, one great and underappreciated method in linear algebra is polarization of forms. The algorithm of de Casteljau for Bezier interpolation and de Boor's algorithm for B-spline interpolation are special cases. It gives you a numerically robust and geometrically intuitive evaluation algorithm that proceeds directly from the control points by iterated linear interpolation. – Per Vognsen Aug 1 2010 at 4:42 An abstract but still elementary application is that every field is a vector space over any of its subfields. In particular, every finite field $F$ is a vector space over its prime field, and so $\|F\| = p^n$ for some $n$ where $p$ is the characteristic of $F$. The same style of reasoning applied to finite extensions of $\mathbb{Q}$ gives negative solutions to the ancient problems of duplicating the cube and trisecting the angle with ruler and compass. Galois theory has plenty of deeper applications of linear algebra to the study of field extensions. But the ones I mentioned are easily enough accessible that they could serve as motivation for the abstract approach to linear algebra. - Since you are going to address undergraduates, there's a book Linear algebra gems that might give you lots of simple, cool stuff to present. It is also available at amazon. - A bit rubbish and easy, but amusing if you haven't seen it before. Let $G$ be a finite group such that $g^2=e$ for all $g \in G$, i.e. every element (except the identity $e$) has order 2. Then $G$ has size $2^n$ for some $n$. This is not too hard to prove directly; but it becomes totally obvious (once you've proved that $G$ is abelian) when you realise that $G$ is a finite-dimensional vector space over $F_2$. - 1 It's already totally obvious by Cauchy's theorem. I usually hear this problem stated differently, although I can't quite remember exactly what is supposed to be shown. – Qiaochu Yuan Aug 1 2010 at 1:41 1 Zen,how exactly is this example obvious to a general audience of beginners with only the barest essentials of linear algebra?Particularly one that hasn't seen group theory yet? Mathematicans often have trouble remembering what it was like to be a rank beginner without many tools yet in thier box.I got the idea those were the kinds of audiences Menny originally had in mind and this one will go way over thier heads. – Andrew L Aug 1 2010 at 4:41 2 Cauchy's theorem is harder than the LA proof, however... – darij grinberg Aug 1 2010 at 19:48 I did say it was rubbish! If not to your taste, just ignore. – Zen Harper Aug 4 2010 at 13:37 This is not rubbish! It is the basis of the analysis of two-level fractional factorial designs in Design of Experiments, in Statistics! – Kjetil B Halvorsen Jul 1 at 19:24 I recommend the use of examples from linear geometry applied to computer graphics. All the basical notions of linear algebra can be easily visualized (in fact I recommend starting with Euclidean affine geometry). See the series "Graphics Gems" for specific examples http://books.google.es/books?id=fvA7zLEFWZgC (there are also a lot of texts books). In my opinion it is the best option to see linear algebra in action. - Here's a fun problem from a recent linear algebra exam at my university. While at university, all students are either in class, in the library, or at the bar. Detailed research by university management has shown that if a student is in class one minute, then after five minutes the student has a 60% chance of still being in class, a 20% chance of being in the library, and a 20% chance of being at the bar. Similarly, if the student is in the library at a certain time, then he or she has a 30% chance of being in class in five minutes' time, a 40% chance of remaining in the library, and a 30% chance of being in the bar. Finally, if the student is in the bar, then there is a 10% chance that he or she will be in class in five minutes' time, a 10% chance of being in the library, and an 80% chance of staying in the bar. What percentage of students do you expect to be in the bar after a long time? So it's a Markov chain problem, which can be used to motivate matrices, vectors, matrix multiplication, eigenvalues and eigenvectors. - Unless the bar/pub/establishment has lock-ins, I would say zero percent... – Yemon Choi Nov 15 2011 at 1:43 1 Do these students ever sleep??? – Victor Protsak Nov 15 2011 at 2:30 9 @Victor, it says the students are in class, it doesn't say they are awake. – Gerry Myerson Nov 15 2011 at 4:24 David Lay's text (which was recommended in the accepted answer to the MO question mathoverflow.net/questions/16994/…) has a whole chapter on application of linear algebra to Markov chains and many problems like this. The bar setting is fun indeed... – Margaret Friedland Nov 15 2011 at 16:23 This article gives a nice connection between linear algebra and calculus, i.e. explains how the fundamental calculus operations of differentiation and integration can be understood instead as a linear transformation. - it should be easy to follow and gives some fascinating insights: http://www.the-idea-shop.com/article/225/the-linear-algebra-view-of-calculus - A beautiful example of applications of linear algebra in linear PDEs is the theory of harmonic functions. With linear algebra and very few analysis one completely characterizes e.g. the space of "spherical harmonics", the eigenfunctions of the spherical Laplacian, which are the n-dimensional analogue of the trigonometric functions sin and cos. - A more elementary application along these lines is the use of linear algebra to express sin and cos of multiple angles as trigonometric polynomials (this can be further adjusted depending on the sophistication level of the audience). – Victor Protsak Jul 31 2010 at 21:44 Rubik's clock can be solved using linear algebra. The only reservation I have about this example is that the Rubik's clock puzzle is unfortunately nowhere near as fun as Rubik's cube. Not only is it obscure, but it's basically impossible to look at both sides of the clock at once. Also, the specimens I've seen are not very well constructed and it's hard to turn the wheels. Despite all that, I personally enjoyed solving Rubik's clock a lot, and a significant part of the fun was discovering that it was a linear algebra problem. (This was back when the puzzle first came out; I was still an undergraduate, and linear algebra was still relatively new to me.) - As important as the Fibonacci sequence and its related sequences are in the overall hierarchy of functions, it just doesn't come naturally to most beginners and the depth of its many interconnections with diverse areas of mathematics will be lost on most of them. Your examples are clever, but I seriously doubt unless your audience is made up of very strong undergraduates with math competition experience and therefore quite bit of familiarity with counting problems, your examples are likely to be met with chirping crickets being heard clearly... Geometry and physics are much more familiar to a general mathematical audience and linear algebra has so many connections with these topics, it’s really much more natural to start with those. These are my favorite examples to give. Describe planes in R3 as linear subspaces of R3, add vectors displaying their parallel lines, give isometries as examples of linear transformations and then construct matrices for them with respect to several possible bases. Show the fundamental theorem of systems of linear equations geometrically (i.e. that the corresponding systems of lines can be parallel, perpendicular or coincident). And then discuss similarities and their corresponding row vectors as eigenvectors of the corresponding eigenspaces. And then you can solve systems of differential equations as your last magic trick. To prepare for the lecture, I'd look at Linear Algebra Through Geometry by Thomas Banchoff and John Wermer as well as the classic Linear Algebra With Applications by Gilbert Strang. Lots of good ideas and examples in these books to guide you in preparing this talk. If you want a lot of very nice specific examples to use in your talk, there’s a terrific discussion and application of convergent sequences of diagonalizable stochastic matrices to solve problems such as the likelihood of graduation of students at a community college and the proportion at any given time of city and rural dwellers in a populated area undergoing mass migrations in the 4th edition of Steven H. Friedberg, Arnold J. Insel and Lawrence E. Spence’s Linear Algebra. It’s in section 5.3. - 11 1. I don't see how on earth you deduced that someone who used the Fibonacci sequence as an example was into combinatorics. 2. The examples you listed are the standard trivial examples you would teach in a beginning undergraduate course in linear algebra. They definitely are not the sorts of fun, novel examples the author is looking for. 3. I don't see how it is useful to list a couple of standard textbooks in linear algebra. – Andy Putman Jul 30 2010 at 20:42 11 I downvoted the response because it led with the first paragraph, which is all unjustified opinion, much of it dismissive of the OP. What followed didn't change my mind. Moreover (since you asked), I find all the spelling and punctuation mistakes unhelpful, and I wonder why you can't take a little extra time to fix these. – Pete L. Clark Jul 31 2010 at 1:04 8 @Andrew: Why don't you fix the spelling and punctuation mistakes this time? By not doing so, you are creating the impression that you don't take the process very seriously. – Pete L. Clark Jul 31 2010 at 5:57 7 @Andrew : Many students see the Fibonacci sequence in high school. I certainly did, and I went to a pretty lousy public school in Georgia. I don't know why you think it is an advanced topic... – Andy Putman Jul 31 2010 at 6:36 6 @Victor P: I had edited many lots of Andrew L's answers to fix his spelling and punctuation. I was disappointed that he was not putting in effort. The authors of the other questions with poor writing habits stop after very few questions and such occasional instances are not worth spending effort for reform. – Anweshi Jul 31 2010 at 20:17 show 15 more comments One of my favorite elementary applications is the classification of projective conics by invoking the spectral theorem on a polarized quadratic form. It makes short work of what would at first glance seem like a messy problem. - Can you give more details? Thanks. – Menny Jul 31 2010 at 17:50 You write the real projective conic via a homogeneous quadratic form and polarize it to get a $3 \times 3$ symmetric matrix. The spectral theorem puts that into the form $Q^T\ D\ Q$ for an orthogonal matrix $Q$ and diagonal matrix $D$. You then absorb the square root of the nonzero diagonal magnitudes into the $Q$ factors (which generally makes them non-orthogonal but still invertible). That leaves diagonal entries that are either -1, 0 or +1. Now you just have to analyze the possible combinations of signs. – Per Vognsen Jul 31 2010 at 18:22 One nice application of linear algebra (mainly dimension theory) is the impossibility of the duplication of the cube (problem that dates back to the greeks and was solved only in 1837 by Wantzel). -	{"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": 139, "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": 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": 0.9490962028503418, "perplexity_flag": "middle"}
http://nrich.maths.org/2292/solution	### Weekly Problem 14 - 2012 Weekly Problem 14 - 2012 ### Lost Can you locate the lost giraffe? Input coordinates to help you search and find the giraffe in the fewest guesses. ### Square Coordinates A tilted square is a square with no horizontal sides. Can you devise a general instruction for the construction of a square when you are given just one of its sides? # Coordinate Patterns ##### Stage: 3 Challenge Level: Wow, we received loads of solutions here! Let's have a look at a few of them: #### Charlie's Squares William and Chris, from Croftlands Junior School, tried to approach Charlie's Squares this way: We found that the best way to organise our information was to draw a table. Here is the table we drew: \| \| \| \| \|--------\|--------------\|--------------\| \| Square \| x-coordinate \| y-coordinate \| \| 1 \| 2 \| 2 \| \| 2 \| 5 \| 3 \| \| 3 \| 8 \| 4 \| \| 4 \| 11 \| 5 \| \| 5 \| 14 \| 6 \| Using the table, we found out that the x-coordinate was going up 3 every time. We added a couple of extra examples in the table. We spotted that the y-coordinate was the number of the square plus 1. Then we tried spotting patterns by working with the y coordinates to find the x coordinates. A great way to start! They then went on to find the general formula correctly. Callum, Elys, Cerys, Elgan, Cullen, Ethan, Ifan and Twm, from Ysgol Llanegryn, jumped straight in with the following: We first found a pattern: (2,2), (5,3), (8,4), and we then discussed how to find any centre square. We turned to algebra. The nth term in the pattern is (3n -1, n + 1) so the coordinates of the 20th centre point would be (60-1, 20+1) which is (59,21). Justin, from the John of Gaunt School, correctly noted: In the case of going to the left, the process is repeated but instead of increasing the horizontal and vertical coordinates, the coordinates decrease by the same amounts. Sam, from Fern Avenue, gave the following answer to Alison's Triangles: I started by comparing triangles 1 and 3, and realised that 'middle' vertices on nearby odd triangles were exactly 8 points horizontally apart. That, with the fact that all odd middle vertices have coordinates of the form (x,10), allows you to find the middle vertex of any odd numbered triangle. The same applies for all even numbered triangles, except that the vertices are on coordinates that are 4 grid squares to the right of the odd triangles. After noticing this, I realised that the even isosceles triangles continued to the middle vertex of the odd numbered triangles, and vice-versa. This meant that they either went down or up 5 squares, and right 2 squares. I now realised that if I knew the coordinates of any triangle, then I knew the middle vertex of the next triangle. Therefore, triangle 23 will have vertices at (88,5), (90,10), and (92,5). Joe and Jack, from Springfield Primary School, gave us their formula for the x-coordinate of the top/bottom vertex of the nth triangle, which was 4n-2. From this they plugged in n = 23, which gave them the correct x-coordinate, and noticed that the y-coordinate alternates between 10 (when n was odd) and 0 (when n was even). Great! #### More Squares from Charlie Penny, Jacob, Patrick and Cameron, from Inter Lakes, submitted solutions to More Squares from Charlie. Cameron wrote: The coordinates of the centre of square 22b are (44,-36). To work this out, first I made a chart for the x-coordinates and the y-coordinates. To get from one 'b' square to the next, add 2 to the x-coordinate and subtract 2 from the y-coordinate. My quick and efficient strategy is knowing the x and y axis rules. Brilliant! #### A recap of rules for the nth term Like some of the students featured above, Ivan Ivanov from the 47th High School in Sofia, Bulgaria put his reasoning into algebraic form. He found formulas for the coordinates of the shapes using the nice notation below. Well done to everyone who found similar formulae. #### Charlie's Squares 1. If $C(n)$ is the centre of square $n$, then the coordinates of $C(n)$ satisfy the equations: $x(n) = 3n - 1$, and $y(n) = n + 1$. 2. If $L(n)$ is the bottom left hand vertex of square $n$, then the coordinates of $L(n)$ satisfy the equations: $x(n) = 3n - 2$, and $y(n) = n - 1$. These formulas both give the correct coordinates even when Charlie goes left with values $n = ....-2, -1, 0, 1, 2, 3....$. #### Alison's Triangles 1. If $C(n)$ is the top or bottom vertex of triangle $n$ (for odd and even $n$ respectively), then the coordinates of $C(n)$ satisfy the equations: $x(n) = 4n - 2$, $y(n) = 10$ when $n$ is odd and $y(n) = 0$ when $n$ is even. 2. If $L(n)$ is the left-most vertex of triangle $n$, then the coordinates of $L(n)$ satisfy the equations: $x(n) = 4n - 4$, $y(n) = 5$. 3. The right-most vertex of triangle $n$ is the same as the left-most vertex of triangle $n-1$. Again, these formulas give the correct coordinates when Alison works left with values $n = ....-2, -1, 0, 1, 2, 3....$. #### More Squares from Charlie If $B(n)$ is the centre of square $nb$, then the coordinates of $B(n)$ satisfy the equations: $x(n) = 2n$, and $y(n)B = -2n + 8$. You might like to think if you can find a formula for squares such as 19c or 199a. What about if Charlie continues the pattern and draws squares such as 1d and 1e, etc? Michael Sena from NSBH sent in a little computer program that could give the coordinates of any of Charlie's first set of squares, even producing a table of the first few. Well done! Thanks again to everyone for their solutions! 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": 33, "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": 0.9065535068511963, "perplexity_flag": "middle"}
http://www.reference.com/browse/caustic-surface	Definitions # BKL singularity A BKL (Belinsky-Khalatnikov-Lifshitz) singularity is a model of the dynamic evolution of the Universe near the initial singularity, described by a non-symmetric, chaotic, vacuum solution to Einstein's field equations of gravitation. According to this model, the Universe is oscillating (expanding and contracting) around a singular point (singularity) in which time and space become equal to zero. This singularity is physically real in the sense that it is a necessary property of the solution, and will appear also in the exact solution of those equations. The singularity is not artificially created by the assumptions and simplifications made by the other well-known special solutions such as the Friedmann-Lemaître-Robertson-Walker, quasi-isotropic, and Kasner solutions. The Mixmaster universe is a solution to general relativity that exhibits properties similar to those discussed by BKL. ## Existence of time singularity The basis of modern cosmology are the special solutions of Einstein's field equations found by Alexander Friedmann in 1922 and 1924 that describe a completely homogeneous and isotropic Universe ("closed" or "open" model, depending on closeness or infiniteness of space). The principal property of these solutions is their non-static nature. The concept of an inflating Universe that arises from Friedmann's solutions is fully supported by astronomical data and the present consensus is that the isotropic model, in general, gives an adequate description of the present state of the Universe. Another important property of the isotropic model is the existence of a time singularity in the spacetime metric. In other words, the existence of such time singularity means finiteness of time. However, the adequacy of the isotropic model in describing the present state of the Universe by itself is not a reason to expect that it is so adequate in describing the early stages of Universe evolution. The problem initially addressed by the BKL paper is whether the existence of such time singularity is a necessary property of relativistic cosmological models. There is the possibilty that the singularity is generated by the simplifying assumptions, made when constructing these models. Independence of singularity on assumptions would mean that time singularity exists not only in the particular but also in the general solutions of the Einstein equations. A criterion for generality of solutions is the number of arbitrary space coordinate functions that they contain. These include only the "physically arbitrary" functions whose number cannot be reduced by any choice of reference frame. In the general solution, the number of such functions must be sufficient for arbitrary definition of initial conditions (distribution and movement of matter, distribution of gravitational field) in some moment of time chosen as initial. This number is four for vacuum and eight for a matter and/or radiation filled space. For a system of non-linear differential equations, such as the Einstein equations, general solution is not unambiguously defined. In principle, there may be multiple general integrals, and each of those may contain only a finite subset of all possible initial conditions. Each of those integrals may contain all required arbitrary functions which, however, may be subject to some conditions (e.g., some inequalities). Existence of a general solution with a singularity, therefore, does not preclude the existence also of other general solutions that do not contain a singularity. For example, there is no reason to doubt the existence of a general solution without singularity that describes an isolated body with a relatively small mass. It is impossible to find a general integral for all space and for all time. However, this is not necessary for resolving the problem: it is sufficient to study the solution near the singularity. This would also resolve another aspect of the problem: the characteristics of spacetime metric evolution in the general solution when it reaches the physical singularity, understood as a point where matter density and invariants of the Riemann curvature tensor become infinite. The BKL paper concerns only the cosmological aspect. This means, that the subject is a time singularity in the whole spacetime and not in some limited region as in a gravitational collapse of a finite body. Previous work by the Landau-Lifshitz group (reviewed in ) led to a conclusion that the general solution does not contain a physical singularity. This search for a broader class of solutions with singularity has been done, essentially, by a trial-and-error method, since a systemic approach to the study of the Einstein equations is lacking. A negative result, obtained in this way, is not convincing by itself; a solution with the necessary degree of generality would invalidate it, and at the same time would confirm any positive results related to the specific solution. It is reasonable to suggest that if a singularity is present in the general solution, there must be some indications that are based only on the most general properties of the Einstein equations, although those indications by themselves might be insufficient for characterizing the singularity. At that time, the only known indication was related to the form of Einstein equations written in a synchronous reference frame, that is, in a frame in which the interval element is $ds^2 = dt^2 - dl^2 , dl^2 = gamma_\left\{alpha beta\right\} dx^\left\{alpha\right\} dx^\left\{beta\right\},$ (eq. 1) where the space distance element dl is separate from the time interval dt, and x0 = t is the proper time synchronized throughout the whole space. The Einstein equation $scriptstyle\left\{R_0^\left\{0\right\}=T_0^\left\{0\right\}-frac\left\{1\right\}\left\{2\right\}T\right\}$ written in synchronous frame gives a result in which the metric determinant g inevitably becomes zero in a finite time irrespective of any assumptions about matter distribution. This indication, however, was dropped after it became clear that it is linked with a specific geometric property of the synchronous frame: crossing of time line coordinates. This crossing takes place on some encircling hypersurfaces which are four-dimensional analogs of the caustic surfaces in geometrical optics; g becomes zero exactly at this crossing. Therefore, although this singularity is general, it is fictitious, and not a physical one; it disappears when the reference frame is changed. This, apparently, stopped the incentive for further investigations. However, the interest in this problem waxed again after Penrose published his theorems that linked the existence of a singularity of unknown character with some very general assumptions that did not have anything in common with a choice of reference frame. Other similar theorems were found later on by Hawking and Geroch (see Penrose-Hawking singularity theorems). It became clear that the search for a general solution with singularity must continue. ## Generalized Kasner solution Further generalization of solutions depended on some solution classes found previously. The Friedmann solution, for example, is a special case of a solution class that contains three physically arbitrary coordinate functions. In this class the space is anisotropic; however, its compression when approaching the singularity has "quasi-isotropic" character: the linear distances in all directions diminish as the same power of time. Like the fully homogeneous and isotropic case, this class of solutions exist only for a matter-filled space. Much more general solutions are obtained by a generalization of an exact particular solution derived by Kasner for a field in vacuum, in which the space is homogeneous and has Euclidean metric that depends on time according to the Kasner metric $dl^2=t^\left\{2p_1\right\}dx^2+t^\left\{2p_2\right\}dy^2+t^\left\{2p_3\right\}dz^2$ (eq. 2) (see ). Here, p1, p2, p3 are any 3 numbers that are related by $p_1+p_2+p_3=p_1^2+p_2^2+p_3^2=1.$ (eq. 3) Because of these relationships, only 1 of the 3 numbers is independent. All 3 numbers are never the same; 2 numbers are the same only in the sets of values $scriptstyle\left\{\left(-frac \left\{1\right\}\left\{3\right\},frac\left\{2\right\}\left\{3\right\},frac \left\{2\right\}\left\{3\right\}\right)\right\}$ and (0, 0, 1). In all other cases the numbers are different, one number is negative and the other two are positive. If the numbers are arranged in increasing order, p1 < p2 < p3, they change in the ranges $-frac \left\{1\right\}\left\{3\right\} le p_1 le 0, 0 le p_2 le frac\left\{2\right\}\left\{3\right\}, frac\left\{2\right\}\left\{3\right\} le p_3 le 1.$ (eq. 4) The numbers p1, p2, p3 can be written parametrically as $p_1\left(u\right)=frac \left\{-u\right\}\left\{1+u+u^2\right\}, p_2\left(u\right)=frac \left\{1+u\right\}\left\{1+u+u^2\right\}, p_3\left(u\right)=frac \left\{u\left(1+u\right)\right\}\left\{1+u+u^2\right\}$ (eq. 5) All different values of p1, p2, p3 ordered as above are obtained by changing the value of the parameter u in the range u ≥ 1. The values u < 1 are brought into this range according to $p_1 left\left(frac \left\{1\right\}\left\{u\right\} right \right)=p_1\left(u\right), p_2 left\left(frac \left\{1\right\}\left\{u\right\} right \right)=p_3\left(u\right), p_3 left\left(frac \left\{1\right\}\left\{u\right\} right \right)=p_2\left(u\right)$ (eq. 6) Figure 1 is a plot of p1, p2, p3 with an argument 1/u. The numbers p1(u) and p3(u) are monotonously increasing while p2(u) is monotonously decreasing function of the parameter u. In the generalized solution, the form corresponding to (eq. 2) applies only to the asymptotic metric (the metric close to the singularity t = 0), respectively, to the major terms of its series expansion by powers of t. In the synchronous reference frame it is written in the form of (eq. 1) with a space distance element $dl^2=\left(a^2l_\left\{alpha\right\}l_\left\{beta\right\}+b^2m_\left\{alpha\right\}m_\left\{beta\right\}+c^2n_\left\{alpha\right\}n_\left\{beta\right\}\right)dx^\left\{alpha\right\}dx^\left\{beta\right\},$ (eq. 7) where $a=t^\left\{p_l\right\}, b=t^\left\{p_m\right\}, c=t^\left\{p_n\right\}$ (eq. 8) The three-dimensional vectors l, m, n define the directions at which space distance changes with time by the power laws (eq. 8). These vectors, as well as the numbers pl, pm, pn which, as before, are related by (eq. 3), are functions of the space coordinates. The powers pl, pm, pn are not arranged in increasing order, reserving the symbols p1, p2, p3 for the numbers in (eq. 5) that remain arranged in increasing order. The determinant of the metric of (eq. 7) is $-g=a^2b^2c^2v^2=t^2v^2 ,$ (eq. 9) where v = l[mn]. It is convenient to introduce the following quantitities $lambda=frac\left\{mathbf\left\{l\right\} mathrm\left\{rot\right\} mathbf\left\{l\right\}\right\}\left\{v\right\}, mu=frac\left\{mathbf\left\{m\right\} mathrm\left\{rot\right\} mathbf\left\{m\right\}\right\}\left\{v\right\}, nu=frac\left\{mathbf\left\{n\right\} mathrm\left\{rot\right\} mathbf\left\{n\right\}\right\}\left\{v\right\}.$ (eq. 10) The space metric in (eq. 7) is anisotropic because the powers of t in (eq. 8) cannot have the same values. On approaching the singularity at t = 0, the linear distances in each space element decrease in two directions and increase in the third direction. The volume of the element decreases in proportion to t. The Einstein equations in vacuum in synchronous reference frame are $R_0^0=-frac\left\{1\right\}\left\{2\right\}frac\left\{partial varkappa_\left\{alpha\right\}^\left\{alpha\right\}\right\}\left\{partial t\right\}-frac\left\{1\right\}\left\{4\right\} varkappa_\left\{alpha\right\}^\left\{beta\right\} varkappa_\left\{beta\right\}^\left\{alpha\right\}=0,$ (eq. 11) $R_\left\{alpha\right\}^\left\{beta\right\}=-left \left(frac\left\{1\right\}\left\{2\right\}sqrt\left\{-g\right\} right \right) frac\left\{partial\right\}\left\{partial t\right\} left \left(sqrt\left\{-g\right\} varkappa_\left\{alpha\right\}^\left\{beta\right\} right \right)-P_\left\{alpha\right\}^\left\{beta\right\}=0,$ (eq. 12) $R_\left\{alpha\right\}^\left\{0\right\}=frac\left\{1\right\}\left\{2\right\} left \left(varkappa_\left\{alpha;beta\right\}^\left\{beta\right\}- varkappa_\left\{beta;alpha\right\}^\left\{beta\right\}right \right)=0,$ (eq. 13) where $scriptstyle\left\{varkappa_\left\{alpha\right\}^\left\{beta\right\}\right\}$ is the 3-dimensional tensor $scriptstyle\left\{varkappa_\left\{alpha\right\}^\left\{beta\right\}=frac\left\{partial gamma_\left\{alpha\right\}^\left\{beta\right\}\right\}\left\{partial t\right\}\right\}$, and Pαβ is the 3-dimensional Ricci tensor, which is expressed by the 3-dimensional metric tensor γαβ in the same way as Rik is expressed by gik; Pαβ contains only the space (but not the time) derivatives of γαβ. The Kasner metric is introduced in the Einstein equations by substituting the respective metric tensor γαβ from (eq. 7) without defining a priori the dependence of a, b, c from t: $varkappa_\left\{alpha\right\}^\left\{beta\right\}=left \left(frac\left\{2 dot a\right\}\left\{a\right\} right \right)l_\left\{alpha\right\}l^\left\{beta\right\}+left \left(frac\left\{2 dot b\right\}\left\{b\right\} right \right)m_\left\{alpha\right\}m^\left\{beta\right\}+left \left(frac\left\{2 dot c\right\}\left\{c\right\} right \right)n_\left\{alpha\right\}n^\left\{beta\right\}$ where the dot above a symbol designates differentiation with respect to time. The Einstein equation (eq. 11) takes the form $-R_0^0=frac\left\{ddot a\right\}\left\{a\right\}+frac\left\{ddot b\right\}\left\{b\right\}+frac\left\{ddot c\right\}\left\{c\right\}=0.$ (eq. 14) All its terms are to a second order for the large (at t → 0) quantity 1/t. In the Einstein equations (eq. 12), terms of such order appear only from terms that are time-differentiated. If the components of Pαβ do not include terms of order higher than 2, then $-R_l^l=frac\left\{\left(dot a b c\right)dot\left\{ \right\}\right\}\left\{abc\right\}=0, -R_m^m=frac\left\{\left(a dot b c\right)dot\left\{ \right\}\right\}\left\{abc\right\}=0, -R_n^n=frac\left\{\left(a b dot c\right)dot\left\{ \right\}\right\}\left\{abc\right\}=0$ (eq. 15) where indices l, m, n designate tensor components in the directions l, m, n. These equations together with (eq. 14) give the expressions (eq. 8) with powers that satisfy (eq. 3). However, the presence of 1 negative power among the 3 powers pl, pm, pn results in appearance of terms from Pαβ with an order greater than t−2. If the negative power is pl (pl = p1 < 0), then Pαβ contains the coordinate function λ and (eq. 12) become $begin\left\{align\right\}$ -R_l^l & =frac{(dot a b c)dot{ }}{abc}+frac{lambda^2 a^2}{2b^2 c^2}=0, -R_m^m & =frac{(a dot b c)dot{ }}{abc}-frac{lambda^2 a^2}{2b^2 c^2}=0, -R_n^n & =frac{(a b dot c)dot{ }}{abc}-frac{lambda^2 a^2}{2b^2 c^2}=0. end{align} (eq. 16) Here, the second terms are of order t−2(pm + pn − pl) whereby pm + pn − pl = 1 + 2 \|pl\| > 1. To remove these terms and restore the metric (eq. 7), it is necessary to impose on the coordinate functions the condition λ = 0. The remaining 3 Einstein equations (eq. 13) contain only first order time derivatives of the metric tensor. They give 3 time-independent relations that must be imposed as necessary conditions on the coordinate functions in (eq. 7). This, together with the condition λ = 0, makes 4 conditions. These conditions bind 10 different coordinate functions: 3 components of each of the vectors l, m, n, and one function in the powers of t (any one of the functions pl, pm, pn, which are bound by the conditions (eq. 3)). When calculating the number of physically arbitrary functions, it must be taken into account that the synchronous system used here allows time-independent arbitrary transformations of the 3 space coordinates. Therefore, the final solution contains overall 10 − 4 − 3 = 3 physically arbitrary functions which is 1 less than what is needed for the general solution in vacuum. The degree of generality reached until now is not lessened by introducing matter; matter is written into the metric (eq. 7) and contributes 4 new coordinate functions necessary to describe the initial distribution of its density and the 3 components of its velocity. This makes possible to determine matter evolution merely from the laws of its movement in an a priori given gravitational field. These movement laws are the hydrodynamic equations $frac\left\{1\right\}\left\{sqrt\left\{-g\right\}\right\}frac\left\{partial\right\}\left\{partial x^i\right\} left \left(sqrt\left\{-g\right\}sigma u^i right \right) = 0,$ (eq. 17) $\left(p+varepsilon\right) u^k left \left\{ frac\left\{partial u_i\right\}\left\{partial x^k\right\}-frac\left\{1\right\}\left\{2\right\} u^l frac\left\{partial g_\left\{kl\right\}\right\}\left\{partial x^i\right\} right rbrace =-frac\left\{partial p\right\}\left\{partial x^i\right\}-u_i u^k frac\left\{partial p\right\}\left\{partial x^k\right\},$ (eq. 18) where u i is the 4-dimensional velocity, ε and σ are the densities of energy and entropy of matter. For the ultrarelativistic equation of state p = ε/3 the entropy σ ~ ε1/4. The major terms in (eq. 17) and (eq. 18) are those that contain time derivatives. From (eq. 17) and the space components of (eq. 18) one has $frac\left\{partial\right\}\left\{partial t\right\} left \left(sqrt\left\{-g\right\} u_0 varepsilon^\left\{frac\left\{3\right\}\left\{4\right\}\right\} right \right) = 0, 4 varepsilon cdot frac\left\{partial u_\left\{alpha\right\}\right\}\left\{partial t\right\}+u_\left\{alpha\right\} cdot frac\left\{partial varepsilon\right\}\left\{partial t\right\} = 0,$ resulting in $abc u_0 varepsilon^\left\{frac\left\{3\right\}\left\{4\right\}\right\}= mathrm\left\{const\right\}, u_\left\{alpha\right\} varepsilon^\left\{frac\left\{1\right\}\left\{4\right\}\right\}= mathrm\left\{const\right\},$ (eq. 19) where 'const' are time-independent quantities. Additionally, from the identity uiui = 1 one has (because all covariant components of uα are to the same order) $u_0^2 approx u_n u^n = frac\left\{u_n^2\right\}\left\{c^2\right\},$ where un is the velocity component along the direction of n that is connected with the highest (positive) power of t (supposing that pn = p3). From the above relations, it follows that $varepsilon sim frac\left\{1\right\}\left\{a^2 b^2\right\}, u_\left\{alpha\right\} sim sqrt\left\{ab\right\}$ (eq. 20) or $varepsilon sim t^\left\{-2\left(p_1+p_2\right)\right\}=t^\left\{-2\left(1-p_3\right)\right\}, u_\left\{alpha\right\} sim t^\left\{frac\left\{\left(1-p_3\right)\right\}\left\{2\right\}\right\}.$ (eq. 21) The above equations can be used to confirm that the components of the matter stress-energy-momentum tensor standing in the right hand side of the equations $R_0^0 = T_0^0 - frac\left\{1\right\}\left\{2\right\}T, R_\left\{alpha\right\}^\left\{beta\right\} = T_\left\{alpha\right\}^\left\{beta\right\}- frac\left\{1\right\}\left\{2\right\}delta_\left\{alpha\right\}^\left\{beta\right\}T,$ are, indeed, to a lower order by 1/t than the major terms in their left hand sides. In the equations $scriptstyle\left\{R_\left\{alpha\right\}^0 = T_\left\{alpha\right\}^0\right\}$ the presence of matter results only in the change of relations imposed on their constituent coordinate functions. The fact that ε becomes infinite by the law (eq. 21) confirms that in the solution to (eq. 7) one deals with a physical singularity at any values of the powers p1, p2, p3 excepting only (0, 0, 1). For these last values, the singularity is non-physical and can be removed by a change of reference frame. The fictional singularity corresponding to the powers (0, 0, 1) arises as a result of time line coordinates crossing over some 2-dimensional "focal surface". As pointed out in , a synchronous reference frame can always be chosen in such way that this inevitable time line crossing occurs exactly on such surface (instead of a 3-dimensional caustic surface). Therefore, a solution with such simultaneous for the whole space fictional singularity must exist with a full set of arbitrary functions needed for the general solution. Close to the point t = 0 it allows a regular expansion by whole powers of t. ## Oscillating mode towards the singularity The four conditions that had to be imposed on the coordinate functions in the solution (eq. 7) are of different types: three conditions that arise from the equations $scriptstyle\left\{R_\left\{alpha\right\}^0\right\}$ = 0 are "natural"; they are a consequence of the structure of Einstein equations. However, the additional condition λ = 0 that causes the loss of one derivative function, is of entirely different type. The general solution by definition is completely stable; otherwise the Universe would not exist. Any perturbation is equivalent to a change in the initial conditions in some moment of time; since the general solution allows arbitrary initial conditions, the perturbation is not able to change its character. In other words, the existence of the limiting condition λ = 0 for the solution of (eq. 7) means instability caused by perturbations that break this condition. The action of such perturbation must bring the model to another mode which thereby will be most general. Such perturbation cannot be considered as small: a transition to a new mode exceeds the range of very small perturbations. The analysis of the behavior of the model under perturbative action, performed by BKL, delineates a complex oscillatory mode on approaching the singularity. They could not give all details of this mode in the broad frame of the general case. However, BKL explained the most important properties and character of the solution on specific models that allow far-reaching analytical study. These models are based on a homogeneous space metric of a particular type. Supposing a homogeneity of space without any additional symmetry leaves a great freedom in choosing the metric. All possible homogeneous (but anisotropic) spaces are classified, according to Bianchi, in 9 classes. BKL investigate only spaces of Bianchi Types VIII and IX. If the metric has the form of (eq. 7), for each type of homogeneous spaces exists some functional relation between the reference vectors l, m, n and the space coordinates. The specific form of this relation is not important. The important fact is that for Type VIII and IX spaces, the quantities λ, μ, ν (eq. 10) are constants while all "mixed" products l rot m, l rot n, m rot l, etc. are zeros. For Type IX spaces, the quantities λ, μ, ν have the same sign and one can write λ = μ = ν = 1 (the simultaneous sign change of the 3 constants does not change anything). For Type VIII spaces, 2 constants have a sign that is opposite to the sign of the third constant; one can write, for example, λ = − 1, μ = ν = 1. The study of the effect of the perturbation on the "Kasner mode" is thus confined to a study on the effect of the λ-containing terms in the Einstein equations. Type VIII and IX spaces are the most suitable models exactly in this connection. Since all 3 quantities λ, μ, ν differ from zero, the condition λ = 0 does not hold irrespective of which direction l, m, n has negative power law time dependence. The Einstein equations for the Type VIII and Type IX space models are $begin\left\{align\right\}$ -R_l^l & =frac{left(dot a b cright)dot{ }}{abc}+frac{1}{2}left (a^2b^2c^2right )left [lambda^2 a^4-left (mu b^2-nu c^2right )^2right ]=0, -R_m^m & =frac{(a dot{b} c)dot{ }}{abc}+frac{1}{2}left(a^2b^2c^2right )left [mu^2 b^4-left(lambda a^2-nu c^2right)^2right]=0, -R_n^n & =frac{left(a b dot cright)dot{ }}{abc}+frac{1}{2}left(a^2b^2c^2right)left[nu^2 c^4-left(lambda a^2-mu b^2right)^2right]=0, end{align} (eq. 22) $-R_0^0=frac\left\{ddot a\right\}\left\{a\right\}+frac\left\{ddot b\right\}\left\{b\right\}+frac\left\{ddot c\right\}\left\{c\right\}=0$ (eq. 23) (the remaining components $scriptstyle\left\{R_l^0\right\}$, $scriptstyle\left\{R_m^0\right\}$, $scriptstyle\left\{R_n^0\right\}$, $scriptstyle\left\{R_l^m\right\}$, $scriptstyle\left\{R_l^n\right\}$, $scriptstyle\left\{R_m^n\right\}$ are identically zeros). These equations contain only functions of time; this is a condition that has to be fulfiled in all homogeneous spaces. Here, the (eq. 22) and (eq. 23) are exact and their validity does not depend on how near one is to the singularity at t = 0. The time derivatives in (eq. 22) and (eq. 23) take a simpler form if а, b, с are substituted by their logarithms α, β, γ: $a=e^alpha, b=e^beta, c=e^gamma,$ (eq. 24) substituting the variable t for τ according to: $dt=abc dtau,$ (eq. 25). Then: $begin\left\{align\right\}$ 2alpha_{tautau} & =left (mu b^2-nu c^2right )^2-lambda^2 a^4=0, 2beta_{tautau} & =left (lambda a^2-nu c^2right )^2-mu^2 b^4=0, 2gamma_{tautau} & =left (lambda a^2-mu b^2right )^2-nu^2 c^4=0, end{align} (eq. 26) $frac\left\{1\right\}\left\{2\right\}left\left(alpha+beta+gamma right\right)_\left\{tautau\right\}=alpha_tau beta_tau +alpha_tau gamma_tau+beta_tau gamma_tau.$ (eq. 27) Adding together equations (eq. 26) and substituting in the left hand side the sum (α + β + γ)τ τ according to (eq. 27), one obtains an equation containing only first derivatives which is the first integral of the system (eq. 26): $alpha_tau beta_tau +alpha_tau gamma_tau+beta_tau gamma_tau = frac\left\{1\right\}\left\{4\right\}left\left(lambda^2a^4+mu^2b^4+nu^2c^4-2lambda mu a^2b^2-2lambda nu a^2c^2-2mu nu b^2c^2 right\right).$ (eq. 28) This equation plays the role of a binding condition imposed on the initial state of (eq. 26). The Kasner mode (eq. 8) is a solution of (eq. 26) when ignoring all terms in the right hand sides. But such situation cannot go on (at t → 0) indefinitely because among those terms there are always some that grow. Thus, if the negative power is in the function a(t) (pl = p1) then the perturbation of the Kasner mode will arise by the terms λ2a4; the rest of the terms will decrease with decreasing t. If only the growing terms are left in the right hand sides of (eq. 26), one obtains the system: $alpha_\left\{tautau\right\}=-frac\left\{1\right\}\left\{2\right\}lambda^2e^\left\{4alpha\right\}, beta_\left\{tautau\right\}=gamma_\left\{tautau\right\}=frac\left\{1\right\}\left\{2\right\}lambda^2e^\left\{4alpha\right\}$ (eq. 29) (compare (eq. 16); below it is substituted λ2 = 1). The solution of these equations must describe the metric evolution from the initial state, in which it is described by (eq. 8) with a given set of powers (with pl < 0); let pl = р1, pm = р2, pn = р3 so that $a sim t^\left\{p_1\right\}, b sim t^\left\{p_2\right\}, c sim t^\left\{p_3\right\}.$ (eq. 30) Then $abc=Lambda t, tau=Lambda^\left\{-1\right\}ln t+mathrm\left\{const\right\}$ (eq. 31) where Λ is constant. Initial conditions for (eq. 29) are redefined as $alpha_tau=Lambda p_1, beta_tau=Lambda p_2, gamma_tau=Lambda p_3 mathrm\left\{at\right\} tau to infty$ (eq. 32) Equations (eq. 29) are easily integrated; the solution that satisfies the condition (eq. 32) is $begin\left\{cases\right\}a^2=frac\left\{2\|p_1\|Lambda\right\}\left\{operatorname\left\{ch\right\}\left(2\|p_1\|Lambdatau\right)\right\}, b^2=b_0^2e^\left\{2Lambda\left(p_2-\|p_1\|\right)tau\right\}operatorname\left\{ch\right\}\left(2\|p_1\|Lambdatau\right),$ c^2=c_0^2e^{2Lambda(p_2-\|p_1\|)tau}operatorname{ch}(2\|p_1\|Lambdatau),end{cases} (eq. 33) where b0 and c0 are two more constants. It can easily be seen that the asymptotic of functions (eq. 33) at t → 0 is (eq. 30). The asymptotic expressions of these functions and the function t(τ) at τ → −∞ is $a sim e^\left\{-Lambda p_1tau\right\}, b sim e^\left\{Lambda\left(p_2+2p_1\right)tau\right\}, c sim e^\left\{Lambda\left(p_3+2p_1\right)tau\right\}, t sim e^\left\{Lambda\left(1+2p_1\right)tau\right\}.$ Expressing a, b, c as functions of t, one has $a sim t^\left\{p\text{'}_l\right\}, b sim t^\left\{p\text{'}_m\right\}, c sim t^\left\{p\text{'}_n\right\}$ (eq. 34) where $p\text{'}_l=frac$ {1-2\|p_1, p'_m=-frac{2\|p_1\|-p_2}{1-2\|p_1, p'_n=frac{p_3-2\|p_1{1-2\|p_1.> (eq. 35) Then $abc=Lambda\text{'} t, Lambda\text{'}=\left(1-2\|p_1\|\right)Lambda.$ (eq. 36) The above shows that perturbation acts in such way that it changes one Kasner mode with another Kasner mode, and in this process the negative power of t flips from direction l to direction m: if before it was pl < 0, now it is p'm < 0. During this change the function a(t) passes through a maximum and b(t) passes through a minimum; b, which before was decreasing, now increases: a from increasing becomes decreasing; and the decreasing c(t) decreases further. The perturbation itself (λ2a4α in (eq. 29)), which before was increasing, now begins to decrease and die away. Further evolution similarly causes an increase in the perturbation from the terms with μ2 (instead of λ2) in (eq. 26), next change of the Kasner mode, and so on. It is convenient to write the power substitution rule (eq. 35) with the help of the parametrization (eq. 5): $begin\left\{matrix\right\}$ mathbf{if} & p_l=p_{1}(u) & p_m=p_{2}(u) & p_n=p_{3}(u) mathbf{then} & p'_l=p_{2}(u-1) & p'_m=p_{1}(u-1) & p'_n=p_{3}(u-1) end{matrix} (eq. 37) The greater of the two positive powers remains positive. BKL call this flip of negative power between directions a Kasner epoch. The key to understanding the character of metric evolution on approaching singularity is exactly this process of Kasner epoch alternation with flipping of powers pl, pm, pn by the rule (eq. 37). The successive alternations (eq. 37) with flipping of the negative power p1 between directions l and m (Kasner epochs) continues by depletion of the whole part of the initial u until the moment at which u < 1. The value u < 1 transforms into u > 1 according to (eq. 6); in this moment the negative power is pl or pm while pn becomes the lesser of two positive numbers (pn = p2). The next series of Kasner epochs then flips the negative power between directions n and l or between n and m. At an arbitrary (irrational) initial value of u this process of alternation continues unlimited. In the exact solution of the Einstein equations, the powers pl, pm, pn lose their original, precise, sense. This circumstance introduces some "fuzziness" in the determination of these numbers (and together with them, to the parameter u) which, although small, makes meaningless the analysis of any definite (for example, rational) values of u. Therefore, only these laws that concern arbitrary irrational values of u have any particular meaning. The larger periods in which the scales of space distances along two axes oscillate while distances along the third axis decrease monotonously, are called eras; volumes decrease by a law close to ~ t. On transition from one era to the next, the direction in which distances decrease monotonously, flips from one axis to another. The order of these transitions acquires the asymptotic character of a random process. The same random order is also characteristic for the alternation of the lengths of successive eras (by era length, BKL understand the number of Kasner epoch that an era contains, and not a time interval). The era series become denser on approaching t = 0. However, the natural variable for describing the time course of this evolution is not the world time t but its logarithm, ln t, by which the whole process of reaching the singularity is extended to −∞. According to (eq. 33), one of the functions a, b, c, that passes through a maximum during a transition between Kasner epochs, at the peak of its maximum is $a_max=sqrt\left\{2Lambda\|p_1\left(u\right)$ (eq. 38) where it is supposed that amax is large compared to b0 and c0; in (eq. 38) u is the value of the parameter in the Kasner epoch before transition. It can be seen from here that the peaks of consecutive maxima during each era are gradually lowered. Indeed, in the next Kasner epoch this parameter has the value u' = u - 1, and Λ is substituted according to (eq. 36) with Λ' = Λ(1 − 2\|p1(u)\|). Therefore, the ratio of 2 consecutive maxima is $frac\left\{a\text{'}_max\right\}\left\{a_max\right\}=left\left[frac\left\{p_1\left(u-1\right)\right\}\left\{p_1\left(u\right)\right\}left\left(1-2\|p_1\left(u\right)\|right\right)right\right]^\left\{frac\left\{1\right\}\left\{2\right\}\right\};$ and finally $frac\left\{a\text{'}_max\right\}\left\{a_max\right\}=sqrt\left\{frac\left\{u-1\right\}\left\{u\right\}\right\}equiv sqrt\left\{frac\left\{u\text{'}\right\}\left\{u\right\}\right\}.$ (eq. 39) The above are solutions to Einstein equations in vacuum. As for the pure Kasner mode, matter does not change the qualitative properties of this solution and can be written into it disregarding its reaction on the field. However, if one does this for the model under discussion, understood as an exact solution of the Einstein equations, the resulting picture of matter evolution would not have a general character and would be specific for the high symmetry imminent to the present model. Mathematically, this specificity is related to the fact that for the homogeneous space geometry discussed here, the Ricci tensor components $scriptstyle\left\{R_alpha^0\right\}$ are identically zeros and therefore the Einstein equations would not allow movement of matter (which gives non-zero stress energy-momentum tensor components $scriptstyle\left\{T_alpha^0\right\}$). This difficulty is avoided if one includes in the model only the major terms of the limiting (at t → 0) metric and writes into it a matter with arbitrary initial distribution of densities and velocities. Then the course of evolution of matter is determined by its general laws of movement (eq. 17) and (eq. 18) that result in (eq. 21). During each Kasner epoch, density increases by the law $varepsilon=t^\left\{-2\left(1-p_3\right)\right\},$ (eq. 40) where p3 is, as above, the greatest of the numbers p1, p2, p3. Matter density increases monotonously during all evolution towards the singularity. To each era (s-th era) correspond a series of values of the parameter u starting from the greatest, $scriptstyle\left\{u_\left\{max\right\}^\left\{\left(s\right)\right\}\right\}$, and through the values $scriptstyle\left\{u_\left\{max\right\}^\left\{\left(s\right)\right\}\right\}$ − 1, $scriptstyle\left\{u_\left\{max\right\}^\left\{\left(s\right)\right\}\right\}$ − 2, ..., reaching to the smallest, $scriptstyle\left\{u_\left\{min\right\}^\left\{\left(s\right)\right\}\right\}$ < 1. Then $u_\left\{min\right\}^\left\{\left(s\right)\right\}=x^\left\{\left(s\right)\right\}, u_\left\{max\right\}^\left\{\left(s\right)\right\}=k^\left\{\left(s\right)\right\}+x^\left\{\left(s\right)\right\},$ (eq. 41) that is, k(s) = [$scriptstyle\left\{u_\left\{max\right\}^\left\{\left(s\right)\right\}\right\}$] where the brackets mean the whole part of the value. The number k(s) is the era length, measured by the number of Kasner epochs that the era contains. For the next era $u_\left\{max\right\}^\left\{\left(s+1\right)\right\}=frac\left\{1\right\}\left\{x^\left\{\left(s\right)\right\}\right\}, k^\left\{\left(s+1\right)\right\}=left\left[frac\left\{1\right\}\left\{x^\left\{\left(s\right)\right\}\right\}right\right].$ (eq. 42) In the limiteless series of numbers u, composed by these rules, there are infinitesimally small (but never zero) values x(s) and correspondingly infinitely large lengths k(s). ## Metric evolution Very large u values correspond to Kasner powers $p_1 approx -frac\left\{1\right\}\left\{u\right\}, p_2 approx frac\left\{1\right\}\left\{u\right\}, p_2 approx 1-frac\left\{1\right\}\left\{u^2\right\},$ (eq. 43) which are close to the values (0, 0, 1). Two values that are close to zero, are also close to each other, and therefore the changes in two out of the three types of "perturbations" (the terms with λ, μ and ν in the right hand sides of (eq. 26)) are also very similar. If in the beginning of such long era these terms are very close in absolute values in the moment of transition between two Kasner epochs (or made artificially such by assigning initial conditions) then they will remain close during the greatest part of the length of the whole era. In this case (BKL call this the case of small oscillations), analysis based on the action of one type of perturbations becomes incorrect; one must take into account the simultaneous effect of two perturbation types. ### Two perturbations Consider a long era, during which 2 out of the 3 functions a, b, c (let them be a and b) undergo small oscillations while the third function (c) decreases monotonously. The latter function quickly becomes small; consider the solution just in the region where one can ignore c in comparison to a and b. The calculations are first done for the Type IX space model by substituting accordingly λ = μ = ν = 1. After ignoring function c, the first 2 equations (eq. 26) give $alpha_\left\{tautau\right\}+beta_\left\{tautau\right\}=0,,$ (eq. 44) $alpha_\left\{tautau\right\}-beta_\left\{tautau\right\}=e^\left\{4beta\right\}-e^\left\{4alpha\right\},,$ (eq. 45) and as a third equation, (eq. 28) can be used, which takes the form $gamma_\left\{tautau\right\}left\left(alpha_\left\{tautau\right\}+beta_\left\{tautau\right\}right\right)=-alpha_taubeta_tau+frac\left\{1\right\}\left\{4\right\}left\left(e^\left\{2alpha\right\}-e^\left\{2beta\right\}right\right)^2.$ (eq. 46) The solution of (eq. 44) is written in the form $alpha+beta=left\left(frac\left\{2a_0^2\right\}\left\{xi_0\right\}right\right)left\left(tau-tau_0right\right)+2ln a_0,$ where α0, ξ0 are positive constants, and τ0 is the upper limit of the era for the variable τ. It is convenient to introduce further a new variable (instead of τ) $xi=xi_0exp left\left[frac\left\{2a_0^2\right\}\left\{xi_0\right\}left\left(tau-tau_0 right\right)right\right].$ (eq. 47) Then $alpha+beta=ln left\left(frac\left\{xi\right\}\left\{xi_0\right\}right\right)+2ln a_0.$ (eq. 48) Equations (eq. 45) and (eq. 46) are transformed by introducing the variable χ = α − β: $chi_\left\{xixi\right\}=frac\left\{chi_xi\right\}\left\{xi\right\}+frac\left\{1\right\}\left\{2\right\}operatorname\left\{sh\right\}2chi=0,$ (eq. 49) $gamma_xi=-frac\left\{1\right\}\left\{4\right\}xi+frac\left\{1\right\}\left\{8\right\}xileft\left(2chi_xi^2+operatorname\left\{ch\right\}2chi-1right\right).$ (eq. 50) Decrease of τ from τ0 to −∞ corresponds to a decrease of ξ from ξ0 to 0. The long era with close a and b (that is, with small χ), considered here, is obtained if ξ0 is a very large quantity. Indeed, at large ξ the solution of (eq. 49) in the first approximation by 1/ξ is $chi=alpha-beta=left\left(frac\left\{2A\right\}\left\{sqrt\left\{xi\right\}\right\}right\right)sin left\left(xi-xi_0right\right),$ (eq. 51) where A is constant; the multiplier $tfrac\left\{1\right\}\left\{sqrt\left\{xi\right\}\right\}$ makes χ a small quantity so it can be substituted in (eq. 49) by sh 2χ ≈ 2χ. From (eq. 50) one obtains $gamma_xi=frac\left\{1\right\}\left\{4\right\}xileft\left(2chi_xi^2+chi^2right\right)=A^2, gamma=A^2left\left(xi-xi_0right\right)+mathrm\left\{const\right\}.$ After determining α and β from (eq. 48) and (eq. 51) and expanding eα and eβ in series according to the above approximation, one obtains finally: $begin\left\{cases\right\}$ `a` `b` end{cases}=a_0sqrt{frac{xi}{xi_0}}left[1pm frac{A}{sqrt{xi}}sin left(xi-xi_0right)right], (eq. 52) $c=c_0 e^\left\{-A^2left\left(xi_0-xiright\right)\right\}.$ (eq. 53) The relation between the variable ξ and time t is obtained by integration of the definition dt = abc dτ which gives $frac\left\{t\right\}\left\{t_0\right\}=e^\left\{-A^2left\left(xi_0-xiright\right)\right\}.$ (eq. 54) The constant c0 (the value of с at ξ = ξ0) should be now c0 $scriptstyle\left\{ll\right\}$ α0· Let us now consider the domain ξ $scriptstyle\left\{ll\right\}$ 1. Here the major terms in the solution of (eq. 49) are: $chi=alpha-beta=kln xi+mathrm\left\{const\right\},,$ where k is a constant in the range − 1 < k < 1; this condition ensures that the last term in (eq. 49) is small (sh 2χ contains ξ2k and ξ−2k). Then, after determining α, β, and t, one obtains $a sim xi^\left\{frac\left\{1+k\right\}\left\{2\right\}\right\}, b sim xi^\left\{frac\left\{1-k\right\}\left\{2\right\}\right\}, c sim xi^\left\{-frac\left\{1-k^2\right\}\left\{4\right\}\right\}, t sim xi^\left\{frac\left\{3+k^2\right\}\left\{4\right\}\right\}.$ (eq. 55) This is again a Kasner mode with the negative t power coming into the function c(t). These results picture an evolution that is qualitatively similar to that, described above. During a long period of time that corresponds to a large decreasing ξ value, the two functions a and b oscillate, remaining close in magnitude $tfrac\left\{a-b\right\}\left\{a\right\} sim tfrac\left\{1\right\}\left\{sqrt\left\{xi\right\}\right\}$; in the same time, both functions a and b slowly ($scriptstyle\left\{sim sqrt\left\{xi\right\}\right\}$) decrease. The period of oscillations is constant by the variable ξ : Δξ = 2π (or, which is the same, with a constant period by logarithmic time: Δ ln t = 2πΑ2). The third function, c, decreases monotonously by a law close to c = c0t/t0. This evolution continues until ξ ~ 1 and formulas (eq. 52) and (eq. 53) are no longer applicable. Its time duration corresponds to change of t from t0 to the value t1, related to ξ0 according to $A^2xi_0=ln frac\left\{t_0\right\}\left\{t_1\right\}.$ (eq. 56) The relationship between ξ and t during this time can be presented in the form $frac\left\{xi\right\}\left\{xi_0\right\}=frac\left\{ln tfrac\left\{t\right\}\left\{t_1\right\}\right\}\left\{ln tfrac\left\{t_0\right\}\left\{t_1\right\}\right\}.$ (eq. 57) After that, as seen from (eq. 55), the decreasing function c starts to increase while functions a and b start to decrease. This Kasner epoch continues until terms c2/a2b2 in (eq. 22) become ~ t2 and a next series of oscillations begins. The law for density change during the long era under discussion is obtained by substitution of (eq. 52) in (eq. 20): $varepsilon sim left\left(frac\left\{xi_0\right\}\left\{xi\right\}right\right)^2.$ (eq. 58) When ξ changes from ξ0 to ξ ~ 1, the density increases $scriptstyle\left\{xi^2_0\right\}$ times. It must be stressed that although the function c(t) changes by a law, close to c ~ t, the metric (eq. 52) does not correspond to a Kasner metric with powers (0, 0, 1). The latter corresponds to an exact solution (found by Taub) which is allowed by eqs. 26-27 and in which $a^2=b^2=frac\left\{p\right\}\left\{2\right\}frac\left\{mathrm\left\{ch\right\}\left(2ptau+delta_1\right)\right\}\left\{mathrm\left\{ch\right\}^2\left(ptau+delta_2\right)\right\}, ; c^2=frac\left\{2p\right\}\left\{mathrm\left\{ch\right\}\left(2ptau+delta_1\right)\right\},$ (eq. 59) where p, δ1, δ2 are constant. In the asymptotic region τ → −∞, one can obtain from here a = b = const, c = const.t after the substitution ерτ = t. In this metric, the singularity at t = 0 is non-physical. Let us now describe the analogous study of the Type VIII model, substituting in eqs. 26-28 λ = −1, μ = ν = 1. If during the long era, the monotonically decreasing function is a, nothing changes in the foregoing analysis: ignoring a2 on the right side of equations (26) and (28), goes back to the same equations (49) and (50) (with altered notation). Some changes occur, however, if the monotonically decreasing function is b or c; let it be c. As before, one has equation (49) with the same symbols, and, therefore, the former expressions (52) for the functions a(ξ) and b(ξ), but equation (50) is replaced by $gamma_\left\{xi\right\} = -frac\left\{1\right\}\left\{4\right\}xi+frac\left\{1\right\}\left\{8\right\}xileft \left(2chi_\left\{xi\right\}^2 + mathrm\left\{ch\right\}2chi + 1right \right).$ (eq. 60) The major term at large ξ now becomes $gamma_\left\{xi\right\} approx frac\left\{1\right\}\left\{8\right\}xi cdot 2, quad gamma approx frac\left\{1\right\}\left\{8\right\} left \left(xi^2-xi_0^2 right \right),$ so that $frac\left\{c\right\}\left\{c_0\right\}=frac\left\{t\right\}\left\{t_0\right\}=e^\left\{-frac\left\{1\right\}\left\{8\right\}left \left(xi_0^2-xi^2 right \right)\right\}.$ (eq. 61) ## References • ; English translation inBelinskii, V.A. (1970). "Oscillatory Approach to a Singular Point in the Relativistic Cosmology". Advances in Physics 19 525–573. . • ; English translation inLifshitz, E.M. (1963). "Problems in the Relativistic Cosmology". Advances in Physics 12 185. . • Vol. 2 of the Course of Theoretical Physics. id="CITEREFBerger2002">Berger, Beverly K. (2002), " Numerical Approaches to Spacetime Singularities", Living Rev. Relativity 5, <http://www.livingreviews.org/lrr-2002-1> (retrieved on 2007-08-04) }}. }}. • . \| accessdate=2003-11-11}}. \| accessdate=2007-02-27}}. Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 96, "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": 0.3, "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": 0.9103150367736816, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/64982-finding-best-fitted-hyperplane-print.html	# Finding the best fitted hyperplane Printable View • December 14th 2008, 03:21 PM TriKri Finding the best fitted hyperplane Given a number of points in $\Bbb{R}^n$, how can you find (one of) the hyperplane(s) from which, according to the least square method, the points is closest located to? That is, minimizing the sum of the square distances from each point to the hyperplane. All times are GMT -8. The time now is 04:13 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": 1, "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": 0.9030493497848511, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/98523/linear-algebra-problem-intersection-of-a-subspace-with-a-cone	# Linear Algebra problem: intersection of a subspace with a cone. In $\mathbb{R}^n$, consider the closed cone $$C^+ = \{ (x_1, \ldots, x_n) : x_i \geq 0,~~i= 1, \ldots, n\}.$$ Let $S \subseteq \mathbb{R}^n$ be a subspace (of any dimension) such that $S \cap C^+ = \{0\}$. Prove that $S^{\perp}$ has non-empty intersection with the interior of $C^+$. The orthogonal complement is taken with respect to the canonical inner product. It's not hard to see why this must be true, but a real proof has eluded me for some time now. - What are the arguments that make you say "it's not hard to see why it must be true"? – Davide Giraudo Jan 12 '12 at 20:53 @Davide My statement, that you quote, is not precise and is not related to the problem. It's just intuition based on a number of particular cases. – student Jan 12 '12 at 22:16	{"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": 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": 0.9281683564186096, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/153607/what-is-the-chance-to-get-a-parking-ticket-in-half-an-hour-if-the-chance-to-get/153629	# What is the chance to get a parking ticket in half an hour if the chance to get a ticket is 80% in 1 hour? This sounds more like a brain teaser, but I had some kink to think it through :( Suppose you're parking at a non-parking zone, the probability to get a parking ticket is 80% in 1 hour, what is the probability to get a ticket in half an hour? Please show how you deduce the answer. Thanks! - 9 As a Frenchman, 80% chance in 1 hour seems really high to me, would be rather 20% chance in 1 day. – Benoit Jun 4 '12 at 13:18 4 80% is about right in Oslo. – stigok Jun 4 '12 at 19:56 3 Where I'm from, between the hours of 3PM - 6PM, M-F, the probability of getting a ticket and towed is 100% within 10 minutes. – user32990 Jun 5 '12 at 8:41 ## 7 Answers It really depends on what model is assumed. However, if the idea is that no matter how long you leave your car there, you have a $20$% chance of getting through any given hour unscathed, you can treat it as an exponential decay problem. Let $p(t)$ be the probability that you do not get a ticket in the first $t$ hours. Then $p(1)=0.2$, $p(2)=0.2^2$ (a $20$% chance of making it through the first hour times a $20$% chance of making it through the second), and in general $p(t)=0.2^t$. The probability of not getting a ticket in the first half hour is then $p(1/2)=0.2^{1/2}=\sqrt{0.2}\approx 0.4472$, and the probability that you do get a ticket in the first half hour is about $1-0.4472=0.5528$. - 1 @Dave: You’re assuming something like full-time checkers with regular routes. Try assuming instead random checks. – Brian M. Scott Jun 4 '12 at 5:34 53 +1 for "It really depends on what model is assumed." Without making assumptions, all we can really say is that the probability of getting a ticket in 20 minutes is less than or equal to 80%. An alternative model is there is an approximately 80% chance that the ticket-writer is working today. If the ticket-writer is working there is an approximately 100% chance that you will get a ticket within 5 minutes. – emory Jun 4 '12 at 6:02 1 @Zarrax, then the challenge is calculating the likelihood of choosing the correct model for determining the probability of getting a ticket. Assume there are N models for determining probability, what is the likelihood of choosing the correct model? You could probably then use it to calculate a margin of error for your model. – smdrager Jun 4 '12 at 18:43 3 @smdrager, I don't understand. There are uncountable infinite number of models to choose from. Anyway you are not provided with any data to calculate likelihood. – emory Jun 4 '12 at 18:49 4 Parking attendants are not memoryless. – Mark Adler Jun 5 '12 at 4:56 show 11 more comments If my probability of getting a parking ticket in a half hour is T, then $$T+(1-T)T$$ are the odds I got one in 1 hour. The first term says that I got it in the 1st half hour (so it doesn't matter what happens after that), and the second term says that I got it in the 2nd half hour (so not in the first). Solving and taking the sensible solution: $$2T-T^2=.8\implies T^2-2T+.8$$ $$T=1-\sqrt{.2}$$ Which are roughly $55.3\%$ odds in a half hour. - 5 @Rock you could instead choose this as your answer. – user23320 Jun 4 '12 at 12:43 @Joshua Now I have it. The accepted answer gives a general solution which I guess is what the question potentially asked for. – Rock Jun 4 '12 at 18:26 3 You've made a Markov assumption, which may or may not be valid. – Ben Voigt Jun 4 '12 at 23:58 1 @BenVoigt fair enough, but it's either this, assume extra conditions that weren't specified, or be unable to do anything at all. You always have to assume some model anyway, and in the absence of other information a Markov assumption seems to be the natural choice. Of course not a statistician though, so I wouldn't be surprised if I'm begging to be corrected here. – Robert Mastragostino Jun 5 '12 at 20:13 – Lucius Sep 11 '12 at 19:01 show 1 more comment Another way of looking at the exponential model: If $P$ is the probability you get lucky and don't get ticketed in the first 30 minutes, then $P^2$ is the chance you luck out twice in a row and don't get ticketed in the first hour. You know $P^2 = 0.2$. So $P$ is the square root of that or $0.447$. So the chance you are unlucky in the first 30 minutes and get a ticket is $1 - 0.447 = 0.553$ or $55.3$ percent. - 1 A very clear explanation! – Simon McKenzie Jun 4 '12 at 7:09 3 For me the smoothest solution ! – berkay Jun 4 '12 at 8:29 Thanks.. but to be clear I'm just explaining the exponential model in layman's terms, and I am not suggesting I agree this model is necessarily the "correct" one for the problem. – Zarrax Jun 4 '12 at 21:18 A simple but not unreasonable model is that the ticket person cycles through her/his round in roughly constant time. Suppose that you park illegally for exactly an hour at a randomly chosen spot in the round, and at a start time uniformly distributed with respect to the cycle. Since your probability of getting a ticket in an hour is $80\%$, the cycle length is $1.25$ hours. Thus your probability of getting a ticket if you park illegally for half an hour is $40\%$. - 5 For the people who wonder how your model can give something different: The other answers assume that getting no ticket in the first half of an hour does not increase the chances of getting one in the second half of the hour. In a place where 80% of cars get a ticket within the hour, this is not a very realistic assumption. So, we can rejoice and expect that the probability is closer to 40% than to 55%. – Phira Jun 4 '12 at 12:35 1 Another explanation to this: Assume there are 100 illegal cars, so the parking warden can handle 80 tickets per hour (we assume linear distribution of tickets per time) and thus 40 tickets half an hour. If you are one of that cars during that half of an hour ... your risk getting caught is 40/100. – Nappy Jun 4 '12 at 14:52 1 IMO from a modeling perspective this is by far the most reasonable answer given. The exponential model seems unreasonable since clearly your hazard (aka force of mortality or failure rate) ought to be increasing over time. – guy Jun 4 '12 at 16:18 1 Agreed, @guy. With an exponential decay model of tickets, no matter how long you leave your car on the street, there is some nonzero chance that it will not get a ticket. That's not the behavior of any conceivable real-world method for writing parking tickets (which generally involve somebody walking down the street and ticketing every illegally parked car). – Larry Gritz Jun 4 '12 at 17:22 There is another way to look at this problem. Most of the answers of ~55.3% (including the currently accepted answer) assume the "meter reader" is checking cars COMPLETELY at random. In this model, there is an extremely small but non-zero chance you could leave a car in the same spot for a month and not get a ticket (although it's up there in "lottery" winner and struck-by-lightning chances). In fact, the chance of getting a ticket using this model is always less than 100% for a finite amount of time. However, meter readers typically work in a linear pattern starting at one point and then looking at ALL the cars in a line down one side of a block. If we consider the realistic pattern that cars are checked in a linear fashion instead of cars being checked randomly, the answer becomes much simpler. Therefore, when they completely cover an area (in a finite amount of time), your car has a 100% chance of getting a ticket. If you are 100% likely to get a ticket within a finite amount of time, then the p(t)=1-0.2^t answer for getting a ticket is invalid. So lets consider linear coverage. If you have an 80% chance of getting a ticket in one hour, the meter readers in your city walk 4 out of 5 blocks per hour (in 60 minutes). Assuming they follow a pattern which has complete coverage and fairly evenly sweeps the whole parking area, they could have 100% coverage of all parking spots in 75 minutes. That means no matter where you park, within 75 minutes you will get a ticket. Under the linear coverage model, p(t)=0.8t. So in half an hour, you have a 40% chance of getting a ticket. This is a case of not trying to get too mathematically clever because the simplest answer could actually be the correct one. The real answer depends whether your meter people ticket cars randomly or sweep areas linearly with even coverage (and gets more complicated with real life where they sweep some areas much more often than others but still get 100% coverage over a finite amount of time). - Not enough data to come up with an answer. You would have to know the pattern that the parking warden takes to get the correct answer (first half hour, second half hour), also this depends on which part of the time unit the parking warden would take up, for example 1pm-2pm, 1.05pm-2.05pm etc 80% can reasonable be assumed for an hour, but for half an hour less so. I would therefore conclude that there is an 80% chance that you will get a ticket as there is no way to distinguish the half hour from the full hour because of lack of data for the movements of the parking warden. - Why be pessimistic? Maybe it's your lucky day and there's a zero percent chance. :) – Zarrax Jun 4 '12 at 21:34 If your a pessimist, there are no unexpected nasty surprises, you will never feel the pain of disappointment, and reality will hold no fear. Abide by the law, there will be a 100% chance of not getting a ticket. :) – WeNeedAnswers Jun 6 '12 at 11:32 You know that there is an 80% chance of getting a ticket in a hour. Since the full hour encompasses the half hour in question, then it is definably an 80% chance through set theory. Without further data about the parking wardens movements, it is impossible to suggest anything better. – WeNeedAnswers Jun 7 '12 at 0:36 This sounds like a Poisson Process to me... so, assuming that: Since we don't have additional data, I will asume that the expected probability of being spotted once per hour is $p=0.8$. This is a Bernoulli experiment (either you get spotted or not). I will model the probability of being spotted by a machine, person, or some other authority that can issue a ticket ;-) If $\lambda$ is the expected times that you get spotted every hour, then $\theta=\frac{1}{\lambda}\$ is the expected time between events. The number of events given in $t$ can be modeled by a random variable $N \sim Poisson(\lambda·t)$ where $t$ is the amount of time (in hours) elapsed, and the time between events can be modeled by a random variable $X\sim Exponential(\theta)$. So, the probability of getting $n$ tickets in $t$ hours, with an expected probability of $\lambda$ that you get a ticket in one hour is: $$Pr\{N=n\ \|\ \lambda, t\}=\frac{e^{-\lambda·t}·(\lambda·t)^{n}}{n!}$$ Since we need to calculate the probability of being spotted at least once, we can calculate $1-Pr\{N=0\}$. Then, for $n=0$, $\lambda=0.8$ and $t=0.5$, then: $$1-Pr\{N=0\ \|\ \lambda=0.8, t=0.5\}=1-\frac{e^{-0.4}·(0.4)^{0}}{0!}=1-e^{-0.4}\approx0.4918$$ You should notice that, even if you "get spotted" more than once per day, this model still holds, since this process is memoryless. Further reference: Ross, Sheldon, "Introduction to probability models", Ninth edition, Chapter 5. (I really hope this is not your homework) - 3 A Markov chain is a much better assumption, since I've never heard of multiple tickets being issued to a car (that remains parked in one place) in the same day. Therefore "got 1 ticket" is a terminal state, there is no chance of moving to two tickets. The problem stated that there is an 80% chance of getting a ticket, not that the mean number of tickets is 0.8. Your answer confuses these two quantities. – Ben Voigt Jun 5 '12 at 0:03 With no further information, the expected value of the tickets per hour is 0.8 (is a Bernoulli experiment). I think the Poisson Process hypothesis holds because all you know is that you may "get caught" with a given probability in a unit of time. Of course you can't be caught more than once a day, but that doesn't implies that you can't be "spotted" n times a day. – Barranka Jun 5 '12 at 18:35 That said, I'm correcting my answer, because I calculated the probability of "being spotted" once per interval, which is not what we are looking for. The probability we're looking for is "what is the probability of 'being spotted' at least one time", which is 1 - Pr{"Not 'being spotted'"}. – Barranka Jun 5 '12 at 18:37 1 @BenVoigt: Unfortunately, in some cities, you can be ticketed multiple times, in the same spot, if your car is still illegally parked. – Dancrumb Jun 5 '12 at 20:48 2 You did it wrong. $\lambda$ is not $0.8$. To get $\lambda$, you need to reverse the calculation that you did get right. If the probability of getting a ticket in an hour is $0.8$, then $1-e^{-\lambda t}=0.8$ where $t$ is one hour, so the rate $\lambda$ is $1.61$ tickets per hour. Then we use $1-e^{-\lambda t}$ with $t$ equal to half an hour. We get $1-e^{\lambda{1\over 2}hr}$ is $0.553$. The same as the exponential examples above, since it is the same memoryless assumption. – Mark Adler Jun 6 '12 at 5:09 show 3 more comments ## protected by Asaf KaragilaJun 5 '12 at 16:34 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": 47, "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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9512406587600708, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/212331-line-equation-vectors.html	# Thread: 1. ## line equation in vectors "find the equation of the line passing through B, with positiion vector b, which is perpendicular to the line r=a + $\lambda$c, c=/=0, given that B is not a point on the line". hi, i dont even get this at all to be honest, how is r=a + $\lambda$c a line? in terms of normal line equations, which is the gradient, and the intercept? because r and c are not axis they are vectors, i dont know what this is asking. 2. ## Re: line equation in vectors A line is essentially a vector (which determines the direction) that is of infinite length, which is then positioned somewhere. So basically, what has happened to get the line $\displaystyle \begin{align} \mathbf{r} \end{align}$ is that $\displaystyle \begin{align} \mathbf{c} \end{align}$ determines the direction of the line, it is multiplied by some parameter $\displaystyle \begin{align} \lambda \end{align}$ to make it of infinite length, and then the point $\displaystyle \begin{align} a \end{align}$ has been added (or rather, each component of $\displaystyle \begin{align} \mathbf{c} \end{align}$ is adjusted numerically by each component of $\displaystyle \begin{align} a \end{align}$) to position it in the correct spot. Do you think you can get a vector-form equation for the line passing through B in the direction of $\displaystyle \begin{align} \mathbf{b} \end{align}$? 3. ## Re: line equation in vectors hi, when you say vector form equation, does this mean i am looking for another line like the r=a+ lc? so it will again be in these terms. but it must be perpendicular. my immediate thought is find to reciprocate and negate the lambda? so it would be r = b -c/l? a is written in the same style as b and r so it is a vector too but is it really just a point? how can a vector (position vector?) be a point, i thought all vectors are lines. 4. ## Re: line equation in vectors Originally Posted by learning hi, when you say vector form equation, does this mean i am looking for another line like the r=a+ lc? so it will again be in these terms. but it must be perpendicular. my immediate thought is find to reciprocate and negate the lambda? so it would be r = b -c/l? a is written in the same style as b and r so it is a vector too but is it really just a point? how can a vector (position vector?) be a point, i thought all vectors are lines. You need to start thinking of lines as VECTORS. How do you know if two VECTORS are perpendicular? 5. ## Re: line equation in vectors hi, i know this measn the dot product is zero for two perpendicular vectors. but what is the vector of r=a + lc? how can i dot this with something? 6. ## Re: line equation in vectors Originally Posted by learning hi, i know this measn the dot product is zero for two perpendicular vectors. but what is the vector of r=a + lc? how can i dot this with something? The vector IS $\displaystyle \begin{align} \mathbf{r} \end{align}$. This is extremely difficult to illustrate without knowing anything about the vectors you've actually been given. Is this part of a larger problem? 7. ## Re: line equation in vectors hi, this is the entirety of this problem there is no 'specific' vector given. i have writen the whole question. one thing i am not sure of is even what should the form be of this perpendicular vector? would it also be r (what is the meaning of the name r?) or should i call it r2 to show its a different vector? so maybe r2 = b + something d? or would it also have c in it? would it also have the lambda or a different parameter? should the parameter be perpendicular? -1/lamba? if i have an equation to solve i can do it no problem at all but i dont know what this is meaning 8. ## Re: line equation in vectors Originally Posted by learning "find the equation of the line passing through B, with positiion vector b, which is perpendicular to the line $\ell_1:~a +\lambda c, c\ne$, given that B is not a point on the line". First write the equation of the plane, $\Pi$ that contains the point $B$ with normal $\lambda$ . Now the given line $\ell_1$ is perpendicular to $\Pi$. Let $\{C\}=\ell_1\cap\Pi$. Now the line $\overleftrightarrow {BC}$ is the answer to your question. 9. ## Re: line equation in vectors you have introducted 4 or 5 symbols and notations which are not in the question. it doesnt help if you go way too technical like that. what you say may be true but its not helpful.. 10. ## Re: line equation in vectors Originally Posted by learning you have introducted 4 or 5 symbols and notations which are not in the question. it doesnt help if you go way too technical like that. what you say may be true but its not helpful.. I am sorry that you find a detailed way to answer the question not helpful. If those symbols confuse you, then perhaps you don't understand the question in the first place. 11. ## Re: line equation in vectors why would i be posting it if i did? :s you may have the technical expertise but you dont have much common sense. being intelligent isnt all thats required of a good teacher.. now i prefer someone to help who actually wants to help rathr than just 'show off' their knowledges.. prove it is giving good hints to steer in the right direction and you just spout out jargon.. if i knew what to do with that information i probably wouldnt have to ask for help on it now would i 12. ## Re: line equation in vectors someone explain the answer please in terms of the vectors for instance, i dont even know what the format required is? something like the r = a + lc, so r = b + md or something? just with different coefficeint and intercept? 13. ## Re: line equation in vectors ya steal need help in this please 15. ## Re: line equation in vectors No threatening learning. Keep it up and no one will want to come anywhere near you. And of course I won't allow it....Rethink how you want to communicate. -Dan Edit: After talking to learning the above comment was not meant as a threat. My apologies. -Dan	{"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": 0.3, "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": 0.9529851675033569, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/255067/using-fermats-little-theorem-prove-if-p-is-prime-prove-1p-2p-3p/255074	# Using Fermat's Little Theorem Prove if $p$ is prime, prove $1^p + 2^p + 3^p +…+(p-1)^p \equiv 0 \bmod{p}$ Using Fermat's Theorem Prove if $p$ is prime, prove $1^p + 2^p + 3^p +...+(p-1)^p \equiv 0 \bmod{p}$ The two definitions of Fermat's Little Theorem is $a^p \equiv a \bmod{p}$ and $a^{p-1} \equiv 1 \bmod{p}$ but I don't know how to use this solve the problem Please help - ## 3 Answers Hint: Use Fermat's theorem for each $k$ ($1\le k\le p-1$) and add all them up. Then use Gauss's formula for the sum $1+2+\cdots +(p-1)$. - $1^p + ... + (p-1)^p \equiv 1 + 2 + ... (p-1) \equiv (1 + (p-1)) + (2 + (p-2)) + ... +(\frac{p-1}{2} + \frac{p+1}{2}) \equiv p + ... + p \equiv 0 (\mod p)$ Edit: Assuming p is an odd prime, of course - I understand the 1+(p-1)+2+(p-2)+.... part but how did you come up with (p-1/2 + p+1/2)? – Cindy Dec 10 '12 at 4:54 Because $p$ is odd, that'll be the last two pairs in the sum. Write out some of these sums for various primes and you'll see where it comes from. – anonymous Dec 10 '12 at 16:39 Since $1^p \equiv 1$, and $2^p \equiv 2$, $\ldots$ $(p-1)^p \equiv p-1$, we have that $$1^p + 2^p + \ldots (p-1)^p \equiv 1 + 2 + \ldots + p-1$$ However, we know this sum: $\sum_{i = 1}^{p-1} i = \frac{p (p-1)}{2}$, which for odd primes would be a multiple of $p$, and we are done by reducing modulo $p$, then we only must check the case for $2$, which doesn't hold, and it's awkward... -	{"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": 19, "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": 0.9412335157394409, "perplexity_flag": "head"}
http://cstheory.stackexchange.com/questions/4928/alphabet-of-single-tape-turing-machine	Alphabet of single-tape Turing machine Can every function $f : \{0,1\}^* \to \{0,1\}$ that is computable in time $t$ on a single-tape Turing machine using an alphabet of size $k = O(1)$ be computed in time $O(t)$ on a single-tape Turing machine using an alphabet of size $3$ (say, $0,1,$ and blank)? (From comments below by the OP) Note the input is written using $0,1$, but the Turing machine using alphabet of size $k$ can overwrite the input symbols with symbols from the larger alphabet. I don't see how to encode symbols in the larger alphabet in the smaller alphabet without having to shift the input around which would cost time $n^2$. - 7 Note the input is written using $0,1$, but the Turing machine using alphabet of size $k$ can overwrite the input symbols with symbols from the larger alphabet. I don't see how to encode symbols in the larger alphabet in the smaller alphabet without having to shift the input around which would cost time $n^2$. – Emanuele Viola Feb 14 '11 at 23:55 4 @Emanuele: You should edit the question and emphasise this aspect; otherwise it sounds exactly like a standard textbook exercise... – Jukka Suomela Feb 15 '11 at 0:11 3 @Tsuyoshi, I think you misunderstood the question. – Suresh Venkat♦ Feb 15 '11 at 0:31 4 @Jukka: On a one-tape Turing machine, everything that can be computed in time $o(n \log n)$ are in fact regular languages. – Kristoffer Arnsfelt Hansen Feb 15 '11 at 6:37 6 @Abel: The result you quote from Arora and Barak gets around the main issue here because in their model (which is fairly standard for multi-tape TMs), they have a separate, read-only input tape. – Joshua Grochow Feb 16 '11 at 5:43 show 11 more comments 2 Answers A partial answer if TM runs in $o( \|x\| \log \|x\|)$ If TM4 is a 4-symbols TM (with alphabet $\Sigma_4 = \{\epsilon,0,1,2 \}$ ) that computes $f:\{0,1\}^* \to \{0,1\}$, i.e. decides language $L = \{ x \| f(x) = 1 \}$ in $(o( \|x\| \log \|x\|))$ One tape deterministic linear-time complexity is $1DLIN = 1DTime(O(n))$ • Hennie proved (1) that $REG = 1DLIN$ • Kobayashi proved (2) that $REG = 1DTime(o(n \log n))$ So $L$ is regular, and is obviously still regular over alphabet $\Sigma_3 = \{\epsilon,0,1\}$ So there is a DFA that decide L and uses only symbols in $\Sigma_3$. A one-tape, 3-symbols TM3 can be built directly from the DFA and it decides L using the same unpadded input of the original TM4. ... you cannot build it directly from TM4, but TM3 exists. If TM4 runs in $\Omega(n^2)$ then you can shift the input and make a direct conversion from TM4 to TM3. As noticed in the comments the difficult case is when TM4 runs in $\Omega(n\log n) \cap o(n^2)$. (1) Hennie, One-tape, off-line Turing machine computations (1965) (2) Kobayashi, On the structure of one-tape nondeterministic Turing machine time hierarchy (1985) - 1 The point about $o(n\log n)$ is already noted by Kristoffer Arnsfelt Hansen in the comments below the question. The really interesting case is $\Omega(n\log n) \cap o(n^2)$. – Kaveh♦ Feb 19 '11 at 5:56 You're right I didn't noticed Kristoffer comment. I badly expressed the interesting case (I don't know how to prove it), so I updated the answer. – Marzio De Biasi Feb 19 '11 at 13:39 1 @Kaveh: What about $o(n \log n)$-time machines for promise problems? Do we already know how to convert, e.g., any machine that solves a promise problem in $O(n)$ time? I don't see how to do it, and the connection to regular languages no longer holds (unless I'm badly mistaken). – Jukka Suomela Feb 19 '11 at 16:35 1 @Kaveh: Can't you simply take a problem $L$ that is not a regular language but can be solved with a Turing machine in, e.g., $O(n^2)$ rounds, and define a promise problem as follows: yes-instances consist of a string $x \in L$ followed by $\|x\|^2$ bits of padding; no-instances consist of a string $x \notin L$ followed by $\|x\|^2$ bits of padding. The promise problem is solvable in $O(n)$ time, and it is not solvable using a finite state machine. – Jukka Suomela Feb 20 '11 at 23:04 1 @Kaveh: I guess the intuitive argument fails because of the following reason: Yes, there is a non-promise problem that is solved by the same machine. However, the running time of the machine may be as high as $\Theta(n^2)$ for certain inputs. (Intuitively, the machine cannot verify that there is enough padding, and hence "to play safe" it must assume that there is enough padding after the prefix $x$. Then it wastes $\Theta(\|x\|^2)$ time to determine if $x \in L$, and this is too much if, e.g., we had only $\Theta(\|x\|)$ bits of padding.) – Jukka Suomela Feb 20 '11 at 23:19 show 3 more comments For all alphabet sizes greater than $1$, runtimes only change by a constant factor since $\log_k(x) \in \Theta(\log_l(x))$ for all $k, l > 1$. Elaboration: In $t$ timesteps, the assumed Turing machine can process at most $t$ positions/bits. Bits are taken from a $k$-nary alphabet, wlog $\{0,1,\dots,k-1\}$. Create a new Turing machine by replacing every transition by $\lceil \log_2(k) \rceil$ transitions; every old bit is encoded by $\lceil \log_2(k) \rceil$ bits in $\{0,1\}$ (blanks are reserved to mark unused cells). Note this is essentially binary coded digits. Obviously, the resulting Turing machine executes at most $\lceil \log_2(k) \rceil \cdot t \in \mathcal{O}(t)$ steps. Addition: Above argumentation breaks because operations that overwrite an input symbol with a bit not in $\{0,1\}$ can not be translated directly; the input has to be shifted. This can be amended by translating the original input before starting computation (essentially padding); this can be done in time $\mathcal{O}(n^2)$, resulting in a total runtime of $\mathcal{O}(n^2) + \lceil \log_2(k) \rceil \cdot t$. Consequently, using only two symbols for encoding intermediate results is of no asymptotic impact if $t(n) \in \Omega(n^2)$, but preprocessing dominates faster algorithms. Since most interesting functions are in $\Omega(n^2)$ (e.g. adding two numbers), one might consider the problem negligable. - 3 Until you convince me why this is supposed to be the case, I shall keep that downvote. – Andrej Bauer Feb 15 '11 at 13:33 1 I would like to hear some evidence for your claim. All of it, it's just one claim. – Andrej Bauer Feb 15 '11 at 18:44 2 Oh, I see what you are referring to. Ok, sorry. However, the question is not about that. It's a slight variation. – Andrej Bauer Feb 16 '11 at 6:15 6 I think that the case with t=Ω(n^2) is the easy case because you can afford the time to shift the input string. The essential case is when t=o(n^2). I do not know how important it is to consider single-tape TM with o(n^2) time, but the question is about that. – Tsuyoshi Ito Feb 16 '11 at 21:38 3 The original question already implies that the case $\Omega(n^2)$ is easy; hence I do not really see how this answer adds anything new... – Jukka Suomela Feb 17 '11 at 8:26 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": 59, "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": 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": 0.902540385723114, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/49084/transforming-a-sequence-using-polynomials/49195	# Transforming a sequence using polynomials Let $$f(x) = \sum_{n \geq 0} a_n x^n = \frac{x-2x^3}{4x^4 - 5x^2 + 1}$$ Now I need to identify a "concrete formula" for $a_n$. This should be done by using the following proposition: For a sequence $a = (a_0,a_1,\cdots)$ with $a_i \in \mathbb{C}$ and a $d$-tuple $(\alpha_1,\cdots,\alpha_d) \in \mathbb{C}^d$ with $\alpha_d \neq 0$ applies: • $f_a(x) = \sum_{n\geq 0} a_n x = \frac{P(x)}{Q(x)}$ with $Q(x) = 1 + \alpha_1 t + \cdots + \alpha_d t^d$ and a polynomial $P(x)$ having degree $< d$. • $a_{n+d} + \alpha_1 a_{n+d-1} + \cdots + \alpha_d a_n = 0$ for $n \geq 0$ • For $n \geq 0$ applies: $$a_n = \sum_{i=?}^k P_i(n) \sigma_i^n$$ with $1 + \alpha_1 x + \cdots + \alpha_d x^d = \prod_{i=1}^k (1- \sigma_i x)^{d_i}$ so that $\sigma_i \neq \sigma_j, 1 \leq i < j \leq k$ and $P_i(t)$ is a polynomial having a degree < $d_i$. Could you please help me to apply this on the initial definition of $f(x)$? How do I start and identify $P(x)$ and $Q(x)$? Is this a method that has its own name (so I could look it up somewhere)? I made an error during the precalculation for the term above, it's correct now. Is this already $P(x)$ and $Q(x)$ as $\deg(P(x)) < \deg (Q(x))$? What is the next step? Thanks in advance! - I guess the term of the sum should be $a_n x^n$, right? – leonbloy Jul 3 '11 at 12:42 @leonbloy yes, sorry – muffel Jul 3 '11 at 21:01 ## 2 Answers Yes, $P(x)$ is $x-2x^3$, and $Q(x)$ is $4x^4-5x^2+1$ (although I note that at one place you have written $Q(x)$ as a polynomial in $t$ when it should be in $x$). The next step is to factor $Q(x)$ as a product of polynomials each of degree 1; that will give you the $\sigma_i$ in the last bullet point. - That would be $4x^4 - 5x^2 + 1 = (-1+x)(1+x)(-1+2x)(1+2x)$. But what's next? What really unsettles me is the index $i$ of P in $a_n = \sum P_i(n) \sigma_i^n$. $a_n$ needs to be like $P_1(n) \cdot (-1+x) + P_2(n) \cdot (1+x) + P_3(n) \cdot (-1+2x) + P_4(n) \cdot (1+2x)$, but I do already have defined P(x), haven't I? – muffel Jul 3 '11 at 21:19 @muffel, the proposition you quoted doesn't tell you how to find $P_i$, but it does tell you that its degree is less than $d_i$. Now from the factorization you have done for $Q(x)$, each $d_i$ is 1, so each $P_i$ is constant. How do you find these 4 constants? One way is to find $a_0,a_1,a_2$, and $a_3$, then you have 4 equations in those 4 unknown constants. – Gerry Myerson Jul 4 '11 at 0:06 thanks again, I now got it! – muffel Jul 4 '11 at 8:49 I see this: $${(1 - 2x)^2 + 2x\over-(2x -1)^3} = {(1 - 2x)^2 + 2x\over(1 - 2x)^3} = {1\over 1 - 2x} + {2x\over (1 - 2x)^3}.$$ Finding a Taylor expansion of this centered at 0 can be done with using the standard tools. - @nemathsadist thank you, but I haven't heart of Taylor expansions yet and need to work it out. Isn't there a way of just defining $P(x)$ and $Q(x)$? – muffel Jul 3 '11 at 8:49 You can just use the Geometric Series Theorem. – ncmathsadist Jul 3 '11 at 13:53	{"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": 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": 0.9504467844963074, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/34231/what-is-the-particle-residence-time-for-given-flow-rates-of-gas-mixture-componen/34300	# What is the particle residence time for given flow rates of gas mixture components? For two different chemical substances there are two open valves with flow rates $$Q_1=a\frac{m^3}{h}\ \text{ and }\ \ Q_2=b\frac{m^3}{h},$$ leading into seperate cables. Next, the cables join, the substances mix perfectly and they flow along together. Then during the route, there is an observational volume $V$ (e.g. a cylinder or length $d$ and cross section $A$). How long is the residence time $\Delta t$ of a particle of the mixture inside the volume? - do you want a resulting flow rate or do you mean by particle a molecule of either of the substances? – Yrogirg Aug 15 '12 at 16:56 @Yrogirg: I was expecting to get a single particle velocity. In that case, if I can compute a single flow rate, I guess the residence time for a particle of the the mixture can be computed. If there is not a single velocity, then a velocity for each of the substances is good too, of course. – Nick Kidman Aug 16 '12 at 8:10 What kind of answer did you expect? My answer is really elementary, what do you want above that (if any)? – Yrogirg Aug 16 '12 at 9:01 @Yrogirg: No, the answer is good, I just don't want to miss something. – Nick Kidman Aug 16 '12 at 9:03 ## 1 Answer First, I assume no chemical reactions take place, as I think you implied. Second, absence of a single velocity for a "perfectly mixed" fluid would meant diffusion. Let's rule it out for the diffusion velocities being small. Anyway, to separate once mixed fluid would require something, like temperature gradient or gravity. Now to the question itself, when we call a speed of the particle a mean speed defined as: $$\boldsymbol v = \frac{\rho_1 \boldsymbol v_1 + \rho_2 \boldsymbol v_2}{\rho_1 + \rho_2}$$ As far as I see, all comes to the mass conservation law $$\rho_1 Q_1 + \rho_2 Q_2 = \rho_o Q_o$$ $o$ for the observational volume. The only thing that hinders from calculating the resulting flow rate is that you don't know the resulting density $\rho_o$. One can do it if the flow is incompressible and one knows the density of the mix. Assuming the incompressible flow, the simplest mix model I guess will yield: $$\rho_o = \frac{\rho_1 Q_1 + \rho_2 Q_2}{Q_1 + Q_2}$$ The residence time itself: $$\Delta t = V / Q_o$$ With the simplest model it is $$\Delta t = \frac{V}{Q_1 + Q_2}$$ So the question is all about compressibility of the flow and the density of the mix. - Okay, so your answer comes down to "The flow rate of the mixed substance is the addition of the two flow rates before (independed on anything, especially independend of geometry)", am I right? – Nick Kidman Aug 16 '12 at 9:01 No, no. First, it is true if the flow is incompressible, just like water or slow gas (< 0.3 Mach) without heating. Second it is true if you take m^3 of the first substance, m^3 of the second, mix and you get 2 m^3 as the result. It is not true for example for water and ammonia (?). I can't remember that gas, water just sucks it almost without changing its volume. – Yrogirg Aug 16 '12 at 9:06 oh it might be true if these two effects compensate each other. For example during the mixing volume increased due to tricky interaction between molecules, but then it decreased as the result of (undersonic) gas being compressed in a tighter pipe. – Yrogirg Aug 16 '12 at 9:22 At least mass flow rates add up :) – Bernhard Aug 16 '12 at 10:51	{"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": 6, "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": 0.3, "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": 0.9292822480201721, "perplexity_flag": "middle"}
http://www.all-science-fair-projects.com/science_fair_projects_encyclopedia/Hill_sphere	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. Hill sphere A Hill sphere approximates the gravitational sphere of influence of one astronomical body in the face of perturbations from another heavier body around which it orbits. It was defined by the American astronomer George William Hill. It is also called the Roche sphere because the French astronomer Édouard Roche independently described it. Considering a central body and a second body in orbit around it (for example the Sun and Jupiter), the Hill sphere is derived from consideration of the following three vector fields: • gravity due to the central body • gravity due to the second body • the centrifugal force in a frame of reference rotating about the central body with the same angular frequency as the second body (Jupiter) The Hill sphere is the largest sphere within which the sum of the three fields is directed towards the second body. A small third body can orbit the second within the Hill sphere, with this resultant force as centripetal force. The Hill sphere extends between the Lagrangian points L1 and L2, which lie along the line of centers of the two bodies. The region of influence of the second body is shortest in that direction, and so it acts as the limiting factor for the size of the Hill sphere. Beyond that distance, a third object in orbit around the second (Jupiter) would spend at least part of its orbit outside the Hill sphere, and would be progressively perturbed by the tidal forces of the central body (the Sun) and would end up orbiting the latter. Formula and examples If the mass of the smaller body is m, and it orbits a heavier body of mass M at a distance a, the radius r of the Hill sphere of the smaller body is $r \approx a \sqrt[3]{\frac{m}{3 M}}$ For example, the Earth (5.97×1024 kg) orbits the Sun (1.99×1030 kg) at a distance of 149.6 Gm. The Hill sphere for Earth thus extends out to about 1.5 Gm (0.01 AU). The Moon's orbit, at a distance of 0.370 Gm from Earth, is comfortably within the gravitational sphere of influence of Earth and is therefore not at risk of being pulled into an independent orbit around the Sun. In terms of orbital period: the Moon has to be within the sphere where the orbital period is not more than 7 months. An astronaut could not orbit the Space Shuttle (mass = 104 tonnes), if the orbit is 300 km above the Earth, since the Hill sphere is only 120 cm in radius, much smaller than the shuttle itself. In fact, in any low Earth orbit, a spherical body must be 800 times denser than lead in order to fit inside its own Hill sphere, or else it will be incapable of supporting an orbit. A spherical geostationary satellite would need to be more than 5 times denser than lead to support satellites of its own; such a satellite would be 2.5 times denser than iridium, the densest naturally-occurring material on Earth. Only at twice the geostationary distance could a lead sphere possibly support its own satellite; the moon itself must be at least 3 times the geostationary distance, or 2/7 its present distance, to make lunar orbits possible. The Hill sphere is but an approximation, and other forces (such as radiation pressure) can make an object deviate from within the sphere. The third object must also be of small enough mass that it introduces no additional complications through its own gravity. Orbits at or just within the Hill sphere are not stable in the long term; from numerical methods it appears that stable satellite orbits are inside 1/2 to 1/3 of the Hill radius (with retrograde orbits being more stable than prograde orbits). Within the solar system, the planet with the largest Hill sphere is Neptune, with 116 Gm, or 0.775 AU; its great distance from the Sun amply compensates for its small mass relative to Jupiter (whose own Hill sphere measures 53 Gm). An asteroid from the main belt will have a Hill sphere that can reach 220 Mm (for 1 Ceres), diminishing rapidly with its mass. In the case of (66391) 1999 KW4, a Mercury-crosser asteroid which has a moon (S/2001 (66391) 1), the Hill sphere varies between 120 and 22 km in radius depending on whether the asteroid is at its aphelion or perihelion! External links 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": 1, "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": 0.9164396524429321, "perplexity_flag": "middle"}
http://mathhelpforum.com/algebra/67123-real-number.html	# Thread: 1. ## Real Number Good evening all of our class is having a online debate with the following question, any feedback would be appreciated: In the Real Number realm, ab = 0 è a = 0 or b = 0 Is the same theorem true in the Complex Number realm? (Why or why not?) 2. Originally Posted by AlgebraicallyChallenged Good evening all of our class is having a online debate with the following question, any feedback would be appreciated: In the Real Number realm, ab = 0 è a = 0 or b = 0 Is the same theorem true in the Complex Number realm? (Why or why not?) In the complex number 'realm', for a complex number to be considered equal to $0$, then both it's real and imaginary parts must be equal to zero! $z_1 = a+ib$ $z_2 = c+id$ $z_1 \times z_2 = ac+adi+cbi-bd = (ac-bd) + i(ad+cb) = 0 + 0i$ Hence $ac-bd = 0$ and $ad+cb=0$ for $z_1 \times z_2=0$ So let's look at the case where $z_1 = a+ib =0+0i$. For this $a=0$ and $b = 0$, which means that $ac-bd = 0(c)-(0)d = 0$. So the first equation is satisfied. $(0)d+(0)c = 0$. So the 2nd is satisfied. Now look at the case where $z_2 = c+id =0+0i$. For this c=0 and d = 0, which means that $ac-bd = 0(a)-(0)b = 0$. So the first equation is satisfied. $(0)b+(0)b = 0$. So the 2nd is satisfied. 3. Ahhh I see what you were asking. But the question remains, are there any combinations of $z_1$ and $z_2$ for which their product is zero, but for which neither of them are 0?! Well. Let's try this then. For a complex number to be zero, then it's modulus must be zero, yes? So let's find the modulus of our product. $\|z_1.z_2\| = \|(ac-bd) + i(ad+bc)\|$ $= \sqrt{(ac-bd)^2 + (ad+bc)^2}$ $= \sqrt{(ac)^2-2abcd+(bd)^2 + (ad)^2+2abcd+(bc)^2}$ $= \sqrt{(ac)^2+(bd)^2 + (ad)^2+(bc)^2} = 0$ Clearly, if this is zero, then the expression inside the square root is zero! $= (ac)^2+(bd)^2 + (ad)^2+(bc)^2 = 0$ $= a^2(c^2+d^2)+b^2(c^2+d^2) = 0$ $= (a^2+b^2)(c^2+d^2) = 0$ Hence, by the logic of REAL numbers (a, b, c and d must all be real, remember!) either: $a^2+b^2 = 0$ OR $c^2+d^2 = 0$ For these to be true: $a = \pm \sqrt{-b^2}$ or $c = \pm \sqrt{-d^2}$ $b^2$ and $d^2$ are always positive numbers, which means that the solutions to these 2 equations for all values of b and d are not real solutions, but purely imaginary solutions. And by the definition of the complex numbers $z_1$ and $z_2$, a, b, c and d must be REAL numbers. Hence there are no two non-zero complex numbers whose product is zero, and hence if $z_1.z_2 = 0 +0i$ then $z_1 = 0+0i$ or $z_2 = 0+0i$	{"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": 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": 0.9367049336433411, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php?title=Parabolic_Reflector&redirect=no	# Parabolic Reflector ### From Math Images Parabolic Reflector Dish Field: Geometry Image Created By: Energy Information Administration Website: [1] Parabolic Reflector Dish Solar Dishes such as the one shown use a parabolic shape to focus the incoming light into a single collector. # Basic Description The geometry of a parabola makes this shape useful for solar dishes. If the dish is facing the sun, beams of light coming from the sun are essentially parallel to each other when they hit the dish. Upon hitting the surface of the dish, the beams are reflected directly towards the focus of the parabola, where a device to absorb the sun's energy would be located. Figure 1: Incoming beams of light perpendicular to the directrix bounce off the dish directly towards the focus. Figure 2: Note that incoming beams reflect 'over' the line perpendicular to the parabola at the point of contact. We can see why beams of light hitting the parabola-shaped dish will reflect towards the same point. A beam of light reflects 'over' the line perpendicular to the parabola at the point of contact. In other words, the angle the light beam makes with the perpendicular when it hits the parabola is equal to the angle it makes with same perpendicular after being reflected, as shown in Figure 2. Near the bottom of the parabola the perpendicular line is nearly vertical, meaning an incoming beam is still nearly vertical after being reflected. Being nearly vertical allows it to reach the focus above the bottom part of the parabola. Further up the parabola the perpendicular becomes more horizontal, allowing a light beam to take on the more horizontal angles needed to reach the focus. Parabolic reflectors are really parabolas rotated about their axis of symmetry (the y axis on this page's diagrams) to form a bowl-like shape known as a paraboloid. This shape is used in many modern application because of its ability to collect incoming information efficiently. For example, TV dishes reflect incoming television signals towards a receiver centered at the focus of the dish. Parabolic reflectors can also work in reverse: if a light emitter is placed at the focus and shined inward towards the parabola, the light will be reflected straight out of the parabola, with the beams of light traveling parallel to each other. Headlights on cars often use this effect to shine light directly forward. # A More Mathematical Explanation Note: understanding of this explanation requires: *Elementary Trigonometry and Calculus. [Click to view A More Mathematical Explanation] The fact that a parabolic reflector can collect light in this way can be proven. We can show that an [...] [Click to hide A More Mathematical Explanation] The fact that a parabolic reflector can collect light in this way can be proven. We can show that any beam of light coming straight down into a parabola will reflect at exactly the angle needed to hit the focus, as follows: Figure 3 Figure 4: A represents equal angles: the line normal to the parabola makes the same angle with the y-axis as the line tangent to the parabola has relative to the x-axis. Step 1 We begin with the equation of a parabola with focus at (0,p). $x^2=4py$ Step 2 We take the derivative with respect to x, giving the slope of the tangent at any point on the parabola: $\frac{x}{2p} = \frac{dy}{dx}$ The slope of this tangent line is relative to the x-axis: when the slope is zero, the tangent line is parallel to the x-axis. The line normal to the parabola makes the same angle with the y-axis as the line tangent to the parabola has relative to the x-axis, as shown in Figure 4. Step 3 We use this slope to find the angle between the normal and the y-axis, which is the same as the angle between the normal and an incoming beam of light. The desired angle $\theta$ can be expressed as: $\tan\theta= \frac{x}{2p}$ (Equation 1) As mentioned previously, a beam of light is reflected 'over' the normal, as shown in Figure 2. This means that the angle the beam of light takes relative to a vertical line is equal to two times the angle the normal makes with the same vertical line. Step 4 We now must show that the direction the light takes after being reflected is exactly the angle needed to hit the focus. Notice from Figure 3 that geometrically, the angle needed to hit the focus is equal to $2\theta$, and satisfies the relationship $\tan2\theta= \frac{x}{p-x^2/4p}$ Step 5 We use a trigonometric identity to rewrite the equation in Step 4: $\tan2\theta= \frac{x}{p-x^2/4p} =\frac{2\tan\theta}{1-\tan^2\theta }$ (Equation 2) Step 6 We now manipulate Equation 1's expression for $\tan\theta$ to show its equivalence to Equation 2 (that is, to show the angle $\theta$ in Equation 1 is the same as the angle $\theta$ in Equation 2). $2tan\theta= 2\frac{x}{2p} = \frac{x}{p}$ , and $1 - \tan^2\theta = 1 - (\frac{x}{2p})^2 = 1 - \frac{x^2}{4p^2}$. Step 7 We combine the two expressions in Step 6. $\frac{2\tan\theta}{1-\tan^2\theta} = \frac{\frac{x}{p}}{1 - \frac{x^2}{4p^2}} = \frac{x}{p-x^2/4p}$ Which is the same as the expression in Equation 2. Therefore, a beam of light will hit the parabola's focus after being reflected. # Teaching Materials There are currently no teaching materials for this page. Add teaching materials. 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": 13, "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": 0.8927724957466125, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/150284-simple-calculus-question.html	Thread: 1. Simple Calculus Question Hi there, There is no forum for basic calculus, so I will have to post in the university-level forum. I have been trying to refresh my maths for the last few months (I had forgotten everything, right down to how to do long division.) and I feel as if I have been making good progress. I have a question about a very simple differentiation exercise: Using the formula $f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{f(x + \delta x) -f(x)}{\delta x}$ find the derivative of $\frac{1}{x^2}$ Here is my attempt at a solution: $f(x + \delta x) = \frac{1}{(x + \delta x)^2}$ $f(x + \delta x) - f(x) = \frac{1}{x^2 + 2x \delta x + (\delta x)^2} - \frac{1}{x^2}$ $= \frac{x^2 - (x^2 + 2x \delta x + (\delta x)^2)}{(x^2 + 2x \delta x + (\delta x)^2)x^2}$ $= \frac{\delta x(- \delta x -2x)}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}$ $\frac{f(x + \delta x) - f(x)}{\delta x} = \frac{\delta x(- \delta x - 2x)}{x^4 + 2x^3 \delta x + x^2(\delta x)^2} \times \frac{1}{\delta x}$ $= \frac{-2x - \delta x}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}$ $f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{-2x - \delta x}{x^4 + 2x^3 \delta x + x^2(\delta x)^2}$ $= \frac{-2x}{x^4}$ $= \frac{-2}{x^3}$ It would be great if someone would tell me where I am going wrong. Regards, Evanator 2. Originally Posted by evanator Hi there, There is no forum for basic calculus, so I will have to post in the university-level forum. I have been trying to refresh my maths for the last few months (I had forgotten everything, right down to how to do long division.) and I feel as if I have been making good progress. I have a question about a very simple differentiation exercise: Using the formula $f'(x) = \displaystyle\lim_{\delta x \to 0} \frac{f(x + \delta x) -f(x)}{\delta x}$ find the derivative of $\frac{1}{x^2}$ Here is my attempt at a solution: $f(x + \delta x) = \frac{1}{(x + \delta x)^2}$ $f(x + \delta x) - f(x) = \frac{1}{x^2 + 2 \delta x + (\delta x)^2} - \frac{1}{x^2}$ .............. you forgot x in second step , $f(x + \delta x) - f(x) = \frac{1}{x^2 + 2x \delta x + (\delta x)^2} - \frac{1}{x^2}$ now continue.... 3. Originally Posted by evanator Hi there, There is no forum for basic calculus, so I will have to post in the university-level forum. [snip] Well done. Thankyou. Someone with some basic common sense. And you even showed your work. 4. I amended the first post. It all comes out now. Thanks for the help, ramiee2010. It is much appreciated. What are you referring to, mr fantastic? 5. Originally Posted by evanator I amended the first post. It all comes out now. Thanks for the help, ramiee2010. It is much appreciated. What are you referring to, mr fantastic? He's saying a lot of people post calculus questions in the pre-calculus forum just because they happen to be in high school and not university. And he's also giving you praise for showing your working, as a lot of people also just post a question and beg for others to do the work for them.	{"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": 16, "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": 0.9599258303642273, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/7566/list	## Return to Answer 3 improved latex formatting Work locally, assume that $f: R\to S$ is a local homomorphism. Let $cmd(R) \operatorname{cmd}(R) =dim R-depth \operatorname{dim} R-\operatorname{depth} R$ (this is the so-called Cohen-Macaulay defect of $R$). Claim: cmd $\operatorname{cmd}$ is preserved by l.c.i maps (easy, essentially because both depth and dimension drop by one when you kill a regular element). Now since the map $\phi: Spec(S) \operatorname{Spec}(S) \to Spec \operatorname{Spec} (R)$ is finite and surjective, $dim \operatorname{dim} R= dim \operatorname{dim} S$, which combines with the last claim to show that $depth \operatorname{depth} R = depth \operatorname{depth} S$. But since l.c.i also implies finite flat dimension, we have $depth \operatorname{depth} R -depth \operatorname{depth} S = pd_RS$, so $S$ is flat over $R$. 2 added 102 characters in body Work locally, assume that $f: R\to S$ is a local homomorphism. Let $cmd(R) =dim R-depth R$ (this is the so-called Cohen-Macaulay defect of $R$). Show that Claim: cmd is preserved by l.c.i mapmaps (easy, essentially because both depth and dimension drop by one when you kill a regular element). Now since the map $\phi: Spec(S) \to Spec (R)$ is finite and surjective, $dim R= dim S$, which combines with the last sentence claim to show that $depth R = depth S$. But since l.c.i also implies finite flat dimension, we have $depth R -depth S = pd_RS$, so $S$ is flat over $R$. 1 Work locally, assume that $f: R\to S$ is a local homomorphism. Let $cmd(R) =dim R-depth R$ (this is the so-called Cohen-Macaulay defect of $R$). Show that cmd is preserved by l.c.i map. Now since the map $\phi: Spec(S) \to Spec (R)$ is finite and surjective, $dim R= dim S$, which combines with the last sentence show that $depth R = depth S$. But since l.c.i also implies finite flat dimension, we have $depth R -depth S = pd_RS$, so $S$ is flat over $R$.	{"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": 28, "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": 0.9542645812034607, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/9022/neutralino-dark-matter-detection?answertab=oldest	# Neutralino Dark Matter Detection Assuming supersymmetry exists and a neutralino is stable, it's often seen as a leading dark matter candidate. What would be expected from the interaction of a neutralino and its anti-particle? Has there been any unambiguous detection of such an interaction? - ## 2 Answers Dear Michael, neutralino carries no conserved charges that can take arbitrarily large values (and no spin) and it is identical with its antiparticle, much like in the case of photons. That still allows two neutralinos to annihilate. Neutralinos only carry a "1" charge under the $Z_2$ symmetry called R-parity. This means that the number of neutralinos must be conserved modulo 2. This is compatible with the annihilation of pairs of neutralinos. The annihilation of neutralino pairs was important when the Universe was young. With the right TeV-like mass and MSSM-like annihilation cross section, models may predict that the dark matter density has been diluted exactly to the observed value (plus minus a few orders of magnitude which is still a nontrivial agreement). The products of this annihilation are most typically two photons (but two gluons or gamma-Z are also possible) and the resulting gamma rays are so rare in the present era that it's hard to observe them. However, people have still tried to argue that they should be observable, at least under various assumptions: http://arxiv.org/abs/astro-ph/9712318 As far as I know, none of those gamma rays has been observed as of today. - Lubos: Dark matter is supposedly 23% of the universe mass/energy. Why are "the resulting gamma rays are so rare in the present era"? Dark matter "out-masses" baryons by 5:1. – Michael Luciuk Apr 23 '11 at 12:44 maybe because dark matter is not made of neutralinos at all? – lurscher Apr 23 '11 at 21:28 2 @Michael: In Lubos' scenario, the neutralino equilibrium period would have ended considerably before the radiation decoupling, so the the annihilation gammas made up until then would never be detected in the present time. Only those made by pairs interacting since the decoupling. – dmckee♦ Apr 24 '11 at 0:55 Thanks dmckee. Good point. – Michael Luciuk Apr 24 '11 at 1:41 1 – stringpheno Apr 24 '11 at 3:56 show 2 more comments It could be that the neutralino is a Majorana fermion. The $Z_2$ structure could be a signature of a Majorana fermion. The Majorana fermion is the charge conjugate fermion of the form $\psi_c~=~\gamma^2\psi^$ with $\psi_c~=~\psi$. This has some interesting statistics. For fermions the exchange of a wave function introduces a negative sign. The additional constraint that $\psi~=~\gamma^2\psi^$, means that the $\gamma^2$ introduces a braid structure into the exchange statistics of Majorana fermions. If this is the case the neutralino is its own anti-particle. The annihilation of the neutralino particle could then be a type of decay process. - 1 Er...perhaps I'm about to expose myself as an ignoramus, but aren't the super-symmetric partners of fermions (like, say, neutrinos) bosons automatically? Or do things change when you get Majorana particles running around the place? – dmckee♦ Apr 24 '11 at 0:51 @dmckee The neutralino isn't related to the neutrino despite the apparent name similarity (similarly, the neutron and neutrino aren't closely related). Neutralinos is actually a mixture of different states of the supersymmetric partners of the Z, photon, and higgs bosons (since they are uncharged and have the same quantum numbers they can mix to form 4 different eigenstates which look like different particles). The least massive such particle is the most likely candidate for the cold dark matter WIMPs. – Wedge Feb 24 at 23:40	{"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": 0.9297683835029602, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/7864/false-beliefs-in-mathematics-conceptual-errors-made-despite-or-because-of-mat?answertab=votes	# False beliefs in mathematics (conceptual errors made despite, or because of, mathematical education) Over on mathoverflow, there is a popular CW question titled: Examples of common false beliefs in mathematics. I thought it would be nice to have a parallel question on this site to serve as a reference for false beliefs within less obscure mathematics. That said, it would be good not get bogged down with misconceptions that are generally assumed to be elementary such as: $(x + y)^{2} = x^{2} + y^{2}$. - @ALL: Should this be community wiki! – anonymous Oct 26 '10 at 4:32 18 I dont really see a need for this question. :/ I think the one on Mathoverflow is welcoming of examples in other areas of math rather than "obscure". I think that it is a cool question, but I really dont think we need this duplicate. – BBischof Oct 26 '10 at 4:34 4 Voted to close. This is a dup, as per the question itself. – Mariano Suárez-Alvarez♦ Oct 26 '10 at 5:05 2 Specifically, the MO discussion focused on mistakes that were (a) conceptual, and (b) known to be made by mathematicians (especially, mistakes the answerers had made). This restriction prevented trivial responses. I suggest revising the question to be "false beliefs YOU -- a presumably mathematically capable math.SE user -- have held" as a separate matter from "review every error that students make!". If the latter is interesting it would be better to explore it in another thread. – T.. Oct 26 '10 at 15:22 1 @T.. Title has been edited as per your suggestion. – Ami Oct 26 '10 at 15:34 show 4 more comments ## 10 Answers Many well-educated people believe that a p-value is the probability that a study conclusion is wrong. For example, they believe that if you get a 0.05 p-value, there's a 95% chance that your conclusion is correct. In fact there may be less than a 50% chance that the conclusion is correct, depending on the context. Read more here. - I recently caught myself thinking that the formula for the determinant of a 2-by-2 matrix also works for a block matrix, i.e. $\det (A B; C D) = \det(A)\det(D) - \det(B)\det(C)$. - 2 It works if e.g. $D$ and $C$ commute. – Plop Oct 27 '10 at 12:55 Every torsion-free Abelian group is free. (This only holds for finitely generated Abelian groups.) - 1 The counterexample which is obvious in hindsight: Q. – Qiaochu Yuan Oct 26 '10 at 15:49 @Qiaochu Yuan: Yes, my problem when I had this false belief was, that I only knew Q, the ring---not Q, the Abelian group. ;) – Rasmus Oct 26 '10 at 17:53 What about Q the omnipotent being? – OghmaOsiris Jul 26 '11 at 21:38 2 @OghmaOsiris: please try to only include relevant comments... – Mariano Suárez-Alvarez♦ Jul 28 '11 at 22:35 3 @Mariano: please try to lighten up. – OghmaOsiris Jul 30 '11 at 21:33 I have seen this one time too many $$\frac{a}{b}+\frac{c}{d}=\frac{a+c}{b+d}$$ - 4 Interestingly, the fraction (a+c)/(b+d) is useful. Of course it's not the sum of a/b and c/d, but it is part of constructing the Farey sequence and is important in finding rational approximations. – John D. Cook Oct 27 '10 at 11:36 – AD. Oct 27 '10 at 12:33 – leonbloy Sep 3 '11 at 17:13 To generalize a few of the answers, for pretty much any function, someone somewhere will make the mistake of treating it as if it is linear in all of its variables. Thus we get: $e^a + e^b = e^{a+b}$, $\sqrt{a + b} = \sqrt{a} + \sqrt{b}$, $a/(b+c) = a/b + a/c$, ... - These are 2 instances which i have seen to happen with my friends. If $A$ and $B$ are 2 matrices, then they believe that $(A+B)^{2}=A^{2}+ 2 \cdot A \cdot B +B^{2}$. Another mistake is if one i asked to solve this equation, $\displaystyle\frac{\sqrt{x}}{2}=-1$, people generally square both the sides and do get $x$ as $4$. - I had posted this on MO, and i am posting this here as well as this appears elementary. – anonymous Oct 26 '10 at 4:35 2 More generally, people simplify X to Y by saying "if X, then Y". They forget that they also need "if Y, then X". In other words, "Sqrt[x] = -2" implies "x=4" (vacuously), but the converse is false. I blame our telling students "you can do anything to both sides of an equation, as long as it's the same thing". That's true, but we should add "most of the time, you want to make sure you can undo it too". – barrycarter Apr 5 '11 at 5:28 1 Actually "you can do anything to both sides of an equation, as long as it's the same thing" is exactly right for deriving new equations from old ones. I think the problem is probably that most students don't realise that solving an equation is the exact opposite procedure. – gfes Jun 6 '11 at 2:37 The question I've heard on many levels (including the grad level): what is the square root of $a^2$? And everyone says: it's $a$! In fact it is $\|a\|$. - 3 It seems they mistook your question for "What is a square root of $a^2$?" – Rahul Narain Oct 27 '10 at 1:12 6 er... in $\mathbb{R}$ I suppose, but surely not in $\mathbb{C}$. – J. M. Oct 27 '10 at 1:14 Both my students and some of my colleagues (!) believe that the graph of a function cannot cross a horizontal asymptote. Obviously this implies that they misunderstand the definition of an asymptote. More worryingly (in my eyes), it also seems to imply that they don't understand why we even care about asymptotes. - So $y=\exp(-x)\cos(x)$ is asymptotic to the horizontal axis, yes? – J. M. Oct 28 '10 at 12:25 I have seen this many times: $$a^2 + a^3 = a^5$$ - 7 it depends on what the meaning of + is.... – a little don Oct 28 '10 at 12:16 I was reading a book on Lie algebras yesterday and found a mistake over something basic. The author stated that two nested exponents equal their product. Let B, x, and y belong to the set of natural numbers and have this equality: $$(B^x)^y = B^{(xy)}$$ Take the $\log_B$ of both sides and we get: $$x^y = xy$$ which is obviously wrong. When I read that I immediately thought of this thread. It's a late addition, I know, but I just thought I'd share my own mathematical pet peeve. - If any one can show me how to represent compound exponents in MathJax? – bwkaplan Sep 3 '11 at 15:40 2 If this post really has a mistake, that is quite ironic. :-) – Srivatsan Sep 3 '11 at 15:54 2 Taking log on the LHS side gives $y\log_B (B^x)==xy\log_B(B)=xy$. – Fredrik Meyer Sep 3 '11 at 16:23 1 @bwkaplan: I'm having trouble parsing your comment. (3. does? I didn't think so!) in particular. "does?" makes no sense to me. In fact, simply directly computing my example shows that it is true. But let's go through your log, and see what happens. $\log_3 (3^4)^5 = 5 \log_3 3^4 = 5\cdot 4 \log_3 3 = 5 \cdot 4 = \log_3 3^{5 \cdot 4}$. So it seems that the log method also does not disagree with my example. But perhaps "does? I didn't think so!" has a completely different meaning? – mixedmath♦ Sep 3 '11 at 16:34 1 @Fredrik Meyer I now see the mistake I made in carrying the exponent outside the logarithm function. Whoops, my bad! – bwkaplan Sep 3 '11 at 16:36 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": 23, "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": 0.3, "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": 0.9585072994232178, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/268071/schroder-functional-equation	# Schröder functional equation I have the following Schröder functional equation: $f(h(s))=c.f(s)$ where $f,h: ℂ→ℂ$, here $f$ is not analytic and $h$ is analytic and $c∈ℝ$. My question is: How we can solve this equation (the form of $f$ and its domain of definition). We can take $h$ as $h(s)=1-s$ or $h(s)=s-1$ and for both cases we can take $c=-1$. Some motivations are available here: http://en.wikipedia.org/wiki/Schr%C3%B6der%27s_equation - ## 1 Answer I solve such questions using the Carleman-matrix-concept (which you also find mentioned in the wikipedia article). Carleman-matrices contain in its rows the coefficients of a function in its power series representation, and of its powers. So the Carleman-matrix, say C contains the coefficients of $f(x)^0$ (which is the constant 1), $f(x)$, $f(x)^2$ ... in its rows. In your case where $f(x)=1 - 1 x$ we have $$C= \begin{bmatrix} 1&.&.&. \\ 1&-1&.&. \\ 1&-2&1&. \\ 1&-3&3&1 \\ \end{bmatrix} \qquad \text{ for } \qquad \begin{array} {} f(x)^0 &=&1 \\ f(x)&=&1-x \\ f(x)^2&=&1-2x+x^2 \\ f(x)^3&=&1-3x+3x^2-x^3 \\ \end{array}$$ such that with a "vandermonde"-type column-vector $V(x) = [1,x,x^2,x^3,...]$ of appropriate size we shall have $$C \cdot V(x) = V(f(x))$$ Then if you find a diagonalization of C such that $M^{-1}\cdot D \cdot M = C$ where $D$ is diagonal and $M$ and $M^{-1}$ are triangular, then $M$ in its second row contains the coefficient of the Schröder-function and that in $M^{-1}$ in its second row its inverse. In our case we find that $$M^{-1} =\begin{bmatrix} 1 & . & . & . \\ 1/2 & 1/2 & . & . \\ 1/6 & 1/2 & 1/3 & . \\ 0 & 1/4 & 1/2 & 1/4 \end{bmatrix}$$ , with $D=\operatorname{diag}([1,-1,1,-1])$ and $$M = \begin{bmatrix} 1 & . & . & . \\ -1 & 2 & . & . \\ 1 & -3 & 3 & . \\ -1 & 4 & -6 & 4 \end{bmatrix}$$ is a possible solution. (Here $M^{-1}$ can be recognized as the set of coefficients of integrals of the bernoulli-polynomials when we extend the dimension of the matrix infinitely) In general, the eigenvector-matrix $M$ can be understood as limit of the n'th power of $C$ scaled by the reciprocal of the n'th power of $f'(0)$ when n goes to infinity, and so the Schröder-function as limit of the n'th iterate of $f(x)$ divided by $f'(0)^n$ where $n \to \infty \qquad$ - but can furtherly be scaled by an arbitrary constant factor $\gamma \ne 0$ Remark: Your example which requires only matrix size of $n \times n= 2 \times 2$ is much easier, but then one wouldn't see the general principle (and the relation to the bernoulli-polynomials, so I used the bigger matrix size with n=4 here. - @ Gottfried Helms : This is a great answer. Thank you very much. But the function $f$ is not analytic. – ZE1 Dec 31 '12 at 11:13 @rh1: hmm, what do you mean? Unfortunately I mixed your given $h(s)$ by my usual $f(x)$ and so this overlaps your $f(s)$, sorry perhaps I should adapt this. By my example, the function $f(s)$ in your sense is $f(s)=-1+2s$ and $f^{-1}(s)=1/2(1+s)$ so $h(s)=f^{-1}(c \cdot f(s))= 1/2(1+(-1(-1+2s)))=1/2(2-2s)=1-s$ for any complex $s$. The $m$'th iterates are then expressible simply by $m$'th powers of $c$ - and, taking care of the problem of non-unique solutions of fractional or complex powers of $c=-1$ one can define even fractional- and complex-valued iterations heights $m$ – Gottfried Helms Dec 31 '12 at 12:55	{"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": 45, "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": 0.9411328434944153, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/21373-use-methods-frobenius-solve.html	# Thread: 1. ## use the methods of frobenius to solve use the methods of frobenius to solve: 2xy"+y'+2y =0 i want to know not just the answer but the steps how to answer it (explanation of the answer) 2. Originally Posted by compufatwa use the methods of frobenius to solve: 2xy"+y'+2y =0 i want to know not just the answer but the steps how to answer it (explanation of the answer) The charachteristic equation is, $2r(r-1) + r+0=0$ thus, $r=0,1/2$. This means to look for solutions of the form: $y=\sum_{n=0}^{\infty} a_n x^n$ and $y=\sum_{n=0}^{\infty} a_n x^{n+1/2}$ Can you take it from there? 3. ur point of view is not clear to me can u tell me more info about it 4. Originally Posted by compufatwa ur point of view is not clear to me can u tell me more info about it Given the equation, $x^2p(x)y''+xq(x)y'+r(x)y=0$ Where $p(x),q(x),r(x)$ are analytic functions* on some open interval $(-R,R)$. The method of Frobenius says that we can look for a solution of the form, $y=\sum_{n=0}^{\infty}a_n x^{n+r}$ for some number $r$. Now to find this $r$ we solve the equation, $r(r-1)p(0) + rq(0) + r(0) = 0$. So I told you what $r$ has to be. *)"Analytic" on $(-R,R)$ means the function can be expanded as its Taylor series about any point, in this case about 0. Yes, I got a question about Islam. Did you know that Halos (the floating circles around your head) were created by Christians. If so why would you use them? 5. i do not think that the question is about islam ,but it's about the floating circles around icon head, it's just an icon i know that it's used in the art as in church and many famous art drawing .	{"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": 13, "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": 0.9638760685920715, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/49737/article-or-book-about-the-history-of-spherical-geometry?answertab=oldest	# Article or book about the history of spherical geometry? I teach a course on non-Euclidean geometry to high schoolers. I'm looking for an article or book that gives a thorough and interesting history of spherical geometry and trigonometry. I'm looking for it for my own learning and possibly to distribute to my students. I'd like something that has some substance to it--not just a couple of facts tossed off for color. A treatment that gives an overview of spherical geometry through the ages would be great, as would a "zoomed in" treatment of some particular episode. Thanks! Edit: Some good resources have been suggested below, especially the book suggested by unclejamil. I still haven't found quite what I'm looking for, though. I'd like something that focuses just on the history of the study of the geometry and trigonometry of the sphere, from ancient times to modernity, focusing on major advances and motivations. Any thoughts? If I can't find such a source ready-made, I'll try to put together a short essay myself. - ## 3 Answers The Mathematics of the Heavens and the Earth: The Early History of Trigonometry will do: Seems to be what you're looking for. Good luck with the kids. ;) - 2 +1 for van Brummelen's book. I heard him give a talk on precisely this subject at the 2010 Joint Mathematics Meetings. His presentation is clear and easy to understand, yet amazingly detailed. – Willie Wong♦ Jul 6 '11 at 13:13 +1 This looks like an awesome resource and an interesting read. I've ordered a copy. Thanks! I'm still hoping to find something that focuses more exclusively on the development of spherical geometry--less interlaced with the corresponding planar story and including more information about non-trigonometric approaches and facts about the geometry of the sphere as well. Clearly all of these things are intertwined, but my hope is that there might be something more streamlined (while remaining substantive) for pedagogic reasons. I'd also like to see it reach at least up to the time of Riemann. – Justin Lanier Jul 6 '11 at 22:20 Try Tristan Needham's Visual Complex Analysis. There is a chapter in there on non-euclidean geometry which includes spherical and hyperbolic geometry. An excellent book that shows the links between different geometries, Möbius Transformations, etc. $\textbf{Edit}$: You don't need to be a genius to see the many beauties of complex analysis in this book; Needham explains many concepts beautifully. However, that being said I know a lecturer who dissed this book asking if it was for physicists. The basis he said that was Needham used things like $ds$ for element of arc-length, this lecturer said that the book was not being rigorous by saying that these are things physicists use without formal definitions. Note: I may or may not support his view but am just conveying that there are some people who dislike this book. - +1 for this amazing book – platinumtucan Jul 6 '11 at 6:11 I, too, once met a lecturer who dissed the book. He was an analyst, and said something along the lines of "well, I'm an analyst, not a geometer, and that's not complex analysis". It is an excellent book, nevertheless. You just have to approach it on its own terms. – Jose L. Lykón Jul 6 '11 at 15:00 I read this book in some depth a couple of summers ago and enjoyed it immensely. The chapter on non-Euclidean geometry is great, but it's not quite what I'm looking for here. – Justin Lanier Jul 7 '11 at 13:51 Today we think of Euler's Polyhedral formula (V+F-E=2) as a result in both geometry, topology, and combinatorics (graph theory) and it is common to give a proof of it using graph theory methods. Euler's "proof" was not correct. The first proof was given by Legendre using methods from spherical geometry! This seems curious but I recently came across a book that makes more sense of what happened in an historical perspective: http://www.springer.com/new+%26+forthcoming+titles+%28default%29/book/978-1-4020-8447-8 The book charts how Legendre pioneered a modern view of symmetry and how his work on solid angles was related to his proof of Euler's polyhedral formula. It is also fascinating how without our modern journals and methods to share ideas among scholars Legendre came to do this work about 1794 after Euler's initial work, about 1750. -	{"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": 0.3, "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": 0.966655969619751, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/61695?sort=newest	## Are there any finitely generated artinian modules that are not notherian? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is well known that for rings, Artinian implies Noetherian (the famous Hopkins–Levitzki theorem) and it is also well known that there are Artinian modules which are not Noetherian. A simple example can be found in http://en.wikipedia.org/wiki/Artinian_module#Relation_to_the_Noetherian_condition Since rings are always finitely generated modules over themselves (all rings considered are unital), it seemed natural to me to ask whether there are finitely generated modules, which are Artinian but not Noetherian (the example given in the reference is clearly not finitely generated). I guess that if the statement "every finitely generated artinian module is noetherian" was true, I would have seen it in any standard text book on algebra, and since I haven't, I guess it's not. But still, I can't find a counter-example for this. Perhaps I'm missing something completely trivial here. I will be happy to see an example of such module or a proof that there are no (just a reference will be much appreciated too of course). - 1 Wouldn't this statement reduce to cyclic modules (one generator)? Then, at least if the ring is commutative (I can't figure out whether you are making this assumption), a cyclic module is a free module over a quotient ring, and so noetherian and artinian properties appear to be equivalent. – t3suji Apr 14 2011 at 13:32 I changed the tag "commutative-algebra", which I put by mistake, to "noncommutative-algebra". The answer about the commutative case was great. so was the comment, that after reading Simon's answer, I understood that it says pretty much the same thing. thanks. I'm still interested in the noncommutative case. – KotelKanim Apr 14 2011 at 13:56 I just followed the wikipedia link that you gave and it gives links to two articles that show that cyclic Artinian modules need not be Noetherian plms.oxfordjournals.org/content/s3-35/1/55 and jlms.oxfordjournals.org/content/55/2/231 I don't have access to either so can't say much more – Simon Wadsley Apr 14 2011 at 14:25 thanks a lot Simon! It settles it completely. I feel stupid for missing the link from the article on wikipedia I linked myself. I promise to do my homework batter for the next question... – KotelKanim Apr 14 2011 at 14:42 ## 2 Answers Suppose you have an Artinian but not Noetherian finitely generated $R$ module $M$. Let $0\leq M_1\leq M_2\leq \cdots \leq M_n=M$ be a finite chain of $R$-modules such that each composition factor $M_i/M_{i-1}$ is cyclic for each $i$. Certainly each composition is Artinian since subquotients of Artinian modules are Artinian. Also one of the composition factors must be non-Noetherian since extensions of Noetherian modules by Noetherian modules are Noetherian. Thus, we may assume that $M$ is a cyclic $R$-module. Now if $R$ is commutative, $M$ is a quotient ring $R/I$ which is Artinian as such and so Noetherian also, as you say. If $R$ is non-commutative then I'm not sure what the answer is. Added: It seems from the wikipedia article linked from the question that Hartley showed that there are cyclic Artinian and non-Noetherian modules over certain non-commutative rings and Cohn gave another construction nearly twenty years later. See the links I give in the comments on the question for precise references. - ### 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. Mistake----Adding a unit makes the radical have lots of submodules. -	{"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": 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": 0.9427419900894165, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/105705/basic-application-of-strong-law-of-large-numbers	# basic application of Strong Law of Large Numbers In $$\sum_{j=0}^q {q\choose j}{1\over n}\sum_{i=1}^n X_i^j(-\bar X)^{q-j} \quad \overrightarrow{a.s.} \quad \sum_{j=0}^q {q\choose j} \mathbb{E}(X^j) (-\mathbb{E}(X))^{q-j}$$ using the Strong Law, why is it, that we can say that $$\frac{1}{n}\sum_{i=1}^n(-\bar X)^{q-j} \quad\overrightarrow{a.s.} \quad (\mathbb{-E}X)^{q-j}$$ The reason I am wondering is, that Slutsky's Theorem would only give me that convergence in probability is preserved under the continuous function $f(x) = x^{q-j}$ and since $\bar{X}$ are not iid it seems that using the Strong law on the complete term would not work ? - 3 There is no dependence on i in $\bar{X}$ so pull it out of the sum. The $\frac{1}{n}$ and the sum cancel (in the second case, or associate with the $X_i^j$ terms in the first case) so you are just looking at the limit as n goes to infinity of $-\bar{X}^{q-j}$. Discarding a set of measure zero, you can assume that $\bar{X}$ converges to $\mathbb{E}$X everywhere at which point that just becomes the statement that if $f$ is continuous and $x_n \to x$ then $f(x_n)\to f(x)$ – Chris Janjigian Feb 4 '12 at 18:21 2 @PeeJay It's not clear to me whether $\bar{X}$ in the sum refers to $\bar{X}_n$ or $\bar{X}_i$. – Ben Derrett Feb 4 '12 at 18:47 @Chris so in the last line of your comment, is that not using Slutsky s Theorem ? ( because if it is, then I could only conclude that convergence in probability follows right ? ) – Beltrame Feb 4 '12 at 18:58 1 @BenDerrett the $\bar{X}$ is supposed to refer to the sample mean, i.e. it s referring to n. – Beltrame Feb 4 '12 at 19:00 1 @PeeJay That is correct, you do not need Slutsky's theorem for that part. Chris's comment above explains it very well. – Byron Schmuland Feb 4 '12 at 20:49 show 3 more comments ## 1 Answer Ok, so the bottom line is that using the strong law we do not need Slutsky or any other result like it, because convergence a.s. is preserved under continuous transformations. This is actually pretty easy to show just using the sequential definition of continuity and applying it to the random variable. Convergence a.e. of the composition follows immediately an no further Results from measure theoretic probability are required. Hopefully this is an adequate summary, if not then comments are most welcome ! -	{"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": 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": 0.9374282956123352, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/1270/how-do-i-estimate-the-joint-probability-of-stock-b-moving-if-stock-a-moves/1294	How do I estimate the joint probability of stock B moving, if stock A moves? I have two stocks, A and B, that are correlated in some way. If I know (hypothetically) that stock A has a 60% chance of rising tomorrow, and I know the joint probability between stocks A and B, how do I calculate the probability of stock B moving tomorrow? For bonus upvotes - do you know of any standard libraries that can calculate the joint probability of stocks A and B, given a time series of historical data? Update: The phrase "conditional probability" is also applicable. - 3 Answers So you want to calculate $\mathbb{P}[B_1 > B_0 + \varepsilon \;\|\; A_1 > A_0 + \varepsilon]$? If you truly have the joint distribution of $A_1$ and $B_1$ and the current prices $A_0$ and $B_0$, this just becomes a simple exercise in integration, by the definition of probability density. Are you asking how to find a conditional probability in general, or is your question about something else? - Interesting. Calculating integrals is not that difficult, thanks or the tip. – Gravitas Jul 9 '11 at 11:38 Why not using the so simple Monte-Carlo estimator $\hat{p}_N =\frac{ \sum_{i=1}^N 1_{\|A_{i+1}-A_i\|>0 \cap \|B_{i+1}-B_i\|>0}} {\sum_{i=1}^N 1_{\|A_{i+1}-A_i\|>0 }}$ where $1_{\|A_{i+1}-A_i\|>0}$ is $1$ if stock $A$ has moved at time $i+1$ - ...do you know of any standard libraries that can calculate the joint probability of stocks A and B, given a time series of historical data? Using R and the LSPM package with the code posted here might be what you are looking for. -	{"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": 10, "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": 0.9378678202629089, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2011/11/24/compact-oriented-manifolds-without-boundary-have-nontrivial-homology/?like=1&source=post_flair&_wpnonce=a34a077b61	# The Unapologetic Mathematician ## Compact Oriented Manifolds without Boundary have Nontrivial Homology If we take $M$ to be a manifold equipped with an orientation given by an orientation form $\omega$. Then $\omega$ is nowhere zero, and $\omega(v_1,\dots,v_n)>0$ for any positively oriented basis $\{v_i\}$ of $\mathcal{T}_pM$ at any point $p\in M$. Next we take $c:[0,1]^n\to M$ to be an orientation-preserving embedding — a singular cube of top dimension. Then the pullback $c^*\omega=fdu^1\wedge\dots\wedge du^n$ for some strictly-positive function $f$. We conclude that $\displaystyle\int\limits_c\omega=\int\limits_{[0,1]^n}fdu^1\wedge\dots\wedge du^n=\int\limits_{[0,1]^n}f(u^1,\dots,u^n)\,d(u^1,\dots,u^n)>0$ The integral of $\omega$ over all of $M$ must surely be even greater than the integral over the image of $c$, since we can cover $M$ by orientation-preserving singular $n$-cubes, and none of them can ever contribute a negative to the integral. If we further suppose that $M$ is compact, we can cover $M$ by finitely many such singular cubes, and the integral on each is well-defined. Using a partition of unity as usual this shows us that the integral over all of $M$ exists and, further, must be strictly positive. In particular it’s not zero. But now suppose that $M$ also has an empty boundary. Since $\omega$ is a top form, we know that $d\omega=0$ — it’s closed in the de Rham cohomology. But we know that it cannot also be exact, for if $\omega=d\eta$ for some $n-1$-form $\eta$ then Stokes’ theorem would tell us that $\displaystyle\int\limits_M\omega=\int\limits_Md\eta=\int\limits_{\partial M}\eta=0$ since $\partial M$ is empty. And so if $M$ is a compact, oriented $n$-manifold without boundary, then there must be some $n$-forms which do not arise from taking the exterior derivatives of $n-1$-forms. If $M$ is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on $M$ which are not the divergence of any vector field on $M$. ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 1 Comment » 1. [...] — we see that any compact, oriented manifold without boundary cannot be contractible, since we know that they have some nontrivial homology! LD_AddCustomAttr("AdOpt", "1"); [...] Pingback by \| December 6, 2011 \| 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": 34, "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": 0.9191231727600098, "perplexity_flag": "head"}
http://nrich.maths.org/5448/clue	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? Factors and Multiples Puzzle Stage: 3 Challenge Level: Can a number be both odd and even? What does this tell you about where the ODD NUMBER and EVEN NUMBER headings have to go? Can you use this idea to position any of the other heading cards? If you have reached a stage where you have placed most of the numbers but have a few that you cannot place, don't panic! Could you swap one of the numbers that you can't place for a number that is already on the grid? You might be able to swap them and then find that the number you have removed can be placed on one of the empty squares. Don't forget: $1$ is not a prime number! Triangle numbers can be represented by a triangular array of squares 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": 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": 0.9454629421234131, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/236624/nabla-times-x-is-given-on-a-surface-can-i-show-that-x-0-on-the-same-surfa	$\nabla \times X$ is given on a surface, can I show that X = 0 on the same surface If I have a volume V enclosed by a surface S, and $\nabla \times X$ is given on the surface, what information does that give me about X on S. Is there a method of showing that X = 0 on S? (in the context of the divergence theorem?) - 1 Answer Note that if $C$ is any constant vector, $\nabla \times (X+C) = \nabla \times X$. If the surface is compact and $X$ is continuous, it is bounded on the surface, so if we take $\|C\|$ large enough there is no point on the surface where $X+C=0$. So there is certainly nothing that $\nabla \times X$ can tell you that would imply that $X=0$ somewhere on $S$. -	{"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": 10, "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": 0.9714235663414001, "perplexity_flag": "head"}
http://mathoverflow.net/questions/2300/what-is-the-field-with-one-element/2306	## What is the field with one element? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've heard of this many times, but I don't know anything about it. What I do know is that it is supposed to solve the problem of the fact that the final object in the category of schemes is one-dimensional, namely `$\mathop{\text{Spec}}\mathbb Z$`. So, what is the field with one element? And, what are typical geometric objects that descend to `$\mathbb F_1$`? - Perhaps `f-1` should have its own tag... but may be `absolute-motives`? – Ilya Nikokoshev Oct 24 2009 at 16:24 1 The problem with tag about `F_1` is that it's not clear how to call it. `f-1` looks like the "Help" button on Windows. And then how to make a non-arbitrary selection between `f-1`, `f-un`, `f-one`, `field-with-one-element`? – Ilya Nikokoshev Oct 24 2009 at 16:26 I wasn't terribly happy with "f-1", but I don't care for "absolute-motives" either--sounds very scary for a phenomenon which has many elementary manifestations. – Reid Barton Oct 24 2009 at 16:35 Good point, you're right. – Ilya Nikokoshev Oct 24 2009 at 16:43 I created the other f-1 tag too, and I have no preference among ilya's other suggestions, so if someone does have a preference, please feel free to change them. – Reid Barton Oct 24 2009 at 17:09 ## 7 Answers As other have mentioned, F_1 does not exist of a field. Tits conjectured the existence of a "field of characteristic one" F_1 for which one would have the equality G(F_1) = W, where G is any Chevalley group scheme and W its corresponding Weyl group. Later on Manin suggested that the "absolute point" proposed in Deninger's program to prove the Riemann Hypothesis might be thought of as "Spec F_1", thus stating the problem of developing an algebraic geometry (and eventually a theory of motives) over it. There are several non-equivalent approaches to F_1 geometry, but a common punchline might be "doing F_1 geometry is finding out the least possible amount of information about an object that still allows to speak about its geometrical properties". A "folkloric" introduction can be found in the paper by Cohn Projective geometry over F_1 and the Gaussian binomial coefficients. It seems that all approaches so far contain a common intersection, consisting on toric varieties which are equivalent to schemes modeled after monoids. In the case of a toric variety, the "descent data" that gives you the F_1 geometry is the fan structure, that can be reinterpreted as a diagram of monoids (cf. some works by Kato). What else are F_1 varieties beyond toric is something that depends a lot on your approach, ranging from Kato-Deitmar (for which toric is all there is) to Durov and Haran's categorical constructions which contains very large families of examples. A somehow in-the-middle viewpoint is Soule's (and its refinement by Connes-Consani) which in the finite type case is not restricted only to toric varieties but to something slightly more general (varieties that can be chopped in pieces that are split tori). No approach is yet conclusive, so the definitions and families of examples are likely to change as the theory develops. Last month Oliver Lorscheid and myself presented an state-of-the-art overview of most of the different approaches to F_1 geometry: Mapping F_1-land: An overview of geometries over the field with one element (sorry for the self promotion). - 4 Also, I forgot to mention, next month there will be a workshop on F_1 geometry for young researchers at Granada. If you are interested check the website: ugr.es/~nc_alg/f1 – javier Oct 24 2009 at 18:23 2 This gets my +1 since it's much better informed than my own answer. – Ilya Nikokoshev Oct 24 2009 at 23:38 Yes, I will be attending that workshop! – Jose Brox Nov 9 2009 at 0:56 ### 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. Update: at the bottom there's a wonderful and fresh reference. There's no field with one element in the literal sense, but there are constructions that work over different fields `$\mathbb F_q$` and which sometimes make sense when `$q=1$`. Examples would include representation theory of $GL_n(\mathbb F_q)$ which, if I'm correct, becomes the representation theory of `$S_n$` under that limit. In particular, indeed, vector spaces — the objects on which `$GL_n(\mathbb F_q)$` act — should become sets, the objects on which `$S_n$` acts. Though I'm not an expert on `$\mathbb F_1$`, I've encountered the viewpoint you're referring to. It's not hard to see why people expect `$\mathbb F_1$` to be the universal base: you usually expect that `$\mathop{\text{Spec}} \mathbb F_n \to \mathop{\text{Spec}} \mathbb F_m$` exists iff `$m$` divides `$n$`, so `$\mathop{\text{Spec}} \mathbb F_1$` should be terminal. Of course, things are not literally that simple, e.g. what `$\mathop{\text{Spec}} \mathbb Z \times \mathop{\text{Spec}} \mathbb Z$` over `$\mathop{\text{Spec}} \mathbb F_1$` is? Also, `$\mathop{\text{Spec}} \mathbb Z$` should be thought of the object of dimension 3 (I think it has dualizing complex in degree 3, and primes are similar to 1-dimensional knots), so there's a pressing need for existence of some smaller scheme. That smaller scheme better "have real dimension 1" so we can claim that `$\mathop{\text{Spec}} \mathbb Z$` is an algebraic curve over it. One then tries to compactify it. When you get stuck geometrically, it helps to lose some information and go to zeta functions. There this approach leads to some direct conjectures. The problem with zeta functions is that everything is hard. Still, search for `absolute zeta` and you'll see some real things being discussed. It's been called `absolute motives` in the works I read, I think by Manin, Kontsevich, and some IHES people. The search for `kontsevich absolute motives` brought me up the article that I found a while ago, math/0702206. This is also somehow related to noncommutative geometry of Connes (`endomotives`), in fact I think `$\mathbb F_1$` is one of his favorite ideas (also owes a lot to Soulé and other people of course) The two articles I found to be of special interest are 0806.2401 and 0809.2926. Also, here's a blog post about `$\mathbb F_1$` with link to the introductory paper. In fact there was a whole blog about `$F_{un}$`, which disappeared (articles, unfortunately, also disappeared from my RSS reader). Here's a paper by Connes and Consani that seems to explain most the topics mentioned above: Schemes over F1 and zeta functions, 0903.2024 - Funnily, the answer was converted to community wiki because I (the author!) edited it too many times... – Ilya Nikokoshev Oct 24 2009 at 18:19 Here's probably the simplest manifestation of the field-with-one-element phenomenon. Define a projective n-space of order q to be a collection of points, lines, planes, etc. satisfying the usual incidence relations with the additional condition that every line has q+1 points on it, every plane has q^2+q+1 points on it, and so forth. For q a prime power, all such spaces come from the usual definition of projective n-space Pn(Fq) over a finite field. But a projective n-space of order 1 is precisely the Boolean algebra of subsets of a set with n elements! (This example is due to Henry Cohn, and it has the virtue that any theorems one wants to prove in this abstract setting don't depend on the value of q.) - Thanks, that's very useful. – Benjamin Antieau Oct 24 2009 at 23:49 There was a great series on this on neverending books. Sadly, the spin off website, "F_un mathematics" seems to have disappeared from the web. - Yes, I've been also looking for it in vain... – Ilya Nikokoshev Oct 24 2009 at 16:06 3 It exists! It's just not a field. – Ben Webster♦ Oct 24 2009 at 16:09 1 ...And so the quest for F_un continues... – Ilya Nikokoshev Oct 24 2009 at 17:05 It appears to have resurfaced: matrix.cmi.ua.ac.be/fun – Jon Paprocki Aug 29 2011 at 14:49 "F_un mathematics" spin off exists, it's just isn't a website – Ostap Chervak Jan 19 at 11:42 Not an expert, but the idea is that there should be some "thing" over which we can define functors which, after base change, are schemes. Over this "thing", rings of integers should be literal curves. The big example of things that I know is that a vector space over F_1 (or F_un, if you're the sort who likes fun notation) is a pointed set. For more info, there have been some "This Week's Finds" posts about them, and also this series at neverendingbooks. - One of the many resources pointed out above linked to the unpublished preprint by Kapranov and Smirnov called Cohomology determinants and reciprocity laws: the number field case. It's posted page by page in jpgs, but it is definitely worth looking at. They work out the details of vector spaces over F_1^n in detail and also relate the classical power residue symbol to determinants of morphisms of these vector spaces. - do you have an alternative link for that paper? It doesn't seem to be working for me. – Jacob Bell Oct 4 at 15:57 @Jason: I don't have another link, but if you find my e-mail from UCLA's website and send me a message, I'll send you the paper. – Benjamin Antieau Oct 4 at 21:06 Charles mentioned "This Week's Finds" posts on the field with one element. There are a number of posts, but the following link is probably the best place to start. -	{"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": 0.9481670260429382, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/147/what-are-some-useful-ways-to-imagine-the-concept-of-spin-as-it-relates-to-subato?answertab=oldest	# What are some useful ways to imagine the concept of spin as it relates to subatomic particles? The answers in this question: What is spin as it relates to subatomic particles? do not address some particular questions regarding the concept of spin: How are some useful ways to imagine a particle without dimensions - like an electron - to spin? How are some useful ways to imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: $\Psi \rightarrow -\Psi$). When spin is not a classical property of elementary particles, is it a purely relativistic property, a purely quantum-mechanical property or a mixture of both? - 1 I've voted to close on this - as the question is stated I don't think there will be a single "right answer". Perhaps you might change it to "What are some useful ways of imagining a particle ..." instead of "How should". – j.c. Nov 3 '10 at 14:05 While answering this question I realized that it was not a real, not argumentative question... You should change the question as @j.c. suggested. – Cedric H. Nov 3 '10 at 14:12 Isn't this similar to, e.g. mathematical dimension, which is a generalization of the common concept? It's not fruitful to ask "How should I imagine the fourth (or the eleventh) dimension. Wikipedia states spin did originate from a classical interpretation: "Ralph Kronig, one of Landé's assistants, suggested in early 1925 that it was produced by the self-rotation of the electron. Pauli heard about the idea, he criticized it severely, noting that the electron's hypothetical surface would have to be moving faster than the speed of light..." – recipriversexclusion Oct 11 '11 at 21:22 ## 5 Answers How should one imagine a particle without dimensions - like an electron - to spin? You don't. If you want to imagine, then you think classically and it is just a particle spinning... Thinking like that doesn't give you any other insight of what spin really is (an intrinsic angular momentum, behaving like an [orbital] angular momentum). How should one imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: Ψ→−Ψ) Just imagine it ... no big deal. Again, classically this is not possible, but quantically it is. When spin is not a classical property of elementary particles, is it a purely relativistic property, a purely quantum-mechanical property or a mixture of both? The spin of elementary particle is a pure quantum mechanical effect. Edit: See @j.c. comment. Relativity also plays a role. Any other interpretation/calculation requires things like commutator, symmetry properties and group theory. The parallel between "real spinning" and "spin" (which is just a name) comes from the fact that the spin operator needed to account for properties of elementary particles behaves (= has the same definition, based on commutators) like orbital angular momentum operator. This again comes from symmetry properties of ... nature. The goal of quantum physics is to provide a way to calculate properties. If you want to calculate or go deeper in the problem, then you don't need this classical interpretation. - 1 It's arguable, but relativity does come into the notion of spin. Consider for instance the fact that photons have only two polarization states despite being "spin 1". I don't know of a way of understanding this without using the (relativistic, I'd argue) fact that the symmetry group of the universe is the Poincaré group. (And you could argue that the rest of the observed spins of particles ultimately arise from this symmetry group too). Of course quantum mechanics is essential for all this as well. – j.c. Nov 3 '10 at 14:16 @j.c. OK, I agree with that. – Cedric H. Nov 3 '10 at 14:17 +1 Trying to "imagine" quantum mechanics too deeply will likely lead you astray! Study the fundamentals, the principles, and the maths, and eventually you'll get a holistic picture. – Noldorin Nov 3 '10 at 14:26 1 @Gerard, it's something that's typically covered at the start of a quantum field theory course (meaning of course, that you should have a solid grasp of both before trying to understand the details of the connection). If I had to make it into a slogan though: relativity is one of the fundamental symmetries of the universe; quantum mechanics tells us how symmetries are reflected in properties of quantum states, in particular, an application of the mathematics of representation theory leads to the concept of spin. – j.c. Nov 3 '10 at 14:52 1 Well, originally "spin" was deduced from experiment... but certainly the current theoretical framework for spin needs the Poincaré group. – j.c. Nov 4 '10 at 9:50 show 3 more comments It's possible to do correct quantum mechanics without believing that particles get altered by 360 degree rotations. Use the "density matrix" form instead of "wave function". http://en.wikipedia.org/wiki/Density_matrix To convert a quantum wave state $\psi(x)$ or $\|a\rangle$ to a density matrix, multiply the ket by the bra as in: $\psi(x) \to \rho(x,x') = \psi(x)\psi^*(x')$ $\|a\rangle \to \|a\rangle \langle a\|$ Since the bras and kets take complex phases, i.e. $e^{+i\alpha}\|a\rangle \equiv e^{-i\alpha}\langle a\|$ the complex phases cancel. This is more general than the -1 you get by a 360 degree rotation, but a factor of -1 is also a phase and so it's also canceled. In short while the state vectors or wave functions take a -1 on 360 degree rotation, the (pure) density matrices are left unchanged. - However, relative phases between different pasts of a system will produce physical results. If you rotate one particle through 360$^\circ$, then you can interfere it against a non-rotated particle to get a shifted pattern. The overall density matrix will reflect this: all the (inter-particle) coherences will change sign. – Emilio Pisanty Jun 6 '12 at 20:15 To have an interference requires that the rotated and non rotated particles begin in a coherent state. Then you can use the density matrix form for them, but it has to be the two particle density matrix. So what you're doing is rotating one particle while not rotating the other. The result is indeed observable, one way of describing it is as a quantum phase, a topological phase, or Berry-Pancharatnam phase. But for a single particle, or a pair of particles, or any number of particles, rotating them all (i.e. rotating the system) by 360 degrees is not detectable. – Carl Brannen Jun 7 '12 at 9:40 How are some useful ways to imagine a particle with spin 1/2 to make a 360° turn without returning to it's original position (the wave function transforms as: Ψ→−Ψ). There is a nice example of such an objects -- "the Dirac scissors": The picture is from the book by Penrose and Rindler "Spinors and space-time." I suggest to read it. - There's a paper$^{1}$ by Battey-Pratt and Racey with an intuitive model of spin 1/2. I'm not sure if it related to reality, but is an interesting read and attempt at an intuitive understanding. -- $^{1}$ E.P.Battey-Pratt and T.J.Racey, Geometric model for fundamental particles, International Journal of Theoretical Physics 19 (1980) 437-475. - I find it useful to think about different spaces having different "sizes", such that one complete rotation through space "A" requires only 360 degrees of turning, but one complete rotation through space "B" requires 720 degrees of turning, making space B in some sense "larger" with respect to complete rotations. Spin 1/2 particles live within the larger space B, where 720 degrees is a full rotation. As objects within that space, the spin 1/2 particles have essentially four "sides" to them, one side per 180 degrees. As observers, we live in space A, and every (classical) object around us in space A reveals all of it's sides in a single 360 degree rotation. But when we try to rotate spin 1/2 particles through their four sides, it takes two full rotations of our space to see one complete rotation of their space. The trick is that a "full rotation" must be a fundamentally different concept than an "amount of rotation". Full rotations depend on the space, amounts of rotation are invariant across different spaces. If this explanation makes sense, it is immediately obvious why we shouldn't think of particles as being "dimensionless". In fact, certain aspects of the particle are more dimension-full than the space we are used to thinking about. -	{"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": 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": 0.9187802672386169, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-math-topics/213163-vectors.html	# Thread: 1. ## Vectors! HI I want to check if my answers are right here, I can't do the last part so I believe my answers in the previous parts must be wrong.. Points A,B,C,D have the following coordinates A1,3,1) B1,2,4) C2,3,6) D5,-2,1) ai.) Evaluate the vector product, giving your answers in terms of the unit vectors i,j,k. I found AB and AC and got -5i, 3j and 1k from doing the cross product. aii.) Find the area of the triangle ABC. I took the magnitude of the cross product result in ai.) and halved it, getting (1/2)sqrt(35)units^2 b.) The plane containing the points A,B, C is denoted by pi and the line passing through D perpendicular to pi is denoted by L. The point of intersection of L and pi is denoted by P. bi.) FInd the cartesian equation of pi. AB is the direction vector of pi, I thought. And then I'll need a point so I got x=1, y=3-t, z=1+3t bii.) FInd the cartesian equation of L. The result of the cross product in ai.) is the direction vector of L, I thought. Dotting it by the direction vector of pi, I will get a zero. So my answer was x=5-5t, y=-2+3t, z=1+t c.) Find the coordinates of P. This is where I know my previous answers are wrong. I though to equate all the x, y and z values of L and pi. I get different values for t each time... Please help me out! Thank you! 2. ## Re: Vectors! Did you get a satisfactory answer to your question? I can help you on this - but I'm not going to if you already have a solution. The post has been here a while and I just joined the site. 3. ## Re: Vectors! Originally Posted by Tutu HI I want to check if my answers are right here, I can't do the last part so I believe my answers in the previous parts must be wrong.. Points A,B,C,D have the following coordinates A1,3,1) B1,2,4) C2,3,6) D5,-2,1) ai.) Evaluate the vector product, giving your answers in terms of the unit vectors i,j,k. I found AB and AC and got -5i, 3j and 1k from doing the cross product. aii.) Find the area of the triangle ABC. I took the magnitude of the cross product result in ai.) and halved it, getting (1/2)sqrt(35)units^2 Yes, this is correct. b.) The plane containing the points A,B, C is denoted by pi and the line passing through D perpendicular to pi is denoted by L. The point of intersection of L and pi is denoted by P. bi.) FInd the cartesian equation of pi. AB is the direction vector of pi, I thought. And then I'll need a point so I got x=1, y=3-t, z=1+3t I don't undertand what you are saying here. You were asked for an equation of the plane, pi. This is the parametric equations for a line, not a plane. bii.) FInd the cartesian equation of L. The result of the cross product in ai.) is the direction vector of L, I thought. Dotting it by the direction vector of pi, I will get a zero. So my answer was x=5-5t, y=-2+3t, z=1+t Yes, this is correct. [qote]c.) Find the coordinates of P. This is where I know my previous answers are wrong. I though to equate all the x, y and z values of L and pi. I get different values for t each time... Please help me out! Thank you![/QUOTE] The equation of a plane, with normal vector <A, B, C> and containing point [tex](x_0, y_0, z_0)[tex] is $A(x- x_0)+ B(y- y_0)+ C(z- z_0)= 0$. You can use that to answer (bi) and then replace x, y, and z in the equation of the plane with the x, y, and z from (bii) to get a single equation to solve for t. #### Search Tags Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"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": 1, "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": 0.9261181354522705, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-math-topics/204781-topologie-question.html	# Thread: 1. ## Topologie question Hello, I've two questions: 1) Given X a set then $\mathcal{P}(X)$ is a topology on X off ( $\mathcal{P}(X)$ being the power set of X). Now if I take c and $c \notin X$ is $( \{c \} \bigcup X ) \bigcup \mathcal{P}(X)$ still a topology on X? ( I'd say yes because the 3 axioms work... I think, I might be wrong...). 2) Let $\{ \mathcal{T}_{\alpha}\}$ be a family of topologies on X. Show that there is a unique smallest topology on X containing all the collections $\mathcal{T}_{\alpha}$. for 2) I thought I could take the subbase $\bigcup_{ \alpha \in A} \mathcal{T}_{\alpha}$. (Where A is a set), but if I'm right in 1) then I think there is a problem bc nothing tells me that none of the $\mathcal{T}_{\alpha}$ is not of the form of what I wrote in 1) so then it wouldn't be a subbase anymore. So can someone give me a hit for 2) (or write the whole thing if you have time). thanks in advance!!	{"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": 8, "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": 0.9452676773071289, "perplexity_flag": "head"}
http://nrich.maths.org/6453/index?nomenu=1	## 'Global Warming' printed from http://nrich.maths.org/ ### Show menu In this investigation you will need to use a variety of pieces of physical data to come up with a reasonable estimate. What data will you need? Where will you find it? What estimates are sensible? What modelling assumptions are you making at each step? These are the sorts of questions that you will need to ask yourself as you undertake your investigation . As a conservative estimate, the average temperature of the atmosphere has increased by $0.4^\circ\mathrm{C}$ over the last thirty years. Estimate how much energy has gone into warming up the planet in this way. Estimate how much burned fuel would be needed to give rise to this increase on the assumption that this were the only cause of changes in the earths temperature. How much fuel burned per week over the last thirty years per person on the planet would this correspond to? What do you think about your answer? How does the amount of burned fuel correspond to the levels of fuel actually used? What other factors would come into a more sophisticated analysis of global warming?	{"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": 0.9595425128936768, "perplexity_flag": "middle"}
http://tcsmath.wordpress.com/tag/integrality-gaps/	# tcs math – some mathematics of theoretical computer science ## February 23, 2011 ### PSD lifting and Unique Games integrality gaps Filed under: Math, Open question — Tags: Integrality gaps, semi-definite programming, unique games — James Lee @ 9:51 am By now, it is known that integrality gaps for the standard Unique Games SDP (see the paper of Khot and Vishnoi or Section 5.2 of this post) can be used to obtain integrality gaps for many other optimization problems, and often for very strong SDPs coming from various methods of SDP tightening; see, for instance, the paper of Raghavendra and Steurer. Problematically, the Khot-Vishnoi gap is rather inefficient: To achieve the optimal gap for Unique Games with alphabet size ${L}$, one needs an instance of size ${\exp(\Omega(L))}$. As far as I know, there is no obstacle to achieving a gap instance where the number of variables is only ${\mathrm{poly}(L)}$. The Walsh-Hadamard code The Khot-Vishnoi construction is based on the Hadamard code. (See Section 5.2 here for a complete description.) If we use ${L^2(\{-1,1\}^k)}$ to denote the Hilbert space of real-valued functions ${f : \{-1,1\}^k \rightarrow \mathbb R}$, then the Walsh-Hadamard basis of ${L^2(\{-1,1\}^k))}$ is the set of functions of the form $\displaystyle u_S(x) = \prod_{i \in S} x_i,$ where ${S \subseteq \{1,2,\ldots,k\}}$. Of course, for two such sets ${S \neq T}$, we have the orthogonality relations, $\displaystyle \langle u_S, u_T \rangle = 0.$ In their construction, the variables are essentially all functions of the form ${f : \{-1,1\}^k \rightarrow \{-1,1\}}$, of which there are ${2^{2^k}}$, while there are only ${2^k}$ basis elements ${\{u_S\}_{S \subseteq [k]}}$ which act as the alphabet for the underlying Unique Games instance. This is what leads to the exponential relationship between the number of variables and the label size. A PSD lifting question In an effort to improve this dependence, one could start with a much larger set of nearly orthogonal vectors, and then somehow lift them to a higher-dimensional space where they would become orthogonal. In order for the value of the SDP not to blow up, it would be necessary that this map has some kind of Lipschitz property. We are thus led to the following (possibly naïve) question. Let ${C(d,\varepsilon)}$ be the smallest number such that the following holds. (Here, ${S^{d-1} \subseteq \mathbb R^d}$ denotes the ${(d-1)}$-dimensional unit sphere and $S(L^2)$ denotes the unit-sphere of $L^2$.) There exists a map ${F : S^{d-1} \rightarrow S(L^2)}$ such that ${\\|F\\|_{\mathrm{Lip}} \leq C(d,\varepsilon)}$ and whenever ${u,v \in \mathbb R^d}$ satisfy ${\|\langle u,v\rangle\| \leq \varepsilon}$, we have ${\langle F(u), F(v)\rangle = 0}$. (Recall that $\\|F\\|_{\mathrm{Lip}} = \sup_{x \neq y \in S^{d-1}} \\|F(x)-F(y)\\|/\\|x-y\\|$.) One can show that $\displaystyle C(d,\varepsilon) \lesssim \frac{\sqrt{d}}{1-\varepsilon}$ by randomly partitioning ${S^{d-1}}$ so that all vectors satisfying ${\|\langle u,v\rangle\| \leq \varepsilon}$ end up in different sets of the partition, and then mapping all the points in a set to a different orthogonal vector. My question is simply: Is a better dependence on ${d}$ possible? Can one rule out that ${C(d,\varepsilon)}$ could be independent of ${d}$? Note that any solution which randomly maps points to orthogonal vectors must incur such a blowup (this is essentially rounding the SDP to an integral solution). Theme: Shocking Blue Green. Blog at WordPress.com.	{"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": 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": 0.9244310855865479, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=97320	Physics Forums ## Diffraction grating I did a physics lab about diffraction grating where we had to determine the angular positions of all spectral lines of the helium discharge tube. I have 3 questions : 1. Why use a discharge tube instead of a light bulb? - I think it might be because you would have to break up the white light of the light bulb with a prism but im not sure if that's right. 2. Define order of diffraction and illustrate it. -I said that the order of diffraction is an integral multiple of wavelengths that allow for constructive interference. ( Im not sure if destructive interference comes in to play here ) 3. If "d" is spacing between 2 slits on a grating, how can we obtain grating constant? - I calculated the values of d but i don't know what grating constant is? I was thinking the number of slits in a certain area but I can't find any info anywhere? Thank you. 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 the helium is used instead of a light bulb because a certain element only emits certain wavelengths of light while the bulb would emit all wavelengths, and even if you were to split this with a prism you would still see all the colors and not specific ones. the order of diffraction is correct by what you said, since the n in the equation $$d sin\theta = n\lambda$$ refers to the order of diffraction, where n=1 is the first, n=2 is the second and so on. the grating constant k, is simply 1/d. d is the distance between slits and k, is the number of slits per unit length. I'd like to add something in regards to #1, if you look at the grating equation, you'll realize that different wavelengths can actually have the same diffraction angle, given thier diffraction orders. For example, a 600nm first order ray will defract in the same angle as a 300nm second order ray. This is a common problem with some gratings when trying to quantify the intensities, though I'm not sure how messy it would look in the experiment. As it has already been stated, the Helium wavelengths are spaced well apart. So in a matter of speaking, the Helium pattern will be more "cleaner" than the lightbulb pattern. Thread Tools Similar Threads for: Diffraction grating Thread Forum Replies Introductory Physics Homework 2 Introductory Physics Homework 1 Introductory Physics Homework 3 Introductory Physics Homework 3 General Physics 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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9348084330558777, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Phase_velocity	# Phase velocity Frequency dispersion in groups of gravity waves on the surface of deep water. The red dot moves with the phase velocity, and the green dots propagate with the group velocity. In this deep-water case, the phase velocity is twice the group velocity. The red dot overtakes two green dots when moving from the left to the right of the figure. New waves seem to emerge at the back of a wave group, grow in amplitude until they are at the center of the group, and vanish at the wave group front. For surface gravity waves, the water particle velocities are much smaller than the phase velocity, in most cases. This shows a wave with the group velocity and phase velocity going in different directions. The group velocity is positive, while the phase velocity is negative. The phase velocity of a wave is the rate at which the phase of the wave propagates in space. This is the velocity at which the phase of any one frequency component of the wave travels. For such a component, any given phase of the wave (for example, the crest) will appear to travel at the phase velocity. The phase velocity is given in terms of the wavelength λ (lambda) and period T as $v_\mathrm{p} = \frac{\lambda}{T}.$ Or, equivalently, in terms of the wave's angular frequency ω, which specifies the number of oscillations per unit of time, and wavenumber k, which specifies the number of oscillations per unit of space, by $v_\mathrm{p} = \frac{\omega}{k}.$ To understand where this equation comes from, imagine a basic sine wave, A cos (kx−ωt). Given time t, the source produces ωt oscillations. At the same time, the initial wave front propagates away from the source through the space to the distance x to fit the same amount of oscillations, kx = ωt. So that the propagation velocity v is v = x/t = ω/k. The wave propagates faster when higher frequency oscillations are distributed less densely in space.[1] Formally, Φ = kx−ωt is the phase. Since ω = −dΦ/dt and k = +dΦ/dx, the wave velocity is v = dx/dt = ω/k. ## Relation to group velocity, refractive index and transmission speed Since a pure sine wave cannot convey any information, some change in amplitude or frequency, known as modulation, is required. By combining two sines with slightly different frequencies and wavelengths, $\cos[(k-\Delta k)x-(\omega-\Delta\omega)t]\; +\; \cos[(k+\Delta k)x-(\omega+\Delta\omega)t] = 2\; \cos(\Delta kx-\Delta\omega t)\; \cos(kx-\omega t),$ the amplitude becomes a sinusoid with phase speed of vg = Δω/Δk. It is this modulation that represents the signal content. Since each amplitude envelope contains a group of internal waves, this speed is usually called the group velocity.[1] In reality, the vp = ω/k and vg = dω/dk ratios are determined by the media. The relation between phase speed, vp, and speed of light, c, is known as refractive index, n = c/vp = ck/ω. Taking the derivative of ω = ck/n, we get the group speed, $\frac{\text{d}\omega}{\text{d}k} = \frac{c}{n} - \frac{ck}{n^2}\cdot\frac{\text{d}n}{\text{d}k}.$ Noting that c/n = vp, this shows that group speed is equal to phase speed only when the refractive index is a constant: dn/dk = 0.[1] Otherwise, when the phase velocity varies with frequency, velocities differ and the medium is called dispersive and the function, $\omega(k)$, from which the group velocity is derived is known as a dispersion relation. The phase velocity of electromagnetic radiation may – under certain circumstances (for example anomalous dispersion) – exceed the speed of light in a vacuum, but this does not indicate any superluminal information or energy transfer. It was theoretically described by physicists such as Arnold Sommerfeld and Léon Brillouin. See dispersion for a full discussion of wave velocities. ## Matter wave phase In quantum mechanics, particles also behave as waves with complex phases. By the de Broglie hypothesis, we see that $v_\mathrm{p} = \frac{\omega}{k} = \frac{E/\hbar}{p/\hbar} = \frac{E}{p}.$ Using relativistic relations for energy and momentum, we have $v_\mathrm{p} = \frac{E}{p} = \frac{\gamma m c^2}{\gamma m v} = \frac{c^2}{v} = \frac{c}{\beta}$ where E is the total energy of the particle (i.e. rest energy plus kinetic energy in kinematic sense), p the momentum, $\gamma$ the Lorentz factor, c the speed of light, and β the speed as a fraction of c. The variable v can either be taken to be the speed of the particle or the group velocity of the corresponding matter wave. Since the particle speed $v < c$ for any particle that has mass (according to special relativity), the phase velocity of matter waves always exceeds c, i.e. $v_\mathrm{p} > c, \,$ and as we can see, it approaches c when the particle speed is in the relativistic range. The superluminal phase velocity does not violate special relativity, as it carries no information. See the article on signal velocity for details. ## References ### Footnotes 1. ^ a b c "Phase, Group, and Signal Velocity". Mathpages.com. Retrieved 2011-07-24. ### Other • Brillouin, Léon (1960), Wave Propagation And Group Velocity, New York and London: Academic Press Inc., ISBN 0-12-134968-3 • Main, Iain G. (1988), Vibrations and Waves in Physics (2nd ed.), New York: Cambridge University Press, pp. 214–216, ISBN 0-521-27846-5 • Tipler, Paul A.; Llewellyn, Ralph A. (2003), Modern Physics (4th ed.), New York: W. H. Freeman and Company, pp. 222–223, ISBN 0-7167-4345-0	{"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": 0, "mathjax_asciimath": 0, "img_math": 10, "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": 0.8781753778457642, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/163532/unique-decomposition-of-primes-in-sums-of-higher-powers-than-2	# Unique Decomposition of Primes in Sums Of Higher Powers than $2$ Primes of the form $p=4k+1\;$ have a unique decomposition as sum of squares $p=a^2+b^2$ with $0<a<b\;$, due to Thue's Lemma. What is known about sums of $n$ higher powers resulting in primes? I tried $a^3+b^3+c^3$, asked Wolfram and found $3,17,27,43,73$, a sequence that I don't know (but Matthew in the comment below). Interestingly, that $251$ has even $2$ decompositions, $1^3+5^3+5^3=2^3+3^3+6^3$. Can anybody help here? - By the solution of Waring's Problem, for any exponent $e$ there is an $n=n(e)$ such that every positive integer is a sum of $n$ (positive) $e$-th powers. The case $e=3$ gets very interesting if we allow some of $a$, $b$, $c$ to be negative. – André Nicolas Jun 26 '12 at 22:10 Huh, I didn't think about this, but I didn't exclude it. – draks ... Jun 26 '12 at 22:11 1 I believe the next case with more than one decomposition is $1009 = 4^3 + 6^3 + 9^3 = 1^3 + 2^3 + 10^3$, and the first with three is $46747=11^3+25^3+31^3 = 4^3 + 27^3 + 30^3 = 3^3 + 4^3 + 36^3$. – Robert Israel Jun 26 '12 at 22:59 1 The most decompositions I have found are 6 each for $3380833$ and $5377373$ – Robert Israel Jun 26 '12 at 23:00 3 There are ${N \choose 3} \approx N^3/6$ triples of positive integers up to $N$, so the expected number of these whose sum of cubes is a given positive integer up to $N^3$ is approximately constant. Heuristically, we might use a Poisson distribution to model the number of decompositions. So there should be examples (increasingly rare as $k$ increases) with any number $k$ of decompositions. – Robert Israel Jun 26 '12 at 23:16 show 5 more comments ## 1 Answer Any cube is congruent to $0$, $1$, or $-1$ modulo $9$. It follows that the sum of three cubes cannot be congruent to $4$ or $5$ modulo $9$. So we have a congruential restriction analogous to the case of two squares. By Dirichlet's Theorem on primes in arithmetic progressions (or undoubtedly by much more elementary means) one can show that there are infinitely many primes congruent to $4$ modulo $9$, and also infinitely many primes congruent to $5$. Thus there are infinitely many primes that cannot be represented as the sum of three cubes. About primes not congruent to $4$ or $5$ modulo $9$, one cannot say much. A long-standing conjecture is that every integer which is not congruent to $4$ or $5$ modulo $9$ is the sum of $3$ cubes. (Here negative cubes are allowed.) There has been a lot of computation on this problem. For other powers, one should go to the vast literature on Waring's Problem. -	{"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": 44, "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": 0.9450239539146423, "perplexity_flag": "head"}
http://mathhelpforum.com/pre-calculus/30419-cordinate-geometry.html	Thread: 1. Coordinate geometry Dear forum members, could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though. Problem Find the values for k for which the lines $2x+ky-11=0$ and $(3-k)x+y+2=0$are perpendicular. My solution $-3x+kx= \frac{k}{2x}$ -3x+kx and 2x have to be the same so $-3x+kx= 2x$ I always keep getting 5 or 1 as the answer for k, is it correct? Thank you in advance! 2. Originally Posted by Coach Dear forum memebers, could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though. Problem Find the values for k for which the lines $2x+ky-11=0$ and $(3-k)x+y+2=0$are perpendicular. My solution $-3x+kx= //frac{k}{2x}$ the bottom numbers have to be the same so $-3x+kx= 2x$ I always keep getting 5 or 1 as the anwer for k, is it correct? Thank you in advance! Two lines are perpendicular if the product of their gradients is equal to -1. $2x+ky-11=0 \Rightarrow y = \frac{-2x+11}{k} \Rightarrow m = -\frac{2}{k}$. $(3-k)x+y+2=0 \Rightarrow y = -(3 - k)x - 2 = (k - 3)x - 2 \Rightarrow m = k - 3$. Therefore you require $\left( -\frac{2}{k} \right) \, (k-3) = -1 \Rightarrow -2(k-3) = -k \Rightarrow k = 6$. 3. Thank you so much! I would not have been able to do that on my own. 4. Originally Posted by Coach Dear forum members, could you please check if my calculations are correct, I have tried several times, I alway keep getting the wrong answer though. Problem Find the values for k for which the lines $2x+ky-11=0$ and $(3-k)x+y+2=0$are perpendicular. I always keep getting 5 or 1 as the answer for k, is it correct? Thank you in advance! I just realised that if you kept getting 5 plus 1 as the answer for k it would be correct 5. Originally Posted by Coach Thank you so much! I would not have been able to do that on my own. I'm sure you would You've just got to work through each link in the chain.	{"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": 13, "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": 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": 0.9599396586418152, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/87167-solved-prove-polynomials-generating-system-vector-spaces.html	# Thread: 1. ## [SOLVED] Prove polynomials are a generating system (vector spaces) Let V be the vector space of the polynomials of degree $\leq$ 2 over $\mathbb{R}$. Prove that the polynomials: $p_1 = 2T^2 + T + 1$ $p_2 = 4T^2 + T$ $p_3 = -2T^2 + 2T + 1$ span, or are a generating system of, V. If I solve the system of linear equations- how do I prove that they span V? Thanks. 2. Originally Posted by bmp05 Let V be the vector space of the polynomials of degree $\leq$ 2 over $\mathbb{R}$. Prove that the polynomials: $p_1 = 2T^2 + T + 1$ $p_2 = 4T^2 + T$ $p_3 = -2T^2 + 2T + 1$ . You just need to show these polynomial are linearly independent. Say that $a_1p_1 + a_2p_2+a_3p_3 = \bold{0}$ where $\bold{0}$ is the zero-polynomial. This means that, $2a_1+4a_2-2a_3=0$ $a_1 + \ \ \ a_2 + 2a_3 = 0$ $a_1 + \ \ \ \ \ \ \ \ \ + a_3 = 0$ Argue that $a_1=a_2=a_3=0$ is the only solution. 3. Write the polynomials as vectors in $\mathbb{R}^{3}$ . $A= \left[\begin{matrix} 1 & 0 & 1\\ 1 & 1 & 2 \\ 2 & 4 & -2 \end{matrix}\right]$ Row operations will show that this matrix A has 3 pivot positions. So it therefore spans $\mathbb{R}^{3}$ and by the isomorphism between $\mathbb{R}^{3} \mbox{ and } \mathbb{P}_{3}$ also $\mathbb{P}_{3}$ .	{"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": 21, "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": 0.9229060411453247, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/111622?sort=votes	## Background for Hejhal’s "The Selberg Trace Formula for $PSL(2, \mathbb{R})$ ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Reposted from math.stackexchange where my question received only five views and no answers... I'm trying to learn the Selberg trace formula, but have very little background in harmonic analysis. I was referred to Dennis Hejhal's The Selberg Trace Formula for $PSL(2, \mathbb{R})$ but just got the book and was dismayed to learn that that the author assumes familiarity with Selberg's original paper (which I don't have access to - would welcome a pointer to an online copy). There's much in the first few pages that I don't know. For example, the author states without proof that the spectrum of the Laplacian on a compact hyperbolic surface is discrete. He gives a reference to a 1912 book by Hilbert, but aside from the fact that I don't read German, it's not clear to me that this is the best place to learn from (in light of the fact that Hejhal's book is from 1976 and many books have been written since). Does anyone have a suggestion as to what to read before Hejhal's book? - publications.ias.edu/selberg – Chandan Singh Dalawat Nov 6 at 6:18 ## 5 Answers Well, the modern viewpoint relays on the interpretation of "fourier transform" (in any generalized fashion you like to define "fourier transform") in representation-theoretic language. As a consequence, there are several approaches today to get the trace formula (either more analytic by Green's functions or the more general representation theoretic manner). A nice introductory account can be found here by Marklof - http://arxiv.org/pdf/math/0407288v2.pdf Another representation-free approach is done in Iwaniec's "spectral methods of automorphic forms". The theorem that Hejhal mentioned is very well known for general (compact, closed) manifolds (follows from Poincare inequality), but in the (cocompact) homogeneous case, one can overcome many analytical complications by just mimicking the proof of the Peter-Weyl theorem in representation theory of compact groups (Hilbert-Schmidt operators and so on). In particular, no Sobolev computations whatsoever, that shows one simple example of the advantages of using representation theory. A more advanced approach (which uses some representation theory) is found in Knapp's article - http://sporadic.stanford.edu/bump/match/trace.pdf Probably a good introduction to this article is Bump's book about automorphic representations (chapters 1-2 I guess, you only need the real part for this article). - ### 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. For books I recommend Audrey Terras' "Harmonic Analysis on Symmetric Spaces and Applications, I." It begins with Fourier analysis on $\mathbb R^m$, includes automorphic forms both Maass and classical, and finishes with the Selberg Trace formula. The orientation is towards number theory, but there are lots of applications and extensive references to the literature. Also lots of exercises. For an expository paper, I recommend H.P. McKean "Selberg's trace formula applied to a compact Riemann surface" in Comm. Pure & Appl. Math., v. 25 (1975), 225-246; with errata in v. 27 (1974) p. 134. Both of these are older than the more modern references in the other answers, but closer perhaps in style to Hejhal's approach. - Deitmar and Echterhoff "Principles in Harmonic Analysis" Chapter 9 and 11 for the cocompact case (Hejhal I). It requires some familiarity with representation theory, but you seem to be more interested in a conceptual approach. It is the only source which really made me understand the underlying concepts of the trace formula. In chapter 9, the main result is as follows: Given a cocompact subgroup $\Gamma$ in a unimodular group $G$, convolution operators $T_\phi$ for functions $\phi \in C_c^\infty(G)$ on $L^2( \Gamma\backslash G)$ are trace class operators. The formula for the trace is given as $$tr T_\phi = \sum\limits_{conj.classes in \Gamma} vol(\Gamma_\gamma \backslash G_\gamma) \int\limits_{G_\gamma \backslash G} \phi(g \gamma g^{-1}) \,d\,g.$$ The space $L^2( \Gamma\backslash G)$ decomposes of course into irreducible reps. It is a nice exercise to deduce the Poisson summation formula from this. In chapter 11, they specialize this to the situation $G=SL(2,\mathbb{R})$ and $\Gamma$ has only hyperbolic elements. The non-compact situation is more difficult (Hejhal II). The main idea (working with $G$) is mostly hidden in Selberg's and Hejhal's presentation, and the underlying computations with special functions can be avoided until a certain point (Chapter 11 is the analogue). This is where these guys start. Iwaniec "Spectral methods" is pretty close to Selberg's Göttingen lecture notes. Hejhal and Selberg are in my opinion a terrible point to enter the subject for an algebraic person. Hejhal remains the most important reference for researchers. Similar route is done for quadratic imaginary fields in Elstrodt, Grunewald, Mennicke "Hyperbolic Groups acting...". It will be probably more useful to understand the Arthur trace formula as presented in Jacquet-Gelbart "Analytic aspects...." given your background. Knightly and Li "Trace of Hecke operators" are useful here for the Hecke eigenvalues, and the lecture notes by Gelbart for the general theory. My thesis might be interesting to you, because I generalize the computations of Hejhal II/Knightly-Li to the number field case and "arbitrary" congruence subgroups by using Arthur's trace formula and the adelic framework: http://webdoc.sub.gwdg.de/diss/2012/palm/. As an example, I derive the Selberg trace formula from the Arthur trace formula there. - To address the narrow question of the discreteness of the spectrum of the laplacian on a compact riemannian manifold: this is part of the elliptic package, which is most of the proof of the Hodge theorem and is proved along the way in many books on differential geometry or topology. I find these treatments too ad hoc. The introduction I greatly prefer is Zimmer's Essential Results of Functional Analysis. This has just enough analytic theory for the geometric applications, which, in turn, helps motivate the theory. The only drawback is that since it really is an analysis book, it leaves the case of compact manifolds as an exercise. - When I tried to read Hejhal's books around 1983, I enjoyed the book $SL_2(\mathbb{R})$ (where $SL$ stands for Serge Lang) for background. -	{"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": 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": 0.932544469833374, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/17366/list	## Return to Answer 2 Added $v\wedge w$. $V \wedge V$ is bad notation when $V$ is a representation, just as $V~ Sym ~V$ would be. $\wedge^2 V$ is less misleading. You could try to define $V \wedge W = \wedge^2 (V \oplus W)$. This For $v\in V, w\in W$, we can naturally identify $v\wedge w$ and $w \wedge v$ with elements of $\wedge^2 (V \oplus W)$, and $v\wedge w = - w \wedge v$.This has some nice properties, perhaps too trivially, but then be careful that $V\wedge V \ne \wedge^2 V.$ $\wedge^2(V\oplus W) = \wedge^2V \oplus \wedge^2 W \oplus V \otimes W.$ $Sym^2(V\oplus W) = Sym^2 V \oplus Sym^2 W \oplus V \otimes W.$ 1 $V \wedge V$ is bad notation when $V$ is a representation, just as $V~ Sym ~V$ would be. $\wedge^2 V$ is less misleading. You could try to define $V \wedge W = \wedge^2 (V \oplus W)$. This has some nice properties, perhaps too trivially, but then $V\wedge V \ne \wedge^2 V.$ $\wedge^2(V\oplus W) = \wedge^2V \oplus \wedge^2 W \oplus V \otimes W.$ $Sym^2(V\oplus W) = Sym^2 V \oplus Sym^2 W \oplus V \otimes W.$	{"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": 22, "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": 0.9829215407371521, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2010/12/06/inducing-the-trivial-representation/?like=1&_wpnonce=34114a6dfa	# The Unapologetic Mathematician ## Inducing the Trivial Representation We really should see an example of inducing a representation. One example we’ll find extremely useful is when we start with the trivial representation. So, let $G$ be a group and $H$ be a subgroup. Since this will be coming up a bunch, let’s just start writing $1$ for the trivial representation that sends each element of $H$ to the $1\times1$ matrix $\begin{pmatrix}1\end{pmatrix}$. We want to consider the induced representation $1\!\!\uparrow_H^G$. Well, we have a matrix representation, so we look at the induced matrix representation. We have to pick a transversal $\{t_i\}$ for the subgroup $H$ in $G$. Then we have the induced matrix in block form: $\displaystyle1\!\!\uparrow_H^G(g)=\begin{pmatrix}1(t_1^{-1}gt_1)&1(t_1^{-1}gt_2)&\cdots&1(t_1^{-1}gt_n)\\1(t_2^{-1}gt_1)&1(t_2^{-1}gt_2)&\cdots&1(t_2^{-1}gt_n)\\\vdots&\vdots&\ddots&\vdots\\1(t_n^{-1}gt_1)&1(t_n^{-1}gt_2)&\cdots&1(t_n^{-1}gt_n)\end{pmatrix}$ In this case, each “block” is just a number, and it’s either $1$ or $0$, depending on whether $t_i^{-1}gt_j$ is in $H$ or not. But if $t_i^{-1}gt_j\in H$, then $t_i^{-1}gt_jH=H, and$latex g(t_jH)=(t_iH)\$. That is, this is exactly the coset representation of $G$ corresponding to $H$. And so all of these coset representations arise as induced representations. ## 1 Comment » 1. [...] now we can define the -module by inducing the trivial representation from the subgroup to all of . Now, the are not all irreducible, but we will see how to identify a [...] Pingback by \| December 7, 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.	{"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": 19, "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": 0.8927940130233765, "perplexity_flag": "head"}
http://mathoverflow.net/questions/86471?sort=votes	## Can every finite poset be realized as divisors of an algebraic curve? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $D_1$, ... , $D_n$ be a finite set of divisor classes on a nonsingular projective irreducible algebraic curve. We say that $D_1\geq D_n$ if the line bundle defined by $D_1-D_n$ has a section. This obviously satisfies the axioms of a partial order. Suppose ${x_1,....,x_n}$ is a finite partially ordered set. Does there exist a (projective, nonsingular) algebraic curve of sufficiently high genus, and a set of divisors on it, that are isomorphic as a partially ordered set to ${x_1,...,x_n}$? - Remark: this is only an order on divisor classes (assuming the curve to be projective and irreducible). – Laurent Moret-Bailly Jan 25 2012 at 8:06 ah, good point. I'll fix the notation. – Will Sawin Jan 25 2012 at 19:43 ## 1 Answer Choosing a general set of $n$ points on a curve of genus at least $n$, you can assume that the divisor they define has a unique global section. Let $P$ denote the poset generated by all the possible sums with coefficients $0,1$ of these $n$ points. This poset is the same as the poset of subsets of an $n$-element set. Since every finite poset seems to be a subposet of the poset of subsets of a finite set [EDIT: this is true-see the comments below], just embed your poset in a "power set poset" and remove the unwanted divisors, to deduce that what you want is true. - 3 To replace "seems to be", note that you can just map each thing to the set of things it's bigger than or equal to. – Will Sawin Jan 23 2012 at 20:23 4 Note, furthermore, that the result to which Will alludes is a special case of the Yoneda lemma (applied to the poset viewed as a category, where $a \le b$ means that there is a single arrow $a \to b$). – Qiaochu Yuan Jan 24 2012 at 16:59	{"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": 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": 0.9190061092376709, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2011/07/15/the-exterior-derivative/?like=1&source=post_flair&_wpnonce=55e0e6033e	# The Unapologetic Mathematician ## The Exterior Derivative The Lie derivative looks sort of familiar as a derivative, but we have another sort of derivative on the algebra of differential forms: the “exterior derivative”. But this one doesn’t really look like a derivative at first, since we’ll define it with some algebraic manipulations. If $\omega$ is a $k$-form then $d\omega$ is a $k+1$-form, defined by $\displaystyle\begin{aligned}d\omega(X_0,\dots,X_k)=&\sum\limits_{i=0}^k(-1)^iX_i\left(\omega(X_0,\dots,\hat{X_i},\dots,X_k)\right)\\&+\sum\limits_{0\leq i<j\leq k}(-1)^{i+j}\omega\left([X_i,X_j],X_0,\dots,\hat{X_i},\dots,\hat{X_j},\dots,X_k\right)\end{aligned}$ where a hat over a vector field means we leave it out of the list. There’s a lot going on here: first we take each vector field $X_i$ out of the list, evaluate $\omega$ on the $k$ remaining vector fields, and apply $X_i$ to the resulting function. Moving $X_i$ to the front entails moving it past $i$ other vector fields in the list, which gives us a factor of $(-1)^i$, so we include that before adding the results all up. Then, for each pair of vector fields $X_i$ and $X_j$, we remove both from the list, take their bracket, and stick that at the head of the list before applying $\omega$. This time we apply a factor of $(-1)^{i+j}$ before adding the results all up, and add this sum to the previous sum. Wow, that’s really sort of odd, and there’s not much reason to believe that this has anything to do with differentiation! Well, the one hint is that we’re applying $X_i$ to a function, which is a sort of differential operator. In fact, let’s look at what happens for a $0$-form — a function $f$: $\displaystyle df(X)=X(f)$ That is, $df$ is the $1$-form that takes a vector field $X$ and evaluates it on the function $f$. And this is just like the differential of a multivariable function: a new function that takes a point and a vector at that point and gives a number out measuring the derivative of the function in that direction through that point. As a more detailed example, what if $\omega$ is a $1$-form? $\displaystyle d\omega(X,Y)=X\left(\omega(Y)\right)-Y\left(\omega(X)\right)-\omega\left([X,Y]\right)$ We’ve got two terms that look like we’re taking some sort of derivative, and one extra term that we can’t quite explain yet. But it will become clear how useful this is soon enough. ### Like this: Posted by John Armstrong \| Differential Topology, Topology ## 9 Comments » 1. [...] further make our case that the exterior derivative deserves its name, I say it’s a derivation of the algebra . But since it takes -forms and [...] Pingback by \| July 16, 2011 \| Reply 2. [...] extremely important property of the exterior derivative is that for all exterior forms . This is only slightly less messy to prove than the fact that is [...] Pingback by \| July 19, 2011 \| Reply 3. [...] turns out that our exterior derivative is uniquely characterized by some of its properties; it is the only derivation of the algebra of [...] Pingback by \| July 19, 2011 \| Reply 4. [...] really important thing about the exterior derivative is that it makes the algebra of differential forms into a “differential graded [...] Pingback by \| July 20, 2011 \| Reply 5. [...] turns out that there is a fantastic relationship between the interior product, the exterior derivative, and the Lie [...] Pingback by \| July 26, 2011 \| Reply 6. [...] formula in hand we can show that the Lie derivative is a chain map . That is, it commutes with the exterior derivative. And indeed, it’s easy to [...] Pingback by \| July 28, 2011 \| Reply 7. [...] . Of course we’ll really start with a -form instead of a vector field, and we already know a differential operator to use on forms. Given a -form we can send it to [...] Pingback by \| October 12, 2011 \| Reply 8. [...] is in terms of differential forms. See, if we take our vector field and consider it as a -form, the exterior derivative is already known to be (essentially) the curl. So what else can we [...] Pingback by \| October 13, 2011 \| Reply 9. [...] not the only 2-category around. The algebra of differential forms — together with the exterior derivative — gives us a chain complex. Since pullbacks of differential forms commute with the exterior [...] Pingback by \| December 2, 2011 \| 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": 27, "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": 0.9456517100334167, "perplexity_flag": "head"}
http://nrich.maths.org/1955	### Logosquares Ten squares form regular rings either with adjacent or opposite vertices touching. Calculate the inner and outer radii of the rings that surround the squares. ### Shape and Territory If for any triangle ABC tan(A - B) + tan(B - C) + tan(C - A) = 0 what can you say about the triangle? ### So Big One side of a triangle is divided into segments of length a and b by the inscribed circle, with radius r. Prove that the area is: abr(a+b)/ab-r^2 # Three by One ##### Stage: 5 Challenge Level: $ABCD$ is a rectangle where $BC$ = $3AB$. $P$ and $Q$ are points on $BC$ such that $BP$ = $PQ$ = $QC$. Show that: angle $DBC$ + angle $DPC$ = angle $DQC$ Generalise this result. N.B. This problem can be tackled in at least 8 different ways using different mathematics learnt in the last two years in school and earlier. The methods are essentially the same when viewed from a more advanced perspective. 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": 12, "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": 0.9102007746696472, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/5143?sort=oldest	## Pushouts in the Category of Schemes ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) When does it make sense to glue schemes together along subschemes? In particular: is there a way to glue two schemes together along a closed point (say we're working over a field)? Can you glue two closed points of the same scheme together? Is it easier to glue in the category of algebraic spaces? - ## 4 Answers Given schemes $X,Y$ and $Z$ such that $Z$ is a closed subscheme of both $X$ and $Y$ the pushout exists in the category of schemes. So in particular one can glue schemes along a closed point. A reference for this (carried out via the category of locally ringed spaces) is given in this paper of Schwede (Corollary 3.9). In general though the pushout in the category of locally ringed spaces need not be a scheme even if one pushes out along a subscheme - see for instance Example 3.3 in Schwede's paper. - Wow.. interesting work there by Schwede! :) – Jose Capco Nov 12 2009 at 7:29 1 Schwede's Example 3.3 doesn't really show that the coproduct doesn't exist in the scheme category; just that the coproduct in the ringed space category isn't a scheme. – Andrew Critch Nov 12 2009 at 7:40 Fixed this. Do you know if the pushout in the category of sheaves is representable in this example? I can think of plenty of non-representable pushouts but none I am sure about along a subscheme. – Greg Stevenson Nov 12 2009 at 7:45 2 I'm sure I'm missing something, but: how does your first sentence imply that "in particular one can ... glue a scheme to itself along a pair of closed points"? Example I have in mind that gives me pause without being a counterexample to your statement: if you try to glue an elliptic curves times the $\mathbb{P}^1$ (over $\mathbb{C}$) to itself, by gluing one fiber over $\mathbb{P}^1$ to another, not by the identity but by a nontorsion element, you just get an algebraic space that is not a scheme. – Ravi Vakil Apr 8 2010 at 5:19 1 Ravi, you are absolutely right. I assume the author meant just choose two closed points, and glue them (but not glue other things), which is true say for quasi-projective schemes I think. You should always be able to do this assuming both points can be represented as points in the same affine patch (if not more generally). Hm, if you have two points which can't live on the same affine patch due to failure of separatedness, I'm not sure what happens if you try to glue them. – Karl Schwede Apr 16 2010 at 16:33 show 3 more comments ### 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. Consider a commutative local ring R, say a valuation domain, with maximal ideal M. Consider the fiber product $R \times_M R$ (I wrote M instead of R/M), coming from the pullback in commutative rings $R\rightarrow R/M$. Then the corresponding prime spectra of this fibered product (in rings) is actually a form of gluing of the same same (affine) scheme Spec R along the closed point M. So this is the case where this happens. So I think you can do such things for affine Schemes. For affine schemes, you can at least reverse the topology (they are sometimes called inverse spectrum) and you can form a sheaf over this topology similar to the canonical structure sheaf, but the closed points becomes the generic points in this topology. I cannot recall correctly, but I think the stalks of this sheaves become integral domains (so it is some form of dual to the affine schemes, local becomes integral and so on) - is there a way to glue two schemes together along a closed point (say we're working over a field)? Is it easier to glue in the category of algebraic spaces? For this particular pushout, the geometric intuition is quite simple: given two algebraic varieties, one of which lives in $\mathbb A^m$, another in $\mathbb A^n$, combine them in two complementary hyperplanes in $\mathbb A^{m+n}$. Algebraically, this easily generalizes to an affine scheme $\mathrm{Spec}\, R_1\times R_2/{\mathrm{relationship}}$ and then you glue everything together. As correctly said above, general pushouts of schemes may not be schemes themselves. - In the affine case, let $X=Spec A$, $Y=Spec B$, and $Z=Spec R$. If you have morphisms $f:Z\rightarrow X$ coming from $\phi:A\rightarrow R$ and $g:Z\rightarrow Y$ coming from $\psi: B\rightarrow R$ (because $Aff$ is anti-equivalent to $CRing$), then the pushout $X \coprod_{Z} Y$, gluing $X$ and $Y$ along $Z$ is given by $Spec D$, where $D=A\times_{ R} B:=$ { $(a,b) \in A\times B \| \phi (a)= \psi (b)$ } . - The coproduct in CRing is given by the tensor product over $\mathbb{Z}$, and the pushout over $R$ is given by the tensor product over $R$. I suspect you meant the pullback of A and B over R, written $A\times_R B$. Also, the "coproduct" that you're referring to is called the pushout, the gluing, or the "fibered coproduct", although this last one is nonstandard. The coproduct of affine schemes is specifically the disjoint union. – Harry Gindi Apr 17 2010 at 5:52 I think I meant something like the "fibered coproduct of schemes". That is, the opposite notion to the actual fibered product in CRings (and the latter corresponds, if I'm not mistaken, to my definition of $D$). – Qfwfq Apr 17 2010 at 6:36 I fixed it for you. – Harry Gindi Apr 17 2010 at 7:16 @Qfwfg: Your pushout is really only the pushout in the category of affine schemes. For non-affine schemes, in general, your pushout does noes have the desired universal property. One has to require, for example, that $\phi$ or $\psi$ is surjective (see the paper by Karl Schwede). – Martin Brandenburg Nov 12 2011 at 11:22 @Martin: yes, my answer was only meant to be a (very) partial answer. – Qfwfq Nov 12 2011 at 13:59	{"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": 37, "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": 0.9297403693199158, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2008/12/31/	# The Unapologetic Mathematician ## The Determinant Let’s look at the dimensions of antisymmetric tensor spaces. We worked out that if $V$ has dimension $d$, then the space of antisymmetric tensors with $n$ tensorands has dimension $\displaystyle\dim\left(A^n(V)\right)=\binom{d}{n}=\frac{d!}{n!(d-n)!}$ One thing should leap out about this: if $n$ is greater than $d$, then the dimension formula breaks down. This is connected with the fact that at that point we can’t pick any $n$-tuples without repetition from $d$ basis vectors. So what happens right before everything breaks down? If $n=d$, then we find $\displaystyle\dim\left(A^d(V)\right)=\binom{d}{d}=\frac{d!}{d!(d-d)!}=\frac{d!}{d!}=1$ There’s only one independent antisymmetric tensor of this type, and so we have a one-dimensional vector space. But remember that this isn’t just a vector space. The tensor power $V^{\otimes d}$ is both a representation of $\mathrm{GL}(V)$ and a representation of $S_d$, which actions commute with each other. Our antisymmetric tensors are the image of a certain action from the symmetric group, which is an intertwiner of the $\mathrm{GL}(V)$ action. Thus we have a one-dimensional representation of $\mathrm{GL}(V)$, which we call the determinant representation. I want to pause here and point out something that’s extremely important. We’ve mentioned a basis for $V$ in the process of calculating the dimension of this space, but the space itself was defined without reference to such a basis. Similarly, the representation of any element of $\mathrm{GL}(V)$ is defined completely without reference to any basis of $V$. It needs only the abstract vector space itself to be defined. Calculating the determinant of a linear transformation, though, is a different story. We’ll use a basis to calculate it, but as we’ve just said the particular choice of a basis won’t matter in the slightest to the answer we get. We’d get the same answer no matter what basis we chose. ## 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": 18, "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": 0.9119871854782104, "perplexity_flag": "head"}

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