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http://mathhelpforum.com/differential-geometry/181491-proving-calculation-integral-definition.html	Thread: 1. Proving and calculation an integral from the definition Hello I have the following problem: given that f : [a,b] > R is integrable, prove that g = f/2 is integrable and find the integral. I'm learning Darboux integrals, so my instinct is to try and calculate the upper and lower sums of g and show that for all epsilon, there exists a partition of [a,b] such that the difference between them is less than epsilon...but I'm not sure how I can break apart the difference in order to use the fact that f is integrable...i'm very confused. 2. It is not necessary to look at upper and lower sums. Given any specific partition, S, of [a, b], and any set of x values, ${x_i}$, consisting of one point in each subset, call that sum A_S. Because f is integrable, the limit, using any partition and any choice of x values, exist and is equal to the integral of f over [a,b]. Using the same partition and x values for g, gives (1/2)A_S, which clearly also converges to 1/2 the integral of f over [a,b].	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9015294313430786, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Quartile	# Quartile In descriptive statistics, the quartiles of a ranked set of data values are the three points that divide the data set into four equal groups, each group comprising a quarter of the data. A quartile is a type of quantile. One definition of the lower quartile is the middle number between the smallest number and the median of the data set. The second quartile is the middle observation, also called the median of the data. The third quartile can be measured as the middle value between the median and highest values of the data set. In applications of statistics such as epidemiology, sociology and finance, the quartiles of a ranked set of data values are the four subsets whose boundaries are the three quartile points. Thus an individual item might be described as being "in the upper quartile". ## Definitions • first quartile (designated Q1) = lower quartile = 25th percentile (splits lowest 25% of data) • second quartile (designated Q2) = median = 50th percentile (cuts data set in half) • third quartile (designated Q3) = upper quartile = 75th percentile (splits highest 25% of data, or lowest 75%) The difference between the upper and lower quartiles is called the interquartile range. ## Computing methods For discrete distributions, there is no universal agreement on choosing the quartile values.[1] ### Method 1 Example 4. Boxplot (with quartiles and an interquartile range) and a probability density function (pdf) of a normal N(0,1σ2) population 1. Use the median to divide the ordered data set into two halves. Do not include the median into the halves. 2. The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data. This rule is employed by the TI-83 calculator boxplot and "1-Var Stats" functions. ### Method 2 1. Use the median to divide the ordered data set into two halves. If the median is a datum (as opposed to being the mean of the middle two data), include the median in both halves. 2. The lower quartile value is the median of the lower half of the data. The upper quartile value is the median of the upper half of the data. ### Method 3 1. If there are an even number of data points, then the method is the same as above. 2. If there are (4n+1) data points, then the lower quartile is 25% of the nth data value plus 75% of the (n+1)th data value; the upper quartile is 75% of the (3n+1)th data point plus 25% of the (3n+2)th data point. 3. If there are (4n+3) data points, then the lower quartile is 75% of the (n+1)th data value plus 25% of the (n+2)th data value; the upper quartile is 25% of the (3n+2)th data point plus 75% of the (3n+3)th data point. This always gives the arithmetic mean of Methods 1 and 2; it ensures that the median value is given its correct weight, and thus quartile values change as smoothly as possible as additional data points are added. ### Example 1 Data Set: 6, 47, 49, 15, 42, 41, 7, 39, 43, 40, 36 Ordered Data Set: 6, 7, 15, 36, 39, 40, 41, 42, 43, 47, 49 Method 1 Method 2 $\begin{cases} Q_1 = 15 \\ Q_2 = 40 \\ Q_3 = 43 \end{cases}$ $\begin{cases} Q_1 = 25.5 \\ Q_2 = 40 \\ Q_3 =42.5 \end{cases}$ ### Example 2 Ordered Data Set: 7, 15, 36, 39, 40, 41 As there are an even number of data points, the two methods give the same results. Method 1 Method 2 $\begin{cases} Q_1 = 15 \\ Q_2 = 37.5 \\ Q_3 = 40 \end{cases}$ $\begin{cases} Q_1 = 15 \\ Q_2 = 37.5 \\ Q_3 = 40 \end{cases}$ ## Outliers There are methods by which to check for outliers in the discipline of statistics and statistical analysis. As is the basic idea of descriptive statistics, when encountered with an outlier, we have to explain this by further analysis of the cause or origin of the outlier. In cases of extreme observations, which are not an infrequent occurrence, the typical values must be analyzed. In the case of quartiles, the Interquartile Range (IQR) may be used to characterize the data when there may be extremeties that skew the data; the interquartile range is a relatively robust statistic (also sometimes called "resistance") compared to the range and standard deviation. There is also a mathematical method to check for outliers and determining "fences", upper and lower limits from which to check for outliers. After determining the first and third quartiles and the interquartile range as outlined above, then determining the fences using the following formula: $\text{Lower fence} = Q_1 - 1.5(\mathrm{IQR}) \,$ $\text{Upper fence} = Q_3 + 1.5(\mathrm{IQR}), \,$ where Q1 and Q3 are the first and third quartiles, respectively. The Lower fence is the "lower limit" and the Upper fence is the "upper limit" of data, and any data lying outside this defined bounds can be considered an outlier. Anything below the Lower fence or above the Upper fence can be considered such a case. The fences provide a guideline by which to define an outlier, which may be defined in other ways. The fences define a "range" outside of which an outlier exists; a way to picture this is a boundary of a fence, outside of which are "outsiders" as opposed to outliers. ## References 1. Hyndman, Rob J; Fan, Yanan (November 1996). "Sample quantiles in statistical packages". American Statistician 50 (4): 361–365. doi:10.2307/2684934.	{"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": 6, "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": 0.9018644094467163, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/11/13/the-coevaluation-on-vector-spaces/?like=1&source=post_flair&_wpnonce=5b12459335	# The Unapologetic Mathematician ## The Coevaluation on Vector Spaces Okay, I noticed that I never really gave the definition of the coevaluation when I introduced categories with duals, because you need some linear algebra. Well, now we have some linear algebra, so let’s do it. Let $V$ be a finite-dimensional vector space with dual space $V^$. Then if we have a basis $\left\{e_i\right\}$ of $V$ we immediately get a dual basis $\left\{\epsilon^i\right\}$ for $V^$ (yet another $\epsilon$ to keep straight), defined by $\epsilon^j(e_i)=\delta_i^j$. We now define a map $\eta:\mathbf{1}\rightarrow V^*\otimes V$ by setting $\eta(1)=\epsilon^i\otimes e_i$. That is, we take the tensor product of each dual basis element with its corresponding basis element, and add them all up (summation convention). But this seems to depend on which basis $\left\{e_i\right\}$ we started with. What if we used a different basis $\left\{f_i\right\}$ and dual basis $\left\{\phi^i\right\}$? We know that there is a change of basis matrix $f_i=t_i^je_j$, so let’s see how this works on the dual basis. The dual basis is defined by the fact that $\phi^j(f_i)=\delta_i^j$. So we use this new expression for $f_i$ to write $\phi^j(t_i^ke_k)=t_i^k\phi^j(e_k)=\delta_i^j$. That is, $\phi^j(e_k)$ must be the inverse matrix to $t_i^k$, which we’ll write as $\left(t^{-1}\right)_k^j$. But now we can check $\left[\left(t^{-1}\right)_i^j\epsilon^i\right](e_k)=\left(t^{-1}\right)_i^j\delta_k^i=\left(t^{-1}\right)_k^j$ And so we find that $\phi^i=\left(t^{-1}\right)_j^i\epsilon^j$ when we change bases. Now we can use the same definition for $\eta$ above with our new basis. We set $\eta(1)=\phi^i\otimes f_i$, and then substitute our expressions in terms of the old bases: $\eta(1)=\left(t^{-1}\right)_j^i\epsilon^j\otimes t_i^ke_k=\left(t^{-1}\right)_j^it_i^k\left(\epsilon^j\otimes e_k\right)=\delta_j^k\left(\epsilon^j\otimes e_k\right)=\epsilon^k\otimes e_k$ which is what we got before. That is, this map actually doesn’t depend on the basis we chose! Okay, now does this coevaluation — along with the evaluation map from before — actually satisfy the conditions for a duality? First, let’s start with a vector written out in terms of a basis: $v=v^ie_i$. Now we use the coevaluation to send it to $v^ie_i\otimes\epsilon^j\otimes e_j$. Next we evaluate on the first two tensorands to find $v^i\delta_i^je_j=v^je_j$. So we do indeed have the identity here. Verifying the other condition is almost the same, starting from an arbitrary covector $\lambda=\lambda_i\epsilon^i$. So now we know that the category $\mathbf{FinVect}(\mathbb{F})$ has duals. Tomorrow we can promote this to a duality on the category of finite-dimensional representations of a Hopf algebra. ### Like this: Posted by John Armstrong \| Algebra, Linear Algebra ## 2 Comments » 1. [...] of Hopf Algebras II Now that we have a coevaluation for vector spaces, let’s make sure that it intertwines the actions of a Hopf algebra. Then we can finish [...] Pingback by \| November 14, 2008 \| Reply 2. [...] we’ve seen it already! When we talked about the coevaluation on vector spaces we worked out how a change of basis affects linear functionals. What we found is that if is our action on , then the action on is by the transpose — the [...] Pingback by \| December 2, 2008 \| Reply « Previous \| Next » ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 30, "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": 0.9184367060661316, "perplexity_flag": "head"}
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The quantum partition function is $Z=tr(e^{-\beta H})$ where H is an operator, the Hamiltonian of the ...	{"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": 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": 0.9260440468788147, "perplexity_flag": "middle"}
http://medlibrary.org/medwiki/Cauchy_sequence	# Cauchy sequence Welcome to MedLibrary.org. For best results, we recommend beginning with the navigation links at the top of the page, which can guide you through our collection of over 14,000 medication labels and package inserts. For additional information on other topics which are not covered by our database of medications, just enter your topic in the search box below: (a) The plot of a Cauchy sequence $(x_n),$ shown in blue, as $x_n$ versus $n$ If the space containing the sequence is complete, the "ultimate destination" of this sequence (that is, the limit) exists. (b) A sequence that is not Cauchy. The elements of the sequence fail to get arbitrarily close to each other as the sequence progresses. In mathematics, a Cauchy sequence (pronounced ), named after Augustin-Louis Cauchy, is a sequence whose elements become arbitrarily close to each other as the sequence progresses. More precisely, given any small positive distance, all but a finite number of elements of the sequence are less than that given distance from each other. The utility of Cauchy sequences lies in the fact that in a complete metric space (one where all such sequences are known to converge to a limit), the criterion for convergence depends only on the terms of the sequence itself. This is often exploited in algorithms, both theoretical and applied, where an iterative process can be shown relatively easily to produce a Cauchy sequence, consisting of the iterates, thus fulfilling a logical condition, such as termination. The notions above are not as unfamiliar as they might at first appear. The customary acceptance of the fact that any real number x has a decimal expansion is an implicit acknowledgment that a particular Cauchy sequence of rational numbers (whose terms are the successive truncations of the decimal expansion of x) has the real limit x. In some cases it may be difficult to describe x independently of such a limiting process involving rational numbers. Generalizations of Cauchy sequences in more abstract uniform spaces exist in the form of Cauchy filters and Cauchy nets. ## In Real numbers[] A sequence $x_1, x_2, x_3, \ldots$ of real numbers is called a Cauchy sequence, if for every positive real number ε, there is a positive integer N such that for all natural numbers m, n > N $\|x_m - x_n\| < \varepsilon,$ where the vertical bars denote the absolute value. In a similar way one can define Cauchy sequences of rational or complex numbers. Cauchy formulated such a condition by requiring $x_m - x_n$ to be infinitesimal for every pair of infinite m, n. ## In a metric space[] To define Cauchy sequences in any metric space X, the absolute value $\|x_m - x_n\|$ is replaced by the distance $d(x_m, x_n)$ (where d : X × X → R with some specific properties, see Metric (mathematics)) between $x_m$ and $x_n$. Formally, given a metric space (X, d), a sequence $x_1, x_2, x_3, \ldots$ is Cauchy, if for every positive real number ε > 0 there is a positive integer N such that for all positive integers m,n > N, the distance $d(x_m, x_n) < \varepsilon.$ Roughly speaking, the terms of the sequence are getting closer and closer together in a way that suggests that the sequence ought to have a limit in X. Nonetheless, such a limit does not always exist within X. ## Completeness[] A metric space X in which every Cauchy sequence converges to an element of X is called complete. ### Examples[] The real numbers are complete under the metric induced by the usual absolute value, and one of the standard constructions of the real numbers involves Cauchy sequences of rational numbers. A rather different type of example is afforded by a metric space X which has the discrete metric (where any two distinct points are at distance 1 from each other). Any Cauchy sequence of elements of X must be constant beyond some fixed point, and converges to the eventually repeating term. ### Counter-example: rational numbers[] The rational numbers Q are not complete (for the usual distance): There are sequences of rationals that converge (in R) to irrational numbers; these are Cauchy sequences having no limit in Q. In fact, if a real number x is irrational, then the sequence (xn), whose n-th term is the truncation to n decimal places of the decimal expansion of x, gives a Cauchy sequence of rational numbers with irrational limit x. Irrational numbers certainly exist, for example: • The sequence defined by $x_0=1, x_{n+1}=\frac{x_n+\frac{2}{x_n}}{2}$ consists of rational numbers (1, 3/2, 17/12,...), which is clear from the definition; however it converges to the irrational square root of two, see Babylonian method of computing square root. • The sequence $x_n = F_n / F_{n-1}\,$ of ratios of consecutive Fibonacci numbers which, if it converges at all, converges to a limit $\phi$ satisfying $\phi^2 = \phi+1$, and no rational number has this property. If one considers this as a sequence of real numbers, however, it converges to the real number $\varphi = (1+\sqrt5)/2$, the Golden ratio, which is irrational. • The values of the exponential, sine and cosine functions, exp(x), sin(x), cos(x), are known to be irrational for any rational value of x≠0, but each can be defined as the limit of a rational Cauchy sequence, using, for instance, the Maclaurin series. ### Counter-example: open interval[] The open interval $X=(0, 2)$ in the set of real numbers with an ordinary distance in R is not a complete space: there is a sequence $x_n=1/n$ in it, which is Cauchy (for arbitrarily small distance bound $d>0$ all terms $x_n$ of $n > 1/d$ fit in the $(0, d)$ interval), however does not converge in X—its 'limit', number 0, does not belong to the space X. ### Other properties[] • Every convergent sequence (with limit s, say) is a Cauchy sequence, since, given any real number ε > 0, beyond some fixed point, every term of sequence is within distance ε/2 of s, so any two terms of the sequence are within distance ε of each other. • Every Cauchy sequence of real (or complex) numbers is bounded (since for some N, all terms of the sequence from the N-th onwards are within distance 1 of each other, and if M is the largest absolute value of the terms up to and including the N-th, then no term of the sequence has absolute value greater than M+1). • In any metric space, a Cauchy sequence which has a convergent subsequence with limit s is itself convergent (with the same limit), since, given any real number r > 0, beyond some fixed point in the original sequence, every term of the subsequence is within distance r/2 of s, and any two terms of the original sequence are within distance r/2 of each other, so every term of the original sequence is within distance r of s. These last two properties, together with a lemma used in the proof of the Bolzano–Weierstrass theorem, yield one standard proof of the completeness of the real numbers, closely related to both the Bolzano–Weierstrass theorem and the Heine–Borel theorem. The lemma in question states that every bounded sequence of real numbers has a convergent monotonic subsequence. Given this fact, every Cauchy sequence of real numbers is bounded, hence has a convergent subsequence, hence is itself convergent. It should be noted, though, that this proof of the completeness of the real numbers implicitly makes use of the least upper bound axiom. The alternative approach, mentioned above, of constructing the real numbers as the completion of the rational numbers, makes the completeness of the real numbers tautological. One of the standard illustrations of the advantage of being able to work with Cauchy sequences and make use of completeness is provided by consideration of the summation of an infinite series of real numbers (or, more generally, of elements of any complete normed linear space, or Banach space). Such a series $\sum_{n=1}^{\infty} x_{n}$ is considered to be convergent if and only if the sequence of partial sums $(s_{m})$ is convergent, where $s_{m} = \sum_{n=1}^{m} x_{n}$. It is a routine matter to determine whether the sequence of partial sums is Cauchy or not, since for positive integers p > q, $s_{p} - s_{q} = \sum_{n=q+1}^{p} x_{n}.$ If $f \colon M \rightarrow N$ is a uniformly continuous map between the metric spaces M and N and (xn) is a Cauchy sequence in M, then $(f(x_n))$ is a Cauchy sequence in N. If $(x_n)$ and $(y_n)$ are two Cauchy sequences in the rational, real or complex numbers, then the sum $(x_n + y_n)$ and the product $(x_n y_n)$ are also Cauchy sequences. ## Generalizations[] ### In topological vector spaces[] There is also a concept of Cauchy sequence for a topological vector space $X$: Pick a local base $B$ for $X$ about 0; then ($x_k$) is a Cauchy sequence if for each member $V\in B$, there is some number $N$ such that whenever $n,m > N, x_n - x_m$ is an element of $V$. If the topology of $X$ is compatible with a translation-invariant metric $d$, the two definitions agree. ### In topological groups[] Since the topological vector space definition of Cauchy sequence requires only that there be a continuous "subtraction" operation, it can just as well be stated in the context of a topological group: A sequence $(x_k)$ in a topological group $G$ is a Cauchy sequence if for every open neighbourhood $U$ of the identity in $G$ there exists some number $N$ such that whenever $m,n>N$ it follows that $x_n x_m^{-1} \in U$. As above, it is sufficient to check this for the neighbourhoods in any local base of the identity in $G$. As in the construction of the completion of a metric space, one can furthermore define the binary relation on Cauchy sequences in $G$ that $(x_k)$ and $(y_k)$ are equivalent if there for every open neighbourhood $U$ of the identity in $G$ exists some number $N$ such that whenever $m,n>N$ it follows that $x_n y_m^{-1} \in U$. This relation is an equivalence relation. More precisely, it is reflexive since the sequences are Cauchy sequences. It is symmetric since $y_n x_m^{-1} = (x_m y_n^{-1})^{-1} \in U^{-1}$ which by continuity of the inverse is another open neighbourhood of the identity. It is transitive since $x_n z_l^{-1} = x_n y_m^{-1} y_m z_l^{-1} \in U' U''$ where $U'$ and $U''$ are open neighbourhoods of the identity such that $U'U'' \subseteq U$; such pairs exist by the continuity of the group operation. ### In groups[] There is also a concept of Cauchy sequence in a group $G$: Let $H=(H_r)$ be a decreasing sequence of normal subgroups of $G$ of finite index. Then a sequence $(x_n)$ in $G$ is said to be Cauchy (w.r.t. $H$) if and only if for any $r$ there is $N$ such that $\forall m,n > N, x_n x_m^{-1} \in H_r$. Technically, this is the same thing as a topological group Cauchy sequence for a particular choice of topology on $G$, namely that for which $H$ is a local base. The set $C$ of such Cauchy sequences forms a group (for the componentwise product), and the set $C_0$ of null sequences (s.th. $\forall r, \exists N, \forall n > N, x_n \in H_r$) is a normal subgroup of $C$. The factor group $C/C_0$ is called the completion of $G$ with respect to $H$. One can then show that this completion is isomorphic to the inverse limit of the sequence $(G/H_r)$. An example of this construction, familiar in number theory and algebraic geometry is the construction of the p-adic completion of the integers with respect to a prime p. In this case, G is the integers under addition, and Hr is the additive subgroup consisting of integer multiples of pr. If $H$ is a cofinal sequence (i.e., any normal subgroup of finite index contains some $H_r$), then this completion is canonical in the sense that it is isomorphic to the inverse limit of $(G/H)_H$, where $H$ varies over all normal subgroups of finite index. For further details, see ch. I.10 in Lang's "Algebra". ### In constructive mathematics[] In constructive mathematics, Cauchy sequences often must be given with a modulus of Cauchy convergence to be useful. If $(x_1, x_2, x_3, ...)$ is a Cauchy sequence in the set $X$, then a modulus of Cauchy convergence for the sequence is a function $\alpha$ from the set of natural numbers to itself, such that $\forall k \forall m, n > \alpha(k), \|x_m - x_n\| < 1/k$. Clearly, any sequence with a modulus of Cauchy convergence is a Cauchy sequence. The converse (that every Cauchy sequence has a modulus) follows from the well-ordering property of the natural numbers (let $\alpha(k)$ be the smallest possible $N$ in the definition of Cauchy sequence, taking $r$ to be $1/k$). However, this well-ordering property does not hold in constructive mathematics (it is equivalent to the principle of excluded middle). On the other hand, this converse also follows (directly) from the principle of dependent choice (in fact, it will follow from the weaker AC00), which is generally accepted by constructive mathematicians. Thus, moduli of Cauchy convergence are needed directly only by constructive mathematicians who (like Fred Richman) do not wish to use any form of choice. That said, using a modulus of Cauchy convergence can simplify both definitions and theorems in constructive analysis. Perhaps even more useful are regular Cauchy sequences, sequences with a given modulus of Cauchy convergence (usually $\alpha(k) = k$ or $\alpha(k) = 2^k$). Any Cauchy sequence with a modulus of Cauchy convergence is equivalent (in the sense used to form the completion of a metric space) to a regular Cauchy sequence; this can be proved without using any form of the axiom of choice. Regular Cauchy sequences were used by Errett Bishop in his Foundations of Constructive Analysis, but they have also been used by Douglas Bridges in a non-constructive textbook (ISBN 978-0-387-98239-7 [Amazon-US \| Amazon-UK]). However, Bridges also works on mathematical constructivism; the concept has not spread far outside of that milieu. ### In a hyperreal continuum[] A real sequence $\langle u_n: n\in \mathbb{N} \rangle$ has a natural hyperreal extension, defined for hypernatural values H of the index n in addition to the usual natural n. The sequence is Cauchy if and only if for every infinite H and K, the values $u_H$ and $u_K$ are infinitely close, or adequal, i.e. $\, \mathrm{st}(u_H-u_K)= 0$ where "st" is the standard part function. ## References[] • Bourbaki, Nicolas (1972). Commutative Algebra (English translation ed.). Addison-Wesley. ISBN 0-201-00644-8 [Amazon-US \| Amazon-UK]. • Lang, Serge (1993), Algebra (Third ed.), Reading, Mass.: Addison-Wesley Pub. Co., ISBN 978-0-201-55540-0 [Amazon-US \| Amazon-UK], Zbl 0848.13001 • Spivak, Michael (1994). Calculus (3rd ed.). Berkeley, CA: Publish or Perish. ISBN 0-914098-89-6 [Amazon-US \| Amazon-UK]. • Troelstra, A. S.; D. van Dalen. Constructivism in Mathematics: An Introduction. (for uses in constructive mathematics) ## [] Content in this section is authored by an open community of volunteers and is not produced by, reviewed by, or in any way affiliated with MedLibrary.org. Licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License, using material from the Wikipedia article on "Cauchy sequence", available in its original form here: http://en.wikipedia.org/w/index.php?title=Cauchy_sequence • ## Finding More You are currently browsing the the MedLibrary.org general encyclopedia supplement. To return to our medication library, please select from the menu above or use our search box at the top of the page. In addition to our search facility, alphabetical listings and a date list can help you find every medication in our library. • ## Questions or Comments? If you have a question or comment about material specifically within the site’s encyclopedia supplement, we encourage you to read and follow the original source URL given near the bottom of each article. You may also get in touch directly with the original material provider. • ## About This site is provided for educational and informational purposes only and is not intended as a substitute for the advice of a medical doctor, nurse, nurse practitioner or other qualified health professional.	{"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": 101, "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": 0.9111930131912231, "perplexity_flag": "head"}
http://mathoverflow.net/questions/66208?sort=oldest	## The cone of positive semidefinite matrices is self-dual? (reference needed) ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm seeking a reference for the following fact. The cone of positive semidefinite matrices is self-dual (a.k.a. self-polar). This result is relatively easy to prove, has been known for a long time, and is fundamental to things like semidefinite programming. Ideally, I would like a reference that reflects all three of those properties. Unfortunately, the properties themselves make it hard to find a good reference to cite. (Many sources I've looked at consider this result elementary and well-known enough to simply state without proof or reference. That was sort of my plan as well, but a referee is now asking for a reference, and seeing as how our paper is outside of optimization theory, I think that's probably reasonable.) By the way, this result is occasionally referred to as Fejer's Trace Theorem, although I have never encountered an actual reference to any publication of Fejer. So if anyone knows the source of this attribution, that would be interesting. Any help would be greatly appreciated! - ## 2 Answers I am pretty sure Boyd's convex optimization (available on his web page as a pdf) talks about this (yes: example 2.24) - Great! I don't know if this is the ideal reference I'm after, but it certainly looks better than any I knew of before. – Louis Deaett Jun 9 2011 at 19:23 ### 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. Perhaps the Notes section of the classic book: Analysis on symmetric cones is of help. In particular, they mention that the following paper of Koecher started the study of symmetric cones. I have not yet read this paper, so cannot say if it was this paper that described the self-duality result that you mention. But I hope the Notes section mentioned above does provide some clues. M. Koecher (1957). Positivitätsbereiche in $R^n$. Amer J. Math., 79. - I'm in the process of trying to get ahold of this book. It sounds promising. – Louis Deaett Jun 9 2011 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": 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": 0.9479028582572937, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/61921/on-closed-totally-disconnected-subgroups-of-connected-real-lie-groups	## On closed totally disconnected subgroups of connected real Lie groups ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) So the following statement seems to be obvious but I don't see how to prove it: Q: How does one prove that a closed totally disconnected subgroup of a connected real Lie group is discrete? Note that it is essential to take into account the group structure since the Cantor set is a closed totally disconnected subset of $\mathbf{R}$ which is not discrete. - Belated comment: adding a tag 'lie-groups' would be appropriate here? – Jim Humphreys Apr 16 2011 at 17:50 and what if we consider a topological group which is not locally euclidean? – Giuseppe Apr 16 2011 at 18:12 2 @Giuseppe: If the group isn't locally Euclidean, then the subgroup needn't be discrete. For example, let $G$ be a direct product of infinitely many circle groups. The circle group has a subgroup of order $2$, so $G$ has a subgroup that is a direct product of infinitely many $2$-element discrete groups. This subgroup is closed and totally disconnected but not discrete. – Stephen S Apr 17 2011 at 8:22 @Stephen: thanks a lot for your attention. – Giuseppe Apr 18 2011 at 6:07 ## 3 Answers Let's just give a quick argument: Let $G$ be a Lie group (with Lie algebra $\mathbb g$) and $H \subset G$ be a closed subgroup. If $H$ is not discrete, then there exists a sequence $(g_n)_{n \in \mathbb N}$ of elements of $H$, which converges to $1_G$. Write $g_n = \exp( \alpha_n \xi_n)$ (for $n$ large enough), where $\xi_n \in \mathbb g$ are unit vectors (with respect to some innerproduct) and $\alpha_n \in \mathbb R_{> 0}$. Since $g_n \to 1_G$, we get $\alpha_n \to 0$. Let $\delta \in \mathbb R$ and find integers $k_n(\delta)$, such that $\|\alpha_n \cdot k_n -\delta\|< \alpha_n$. Let $\xi$ be some accumulation point of the set $\lbrace \xi_n \mid n \in \mathbb N \rbrace$. Now, it is easy to see that $$g_n^{k_n(\delta)} = \exp(\alpha_n k_n(\delta) \xi_n) \to \exp(\delta \xi)$$ on a subsequence. This implies that the whole 1-parameter subgroup generated by $\xi$ lies in $H$. In particular, $H$ is not totally disconnected. - Thanks a lot Andreas, this is exactly what I was looking for. – Hugo Chapdelaine Apr 16 2011 at 17:36 Corollary: Any continuous Galois representation on Euclidean space has finite image. – Kevin Ventullo Apr 16 2011 at 21:10 ### 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. 0)The solution by Gleason, Montgomery, and Zippin of the fifth Hilbert's problem says that any òpcally euclidean topological group admits a compatible analytical structure. 1)By Cartan--Von Neumann theorem on closed subgroups, if $G$ is a Lie group (i.e. smooth group manifold) and $H$ is a closed subgroup of $G$, then $H$ is an embedded Lie subgroup of $G$, so its connected components w.r.t. the subspace topology and its connected components w.r.t. this Lie group structure are the same. 2)The connected components of a topological manifold are open. Now 0),1) and 2) should be conclusive for more than your question, i.e.: If $G$ is a locally euclidean topological group and $H$ is a totally disconnected, closed subgroup of $G$ then $H$ is discrete topological subgroup of $G$. - Hi Giuseppe, but $H$ is not necessarily a topological manifold. So are you saying that $H$ is a $0$-dimensional manifold so therefore discrete? – Hugo Chapdelaine Apr 16 2011 at 16:22 1 Cartan's theorem implies that $H$ is necessarily a smooth submanifold of $G$. So if it's totally disconnected, it must be 0-dimensional, i.e. discrete. – Brad Hannigan-Daley Apr 16 2011 at 16:42 So what is the definition of a 0-dimensional manifold? – Hugo Chapdelaine Apr 16 2011 at 16:55 1 @Hugo: exactly what it sounds like: a (second-countable Hausdorff etc.) space which is locally homeomorphic to a point, hence discrete. – Qiaochu Yuan Apr 16 2011 at 17:33 Well take $G=H=\mathbf{Z}_p$, then $H$ is not discrete. The result that you claim in $0)$ probably applies to topological groups which have a topological real manifold structure. – Hugo Chapdelaine Apr 16 2011 at 17:35 show 4 more comments To complement what has been said earlier: a Lie group $G$ has no small subgroup, i.e. there exists a neighborhood $U$ of the identity $e\in G$ such that the only subgroup of $G$ contained in $U$, is ${e}$. [To see it, let $B$ be a small ball around $0$ in the Lie algebra of $G$, such that the exponential map $\exp$ induces a diffeomorphism between $B$ and its image. Set $U=\exp(B/2)$. Then for every $x\in U\backslash{e}$, there exists $p>1$ such that $x^p\notin U$.] Let then $H$ be a closed subgroup of $G$. If $H$ is totally disconnected, then there is a basis of neighborhoods of $e$ in $H$, consisting of open subgroups of $H$. So $H\cap U={e}$, meaning that $H$ is discrete. - So how do you show the existence of such a basis of neighborhoods at $e$ in $H$. Definitely you need $H$ to be closed but I don't see how to use it. – Hugo Chapdelaine Apr 26 2011 at 14:54 Well, I suppose I'm appealing to van Dantzig's theorem... – Alain Valette Apr 26 2011 at 16:11	{"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": 70, "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": 0.9316482543945312, "perplexity_flag": "head"}
http://mathoverflow.net/questions/29384/convergence-of-orthogonal-polynomial-expansions/29390	## Convergence of orthogonal polynomial expansions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) "Everyone" knows that for a general $f\in L^2[0,1]$, the Fourier series of $f$ converges to $f$ in the $L^2$ norm but not necessarily in most other senses one might be interested in; but if $f$ is reasonably nice, then its Fourier series converges to $f$, say, uniformly. I'm looking for similar results about orthogonal polynomial expansions for functions on the whole real line. What I specifically want at the moment is sufficient conditions on a bounded function `$f:\mathbb{R} \to \mathbb{R}$` so that the partial sums of its Hermite polynomial expansion are uniformly bounded on compact sets, but I'm also interested in learning what's known about pointwise/uniform/etc. convergence results for Hermite and other classical orthogonal polynomials. Possibly such results follow trivially from well-known basic facts about Hermite polynomials, but I'm not familiar with that literature and I'm having trouble navigating it. So in addition to precise answers, I'd appreciate literature tips (but please don't just tell me to look at Szegő's book unless you have a specific section to recommend). - 1 I think you should add a weight to your question. The Hermite polynomials are an orthonormal basis of $L^2(\exp(-x^2/2) dx)$ and not of $L^2$. So assuming that $f \in L^2(\exp(-x^2/2) dx)$ would be sufficient to have an expansion in Hermite polynomials converging in $L^2$ (with respect to that weight). So e.g. the assumption $f(x) \exp(x^2/2)$ is bounded would be more naturall than just $f$ bounded... (in some sense ...) – Helge Jun 24 2010 at 15:08 Your last sentence doesn't make sense to me since you're proposing a much stronger condition than boundedness. In any case, I didn't mention the weight because (1) for reasons external to this question, bounded functions are what I'm really interested in; and (2) I wanted to keep the statement of my question simple. – Mark Meckes Jun 24 2010 at 15:28 Now, I'm confused. Do you want to approximate with the Hermite polynomials $H_n(x)$ or with $H_n(x) \exp(- x^2/2)$ ? In the first case uniform convergence is trivally wrong! One might hope for uniform convergence on compact sets. In the second case, the condition $\liminf_{\|x\|\to\infty} \|f(x)\| > 0$ should imply that the expansion coefficients are unbounded. (This statement is a feeling, I didn't try proving it). This again destroys all hopes for a nice sense of convergence... – Helge Jun 24 2010 at 15:39 Ack, what I really wanted was uniform boundedness on compact sets, and I forgot to include that crucial last point. I've edited to correct that. I did want to approximate $f$ by $\sum c_n H_n$, but now you've got me thinking that maybe it would work better to approximate $f(x)\exp(-x^2/2)$ by $\sum c_n H_n(x) \exp(-x^2/2)$; I'll think about that some more. – Mark Meckes Jun 24 2010 at 15:54 1 There is a chapter (9.1 ff p368) in the book of G. Sansone "Orthogonal functions" about uniform convergence of series of Hermite polynomials. There is a preview of the book here books.google.com/…. – coudy Jun 24 2010 at 18:07 show 2 more comments ## 2 Answers Define $\psi_n(x) = c_n H_n(x) e^{-x^2/2}$ as in http://en.wikipedia.org/wiki/Hermite_polynomials . Also define the differential operator $H u = - u'' + x^2 u$. Then the $\psi_n$ form an othonormal basis of $L^2$ and $H \psi_n = (2n + 1) \psi_n$. Warning: As coudy points out below: one needs $\\|H f\\| < \infty$ and not just $\langle f, Hf\rangle < \infty$. So the computations below need to be changed. Rest of original post Given now $f$ such that $$A = \langle f,Hf \rangle =\int \overline{f(x)} (Hf)(x) dx$$ is finite. Then by writing $f(X) = \sum_{n \geq 0} f_n \psi_n(X)$, we obtain $$A = \langle f,Hf \rangle =\langle f, \sum_{n \geq 0} f_n H\psi_n(X) \rangle = \langle \sum_{n \geq 0} f_n \psi_n(X) , \sum_{n \geq 0} f_n (2 n + 1)\psi_n(X) \rangle$$ Now using orthonormality of the $psi_n$, we conclude that $$A = \sum_{n \geq 0} \|f_n\|^2 (2n + 1).$$ Now using that the $\psi_n(x)$ are all bounded by $2$ it follows that the sequence converges uniformly! Now, what does $\langle f,Hf \rangle < \infty$ mean for $f(x) = e^{-x^2/2} g(x)$. This can be computed to mean $$\int \|g'(x) + \frac{x}{2} g(x)\|^2 e^{-x^2} dx.$$ On a philosophical level, this is not about the $H_n$ being orthogonal polynomials, but about them being eigenfunctions of a self-adjoint operator. (well the $\psi_n$ are). - Am I missing something ? Shouldn't the condition be Hf in L^2 ? With $f_n=1/(n \log n)$, I don't see how to get the uniform convergence of $\sum f_n \psi_n$ from the finiteness of <f,Hf>. – coudy Jun 24 2010 at 20:58 I think you are right, and one indeed needs $\\| H f\\|$ to be finite. I'll add a comment ... – Helge Jun 24 2010 at 22:11 Well, yeah, clearly. But I accepted this answer because it gives me all the tools I need to prove the kinds of results I can use. – Mark Meckes Jun 24 2010 at 23:30 ### 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. Let me recall a quick $L^2$ proof of the uniform convergence of the Fourier series of a $C^1$ function $f$. Let $c_n$ be its Fourier coefficients. Then $$\|f(x)\| \leq \|c_0\|+ \Sigma\|c_n\|\ n \ {1\over n}\ \ \leq \|c_0\| + \ \ \sqrt{\Sigma \ n^2 \|c_n\|^2}\ \ \sqrt{\Sigma\ 1/n^2}$$ Replacing f by f minus its partial sum, and noting that $\Sigma \ n^2\|c_n\|^2 = \|\|f'\|\|_2^2 \$ is finite, you get uniform convergence. So maybe you can use a similar computation in case of a family of orthogonal polynomials ? - Quite possibly something along the same lines, but necessarily more complicated, works. I'd need to know how to simply relate the coefficients for $f$ to those for $f'$, and also uniform bounds on the Hermite polynomials themselves. These are the kinds of "well-known basic facts about Hermite polynomials" that I'm missing. – Mark Meckes Jun 24 2010 at 15:32 That is, assuming these are "well-known basic facts" at all. – Mark Meckes Jun 24 2010 at 15:34	{"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": 45, "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": 0.9454056620597839, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/182459/an-infinite-set-having-one-more-element-than-another-infinite-set	# An infinite set having “one more element” than another infinite set A classic example of homeomorphism is between a sphere missing one point and a plane To see this, place a sphere on the plane so that the sphere is tangent to the plane. Given any point in the plane, construct a line through that point from the "north" pole (assuming the point of tangency is the south pole); this necessarily intersects the sphere in a unique point other than the north pole. Similarly, given any point on the sphere other than the north pole, the line through the north pole and that point will intersect the plane at a unique point. Thus, the homeomorphism is established. Since there is a homeomorphism between a sphere missing one point and a plane, does it make sense to say that a sphere has one more point than a plane, even though both sets are infinite and uncountable? - 1 – tomasz Aug 14 '12 at 14:17 1 In fact, a set is infinite if and only if it can be placed in bijective correspondence with a proper subset of itself. – Charles Staats Aug 14 '12 at 14:23 3 @CharlesStaats: Assuming axiom of choice, that is. ;) – tomasz Aug 14 '12 at 14:40 ## 4 Answers Since there is a homeomorphism between a sphere missing one point and a plane, does it make sense to say that a sphere has one more point than a plane, even though both sets are infinite and uncountable? It can certainly be given meaning. For example, given any inclusion of subsets $A \subseteq B$, we can define the phrase "$B$ has $\alpha$ more elements than $A$" to mean that $\|B \setminus A\| = \alpha$. More generally, given any injection $i: A \to B$, we can define "$B$ has $\alpha$ more elements than the subset defined by $i$" or "$B$ has $\alpha$ more elements than $A$ relative to the inclusion $i$" to mean $\|B \setminus i(A)\| = \alpha$. These notions can even be useful. For example, the notion of a cofinite subset often comes up: a subset $S$ of a set $X$ such that $\|X \setminus S\| < \aleph_0$. - 3 It should be stressed that this really measures the size with regards to the inclusion map. Different inclusions give rise to different sizes – Thomas Rot Aug 14 '12 at 14:56 Yes, and no, it really just depends on how you're defining the "size" of sets. In each case you're applying a metric, or something close to it (since not every measure of size inherently fits the definition of a metric in a more formal sense). I can easily say that, in the real numbers, the size of a set is it's lebsgue measure, or even the maximal distance of it's supremum and infimum; in the same way you can define one set to be larger than another if the removal of an element allows a homeomorphism to exist between them. This can then be used to create an equivalence relation of the sets in question, and an ordering between those equivalence classes. Ordinarily cardinality is used in this sense, but it's not the only way to define a set's size. - Potential problems: 1) There exist spaces that are homeomorphic to the same space with a point removed. 2) The order relation you have defined gives no relation at all between, say, $\mathbb N$ and $\mathbb R$. – Charles Staats Aug 14 '12 at 14:55 • Is it true that there is a one-to-one correspondence between $B$ and $A\setminus\{a\}$, where $a\in A$, and that there is no one-to-one correspondence between $B$ and $A$? • Is it true that there is a well-behaved one-to-one correspondence between $B$ and $A\setminus\{a\}$, where $a\in A$, and that there is no well-behaved one-to-one correspondence between $B$ and $A$? The first can certainly be true if $A$ and $B$ are finite. In conventional set theory, it cannot be true if they're infinite. The second depends on which notion of "well-behaved" is invoked, and there you have a vast menu to choose from. Sometimes "well-behaved" means "linear", and then a three-dimensional space is bigger than a two-dimensional space. Sometimes well-behaved means a homeomorphism, and then you've got your plane-and-sphere example. Sometimes well-behaved means measure-preserving. Sometimes it means an isometry. Here's an example that I like. In non-standard analysis one has "infinite integers". Let $n$ be one such. Consider the sets $\{1,2,3,\ldots,n\}$ and $\{1,2,3,\ldots,n,n+1,n+2\}$. If "well-behaved" has no content, so that all bijections are well-behaved, then these two sets have the same cardinality and that's that. But in non-standard analysis one speaks of internal and external objects. Every internal (i.e. well-behaved) one-to-one function from $\{1,2,3,\ldots,n\}$ to $\{1,2,3,\ldots,n,n+1,n+2\}$ omits exactly two members of the latter set from its image. Take your pick as to which two they will be. - Size of sets is measured by cardinality. A set $A$ is said to have a smaller or equal cardinality than $B$ if there exists an injective function $f:A\rightarrow B$. If there exists a bijective function the sets have the same cardinality. A nonempty set is finite if it has the same cardinality as the set $\{1,\ldots,n\}$ for some $n\in\mathbb{N}$. Otherwise it is infinity. It is countable if it is in bijective correspondence with $\mathbb{N}$ The following sets have the same cardinality, even though "intuitively" one would expect this not to be the case Natural numbers $\mathbb{N}$, The even natural numbers $2\mathbb{N}$, the set of prime numbers $P$, The set of all integers $\mathbb{Z}$. The rationals $\mathbb{Q}$. The set of reals $\mathbb{R}$ has a (strictly) larger cardinality than the set of natural numbers though. This is Cantors famous diagonal argument. In the example you describe, you give an injection (but not bijection) of the plane into the sphere. But there also exists an injection (not a bijection) of the sphere into the plane. -	{"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": 48, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9363235831260681, "perplexity_flag": "head"}
http://mathoverflow.net/questions/4442/is-there-a-theorem-that-says-that-there-is-always-more-than-one-way-to-continue/75260	## Is there a theorem that says that there is always more than one way to “continue a finite sequence”? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I have come across a bit of folklore(?) which goes something like "given any finite sequence of numbers, there is more than one 'valid' way of continuing the sequence". For example see here. I would like to know if this is actually stated and proved rigorously, and if so, where can I find a statement and proof? EDIT: The first couple of answers reflect my concerns exactly. But to quote from the book review I linked to, Wittgenstein's Finite Rule Paradox implies that any finite sequence of numbers can be a continued in a variety of different ways - some natural, others unexpected and surprising but equally valid. I didn't use the terms "Wittgenstein's Finite Rule Paradox" and "Wittgenstein on rule-following" before as googling them turns up results which look more like philosophy and linguistics than mathematics. My background in logic is nonexistent, I'm looking for any logicians out there who may have seen this before. - 4 In response to your edit, this is philosophy and linguistics, not logic. What is natural or unexpected depends entirely on what kind of patterns you prefer. – Qiaochu Yuan Nov 6 2009 at 22:00 16 -1, this isn't mathematics, and we shouldn't feed the people who are inevitably (and have already) going to come and ask us "what's the next number in ..." – Scott Morrison♦ Nov 7 2009 at 6:08 8 I like the question and it led to 8 nice answers – Gil Kalai Nov 27 2009 at 18:31 2 @Scott: wait, wouldn't an affirmative answer to this question be more like starving people who come asking us to continue a sequence? – aorq Jan 28 2010 at 16:31 3 @Charles Roque: Scott posted his comment when this question was rather stupidly named "Continuing a finite sequence." I just changed it to the current title today. Orthogonally, it's interesting how a title-change translates into interest in a question terms of views and votes. (I got 3 upvotes within a couple of hours.) – Sonia Balagopalan Jan 28 2010 at 18:06 show 4 more comments ## 11 Answers Since you mentioned Wittgenstein's paradox, I thought someone should explain what that is. Warning: As you suspected, what follows is really philosophy and not mathematics. Nevertheless, it seems worth clarifying what people mean by Wittgenstein's paradox so that you don't go chasing wild geese. In Wittgenstein's Philosophical Investigations, he makes an argument about "private languages" that Saul Kripke later interpreted in a certain way. The basic point is that it is very difficult, if not impossible, to pin down what a "rule" is. Imagine that you are trying to teach a Martian the syntactic rule, "append a 1 to the end of a string." The Martian looks puzzled so you give some examples: 0 $\to$ 01 101 $\to$ 1011 0010 $\to$ 00101 and so forth. The Martian seems to get the idea, and does a few examples to confirm with you. The first few examples look good, but then all of a sudden the Martian comes up with 1111111 $\to$ 111111110 Say what? Somehow it seems that the Martian hasn't gotten the rule after all. Or maybe the Martian has extrapolated from your examples to a different rule? How do you make sure you communicate the rule you intend? If you have previously already agreed on some basic rules then you can build on those to define new rules, but how do you get started? It's hard to get more basic than "append a 1." Perhaps you could try building a physical device that optically scans its input and writes a 1 next to it. But any physical device will eventually fail to implement your intended "append a 1" rule when it reaches a certain physical limit, so the device doesn't unambiguously communicate your intended rule to the Martian either. No matter how you slice it, it seems that you can't guarantee that you have communicated your rule to the Martian, since any finite amount of interaction is consistent with infinitely many rules. Once we see this, we could take a more radical step and wonder, maybe I'm the Martian. Maybe all these years I've been assuming that I know what people mean when they specify syntactic rules, but actually I've just been lucky and haven't discovered the discrepancy between my understanding of what "append a 1" means and what everyone else means by it. (Here you can get a glimpse of where Wittgenstein's term "private language" comes into the discussion.) Even more radically, we could wonder whether the notion of a "rule" is incoherent. Perhaps there really is no such thing as a "rule" in the sense of some unambiguous finite description of something that applies to an infinite number of cases. Anyway, it is not my intent to start a debate here on MO about this issue, about which many philosophical papers have been written. It is just to confirm that Wittgenstein isn't who you're looking for, if you want a mathematical answer to your question. - ### 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. As others have mentioned, there can be many ways to continue a sequence, but it can be difficult to argue that a particular way is best. Kolmogorov complexity gives us a quantitative method to say that one way to continue a sequence is better than another. Namely, the "better" method is the one that can be specified with a smaller ruleset. However, this does not necessarily yield a unique answer, and your answer can depend on the language you use to specify your rules. - 1 Also, for sequences with any non-trivial algorithm, like counting trees, the program that just prints the first 10 sequence numbers and then repeats will be in most languages shorter than the honest realization of algorithm. – Ilya Nikokoshev Nov 6 2009 at 23:07 1 I think it is not a real problem that the program doing the "right" thing may be longer than one doing the silly thing: in this case the number of given terms is to small. Thus, one could ask for a program that is shorter than the one that simply lists the terms given and continues with zero. As an aside, Section 2.4 of an article by Kauers and Bostan, arxiv.org/abs/0811.2899 lists some ways to "check" whether a guessed formula is "correct". – Martin Rubey Nov 24 2010 at 14:20 @S. Carnahan: Harvey Friedman has made an attempt to get around the problem that you mention, of the non-canonical choice of language. See cs.nyu.edu/pipermail/fom/2004-January/007798.html and for a clarification of how his proposal differs from Kolmogorov complexity, see cs.nyu.edu/pipermail/fom/2004-January/007805.html Friedman's idea looks promising to me, but it's very sketchy, and I don't think he has developed it further. – Timothy Chow Sep 13 2011 at 1:06 @Timothy: Thank you for letting me know. However, from the first link, it looks like Friedman is avoiding the non-canonical choice of language by choosing a language once and for all. I don't understand some of the vocabulary well enough to tell if my interpretation is correct, though. – S. Carnahan♦ Sep 13 2011 at 5:12 @S. Carnahan: Friedman picks one language to start with, but aims to show later that the resulting concept of simplicity is robust to "reasonable" changes in the choice of language. In the Kolmogorov complexity literature, people generally prove uniqueness up to some unspecified constant c, and they make no effort to investigate c further since it disappears asymptotically. So, e.g., languages containing arbitrarily large but finite lookup tables are allowed. But intuitively, those languages are "unreasonable." (continued) – Timothy Chow Sep 13 2011 at 14:11 show 2 more comments That depends entirely on your definition of "valid." If "valid" means "describable by a Turing machine," then the answer is trivially yes. For example, here are three: • The unique interpolating polynomial of minimal degree. • The periodic sequence given by repeating what was given. • The sequence which is constant after what was given. Edit: Sequence-continuation problems happen to be one of my pet peeves. I do not consider them mathematical exercises, but psychological and cultural: the real goal of any sequence-continuation exercise is to understand what kind of sequences would interest the person who came up with the exercise and what they would consider meaningful. And that is not mathematics. My favorite example of this is the following sequence: $1, 1, \infty, 5, 6, 3, ...$. How does it continue? (I won't spoil the answer; try looking it up on the OEIS if you're stuck. They write $\infty$ as $-1$.) - 7 People who say that sequence-continuation problems are one of their pet peeves are one of my pet peeves. – Michael Lugo Nov 6 2009 at 21:59 1 "Sequence-continuation problems happen to be one of my pet peeves." I totally agree! Psychologically, the reason I'm asking this question is that I want to see a 'proof' that sequence-continuation is not a mathematical exercise. – Sonia Balagopalan Nov 6 2009 at 22:01 What would you consider a "proof"? – Qiaochu Yuan Nov 6 2009 at 22:06 1 And all this time I've been trying to figure out what to feed my peeves... – GS Jan 28 2010 at 13:55 4 Just so I can make my point better, the sequence I gave above is the number of regular polytopes in n dimensions (starting from zero). I cannot imagine any conceivable problem-solving method that could have deduced this from the sequence alone unless you were already familiar with it. – Qiaochu Yuan Jan 28 2010 at 14:01 show 2 more comments There is an area of computer science called grammatical inference. Let $\Sigma$ be a finite alphabet, $\Sigma^*$ be the set of finite length strings over $\Sigma$ and a language be a set of strings. Let $F$ be a family of languages. The data for the problem is a sequence of words $w_0, w_1, \ldots$ from a language in $F$. A learner receives the data and generates a sequence of languages $H_0, H_1, \ldots$ called hypotheses. Typically the hypothesis $H_i$ includes all words provided till that point. The learner stabilizes if the hypotheses do not change after a point and is successful if the stable hypothesis is the language from which the words were drawn. A family of languages can be learnt in the limit if there is an algorithm that can successfully learn the source language. There is a theorem due to E. Mark Gold (Language Identification in the Limit, 1967) stating that if the family $F$ contains all languages of finite cardinality and at least one language of infinite cardinality, it cannot be learnt in the limit. This result may be one formalisation related to the original question, but does not tell the whole story. Dana Angluin (Inference of Reversible Languages, 1982) showed that there are families containing infinitely many languages that can be learnt in the limit. There is a recent text Grammatical Inference: Learning Automata and Grammars by Colin de la Higuera devoted to this area. - It depends on the context in which the sequence is viewed. If some sort of meaningful information is compressed the better the compression the more the digits representing the compressed information resemble a random number and the more the remainder of the sequence is unpredictable and hence has more possible continuations. In some mathematical contexts a random number is desired predictability is bad for a random number. So again in this case there would be many valid continuations. In the case of a random number you do not want low Kolmogorov complexity. There are problems in which randomness helps see the section entitled "When randomness helps" at the wiki article at: [http://en.wikipedia.org/wiki/Randomized_algorithm] - I believe you mean "describable by a polynomial formula", in which case the answer is "yes". Create a system of n equations such that for each polynomial where x = 0, .. , n-1 the polynomial is set equal to . Solve the system of equations for all terms and voila, a formula. Now repeat the same thing for a polynomial of one higher degree, dropping the term so there are still n terms in total. Voila, a second formula. - Nice characterisation of a similar point here: qntm.org/1111 – Seamus Mar 21 2010 at 17:59 Normally, "valid" means "defined by a clear and short rule". Formally we may talk about having a generating program on some abstract machine that is much shorter than the sequence itself. The formal unique continuation statement would be "for each $n$ there is $m$ such that if two infinite sequences generated by valid programs of length $n$ coincide in the first $m$ positions, they coincide in all positions. That is trivially true because there are only finitely many programs of length $n$ but the real mathematical question is whether there is any way to effectively estimate $m$ given $n$? My knowledge of logic is too shaky to answer that but, I hope, somebody more knowledgeable will enlighten us. - Here's how I think about this. Suppose you're given n terms Define , where the denominator skips the factor . Consider the function f(x) = . Then for i an integer between 1 and n, , except that you skip the factor (i-i). Thus, f(i) = ai. But by changing c, you can make the next term f(n+1) whatever you want. - Sonia, you need to define what you mean by "valid". Otherwise, I can continue in any way I want. For instance, here are two different continuations of the sequence "3,1,4,1": 3,1,4,1,0,0,0,0,0,0,0,0... 3,1,4,1,1,1,1,1,1,1,1,1... (There's also a more famous way of continuing this sequence.) - 4 3.1.4.99999999999999999999999999999999999999999999999? – Harry Gindi Jan 28 2010 at 14:38 There is no "a priori" valid way of continuing a sequence... there are many of them! (In fact, uncountable infinitely many). Think of the first terms of your sequence as the integer part of a real number; for example, if you start with $\{2,4,6,8\}$ think you have the number $2468,...$. In how many ways can you continue this real number by its decimal part? You see that we have $2468,0000...$ and $2468,1000...$, and $2468,01000...$, and $2468,101001000...$ and... that is, we can take any real number in the interval $[2468,2469]$. But for any of these reals we have an obvious way to continue your sequence, namely by taking every next decimal as the next element of the sequence (e.g., $2468,10111213...$ gives $\{2,4,6,8,1,0,1,1,1,2,1,3,...\}$. Thus, there are at least as many ways of continuing your sequence as there are real numbers in $[2468,2469]$, i. e., uncountably many. (Note that for every of these reals, we could also take every TWO (or three, or four...) decimals as the next element of the sequence, or even take an arbitrary number of decimals for every new element; whence there are in fact "more" ways to continue the sequence than real numbers in that interval). You would have a different case if you had a good criterion for determining "a posteriori" which sequence continuations are accepted as 'natural' and which ones not; but then, the number of 'valid' continuations will completely depend on that criterion (and should be computed in every case), and not in any axiom or theorem of general mathematics. - (Most of the things I post have already been covered in the previous answers, but not combined this way). Here is my view of the problem: Given a finite sequence of numbers $a_1,...,a_n$ we can use the Lagrange interpolation formula to get a polynomial $P(x)$ so that $a_k=P(k) \forall k \leq n$. As yuan mentions this is an unique answer in the following two (equivalent) ways: - The only polynomial of degree at most $n-1$ which fits this data - The polynomial of smallest degree which fits this data. Now pick a function $f : R \rightarrow R$ {\bf at random}. Let $g(x)= P(x)+ (x-1)(x-2)...(x-n)f(x)$. Then $g(1)=a_, ..., g(n)=a_n$, and this provides uncountably many different ways of continuing $a_1,...,a_n$. Second point: Pick an $x$ at random. Isn't then any "natural" way of continuing the sequence $a_1,...,a_n, a_{n+1}=x$ also in some sense a "natural" way of continuing $a_1,...,a_n$? But each choice of $x \in R$ leads to a different 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": 45, "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": 0.9467251300811768, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/32506/counting-points-on-lattices/32520	## Counting points on lattices ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I expect that the following is a standard problem from analytic number theory, but I don't know where exactly to look for an answer. Let f: ℤr→ H be a surjective homomorphism into a finite group. Let `$S(N)= \frac{1}{N^r}\#\{(x_1,\dots,x_r)\in \ker f\colon 0\leq x_i <N\}$`. One expects that S(N) is roughly \|H\|-1. My question is: What is the best known estimate of the error term S(N)-\|H\|-1 in terms of N and the structure of H? I am especially interested in the case, when H=(ℤ/pnℤ)d, for some d<r. To give some idea of what kind of results I am looking for, I will give the estimate, that I managed to find myself. If "e" is the exponent of the group H, and "h" is its size, then by estimating character sums one gets `$S(N) - h^{-1} \ll h^{-1}(\log \min\{h,N\})^r \max\{eN^{-1},e^rN^{-r}\}$`, where the implicit constant depends only on r. I think, that this can be improved at least when N is small with respect to e. - ## 2 Answers Note that $ker(f)$ is a sublattice of the integer lattice `$\mathbb{Z}^r$`. And conversely any such integer sublattice will give you such a homomorphism $f$. The index of this sublattice is exactly what you call $h$ (since the homomorphism is surjective). If $C$ is a symmetric convex body in `$\mathbb{R}^r$` and $L$ a full rank lattice then the number of lattice points in $t C$ (the dilation of $C$ by a real factor of $t$) is asymptotic to $t^r vol(C)/D$, where $D$ is the volume of the fundamental domain of $L$. In your case $D=h$ and $C$ is the unit hypercube. Getting a good remainder is usually a much harder job. In general the remainder is bounded by something proportional to the area of $C$. When $C$ has a smooth boundary (which alas, the hypercube doesn't) you can get better estimates (but usually have to work quite hard). When you're working in a high dimension a large fraction of the volume is close to the vertices -- making things much more difficult. In particular you might look at the work of Martin Huxley. For example his book "Lattice Points, Area and Exponential Sums" or a number of his papers on this subject. - ### 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. Not a full answer, but here is a standard approach to questions of this kind. [Added Upon a further reflection, I am not sure whether this approach is the best or robust enough.] The set `$$\Gamma_f=\{(x,y)\in\mathbb{Z}^r\oplus\mathbb{Z}: x\in\operatorname{Ker} f,\ 0\leq x_i\leq y \text{ for } 1\leq i\leq r\}$$` is an affine semigroup whose semigroup algebra is the homogeneous coordinate ring of a projective toric variety with Hilbert polynomial $H_{\Gamma_f}(t)$ with respect to a positive $\mathbb{Z}$-grading by $\deg y.$ For a large enough $N$, the number that you are interested in is $\frac{1}{N^r}H_{\Gamma_f}(N-1)$ and the error term you need is determined by the subleading coefficient of $H_{\Gamma_f}(t).$ It can probably be extracted from the information in Miller and Sturmfels, Combinatorial Commutative Algebra. - To supplement this -- by applying the inverse of the matrix formed by the basis of the lattice $ker(f)$ we reduce to the problem of counting integral lattice points in a polytope. This paper math.ucdavis.edu/~deloera/RECENT_WORK/… contains a nice survey. In particular using the Hilbert Polynomial (as Victor P. suggested) will be helpful. – Victor Miller Jul 21 2010 at 13:27	{"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": 0.9357429146766663, "perplexity_flag": "head"}
http://mathoverflow.net/questions/4361/cohomology-of-fibrations-over-the-circle/4428	Cohomology of fibrations over the circle Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Are there any general results on the (integral) cohomology of manifolds that are fibrations over the circle? Any literature references much appreciated. - 1 There's all kinds of general results as the problem is pretty general and fits into a standard formalism where many of the details are already worked out. Is there a more specific issue you're interested in? – Ryan Budney Nov 6 2009 at 18:50 6 Answers The above Mayer-Vietoris argument gives the cohomology of a fiber bundle over a circle in a concrete fashion. For sake of "mathematical culture", I thought I'd mention what happens for fiber bundles over a higher dimensional sphere (this is also a good excuse for me to test drive the new latex support). For fiber bundles $F \hookrightarrow E \rightarrow S^n$ with $F$ connected and $n \geq 2$, the Serre spectral sequence degenerates in a very simple fashion into what is known as the Wang exact sequence. Namely, we have a long exact sequence of the form $$\cdots \rightarrow H^k(E) \rightarrow H^k(F) \rightarrow H^{k-n+1}(F) \rightarrow H^{k+1}(E) \rightarrow \cdots$$ The proof of this is completely analogous to the proof of the better known Gysin exact sequence, which tells you what happens for fiber bundles whose fibers are spheres. A reference for this material is McCleary's "User's guide to spectral sequences", page 145. - 2 This is also a Mayer-Vietoris argument, decomposing E as the pull-back of the decomposition of the sphere into two discs. But you also need the computation of the homology of F x S^{n-1} so it's a smidgen more complicated. This is also why the map H^k(F) --> H^{k-n+1} does not preserve dimension, the the homology of S^{n-1} introduces a degree shift via the Kunneth Theorem. – Ryan Budney Nov 6 2009 at 20:54 An excellent point. One thing to remark about the Wang exact sequence, though, is that it also works for simply connected homology spheres. – Andy Putman Nov 6 2009 at 21:13 6 The meta-point of bringing up the Wang/Gysin exact sequence is that they are illustrations of the fact that spectral sequences with only two nonzero rows/cols degenerate nicely into long exact sequences. This bit of algebra shows up a lot, so it is useful to internalize it. – Andy Putman Nov 6 2009 at 21:22 1 The Mayer-Vietoris argument also works for simply-connected homology spheres. You decompose the homology sphere into the union of a disc and a homology ball, and then the remainder follows analogously. Basically, this boils down to the observation that the Serre spectral sequence is just Mayer-Vietoris + Kunneth applied to a cell decomposition of the base space. – Ryan Budney Nov 6 2009 at 21:39 3 Ah, okay. But I like cheating. :) – Ryan Budney Nov 7 2009 at 1:31 show 1 more comment 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. This began as a comment, but it is interesting enough that I decided to make it an answer instead. Let's consider bundles over the circle whose fibers are closed genus $g$ surfaces $\Sigma_g$. The diffeomorphism type of the total space of our bundle only depends on the isotopy class of the monodromy map. Denote by $M_g$ the mapping class group, ie the group of isotopy classes of orientation-preserving diffeomorphisms of $\Sigma_g$. For $f \in M_g$, denote by $B_f$ the surface bundle over the circle determined by $f$. In a comment, Tom Church observed that the homology of $B_f$ will be the same as the homology of the trivial bundle if and only if $f$ acts trivially on $H_1(\Sigma_g)$. The group of mapping classes that act trivially on $H_1(\Sigma_g)$ is known as the Torelli group. One could demand more. Namely, we could require that the cup-product structure on $H^{\ast}(B_f)$ be the same as the cup product structure on the trivial bundle. As Tom observed, a beautiful theorem of Dennis Johnson gives a precise characterization of the subgroup of $I_g$ consisting of monodromies with this property. One easy way of describing it is that it is the kernel of the (outer) action of $M_g$ on the second nilpotent truncation of $\pi_1(\Sigma_g)$ (the group $H_1(\Sigma_g)$ is the first nilpotent, ie abelian, trunctation). The story does not end here. A topological space has a "higher-order" intersection theory given by the so-called Massey products. They are sort of like generalized cup products. Anyway, Kitano generalized Johnson's work and gave a precise and beautiful description of the monodromies of surface bundles over the circle in which these higher intersection products (up to a certain level) are trivial. Namely, all the degree at most $k$ Massey products of $B_f$ will be trivial if and only if $f$ acts trivially on the (k+1)st nilpotent truncation of $\pi_1(\Sigma_g)$. For the details of this plus references to Johnson's papers, see the following paper: MR1381688 (97f:57014) Kitano, Teruaki(J-TOKYTE) Johnson's homomorphisms of subgroups of the mapping class group, the Magnus expansion and Massey higher products of mapping tori. (English summary) Topology Appl. 69 (1996), no. 2, 165--172. - For 3-manifolds, the cohomology could be almost anything surjecting $\mathbb{Z}$. In this case, the fiber is a surface, and since $\mathrm{Mod}(S) \to \mathrm{Sp}(2g, \mathbb{Z})$ is surjective, you can make the action of the monodromy on the homology of the fiber anything you like (as long as it preserves the intersection form). - 2 If the 3-manifold is the complement of a fibered knot K in S^3, so the fiber is a punctured surface of genus g(K), then H^1(S^3-K) is exactly Z and the characteristic polynomial of that monodromy has to be the Alexander polynomial of K. – Steven Sivek Nov 6 2009 at 13:42 Thanks for these interesting examples. Is there anything like the Kunneth formula in these cases? – Dan Nov 6 2009 at 14:33 In general you have the Leray-Hirsch condition; from this you can show that for a closed 3-manifold fibered over the circle, H(M) = H(S) tensor H*(S^1) as groups iff the monodromy action on the homology of the fiber is trivial. Work of Johnson implies that for this to be an isomorphism of rings (i.e. cup product behaves like in Kunneth) you need the monodromy to lie in the Johnson kernel (a certain subgroup of the mapping class group). This should tell you that Kunneth-like behavior in higher dimensions will be very difficult to guarantee. – Tom Church Nov 6 2009 at 23:57 Given a bundle $F \to M \to S^1$, the Mayer-Vietoris sequence corresponding to the decomposition of $M$ coming from writing $S^1$ as the union of two intervals tells you there's a short exact sequence: $$0 \to coker( f_n - I ) \to H_n(M) \to ker( f_{n-1} - I ) \to 0$$ Here $f_n : H_n F \to H_n F$ is the induced map from the monodromy of the bundle, ie: you think of the bundle as $R \times_f F, f: F --> F$ a homeomorphism / diffeomorphism / whatever. And $I$ is the identity map on $H_n(M)$ and $H_{n-1} M$ respectively. There's a similar decomposition for cohomology, and this is what the Serre spectral sequence gives you, too. The short-exact sequence basically encodes the extension problem from the spectral sequence. - Not very exciting, but the Euler characteristic is zero. I assume that you are aware of the Serre spectral sequence? - If the manifold happens to be aspherial manifold(that means the universal covering spacec is contractible), then the cohomology groups is just cohomology of the fundamental groups, which can be detecked by the fundamental group of the fiber and S^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": 33, "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": 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": 0.9230321049690247, "perplexity_flag": "head"}
http://faculty.smu.edu/reynolds/arkode/	# The ARKode Solver¶ ARKode is a solver for stiff and nonstiff systems of ordinary differential equations (ODEs) given in explicit form, $M \dot{y} = f_E(t,y) + f_I(t,y).$ Here, • $$t$$ is the independent variable, e.g. time, • $$y$$ is the set of dependent variables (in $$\Re^N$$), • $$M$$ is a user-specified, non-singular operator from $$\Re^N \to \Re^N$$ (the “mass matrix”, possibly time dependent, but independent of $$y$$), • $$f_E(t,y)$$ is the portion of the right-hand side containing the “slow” time scale components in the system that should be integrated explicitly, and • $$f_I(t,y)$$ is the portion of the right-hand side containing the “fast” time scale components to be integrated implicitly. Either of the $$f_E$$ or $$f_I$$ operators may be disabled, allowing for fully explicit, fully implicit, or combination implicit-explicit (IMEX) time integration. ARKode is a component of the SUNDIALS suite of nonlinear and differential/algebraic equation solvers. ARKode is written in C, with C++ and Fortran interfaces. ## News:¶ 23 January 2013: ARKode is still in development. An official release as part of the SUNDIALS suite is planned for summer 2013, but early access to the development repository will be opened up much sooner. Both events will be posted to this page when they occur. ## Contact¶ Daniel Reynolds, Department of Mathematics, Southern Methodist University ## Support¶ This work is supported by the U.S. Department of Energy through the FASTMath SciDAC Institute, under subcontract B598130 from Lawrence Livermore National Laboratory.	{"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": 10, "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": 0.8919576406478882, "perplexity_flag": "middle"}
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However, the bare masses of the up and down quarks are only around 5 MeV, quite a bit ...	{"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": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 20, "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": 0.9276655912399292, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/39243/enumerating-triangulations-of-quadrangulations-in-cycles	## Enumerating triangulations of quadrangulations in cycles ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Consider a cycle of length $(2n+2)$. Now we quadrangulate this cycle into $n$ quadrants. We want to enumerate the number of quadrangulations, and we denote this number by $q_n$. Now we triangulate this quadrangulation by triangulating each quadrant. We denote the number of triangulations $t_n$. It is clear that $t_n = 2^nq_n$. Do I count some triangulations multiple times? That is, will some triangulation be the result of multiple different quadrangulations? - When I talk about quadrangulations, I mean planar quadrangulations. – utdiscant Sep 18 2010 at 20:21 ## 2 Answers No triangulation occurs multiple times. At least one of the triangles $T$ of the triangulation has two edges on the boundary. There is a unique quadrant (I would say quadrilateral) $Q$ made up of two triangles of the triangulation, one of which is $T$. Remove the three edges of $T$, obtaining two quadrangulated cycles (possibly degenerate), and induct. Note that not every triangulation of a $(2n+2)$-cycle is obtained by bisecting the quadrilaterals of a quadrangulation. Also, it is well-known that the number of quadrangulations of a $(2n+2)$-gon is $\frac{1}{2n+1}\binom{3n}{n}$. - ### 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. Another way to think about this is that each triangulation gives you a binary tree, while each quadrangulation gives you a ternary tree (consider the dual graphs). If a triangulation comes from a quadrangulation then the correspondence gives you a perfect matching on the binary tree. Now show that every tree has at most one perfect matching. This translates to: every triangulation comes from at most one quadrangulation. - This approach seems easiest to prove. – utdiscant Sep 18 2010 at 23:11	{"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": 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": 0.9274721145629883, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/4994/fundamental-examples/24017	## Fundamental Examples ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It is not unusual that a single example or a very few shape an entire mathematical discipline. Can you give examples for such examples? (One example, or few, per post, please) I'd love to learn about further basic or central examples and I think such examples serve as good invitations to various areas. (Which is why a bounty was offered.) Related MO questions: What-are-your-favorite-instructional-counterexamples, Cannonical examples of algebraic structures, Counterexamples-in-algebra, individual-mathematical-objects-whose-study-amounts-to-a-subdiscipline, most-intricate-and-most-beautiful-structures-in-mathematics, counterexamples-in-algebraic-topology, algebraic-geometry-examples, what-could-be-some-potentially-useful-mathematical-databases, what-is-your-favorite-strange-function To make this question and the various examples a more useful source there is a designated answer to point out connections between the various examples we collected. In order to make it a more useful source, I list all the answers in categories, and added (for most) a date and (for 2/5) a link to the answer which often offers more details. (~year means approximate year, year means a year when an older example becomes central in view of some discovery, year? means that I am not sure if this is the correct year and ? means that I do not know the date. Please edit and correct.) Of course, if you see some important example missing, add it! Logic and foundations: $\aleph_\omega$ (~1890), Russell's paradox (1901), Halting problem (1936), Goedel constructible universe L (1938), McKinsey formula in modal logic (~1941), 3SAT (1970), The theory of Algebraically closed fields (ACF) (?), Physics: Brachistochrone problem (1696), Ising model (1925), The harmonic oscillator,(?) Dirac's delta function (1927), Heisenberg model of 1-D chain of spin 1/2 atoms, (~1928), Feynman path integral (1948), Real and Complex Analysis: Harmonic series (14th Cen.) {and Riemann zeta function (1859)}, the Gamma function (1720), li(x), The elliptic integral that launched Riemann surfaces (1854?), Chebyshev polynomials (?1854) punctured open set in C^n (Hartog's theorem 1906 ?) Partial differential equations: Laplace equation (1773), the heat equation, wave equation, Navier-Stokes equation (1822),KdV equations (1877), Functional analysis: Unilateral shift, The spaces $\ell_p$, $L_p$ and $C(k)$, Tsirelson spaces (1974), Cuntz algebra, Algebra: Polynomials (ancient?), Z (ancient?) and Z/6Z (Middle Ages?), symmetric and alternating groups (1832), Gaussian integers ($Z[\sqrt -1]$) (1832), $Z[\sqrt(-5)]$,$su_3$ ($su_2)$, full matrix ring over a ring, $\operatorname{SL}_2(\mathbb{Z})$ and SU(2), quaternions (1843), p-adic numbers (1897), Young tableaux (1900) and Schur polynomials, cyclotomic fields, Hopf algebras (1941) Fischer-Griess monster (1973), Heisenberg group, ADE-classification (and Dynkin diagrams), Prufer p-groups, Number Theory: conics and pythagorean triples (ancient), Fermat equation (1637), Riemann zeta function (1859) eliptic curves, transendental numbers, Fermat hypersurfaces, Probability: Normal distribution (1733), Brownian motion (1827), The percolation model (1957), The Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble, SLE (1999), Dynamics: Logistic map (1845?), Smale's horseshoe map(1960). Mandelbrot set (1978/80) (Julia set), cat map, (Anosov diffeomorphism) Geometry: Platonic solids (ancient), the Euclidean ball (ancient), The configuration of 27 lines on a cubic surface, The configurations of Desrague and Pappus, construction of regular heptadecagon (1796), Hyperbolic geometry (1830), Reuleaux triangle (19th century), Fano plane (early 20th century ??), cyclic polytopes (1902), Delaunay triangulation (1934) Leech lattice (1965), Penrose tiling (1974), noncommutative torus, cone of positive semidefinite matrices, the associahedron (1961) Topology: Spheres, Figure-eight knot (ancient), trefoil knot (ancient?) (Borromean rings (ancient?)), the torus (ancient?), Mobius strip (1858), Cantor set (1883), Projective spaces (complex, real, quanterionic..), Poincare dodecahedral sphere (1904), Homotopy group of spheres, Alexander polynomial (1923), Hopf fibration (1931), The standard embedding of the torus in R^3 (1934 in Morse theory), pseudo-arcs (1948), Discrete metric spaces, Sorgenfrey line, Complex projective space, the cotangent bundle (?), The Grassmannian variety,homotopy group of spheres (1951), Milnor exotic spheres (1965) Graph theory: The seven bridges of Koenigsberg (1735), Petersen Graph (1886), two edge-colorings of K_6 (Ramsey's theorem 1930), K_33 and K_5 (Kuratowski's theorem 1930), Tutte graph (1946), Margulis's expanders (1973) and Ramanujan graphs (1986), Combinatorics: tic-tac-toe (ancient Egypt(?)) (The game of nim (ancient China(?))), Pascal's triangle (China and Europe 17th), Catalan numbers (mid 19th century), (Fibonacci sequence (12th century; probably ancient), Kirkman's schoolgirl problem (1850), surreal numbers (1969), alternating sign matrices (1982) Algorithms and Computer Science: Newton Raphson method (17th century), Turing machine (1937), RSA (1977), universal quantum computer (1985) Social Science: Prisoner's dilemma (1950) (and also the chicken game, chain store game, and centipede game), the model of exchange economy, second price auction (1961) - 15 I think that this should be community wiki. – Andrew Stacey Nov 11 2009 at 7:55 12 @Jose: Hard to say exactly. My instinct is that the kind of answers that this question will garner are those that didn't involve much actual thought, and the votes up or down will be more an assessment of whether the voter liked the example rather than whether the voter liked the answer (which, ideally, should contain an explanation of why that example shaped the discipline); both of these indicate that the answerers should not gain reputation for their answers, hence community wiki. – Andrew Stacey Nov 11 2009 at 9:50 4 I've hit this with the wiki hammer. – Scott Morrison♦ Nov 11 2009 at 19:34 7 I can't imagine a counterexample to the following rule: Any question whose purpose is to produce a sorted list of resources (i.e. the question includes, or should include, "one per post please") should be community wiki. – Anton Geraschenko♦ Nov 12 2009 at 8:03 8 Why does this question have a bounty anyway? – Kevin Lin Nov 21 2009 at 17:33 show 17 more comments ## 129 Answers Three related fundamental examples in random matrix theory (in mathematical physics and probability) are the Gaussian Orthogonal Ensemble, the Gaussian Unitary Ensemble, and the Gaussian Symplectic Ensemble. Much of random matrix theory has been devoted to determining the properties of these families of random matrices and proving that other families exhibit the same behaviors. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The discovery of Transcendental numbers, or numbers that are not the root of any finite polynomial with rational coefficients. Also, the proof that e and π were transcendental, the latter via the proof that ea is only algebraic for transcendental values of a (and eiπ = -1 is algebraic, as is i, so therefore π must be transcendental). Their discovery, as well as the first explicitly created example, the Liouville number, sparked what's called "Transcendence theory". And as it turns out, any randomly chosen real number is "almost surely" transcendental. In other words, the density of transcendental numbers among the real numbers is 1!! - The Sorgenfrey line is an example that has motivated a lot of research in general topology, mostly generalized metric properties and ordered space theory. It's an example of a hereditary normal space with non-normal square, it is separable, Lindelöf, first countable, but not second countable; a generalized ordered space that is not orderable, and many more. - The rational parametrization of the locus of the equation $X^2+Y^2=1$ by $(\frac{t^2-1}{t^2+1},\frac{2t}{t^2+1})$. It can be viewed geometrically by taking a line that intersects the unit circle at one rational point and then considering all possible (rational) slopes of the line (including infinity), which are in correspondence with (rational) points of the circle. This is the most basic example of using a geometric idea to find solutions to a diophantine equation, and it leads to very deep mathematics. - Motivated by Amit Kumar Gupta's answer about the continuum hypothesis, let me add an example that is less natural but has inspired an amazing amount of set theory, namely Suslin's Hypothesis. This conjecture, proposed in 1920 and now known to be independent of ZFC, says that the real line with its usual ordering relation is characterized up to isomorphism by the following properties: • dense linear order without endpoints • Dedekind-complete • No uncountable family of pairwise disjoint open intervals. The point of the conjecture is that it was proved much earlier by Cantor that one gets a characterization of $\mathbb R$ if one puts in place of the last property the stronger statement that there is a countable dense set. So Suslin is simply asking whether one can weaken this separability assumption to the third property in the list above (often called the "countable chain condition"). I can't claim that this question is anywhere near as natural as the continuum hypothesis, but what makes it important (in my opinion) is its impact on the development of set theory. The fact that Suslin's hypothesis is false in Gödel's constructible universe $L$ was one of the first applications (and probably a major motivation, though I don't actually know that) for Jensen's theory of the fine structure of $L$, a theory that has grown tremendously as a component of the inner model program in contemporary set theory. The fact that Suslin's hypothesis is consistent with ZFC was the initial application and the motivation for the theory of iterated forcing, now a central tool in set theory. It also provided the occasion for the invention of Martin's axiom. That axiom and the combinatorial principles isolated by Jensen from the fine structure of $L$ have become standard tools for proving independence results without explicitly referring to forcing or to $L$. - The irrational rotation on the torus $\mathbb{T^2}$: $T_t (z,w) = (e^{2 \pi i \alpha }z, e^{2 \pi i \beta} w)$, where $\alpha, \beta \in \mathbb{R}, \frac{\alpha}{\beta} \notin \mathbb{Q}$. It is perhaps the first nontrivial example of an ergodic dynamical system. By considering its orbits $t \mapsto T_t (z,w)$, one gets examples of one-to-one immersions (with dense image) which are not embeddings, immersed submanifolds which are not closed/regular submanifolds, Lie subgroups which are not closed subgroups, non-Hausdorff quotient space and similar phenomena. - In combinatorics there are very simple basic graphs from which a whole lot of theory came. For example the complete graphs `K_5` and `K_{3,3}` which alone provide the ground level for any non-planar graph according to Kuratowski's theorem. Another simple graph that gave rise to a huge amount of theory is Petersen's graph, which I like to think as the graph whose vertices are the ten two-element subsets of `{1,2,3,4,5}`, and for which two such vertices are connected iff they are disjoint. A link for Kuratowski's theorem is http://en.wikipedia.org/wiki/Kuratowski's_theorem - Second price auction (or Vickrey auction): an auction in which the bidder who submitted the highest bid is awarded the object being sold and pays a price equal to the second highest amount bid. - 1 It seems close to the classic pie division procedure. The cutter cuts and the other chose first the part he wants. – ogerard May 13 2010 at 17:12 Theorem on Friends and strangers in Ramsey Theory. - Dear Gil, I will quibble with your listing the Fano Plane under graph theory rather than under geometry or perhaps combinatorics. Now I know people who do not think the theory of finite planes is very geometric, and I agree there is much truth here. In some cases it requires algebraic rather than geometrical work to make progress, and in other cases combinatorial ideas. Yet, trying to "imitate" in the finite plane world interesting geometric phenomenon in the Euclidean, projective, or hyperbolic planes I think has proved very fruitful. I don't really see that the Fano Plane leads to graph theory questions that are of great interest, and that would not be raised from some other point of view. Best, Joe - 3 Shouldn't this be a comment, rather than an answer? – Grétar Amazeen Dec 12 2009 at 22:24 1 Based on the tiny amount I know, I think your comment makes sense. But it's still a comment, not an answer. – Darsh Ranjan Dec 13 2009 at 0:46 1 I tried to do this as a comment but the system did not seem to want to allow me to do this. – Joseph Malkevitch Dec 13 2009 at 12:56 1 Now that this has been changed, shouldn't this answer be deleted from the list of examples? – Douglas Zare Jan 13 2010 at 15:51 show 1 more comment The Delaunay triangulation is fundamental in computational (Euclidean) geometry. For a finite point set S in general position, it can be defined in several ways: (1) as the unique triangulation in which every simplicial cell is Delaunay (i. e., its circumsphere does not contain any points of S in its interior), (2) as the uniqe triangulation in which every facet (of any dimension) of every simplicial cell is Delaunay (meaning it has some empty circumsphere), or (3) as the dual of the Voronoi diagram (which is also fundamental). In the plane, the Delaunay triangulation has the additional property of maximizing the smallest angle of all its triangles, among all triangulations. The Delaunay triangulation is usually the most obvious candidate for "the right" triangulation of a given point set, and most simplicial mesh-generating methods seem to be based on it. It doesn't hurt that there are reasonably fast and elegant algorithms for constructing it (very fast in the plane, but unfortunately (and necessarily) exponential in the dimension in the worst case). - The solution of Kirkman's Schoolgirl Problem is the archetypal example of a resolvable triple system. This example essential shaped the entirety of Design Theory. We might also consider Euler's 36 Officers Problem to be one of the fundamental counter-examples within this field. Answer by Disonnant - Taking introductory topology, I got the impression that the real line is the fundamental example of a topological space. I wouldn't be surprised if the open and closed intervals of $\mathbb{R}$ were the prototypical examples of open and closed sets, and I think many important topological properties---including compactness, connectedness, and Hausdorffness---first arose because you need them to prove obvious facts about $\mathbb{R}$ and its subsets. - 1 There are several examples that I would regard as "too fundamental" for the list like: 0, 1,2, $\sqrt 2$, the real numbers, the natural numbers, the prime numbers, the triangle. I also consider Alef_0 and Alef as "too fundamental" and chose Alef_\omega to start the set theory examples. – Gil Kalai Feb 20 2010 at 7:33 show 4 more comments $\mathbb{Q}_{p}$. The field of p-adic numbers brings the study of local methods. Hensel's lemma is a great example. It is also interesting that p-adic integers is the projective limit of the rings $\mathbb{Z}/p^{n}\mathbb{Z}$. - To make this question and the various examples a more useful source this is a designated answer to point out connections between the various examples we collected. please indicate only strong, definite, nontrivial, and clear connections. 1) The Petersen graph is obtained by identifying antipodal vertices and edges in the graph of the dodecahedron - one of the five platonic solids. Such an identification gives a polyhedral complex realizing the real projective plane. Applying this operation to the icosahedron leads to a 6-vertex triangulation of the real projective plane. - The Möbius strip or Möbius band (a surface with only one side and only one boundary component). - I think no one's pointed Lorenz equations. - The cyclotomic field $\mathbb{Q}(\zeta_p)/\mathbb{Q}$ is the most basic example of a field extension in which splitting of primes depends on an obvious congruence condition. Specifically, if $\ell$ is another prime, then the Frobenius of $\ell$ is $\ell \mod p \in (\mathbb{Z}/p)^\times = \mathrm{Gal}(\mathbb{Q}(\zeta_p)/\mathbb{Q})$. In particular, $\ell$ splits in the field iff its Frobenius is trivial, and this is true iff $\ell \equiv 1 \mod p$. We can then relate other congruences to splitting in subfields of $\mathbb{Q}(\zeta_p)$, etc. The theorems of global class field theory show that this basic concept holds in a very general case, although the general case is much harder to prove. This basic example, does, however, motivate the ideas in class field theory, which have greatly influenced modern number theory and related areas. (As an added note, the fact that the Artin reciprocity law is true for cyclotomic fields is actually a key ingredient in the proof for general abelian extensions!) Answered by Davidac897 - Heisenberg model of 1-D chain of spin 1/2 atoms, solved exactly by Bethe in 1931, is where Bethe Ansatz was born, and with it the field of integrable models in statistical and quantum mechanics. Answered by Mio - The gamma function is a fundamental example of an interesting function defined only on the integers which has an analytic (meromorphic) continuation to the whole complex plane. This ability to extend an interesting, seemingly discrete function to a complex differentiable function motivates a lot of later material. Answered by Davidac897 - The semicircular law and the Marchenko-Pastur distribution are fundamental examples of probability distributions in random matrix theory. - The free group factors $L(\mathbb{F}_{n})$, which are the closer in the weak operator topology of the left regular representation of the free group $\mathbb{F}_n$, are fundamental examples in von Neumann algebras. The isomorphism question is the root of the so important Free Probability theory of Voiculescu. - In geometric/combinatorial group theory, Grigorchuk's 2-group is a fundamental example. It originally was constructed as a particularly elegant infinite finitely generated torsion group. Then Grigorchuk showed it had intermediate growth, answering Milnor's problem. For a time it was seen as the universal counterexample in group theory. But now it has spawned a theory of groups acting on rooted trees, self-similar groups and branch groups. Pierre de la Harper has an entire chapter of his book devoted to this group. - The Lorenz system of ordinary differential equations: $$\dot x=\sigma(y-x)$$ $$\dot y = rx-y-xz$$ $$\dot z = xy-bz$$ ($\sigma$, $r$, $b$ are parameters) is a good example in dynamical system. It is an example of a deterministic system displaying chaotic behaviour. Also the Lorenz attractor. Date: 1963. - Fermat hypersurfaces and their Zeta functions. Udi de Shalit proposed: One example that comes to my mind is the Zeta functions of the Fermat hypersurfaces, which were studied by Weil and are known to have helped him in formulating the Weil conjectures, later on proved by Grothendieck and Deligne. - I suppose the discrete metric space is a crucial example in the metric spaces theory and in the introductory mathematical analysis. It shows many aspects and pathological behavior of metric spaces in general. - Dirichlet's theorem is the first use of analysis to prove a number theoretic result which does necessarily seem analytic. His proof leads to a lot of ideas about distributions of primes, many of which used analysis. It even leads to an analytic proof of one of the inequalities in class field theory, a result which can also be proved using a good deal of cohomology and which is therefore not exclusively analytic. (I am not counting the prime number theorem, since that is an asymptotic result and thus reeks of analysis as soon as it is conjectured. It was also proven later.) - The (complex analytic) proof of the Prime Number Theorem is the first major use of complex analysis to prove results about asymptotic behavior of prime numbers, which, at first glance, do not at all seem to be tied to complex numbers. This has led to an enormous amount of mathematics. - The field extension $\mathbb{Q}(\sqrt[3]{2})/\mathbb{Q}$ is the most basic example of an algebraic extension which is not Galois. In particular, one notices that it has no non-trivial automorphisms, and that this is related to the fact that not enough of the roots of $x^3-2$ are in this field. This leads to the key concept of Galois extensions and the relation between automorphisms and roots. It also led Galois to develop the concept of normal subgroup. - The pseudo-arc in continuum 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": 39, "mathjax_display_tex": 3, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9140534996986389, "perplexity_flag": "middle"}
http://scicomp.stackexchange.com/questions/2788/quickly-computing-inversion-of-a-large-sparse-partial-stochastic-matrix	# Quickly computing inversion of a large sparse partial stochastic matrix Suppose I have a sparse stochastic matrix $M$ (with thousands or millions of stochastic column vectors), possibly encoding some links in a web graph. Now I split it into two matrices: $D$ containing only the diagonal entries of $M$, and $R$ containing the remaining entries of $M$. What would be a fast way to compute $D(I−R)^{−1}$ (or a good approximation)? Instead of low computational complexity, I'm looking for fast practical performance. What I actually care about is $D(I−R)^{−1}$ for a continuous stream of vectors $x$'s, though $M$ may also change (both nodes and edges), but less often. Looks similar to PageRank, but still quite different. What would be a fast implementation? - Thanks, Michelle - Hello Michelle and welcome to Scicomp! Is the matrix $(I-R)$ well-conditioned, in general? This can have a big impact on whether approximating its inverse is practical or not. – Paul♦ Jul 12 '12 at 15:23 What do you know about the entries of $M$? You say they are sparse, so the vectors that make up $M$ are sparse as well. Are the few non-zero entries simply ones, or are they distributed real numbers? Also, are the diagonal entries of $M$ (i.e., the matrix $D$) always nonzero? – Wolfgang Bangerth Jul 12 '12 at 17:25 Hello Paul, thanks. I don't know whether $(I-R)$ is technically well-conditioned, but intuitively the answer varies gradually with gradual changes in $M$. – Michelle Dolly Jul 12 '12 at 18:12 Wolfgang, The non-zero entries are usually not 1, but distributed reals (< 1). Some diagonal entries may be zero, but most will be non-zero. Thanks. – Michelle Dolly Jul 12 '12 at 18:15 1 Too bad. I had hoped that the diagonal entries are as sparse as the rest which would mean that $D$ would have only few nonzeros on the diagonal -- making the problem substantially lower-dimensional. – Wolfgang Bangerth Jul 12 '12 at 23:41 show 4 more comments ## 1 Answer Here is an idea. How about developing $(I-R)^{-1}$ in a geometric series? It only converges if all the eigenvalues of R are smaller than 1. But if it does then $D(I-R)^{-1} = D(I+R+R^2+\ldots)$ The problem would be reduced to (hopefully) not many sparse matrix multiplications. -	{"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": 0.9328204393386841, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/207344/how-to-find-the-numbers-that-sum-up-to-a-given-number?answertab=votes	# How to find the numbers that sum up to a given number? I have a list of numbers, finite, about 50 and I want to know which permutations with subsets of that set sum up to a given number. I found a formula for the number of ways but I don't know how to find the actual numbers. Can you help me? - 1 "which permutations with subsets of that set"? – Michael Zhao Oct 4 '12 at 19:05 1 – ziyuang Oct 4 '12 at 19:14 1 So, just to be clear, for example, you have {1, 2, 3, 4, 9, 10, 13, 18}, and you want to know how many ways you can add some subset together to get 23? And, you want to know the numbers involved as well, not just the number of ways to do it? – Graphth Oct 4 '12 at 19:15 It's similar to the subset-sum problem and might be transformed to that. As far as I can tell you have understood the question correctly. I think it is a nice question because it is easy to understand. – Nick Rosencrantz Oct 4 '12 at 21:10 ## 1 Answer I assume you mean you want to find the subsets of your list whose sum is a given number $S$. You can do it with dynamic programming. Suppose your numbers $x_1, \ldots, x_n$ are positive integers and the target sum is $s$. If $A_{k,m}$ for $1 \le k \le n$, $0 \le m \le s$, is the number of subsets of $\{x_1, \ldots, x_k\}$ whose sum is $m$, we have $A_{1,0} = A_{1,x_1} = 1$, $A_{1,m} = 0$ otherwise, and for $k \ge 2$, $A_{k,m} = A_{k-1,m} + A_{k-1,m-x_k}$ if $m \ge x_k$, $A_{k-1,m}$ if $m < x_k$. Now if $A_{k,m} > 0$ we enumerate the subsets of $\{x_1,\ldots,x_k\}$ with sum $m$ recursively as follows: if $A_{k-1,m} > 0$, first enumerate the subsets of $\{x_1,\ldots,x_{k-1}\}$ with sum $m$; then if $m \ge x_k$ and $A_{k-1,m-x_k} > 0$ enumerate the subsets of $\{x_1,\ldots,x_{k-1}\}$ with sum $m-x_k$ and adjoin $x_k$ to each. - Thank you for the answer. As far as I can tell you've understood the question correctly. – Nick Rosencrantz Oct 4 '12 at 21:10	{"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": 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": 0.9436733722686768, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/55278/distance-traveled-in-a-simple-two-body-problem?answertab=active	# Distance traveled in a simple two body problem [duplicate] This question already has an answer here: • Radial fall in a Newtonian gravitational field [duplicate] 1 answer I'm trying to program an $N$-body simulation and I'd like to be able to test it with a known solution to a simple, two-body problem. I've looked at multiple sources, but I just don't know how to apply it to my simple test case. Two objects at rest placed 10 meters apart with mass of 1. The force between them is a modified gravitational force of F = 10 * m1 * m2 / r^2. How long will it take for each object to travel 4 meters? - – Qmechanic♦ Feb 27 at 10:26 ## marked as duplicate by Qmechanic♦Feb 27 at 10:26 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 First of all, this is by no means a trivial problem. The usual method goes something like the following. The force from mass 2 on mass 1 is: $$F_{21} = G\frac{m_1 m_2}{(x_2 - x_1)^2} = m_1 \ddot{x}_1$$ Similarly: $$F_{12} = -F_{21} = -G\frac{m_1 m_2}{(x_2 - x_1)^2} = m_2 \ddot{x}_2$$ Canceling masses and subtracting the equations from each other gives: $$\ddot{x}_2 - \ddot{x}_1 = \frac{d^2}{dt^2}(x_2 - x_1) = -G\frac{m_1 + m_2}{(x_2 - x_1)^2}$$ If we define $~r=x_2-x_1$ as the separation between the masses, then our equation becomes: $$\ddot{r} = -G\frac{m_1 + m_2}{r^2}$$ Now it gets a bit trickier. We use the fact that $\ddot{r}=\dot{r}~d\dot{r}/dr$ to separate the differential equation: $$\dot{r}~d\dot{r}=-G\frac{m_1 + m_2}{r^2}~dr$$ For $\dot{r}=0$ at $r_0$ (they're initially at rest), the integral of the above yields: $$\frac{dr}{dt} = \sqrt{ \frac{2 G (m_1 + m_2)}{r} - \frac{2 G (m_1 + m_2)}{r_0}} = \sqrt{ \frac{2 G r_0 (m_1 + m_2) - 2 G r (m_1 + m_2)} {r\ r_0}}$$ So then: $$\Delta t=\sqrt{\frac{r_0}{2 G (m_1 + m_2)}}~\int_{r_0}^{r} \sqrt{\frac{r}{r_0 - r}}~dr$$ In your case you've set $G=10$, $r_0=10$, and the masses each to one. When they've each traveled four meters, $r=2$. So you need to integrate the above from 10 to 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": 8, "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": 0.9466270208358765, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/141627-divergence-flux-volume-independant-limiting-volume-print.html	# divergence = flux / volume is independant of the limiting volume Printable View • April 26th 2010, 07:31 PM maddas divergence = flux / volume is independant of the limiting volume In physics books, one often finds the coordinate-free definiton $\mathrm{div}\; \mathbf{f} = \lim \frac1{\mathrm{d}V} \oint \mathbf{f} \cdot \mathrm{d}\mathbf{S}$, where the surface integral is taken over the surface of a small volume element with volume $\mathrm{d}V$ as this volume goes to zero, together with the claim that the limit is "independant of the shape of the volume element" (which my book dismisses with "this claim is somewhat troublesome to prove"). What is the precise statement for the limit being independant of the "shape of the volume element"? All times are GMT -8. The time now is 02:59 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": 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": 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": 0.8805397152900696, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/circle+coordinate-systems	# Tagged Questions 1answer 85 views ### Find parametric expression of an arc given its start point, end point and central angle in 3D cartesian coordinate system In a 3D cartesian coordinate system, the coordinates of start point and end point have been given as $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$. If the central angle of the two points (the one smaller ... 1answer 516 views ### How to find an end point of an arc given another end point, radius, and arc direction? Given an arbitrary arc, where you know the following values: end point (x1,y1), radius (r) and arc direction (e.g. clockwise or counterclockwise from start to end), how can I calculate the other ... 1answer 226 views ### Find the coordinates of a point on a circle given 2 points and an angle I have a circle with A as a center, B and C 2 points on the circle. I have the coordinates of A (the center) and B (the point on the circle). How can I find the coordinates of C (another point on the ... 1answer 192 views ### Finding points relating to the edge of a circle in an x,y coordinate system My question is a bit hard for me to express, so please bear with me. I never got far in trig, and haven't done much on the subject in years; trying to get back into it as it's a pretty major part of ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8937259912490845, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/119749/significance-of-eigenvalue?answertab=oldest	Significance of eigenvalue When I represent a graph with a matrix and calculate its eigenvalues what does it signify? I mean, what will spectral analysis of a graph tell me? - 3 Answers Here are some things you can learn from eigenvalues. If there are $d$ distinct eigenvalues, then the diameter of the graph is at least $d+1$. If a graph is $k$-regular, then the multiplicity of $k$ as an eigenvalue gives you the number of connected components in the graph. In general, if a graph is connected, then the largest eigenvalue must have multiplicity 1. For strongly regular graphs, you can use information about the eigenvalues and their multiplicities to determine if certain parameter values are even possible. That is, if we have a strongly regular graph with parameters (n, k, a, c), then there exists one eigenvalue of multiplicity 1 and two other eigenvalues, $\theta$ and $\tau$. These eigenvalues have multiplicities: $m_\theta = \frac{1}{2} \left( (n-1) - \frac{2k+(n-1)(a-c)}{\sqrt{\Delta}}\right)$ and $m_\tau = \frac{1}{2} \left( (n-1) + \frac{2k+(n-1)(a-c)}{\sqrt{\Delta}}\right)$ where $\Delta = (a - c)^2 + 4(k - c)$. Since these represent eigenvalue multiplicities, they must be integers. Therefore, any parameter values (n, k, a, c) which do not give integer values for these multiplicities, are not possible. See Algebraic Graph Theory by Godsil and Royle, chapter 10, for more on this. If you are interested in learning about the connection between eigenvalues and graphs, then read a book. You already know the term spectral analysis. Read Spectra of Graphs, for instance: http://homepages.cwi.nl/~aeb/math/ipm/ipm.pdf P14 of this book gives the first result I gave, plus the proof. - The first eigenvalue also bounds the Cheeger constant of the graph, which gives you information about whether it has any bottlenecks. – Neal Mar 13 '12 at 17:42 But how can you connect notion of eigen values and graph theory?i'm not getting this point – user997704 Mar 13 '12 at 17:43 @user997704 The diameter is a notion of graph theory. I explained one connection between graphs and their eigenvalues, which involves the diameter. I'm not sure what you mean? I'm not saying my answer gives you a complete picture. I'm giving you a bit of information. – Graphth Mar 13 '12 at 17:44 I got wht u r saying but is there any proof for this statement?How did u come to this statement?Please post some mathematical proof – user997704 Mar 13 '12 at 17:46 @user997704 See Algebraic Graph Theory by Biggs. It has the proof in it. – Graphth Mar 13 '12 at 17:48 show 2 more comments The dominant (largest) eigenvalue can be used to tell you which node(s) are the most connected. Since the matrix is symmetric, it is diagonalizable (in fact, orthogonally diagonalizable, but diagonalizable is enough for what follows). So all $n$-vectors can be written as a linear combination of eigenvectors. Say $$\vec{x}=\sum_{i=1}^nc_i\vec{v}_i$$ where $\vec{v}_i$ is an eigenvector for $\lambda_i$ and $\lambda_1$ is the dominant eigenvalue. In particular, apply this to the columns of the matrix $\vec{a_j}$: $$\vec{a_j}=\sum_{i=1}^nc_{ij}\vec{v}_i$$ I assume you understand that $A^n$ shows you how many paths there are of length $n$ from each vertex $i$ to each vertex $j$. But $$A^n=A^{n-1}A=\begin{bmatrix}A^{n-1}\vec{a}_j\end{bmatrix}=\begin{bmatrix}\sum_{i=1}^nc_{ij}\lambda_i^{n-1}\vec{v}_i\end{bmatrix}$$ For large $n$, $\lambda_1^{n-1}$ is relatively much larger than all other such terms. So $$A^n\approx\begin{bmatrix}c_{1j}\lambda_1^{n-1}\vec{v}_1\end{bmatrix}$$ where the $\approx$ comparison is made entry by entry and compares entires by looking for ratios close to $1$. (I'm being informal, for ease of reading). Thus the columns of $A^n$ are proportional to $\vec{v}_1$, so for each column of $A^n$, the entries of $\vec{v}_1$ tell you which entries are largest, which in turn tells you which nodes have the most paths of length $n$ to each of the other nodes. As $n$ gets larger, this feature of $A^n$'s columns won't change. So in a nice sense, you will understand which nodes are the most connected to other nodes through paths of long-enough lengths. - – user997704 Mar 13 '12 at 18:08 @user997704 Read chapter 1. That's not that complicated and that would be quick and you'd learn a lot. – Graphth Mar 13 '12 at 18:15 The ratio between the largest and smallest eigenvalue can be used to estimate the chromatic number: $$\chi(G)\ge1+\frac{\lambda_{\min}}{\lambda_\max}.$$ A proof can be found here (Theorem 2.4). -	{"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": 29, "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": 0.906960129737854, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/32918/interactions-between-the-poles-of-the-same-magnet/32961	# Interactions Between the poles of the same magnet Suppose a bar magnet like so : ````[(+m)----------------(-m)] ```` Do the two individual poles interact with each other ? If yes then how do you explain it if no then how? - ## 2 Answers There aren't two distinct poles. The battery is made up of lots of tiny magnets (e.g. iron atoms), and all those tiny magnets do indeed interact with other. The overall field of the magnet is the result of these interactions. The poles are just where the field leaves the end of the magnet. - So they interact or not? – The-Ever-Kid Jul 26 '12 at 18:07 All the poles of the magnet (all the little poles in John Rennie's answer) interact with each other. The total magnetic energy is the field energy in the magnetic field everywhere in space: $$\int {1\over 2} B^2 d^dx$$ in appropriate units. Increases and decreases in this magnetic field energy also give you the energy changes when you bring two magnets together. The identification of the interaction potential energy with field energy removes any question of whether two objects interact--- the objects interact with the field, and the change in their energy as you change their configuration is the change in the field energy of this configuration. You don't have to do a pairwise sum over all poles to get the interaction energy. -	{"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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9072219133377075, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/194177/algorithmic-complexity-of-i2?answertab=votes	# Algorithmic Complexity of $i^2$ I am new to the Big O notation in regards to algorithm design. I have had some exposure to it but I am not sure how to find the algorithmic complexity of a given function for a summation. If someone can point me in the right direction to solve this problem that would be great. The Problem: Prove that $\sum_{i=1}^n i^2$ is $O(n^3)$. I am not sure how this complexity is derived because in my own mind I feel that the complexity is closer to $n$ for each value in the list of $i$ values. - Is this the problem in its entirety? – Arkamis Sep 11 '12 at 15:06 4 I think the most important thing is to understand that what you're trying to find is not the complexity of computing the value; rather, you're trying to figure out how the value of the sum itself, as a function of $n$, behaves as $n\rightarrow\infty$. – Steven Stadnicki Sep 11 '12 at 15:06 Yes this is the problem in its entirety. @StevenStadnicki Yes that is basically what I am trying to figure out. Once I can find out the behavior of f(n) and how that relates to the complexity I feel that I would be all set. – Michael Guantonio Sep 11 '12 at 15:12 2 @MichaelGuantonio The point is that complexity doesn't actually factor into this problem at all; it may motivate it, but the $O()$ notation is much broader than just complexity and it's really worthwhile to try and divorce the two in your mind. Once you understand the behavior of this sum you can relate it back to complexity, but I'd suggest not thinking about complexity at all while you work this problem. – Steven Stadnicki Sep 11 '12 at 15:13 @StevenStadnicki Sorry, the computer science student in me is trying to make since of it all. This means creating relations that are not there. – Michael Guantonio Sep 11 '12 at 19:58 ## 2 Answers You seem to be confused by what $O(\cdot)$ really means. It doesn't inherently have anything to do with algorithms; the definition is as follows: Given two functions, $f$ and $g$, we say that $g$ is $O(f)$ if there exists some constants $c$ and $n_0$ such that $g(n) \le c \cdot f(n)$ for all $n \ge n_0$. For your problem, \begin{align} \sum_{i=1}^n i^2 \le \sum_{i=1}^n n^2 = n^3 \text{$\quad$for $n \ge 0$}\\ \end{align} Thus by definition, $\sum_{i=1}^n i^2$ is $O(n^3)$. - Still not quite sure how you got this. I can see that you took the general i case and said that it was less than or equal to the n max case. But how you determined because of that the complexity is $O(n^{3})$ baffles me. – Michael Guantonio Sep 11 '12 at 20:16 1 Read the definition again. We just need to find $c$ and $n_0$ such that $\sum_{i=1}^n i^2 \le cn^3$ for all $n \ge n_0$. The steps I took above show that $c=1$ and $n_0 = 0$ are valid witnesses. That is, $\sum_{i=1}^n i^2$ is always less than $n^3$ and therefore by definition is $O(n^3)$. – Zach Langley Sep 11 '12 at 22:34 @MichaelGuantonio (The word "complexity" does not apply here. There is no computation, only a comparison of functions.) – Zach Langley Sep 14 '12 at 14:55 How can I deal with a more general function for this problem. Like proving for the general case that $$\sum_{i=1}^n i^k is O(n^{k-1})$$ – Michael Guantonio Sep 14 '12 at 22:48 @MichaelGuantonio You won't be able to prove that because it's not true. $\sum_{i=1}^n i^k \ge n^k$ so the function is $\omega(n^{k-1})$. If you meant to switch the exponents, the exact same techniques I used above will work. – Zach Langley Sep 15 '12 at 14:49 By the very definition of $O$, you have to prove that there is an $N \in \mathbb N$ and a $C \ge 0$ such that $$\sum_{i=1}^n i^2 \le C n^3, \quad n \ge N,$$ right? But now for each $n$ and each $1 \le i \le n$ we have $i^2 \le n^2$, so $$\sum_{i=1}^n i^2 \le \sum_{i=1}^n n^2 = n^3,$$ and we are done with $N = 1$, $C = 1$. -	{"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": 41, "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": false, "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": 0.9591408371925354, "perplexity_flag": "head"}
http://mathoverflow.net/questions/7114?sort=newest	## Why are planar graphs so exceptional? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) As compared to classes of graphs embeddable in other surfaces. Some ways in which they're exceptional: 1. Mac Lane's and Whitney's criteria are algebraic characterizations of planar graphs. (Well, mostly algebraic in the former case.) Before writing this question, I didn't know whether generalizations to graphs embedded in other surfaces existed, but some lucky Google-fu turned up some references -- in particular there seems to be a generalization of Whitney's criterion due to Jack Edmonds for general surfaces, although frustratingly I can't find the paper, and the main reference I found implies that there might be a small problem on the Klein bottle. Anyone know if Edmonds' result is as easy to prove as Whitney's? 2. Kuratowski's classic characterization of planar graphs by forbidden minors. Of course this does generalize to other surfaces, but this result is both incredibly deep and difficult (as opposed to the proof of Kuratowski, which is by no means trivial but is obtainable by a sufficiently dedicated undergraduate -- actually my working it as an exercise is largely what motivated the question) and is in some sense "essentially combinatorial" in that it applies to a wider class of families that aren't inherently topologically defined. 3. In the other direction of difficulty, the four-color theorem. It's apparently not difficult to show (except for the plane) that what turns out to be the tight upper bound on the chromatic number of a graph embeddable on a surface (other than, for whatever reason, the Klein bottle) is, in fact, an upper bound -- the problem is showing tightness! Whereas it's pretty much trivial to show that $K_4$ is planar (to be fair, though, tightness is easy to check for surfaces of small genus -- the problem's in the general case), but the four-color theorem requires inhuman amounts of calculation and very different, essentially ad-hoc methods. I realize that the sphere has genus 0, which makes it unique, and has trivial fundamental group, which ditto, but while I imagine this information is related to the exceptionalness of the plane/sphere, it's not really all that satisfying as an answer. So, why is it that methods that work everywhere else tend to fail on the sphere? Related questions: reasons-for-the-importance-of-planarity-and-colorability - 2 While you're collecting ways in which planar graphs are exceptional, there's also the Colin de Verdière number, which defines a hierarchy: trees, outerplanar graphs, planar graphs, ... – David Eppstein Nov 29 2009 at 7:48 @David: Ooh, good one! I knew I'd seen more examples, but I couldn't think of others... – Harrison Brown Nov 29 2009 at 8:01 A follow up question which gives also a wide context for Harrison's question is here mathoverflow.net/questions/7650/… A follow up blog discussion is here gilkalai.wordpress.com/2009/12/03/… – Gil Kalai Dec 4 2009 at 9:43 ## 3 Answers (I think that the question of why planar graphs are exceptional is important. It can be asked not only in the context of graphs embeddable on other surfaces. Let me edit and elaborate, also borrowing from the remarks.) Duality: Perhaps duality is the crucial property of planar graphs. There is a theorem asserting that the dual of a graphic matroid M is a graphic matroid if and only if M is the matroid of a planar graph. In this case, the dual of M is the matroid of the dual graph of G. (See this wikipedia article). This means that the circuits of a planar graph are in one to one correspondence with cuts of the dual graph. One important manifestation of the uniqueness of planar graphs (which I believe is related to duality) is Kasteleyn's formula for the number of perfect matchings and the connection with counting trees. Robust geometric descriptions: Another conceptual difference is that (3-connected or maximal) planar graphs are graphs of convex 3-dimensional polytopes and thus have extra geometric properties that graphs on surfaces do not share. The geometric definition of planar graphs (unlike various generalizations) is very robust. A graph is planar if it can be drawn in the plane such that the edges do not intersect in their interiors and are represented by Jordan curves; The class of planar graphs is also what we get if we replace "Jordan curves" by "line intervals," or if we replace "no intersection" by "even number of crossings". The Koebe-Andreev-Thurston theorem allows to represent every planar graph by the "touching graph" of nonoverlapping circles. Both (related) representations via convex polytopes and by circle packings, can respect the group of automorphisms of the graph and its dual. Simple inductive constructions. Another exceptional property of the class of planar graphs is that planar graphs can be constructed by simple inductive constructions. (In this respect they are similar to the class of trees, although the inductive constructions are not so simple as for trees.) This fails for most generalizations of planar graphs. A related important property of planar graphs, maps, and triangulations (with labeled vertices) is that they can be enumerated very nicely. This is Tutte theory. (It has deep extensions to surfaces.) It is often the case that results about planar graphs extend to other classes. As I mentioned, Tutte theory extends to triangulations of other surfaces. Another example is the fundamental Lipton-Tarjan separator theorem, which extends to all graphs with a forbidden minor. The study of planar graphs have led to important graph theoretic concepts Another reason (of a different nature) why planar graphs are exceptional is that several important graph-theoretic concepts were disvovered by looking at planar graphs (or special planar graphs.) The notion of vertex coloring of graphs came (to the best of my knowledge) from the four color conjecture about planar graphs. Similarly, Hamiltonian paths and cycles were first studied for planar graphs. Graphs on surfaces and other notions generalizing planarity. Considering the class of all graphs that can be embedded in a given surface is a natural and important extension of planarity. But, in fact, for various questions, graphs embeddable on surfaces may not be the right generalization of planar graphs. David Eppstein mentioned another generalization via the colin de Verdier invariant. This describes a hiearachy of graphs where the next class after planar graphs are "linklessly embeddable graphs". Those are graphs that can be embedded in space without having two disjoint cyles geometrically link. As it turned out this is also a very robust notion and it leads to a beautiful class of graphs. (They all have at most 4v-10 edges where v is the number of vertices; The known case of Hadwiger's conjecture for graphs not having K_6 minor implies that they are all 5 colorable.) Further classes in this hierarchy are still very mysterious. Other extensions of planarity are: 3) (not literally) Graphs not having K_r as a minor; 4;5) (Both very problematic) As Joe mentioned, graphs of d-polytopes, and also graphs obtained from sphere packings in higher dimensions; 6) (not graphs) r-dimensional simplicial complexes that cannot be embedded in twice the dimension, 7) [A notion that I promoted over the years:] graphs (and skeleta) of d-polytope with vanishing second (toric) g-number, and many more. Forbidden minors and coloring. As for the second and third items in the question. About coloring I am not sure if we should consider 4-coloring planar graphs and coloring graphs on other surfaces as very related phenomena. Regarding forbidden minors. Kuratowski's theorem on surfaces is a special case (and also an important step of the proof) of a much more general result (Wagner's conjecture proved by Robertson and Seymour) about any minor-closed class of graphs. This result can be regarded as extending Kuratowski theorem and also (and perhaps more importantly) extending Kruskal and Nash-Williams theorem on trees. Indeed Kuratoski's theorem is related nicely to the more general picture of topological obstruction to embeddibility. If you would like to propose a different (perhaps topological) understanding of the extension of Kuratowski's theorem for surfaces, then maybe you should start by the well-quasi ordering theorem for trees. - Ah, I posted a comment before I saw your edit -- the thing about duality makes more sense now, although I'm still suspicious of duality being behind everything, considering how naturally it generalizes in the topological setting. I hadn't considered your point about polytopes, though, and it's a good one. – Harrison Brown Nov 29 2009 at 8:10 2 One important manifestation of the uniqueness of planar graphs is Kasteleyn's formulas for the number of perfect matching and the connection with counting trees. Another geometric description (related to the relation with polytopes) is the Koebe-Andreev-Thurston theorem that allows to represent every planar graph by the "touching graph" of nonoverlapping circles. – Gil Kalai Nov 29 2009 at 8:14 1 I dont know if duality is behind everything but it is certainly behind many things. For example, consider 3-connected graphs. When we have such a planar graph the 2-cells are determined uniquely. (A theorem of Whiteney asserts that an induced cycle is a 2-face iff it is non seperating. On the other hand consider for example the polyhedral cell complex realizing the real projectice plane obtained by identifying opposite faces of the dodecahedron. The graph is the Petersen graph. The cells are not determined by the graph: Comparing the automorphism groups of the graph and of the 2-dim complex. – Gil Kalai Nov 29 2009 at 8:28 3 The (Koebe-Andreev)-Thurston theorem is true (and was actually proved first in this generality) for graphs on arbitrary surfaces (the question explicitly asks about planar graphs as compared to graphs embedded in other surfaces, so I thought I'd mention it). This was one of the steps in Thurston's original proof of geometrization of Haken manifolds (a later simplification, due to Otal I think, showed that it sufficed to prove the theorem for planar graphs). – Danny Calegari Dec 1 2009 at 16:38 That's interesting. I was not aware of it; So is the statement that every graph can be realized by a circle packing on arbitrary surface? (with a Riemannian metric?) Is there also uniqueness results like in the planar case? – Gil Kalai Dec 1 2009 at 17:05 ### 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. Another (less known) characterization of planar graphs is Schnyder's theorem, which characterizes planar graphs according to order dimension. That is, a graph is planar if and only if its incidence poset has order dimension at most 3. Also, I would be remiss to not mention the beautiful (strong) Hanani-Tutte theorem: A graph is planar if and only if it has a drawing in the plane such that every two non-adjacent edges cross an even number of times. - As Gil Kalai notes Steinitz's Theorem (A graph G is the vertex-edge graph of a convex 3-dimensional polyhedron, if and only if G is planar and 3-connected) is very powerful. It means that the combinatorial properties (hamiltonian circuits, matchings, etc.) of something 3-dimensional can be studied by looking at a special kind of graph in 2-dimensions. Nothing completely like this happens for higher dimensional convex polytopes. Similarly, we do not know how to study all "toroidal polytopes" using a special class of graphs on the torus. What seems intriguing is the interface between capturing "geometric" properties of objects from combinatorial (graph theory) properties. For example, for some time I have been interested in which 3-dimensional convex polyhedra with all of its faces being triangles can be realized with isosceles triangles as faces. One way to attack this is to look at the planar 3-connected graphs graphs with at least 4 vertices all of whose faces are triangles (maximal planar graphs). It turns out one can color the edges of such a graph with two colors a and b so that each face gets two a edges and one b edge. Now one can ask the question of whether these colorings can be realized with congruent isosceles triangles, the colors a and b being used to code the two lengths of edges of the congruent isosceles triangles. It is not hard to show that some maximal planar graphs can not be realized with congruent isosceles triangles but telling exactly which graphs can be so realized seems much harder. Also, the graph theory made it possible for me to see many ways to realize such polyhedra geometrically, when such a realization is possible - in other words the same combinatorial type often can be realized with congruent isosceles triangles in many ways) other than ways which had lots of symmetry. If 4 divides the number of faces of a maximal planar graph one can color its edges with two colors a and b so that the number of a, a, b triangles and b, b, a triangles are equal. However, I have been unable to explicitly tell when such colorings translate into geometrical realizations. Loosely speaking, the combinatorial questions seem "easy" while the geometric ones seem "hard." I do not know if the maximal planar graphs with at least 4 vertices can be realized by convex polyhedra with isosceles triangles. (My guess is yes.) There are also interesting questions when one allows mixtures of isosceles triangles and equilateral triangles. (It is known there are exactly 8 convex polyhedra with only equilateral triangles for faces.) One of the sticking points of these questions is trying to tell when two triangles which share an edge in the graph flatten out and lie in one plane in the realization in 3-space. My point though is that Steinitz's Theorem allowed progress on many questions and encouraged me to formulate new geometrical questions that would not have occurred to me otherwise. Best, Joe Malkevitch -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9454983472824097, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/117406/finding-probability-distribution-of-a-random-variable-from-the-others	# Finding probability distribution of a random variable from the others Let $X_1,...,X_n$ be independent exponential distribution.and now i want to calculate the density function of $\sum_{i=1}^{n} X_i$,i tried to find its distributing function,$F(T(X)\le x)$ but then i don't know what to do. - Got something from the answer below? – Did Aug 15 '12 at 15:42 ## 1 Answer Suppose your random variables are i.i.d. exponential with parameter $\lambda$. Then the sum $\sum X_i$ is a gamma random variable with parameter $(n, \lambda)$. Here's a sketch of the proof: (you should fill in the details) 1. Let $X$ and $Y$ be gamma random variables with parameters $(s,\lambda)$ and $(t, \lambda)$, respectively. (Look up the gamma density function, if necessary). 2. Prove that $X+Y$ is a gamma random variable with parameter $(s+t, \lambda)$. 3. Prove by induction that if $X_1, \dots, X_n$ are gamma$(t_i, \lambda)$ random variables, then $\sum X_i$ is a gamma$(\sum t_i, \lambda)$, random variable. 4. Note that an exponential rv with parameter $\lambda$ is a gamma$(1,\lambda)$ random variable. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.7924346923828125, "perplexity_flag": "head"}
http://en.wikisource.org/wiki/Notes_on_the_Principle_of_Action_and_Reaction_in_General_Dynamics	# Notes on the Principle of Action and Reaction in General Dynamics From Wikisource Notes on the Principle of Action and Reaction in General Dynamics (1908) by Max Planck, translated by Wikisource In German: Bemerkungen zum Prinzip der Aktion und Reaktion in der allgemeinen Dynamik, Physikalische Zeitschrift 9 (23): 828-830, (Session from September 23, 1908.) Notes on the Principle of Action and Reaction in General Dynamics Max Planck Wikisource 1908 Max Planck (Berlin-Grunewald) Notes on the principle of action and reaction in general dynamics. The Newtonian principle of equality of action and reaction is well known to consist mainly of the theorem of the constancy of momentum or impulse of motion; and I would therefore like to speak of such a principle only in the meaning of this theorem, especially in its importance for the general dynamics that not only covers mechanics in the narrower sense, but also electrodynamics and thermodynamics. Many of us probably still remember the stir that was caused, when H.A. Lorentz in his construction of atomistic electrodynamics on the basis of a stationary aether, denied the universal validity of the third Newtonian axiom, and it was inevitable that this circumstance was considered (for example by H. Poincaré) as a serious objection to Lorentz's theory. A kind of stabilization only came back as it turned out, especially by the investigations of M. Abraham, that the reaction principle can still be saved in its full generality, [ 829 ] if we introduce a new momentum in addition to the mechanical momentum only known so far, i.e. the electromagnetic momentum. Abraham has made this even more plausible, by bringing the conservation of momentum in comparison with the conservation of energy. Similarly, as the energy principle is violated when we don't consider the electromagnetic energy, and fulfilled when we introduce this type of energy, the reaction principle is violated if we only consider the mechanical momentum, and is satisfied on the other hand, as soon as we have considered the electromagnetic momentum. However, this certainly indisputable comparison doesn't affect a still substantial difference. As to the energy we know already a number of different forms: the kinetic energy, gravitation, the elastic deformation energy, heat, chemical energy, and it is therefore not a principally new feature when to these different forms, the electromagnetic energy is attached as another form. Conversely, until now only a single form was known as momentum: mechanical momentum. While the energy from the very outset represented a universal physical expression, the momentum was so far a special mechanical expression, the reaction principle a special mechanical theorem, and therefore the extension recognized as necessary was perceived as a revolution of principle, by which the previous relatively simple and uniform concept of momentum becomes considerably more complicated in nature. Is it not possible, even from the standpoint of general dynamics, to form the definition of momentum (although it now includes both the mechanical and the electromagnetic form) in the same uniform way as it was earlier done in mechanics? A positive answer to this question would certainly lead to an advance in the understanding of the true meaning of the reaction principle. In fact, such a uniform definition of momentum seems to be possible and feasible, at least if we also admit of Einstein's theory of relativity.[1] However, it must now be emphasized that this theory is regarded nowadays by no means assured. But as their deviations from other eligible theories are only limited to very small terms, one may say however, that they can be considered as correct just up to those differences, and to that extent, therefore, the following considerations keep in all circumstances a certain importance. In the theory of relativity, the momentum can now be generally attributed to that vector which expresses the energy flux, not only the Poynting electromagnetic energy flux, but the energy flux in general. From the standpoint of direct action theory all types of energy can change their location within space only by continuous propagation, not by changes in the form of leaps. Therefore, the energy principle generally requires that the change of total energy contained in a certain space is like a surface integral, namely, the algebraic sum of the entire incoming energy through the surface of that space. The flux can be submitted by radiation, such as the Poynting vector, by conduction as with pressure or with pushes and with heat conduction, and by convection as with the admission of ponderable atoms or electrons through the considered surface. In any case, the entire energy flux at every point in space, with respect to unit surface and unit time, is a certain finite vector, and the ratio of this vector by the square of the speed of light c is in general the momentum in relation to unit volume. Take for example a ponderable fluid moving with velocity q under the pressure p. Through a surface element df of a stationary area perpendicularly directed to q, the mechanical work is: p · df · qdt. The entrained energy is: df · ε · qdt, where ε denotes the energy density. Accordingly, by that definition the momentum of unit volume is: $\frac{(\epsilon+p)q}{c^{2}}$. Comparing this expression with the usual mechanical momentum kq, where k is the density of the liquid, we find: $k=\frac{\epsilon+p}{c^{2}}$, a known relationship of the theory of relativity[2]. [ 830 ] From the point of view described we can generally denote the principle of equality of action and reaction as the "theorem of inertia of energy". But we can go a step further. As the constancy of energy implies the concept of energy flux, the constancy of momentum also implies the concept of "flux of the momentum", or briefly spoken: the "momentum flux". As the momentum located in a particular area can only change due to external effects, i.e. according to the theory of direct action only by processes at the surface of that space, the amount of change in unit time is a surface integral which can be described as the total momentum flux in the interior of that space. A major difference, however, to the energy flux is the fact that the energy is a scalar, yet the momentum is a vector. Therefore, the incoming energy in that space is expressed by a single surface integral, and the energy flux is a vector. However, the momentum flux into that space is expressed by three surface integrals, corresponding to the three components of momentum, and the momentum flux in one place is a tensor triple, in the notation of W. Voigt[3], which is characterized by six components To get an idea of the significance of these tensor triples, we first consider the mechanical momentum and its corresponding mechanical momentum flux. The total momentum flux into the interior of a room, i.e. the increase in momentum in the interior per unit time, is equal to the resulting mechanical force which acts on the entire mass located within that space. Consequently, the momentum flux through a surface element is nothing else than the mechanical pressure on the surface element, and its components have the form: $X_{n}=X_{x}\cos(nx)+X_{y}\cos(ny)+X_{z}\cos(nz)$ $Y_{n}=Y_{x}\cos(nx)+Y_{y}\cos(ny)+Y_{z}\cos(nz)$ $Z_{n}=Z_{x}\cos(nx)+Z_{y}\cos(ny)+Z_{z}\cos(nz)$, where n denotes the inner normal of the surface element. $X_{x},\ Y_{y},\ Z_{z},\ X_{y}=Y_{x},\ Y_{z}=Z_{y},\ Z_{x}=X_{z}$ are the six components of the tensor triples representing the momentum flux. The same is true for the electromagnetic momentum flux in a vacuum. The components of this tensor triple are nothing else than the known Maxwell stresses. Its integration over a closed surface gives the total momentum flux into the interior, and thus the increase of all mechanical and electromagnetic momentum contained within the enclosed space. It is remarkable, how by this theorem the physical meaning of the Maxwell stresses is increased for the stationary aether theory. Because, as a force of pressure these stresses make not really sense in this theory, since we may not attribute a meaning to a force acting on something absolutely immovable.[4] The fact that the Maxwell stresses, although they were officially abolished, have maintained themselves within the theory of a stationary aether, because they often proved so convenient as mathematical tools for certain calculations, could already suggest the idea, that they play some special physical role by which they are legitimate also for the stationary aether. It is tempting to transfer the concept of momentum flux to the gravitational field, where, apart from the fatal sign, there are given a remarkable number of analogies; but a more detailed discussion of this problem would take too far at this place. ## Discussion. Minkowski: The theorems on momentum can be derived directly from the energy theorem in my point of view. Namely, the energy equation in Lorentz's theory depends on the reference frame for space and time. If we write the energy theorem for each possible reference frame, we have some equations, and in those the theorems on momentum are included. Planck: Certainly. But I consider the independence from the reference frame not as a solid physical result, but more as a hypothesis, which I consider as promising but not to be established yet. However, we just need to consider whether these relationships actually exist in nature. This we can only learn by experiment, and hopefully the time is not far off when we experience it. 1. See especially F. Hasenöhrl (Sitzungsbericht de. Akad. d. Wiss. zu Wien vom 31. Oktober 1907, S. 1400), although he doesn't directly proceed from the theory of relativity, but as far as I can see he get quite the same results. 2. See for example M. Planck, Ann. d. Phys. (4), 25, 27, 1908. equation (48). 3. See M. Abraham, Enzyklopädie d. math. Wiss. IV, 14, p. 28 This is a translation and has a separate copyright status from the original text. The license for the translation applies to this edition only. Original: This work is in the public domain in the United States because it was published before January 1, 1923. The author died in 1947, so this work is also in the public domain in countries and areas where the copyright term is the author's life plus 60 years or less. This work may also be in the public domain in countries and areas with longer native copyright terms that apply the rule of the shorter term to foreign works. Translation: This work is released under the Creative Commons Attribution-ShareAlike 3.0 Unported license, which allows free use, distribution, and creation of derivatives, so long as the license is unchanged and clearly noted, and the original author is attributed.	{"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": 6, "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": 0.9451974630355835, "perplexity_flag": "head"}
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I know that black holes are "black" because nothing can escape it due to the massive gravity, but I am wondering if there are any theories as to what happens to the light or mass that enters a black ... 2answers 90 views ### What happens to things when things get crushed in a blackhole [duplicate] When a black hole destroys things until they are smaller than molecules, where does it go and what happens when it clogs up? 1answer 72 views ### The effects of heat on gravitational fields In boiling soapy water, globs of soap coalesce as the temperature increases to boiling. Does this mean that temperature increases the gravitational pull of bodies? 5answers 937 views ### The square in the Newton's law of universal gravitation is really a square? 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http://mathoverflow.net/revisions/116580/list	Return to Question 2 added 14 characters in body Hello, While studying Sobolev spaces, the following question came to my mind. Any help in this direction is appreciated. QUESTION Let $U\subseteq\mathbb{R}^n$ be open. Does there exist a function $f\in L^1_{\text{loc}}(U)$ such that 1) the classical derivative $Df$ exists everywhere in $U$. 2) $f$ is weakly differentiable in $U$. Let us write $D_w f$ to denote the weak derivative of $f$. 3) $Df\neq D_w f$, on a set of positive measure. Note that, we are assuming the existence of both the derivatives. I'm aware of examples where one exists while other one does not. The problem seems to be related to the question of validity of integration by parts for differentiable functions that are only differentiable. Thank you. 1 Classical Derivative, Weak Derivative and Integration by Parts Hello, While studying Sobolev spaces, the following question came to my mind. Any help in this direction is appreciated. QUESTION Let $U\subseteq\mathbb{R}^n$ be open. Does there exist a function $f\in L^1_{\text{loc}}(U)$ such that 1) the classical derivative $Df$ exists everywhere in $U$. 2) $f$ is weakly differentiable in $U$. Let us write $D_w f$ to denote the weak derivative of $f$. 3) $Df\neq D_w f$, on a set of positive measure. Note that, we are assuming the existence of both the derivatives. I'm aware of examples where one exists while other one does not. The problem seems to be related to the question of validity of integration by parts for differentiable functions. Thank you.	{"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": 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": 0.9397565126419067, "perplexity_flag": "head"}
http://mathoverflow.net/questions/20445/coend-computation	Coend computation Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $F:A^{\mbox{op}} \to \mbox{Set}$ and define `$G_a:A\times A^{\mbox{op}} \to \mbox{Set}$` `$G_a(b,c) = \mbox{hom}(a,b) \times F(c)$`. I think the coend of `$G_a$`, `$\int^AG_a$`, ought to be $F(a)$--it's certainly true when A is discrete, since then hom is a delta function. But my colimit-fu isn't good enough to actually compute the thing and verify it's true. Can someone walk me through the computation, please? - is this a category-theoretic analogue of the main theorem of the fundamental theorem of calculus? – Martin Brandenburg Apr 6 2010 at 0:13 2 @Martin: an FToC analogue would require a 'boundary' computation of some sort. This seems more like a distribution computation, where the hom works like $\delta(a-b)$, so the integral over the whole space is just evaluation. – Jacques Carette Apr 6 2010 at 3:20 1 Answer Hi Mike. This is what's often called the Density Formula, or (at the n-Lab) the coYoneda Lemma (I think), or (by Australian ninja category theorists) simply the Yoneda Lemma. (But Australian ninja category theorists call everything the Yoneda Lemma.) In any case, it's a kind of dual to the ordinary Yoneda Lemma. But you asked to be walked through it. First: yes, it is $F(a)$. Another way of writing your coend $$\int^A G_a$$ is as $$\int^{b \in A} G_a(b, b) = \int^b \mathrm{hom}(a,b) \times F(b).$$ I claim this is canonically isomorphic to $F(a)$. I'll prove this by showing that for an arbitrary set $S$, the homset $\mathrm{hom}(\mathrm{this}, S)$ is canonically isomorphic to $\mathrm{hom}(F(a), S)$. The claim will then follow from the ordinary Yoneda Lemma. So, let $S$ be a set. Then ```$$ \begin{align} \mathrm{Set}(\int^b \mathrm{hom}(a, b) \times F(b), S) & \cong \int_b \mathrm{Set}(\mathrm{hom}(a, b) \times F(b), S) \\ &\cong \int_b \mathrm{Set}(\mathrm{hom}(a, b), \mathrm{Set}(F(b), S)) \\ &\cong \mathrm{Nat}(\hom(a, -), \mathrm{Set}(F(-), S)) \\ &\cong \mathrm{Set}(F(a), S) \end{align} $$``` I don't know how much of this you'll want explaining, so I'll just say it briefly for now. If you want further explanation, just ask. The first isomorphism is kinda the definition of colimit. The second is the usual exponential transpose/currying operation. The third is maybe the most important: it's a fundamental fact about ends that if $F, G: C \to D$ are functors then $$\mathrm{Nat}(F, G) = \int_c D(F(c), G(c)).$$ The fourth and final isomorphism is the ordinary Yoneda Lemma applied to the functor $\mathrm{Set}(F(-), S)$. - I'm a little confused by your notation: specifically, what is the category $\mathrm{Nat}$? From the types its objects are functors $A^{op} \to \mathrm{Set}$, so it's the functor category whose objects are the natural transformations between those functors? – Neel Krishnaswami Apr 6 2010 at 14:14 FWIW You can formalise the steps in the construction of this isomorphism as a Haskell program. Here's one direction, with `f` being the relevant morphism: hpaste.org/fastcgi/hpaste.fcgi/view?id=24722 I suspect that by interpreting Haskell code as the internal language of some family of categories then this becomes a perfectly good definition of the isomorphisms for mathematical purposes too, and not just a statement about some Haskell functions. – Dan Piponi Apr 6 2010 at 16:53 Thanks! The third step is really the one I need to understand, so I'll think about that for a while and come back if I need more help. – Mike Stay Apr 6 2010 at 18:30 1 Neel, I guess Nat is a slightly unspecific notation, like hom. The first occurrence of Nat meant "hom in $[A^{op}, \mathrm{Set}]$". Here $[A^{op}, \mathrm{Set}]$ is the category whose objects are functors $A^{op} \to \mathrm{Set}$. So: Nat is not a category; it means "hom" in that functor category. I might equally well have written "$[A^{op}, \mathrm{Set}]$" in place of "Nat". (In fact I prefer to; I was making a perhaps misguided effort to be more widely comprehensible.) Similarly, the second occurrence of "Nat" could be replaced by "$[C, D]$". – Tom Leinster Apr 6 2010 at 20:08 1 Ah, thanks -- I've just never seen Nat used like that before. Amusingly, the $[A^{op}, \mathrm{Set}]$ notation is the one I've seen before. :) – Neel Krishnaswami Apr 6 2010 at 22:43 show 1 more comment	{"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": 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": 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": 0.9310299754142761, "perplexity_flag": "middle"}
http://bakingandmath.wordpress.com/2012/12/10/so-tubular-err-cubular/	# So tubular! Err… cubular 10 Dec What I’m reading right now: Special Cube Complexes, a 2008 paper by Frederic Haglund and Dani Wise. Recently Ian Agol proved the Virtual Haken Conjecture , which was a Big Deal in math (this link is LONG but a very well written non-math-person friendly summary of 30 years of math). In fact, one of my professors from undergraduate, Jesse Johnson, wrote a nice little blog post on what it might mean for the future of low dimensional topology. Basically, Agol used this special cube complex stuff to prove this Big Deal, which means that we might be able to use these to prove Lots of Big Deal and Little Deal theorems. So let’s get into what these guys are. Update: I just found out that it’s my turn to give the talk in our little colloquium this week. There’s seven of us, four are my advisor’s students (I guess he’s not technically my adviser yet) and three are in related fields. So every two months or so, you have to give a half hour talk on some math you’re learning about. We aren’t supposed to talk about our research, but I think I get a pass since it’s still my first year. So this post is a prelude to my talk! A cube complex is an object built by gluing a whole bunch of Euclidean cubes together. So a one-dimensional cube complex is built by gluing a bunch of lines together; that is, it’s a (mathematical) graph. And a two-dimensional cube complex is built by gluing a bunch of squares together. The gluings can happen in funky ways though, and special cube complexes are objects where these pathologies don’t happen. We’ll define these topologies in terms of hyperplanes. So if I have a square $[-1,1]\times [-1,1]$, I’ll have two hyperplanes running through it: one at $0\times [-1,1]$ and one at \$[-1,1]\times 0\$. In the picture below, which is taken from the 11th page of this paper, the red lines are hyperplanes, and the gray lines represent the cubes they’re cutting through. To be special, we need our cube complex to a) not self-intersect, b) have no one-sided hyperplanes, c) not directly self-osculate, and e) have no two hyperplanes that inter-osculate. Turns out that case d) hyperplanes indirectly self-osculate is OK. Pathologies of cube complexes (these guys are not special, but don’t tell their parents I said that). Really quick, notice that a cube complex is special if and only if its two skeleton is (the part made up of filled-in squares). That’s why we can just use this picture. So what’s so special about special cube complexes? The ultimate idea is that given a cube complex, if none of these funky things happens, I’ll be able to cut along the hyperplanes and have nice things happen. That’s how we get to the Big Deal. But that’s neither here nor there; this post is about a Smaller Deal: that a cube complex is special if and only if it corresponds to a right angled Artin group, that is, that there’s some graph so that our cube complex has an isometry into the Salvetti complex of that graph. Turns out google image searching “Salvetti complex” is utterly useless, so I’ll just describe it. Start with a single vertex. Now given a graph $\Gamma$, we add one loop to this vertex for each vertex $x_i$ in $\Gamma$. For any edge in $\Gamma$, say it’s $(x_i,x_j)$, we attach a square that looks like this: Pub crawl that winds up back at the first bar invite idea: be there or be a torus! So far we have the standard 2-complex for the group $A(\Gamma) = \langle x_i: i\in\text{Vertices}(\Gamma): [x_i,x_j]: (x_i,x_j)\in\text{Edges}(\Gamma)\rangle$, which is the definition of a right angled Artin group. Now to make the Salvetti complex, we attach an n-torus for every n-cycle in our graph. So if there was a triangle in our graph, we get a corresponding 3-torus in the Salvetti complex. And the fundamental group of our Salvetti complex is that right angled Artin group. To restate our deal, we’re saying that special cube complexes always have some graph $\Gamma$ so that we can see our cube complex somewhere in the corresponding Salvetti complex, and everything still looks nice. More formally and rigidly, we’re saying that X is special if and only if there’s an immersion from X into some Salvetti complex, which is a local isometry on the 2-skeleton. Proof in one direction: Suppose I’ve got a local isometry from the 2-skeleton of my cube complex X to a Salvetti complex. Since the Salvetti complex is special from how we built it, and local isometries keep things tidy (you can’t uncross intersecting hyperplanes, for instance), that means my 2-skeleton is also special. So from our remark above, X is special. Proof in the other direction: Say my cube complex is special. Then make a graph with vertices being the hyperplanes of X, and edges connecting intersecting hyperplanes. Now make the Salvetti complex of this graph. We can map X into the complex by sending an edge to the vertex of the hyperplane that crosses it, and then extend the rest of the map. It’s a hop and skip (no jumping allowed) that this map is, in fact, a local isometry on the level of 2-skeleta. Right angled Artin groups have really nice properties and are fun stuff, so this little theorem can lead to a whole bunch of conclusions. Phew, first math blog post. I’ll get better at these, I promise. I have to figure out what audience I’m writing 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 9, "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": 0.9030659794807434, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/81475/is-it-possible-to-solve-the-argument-maximization-problem-arg-max-x-langle-x	# Is it possible to solve the argument maximization problem $\arg\max_x \langle x,l\rangle - f_1(x) - f_2(x)$ via convex duality? I am attemping to solve the argument maximization problem $$\arg\sup_x \{\langle x,l\rangle - f_1(x) - f_2(x)\}\qquad\qquad\qquad\qquad (1)$$ where the functions $f_1$ and $f_2$ are concave but difficult to evaluate but their convex conjugates $f_1^$ and $f_2^$ are easy to evaluate. Since the sum operation is dual to the infimal convolution (or epi-sum) operation $$(g\#h)(x) = \inf_w \{g(x-w)+h(w)\},$$ the standard maximization problem is easy to compute by duality using the identity $$\sup_x \{\langle x,l\rangle - f_1(x) - f_2(x)\} = \inf_w \{f_1^(l-w)+f_2^(w)\}.$$ Is it possible to compute the solution to problem (1) is an analogous manner, making only calls to the conjugate functions $f_1^$ and $f_2^$? - To get the correct spacing, use `\langle` and `\rangle`, which produce, respectively, $\langle$ and $\rangle$, instead of `<` and `>`. – Arturo Magidin Nov 12 '11 at 23:32	{"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": 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": 0.8815906047821045, "perplexity_flag": "head"}
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http://math.stackexchange.com/questions/173860/why-is-choosing-elements-in-equivalence-classes-not-a-choice	# Why is choosing elements in equivalence classes not a choice? This is Asaf's answer from this link: How do we know an $\aleph_1$ exists at all? I don't understand this sentence that is; From each equivalence class choose the representative which is an ordinal (which does not require any form of choice, as the equivalence classes can be described without the axiom of choice, as well as being an ordinal). The set of representatives is a set of ordinals, we take its union. Here's what i think this means. Please tell me i'm following this argument correctly. Let $X$ be the class of all the well orderings of $\omega$ Let $[G]$={$F \in X$\|$F$ is isomorphic with $G$} for every $G\in X$. Then we 'choose' representatives from each $[G]$ and take a union. I see this is definitely a choice since there might be infinitely many [G]'s. Why is this not a choice?? - I think I will revise my answer on that thread to make this construction clearer. – Asaf Karagila Jul 22 '12 at 12:53 @Asaf Thank you. Excuse me keep asking different questions, but it's hard for me to understand the part deriving a contradiction too.(that is, i know that $\alpha$ is a supremum but why ordinals less than $\alpha$ is isomorphic to some order relation that well orders $\omega$? And why that $\alpha + 1$ is countable is a contradiction?) – Katlus Jul 22 '12 at 13:09 @Asaf Or did you mean this? Let $X$={$\beta \in OR$\|$\alpha$ is equipotent with $\omega$}. Let $\gamma$ be the least element of $X$. Is this $\gamma$ same as $\alpha$ in your argument? – Katlus Jul 22 '12 at 13:24 I dunno why edit button is unavailable to me now.. I meant 'union' not 'least' above. – Katlus Jul 22 '12 at 13:26 ## 2 Answers Since your choice comes from the collection of ordinals, you can specify that you want the least such ordinal that satisfies the condition. This ordinal exists and is unique, so there is no choice involved. (Of course as Asaf's answer says it turns out that there is only one ordinal that satisfies the property, but you can also make this selection without knowing that fact) - Now i got it. Thank you. Would you please answer my comment too please? – Katlus Jul 22 '12 at 13:12 @Katlus To which comment are you referring? – Francis Adams Jul 22 '12 at 16:15 There is a theorem which states that every well-ordered set is isomorphic to a unique ordinal. Since the equivalence relation is that of order isomorphism all the elements in a given isomorphism class are isomorphic to a unique ordinal. We thus have eliminated the need to choose, since there is only one possible choice. The full poin I the sentence was that the statements saying that two ordered sets are isomorphic, and that a certain set is an ordinal do not require the axiom of choice. Thus this makes sense to do these condtructions in ZF. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9486261606216431, "perplexity_flag": "head"}
http://ams.org/bookstore?fn=20&arg1=amsipseries&ikey=AMSIP-45	New Titles \| FAQ \| Keep Informed \| Review Cart \| Contact Us Quick Search (Advanced Search ) Browse by Subject General Interest Logic & Foundations Number Theory Algebra & Algebraic Geometry Discrete Math & Combinatorics Analysis Differential Equations Geometry & Topology Probability & Statistics Applications Mathematical Physics Math Education Return to List Extensions of the Stability Theorem of the Minkowski Space in General Relativity Lydia Bieri and Nina Zipser, Harvard University, Cambridge, MA A co-publication of the AMS and International Press. SEARCH THIS BOOK: AMS/IP Studies in Advanced Mathematics 2009; 491 pp; hardcover Volume: 45 ISBN-10: 0-8218-4823-2 ISBN-13: 978-0-8218-4823-4 List Price: US\$119 Member Price: US\$95.20 Order Code: AMSIP/45 See also: Advances in Lorentzian Geometry: Proceedings of the Lorentzian Geometry Conference in Berlin - Matthias Plaue, Alan Rendall and Mike Scherfner This book consists of two independent works: Part I is "Solutions of the Einstein Vacuum Equations", by Lydia Bieri. Part II is "Solutions of the Einstein-Maxwell Equations", by Nina Zipser. A famous result of Christodoulou and Klainerman is the global nonlinear stability of Minkowski spacetime. In this book, Bieri and Zipser provide two extensions to this result. In the first part, Bieri solves the Cauchy problem for the Einstein vacuum equations with more general, asymptotically flat initial data, and describes precisely the asymptotic behavior. In particular, she assumes less decay in the power of $$r$$ and one less derivative than in the Christodoulou-Klainerman result. She proves that in this case, too, the initial data, being globally close to the trivial data, yields a solution which is a complete spacetime, tending to the Minkowski spacetime at infinity along any geodesic. In contrast to the original situation, certain estimates in this proof are borderline in view of decay, indicating that the conditions in the main theorem on the decay at infinity on the initial data are sharp. In the second part, Zipser proves the existence of smooth, global solutions to the Einstein-Maxwell equations. A nontrivial solution of these equations is a curved spacetime with an electromagnetic field. To prove the existence of solutions to the Einstein-Maxwell equations, Zipser follows the argument and methodology introduced by Christodoulou and Klainerman. To generalize the original results, she needs to contend with the additional curvature terms that arise due to the presence of the electromagnetic field $$F$$; in her case the Ricci curvature of the spacetime is not identically zero but rather represented by a quadratic in the components of $$F$$. In particular the Ricci curvature is a constant multiple of the stress-energy tensor for $$F$$. Furthermore, the traceless part of the Riemann curvature tensor no longer satisfies the homogeneous Bianchi equations but rather inhomogeneous equations including components of the spacetime Ricci curvature. Therefore, the second part of this book focuses primarily on the derivation of estimates for the new terms that arise due to the presence of the electromagnetic field. Titles in this series are co-published with International Press, Cambridge, MA. Readership Graduate students and research mathematicians interested in general relativity. Reviews "Both parts are well written. ...the book should be of interest to anyone who is doing research in mathematical relativity." -- Mathematical Reviews AMS Home \| Comments: [email protected] © Copyright 2012, American Mathematical Society Privacy Statement	{"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": 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": 0.8356271982192993, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/296551/what-is-a-minimal-polynomial-of-a-group-element-and-why-would-we-care-if-it-was?answertab=votes	What is a minimal polynomial of a group element, and why would we care if it was quadratic? EDIT: the $p$-stable definition I give below is incorrect. I have included the correct definition as an answer to this question. I am trying to understand the definition of a p-stable group. The first part of the definition is A faithful representation of a finite group $G$ on a vector space over a field of characteristic $p\not= 2$ is called $p$-stable if no $p$-element of $G$ has a quadratic minimal polynomial. What does it mean for a group element to have a minimal polynomial? Additionally, any intuition on a meaningful interpretation of this definition would be much appreciated. What is special about elements which have quadratic minimal polynomials? Why would we want to get rid of them? What's wrong with $p=2$? After this, 1. If $G$ has no nontrivial $p$-subgroups, $G$ is $p$-stable if every faithful characteristic $p$ representation is $p$-stable. 2. If $1<O_p(G)$ and $1=O_{p'}(G)$ then $G$ is $p$-stable if for all normal nontrivial $p$-subgroups $P$, for every $p$-element $x$ such that $[[x,P],x]=1$, the image $\overline{x}$ in $G/C_G(P)$ is contained in a normal $p$-subgroup. 3. If $1<O_p(G)$ and $1<O_{p'}(G)$, then $G$ is $p$-stable if $G/O_{p'}(G)$ is $p$-stable. What, mainly, is the connection between the $p$-stable representation definition and $\#2$? Are these somehow the same, but in a different light? (I see that $[[x,P],x]$ are elements of the form $p^{-1}x^{-1}pxx^{-1}x^{-1}p^{-1}xpx=(x^{-1})^p(p^{-1})^xpx$, so if that is equal to $1$ then $px=p^xx^p$. So there's sort of a "double twist" happening, which must be important in some way; but I don't see immediately any connection to minimal polynomials.) Sorry if these are basic questions on advanced material. I am sure the answer to this part is, to some extent, because this is a technical definition which is made to prove things with, but even the broadest intuition on this would help. - 1 I don't have any intuition on the definition. But presumably minimum polynomial of a group element $g \in G$ with respect to a representation $\phi$ refers to the minimum polynomial of $\phi(g)$. – JSchlather Feb 6 at 20:31 I think, from the phrase "satisfying a technical condition introduced by Gorenstein and Walter ... in order to extend Thompson's uniqueness results in the odd order theorem" this definition came backwards - it was defined to be useful based on an existing argument that worked in other cases. I can't be sure of that, but that feels likely. – Thomas Andrews Feb 6 at 20:54 @ThomasAndrews Maybe somebody who knows how these things were used in the classification theorems can help. If it is backwards like that, it would at least be useful to know what they were trying to do with it. – Samuel Handwich Feb 6 at 20:58 That wikipedia page is also somewhat suspect - it has no links to external definitions. (I thought $p$-element would be an element of order $p$, but later definitions would be meaninngless if that was true .) – Thomas Andrews Feb 6 at 21:05 @ThomasAndrews I think a $p$-element is supposed to be an element of $p$-power order. – Samuel Handwich Feb 6 at 21:08 show 3 more comments 4 Answers ````What is special about elements which have quadratic minimal polynomials? ```` As the action of a $p$-element $x\in G$ on a (finite dimensional) vector space $V$ over a field of characteristic $p$ is nilpotent, defining $V_0 = V, V_{i+1} := [V_i, x]$ you get $V_n = 0$ for some $n\ge 0$. As $x$ acts trivially if $n\le 1$, the minimal nontrivial case is $n=2$, i.e., $x$ acts quadratically. ```` Why would we want to get rid of them? ```` An action of $G$ on $V$ is $p$-stable if for all $a \in G$ holds $$[V, a, a] = 1 \implies a\mathrm{C}_G(V) \in \mathrm{O}_p(G/\mathrm{C}_G(V)).$$ The purpose of this condition is to exclude sections of $G/\mathrm{C}_G(V)$ that are isomorphic to $\mathrm{SL}_2(p)$ (and act "naturally" like $\mathrm{SL}_2(p)$, see for example the remark after Theorem 9.1.4 in [KS]). If you have a group $G$ with $\mathrm{C}_G(\mathrm{O}_p(G)) \le \mathrm{O}_p(G)$ and the action of $G$ on the chief factors of $G$ in $\mathrm{O}_p(G)$ is $p$-stable then Glauberman's ZJ-Theorem states that $Z(J(S))$ is normal $G$ for every $p$-Sylow subgroup $S$ of $G$ (see beginning of section 9.4 in [KS]). ````What's wrong with p=2? ```` Every element of order $2$ acting non-trivially on a vector space over a field of characteristic $2$ acts quadratically (according to the remark after 9.4.5 in [KS] you can replace $p$-stability by excluding sections of $G$ isomorphic to $\mathrm{SL}_2(2) = S_3$ directly to get meaningful results for $p=2$). [KS] Kurzweil, Stellmacher: The Theory of Finite Groups - I don't study these things so I can't give you any intuition, but here is the definition of the minimal polynomial: Given a representation $\phi\colon G \to \mathrm{GL}_n(k)$ and an element $g \in G$ the matrix $A = \phi(g)$ is square so given any polynomial $$f(x) = c_nx^n + c_{n-1}x^{n-1} + \cdots c_1x + c_0$$ it is perfectly well defined to set $$f(A) = c_nA^n + c_{n-1}A^{n - 1} + \cdots + c_1A + c_0I$$ where $I$ is the $n \times n$ identity matrix. The minimal polynomial of $g$ with respect to $\phi$ is then the monic polynomial $f$ of minimal degree such that $f(A) = 0$ (the $n \times n$ matrix whose entries are all zero). Also: we can define the ideal $\mathrm{ann}(A)$ of all polynomials $p$ such that $p(A) = 0$. The minimal polynomial $f$ is the monic polynomial that generates this ideal. So if $p(A) = 0$ then you can write $p = fh$ for some other polynomial $h$. - If there are no $p$-subgroups of $G$, then there are no $p$-elements of $G$, and therefore, by the definition of $p$-stable representation, all faithful representations are $p$-stable, since a representation being $p$-stable is defined only in terms of how it affects $p$-elements, and there are none. There are lots of indications that this Wikipedia page is not "mature." Specifically, there are no links to external articles defining the terms, and there are cases of mangled grammar. I would not trust that page. Probably best to go to the primary source. - I have found a paper in my library which includes the definition, which has some differences. Should I rewrite this question or start a new one? – Samuel Handwich Feb 6 at 21:17 It depends on your intent. I think you have the answer to your "technical" question: What does it mean by the minimal polynomial? The bigger question of "why this definition?" is still open, and it does seem like a separate question. – Thomas Andrews Feb 6 at 21:19 As @ThomasAndrews suggested, the definition given on Wikipedia was not correct. I have located the definition in an article by Glauberman: 1. Let $p$ be an odd prime and $G$ be a finite group with $O_p(G)\not= 1$. Then $G$ is $p$-stable if it satisfies the following condition: Let $P$ be an arbitrary $p$-subgroup of $G$ such that $O_{p'}(G)P$ is a normal subgroup of $G$. Suppose that $x\in N_G(P)$ and $\overline{x}$ is the coset of $C_G(P)$ containing $x$. If $[P,x,x]=1$, then $\overline{x}\in O_n(N_G(P)/C_G(P))$. 2. Define $\mathcal{M}_p(G)$ as the set of all $p$-subgroups of $G$ maximal with respect to the property that $O_p(M)\not= 1$. 3. Let $G$ be a finite group and $p$ an odd prime. Then $G$ is called $p$-stable if every element of $\mathcal{M}_p(G)$ is $p$-stable. (I am not sure what little $n$ is in (1). I am guessing just any $n\in \mathbb{N}$?) So I guess that $[P,x,x]$ is a commutator condition that no critical $p$-subgroups are allowed to have when we want $G$ to be $p$-solvable. (Evidently, the case $p\not= 2$ is degenerate with respect to the $[P,x,x]$ condition, which is why it is not included in this.) As a side note, excluding $SL(2,p)$ from a group implies that it is $p$-stable - I suspect because the $[P,x,x]$ relation gives rise to that sort of subgroup. I am still not sure what to think intuitively about what makes these groups "stable" with respect to $p$. Perhaps I will find and read the papers in the bibliography. - If you can get the book The Theory of Finite Groups by Kurzweil and Stellmacher, take a look at chapter 9 "Quadratic Action". – jug Feb 6 at 21:50	{"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": 142, "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": 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": 0.9515759944915771, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/273155/where-have-i-made-my-mistake-in-calculating-p-1ap	# Where have I made my mistake in calculating $P^{-1}AP$? I have $$Q = \begin{pmatrix} -\mu & \mu \\ \lambda & -\lambda \end{pmatrix}$$ and I want to work out the value of $\mathbb{P}(t) = \exp(Qt)$ So I diagonalised $Q$ and then worked out the exponential of the diagonal matrix. I got this to be: $${Q}t = \pmatrix{-\mu t &\mu t \\ \lambda t & -\lambda t} = \pmatrix{1 & -\frac{\mu }{\lambda } \\ 1 & 1}^{-1} \cdot \pmatrix{0 & 0 \\ 0 & -t(\lambda + \mu)} \cdot \pmatrix{1 & -\frac{\mu }{\lambda } \\ 1 & 1}.$$ So using the middle matrix, I got $$\exp (Qt) = \pmatrix{1 & 0 \\ 0 & \exp(-t(\lambda + \mu))} = \pmatrix{1 & 0 \\ 0 & \exp(T)},$$ where $T = -t(\lambda + \mu)$. Then, using $\exp (P^{-1}AP) = P^{-1}e^AP$, I was supposed to get $$\mathbb{P}(t) = \exp({Q} t) = \frac{1}{\lambda + \mu}\pmatrix{\lambda + \mu \exp(T) & \mu - \mu \exp(T) \\ \lambda - \lambda\exp(T) & \mu + \lambda\exp(T)}.$$ This is what I did. To first work out $P^{-1}$ I got $$P^{-1} = \frac{1}{1 + \frac{\mu}{\lambda}} \pmatrix{1 & \frac{\mu}{\lambda} \\ -1 & 1} = \frac{\lambda}{\lambda + \mu} \pmatrix{1 & \frac{\mu}{\lambda} \\ -1 & 1}$$ Then doing $P^{-1}e^A$ gave me $$\frac{\lambda}{\lambda + \mu} \pmatrix{1 & \frac{\mu}{\lambda} \exp (T) \\ -1 & \exp (T)}$$ Then doing this times $P$ gave me $$\frac{\lambda}{\lambda + \mu} \pmatrix{1 + \frac{\mu}{\lambda} \exp (T) & -\frac{\mu}{\lambda} + \frac{\mu}{\lambda} \exp (T) \\ -1 + \exp (T) & \frac{\mu}{\lambda} + \exp (T)}$$ Multiplying through by $\lambda$ gives me $$\frac{\lambda}{\lambda + \mu} \pmatrix{\lambda + \mu \exp (T) & - \mu - \mu \exp (T) \\ - \lambda + \exp(T) & \mu + \lambda \exp (T)}$$ Clearly it's started going wrong in the matrix before this but I can't see where I've made my mistakes. Can someone help please? Thank you - 1 Are you sure about the diagonalization? I think your $P$ should be transposed. – leonbloy Jan 8 at 23:17 @leonbloy If I'm solving $det(Qt - AI) = 0$, I get my complementary (if thats what it's called) equation to be $A^2 + A(\mu t + \lambda t) + 2\mu \lambda t^2 = 0$. Is that correct? I'm having difficulty solving that now... – Kaish Jan 8 at 23:36 ## 2 Answers When you calculate $\exp(tQ)$, you don't need to diagonalize $tQ$. Instead, you can diagonalize $Q=V\Lambda V^{-1}$, where the column vectors of $V$ are the eigenvectors of $Q$. (When you don't remember whether $Q=V\Lambda V^{-1}$ or $Q=V^{-1}\Lambda V$ is correct, think about $QV=V\Lambda$.) Once you have $Q=V\Lambda V^{-1}$, $tQ=V(t\Lambda) V^{-1}$. Then you would get $$\exp(tQ)=V\exp(t\Lambda)V^{-1}$$ where $\exp(t\Lambda)$ is easy to get. For a degenerate $2\times 2$ matrix $Q$, you don't need to calculate the eigenvalues by the characteristic polynomial. One of them must be $\lambda_1=0$, and another one is $\lambda_2=\operatorname{trace}(Q)$. - What does trace mean? – Kaish Jan 9 at 11:29 – Jack Jan 9 at 13:02 Kaish in in first post you write correct, now in this post not. $Qt=PAP^{-1}$. So my hint must be: you can use $e^{PAP^{-1}}=Pe^AP^{-1}$ Calculating the determinant we have $(-\mu-A)(-\lambda-A)-\mu\lambda=0$ $\implies$ $A^2+(\mu+\lambda)A=0$ $\implies$ A=0 or $A=-(\mu+\lambda)$ can you find the eigenvector from here? - Oh ok, thanks. I'll try that again then. But still, I can't get the right eigenvectors and values to work out my $PAP^{-1}$, how do I do this? – Kaish Jan 8 at 23:43 @Kaish what you want is caculate eingevectors of \begin{pmatrix} -\mu & \mu \\ \lambda & -\lambda \end{pmatrix}? – user52188 Jan 8 at 23:49 @Kaish see above with $t$ is the same – user52188 Jan 9 at 0:16	{"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": 36, "mathjax_display_tex": 9, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9355655908584595, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-software/174764-reduce-precision-mathematica-print.html	# Reduce Precision in Mathematica Printable View • March 16th 2011, 08:39 AM mattfred Reduce Precision in Mathematica I've spent quite a bit of time searching for an answer to my problem, but one of the main problems is that I don't exactly know how to phrase it correctly. I have an equation in two variable that I want to find the roots in one variable (E) as a function of the other(r). I can plot f(E) with a known r relatively fast, but I cannot find the roots at all. I've tried Solve, FindRoot, and FindInstance with various accuracy and precision goals and damping factors. I think one of the main problems is that the numbers involved are quite large. I only care about two or three sig figs, but even at the lowest precision, mathematica wants the accuracy to be to the right of the decimal place, which gives about 15 significant figures. I also think there is a small imaginary component to the solution that I don't care about, but may be preventing the calculation from converging. Basically, I want a rough, fast estimate of the zeros. I could do it by hand, but it seems to me that if the computer can plot the function as rapidly as it does, I should be able to set the tolerances where I want and get an idea of where the zero is. Thanks! • March 16th 2011, 08:46 AM Ackbeet • March 16th 2011, 10:13 AM mattfred 1 Attachment(s) I've attached a screenshot of the input. I don't really know how tractable it will be to see the input. Every variable except Eh and r is a defined constant. I never took a course in linear algebra, so I don't really know much about the matrix operations. The problem is initially solving for the coefficients of $\psi$ (line 45) with a certain set of boundary conditions. I'm positive that that is all handled correctly as when I evaluate Ehole with the proper r and Eh I get the values I expect. I can easily plot the argument in Ehole. I just want to relax the convergence criteria so that I can rapidly model Ehole as a function of R for various changes in the other input parameters. Thanks! Attachment 21162 • March 17th 2011, 07:06 AM Ackbeet Hmm. That's extraordinarily complicated. Could you please send me the Mathematica notebook that has all that? I'm going to have to play around a bit in order to make anything of it. If MHF balks at uploading the file, just change the extension to pdf or jpg or something like that. Thanks! • March 17th 2011, 08:07 AM mattfred 1 Attachment(s) Haha yeah it is quite complicated. I posted about solving the system earlier: http://www.mathhelpforum.com/math-he...tml#post621430 but eventually got help from a colleague. Everything up until "Determination of Energy" is definitions of constants and solving for the coefficients with the appropriate boundary conditions. I'm certain that that portion is handled correctly. Attachment 21171 • March 17th 2011, 08:17 AM Ackbeet Well, I tried opening the file, but I have Mathematica 4, not 7. Is there any way you can save the notebook in a backwards-compatible way? • March 17th 2011, 10:19 AM mattfred 1 Attachment(s) The best option I can come up with is converting to Mathematica Player. It's a free download and claims that it retains all the features of Mathematica with certain licensing agreements that I believe this fails under: Wolfram Mathematica Player: Free Interactive Player for Mathematica Notebook Documents When I open the file, it looks indistinguishable to me. You'll have to change the file extension to ".nbp" Attachment 21174 • March 19th 2011, 04:13 PM Ackbeet I hate to break it to you, but I don't think I'm going to be of much help here. In order for me to really do anything useful, I need to be able to evaluate cells and play around. I successfully installed the Wolfram software you linked to, but it doesn't let me edit or even evaluate cells (unless I'm missing something). I could try to type out the file in Mathematica 4 format, but that seems like an enormous amount of work.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.9507363438606262, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/214681-explicit-description-line.html	Thread: 1. explicit description of a line? Find an explicit description of the line 3x+2y = 0 in R2 through the origin. the answer is: span {[2, -3]} , but how? 2. Re: Explicit description of a line? The line $3x+2y=0$ in $\mathbb R^2$ is spanned by any nonzero vector $\begin{pmatrix}x \\ y\end{pmatrix}$ such that $(x,y)$ lies on the line. $\begin{pmatrix}2 \\ -3\end{pmatrix}$ is one such vector; another would be $\begin{pmatrix}-1 \\ \frac32\end{pmatrix}$. 3. Re: explicit description of a line? Originally Posted by yangx Find an explicit description of the line 3x+2y = 0 in R2 through the origin. the answer is: span {[2, -3]} , but how? I do not necessarily disagree with reply #2. However, the point $(-2,3)$ is on the line $3x+2y=0~.$ That line contains $(0,0)$, so its direction vector is $<-2,3>$. Thus in traditional vector geometry its equation is $<0,0>+t<-2,3>~.$ So notation is everything here. Unless you tell us about the notation in use in your notes, we cannot really 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": 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": 0.9257779121398926, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/tagged/complexity+encryption	# Tagged Questions 1answer 389 views ### How do I derive the time complexity of encryption and decryption based on modular arithmetic? I want to calculate the time complexity of two encryption and decryption algorithms. The first one (RSA-like) has the encryption $$C := M^e \bmod N$$ and decryption $$M_P := C^d \bmod N.$$ ... 2answers 298 views ### What is the time complexity of the RC4 encryption & decryption algorithms? I'm trying to figure out what the time complexity of RC4 encryption & decryption algorithms is, in big-O-notation. 1answer 645 views ### Big-O Notation: Encryption Algorithms I am currently completing a dissertation concerning the encryption of data through a variety of cryptographic algorithms. I have spent much time reading journals and papers but as yet have been ...	{"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": 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": 0.9268044233322144, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/12100?sort=oldest	## Why Weil group and not Absolute Galois group? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In many formulation of Class Field theory, the Weil group is favored as compared to the Absolute Galois group. May I asked why it is so? I know that Weil group can be generalized better to Langlands program but is there a more natural answer? Also we know that the abelian Weil Group is the isomorphic image of the reciprocity map of the multiplicative group (in the local case) and of the idele-class group (in the global case). Is there any sense in which the "right" direction of the arrow is the inverse of the reciprocity map? Please feel free to edit the question into a form that you think might be better. - I have accepted the answer by Johnson but I still hope to receive the answer for the second part of the question. – Tran Chieu Minh Jan 17 2010 at 19:28 ## 3 Answers One reason we prefer the Weil group over the Galois group (at least in the local case) is that the Weil group is locally compact, thus it has "more" representations (over $\bf C$). In fact, all $\bf C$-valued characters of $Gal(\bar{\bf Q_p} / \bf Q_p)$ have finite image, where as that of $W_{\bf Q_p}$ can very well have infinite image. The same goes for general representations of these groups (recall that $\bf{GL}_n(\bf C)$ has no small subgroups.) The global Weil group (which is much more complicated than the local one), on the other hand, is a rather mysterious object that is pretty much untouched in modern number theory as far as I can tell. Supposedly the global Langlands group used in the global Langlands correspondence should be the extension of the global Weil group by a compact group, but this is still largely conjectural. The standard reference is Tate's "Number Theory Background" in the Corvallis volumes (available for free at ams.org). Also Brooks Roberts has notes on Weil representations available at his website. - 1 Thank you for the reply. Do you think it is possible to have a natural reason beside utility for which we prefer Weil group? – Tran Chieu Minh Jan 17 2010 at 19:17 1 Tran: Isn't utility one of the best motivators to do anything? – Adam Hughes Mar 22 2011 at 13:21 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The Weil group appears for several reasons. Firstly: if $K$ is a non-archimedean local field with residue field $k$, the local reciprocity law induces an embedding $K^{\times} \hookrightarrow G_K^{ab}.$ The image consists of all elements in $G_k$ whose image is an integral power of Frobenius. This is the abelianized Weil group; it just appears naturally. Secondly: suppose that $K$ is a global field of positive characteristic, i.e. the function field of a curve over a finite field $k$. Then the global reciprocity map identifies the idele class group of $K$ with a subgroup of $G_K^{ab}$ consisting of elements which act on $k$ by integral powers of Frobenius. So again, it is the abelianized Weil group that appears. Thirdly: suppose that $E$ is an elliptic curve over a quadratic imaginary field $K$ with complex multipliction by $\mathcal O$, the ring of integers in $K$. (Thus I am implicity fixing $K$ to have class number one, but this is not so important for what I am going to say next.) If $\ell$ is a prime, then the $\ell$-adic Tate module is then free of rank one over $\mathcal O_{\ell}$ (the $\ell$-adic completion of $\mathcal O$), and the $G_K$-action on this Tate module induces a character $\psi_{\ell}:G_K^{ab} \rightarrow \mathcal O_{\ell}^{\times}$. There is a sense in which the various $\psi_{\ell}$ are indepenent of $\ell$, but what is that sense? Well, suppose that $\wp$ is a prime of $K$, not dividing $\ell$ and at which $E$ has good reduction. Then the value of $\psi_{\ell}$ on $Frob_{\wp}$ is indepenent of $\ell$, in the sense that its value is an element of $\mathcal O$, and this value is independent of $\ell$. More generally, provided that $\wp$ is prime to $\ell$, the restriction of $\psi_{\ell}$ to the local Weil group at $\wp$ is independent of $\ell$ (in the sense that the value at a lift of Frobenius will be an algebraic integer that is independent of $\ell$, and its restriction to inertia at $\wp$ will be a finite image representation, hence defined over algebraic integers, which again is then independent of $\ell$). Note that independence of $\ell$ doesn't make sense for $\psi_{\ell}$ on the full local Galois group at $\wp$, since on this whole group it will certainly take values that are not algebraic, but rather just some $\ell$-adic integers, which can't be compared with one another as $\ell$ changes. Now there is also a sense in which the $\psi_\ell$, as global Galois characters, are indepdendent of $\ell$. Indeed, we can glue together the various local Weil group representations to get a representation $\psi$ of the global Weil group $W_K$. Since it is abelian, this will just be an idele class character $\psi$, or what is also called a Hecke character or Grossencharacter. It will take values in complex numbers. (At the finite places it even takes algebraic number values, but when we organize things properly at the infinite places, we are forced to think of it as complex valued.) Note that $\psi$ won't factor through the connected component group, i.e. it won't be a character of $G_K^{ab}$. It is not a Galois character, but a Weil group character. It stores in one object the information contained in a whole collection of $\ell$-adic Galois characters, and gives a precise sense to the idea that these various $\ell$-adic characters are independent of $\ell$. This is an important general role of Weil groups. Fourthly: The Hecke character $\psi$ above will be an algebraic Hecke character, i.e. at the infinite places, it will involve raising to integral powers. But we can also raise real numbers to an arbitrary complex power $s$, and so there are Hecke characters that do not come from the preceding construction (or ones like it); in other words, there are non-algebraic, or non-motivic, Hecke characters. But they are abelian characters of the global Weil group, and they have a meaning; the variable $s$ to which we can raise real numbers is the same variable $s$ as appears in $\zeta$- or $L$-functions. In summary: Because Weil groups are "less completed", or "less profinite", then Galois groups, they play an important role in describing how a system of $\ell$-adic representations can be independent of $\ell$. Also, they allow one to describe phenomena which are automorphic, but not motivic (i.e. which correspond to non-integral values of the $L$-function variable $s$). (They don't describe all automorphic phenomena, though --- one would need the entire Langlands group for that.) - Excellent answer ! By the way, the second section of Gross-Reeder paper on arithmetic invariants (Duke 2010) is also nice for Weil groups – unknown (google) Jun 24 at 16:56 Perhaps something worth pointing out, relating to how the Weil group appears naturally: it arises from a compatible system of group extensions at finite levels. Indeed, one of the "axioms" of class field theory, is the existence of a "fundamental class" uL/K in $H^2(\operatorname{Gal}(L/K),C_L)$ for each finite Galois extension $L/K$ (where $C_L$ is the class module). Each of these gives a group extension $$1\rightarrow C_L\rightarrow W_{L/K}\rightarrow \operatorname{Gal}(L/K)\rightarrow 1.$$ The projective limit of these gives the absolute Weil group fitting in $$1\rightarrow C\rightarrow W_K\rightarrow G_K$$ with the rightmost map having dense image (and $C$ is the formation module of the class formation). Thus, you can think of the Weil group as arising canonically out of the results of class field theory, thus making it a natural replacement of $G_K$ in questions related to CFT. I like section 1 of chapter III of Neukirch–Schmidt–Wingberg's Cohomology of number fields and the last chapter of Artin–Tate for this material. -	{"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": 75, "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": 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": 0.9334783554077148, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/206679/integer-solutions-of-x-y-z/206711	# Integer solutions of $x! = y! + z!$ There was an interesting problem asked about triples $(x,y,z)$ which are solutions of $$x! = y! + z!.$$ Here $(2,1,1)$ is a solution because $2! = 1! + 1!$, as are $(2,1,0)$ and $(2,0,1)$. Now I wanted to analyze this a bit further and thought of using the gamma function definition of a factorial, to see where it led and this is what I got: $x! = y! + z!$ $\Gamma(x) = \Gamma(y) + \Gamma(z)$ $\int_{-\infty}^{\infty}t^xe^{-t}dt = \int_{-\infty}^{\infty}t^ye^{-t}dt + \int_{-\infty}^{\infty}t^ze^{-t}dt$ $\int_{-\infty}^{\infty}t^xe^{-t}dt - \int_{-\infty}^{\infty}t^ye^{-t}dt - \int_{-\infty}^{\infty}t^ze^{-t}dt = 0$ $\int_{-\infty}^{\infty}(t^x - t^y - t^z)e^{-t}dt = 0$ Now my line of thinking was that, in a manner similar to the fundamental lemma of the calculus of variations, this should imply that the above integral can only be true for arbitrary values of $x, y$ and $z$ if $t^x - t^y - t^z = 0$. I'm not really sure if that is justifiable so I'd appreciate some comment on this. The reason I continued on despite the uncertainty is that when you're left with the polynomial $t^x - t^y - t^z = 0$ you encounter a strange fact. First off, plugging in a triple like $(2,1,1)$ results in $t^2 - t^1 - t^1 = 0$ which implies $t^2 = 2t$ thus $t = 2$. Now plugging in $t = 2$ gives $2^2 = 1^2 + 1^2$. In other words you get what you'd expect. However plugging in a triple you know is inconsistent, like $(0,0,0)$, gives $t^0 = t^0 + t^0$ $1 = 1 + 1$ $1 = 2$. You get absurdities when you plug in inconsistent triples. So while $(2,1,0)$, which works, gives $t^2 - t^1 - t^0 = 0$ $t^2 - t - 1 = 0$ whose roots are $\frac{1}{2} + \frac{\sqrt{5}}{2}$ $\frac{1}{2} - \frac{\sqrt{5}}{2}$ and on plugging these into $t^x - t^y - t^z = 0$ gives a consistent equality for both roots, an inconsistent triple like $(3,2,1)$ leaves us with $t^3 - t^2 - t = 0$, one of whose roots are $t = 0$ (which gives a consistent equality $0^3 - 0^2 - 0 = 0$ whereas another one of it's roots, $t = \frac{1}{2} + \frac{\sqrt{5}}{2}$ does not give a consistent equality $(\frac{1}{2} + \frac{\sqrt{5}}{2})^3 - (\frac{1}{2} + \frac{\sqrt{5}}{2})^2 - (\frac{1}{2} + \frac{\sqrt{5}}{2}) = 0$. If all of the above is magically okay you can use this idea to show that no triples $(n, n - 1, n - 2)$ will work for $n$ greater than $2$. I'd like to go further with it but I'm afraid it's all just conceptually flawed, is it? A more interesting question stems from something a great lecturer told me, she basically analyzed the whole question geometrically and said something about certain triples having to do with the area of those gamma integrals canceling out. I thought I understood what she was saying but I actually don't so I'd appreciate any comment on what this could mean. I hope I was clear enough, thanks for your time. - 4 The reasoning after and including calculus of variations doesn't make sense to me. Also I would note: this isn't the usual definition of the gamma function, which interpolates as $\Gamma(x) = (x-1)!$. – Cocopuffs Oct 3 '12 at 17:23 1 Wow, I'm such an idiot. Based on forgetting the - 1 term I went in circles doing a lot of nonsense. Just delete this post, no marks, complete & utter failure... Thanks for your time. – sponsoredwalk Oct 3 '12 at 18:13 5 no! upvote for you for forging ahead in muddy waters! – AK4749 Oct 3 '12 at 20:37 ## 1 Answer $x! = y! + z!$ does not have any solutions in integers with $x \geq 3.$ As soon as $x \geq 3,$ we have $(x-1)! \leq x! / 3.$ With the necessary $y,z < x,$ we get $y! \leq x! / 3, \; \; z! \leq x! / 3,$ so $y! + z! \leq 2 x! / 3$ and $y! + z! \neq x!$ - Neat solution. You can rewrite the solution this way: $y,z <x$ hence $y,z \leq x-1$, and thus $x!=y!+z! \leq (x-1)!+(x-1)!$ which implies $x \leq 2$. – N. S. Oct 3 '12 at 17:46 2 The more interesting problem is $x! = y! \; z!.$ For any $n,$ we know $n! = (n-1)! \; n.$ Taking $n = z!,$ we get the infinite family $(z!)! = (z! - 1)! \; z!.$ I think there is at least one other solution, the question is to find others. It's on MO somewhere. – Will Jagy Oct 3 '12 at 18:07 – Rick Decker Oct 3 '12 at 18:21 @RickDecker thanks, I think that is actually where I saw it. Also I had not remembered $6! \; 7! = 10!$ – Will Jagy Oct 3 '12 at 18:34	{"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": 59, "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": 0.9659449458122253, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/14963/bi-invariant-metrics-on-sl-n-mathbbr	# Bi invariant metrics on $SL_n(\mathbb{R})$ Does there exist a bi-Invariant metric on $SL_n(\mathbb{R})$. I tried to google a bit but I didn't find anything helpful. - Doesn' t the negative of the Killing form do? – Mariano Suárez-Alvarez♦ Dec 20 '10 at 16:04 Ah: That gives a biinvariant semiRiemannian metric, rather. – Mariano Suárez-Alvarez♦ Dec 20 '10 at 16:06 ## 2 Answers A connected Lie group has a biinvariant metric iff it is isomorphic to the dirtect product of a compact one and a vector space---see Milnor's Curvatures of left invariant metrics. Such a factorization would carry over to the Lie algebras, etc. - It's enough to show the answer is "no" for $SL_2(\mathbb{R})$, because $SL_2(\mathbb{R})$ is naturally embedded in $SL_n(\mathbb{R})$, and if $SL_n(\mathbb{R})$ has a biinvariant metric, then the induced metric on $SL_2(\mathbb{R})$ would be biinvariant. So let's focus on $SL_2(\mathbb{R})$. Finding a biinvariant metric on $SL_2(\mathbb{R})$ is equivalent to finding an $Ad(SL_2(\mathbb{R})$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, so I'll show there is no such $Ad$ invariant metric. Now, $\mathfrak{sl_2(\mathbb{R})}$ is 3 dimensional with basis $E_{12}, E_{21}, E_{11} - E_{22}$ where $E_{ij}$ denotes the matrix with a 1 in the $ij$ slot and a 0 elsewhere. Consider $A = diag(2,1/2)$ ,the diagonal matrix with diagonal entries 2 and 1/2. This is clearly in $SL_2(\mathbb{R})$. Note then that $AE_{12}A^{-1} = 4 E_{12}$. But if an inner product were $Ad$ invariant, it would have to preserve the length of $E_{12}$, giving a contradiction. Thus, there is no $Ad$ invariant inner product on $\mathfrak{sl_2(\mathbb{R})}$, and hence there is no biinvariant metric on $SL_n(\mathbb{R})$. -	{"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": 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": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8921047449111938, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/20306/joining-the-definitions-of-entropy	Joining the definitions of entropy $\int \frac{Q_{rev}}{T} = \Delta(k_B\ln\Omega)=\Delta S$ Could anyone give some definite proof for this? I was able to prove that the two definitions of change in entropy are equivalent for an isothermal process carried out on a gas (by quantizing space and then limiting the quantization to infinity), but my proof makes the absolute entropy of the gas infinite. If the process is not isothermal, the particle's velocities come into the picture and I don't know how to deal with that. I tried making various assumptions (quantizing time, etc), but it didn't work. I know that once I prove it for another process, it will be proven for any process carried out on ideal gases(as I can write any process as the combination of isothermal and another process). Could someone please nudge me in the right direction/give a proof? - In what framework do you work? If you know define the temperature as $\frac{1}{T}=\frac{\partial S}{\partial U}$, then you can just solve the first law of thermodynamics $\delta Q=PV+dU(S,V)$ for $dS$. – Nick Kidman Jan 31 '12 at 14:35 @NickKidman I'm defining everything on the basis of (P,V,T). Then I define $\Delta S$ as $\int\frac{Q}{T}$, and try to prove that this is equal to change in %k_Bln\Omega\$ . How does your approach merge these two definitions of entropy? It seems that you are only proving a general formula for entropy (I'm not sure, though) – Manishearth♦ Feb 3 '12 at 2:37 In classical thermodynamics, there are only macroscopical quantities. For every $(P,V,T)$, there are many many microstates. The introduction of microstates only comes with statistical mechanics. In that framework the volume $\Omega(E,V)$ is a natural object in phase space and $S(E,V)=log(\Omega(E,V))$ is the definition of the entropy. You can now compute the derivative $dS$ (where $dE$, i.e. $dU$, is related to $\delta Q$ via first law of theremodynamics, if you will). Both $T$ and $P$ are abstractly defined in terms of derivatives, although $P$ is easy to motivate. – Nick Kidman Feb 3 '12 at 8:17 Yeah, but I want to calculate $\Omega$ as a function of P/V/T/U/whatever. Omega as a func of V is easy at a constant temperature, but when the temperature changes, the speegs of the molecules add to the microstates in a way which I don't know how to calculate. – Manishearth♦ Feb 3 '12 at 10:36 I don't understand. So you're doing calculations within the framework of statistical mechanics after all? If you have a potential like $U$ or $S$, then you can compute the equation of state, and if you like you can express $V$ as function of, for example $T$. – Nick Kidman Feb 3 '12 at 12:26 show 10 more comments 1 Answer The fact that $k_B \ln \Omega$ coincides with entropy $S$ defined in thermodynamics comes from microcanonical ensemble. There are many resources out there on microcanonical ensemble, for example, this. After you come to the conclusion that $$\beta=\left( \frac{ \partial \ln \Omega }{ \partial U }\right)_{N,V}$$ fully characterizes thermal equilibrium, you know that it must be a function of thermodynamic temperature, and thermodynamic temperature alone, by virtue of zeroth law of thermodynamics. So $\beta=f(T)$. Compare this to $$\frac{1}{T}=\left( \frac{ \partial S }{ \partial U }\right)_{N,V}$$ and you get that S must be the function of $\ln \Omega$. The remaining question is the exact form of this function, and you already derive it from special cases. - Could you please provide a more detailed explanation? I'm not too familiar with the statistical half of thermodynamics. I want to try to prove it by counting microstates in a system given P,V,T. If this is not possible, do you have a link/etc to the proof? – Manishearth♦ Feb 3 '12 at 2:42 By the way, S as a function of $\ln\Omega$ can be more easily proved from the multiplicativeness of probability and from the fact that S is extensive. Not exactly the type of answer I was looking for, but it'll do. Nick Kidman gave a nice explanation above that clarified my main confusion. – Manishearth♦ Feb 6 '12 at 4:10 @Manishearth: How do you prove S as a function of $\ln \Omega$ from the multiplicativeness of probability and from the fact that S is extensive? – C.R. Feb 6 '12 at 9:47 If I mix two systems with $\Omega_1$, $\Omega_2$, the resultant $\Omega=\Omega_1\times\Omega_2$. But, Entropy is extensive, so $S=S_1+S_2$. If $S_r=f(\Omega_r)$, then $f(x)=k\ln x$ (only function which turns multiplication into addition), thus $S=k\ln\Omega$. We can later set the constant $k=k_B$. – Manishearth♦ Feb 6 '12 at 12:05 @Manishearth: But you need to prove that S is indeed related to Ω, and not to any other quantities. U is also extensive, but you don't get $U=k_B \ln \Omega$. – C.R. Feb 6 '12 at 12:40 show 1 more comment	{"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": 36, "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": 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": 0.9381547570228577, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-equations/190618-existence-uniqueness-theorem-print.html	# Existence and Uniqueness Theorem Printable View • October 17th 2011, 12:32 PM JohnDoe Existence and Uniqueness Theorem Greetings I have a question about the basic existence and uniqueness theorem for nth order homogeneous linear differential equations on a certain interval.If $f(x_0)=0,f'(x_0)=0,...,f^{(n-1)}(x_0)$ where x, then $f(x)=0$ on the whole interval. I did not understand the logic behind this, can someone explain it to me? I would be grateful. I think this is also valid for first order differential equations if $f(x_0)=0$ then $f(x)=0$ if I am not mistaken? Thanks • October 17th 2011, 04:36 PM Ackbeet Re: Existence and Uniqueness Theorem For the linear nth order homogeneous differential equation on an interval, you know that if the coefficients are continuous functions, and the coefficient of the highest-order derivative is nonzero on the interval, then there exists a unique solution to the initial value problem. Does f(x) = 0 solve the initial value problem? What could you say then? • October 18th 2011, 12:46 AM JohnDoe 1 Attachment(s) Re: Existence and Uniqueness Theorem Thanks for your reply.The attached pdf is from Shepley, Ross Differential Equations 3rd Edition and the theorem I did not understand is theorem 4.1 and especially its corollary. Can you shed some light on that ? • October 18th 2011, 10:47 AM HallsofIvy Re: Existence and Uniqueness Theorem What exactly is your confusion? The theorem and it corollary state that the nth order linear differential equation, with continuous coefficients in some neighborhood of $x_0$, has unique solution to the "initial value" problem- that is finding a function, f, that satisfies the given differential equation and has specified values for the function and its derivatives, up to the n-1 derivative, at $x_0$. Specifically, the "theorem" says there is a solution and the corollary says there is only one solution. • October 18th 2011, 01:03 PM JohnDoe Re: Existence and Uniqueness Theorem I did not understand the corollory in the special case $f(x_0)=0 ... f^{n-1}(x_0)=0$. Why does it imply that $f(x)=0$ on the whole interval? • October 18th 2011, 04:18 PM Ackbeet Re: Existence and Uniqueness Theorem First answer this question: is f(x) = 0 a solution of the initial value problem? • October 19th 2011, 11:28 AM JohnDoe Re: Existence and Uniqueness Theorem F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if $f(x_0)=0$ and $x_0$ belongs to aforementioned certain interval I, then f(x)=0 for every x that is a member of this interval. How is this possible ? • October 19th 2011, 11:39 AM Ackbeet Re: Existence and Uniqueness Theorem Quote: Originally Posted by JohnDoe F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if $f(x_0)=0$ and $x_0$ belongs to aforementioned certain interval I, then f(x)=0 for every x that is a member of this interval. How is this possible ? You don't have to assume anything. Tell me this: 1. Does f(x) = 0 satisfy the differential equation? 2. Does it satisfy the initial conditions? • October 19th 2011, 11:43 AM JohnDoe Re: Existence and Uniqueness Theorem Quote: Originally Posted by Ackbeet You don't have to assume anything. Tell me this: 1. Does f(x) = 0 satisfy the differential equation? 2. Does it satisfy the initial conditions? It satisfies the differential equation and the initial conditions. • October 19th 2011, 11:46 AM Ackbeet Re: Existence and Uniqueness Theorem Quote: Originally Posted by JohnDoe It satisfies the differential equation and the initial conditions. Right. So that tells you that f(x) = 0 is a solution. The question is, is it the only solution? Now, you tell me: does the existence and uniqueness theorem apply to this problem? • October 19th 2011, 11:47 AM JohnDoe Re: Existence and Uniqueness Theorem Yes it does apply then I think I answered my own question. I think it is the trivial solution that is unique if I am not mistaken ? • October 19th 2011, 11:48 AM Ackbeet Re: Existence and Uniqueness Theorem Quote: Originally Posted by JohnDoe Yes it does apply then I think I answered my own question. (Clapping) • October 19th 2011, 01:04 PM HallsofIvy Re: Existence and Uniqueness Theorem Quote: Originally Posted by JohnDoe Yes it does apply then I think I answered my own question. I think it is the trivial solution that is unique if I am not mistaken ? Your wording implies that you think this solution is unique because it is the trivial solution. In fact, for the conditions given on the coefficients, there exist a unique solution with $f(x_0)$, $f'(x_0)$, ..., $f^{n-1}(x_0)$ any given numbers. Since the trival function, f(x)= 0, satisfies both the differential equation and the conditions $f(x_0)= 0$, $f'(x_0)= 0$, ..., $f^{n-1}(x_0)= 0$, it is the only solution to that problem. By the way, this is not true for "boundary value problems". The functions f(x)= C sin(x) satisfies the differential equation $\frac{d^2f}{dt^2}= -f$ and the conditions $y(0)= 0$, $y(\pi)= 0$ for any value of C. • October 20th 2011, 05:49 AM JohnDoe Re: Existence and Uniqueness Theorem I concluded that this is only valid for an initial value problem and understood what is told in the book many thanks again for your help. All times are GMT -8. The time now is 06:18 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": 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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9337546825408936, "perplexity_flag": "head"}
http://www.physicsforums.com/showpost.php?p=2689902&postcount=5	View Single Post My question is then: Does the time dilation due to gravity need to be added to the time dilation due to velocity? IE, GPS satellites - do they have to account for both? IE: $$T_{1}=T_{2}\sqrt{1-{Gm_{1}}/r^2} + T_{2}\sqrt{1-{2Gm_{1}}/r^2}$$ Where the first term is the time dilation due to the velocity of the circular orbit, and the 2nd due to gravity. I still think there is a problem in my circular orbit formula, and that the two terms should match. Or am I just way to far off, and I should go RTFM some more	{"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": 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": 0.9698732495307922, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/54909/geodesic-versus-geodesic-loop	# Geodesic versus geodesic loop Let (M,g) be a closed manifold and let $\alpha$ be an element of $G=\pi_1(M,p)$ we can define the norm of $\alpha$ with respect to p as the infinimum riemannian length of a representative of $\alpha$ . people say this norm is realised by a geodesic loop at p my question is why it is realised by a geodesic loop and not by a geodesic it must be something with the regularity here is a simple proof for why i am saying geodesic and not geodesic loop. fix $\tilde{p}$ in the universal cover of $(M,g)$ let c be a representitve of $\alpha$ , $\tilde{c}$ be the lift of c to $\tilde{M}$ with base point $\tilde{p}$ and let $c\prime$ be the unique minimizing geodsic joinging $\tilde{p}$ to $\alpha(\tilde{p})=\tilde{c}(\tilde{p})$ . obviously $\tilde{c}$ and $c\prime$ are homotopic and hence $\pi oc\prime$ and $\pi o\tilde{c}=c$ are homotopic where $\pi : \tilde{M}\rightarrow M$. and I know the the image of a geodesic by a local isometry is a geodesic hence $\pi oc\prime$ is a closed geodesic at p and sure it minimise the length otherwise we will have a contradiction cause it lift $c\prime$ is minimizing . - 1 Hi, welcome to this site. I'm sorry to ask, but: What exactly is your question? It would be great if you could insert a few more periods and make this one block of text into two or three paragraphs. – t.b. Aug 1 '11 at 10:05 my question is the length of alpha is realised by a closed geodesic or closed geodesic loop ? – Alfie Aug 1 '11 at 10:17 1 On the other hand, if instead of looking at homotopy classes of based maps from $S^1$, one just considers the homotopy classes of maps from $S^1$ into a compact manifold $M$, then such classes are represented by closed geodesics. – Jason DeVito Aug 1 '11 at 16:02 ## 2 Answers Are you sure you have the definition of a "closed geodesic" and "geodesic loop" correct? Usually we have Definition A closed geodesic $\gamma$ on $(M,g)$ is a smooth image of $\mathbb{S}^1$ that is geodesic. Definition A geodesic loop $\gamma$ on $(M,g)$ is a smooth image of $[0,1]$, that is geodesic, and such that $\gamma(0) = \gamma(1)$. See, for example, the Springer EOM. Then by the argument given in your question statement, you have that a minimizing object is automatically a geodesic loop. But it doesn't have to be a closed geodesic as there may be an angle. Example Imagine a T-shaped pipe formed by joining two cyclinders at right angles. (To make it closed you can glue the ends of the pipe to a big sphere.) Smooth out the junctions. Let your base point $p$ be the point on the horizontal part of the T that sits directly opposite the vertical leg of the T. - $\pi \circ c′$ is not smooth at its endpoints. That is, it is not smooth at $p$. Thus $\pi \circ c′$ is not a geodesic. Instead it is a geodesic loop. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 34, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9476423859596252, "perplexity_flag": "head"}
http://regularize.wordpress.com/2013/01/16/comparing-shapes-from-prokhorovwasserstein-over-hausdorff-to-gromov-hausdorff-metrics/	# regularize Trying to keep track of what I stumble upon January 16, 2013 ## Comparing shapes: From Prokhorov/Wasserstein over Hausdorff to Gromov-Hausdorff metrics Posted by Dirk under Imaging, Math \| Tags: functional analysis, gromov-hausdorff metric, hausdorff metric, image processing, metrics spaces, prokhorov metric, shapes, sparsity, wasserstein metric \| 1 Comment With this post I delve into a topic which is somehow new to me, although I planned to look deeper into this for quite some time already. I stumbled upon the paper Gromov-Wasserstein distances and the metric approach to object matching by Facundo Mémoli which was a pleasure to read and motivated this post. 1. Comparing measures with norms and metrics There are different notions in mathematics to compare two objects, think of the size of real numbers, the cardinality of sets or the length of the difference of two vectors. Here we will deal with not only comparison of objects but with “measures of similarity”. Two fundamental notions for this are norms in vector spaces and metrics. The norm is the stronger concept in that it uses more structure than a metric and also, every norm induces a metric but not the other way round. There are occasions in which both a norm and a metric are available but lead to different concepts of similarity. One of these instances occurs in sparse recovery, especially in the continuous formulation, e.g. as described in a previous post. Consider the unit interval ${I = [0,1]}$ and two Radon measures ${\mu_1}$ and ${\mu_2}$ on ${I}$ (${I}$ could also be an aritrary metric space). On the space of Radon measures ${\mathfrak{M}(I)}$ there is the variation norm $\displaystyle \\|\mu\\|_{\mathfrak{M}}= \sup_\Pi\sum_{A\in\Pi}\|\mu(A)\|$ where the supremum is taken over all partitions ${\Pi}$ of ${I}$ into a finite number of measurable sets. Moreover, there are different metrics one can put on the space of Radon measures, e.g. the Prokhorov metric which is defined for two probability measures (e.g. non-negative ones with unit total mass) $\displaystyle \begin{array}{rcl} d_P(\mu_1,\mu_2) & = & \inf\{\epsilon>0\ :\ \mu_1(A)\leq \mu_2(A^\epsilon) + \epsilon,\nonumber\\ & & \qquad \mu_2(A)\leq \mu_1(A^\epsilon) + \epsilon\ \text{for all measurable}\ A\} \end{array}$ where ${A^\epsilon}$ denotes the ${\epsilon}$-neighborhood of ${A}$. Another familiy of metrics are the Wasserstein metrics: For ${p\geq 1}$ define $\displaystyle d_{W,p}(\mu_1,\mu_2) = \Big(\inf_\nu\int_{I\times I} \|x-y\|^p d\nu(x,y)\Big)^{1/p} \ \ \ \ \ (1)$ where the infimum is taken over all measure couplings of ${\mu_1}$ and ${\mu_2}$, that is, all measures ${\nu}$ on ${I\times I}$ such that for measurable ${A}$ it holds that $\displaystyle \nu(A\times I) = \mu_1(A)\ \text{and}\ \nu(I\times A) = \mu_2(A).$ Example 1 We compare two Dirac measures ${\mu_1 = \delta_{x_1}}$ and ${\mu_2 = \delta_{x_2}}$ located at distinct points ${x_1\neq x_2}$ in ${I}$ as seen here: The variation norm measures their distance as $\displaystyle \\|\mu_1-\mu_2\\|_{\mathfrak{M}} = \sup_\Pi\sum_{A\in\Pi}\|\delta_{x_1}(A) - \delta_{x_2}(A)\| = 2$ (choose ${\Pi}$ such that it contains ${A_1}$ and ${A_2}$ small enough that ${x_1\in A_1}$, ${x_2\in A_2}$ but ${x_1\notin A_2}$ and ${x_2\notin A_1}$). The calculate the Prokhorov metric note that you only need to consider ${A}$‘s which contain only one of the points ${x_{1/2}}$ and hence, it evaluates to $\displaystyle d_P(\mu_1,\mu_2) = \|x_1-x_2\|.$ For the Wasserstein metric we observe that there is only one possible measure coupling of ${\delta_{x_1}}$ and ${\delta_{x_2}}$, namely the measure ${\nu = \delta_{(x_1,x_2)}}$. Hence, we have $\displaystyle d_{W,p}(\mu_1,\mu_2) = \Big(\int_{I\times I}\|x-y\|^pd\delta_{(x_1,x_2)}(x,y)\Big)^{1/p} = \|x_1-x_2\|.$ The variation norm distinguishes the two Diracs but is not able to grasp the distance of their supports. On the other hand, both metrics return the geometric distance of the supports in the underlying space ${I}$ as distance of the Diracs. Put in pictures: The variation norm of the difference measures the size ob this object while both metrics capture the distance of the measures like here It should not stay unnoted that convergence in both the Prokhorov metric and the Wasserstein metrics is exactly the weak convergence of probability measures. The above example provides a motivation to study metric structures on spaces, even if they are also equipped with a norm. Another reason to shift attention from normed spaces to metric spaces is the fact that there has emerged a body of work to build a theory of analysis in metric spaces (see, e.g. this answer on mathoverflow or the book Gradient Flows: In Metric Spaces And In The Space Of Probability Measures by Ambrosio, Gigli and Savaré (which puts special emphasis on the space of probability measures)). Yet another motivation for the study of metrics in this way is the problem of comparing shapes (without being precisely defined yet): Which of these shapes look most alike? (Note that shapes need not to be two dimensional figures, you may also think of more complex objects like surfaces in three dimensions or Riemannian manifolds.) One may also ask the question how two compare different images defined on different shapes, i.e. different “distributions of colour” on two different shapes. 2. Comparing shapes: Metric spaces Up to now we tried to compare different measures, defined on the same set. At least to me it seems that both the Prokhorov and the Wasserstein metrics are suited to measure the similarity of measures and in fact, they do so somehow finer than the usual norm does. Let’s try to go one step further and ask ourselves, how we could compare two measures ${\mu_1}$ and ${\mu_2}$ which are defined on two different sets? While thinking about an answer one need to balance several things: • The setup should be general enough to allow for the comparison of a wide range of objects. • It should include enough structure to allow meaningful statements. • It should lead to a measure which is easy enough to handle both analytically and computationally. For the first and second bullet: We are going to work with measures not on arbitrary sets but on metric spaces. This will allow to measure distances between points in the sets and, as you probably know, does not pose a severe restriction. Although metric spaces are much more specific than topological spaces, we still aim at quantitative measures which are not provided by topologies. With respect to the last bullet: Note that both the Prokhorov and the Wasserstein metric are defined as infimums over fairly large and not too well structured sets (for the Prokhorov metric and need to consider all measurable sets and their ${\epsilon}$-neighborhoods, for the Wasserstein metric, one need to consider all measure couplings). While they can be handled quite well theoretically, their computational realization can be cumbersome. In a similar spirit than Facundo Memoli’s paper we work our way up from comparing subsets of metric spaces up to comparing two different metric spaces with two measures defined on them. 2.1. Comparing compact subsets of a metric space: Hausdorff Let ${(X,d)}$ be a compact metric space. Almost hundred years ago Hausdorff introduced a metric on the family of all non-empty compact subsets of a metric space as follows: The Hausdorff metric of two compact subsets ${A}$ and ${B}$ of ${X}$ is defined as $\displaystyle d_H(A,B) = \inf\{\epsilon>0 \ :\ A\subset B_\epsilon,\ B \subset A_\epsilon\}$ (again, using the notion of ${\epsilon}$-neighborhood). This definition seems to be much in the spirit of the Prokhorov metric. Proposition 2.1 in Facundo Memolis paper shows that the Hausdorff metric has an equivalent description as $\displaystyle d_H(A,B) = \inf_R \sup_{(a,b) \in R} d(a,b)$ where the infimum is taken over all correspondences ${R}$ of ${A}$ and ${B}$, i.e., all subset ${R\subset A\times B}$ such that for all ${a\in A}$ there is ${b\in B}$ such that ${(a,b) \in R}$ and for all ${b\in B}$ there ${a\in A}$ such that ${(a,b)\in R}$. One may also say set coupling of ${A}$ and ${B}$ instead of correspondence. Example 2 There is always the full coupling ${R = A\times B}$. Three different set couplings of two subsets ${A}$ and ${B}$ of the unit interval are shown here: the “full one” ${A\times B}$ in green and two “slim” ones in red and orange. Other “slim” couplings can be obtained from surjective mappings ${f:A\rightarrow B}$ by ${R = \{(a,f(a))\ :\ a\in A\}}$ (or with the roles of ${A}$ and ${B}$ swapped): If you couple a set ${A}$ with itself, there is also the trivial coupling $\displaystyle R = \{(a,a)\ : \ a\in A\}$ which is just the diagonal of ${A\times A}$ Note that the alternative definition of the Hausdorff metric is more in the spirit of the Wasserstein metric: It does not use enlarged objects (by ${\epsilon}$-neighborhoods) but couplings. The Hausdorff metric is indeed a metric on the set ${\mathfrak{C}(X)}$ of all non-empty compact subsets of a metric space ${X}$ and if ${X}$ itself is compact it even holds that ${(\mathfrak{C}(X),d_H)}$ is a compact metric space (a result, known as Blaschke Selection Theorem). One may say that we went up an abstraction ladder one step by moving from ${(X,d)}$ to ${(\mathfrak{C}(X),d_H)}$. 2.2. Comparing compact metric spaces: Gromov-Hausdorff In the previous subsection we worked within one metric space ${X}$. In the book “Metric Structures for Riemannian and Non-Riemannian Spaces” Misha Gromov introduced a notion to compare two different metric spaces. For compact metric space ${X}$ and ${Y}$ the Gromov-Hausdorff metric is defined as $\displaystyle d_{GH}(X,Y) = \inf_{Z,f,g} d_H(f(X),g(Y)) \ \ \ \ \ (2)$ where the infimum is taken over • all metric spaces ${Z}$ and • all isometric embeddings ${f}$ and ${g}$ which embed ${X}$ and ${Y}$ into ${Z}$ respectively. In words: To compute the Gromov-Hausdorff metric, you try embed both ${X}$ and ${Y}$ into a common larger space isometrically such that they are as close as possible according to the Hausdorff metric in that space. Strictly speaking, the above definition is not well stated as one can not form an infimum over all metric spaces since this collection does not form a set according to the rules of set theory. More precisely one should write that ${d_{GH}(X,Y)}$ is the infimum over all ${r>0}$ such that there exists a metric space ${Z}$ and isometric embeddings ${f}$ and ${g}$ of ${X}$ and ${Y}$, respectively, such that ${d_H(f(X),g(Y))<r}$. As the Hausdorff metric could be reformulated with set couplings there is a reformulation of the Gromov-Hausdorff metric based on metric couplings: A metric coupling of two metric spaces ${(X,d_X)}$ and ${(Y,d_Y)}$ is a metric ${d}$ on the disjoint union ${X\sqcup Y}$ of ${X}$ and ${Y}$ such that for all ${x,x'\in X}$ and ${y,y'\in Y}$ it holds that ${d(x,x') = d_X(x,x')}$ and ${d(y,y') = d_Y(y,y')}$. Example 3 We couple a metric space ${(X,d)}$ with itself. We denote with ${(X',d')}$ an identical copy of ${(X,d)}$ and look for a metric ${D}$ on ${X\times X'}$ that respects the metrics ${d}$ and ${d'}$ in the way a metric coupling has to. To distinguish elements from ${X}$ and ${X'}$ we put a ${'}$ on all quantities from ${X'}$. Moreover, for ${x\in X}$ we denote by ${x'}$ its identical copy in ${X'}$ (and similarly for ${x'\in X'}$, ${x}$ is its identical twin). Then, for any ${\epsilon>0}$ we can define ${D_\epsilon(x,x') = D_\epsilon(x',x) = \epsilon}$ (i.e. the distance between any two identical twins is ${\epsilon}$. By the triangle inequality we get for ${x\in X}$ and ${y'\in X'}$ that ${D_\epsilon(x,y')}$ should fulfill $\displaystyle D_\epsilon(x',y') - D_\epsilon(x',x) \leq D_\epsilon(x,y') \leq D_\epsilon(x,y) + D_\epsilon(y,y')$ and hence $\displaystyle d(x,y) - \epsilon \leq D_\epsilon(x,y') \leq d(x,y) + \epsilon.$ Indeed we can choose ${D_\epsilon(x,y') = d(x,y)}$ if ${x\in X}$ and ${y'\in Y}$ leading to one specific metric coupling for any ${\epsilon}$. This couplings allow to distinguish identical twins and behave as a metric on the whole disjoint union. In the limiting case ${\epsilon\rightarrow 0}$ we do not obtain a metric but a semi-metric or pseudo-metric which is just the same as a metric but without the assumption that ${d(x,y) = 0}$ implies that ${x=y}$. Example 4 The above example of a metric coupling of a metric space with itself was somehow “reproducing” the given metric as accurate as possible. There are also other couplings that put very different distances to points ${D(x,y')}$ and there is also a way to visualize metric couplings: When building the disjoint union of two metric spaces ${X}$ and ${Y}$, you can imagine this as isometrically embedding both in a larger metric space ${Z}$ in a non-overlapping way and obtain the metric coupling ${D}$ as the restriction of the metric on ${Z}$ to ${X\sqcup Y}$. For ${X=Y=[0,1]}$ you can embed both into ${Z = {\mathbb R}^2}$. A metric coupling which is similar (but not equal) to the coupling of the previous example is obtained by putting ${X}$ and ${Y}$ side by side at distance ${\epsilon}$ as here (one space in green, the other in blue). A quite different coupling is obtained by putting ${X}$ and ${Y}$ side by side, but in a reversed way as here: You may even embed them in a more weired way as here: but remember that the embeddings has to be isometric, hence, distortions like here are not allowed. This example illustrate that the idea of metric coupling is in similar spirit as of “embedding two spaces in a common larger one”. With the notion of metric coupling, the Gromov-Hausdorff metric can be written as $\displaystyle d_{GH}(X,Y) = \inf_{R,d} \sup_{(x,y)\in R} d(x,y) \ \ \ \ \ (3)$ where the infimum is taken over all set couplings ${R}$ of ${X}$ and ${Y}$ and all metric couplings ${d}$ of ${(X,d_X)}$ and ${(Y,d_Y)}$. In words: To compute the Gromov-Hausdorff metric this way, you look for a set coupling of the base sets ${X}$ and ${Y}$ and a metric coupling ${d}$ of the metrics ${d_X}$ and ${d_Y}$ such that the maximal distance of two coupled points ${x}$ and ${y}$ is as small as possible. While this may look more complicated than the original definition from~(2), note that the original definition uses all metric spaces ${Z}$ in which you can embed ${X}$ and ${Y}$ isometrically, which seems barely impossible to realize. Granted, the new definition also considers a lot of quantities. Also note that this definition is in spirit of the Wasserstein metric from~(1): If there were natural measures ${\mu_R}$ on the set couplings ${R}$ we could write \begin{equation} d_{GH}(X,Y) = \inf_{R,d} \Big(\int d(x,y)^pd\mu_R\Big)^{1/p} \end{equation} and in the limit ${p\rightarrow\infty}$ we would recover definition~(3). Example 5 The Gromov-Hausdorff distance of a metric space ${(X,d_X)}$ to itself is easily seen to be zero: Consider the trivial coupling ${R = \{(x,x)\ :\ x\in X\}}$ from Example~2 and the family ${D_\epsilon}$ of metric couplings from Example~3. Then we have ${d_{GH}(X,X) \leq \epsilon}$ for any ${\epsilon >0}$ showing ${d_{GH}(X,X) = 0}$. Let’s take one of the next-complicated examples and compute the distance of ${X = [0,1]}$ and ${Y=[0,2]}$, both equipped with the euclidean metric. We couple the sets ${X}$ and ${Y}$ by ${R = \{(x,2x)\ : \ x\in X\}}$ and the respective metrics by embedding ${X}$ and ${Y}$ into ${{\mathbb R}^2}$ as follows: Put ${Y}$ at the line from ${(0,0)}$ to ${(2,0)}$ and ${X}$ at the line from ${(\tfrac12,\epsilon)}$ to ${(1\tfrac12,\epsilon)}$: This shows that ${d_{GH}(X,Y) \leq \tfrac12}$ and actually, we have equality here. There is another reformulation of the Gromov-Hausdorff metric, the equivalence of which is shown in Theorem 7.3.25 in the book “A Course in Metric Geometry” by Dmitri Burago, Yuri Burago and Sergei Ivanov: $\displaystyle d_{GH}(X,Y) = \tfrac12\inf_R \sup_{\overset{\overset{x_{1/2}\in X}{y_{1/2}\in Y}}{(x_i,y_i)\in R}}\big\| d_X(x_1,x_2) - d_Y(y_1,y_2)\big\| \ \ \ \ \ (4)$ where the infimum is taken over all set couplings ${R}$ of ${X}$ and ${Y}$. In words: Look for a set coupling such that any two coupled pairs ${(x_1,y_1)}$ and ${(x_2,y_2)}$ have the “most equal” distance. This reformulation may have the advantage over the form (3) in that is only considers the set couplings and the given metrics ${d_X}$ and ${d_Y}$ and no metric coupling is needed. Note that, as the previous reformulation~(3), it is also in the spirit of the Wasserstein metric: If there were natural measures ${\mu_R}$ in the set couplings ${R}$, we could write $\displaystyle d_{GH}(X,Y) = \tfrac12\inf_R \Big(\int_{R\times R}\big\| d_X(x_1,x_2) - d_Y(y_1,y_2)\big\|^p d\mu_R(x_1,y_1)d\mu_R(x_2,y_2)\Big)^{1/p}.$ and recover the formulation~(4) in the limit ${p\rightarrow\infty}$. One may say that we went up an abstraction ladder one step further by moving from ${(X,d)}$ to ${(\mathfrak{C}(X),d_H)}$ to ${(\text{All compact metric spaces},d_{GH})}$. Since this post has been grown pretty long already, I decided to do the next step (which is the already announced metric on metric spaces which additionally carry some measure on them – so-called metric measure spaces) in a later post. ### One Response to “Comparing shapes: From Prokhorov/Wasserstein over Hausdorff to Gromov-Hausdorff metrics” 1. March 19, 2013 at 11:52 am [...] I continue my previous post on methods to compare shapes. In that post we started with different metrics between two probability measures and defined on [...] Cancel	{"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": 212, "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": 0.9366541504859924, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/122384/venn-diagram-3-set?answertab=active	# Venn diagram 3 set I understand Venn diagram equations of 2 sets. It has been perhaps 3 years since I have done Venn diagram so I have gotten really rusty at this. I don't understand in the equation for 3 sets why in the end they add the intersection of 3 sets... I have such a hard time visualising these diagrams and knowing what to subtract from intersection etc... I know this is a super easy question but it is one of those questions you either get or you don't and right now I cant figure out why they add the intersection at the end.. shouldn't it be subtract the intersection? This is how I looked at the problem to solve it. You look at the commonality between A and B at first so you look at the part where A = B or A and B...You subtract that because of overlay. Same concept as 2 set Venn diagram. Then you look at commonality between B and C or simply when B = C, subtract overlay. C and A or C=A subtract overlay...Now you have a hole left in the center which you can fill by adding A and B and C or simply the commonality between A and B and C or A = B = C I'm lost once more... this time its 4 sets or 5 or 6 or 7 or 8.... I don't understand the general formula given below. - ## 3 Answers Let us assume one wants to evade Venn diagrams, then indicator functions are an alternative. Recall that for every set $A$ and every element $x$, $\mathbf 1_A(x)$ is $1$ if $x\in A$ and $0$ if $x\notin A$. Then, for every subset $A$ of a given set $E$, this defines a function $\mathbf 1_A:E\to\mathbb R$, called the indicator function of $A$ (in $E$), such that $$\|A\|=\sum\limits_{x\in E}\mathrm 1_A(x).\tag{1}$$ This formulation might seem to only complicate things. Let us show the opposite is true, for the reason that now one deals with sums on a fixed set $E$. It happens that indicator functions are especially well suited to intersections. To wit, for every collection of sets $A_k$ with intersection $C$, $$\mathbf 1_C=\prod\limits_k\mathbf 1_{A_k}.$$ Let us now put indicator functions and intersections of sets together. First consider two sets $A$ and $B$. Then one might remember that $\mathbf 1_{A\cup B}=\mathbf 1_A+\mathbf 1_B-\mathbf 1_{A\cap B}$ and proceed. So far so good, except this is not a new formula at all. To see this, consider that unions of complementary sets are complementary sets of intersections so let us find the intersection hidden here. One gets $$E\setminus(A\cup B)=(E\setminus A)\cap(E\setminus B),$$ hence $$\mathbf 1_{E\setminus(A\cup B)}=\mathbf 1_{E\setminus A}\cdot\mathbf 1_{E\setminus B}.\tag{2}$$ One last tool before we proceed. To enumerate a complementary set is easy, since $$\mathbf 1_{E\setminus A}=1-\mathbf 1_A.\tag{3}$$ Note that, using (1), (3) yields $$\|E\setminus A\|=\sum\limits_{x\in E}(1-\mathrm 1_A(x))=\sum\limits_{x\in E}1-\sum\limits_{x\in E}\mathrm 1_A(x)=\|E\|-\|A\|.$$ Coming back to the union, (2) and (3) yield $$1-\mathbf 1_{A\cup B}=(1-\mathbf 1_{A})\cdot(1-\mathbf 1_{B})=1-\mathbf 1_A-\mathbf 1_B+\mathbf 1_{A\cap B},$$ which can be rewritten as $$\mathbf 1_{A\cup B}=\mathbf 1_A+\mathbf 1_B-\mathbf 1_{A\cap B}.$$ For two sets, this was complicating well known arguments for nothing but now, let us consider $n$ subsets $A_k$ of $E$ and let us try to imitate the arguments above. Calling $A$ the union of the sets $A_k$, $B_k=E\setminus A_k$ and $B$ the intersection of the sets $B_k$, one gets $$E\setminus A=B,\quad \mathbf 1_B=\prod\limits_{k=1}^n\mathbf 1_{B_k}, \quad \mathbf 1_B=1-\mathbf 1_A, \quad \mathbf 1_{B_k}=1-\mathbf 1_{A_k}.$$ To summarize, the key formula here is $$1-\mathbf 1_A=\prod\limits_{k=1}^n(1-\mathbf 1_{A_k}).\tag{4}$$ The rest is algebra. Expanding the RHS of (4) yields a sum over every sequence of choices of $1$ or $(-\mathbf 1_{A_k})$, from $k=1$ to $n$. One can choose $(-\mathbf 1_{A_k})$ for every $k$ in $K$ and $1$ for every $k$ not in $K$, and the product is the sum of the results over every possible subset $K$ of $\{1,2,\ldots,n\}$, that is, $$\prod\limits_{k=1}^n(1-\mathbf 1_{A_k})=\sum\limits_K\prod\limits_{k\in K}(-\mathbf 1_{A_k})=\sum\limits_K(-1)^{\|K\|}\mathbf 1_{A_K}, \qquad A_K=\bigcap\limits_{k\in K}A_k.$$ The $K=\varnothing$ term is $1$, which can be cancelled with $1$ on the LHS of (4), hence, changing the signs, one is left with $$\mathbf 1_A=\sum\limits_{K\ne\varnothing}(-1)^{\|K\|+1}\mathbf 1_{A_K}, \quad\text{with}\quad A=\bigcup\limits_{k=1}^nA_k, \quad A_K=\bigcap\limits_{k\in K}A_k.$$ This yields finally the size of $A$ by summation over $x$ in $E$, namely, $$\|A\|=\sum\limits_{K\ne\varnothing}(-1)^{\|K\|+1}\|A_K\|.$$ Note that these identities are valid for every number $n$ of sets $A_k$. Example When $n=3$, the sum over the sets $K$ such that $\|K\|=1$ is $\|A_1\|+\|A_2\|+\|A_3\|$, the sum over the sets $K$ such that $\|K\|=2$ is $-1$ times $\|A_1\cap A_2\|+\|A_2\cap A_3\|+\|A_3\cap A_1\|$, and the sum over the sets $K$ such that $\|K\|=3$ is $\|A_1\cap A_2\cap A_3\|$, which is the formula in your post. - The numbers in the diagram indicate how many times the given formula counts that particular portion of the Venn Diagram. In the first picture, for example, an element in the intersection of A, B, and C will be counted 3 times. Each term of the inclusion-exclusion formula gradually refines these numbers until each portion of the Venn Diagram is counted exactly once. - thank you that helps a lot in understand now!!! +1 – Raynos Mar 21 '12 at 16:43 Take a point in the intersection and see how many times it gets counted on the left side, and how many times on the right - then you'll see why it has to be added. - thank you gary. I think I understand where the intersection at the end is coming from. I can do each set and its intersection with the other set separately and then I have a hold left in the center and to cover for that hole I can do intersection of the 3 sets....so its kind of like saying A or B or C is equal to the commonality between A and B, B and C C and A, then you also want to add the commonality that all 3 are the same so you do it like the above equation for that reason – Raynos Mar 20 '12 at 5:26 I edited my post. The new diagram represents what I just said in comments. I hope this helps other people confused, there are ways to solve complex problems in many ways, you can solve them in complex methods or with super simple and easy methods. How you look at the problem is what determines whether you will find a simple easy solution or a complex one. This I have now learned. It is much easier I find to visualize the hole in center as oppose to overlay of the 3 sets intersection. – Raynos Mar 20 '12 at 5:31 darn my "trick" fails if you go for 4 venn diagrams... – Raynos Mar 20 '12 at 5:42 But my explanation succeeds. You want to count each element in the union exactly once. If an element is in, say, $A_1$ only, it just gets counted once, as an element of $A_1$. If it's in just $A_1$ and $A_2$, it gets counted twice, but then subtracted as an element of the intersection, so the net effect is to count it once. If it's in just $A_1,A_2,A_3$, it gets counted 3 times, then subtracted three times, then added back in; the net effect is to count it once. If it's in $A_i$, $i=1,2,3,4$, it gets counted 4 times, subtracted 6 times, added back 4 times, subtracted once; net, counted once. – Gerry Myerson Mar 20 '12 at 6:41	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 64, "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": 0.18, "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": 0.9491885900497437, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/145142-finding-eigenvalues.html	# Thread: 1. ## Finding eigenvalues Let $T: P_2 \mapsto P_2$ be defined by T(p)=xp'. Find eigenvalues of T. I'm not sure what the notation of T(p)=xp' means and how to go about finding the eigenvalues. Thanks. 2. Originally Posted by wattkow Let $T: P_2 \mapsto P_2$ be defined by T(p)=xp'. Find eigenvalues of T. I'm not sure what the notation of T(p)=xp' means and how to go about finding the eigenvalues. Thanks. That is the linear transformation which is x times the derivative of the polynomial. 3. Originally Posted by dwsmith That is the linear transformation which is x times the derivative of the polynomial. Ok, so p is a polynomial? How would I go about finding eigenvalues for said polynomial? 4. Originally Posted by wattkow Ok, so p is a polynomial? How would I go about finding eigenvalues for said polynomial? I am assuming P is a polynomial since we are mapping $P_2$ on $P_2$. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate $P_2$ differently. For instance, mine says $P_2$ are polynomial of degree 1 or less. Thus, the 2 in $P_2$ is the amount of elements {1, x} but some books refer to $P_2$ as degree two or less, { $1$, $x$, $x^2$}. What does your book say? 5. Originally Posted by dwsmith I am assuming P is a polynomial since we are mapping $P_2$ on $P_2$. Unless your book notates this differently, I am almost positive these are polynomials; however, different books notate $P_2$ differently. For instance, mine says $P_2$ are polynomial of degree 1 or less. Thus, the 2 in $P_2$ is the amount of elements {1, x} but some books refer to $P_2$ as degree two or less, { $1$, $x$, $x^2$}. What does your book say? I believe in my case it's the 2nd degree polynomial. 6. Originally Posted by wattkow I believe in my case it's the 2nd degree polynomial. Ok so your linear transformation goes from $ax^2+bx+c\rightarrow xp'(x)=x(2ax+b)=2ax^2+bx$. Now you need to find the eigenvalues of the linear transformation. 7. I would do the following: A basis for $P_{2}$ is $\{1,x,x^{2} \}$ $T(1) = 0$ $T(x) = x$ and $T(x^{2}) = 2x^{2}$ so the transformation matrix is $\begin{bmatrix}<br /> 0 & 0 & 0 \\<br /> 0 & 1 & 0\\<br /> 0 & 0 & 2<br /> \end{bmatrix}$ which has eigenvalues 0,1, and 2 8. Originally Posted by wattkow I believe in my case it's the 2nd degree polynomial. The "set of all 2nd degree polynomials" does not form a vector space. As dwsmith said, $P_2$ is the set of all polynomials of degree 2 or less. Random Variable showed how to represent this transformation as a matrix- worth knowing for itself. But for this problem, you don't have to do that. As dwsmith said, if $p= ax^2+ bx+ c$, then $T(p)= xp'= x(2ax+ b)= 2ax^2+ bx$. If $\lambda$ is an eigenvalue for T, we must have some a, b, c, not all 0, such that $T(p)= 2ax^2+ bx= \lambda(ax^2+ bx+ c)$. Since that must be true for all x, we can set corresponding coefficients equal: $2a= \lambda a$, $b= \lambda b$, and $0= \lambda c$. If a is not 0, then $\lambda= 2$, if b is not 0, then $\lambda= 1$, and if c is not 0, then $\lambda= 0$. That is, 0 is an eigenvalue with eigenvectors any multiple of 1, 1 is an eigenvalue with eigenvectors any multiple of x, and 2 is an eigenvalue with eigenvectors any multiple of $x^2$, exactly what Random Variable got. 9. Thanks fellas.	{"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": 39, "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": 0.9436976313591003, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/83042/how-to-calculate-a-specific-area-inside-a-circle?answertab=active	# How to calculate a specific area inside a circle? I want to calculate the area displayed in yellow in the following picture: The red square has an area of `1`. For any given square, I'm looking for the simplest formula to compute the yellow area in this picture. (Squares are actually pixels in an image processing software, and I need to compute the opacity of the given pixel based on the relative area of the circle compared to the area of the pixel, to create smooth rounded corners.) Update: I need to be able to do that for any individual square in the image: also the orange and the green one, for example. - I think what you really want is just an algorithm for drawing an anti-aliased circle. I don't think Wu's algorithm actually computes exact areas, and I also don't think you need to. Remember, a pixel is not a little square. – Rahul Narain Nov 18 '11 at 5:33 ## 2 Answers Well, I guess I'll take a stab at this. This is definitely a calculus problem. To take the area between two curves, you want to take the integral of the greater function minus the lesser function. For the yellow area, the greater function is $y=1$. The lesser function will take some manipulation. The formula for the circle is: $(x-4)^2+(y-4)^2=16$ $(y-4)^2=16-(x-4)^2$ $y-4=-\sqrt{16-(x-4)^2}$ $y=4-\sqrt{16-(x-4)^2}$ So our integral is $\int^3_2[1-(4-\sqrt{16-(x-4)^2}]dx$=$\int^3_2-3dx+\int^3_2\sqrt{16-(x-4)^2}dx$. The first integral is $-3x$ evaluated from 2 to 3, or in other words, $-3(3)-[-3(2)]=-9+6=-3$. For the second half of that integral, we'll use the info from Andreas's comment. We'll perform a change of variable $u=x-4,du=dx$ $\int^3_2\sqrt{16-(x-4)^2}dx=\int^{-1}_{-2}\sqrt{16-u^2}du=\frac12[u\sqrt{16-u^2}+16sin^{-1}\frac u4]^{-1}_{-2}=\frac12[(x-4)\sqrt{16-(x-4)^2}+16sin^{-1}\frac{x-4}4]^3_2$ That solves the yellow area. For the other 2, you'll want to know where the 2 functions cross. $(x-4)^2+(1-4)^2=16$ $(x-4)^2=7$ $x=4-\sqrt7$ For the green area, the 2 functions are the same, but it's evaluated from $4-\sqrt7$ to 2. For the orange area, the greater function is $y=2$, but the lesser function changes. It should be easy to see, though, that the right half is a rectangle. The left half is integrated from 1 to $4-\sqrt7$. Also, the first half of the integral has changed from $\int-3dx$ to $\int-2dx=-2x$. So the total orange area is $2x-\frac12[(x-4)\sqrt{16-(x-4)^2}+16sin^{-1}\frac{x-4}4]$ evaluated from 1 to $4-\sqrt7$ plus $1[2-(4-\sqrt7)]$, or $\sqrt7-2$. - ## Did you find this question interesting? Try our newsletter take $\int_2^3\sqrt{4^2-x^2}dx$ (upper part of the circle) and subract 3, for the boxes below. Is it possible to break that down to simple functions like `cos()` & `sin()` ? – Benjamin Nov 17 '11 at 13:43 by integrating from 1 to 2 you'll get $(green+orange)$ if you subtract 2. now you calculate yellow' from 2 to 4. next you rotate your cirlce by $90^\circ$ clockwise. $green + yellow' =\int_7^8 ...dx$. now you get $green = (green+yellow')-yellow'$ and then orange in the same manner. continue like this to the end. – draks ... Nov 17 '11 at 14:41 It's a bit strange looking at it with the y axis going down instead of up. I think you have the formula slightly off though. The equation of the circle is $(x-4)^2+(y-4)^2=16$. You could probably perform a change of variable to recenter the circle, but the limits of integration would be different, wouldn't they? – Mike Nov 17 '11 at 14:54	{"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": 29, "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": 0.9373422265052795, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/20156/is-there-a-point-of-no-hearing-if-two-speakers-are-emitting-the-same-sounds-wa	# Is there a point of no hearing if two speakers are emitting the same sounds - Wave Superposition I'm talking bout this phenomenon (The third one): I would like to know if with sound waves, happen the same as with (water) waves. So, If two speakers were emitting the same sound, would there technically be a point where you wouldn't hear. (I guess this would only happen if you heard with only one ear). Is this true for sound wave too? Does this happen in real life, say, in car? Thanks in advance! - ## 2 Answers Since you've gotten one 'yes' answer, let me give you a 'no' answer. :-) More precisely, there are two different questions here: Brian's 'yes' answer handles the question 'for any sound wave, is there some other wave that will cancel that wave out?' The question you asked, on the other hand, was closer to 'for any sound wave, is there some shift of that wave that will cancel the original out' - in other words, does every wave have a set of peaks and troughs that can be shifted in time to cancel it out? In general, the answer is no: there are wave functions for which no shift can possibly cancel out the function. What does it mean to be self-cancellable? It means there's some shift of your wavefunction $f(t)$ - say, by $t_0$ - such that the sum of the two equals zero: $f(t)+f(t-t_0) = 0$. But then, this means that $f(t) = -f(t-t_0)$, and $f(t-t_0) = -f(t-t0-t0) = -f(t-2t_0)$ - so $f(t) = -f(t-t_0) = -(-f(t-2t_0)) = f(t-2t_0)$ - in other words, $f(t)$ is periodic with period $2t_0$. So since every self-cancellable wavefunction is periodic, no non-periodic function can be self-cancellable; as a simple example, any glissando (rising tone) won't be self-cancellable. On the other hand, being periodic isn't sufficient to be self-cancellable, either. The simplest example is probably the Sawtooth wave: $f(t) = t - \lfloor t\rfloor$. Note that this function has derivative (i.e. slope) 1 everywhere its derivative is defined, and so every shift of it (by, say, $t_0$) also has derivative 1 everywhere its derivative is defined. But then $f(t) + f(t-t_0)$ has derivative 2 everywhere it's defined, whereas to cancel out they would need to add to the zero function - whose derivative is also zero. - Very interesting answer, but I'm having some trouble following (No expert). What I did not get is why some functions can't be self cancelable (couldn't quite follow the equations). Shouldn't each wave have it's "anti-wave" even if it's a rising tone, by any chance do you have any visual representation of this phenomenon? (helps too see sometimes). Thank you very much for your answer! – Trufa Feb 3 '11 at 0:56 Unfortunately, I'm lousy at drawing pictures. :-) You're right that each wave has an 'anti-wave' - some other wave that you could add to it to come up with cancellation; that, just like Brian said, is how noise-cancelling headphones worked. But what you asked was 'for two speakers playing the same sound, is there a point where you wouldn't hear?' If two speakers are playing the same sound, then they're going to send the exact same wavefunction out; the only thing that changing your position can do then is to shift the time difference between them. – Steven Stadnicki Feb 3 '11 at 1:00 1 For instance, if you're 10m from one speaker and 20m from another, then you would have a delay of about .03 seconds between the sound from one speaker and the sound from the other. That's why I talk about self-cancellation specifically. Imagine taking the graph of the sawtooth and sliding it left and right along itself to find some way of 'cancelling' the notches; you can see that this is impossible, whereas doing it for (for instance) a square wave or a sine wave is easy. – Steven Stadnicki Feb 3 '11 at 1:03 @Steven (sorry for the delay I got no notification) Ok I think I "got it" now with the moving around of the graph, (I'm and image kind of graph). So then in a car (emitting the same sounds) is a definite no go. If the waves are no inverse, there is no way they could cancel each other, this makes sense! is this true? But the this leads me to this question, what does an "inverse" sound, sound like, I mean I get it is obviously not playing the tape backwards :) what would you actually have to do to, let's say a song to cancel the original song? – Trufa Feb 3 '11 at 1:27 Trufa: it turns out that it sounds almost exactly like the original sound! The short reason for this is that your ears/brain can't distinguish the 'shape' of a waveform, only how much energy is in different frequency bands (roughly, think of it as how hard the individual notes are being hit in a piano chord), and 'flipping' the wave upside-down doesn't change its energy profile. – Steven Stadnicki Feb 3 '11 at 2:00 show 1 more comment ## Did you find this question interesting? Try our newsletter email address Yes. Take a look at this: Noise Cancelling Headphones - – Brian Feb 3 '11 at 0:14 Thank you and interesting references! – Trufa Feb 3 '11 at 0:28 2 But does this really answer the question? The headphones answer the question, given a noise source, is there another noise that if we emit it at known distance, then it will cancell the given noise. That answer is clearly yes, essentially by the third diagram in the question. But the original question is: given two sources emitting the same noise, is there a position where they are exactly cancelled? – JDH Feb 3 '11 at 0:36 – Brian Feb 3 '11 at 0:38 – Brian Feb 3 '11 at 0:42 show 2 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "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": 0.959145724773407, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/32292/what-are-some-of-the-big-open-problems-in-3-manifold-theory/36594	## What are some of the big open problems in 3-manifold theory? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) From what I understand, the geometrization theorem and its proof helped to settle a lot of outstanding questions about the geometry and topology of 3-manifolds, but there still seems to be quite a lot of activity. I am not prepared to make a full push to familiarize myself with the literature in the near future, but I am still curious to know what people are working on, what techniques are being developed, etc. So I was hoping people could briefly explain some of the main open questions and programs that are motivating current research on 3-manifolds. I'll let the community decide if this undertaking is too broad, but I'm hoping it is possible to give a rough impression of what is going on. References to survey articles are appreciated, especially if they are accessible to non-experts like myself. It seems like the community wiki designation is appropriate for this question, and the usual rules ought apply. - 6 See my talk for some discussion of classification of geometric structures. math.auckland.ac.nz/wiki/… – Agol Jul 17 2010 at 17:21 3 It would be beneficial to the mathematics community if someone knowledgeable updates the relevant Wikipedia entries, which do not seem to be as detailed as the union of the answers to this (good!) question. – Joseph O'Rourke Aug 24 2010 at 22:50 I added tag open-problems-list, it implies poping-up this nice question -- hope this serves the community. – Alexander Chervov Dec 8 at 13:30 ## 12 Answers Let $M$ be a finite-volume hyperbolic 3-manifold. (Some of these extend, suitably restated, to larger classes of 3-manifolds. But it follows from Geometrisation that the hyperbolic case is often the most interesting. These are all trivial or trivially false in the elliptic case, for example.) The Surface Subgroup Conjecture (SSC). $\pi_1M$ contains a subgroup isomorphic to the fundamental group of a closed surface. (Recently proved by Kahn and Markovic.) The Virtually Haken Conjecture (VHC). $M$ has a finite-sheeted covering space with an embedded incompressible subsurface. Virtually positive first Betti number (VPFB). $M$ has a finite-sheeted covering space $\widehat{M}$ with $b_1(\widehat{M})\geq 1$. Virtually infinite first Betti number (VIFB). $M$ has finite-sheeted covering spaces $\widehat{M}_k$ with $b_1(\widehat{M}_k)$ arbitrarily large. Largeness (L). $\pi_1(M)$ has a finite-index subgroup that surjects a non-abelian free group. The Virtually Fibred Conjecture (VFC). $M$ has a finite sheeted cover that is homeomorphic to the mapping torus of a (necessarily pseudo-Anosov) surface automorphism. This is false for graph manifolds. There are fairly easy implications $L\Rightarrow VIFB \Rightarrow VPFB \Rightarrow VHC \Rightarrow SSC$. Also, a fortiori, $VFC\Rightarrow VPFB$. Recently, Daniel Wise announced a proof that $VHC\Rightarrow VFC$. His proof also shows that, if $M$ has an embedded geometrically finite subsurface, then we get $L$ and other nice properties. This list is similar to the one that Agol links to in the comments. Also, I suppose it's exactly what Daniel Moskovich meant by 'The Virtually Fibred Conjecture, and related problems'. I thought some people might be interested in a little more detail. Paul Siegel asks in comments: 'Would it be correct to guess that the "virtually _ conjecture" problems can be translated into a question about the large scale geometry of the fundamental group?' Certainly, it's true that most of these can be translated into an assertion about how (some finite-index subgroup of) $\pi_1M$ splits as an amalgamated product, HNN extension or, more generally, as a graph of groups. The equivalence uses the Seifert--van Kampen Theorem in one direction, and something like Proposition 2.3.1 of Culler--Shalen in the other. Rephrased like this, some of the above conjectures turn out as follows. The Virtually Haken Conjecture (VHC). $M$ has a finite-sheeted covering space $\widehat{M}$ such that $\pi_1(\widehat{M})$ splits. Virtually positive first Betti number (VPFB). $M$ has a finite-sheeted covering space $\widehat{M}$ such that $\pi_1(\widehat{M})$ splits as an HNN extension. Largeness (L). $M$ has a finite-sheeted covering space $\widehat{M}$ such that $\pi_1(\widehat{M})$ splits as a graph of groups with underlying graph of negative Euler characteristic. The Virtually Fibred Conjecture (VFC). $M$ has a finite-sheeted covering space $\widehat{M}$ such that $\pi_1(\widehat{M})$ splits can be written as a semi-direct product $\pi_1(\widehat{M}) \cong K\rtimes\mathbb{Z}$ with $K$ finitely generated. (Here we invoke Stallings' theorem that a 3-manifold whose fundamental group has finitely generated commutator subgroup is fibred.) I don't think I know a way to rephrase $VIFB$ in terms of splittings of $\pi_1$. Often, when people say 'the large scale geometry of $\pi_1$' they're talking about properties that are invariant under quasi-isometry. I'm really not sure whether these splitting properties (or, more exactly, 'virtually having these splitting properties') are invariant under quasi-isometry. Perhaps something like the work of Mosher--Sageev--Whyte does the trick? - At the very least I was interested in a little more detail - thanks for providing it! Would it be correct to guess that the "virtually _ conjecture" problems can be translated into a question about the large scale geometry of the fundamental group? – Paul Siegel Aug 24 2010 at 16:54 1 The question whether "there splitting properties are quasi-isometry invariant" is answered by a theorem of Richard Schwartz: quasi-isometry is equivalent to commensurability for non-uniform lattices in the isometry group of the hyperbolic 3-space. And all uniform lattices are qi to the hyperbolic space. So there seems to be no qi versions of the above conjectures. – Igor Belegradek Aug 25 2010 at 15:32 Igor, you're right to point that out. But I was wondering more about the other side of the question, namely whether `virtually acting on a tree' is invariant under quasi-isometry. I would guess that it isn't, but that it might be under some fairly mild hypotheses like being $PD_3$. – HW Aug 27 2010 at 20:49 Sorry, I should have said $PD_n$. As you point out below, $PD_3$ could be the same as being a 3-manifold group! – HW Sep 1 2010 at 16:12 ### 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. One of the unresolved questions about 3-manifolds is the generalized Smale conjecture, which roughly interpreted asks for the homotopy type of the space of diffeomorphisms of a 3-manifold. Smale originally conjectured that $Diff(S^3)\simeq O(4)$, and this was proven by Hatcher. He also worked out the homotopy type of diffeomorphisms of Haken 3-manifolds. Another interpretation of Smale's question is that the space of round (constant sectional curvature $=1$) metrics on $S^3$ is contractible. Gabai proved the analogous statement that the space of hyperbolic metrics on a hyperbolic 3-manifold is contractible, and recently McCullough and Soma have dealt many small (non-Haken) Seifert-fibered spaces. However, the case of the generalized Smale conjecture for elliptic manifolds is still open (see however the work of Hong et. al.). I think this is an important open question, and it would be useful to have a unified proof of these results (in particular, Gabai's results makes use of a computer-aided proof of the existence of "non-coalescable insulator families"). One possible approach is to try to prove that the space of metrics is contractible (on a constant curvature manifold) by showing that all the homotopy groups vanish (it is known to be of the homotopy type of a CW-complex, so this suffices). This was the approach that Gabai took. You can fill in a sphere of constant curvature metrics with a ball of Riemannian metrics, since the space of Riemannian metrics is convex. Then you could try to "flow" towards a ball of constant curvature metrics using Ricci flow (which would stay fixed on the boundary of the ball). The issue is that under Ricci flow, singularities may occur. However, what I hope is that some sort of canonical Ricci-flow with surgery may be used to fill in the sphere with a ball of constant curvature metrics. Thus, I see it as an important question for 3-manifold topology to obtain an understanding of a version of Ricci flow-with-surgery and Perelman's proof of geometrization for families of Riemannian metrics. This approach for more general Seifert fibered spaces would be trickier, since one would probably have to get a very good idea of how the collapsing occurs at infinite time under Ricci flow, and prove finiteness of surgeries. - 1 Related, there recently was a new proof of Cerf's theorem $\pi_0 Diff(S^3)≃\mathbb Z_2$ put up on the arXiv: front.math.ucdavis.edu/1007.3606 – Ryan Budney Sep 1 2010 at 17:48 Rob Kirby has a huge collection of open problems: http://math.berkeley.edu/~kirby/problems.ps.gz - The volume conjecture. See e.g. H. Murakami's survey http://arxiv.org/abs/1002.0126 and references therein. - 3 Everytime someone tells to "have a look at [...] and the references therein" I suddenly think of some C++ programs and pointers to pointers to pointers to ... – Johannes Hahn Jul 18 2010 at 0:54 Johannes -- sorry, I know exactly how you feel. What I meant to say was that this is a very good introduction containing the statement of the conjecture and its variants along with the references to the original papers, settled cases etc. – algori Jul 18 2010 at 3:55 1 This is a very nice paper (and quite readable) - thanks! – Paul Siegel Aug 24 2010 at 16:51 Paul -- welcome! – algori Aug 24 2010 at 22:55 I suppose I'm a little contrary but I don't consider the virtual fibering conjecture to really be a big problem in 3-manifold theory. In an earlier era when it might have been an approach to proving geometrization, sure, but nowadays with geometrization a fixture of the landscape, the problem is far less important. Still quite significant, but no longer vital, and I'd rank it well below these problems: • Find an algorithmic formalism for the Ricci flow (with surgery). i.e. find a combinatorial formalism for curvature on a manifold and the resulting flow. This should be compatible with means for representing surfaces in the 3-manifold so that surgery can be implemented, for example, a formalism using triangulations of the manifold so that it would be compatible with normal surface theory. Likely you would want a suitable notion of Pachner complex to get this formalism off the ground. • Build stronger connections between the geometric perspective on 3-manifolds and other perspectives on 3-manifolds. I would put problems like understanding the properties of the Gordian graph of knots in here. Or the volume conjecture. 4-manifold theory enters the picture here because the question of how geometrization relates to surgery is a big one. Questions like which (rational) homology spheres bound (rational) homology balls, embedding 3-manifolds in 4-manifolds, etc. - 2 Ryan, I think your distinction is a little artificial. For instance, I would rate Friedl and Vidussi's proof that a 3-manifold $N$ is fibred if and only if $S^1 \times N$ is symplectic as one of the most attractive examples of 'a strong connection between the geometric perspective on 3-manifolds and other perspectives'. Their proof made crucial use of Agol's work on the Virtual Fibring Conjecture. – HW Aug 25 2010 at 5:23 1 Somehow an argument making the point of the interconnectedness of things in a thread which has the intent to artificially single-out individual "big" problems seems contrary to the point. :) My point being largely that 3-manifold theory's future growth should largely be outward to subjects bordering-on 3-manifold theory. And refining geometrization to make it more easily applicable and useful. – Ryan Budney Aug 25 2010 at 7:04 On the other hand, the question didn't ask for opinions as to which problem is 'the biggest'! Actually, I'm interested in your opinion and I don't disagree that (the subject of) 3-manifolds needs to look outwards. Which is why I wanted to point out that some of the problems already mentioned are indeed 'applicable'. – HW Aug 25 2010 at 16:19 I'd argue you made the point that some of Agol's work is applicable. VFC may have been part of his interest but it's not VFC that's actually been applied. But that's a small point. – Ryan Budney Aug 25 2010 at 17:14 1 I must respectfully disagree. In light of the Kahn-Markovic theorem, the importance of VFC has been heightened even further. Combining K-M with Thurston's dichotomy for surface subgroups (every surface subgroup of an $H^3$-manifold group is either undistorted or $\pi_1$ of a virtual fiber) VFC is now known to be equivalent to the following: The fundamental group of every closed hyperbolic 3-manifold contains a distorted surface group. I believe it is true (someone correct me if I'm wrong) that the surface subgroups constructed by K-M are all undistorted. – Lee Mosher Feb 27 2012 at 21:06 show 2 more comments Here are two problems on 3-manifold groups (i.e. fundamental groups of compact 3-manifolds) that I find important. a. Are 3-manifold groups linear? Comments: A group is called linear if it is isomorphic to a subgroup of $GL(n,\mathbb C)$. One can also ask this over other fields but let's focus on $\mathbb C$. Thurston conjectured that 3-manifold groups are linear, because the geometrization implies that they are residually finite (which is weaker than linearity for finitely generated groups). Aschenbrenner-Friedl recently showed that 3-manifold groups are virtually residually $p$ for all but finitely many $p$'s, which again is known for fg linear groups. b. Is it true that every 3-dimensional Poincare duality group is a 3-manifold group? Comments: This is wide open, but see e.g. this survey of Wall, and this list of questions by Hillmann. - The Virtually Fibred Conjecture, and related problems. For a weaker definition of 3-manifold topology, I think the Andrews-Curtis conjecture is a key problem. Also, anything which relates to the classification of non-simply-connected topological 4-manifolds, for instance problems related to knot and link concordance. - 1 Can you, please, comment on the precise relation of the VFC to the virtual positive Betti number conjecture? – Victor Protsak Jul 18 2010 at 16:28 The rank versus Heegaard genus conjecture: It states that given a closed (compact works as well, I think) hyperbolic 3-manifold $M$, the conjecture states that the Heegaard genus of $M$ is equal to the rank of $\pi_1(M)$. It's is relatively easy to see that the Heegaard genus is always greater than or equal to the rank just by looking at the definition of a Heegaard splitting, but the other inequality is not known. For non-hyperbolic 3-manifolds, there are examples where the genus is strictly greater than the rank (I don't have a reference for this off the top of my head). - The geometization conjecture shows that every 3-manifold decomposes into geometric pieces. In some sense, the non-hyperbolic pieces are "well-known" since many decades, whereas the hyperbolic pieces are not. Therefore the current research focuses mainly in "understanding" hyperbolic 3-manifolds. Of course, "understanding" is not a well-defined mathematical problem: however, I think that researchers in the field mostly agree that we are still far from reaching this goal. For instance, as opposite to Seifert manifolds, hyperbolic manifolds are not classified in a strict sense: every Seifert manifold has a standard unique "name" which tells many things about its geometry and topology, but hyperbolic manifolds do not have such univoque names. Volumes of hyperbolic manifolds are still poorly understood, and even a simple relationship between manifolds like a "topological covering" is far from being understood, all the conjectures listed by Wilton above show. - 4 In some sense hyperbolic $3$-manifolds do have names. Their fundamental group functions as a name, since the isomorphism problem is decidable for them. I think in some way the mystery is more the other-way around -- instead of assigning names we want a "phone book". Like: I want a non-repeating list of all hyperbolic 3-manifolds with volume less than 20, how do I generate that list efficently? Similarly, "is $\sqrt{71\sqrt{11}}$ the volume of a hyperbolic 3-manifold?" – Ryan Budney Aug 31 2010 at 23:51 Cannon's Conjecture: Every finitely generated word hyperbolic group with Gromov boundary $S^2$ has a finite normal subgroup whose quotient is the fundamental group of a closed hyperbolic 3-orbifold. - Actually, this is a special case of Wall's conjecture mentioned by Belegradek (PD(3) groups are manifold groups). Of course, Cannon's conjecture might be easier to settle, at least there are some tools available in hyperbolic setting. – Misha Dec 8 at 16:28 If the 3-manifold theory is understood broadly enough, then one should mention the Vassiliev conjecture and in general, the problem of computing the cohomology of the spaces of knots in 3-manifolds. Note that for manifolds of dimension 4 or more this has been completely solved. For an introduction to all this see Vassiliev's ICM 1994 talk (MR1403923) and for more details see his Complements of discriminants of smooth maps (MR1168473). - 1 FYI, in my papers I reduce the computation of the homology of $Emb(S^1,S^3)$ to a question about certain canonical representations of certain subgroups of the group of hyperbolic isometries of certain hyperbolic link complements (in $S^3$). Here "certain" means the qualifiers are simple to state, just too long for this comment. Your "completely solved" statement needs more qualifiers. Perhaps you're talking about convergence of an appropriate spectral sequence provided the co-dimension of the embeddings is 3 or larger? – Ryan Budney Aug 25 2010 at 3:58 Ryan -- yes, I meant to say the spectral sequences degenerate and one can compute each $E^{pq}_\infty$; these will be finite dimensional and one can in principle construct explicit cohomology classes and evaluate them on each knot. In general (that is if one considers embeddings whose source is not necessarily the circle) codimension 3 does not suffice: one needs the discriminant to be of real codimension $>1$ in order for the spectral sequence to degenerate, so by counting parameters if $n$ is the dimension of the source, the dimension of the target must be at least $2n+2$. – algori Aug 25 2010 at 5:04 .. that is one can compute each $E^{pq}_{\infty}$... (the formula disappears when I press add comment, not sure why) – algori Aug 25 2010 at 5:08 @algori - do you know of a proof that these spectral sequences in dimensions 4 and higher degenerate over the integers? I only know of one for rational cohomology and homotopy, for embeddings of spheres in Euclidean spaces. These degeneration results are by Arone-Lambrechts-Turchin-Volic. Also, explicit cohomology classes are only known (to me) over the reals using de Rham theory (by Cattaneo-Cotta Ramussino-Longoni) and not even over the rationals much less the integers. – Dev Sinha Sep 1 2010 at 6:59 1 Yes, there is a description of the E_1 in those terms - see arxiv.org/pdf/math.AT/0407039.pdf and arxiv.org/pdf/math/0202287.pdf But there is a big difference between convergence and collapse. In both cases "there exists an algorithm" but if all you have is convergence the algorithm is hardly an algorithm. The collapse results of Lambrechts-Turchin-Volic say much more - I suggest you take a look at them if you have an interest in this area. – Dev Sinha Sep 2 2010 at 4:38 show 6 more comments The simple loop conjecture. The statement 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": 62, "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": 0.9405860900878906, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/242291/curious-facts-about-ordinal-numbers?answertab=oldest	# Curious facts about ordinal numbers I have some notes about curious facts about ordinal numbers, for example that their addition is not commutative, multiplication is not distributive from the right hand side and that the exponent rule doesn't always hold. Also that some things that are undefined in analysis like $0^0, \infty^0, 1^\infty$ are actually defined for ordinal numbers. I know that there's been some investigations devoted to pursuing ordinal arithmetic along the lines of classical results in number theory for example. Do you happen to know other curious facts about ordinal numbers (compared to facts in analysis or other)? - Nice list, but the notation is so small (at least in my browser!) – amWhy Nov 21 '12 at 22:32 I don't think there are many fun facts on ordinals, or maybe I just don't know what you want to hear... – Asaf Karagila Nov 21 '12 at 22:36 John Baez (math.ucr.edu/home/baez) has been writing a series of posts on Google+ (plus.google.com/117663015413546257905/posts) on countable ordinals, which you may find interesting. He is a very engaging expositor. The latest one is on the Hercules-Hydra game: plus.google.com/117663015413546257905/posts/HSS2PMVENe3 (and see also plus.google.com/117663015413546257905/posts/PjyerUy3AeH ) – Andres Caicedo Nov 21 '12 at 22:38 2 Well for example other fun facts I know of is that Fermat's Last Theorem and Goldbach's Hypothesis are known to be false in ordinal number theory (Sierpinski [1950]). – Nicolas Nov 21 '12 at 22:48 1 As a sanity check, by "$\infty^0$" and "$1^\infty$", do you really mean for $\infty$ as a variable whose values range over the infinite ordinals? As opposed, e.g. to suggesting there is an ordinal named $\infty$ or that it is the only infinite ordinal? – Hurkyl Nov 21 '12 at 23:55 show 1 more comment ## 3 Answers 1) Every countable ordinal can be written in a unique, canonical way, called the Cantor Normal form. It is basically like writing the ordinal "in base omega." 2) There is a fact in number theory that has a natural proof using (necessarily) infinite ordinals. The question is does every "Goldstein sequence" eventually converge to 0. A Goldstein sequence is, basically, a process where you take a number, write it in base 2, replace all of the 2's with 3's, then subtract 1. Then do this for base 3 into 4, etc. The answer is yes they all converge, as you can see here, and the proof is rather shocking for a number theorist. - Welcome Mike! :-) – Asaf Karagila Nov 22 '12 at 5:16 Why do you restrict yourself to countable ordinals? Every ordinal has a Cantor normal form. – Chris Eagle Nov 22 '12 at 18:14 Every countable ordinal is order isomorphic to some closed subset of (the rationals in) the closed unit interval. Not really related to ordinals, but it's might be interesting: $\mathbb Q$ is universal for all countable order types, i.e. any countable totally ordered set embeds into $(\mathbb Q,\leq)$. - Given any mapping $f : \mathbf{Ord} \to \mathbf{Ord}$ which is strictly increasing and continuous at limit ordinals (such a mapping is called a normal mapping), there are arbitrarily large ordinals $\alpha$ such that $f ( \alpha ) = \alpha$. • For example, the mapping $\alpha \mapsto \omega^\alpha$ is normal. The least fixed point of this function is commonly denoted $\epsilon_0$, and has great importance to the proof theory of Peano arithmetic. (Of course, as $\alpha \mapsto \omega^\alpha$ has arbitrarily large fixed points, we can define a new mapping $$\alpha \mapsto \epsilon_\alpha = \text{the } \alpha^{\text{th}} \text{ fixed point of the above mapping}.$$ This new mapping is also normal, and thus has arbitrarily large fixed points. And by a diagonalisation process at limit ordinals we can continue this indefinitely.) • The mapping $\alpha \mapsto \aleph_\alpha$ is also normal, and thus has arbitrarily large fixed points. That is, there are infinite cardinals $\kappa$ that have $\kappa$-many (infinite) cardinals below them. Even more, it is consistent with ZFC that $2^{\aleph_0} = \| \mathbb{R} \|$ is such a cardinal. - Haven't I told you about that last one? – Asaf Karagila Nov 25 '12 at 20:28	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 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": 0.9384346604347229, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/207543/how-to-programmatically-find-the-inverse-of-a-2x2-matrix-mod-26	# How to programmatically find the inverse of a 2x2 matrix (mod 26). I'm trying to create a hill cipher utility. One feature I want is to be able to compute the key if you have the plaintext and ciphertext. $C$ = ciphertext matrix ($2\times 2$), $P$ = plaintext matrix $\left(2\times\frac{N}{2}\right)$, $K$ = key ($2\times 2$). $$PK = C$$ So if I have $C$ and $P$, then $K = P^{-1}C$. How do I best find $P^{-1}$? EDIT: It is important to note that this is all mod $26$. Sorry. - 1 – Douglas S. Stones Oct 5 '12 at 2:16 ## 1 Answer You could use the direct formula for $2 \times 2$: $$\pmatrix{a & b\\ c & d}^{-1} = \dfrac{1}{ad-bc}\pmatrix{d & -b\\-c & a}$$ - The problem is that doesn't work out in with the modular. I don't know how to format this nicely, sorry. but lets say our plaintext is b,a,z,z. {{b,a}, {z,z}} = {{1,0}, {25,25}} then if we try multiplying by 1/(ad-ab) we get a fraction. That doesn't work when you are working with only discrete values. Sorry if I was not clear enough. – Alan Oct 5 '12 at 1:55 1 Yes, it would have to be the modular inverse, the extended gcd algorithm would be one way to do it. In your example, 25 is its own inverse, another example is $3^{-1} \equiv 9 (mod 26)$ since $3 \times 9 = 27 \equiv 1 (mod 26)$ – adam W Oct 5 '12 at 2:09 Okay, I think I got it. So {{1,0},{25,25}}^(-1) actually works out to {{1,0},{25,25}}, right? – Alan Oct 5 '12 at 2:15 Yes! Also, consider that $2$ and $13$ and any multiples (basically just even numbers and the number $13$) do not have modular inverses. – adam W Oct 5 '12 at 2:19	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 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": 0.9336446523666382, "perplexity_flag": "middle"}
http://mathhelpforum.com/math-challenge-problems/137094-classic-geometry-problem.html	# Thread: 1. ## A classic geometry problem Challenge Question: Triangle $ABC$ is an isosceles triangle with base $BC$ , the included angle of $A$ is $20$ degree . A point $D$is marked off on side $AC$ such that $AD = BC$ . What is the angle of $ABD$ ? Moderator edit: This is now an approved challenge question. 2. Originally Posted by simplependulum Triangle $ABC$ is an isosceles triangle with base $BC$ , the included angle of $A$ is $20$ degree . A point $D$is marked off on side $AC$ such that $AD = BC$ . What is the angle of $ABD$ ? Dear simplependulum, If you draw the traingle as indicated you will see that, Take, $A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x$ By the law of sines, $\frac{BC}{Sinx}=\frac{a}{Sin(160-x)}$ $BC=2aSin10$ $\frac{2aSin10}{Sinx}=\frac{a}{Sin(160-x)}$ Can you do it from here??? 3. Originally Posted by Sudharaka Dear simplependulum, If you draw the traingle as indicated you will see that, Take, $A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x$ By the law of sines, $\frac{BC}{Sinx}=\frac{a}{Sin(160-x)}$ $BC=2aSin10$ $\frac{2aSin10}{Sinx}=\frac{a}{Sin(160-x)}$ Can you do it from here??? Hmmm... I can't see clearly what the continuation is without getting into heavy stuff (probably power series or something: I'm not sure), but it perhaps doesn't matter: the problem's title says geometry and, at least over here, this means the problem must be solved using exclusively tools from geometry, without any trigonometry, calculus, etc. I wonder what the OP originally meant. Tonio 4. Originally Posted by Sudharaka Dear simplependulum, If you draw the traingle as indicated you will see that, Take, $A\hat{B}D=x\Rightarrow{D\hat{B}C}=80-x$ By the law of sines, $\frac{BC}{Sinx}=\frac{a}{Sin(160-x)}$ $BC=2aSin10$ $\frac{2aSin10}{Sinx}=\frac{a}{Sin(160-x)}$ Can you do it from here??? I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines . Below is the trigonometric method which is similar to the contest 's . Spoiler: we have $\frac{ \sin(x+20) }{\sin(x) } = \frac{1}{2\sin(10)}$ $\cos(20) + \cot(x) \sin(20) = \frac{1}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{1- 2\sin(10)\cos(20)}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{1- (\sin(30) - \sin(10))}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{\sin(30) + \sin(10)}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{2 \sin(20)\cos(10)}{2\sin(10)}$ $\tan(x) = \tan(10)$ $x = 10$ 5. Originally Posted by simplependulum I first saw this problem in a paper of IMO preliminary selection contest but the suggested solution uses trigonometry . Of course , as i said it classic , there is actually a solution ( collected from a book )which only uses the properties of isosceles , equilateral triangle , similar triangles and parallel lines . Below is the trigonometric method which is similar to the contest 's . Spoiler: we have $\frac{ \sin(x+20) }{\sin(x) } = \frac{1}{2\sin(10)}$ $\cos(20) + \cot(x) \sin(20) = \frac{1}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{1- 2\sin(10)\cos(20)}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{1- (\sin(30) - \sin(10))}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{\sin(30) + \sin(10)}{2\sin(10)}$ $\cot(x) \sin(20) = \frac{2 \sin(20)\cos(10)}{2\sin(10)}$ $\tan(x) = \tan(10)$ $x = 10$ Ok, and what's the pure-geometric solution, please? Tonio 6. Originally Posted by tonio Ok, and what's the pure-geometric solution, please? Tonio It's a CHALLENGE question, Tonio. See edit on initial post. 7. Originally Posted by Wilmer It's a CHALLENGE question, Tonio. See edit on initial post. Yes, I know but the OP said there exists a pure-geometric solution though his uses trigonometry. I'm asking for the pure-geometric one. Tonio 8. Originally Posted by tonio Yes, I know but the OP said there exists a pure-geometric solution though his uses trigonometry. I'm asking for the pure-geometric one. Tonio Don't think so: he gave the trig solution because it's evident; BUT the challenge is to find the geometric solution; right, simplep ? 9. I shouldn't give the solution now but if you really want the pure-geometric solution , i think it is okay to give the hints : The solution has three additional lines but i am going to give you only one of the lines because this is exactly the key to this problem while the others are not very important . Hints: Spoiler: Draw $DE \parallel BC$ such that $B,E$ are on the same side of $AC$ and $DE = AB$ . Think of why $20^o$ is used but not the others . 10. Originally Posted by Wilmer Don't think so: he gave the trig solution because it's evident; BUT the challenge is to find the geometric solution; right, simplep ? Any method is welcome . The geometric solution just brings us more happiness . BTW , my initial solution is the use of complex number . haha . 11. A figure is added in the attachment to make the proof easier to read. On the given triangle $\triangle ABC$ build a regular $18$-gon such that $A$ is the center of it (which is possible due to the fact that $AB=AC$, $\angle BAC = 20^{\circ}$). Now, we are not assuming $AD=BC$. We'll come back to it later. Continue AB until it gets to another vertex, call it $J$, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line $BS$ as shown in the attached figure. Next, draw $JT$ as shown, and call $M$ the intersection of $JT$ with $BS$. All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle $\triangle JMS$ it follows that $[JM]=2[JS]$, and in triangle $\triangle JMB$ we have $[JM]=2[AD]$. Therefore $[JS]=[AD]$, but $[JS]=[BC]$, so $[AD]=[BC]$. Next notice that $\angle JBS=10^\circ$ (again, by considering the circumscribing circle). We have thus proven that the conditions $\angle JBS=10^\circ$ and $[AD]=[BC]$ are obtained simultaneously, from which the result follows. Attached Thumbnails 12. Originally Posted by Unbeatable0 A figure is added in the attachment to make the proof easier to read. On the given triangle $\triangle ABC$ build a regular $18$-gon such that $A$ is the center of it (which is possible due to the fact that $AB=AC$, $\angle BAC = 20^{\circ}$). Now, we are not assuming $AD=BC$. We'll come back to it later. Continue AB until it gets to another vertex, call it $J$, of the regular polygon (I'll leave it to you to verify the fact that it is indeed getting there), and draw a line $BS$ as shown in the attached figure. Next, draw $JT$ as shown, and call $M$ the intersection of $JT$ with $BS$. All the angles shown in the figure can easily be calculated by considering equal arcs on the circumscribing circle of the regular polygon. Thus, from triangle $\triangle JMS$ it follows that $[JM]=2[JS]$, and in triangle $\triangle JMB$ we have $[JM]=2[AD]$. Therefore $[JS]=[AD]$, but $[JS]=[BC]$, so $[AD]=[BC]$. Next notice that $\angle JBS=10^\circ$ (again, by considering the circumscribing circle). We have thus proven that the conditions $\angle JBS=10^\circ$ and $[AD]=[BC]$ are obtained simultaneously, from which the result follows. Wow , again ! A unbeatable method ! The key to this problem is : this isosceles triangle has the property that the difference of its base angle and the other angle is equal to 60 : $80^o - 20^o = 60^o$ I would like to share my three solutions : Spoiler: Note that Draw such that are on the same side of and . Since , , , is congruent to and . Consider and , therefore , is an equilateral triangle . Then , is an isosceles triangle with base angle , so . Spoiler: Let $E$ be a point outside the triangle such that $\Delta ADE$ is an equilateral triangle so $\angle BAE = 20^o + 60^o = 80^o$ . It is easy to find that $\Delta ABE$ is congruent to $\Delta ABC$ , $AB = BE$ . Note that $D$ is inside the $\Delta ABE$ . Now consider $\Delta ABD$ and $\Delta EBD$ , we have $AD = DE$ , $AB = BE$ and $\angle BAC = \angle BED = 20^o$ . Therefore , $\Delta ABD$ is congruent to $\Delta ABE$ . We can see that $AD$ bisects $\angle ABE$ so $\angle ABD = \frac{20^o}{2} = 10^o$ Spoiler: Let $E$ be a point inside the $\Delta ABC$ such that $\Delta BCE$ is an equilateral triangle , $\angle ABE = 80^o - 60^o = 20^o$ . It is easy to prove that $ABED$ is an isosceles trapezium . $\angle ABD = \angle ABE = \frac{20^o}{2} = 10^o$.	{"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": 118, "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": 0.9402906894683838, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-algebra/205168-circle-group-related-groups-print.html	The circle group and related groups Printable View • October 12th 2012, 01:12 AM tnsten The circle group and related groups Hello! My problem is as follows: Circle group $T$ = \{z \in C: \|z\| = 1\}$, $G_n = \{z \in T: z^n = 1\}, G = \cup_{i = 1}^\infty(G_n)$$ First of all, which are cyclic? T is not as it is not countable. Gn is and G should not be since it is here described as a union of proper cyclic subgroups. But I am afraid that for some large n Gn becomes G or something weird like that. How can I assure myself this is not the case, and that Gn for every n is a proper subgroup of G? Further I'm asked to describe the cyclic subgroups of these 3. In particular does T have a subgroup isomorphic to Z x Z? As I understand it T only has Gn as proper subgroups, but how do I prove non-existance of others? Gn is also a subgroup of G by definition, but does G have any other subgroups? This one in particular I do not know where to start with. Further I am to show that T is generated by every neighbourhood of the identity. So basicly if we have an interval ]a,b[, where a and b are the arguments of the complex numbers in the original neighbourhood I of e (identity). So I'm thinking I^2 will be mapped onto something that has arguments in the range ]2a, 2b[. However, it feels they are kind of "thinned out" in the edges as it is only half of the numbers that map to something "new" and the rest end up in the previous interval. I know this is very handwavy etc, but I feel my knowledge of the real numbers is terribly insufficient so I can not really make a clear argument. How should I view this, and reassure myself nothing bad happens by just covering more and more of the unit circle? Also is T countably generated? An interval of reals should not be countable right? But I have so far only shown it can be generated by an uncountable set, not non-existance of countable generating sets. Cheers and thanks in advance, -Sten • October 12th 2012, 04:05 AM chiro Re: The circle group and related groups Hey tnsten. You should consider sub-groups where you have z^n = 1 where z is a complex number and n is a natural number. Also, based on what you've said, consider Lie Groups (and there are tonnes of books and resources out there for this): Lie group - Wikipedia, the free encyclopedia All times are GMT -8. The time now is 04:09 PM. Copyright © 2005-2013 Math Help Forum. All rights reserved. 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": 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": 0.9567133784294128, "perplexity_flag": "middle"}
http://crypto.stackexchange.com/questions/3334/can-i-build-a-secure-tweakable-block-cipher-from-a-normal-one-by-adding-key-and?answertab=votes	Can I build a secure tweakable block cipher from a normal one by adding key and tweak? Let `(E,D)` be a secure block cipher. Consider the following tweakable block cipher: ````E'(k,t,x) = E( (k+t) mod \|K\|, x) D'(k,t,c) = D( (k+t) mod \|K\|, c) ```` Is `(E',D')` secure? - 2 Answers Well, whether it is a secure tweakable block cipher depends on how resistant `(E,D)` are to related key attacks; that's not a standard assumption for block ciphers. For example, this would not be a secure tweakable block cipher with 3DES; because every 8th bit is ignored, the attacker can effectively test the value of the 7 adjacent bits (except for the 7 MSBits) by a choosen tweak attack. For example, if he can find a value t with (for an arbitrary block x): ````E'(k,t,x) = E'(k,t+256,x) E'(k,t+2,x) != E'(k,t+258,x) ```` then he can deduce that the addition of k and t+258 was just enough to propagate a carry past bit 8 (which doesn't affect the operation of 3DES) into bit 9 (which does). This allows him to immediate deduce the lowest 7 used bits of the key. Repeating this observation for the rest of the key bits allows us to recover all but 7 bits of the key with about 400 adaptively chosen tweaks; the last 7 is easy to brute force. - I never understood why DES was treated as having a 64 bit key, and not a 56 bit key. – CodesInChaos Aug 30 '12 at 22:32 To expand / generalize @poncho's reply, given a block cipher $(E,D)$ with keylength $n$, you can make a new one $(E',D')$ with key length $n+1$, which ignores the last bit of the key and just runs $(E,D)$. If $(E,D)$ is a secure PRP, then so is $(E',D')$. But plugging $(E',D')$ into the OP's construction does not yield a secure tweakable block cipher. To see this, ask for encryptions of the same block under tweaks 0, 1, and 2. Two of the three resulting blocks will be identical (depending on whether the secret key was chosen to be even or odd), whereas they should all look independently random. So the construction requires more than just standard [s]PRP security of the underlying block cipher. Note that the construction you suggest $E'(k,t,x) = E( (k+t) \bmod \|K\|, x)$ is a special case of a related-key secure PRP (see A Theoretical Treatment of Related-Key Attacks: RKA-PRPs, RKA-PRFs, and Applications by Bellare & Kohno). In their terminology, you are asking for the block cipher to be secure against a group-induced key relation -- in this case, the group operation is addition mod $N$. Some PRPs were shown secure against these kinds of attacks in two papers (Pseudorandom Functions and Permutations Provably Secure Against Related-Key Attacks by Bellare & Cash, Fast Cryptographic Primitives Based on the Hardness of Decoding Random Linear Codes by Applebaum). Neither of these uses addition mod N as the underlying group operation, I think. Instead they use the group operation in a DDH group and XOR. -	{"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": 0.9188239574432373, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/62761-calculus-volume-problem.html	# Thread: 1. ## calculus volume problem From Calculus 1: A ball of radius 13 has a round hole of radius 7 drilled through its center. Find the volume of the resulting solid. Okay, I have no idea what I am doing wrong here: General plan is to find the volume of the ball and then subtract the volume of the part drilled out. The volume of the ball: V = 4/3pi(13^3) = 9202.77208 volume of the part drilled: V = Hpi(7^2) = 26 * pi * 49 = 4002.389041 So I get a final answer of about : 5200.38 But it's the wrong answer. Help?? 2. Well I would recomend coming up with some system of equations and geting to a single variable. From there try performing some calculus. Also consider that you are not actually removing a right cylinder from the sphere - it does not have flat ends. I hope that helps get you going. 3. ## volume Take the sphere as its centre at the origin. The volume you found for the sphere is correct. The area of the cylinder in the middle needs to be subtracted. Because the hole drilled has a radius of 7, this line is x=7. Rotating the line x=7 about the y axis gives $V = \pi \int \limits_{-6.5}^{6.5} 7^2 dy$ Integrating you get $\pi [49y]$ with limits -6.5 and 6.5 $\pi * [(49 * 6.5) - (49 * -6.5)] = 13 * 49 \pi = 2001.19452<br />$ Now subtract this from the sphere volume and you should end up with the correct answer. 4. Why did you use 13 as the height for the cylinder and not 26? I thought if the radius of a sphere is 13 then the height would be the same as the diameter. why is my sphere volume wrong? I thought the formula for it is V = pi r^3 The only thing I can think of is to find it with integration... something like... pi r^2 integrated from 0 to 13? pi [r^3/3] with 13 as r ... V = 2300.69302 but that doesn't give me the right answer either... I'm confused. help? 5. ## volume littlejodo, sorry about that mistake I made. My approach would be: The sphere should be $4/3 * \pi * 13^3$ The cylinder should be $\pi * 7^2 * 26$ Subtract the cylinder volume from the sphere volume which gives 5200.38. Which is apparently wrong? Could you post the correct answer please? 6. I don't have the correct answer. I do submissions online and it just tells me when I'm wrong. I think you were on to something when you used the integration, I am just not sure how to accomplish that for the sphere as well. 7. Does anybody think this could be set up as a solid of revolution problem? Would I rotate around the x or y axis, if so? Thanks!	{"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": 5, "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": 0.9493085741996765, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/tagged/elliptic-surfaces	## Tagged Questions 0answers 127 views ### Topology of K3 as a sum of two abelian fibrations. Let $E$ be a blow-up of $\mathbb{P}^2$ at 9-points in the bases locus of a pencil of elliptic curves (A $T^2$ fibration over $S^2$). K3 surfaces is obtained by removing a fiber fr … 0answers 190 views ### Singular fibers of an elliptic fibered K3 surface. Let $f:S\rightarrow \mathbb{P}^1$ be an elliptic K3 surface. Assume that $\mathrm{Pic}(S)\cong U$, where $U$ stands for the hyperbolic lattice. I think that the elliptic fibration … 2answers 203 views ### Singular fibres in the definition of an elliptic surface I have a question about the definition of an elliptic surface. One defines an elliptic surface $S$ over a base curve $C$ (over some field $k$) as a surjective morphism $f: S \to C$ … 3answers 607 views ### A K3 over $P^1$ with six singular $A_1$- fibers? Hirzebruch, in the paper 'Arrangements of Lines and Algebraic Surfaces' constructs a special $K3$ surface out of a 'complete quadrilateral' in $CP^2$. A complete quadritlateral … 1answer 426 views ### Dodecahedral K3? In pondering this MO question and in particularly its 1st answer, and answers to this one recently posed, I realized there ought to be a dodecahedral K3 surface $X$. This $X$ woul …	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9006113409996033, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/tagged/kullback-leibler	# Tagged Questions The kullback-leibler tag has no wiki summary. 0answers 22 views ### Symmetric Kullback-Leibler divergence OR Mutual Information as a metric of distance between two distributions? I need some metric of divergence of two distributions. (They are complex and don't fit with exponential family, normal, log-normal, power-law. Maybe some mixture of that, but I'm not feeling right ... 1answer 82 views ### Is the square root of the symmetric Kullback-Leibler divergence a metric? It is well known that the square root of the Jensen-Shannon divergence is a true metric, but how about the square root of symmetric KL: D(P\|\|Q)+D(Q\|\|P)? I have reasons to believe that it also is a ... 0answers 13 views ### Is there any meta-approach for variable selection based of measures of similarity between each two variables? 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Define $$L(q)=D(q\|\|p_1)-D(q\|\|p_2)$$ where $D$ is the usual Kullback-Leibler divergence. Assume the support of $p_2$ is included in ... 1answer 286 views ### Kullback–Leibler divergence between two wishart distributions The result is shown in: [1] W.D. Penny, KL-Divergences of Normal, Gamma, Dirichlet, and Wishart densities, Available at: www.fil.ion.ucl.ac.uk/~wpenny/publications/densities.ps But could anyone help ... 1answer 74 views ### How does one express the decrease in minimal type II error bound for each observation added? 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Quite simple, I have some probability distribution p(x), how can I measure whether one empirical density (set of delta masses) is a better approximation than another. I know that KL-divergence is a ... 0answers 122 views ### How to to calculate the topic distribution of a document I have a simple (may be stupid) question. I want to calculate Kullback–Leibler divergence on two documents. It requires probability distribution of each document. I do not know how to calculate ... 3answers 678 views ### Kullback–Leibler divergence between two gamma distributions Choosing to parameterize the gamma distribution $\Gamma(b,c)$ by the pdf $g(x;b,c) = \frac{1}{\Gamma(c)}\frac{x^{c-1}}{b^c}e^{-x/b}$ The Kullback-Leibler divergence between $\Gamma(b_q,c_q)$ and ... 1answer 1k views ### Kullback–Leibler vs Kolmogorov-Smirnov distance I can see that there are a lot of formal differences between Kullback–Leibler vs Kolmogorov-Smirnov distance measures. However, both are used to measure the distance between distributions. Is there ... 2answers 3k views ### KL divergence between two univariate Gaussians I need to determine the KL-divergence between two Gaussians. I am comparing my results to these, but I can't reproduce their result. My result is obviously wrong, because the KL is not 0 for KL(p, p). ... 4answers 2k views ### An adaptation of the Kullback-Leibler distance? Look at this picture: If we draw a sample from the red density then some values are expected to be less than 0.25 whereas it is impossible to generate such a sample from the blue distribution. As a ... 1answer 734 views ### Estimate the Kullback-Leibler divergence I would like to be sure I am able to compute the KL divergence based on a sample. Assume the data come from a Gamma distribution with shape=1/.85 and scale=.85. set.seed(937) theta <- ... 2answers 2k views ### Kullback-Leibler divergence - interpretation I have a question about the Kullback-Leibler divergence. Can someone explain why the "distance" between the blue density and the "red" density is smaller than the distance between the "green" curve ... 1answer 244 views ### Properties of Battacharyya distance vs Kullback-Leibler divergence What properties do these measures have and how can I determine which one is better for a given purpose? What are extreme cases where they differ a lot? 3answers 3k views ### Questions about KL divergence? I am comparing two distributions with KL divergence which returns me a non-standardized number that, according to what I read about this measure, is the amount of information that is required to ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 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": 0.8921208381652832, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/61763/zero-cycles-on-an-arithmetic-surface/61771	Zero-cycles on an arithmetic surface Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Could anyone give a reference for the following statement, which I believe is true. "Let X be a regular scheme, flat over $Spec( \mathbb{Z})$, with fiber dimension $1$. Then the Chow group $CH^2(X)$ is finite." - 1 Answer The finiteness is known; see Szamuely's Seminaire Bourbaki expose and Remarque 3.4 (5) on page 11 is a precise reference. - A similar result is valid for any regular scheme $X$ of finite type over $\mathop{\rm Spec}(\mathbb Z)$: the Chow group of zero cycles is finite, with the exceptions of schemes which are proper over a finite field~$\mathbb F_p$ for which the Chow group of zero cycles has the form $\mathbb Z\oplus G$, for some finite group $G$. This is a part of geometric class field theory, due to Rosenlicht, Lang (50s), Bloch (1981), Kato-Saito (1986)... Szamuely's Seminaire Bourbaki explains the new point of view given by work of Wiesend. – ACL Apr 15 2011 at 6:43 Yes, indeed! Thanks! – SGP Apr 15 2011 at 21:06	{"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": 8, "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": 0.8700210452079773, "perplexity_flag": "head"}
http://mathoverflow.net/questions/76412?sort=oldest	## reference for the slice theorem for Banach Lie group actions on Banach manifolds ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am looking for a reference treating the slice theorem for Banach Lie group actions on Banach manifolds, i.e. proving that a smooth, free and proper action of a Banach Lie group $G$ on a Banach manifold $M$ with embedded orbits ensures that the quotient $M/G$ inherits the structure of a Banach manifold, s.t. $M\rightarrow M/G$ becomes a $G$-principal bundle. I know this is briefly treated in Bourbaki's "Lie groups and Lie algebras". Unfortunately, in the proof they refer to the Bourbaki book "differentiable and analytic manifolds" which is not available to me. Could anyone please provide me with a reference? - ## 1 Answer Dear Orbicular the theorem on the existence of slices is stated without proof as Theorem 5.2.6 in Critical Point Theory and Submanifold Geometry, LNM 1353, of Palais and Terng (for example see here). The proof should be adapted without difficulty from that in the finite-dimensional case. For this case you can look at On the existence of slices for actions of non-compact groups'' by Palais (for example see 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": 5, "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": 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": 0.9023267030715942, "perplexity_flag": "head"}
http://mathoverflow.net/questions/119981/are-all-anabelian-galois-actions-faithful/119984	## Are all anabelian Galois actions faithful? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $C/\mathbb Q$ be a smooth projective curve of genus $g\geq 2$ or a smooth affine curve of genus $g \geq 1$. The exact sequence $1 \to \pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q}) \to \pi_1^{et}(C) \to \operatorname{Gal}(\bar{\mathbb Q}\|\mathbb Q) \to 1$ gives a homomorphism from $\operatorname{Gal}(\bar{\mathbb Q}\|\mathbb Q)$ to the outer automorphism group of $\pi_1^{et}(C \otimes_\mathbb Q \bar{\mathbb Q})$. Is this homomorphism always injective? If one instead takes $C$ to be a curve of genus $0$ with $3$ points removed, it is injective, by Belyi's Theorem. - ## 2 Answers The answer is "yes" I think, even if you replace $\mathbb Q$ with a number field. In the affine case this is a result of Matsumoto, as pointed out by Felipe Voloch, see Matsumoto, Makoto Galois representations on profinite braid groups on curves. J. Reine Angew. Math. 474 (1996), 169–219. In the proper case this is a more recent result of Hoshi and Mochizuki, see Hoshi, Yuichiro; Mochizuki, Shinichi On the combinatorial anabelian geometry of nodally nondegenerate outer representations. Hiroshima Math. J. 41 (2011), no. 3, 275–342. See also Tamás Szamuely's nice survey: Heidelberg lectures on fundamental groups, in J. Stix (ed.) The Arithmetic of Fundamental Groups (PIA 2010), Contributions in Mathematical and Computational Sciences, Vol. 2, Springer-Verlag, 2012, 53--73 available here http://www.renyi.hu/~szamuely/pia.pdf - ### 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. In the affine case, when the divisor at infinity has at least one rational point, this follows from theorems 2.1 and 2.2 of A. Matsumoto, A. Tamagawa, Mapping-class-group action versus Galois action on profinite fundamental groups. Amer. J. Math. 122 (2000), 1017--1026. I don't know what happens in the general case. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.8393454551696777, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/20904/two-way-mixed-effects-anova-model	# Two-way mixed effects ANOVA model Consider the two-way ANOVA model with mixed effects : $$Y_{i,j,k} = \underset{M_{i,j}}{\underbrace{\mu + \alpha_i + B_j + C_{i,j}}} + \epsilon_{i,j,k},$$ with $\textbf{(1)}$ : $\sum \alpha_i = 0$, the random terms $B_j$, $C_{i,j}$ and $\epsilon_{i,j,k}$ are independent, $B_j \sim_{\text{iid}} {\cal N}(0, \sigma_\beta^2)$, $\epsilon_{i,j,k} \sim_{\text{iid}} {\cal N}(0, \sigma^2)$ ; and there are two possibilities for the random interactions : $\textbf{(2a)}$ : $C_{i,j} \sim_{\text{iid}} {\cal N}(0, \sigma_\gamma^2)$ or $\textbf{(2b)}$ : $C_{i,j} \sim {\cal N}(0, \sigma_\gamma^2)$ for all $i,j$, the random vectors $C_{(1:I), 1}$, $C_{(1:I), 2}$, $\ldots$, $C_{(1:I), J}$ are independent, and $C_{\bullet j}=0$ for all $j$ (which means that mean of each random vector $C_{(1:I), j}$ is zero). Model $\textbf{(1)}$ + $\textbf{(2a)}$ is the one which is treated by the nlme/lme4 package in R or the PROC MIXED statement in SAS. Model $\textbf{(1)}$ + $\textbf{(2b)}$ is called the "restricted model", it satisfies in particular $M_{\bullet j} = \mu + B_j$. Do you think one of these two models is "better" (in which sense) or more appropriate than the other one ? Do you know whether it is possible to perform the fitting of the restricted model in R or SAS ? Thanks. - Is the question not clear ?.. – Stéphane Laurent Jan 12 '12 at 9:50 According to the book "Design and Analysis of Gauge R&R Studies" (Richard K. Burdick, Connie M. Borror, Douglas C. Montgomery) this question is discussed in Section 7.2 of the book "Confidence intervals on variance components" (Richard K. Burdick, Franklin A. Graybill). – Stéphane Laurent Apr 23 '12 at 19:19 So which model is better? – Henrik Apr 23 '12 at 21:01 I do not have the book ! – Stéphane Laurent Apr 24 '12 at 4:52 ## 1 Answer I will try to give an answer, but I am not sure if I understood your question correctly. Hence, first some clarification on what I tried to answer (as you will see, I am not mathematician/statistician). We are talking about a classical split-plot design with the following factors: experimental unit $B$, repeated-measures factor $C$ (each experimental unit is observed under all levels of $C$), and fixed-effect factor $\alpha$ (each experimental unit is observed under only one level of $\alpha$; I am not sure why $\sum \alpha_i = 0$, but as there needs to be a fixed factor, it seems to be $\alpha$). Model $\textbf{(1)}$ + $\textbf{(2a)}$ is the standard mixed-model with crossed-random effects of $B$ and $C$ and fixed effect $\alpha$. Model $\textbf{(1)}$ + $\textbf{(2b)}$ is the standard split-plot ANOVA with a random effects for $B$, the repeated-measures factor $C$ and fixed effect $\alpha$. That is, $\textbf{(1)}$ + $\textbf{(2a)}$ does not enforce/assumes a specific error strata, whereas $\textbf{(1)}$ + $\textbf{(2b)}$ enforces/assumes variance homogeneity and sphericity. You could fit $\textbf{(1)}$ + $\textbf{(2a)}$ using `lme4`: ````m1 <- lmer(y ~ alpha + (1\|B) + (1\|C)) ```` You could fit $\textbf{(1)}$ + $\textbf{(2b)}$ using `nlme`: ````m2 <- lmer(y ~ alpha * C, random = ~1\|C, correlation = corCompSymm(form = ~1\|C)) ```` Notes: • Note that there is one difference between the two models namely that `m1` does not have the $B \times C$ interactions. Experts on `lme4` will probably be able to help you with it. • To enforce the sphericity for $\textbf{(2b)}$ when using `lme` I use a compound correlation structure. See my answer to another question for more practical stuff on this use of `lme`. As far as I get it, it is kind of difficult/mpossible to extend this approach to more than one repeated-measures factor. • You will need a `data` argument in both calls. - I don't believe your proposal for (1)+(2b) is correct. Nothing looks like the constraint $C_{\bullet j}=0$ in your model m2. – Stéphane Laurent Apr 24 '12 at 5:00 1 Hmm, I have to say that then, I dont get it. Can you clarify in a little less mathematical terms, what this constraint means? – Henrik Apr 24 '12 at 6:31	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "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": 0.9071180820465088, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/56451/list	## Return to Answer 3 correct hypothesis I realized that I completely missed the second part of the question (the example). Note that ZMT implies that $f$ is a quasi-affine morphism. Then $X\to \mathit{Spec}(f_\mathcal O_X)$ is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if $f_\mathcal O_X$ is finite over $\mathcal O_Y$. Some comments: one should note that in general, the quasi-coherent algebra $f_\mathcal O_X$ is not finite over $\mathcal O_Y$ and even worse, the morphism $\mathit{Spec}(f_\mathcal O_X)\to Y$ may not be of finite type (take $Y$ an algebraic variety and $f$ an open immersion. Then $\mathcal O(X)$ is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation $X\to Z\to Y$. If the complementary of $X$ in $Z$ only consists in points of depth at least 2 (see discussions here), then $f_\mathcal O_X=h_\mathcal O_Z$ is finite and we are happy. This happens when $X$ is normal (or with non-normal locus finite over $Y$) and surjective to $Y$. Y$with complementary in$Z\$ of codimension at least 2. But I don't have a general criterion. 2 typo in URL I realized that I completely missed the second part of the question (the example). Note that ZMT implies that $f$ is a quasi-affine morphism. Then $X\to \mathit{Spec}(f_\mathcal O_X)$ is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if $f_\mathcal O_X$ is finite over $\mathcal O_Y$. Some comments: one should note that in general, the quasi-coherent algebra $f_\mathcal O_X$ is not finite over $\mathcal O_Y$ and even worse, the morphism $\mathit{Spec}(f_\mathcal O_X)\to Y$ may not be of finite type (take $Y$ an algebraic variety and $f$ an open immersion. Then $\mathcal O(X)$ is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation $X\to Z\to Y$. If the complementary of $X$ in $Z$ only consists in points of depth at least 2 (see discussions here), then $f_\mathcal O_X=h_\mathcal O_Z$ is finite and we are happy. This happens when $X$ is normal (or with non-normal locus finite over $Y$) and surjective to $Y$. But I don't have a general criterion. 1 I realized that I completely missed the second part of the question (the example). Note that ZMT implies that $f$ is a quasi-affine morphism. Then $X\to \mathit{Spec}(f_\mathcal O_X)$ is always an open immersion (see stack project, chapter 21, Lemma 12.3). So the Stein factorization witness ZMT if and only if $f_\mathcal O_X$ is finite over $\mathcal O_Y$. Some comments: one should note that in general, the quasi-coherent algebra $f_\mathcal O_X$ is not finite over $\mathcal O_Y$ and even worse, the morphism $\mathit{Spec}(f_\mathcal O_X)\to Y$ may not be of finite type (take $Y$ an algebraic variety and $f$ an open immersion. Then $\mathcal O(X)$ is or not finitely generated is related to Hilbert's 14th problem). Now consider a ZMT factorisation $X\to Z\to Y$. If the complementary of $X$ in $Z$ only consists in points of depth at least 2 (see discussions here), then $f_\mathcal O_X=h_\mathcal O_Z$ is finite and we are happy. This happens when $X$ is normal (or with non-normal locus finite over $Y$) and surjective to $Y$. But I don't have a general criterion.	{"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": 52, "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": 0.929092526435852, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/174646/approximation-of-function-on-interval?answertab=votes	# Approximation of function on interval I'm looking for an accurate but as simple as possible approximation of $S(x,\lambda) = \frac{1}{(1-x) [x-(1-\lambda )]}\left((1+\lambda ) \left(\frac{x(1+\lambda)}{1-\lambda }\right)^{\frac{x(1-\lambda )-(1-\lambda )}{x(1+\lambda )-(1-\lambda )}}-\lambda -x\right)$ for $x \in (0,1), \lambda \in (0,1)$. Does such a thing exist, and how can one find it? From what I've tried, Taylor expansion does not seem very effective in approximating the function in question... any other ideas? Many thanks! - 1 Oh dear. How exactly did you end up with this beast? – J. M. Jul 24 '12 at 13:17 How accurate/fast does the approximation have to be? What is the application for it? – Shaktal Jul 24 '12 at 13:22 It's an expression for "the gains of waiting" in an economics article. To be precise, it measures how much it pays for a consumer to delay his purchase, given that firms sample prices from an equilibrium distribution function $G(p)$. $\lambda$ is a parameter of the model and $x$ a variable I'd like to solve for. – Martin Jul 24 '12 at 13:26 Ideally I'd just like some kind of low degree polynomial, algebraic fraction or logarithmic function such that I can solve $S(x,\lambda) = W(x)$ for nice sample functions $W(x)$. I have not an extensive math background. – Martin Jul 24 '12 at 13:28 ## 1 Answer The graph of $S$ as a function of $x$ looks very much like a hyperbola, which suggests using the approximation $S(x,\lambda)\approx a+\frac{b}{x+c}$ where $a,b,c$ depend on $\lambda$ in some way. I first tried to fit a hyperbola to the data $$S(0)=\frac{\lambda}{1-\lambda},\quad S'(0)=-\frac{\lambda(1+2\lambda)}{(1-\lambda)^2}, \quad S(1)=\frac{1}{2\lambda^2}\left(2\lambda-(1-\lambda^2)\ln\frac{1+\lambda}{1-\lambda}\right)$$ by solving the system $a+b/c=S(0)$, $-b/c^2=S'(0)$, and $a+b/(1+c)=S(1)$. Unfortunately this results in a substantial underestimate. The reason is that $S'(0)$ is a poor indicator of the behavior of the curve: $S'$ changes rapidly near $0$. Then I fit the model $a+b/(x+c)$ to three points $S(0), S(1/3), S(1)$, and this worked much better. Here is the comparison for $\lambda=0.17$ ($S$ in red, approximation in blue): For $\lambda=0.37$ For $\lambda=0.87$ The coefficients $a,b,c$ are ugly, so I did not reproduce them here. The ugliness does not matter much if you work with specific $\lambda$. If you also need reasonably simple dependency on $\lambda$, let us know: maybe someone will come up with a better way to get $a,b,c$. (Added) Here is an approximation with simple dependency on $\lambda$. I found it more convenient to work with $\tilde S=1-\frac{1-\lambda}{\lambda}S$, since this function is increasing from $0$ to something in the range 0.3..1, which is to say it has a milder dependency on $\lambda$ than the original one. I picked the model $\tilde S\approx \alpha x/(\beta x+\gamma)$ where $\alpha,\beta,\gamma$ are linear functions of $\lambda$. Here is one possibility: $$\tilde S\approx \frac{(1+2\lambda)x}{(2+\lambda)x+1-\lambda}$$ This approximation is not great, but you can tweak the six (well, actually five) numeric parameters to make it look better. - I very much like the idea of fitting S(x,.) to specific, simpler functions of x. Thank you very much for putting in the effort!A simple dependency on $\lambda$ would of course be nice, but what really matters is a simple dependency on $x$ which is robust to changes in $\lambda$. In particular, a good approximation for $\lambda \approx 0$ and $\lambda \approx 1$ would be nice. I will toy around some more based on your approach! – Martin Jul 27 '12 at 9:52 @Martin Added a pattern for simpler approximation. – user31373 Jul 27 '12 at 20:08 Thanks Leonid. I tried your suggestion but it only seems to give a good approximation for $\lambda$ close to zero (assuming that $S = \frac{\lambda}{1-\lambda}(1-\tilde{S})$). I think I found a reasonably good approximation (with a complicated dependency on $\lambda$) to be $S(x,\lambda) = \frac{a}{x^{0.8}+b}$, with $b = \frac{2\lambda - (1-\lambda^2)Log\left(\frac{1+\lambda}{1-\lambda}\right)}{\frac{2\lambda^3}{1-\l‌ambda}-2\lambda - (1-\lambda^2)Log\left(\frac{1+\lambda}{1-\lambda}\right)}$ and $a = \lambda b$. This function has only an exact fit for $S(0;\lambda)$ and $S(1;\lambda)$. – Martin Jul 30 '12 at 9:48 There is a typo in the denominator and the website doesn't let me change it. It should be $\frac{2\lambda^3}{1-\lambda}-[2\lambda-(1-\lambda^2)Log\left(\frac{1+\lambda}{1‌-\lambda}\right)].$ – Martin Jul 30 '12 at 10:12	{"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": 49, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9547688961029053, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4267337	Physics Forums Recognitions: Gold Member ## Pisano Periods - Fibonacci Numbers mod p Let Fn be the nth number of a Fibonacci sequence. We know that Fnmod(p) forms a periodic sequence (http://en.wikipedia.org/wiki/Pisano_period) called the Pisano Period. Let p = a prime such that p$\equiv${2,3}mod 5 so that h(p)$\mid$ 2 p + 2. Let h(p) denote of the length of the Pisano period. If D = {d1,d2,d3$\cdots$dk} is the non-empty set of k divisors of 2 p + 2 Then: h(p) = min[di] such that Fd(i + 1)$\equiv$ 1 mod p and 1. di ~$\mid$$\frac{1}{2}$ p (p + 1) 2. di ~$\mid$ p + 1 3. di ~$\mid$ 3 (p - 1) Now let p = a prime such that p$\equiv${1,4}mod 5 so that h(p)$\mid$ p - 1. If p has a primitive root such that g2$\equiv$ g + 1 mod(p) then h(p) = p - 1. Note that g2$\equiv$ g + 1 mod(p) has two roots: 1.618033988 and -0.618033988 - variants of the Golden Ratio. If p has no primitive root then D = {d1,d2,d3$\cdots$dk} is the non-empty set of k divisors of p - 1. Let h(p) = min[di] such that Fd(i + 1)$\equiv$ 1 mod p and di ~$\mid$ p + 1 and di ~$\mid$ floor [ p/2]]. If m is any positive integer > 3 we can write Fn mod Fm where h(Fm) is given by 1. h(Fm) = 2m ↔ m is even 2. h(Fm) = 4m ↔ m is odd PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Blog Entries: 2 Quote by LDP Let Fn be the nth number of a Fibonacci sequence. We know that Fnmod(p) forms a periodic sequence (http://en.wikipedia.org/wiki/Pisano_period) called the Pisano Period. Let p = a prime such that p$\equiv${2,3}mod 5 so that h(p)$\mid$ 2 p + 2. Let h(p) denote of the length of the Pisano period. If D = {d1,d2,d3$\cdots$dk} is the non-empty set of k divisors of 2 p + 2 Then: h(p) = min[di] such that Fd(i + 1)$\equiv$ 1 mod p anddi ~$\mid$$\frac{1}{2}$ p (p + 1) di ~$\mid$ p + 1 di ~$\mid$ 3 (p - 1) Now let p = a prime such that p$\equiv${1,4}mod 5 so that h(p)$\mid$ p - 1. If p has a primitive root such that g2$\equiv$ g + 1 mod(p) then h(p) = p - 1. Note that g2$\equiv$ g + 1 mod(p) has two roots: 1.618033988 and -0.618033988 - variants of the Golden Ratio. If p has no primitive root then D = {d1,d2,d3$\cdots$dk} is the non-empty set of k divisors of p - 1. Let h(p) = min[di] such that Fd(i + 1)$\equiv$ 1 mod p and di ~$\mid$ p + 1 and di ~$\mid$ floor [ p/2]]. If m is any positive integer > 3 we can write Fn mod Fm where h(Fm) is given by h(Fm) = 2m ↔ m is even h(Fm) = 4m ↔ m is odd Interesting to say the least. I am curious as to what is meant by 1. di ~$\mid$$\frac{1}{2}$ p (p + 1) 2. di ~$\mid$ p + 1 3. di ~$\mid$ 3 (p - 1) I think you are saying di approximately divides the expressions on the right, but I dont know what that means. Can you give an example? Recognitions: Gold Member Thanks. No, I was trying to find - does not divide - but couldn't so I sort of made that up. But perhaps I should clarify it, because it is not the standard notation. Blog Entries: 2 ## Pisano Periods - Fibonacci Numbers mod p Quote by LDP Thanks. No, I was trying to find - does not divide - but couldn't so I sort of made that up. But perhaps I should clarify it, because it is not the standard notation. Clarification noted. Only problem that I see is for p = 2 (for which you say h(p) = 2p + 2 = 6). Since h(2) = 3, I guess you meant odd primes that = 2,3, mod 5 (whose last digit is either a 3 or 7). PS, I noted that for the composites ending in 3 or 7 which I checked, that h(p) <> 2p + 2. Could this be a test for primes ending in 3 or 7? Recognitions: Gold Member Quote by ramsey2879 Clarification noted. Only problem that I see is for p = 2 (for which you say h(p) = 2p + 2 = 6). Since h(2) = 3, I guess you meant odd primes that = 2,3, mod 5 (whose last digit is either a 3 or 7). PS, I noted that for the composites ending in 3 or 7 which I checked, that h(p) <> 2p + 2. Could this be a test for primes ending in 3 or 7? Indeed, I should have said "only odd primes". Good catch. I will add some more observations on this topic as time permits. Blog Entries: 2 Quote by ramsey2879 Clarification noted. Only problem that I see is for p = 2 (for which you say h(p) = 2p + 2 = 6). Since h(2) = 3, I guess you meant odd primes that = 2,3, mod 5 (whose last digit is either a 3 or 7). PS, I noted that for the composites ending in 3 or 7 which I checked, that h(p) <> 2p + 2. Could this be a test for primes ending in 3 or 7? I checked and found that the following composites ending in 7 have Pisano periods that divide 2p + 2 and do not divide (1/2)*p(p+1) or p+1 or 3p-3. (All primes ending in 7 less than 100,000 meet this test also). The composites ending in 7 and less than 100,000 meeting the test are 377, 3827, 5777, 10877, 25877, 60377, 75077 and 90287. Thread Tools \| \| \| \| \|---------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Pisano Periods - Fibonacci Numbers mod p \| \| \| \| Thread \| Forum \| Replies \| \| \| Programming & Comp Sci \| 4 \| \| \| Calculus & Beyond Homework \| 3 \| \| \| Linear & Abstract Algebra \| 4 \| \| \| Introductory Physics Homework \| 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": 36, "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": 0.8820688724517822, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/247795/what-is-the-covariance-of-mixture-bernoulli-distribution/247819	# What is the covariance of mixture Bernoulli distribution? For mixture of multivariate Bernoulli distribution we have that, $$p(x\|\mu,\pi) =\Sigma_{k=1}^{K}\pi_kp(x\|\mu_k)$$ where $$p(x\|\mu_k) = \prod_{i=1}^{D}\mu_{ki}^{x_i}(1-\mu_{ki})^{1-x_i}$$ I read it from the book that $$E[x] = \Sigma_{i=1}^{K}\pi_k\mu_k$$ $$cov[x] = \Sigma_{k=1}^{K}\pi_k(\Sigma_k+\mu_k\mu_k^T) - E[x]E[x]^T$$ The mean is trivial to prove, however I can't find proof for the covariance and I don't know how to prove it. Can anyone help? - ## 1 Answer You know that, in general, $cov[x] = E[x x^T] - E[x]E[x]^T$ and you want to compute the first term. We can use the property: $E[g(X)] = E [ E(g(X)\|k]]$ But $E(x x^T \| k)$ (i.e., fixing the population 'k' index of the mixing), is $\Sigma_k+\mu_k\mu_k^T$ (don't let the $k$ subindexes confuse you: they are the first and seconds moments of $x$, for a fixed population index $k$). Now, the above is a function ok $k$, and its expectation is $\sum_{k=1}^K \pi_k (\Sigma_k+\mu_k\mu_k^T)$ Notice that this result is not really connected to the Bernoulli distribution, it's valid in general. - Hi, can I know why $$E(xx^T\|k) = \Sigma_k+\mu_k\mu_k^T$$ I think that's the thing that confuses me – Jing Nov 30 '12 at 4:55 @Jing: It's just the same first equation of my answer, rearranged, and applied not to $x$ (the mixed variable) but to $x_k$ (one of the $K$ components of the mixture) – leonbloy Nov 30 '12 at 11:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 12, "mathjax_display_tex": 5, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9498900771141052, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/228972/solving-an-equation-with-different-base/228975	solving an equation with different base I am trying to solve a equation $(2^x)(3^{x-2}) = 4$ (the solution is $x = 2$) My approach is to use the natural logarithm $\ln$ (that's suggested when googleing the problem) So this is how it goes for me: $$\ln{2^x}\cdot\ln{3^{x-2}} = \ln{2^2}$$ but I am unsure if using $\ln$ is correct or where to go from here. - Thank you for all of the replies i really appreciate the help! I can see two types of approach one using logarithms and the other is just using exponents... which is more correct ? – 54N1 Nov 4 '12 at 17:38 All of them works, it depends on what you can use and what you feel more confortable with – Jean-Sébastien Nov 4 '12 at 18:27 3 Answers The correct way to use the logarithm is to take the logarithm of both sides: $$\ln{(2^{x}\cdot3^{x-2})}=\ln{4}$$ Using the logarithm law: $\ln{a\cdot b}=\ln{a}+\ln{b}$, we now have: $$\ln{2^{x}}+\ln{3^{x-2}}=\ln{4}$$ We now use another law of logarithms: $\ln{a^{b}}=b\ln{a}$: $$x\ln{2}+(x-2)\ln{3}=2\ln{2}$$ Re-arranging we get: $$x(\ln{2}+\ln{3})=2\ln{2}+2\ln{3}$$ We can solve for $x$ to get: $$x=\frac{2(\ln{2}+\ln{3})}{\ln{2}+\ln{3}}=2$$ - Note that $2^x 3^{x-2} = \dfrac{6^x}9$. This gives us that $6^x = 36 \implies x = 2$. - $(2^x)(3^{x-2}) = 4$ $2^x 3^x3^{-2} = 4$ $6^x/3^2 = 4$ $6^x =36$ $6^x =6^2$ $x=2$ -	{"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": 15, "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": 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": 0.9114459753036499, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/18280/why-do-quantum-physical-properties-come-in-pairs/18283	# Why do quantum physical properties come in pairs? Why do quantum physical properties come in pairs, governed by the uncertainty principle (that is, position and momentum?) Why not in groups of three, four, etc.? - ## 8 Answers The duality is a duality because of the notion of canonical conjugation in classical mechanics. The reason people say that they come in pairs has nothing to do with quantum mechanics, but with the structure of classical mechanics. In classical mechanics, to give you the initial conditions for a system, you need to give the initial position of everything, and also the initial momentum. The classical variables come in pairs. These pairs are called canonically conjugate, because they have the property that their time rate of change of one is given by the derivative of the energy with respect to the other one. The quantum mechanical description is only for wavefunctions which vary over values of one of the two canonically conjugate pairs. The other one is not freely specified, the wavefunction which gives its quantum description can be derived from the first. People express the failure of classical mechanics by saying that half of all the initial data is related by uncertainty to the other half. This is the quantum duality. This idea was important historically, becuase it could explain how the classical equations could be taken up in quantum mechanics unchanged, while the predictions became probabilistic. People made analogies with the case where you have a classical particle whose position and momentum are unknown, and obey the Heisenberg relation. Quantum mechanics is completely different in that the presence at different position states is parametrized by probability amplitudes, not probabilities. But the description is always on half the phase space variables. The uncertainty in position/momentum is directly analogous to the uncertainty in angular position/angular momentum, in the uncertainty of phase of a field mode and particle number of that mode, and in every other canonically cojugate pair. States of definite position are also uncertain energy, because energy and position do not commute, but nobody calls this a duality, because energy and position are not canonically conjugate. I should also point out the energy/time uncertainty principle, which is hard to think of in terms of canonical pairs in the usual formulations of QM, because particles don't have a time associated to them, but all particles have a global time. In Schwinger/Feynman particle formalisms this is taken care of, but the uncertainty relation can be worked out in any formalism of course. This uncertainty relation might not be called a duality by some, I don't know. - Good question! For the properties related by the uncertainty principle, there are two reasons why they come in pairs: 1. Intuitively, the uncertainty principle relates the variance of a function to the variance of its Fourier transform. And, up to a couple of numerical factors, the Fourier transform of a Fourier transform is the original function. (Mathematicians will balk at this statement because it's not technically true, but conceptually it's accurate enough for my purposes here.) So the process of Fourier conjugation leads you through a cycle of two functions. 2. Mathematically, the uncertainty principle is based on commutators, namely that $$\sigma_A \sigma_B \ge \biggl\|\frac{1}{2i}\langle[A,B]\rangle\biggr\|$$ and it doesn't really make as much sense to compute the commutator of three or more operators since there are multiple ways in which you can rearrange them. - 1 indeed. I agree mostly with you, but still one can avoid to think that for instance, we also have vector spaces and their duals, but have not more complex tuples of vector spaces. What is called again an algebraic object with ternary operators beside the usual binary operators that define groups? we don't have it because we don't think in those terms, not necessarily because it might not make sense – lurscher Dec 14 '11 at 19:32 I agree with the first of the arguments. As for the second, one could easily generalize commutators to n tuples of operators in a manner similar to antisymmetric tensors (i.e. sum over all permutations employing permutation sign). See Levi-Civita symbol. – Adam Zalcman Dec 14 '11 at 20:10 exactly, maybe there is some generalization out there for 'Lie thing of ternary operators' using such symbols as infinitesimal generators, but maybe such math is too weird to fit in our thinking models – lurscher Dec 14 '11 at 20:53 Yeah, I've seen things like this on occasion (like the Nambu bracket Qmechanic mentioned), but they're not as generally useful as the commutator. Albeit perhaps only because the math is more complicated and less intuitive. – David Zaslavsky♦ Dec 16 '11 at 20:35 1) There are already many good answers explaining the conventional theory and observations. Nevertheless, related to comments by lurscher and Adam Zalcman, it seems appropriate to mention the Nambu bracket, which is a Poisson-like bracket $$\{ f,g,h \}$$ with 3 function entries, originally invented by Nambu in 1973, purportedly in a failed attempt to explain the $SU(3)$ symmetry of quarks. Already Nambu discusses an operator 3-bracket $$[\hat{f},\hat{g},\hat{h}],$$ and one can imagine some kind of uncertainty relation associated to this, where canonical variables come in triples. Unfortunately, the subject so far has remained just theoretical speculations. [Authors even don't agree what should replace the Jacobi identity for the Poisson bracket $\{ f,g\}$, although most think it should be the so-called Filippov fundamental Identity (FI).] 2) Recently in 2008, the Nambu-bracket has been used in the Bagger–Lambert–Gustavsson M2 brane proposal. 3) The exist various generalizations to higher Nambu brackets $$\{ f_1,\ldots ,f_n \}$$ with $n$ entries. - I think this has to do with the fact that one is the infinitesimal generator of the other (i.e p is the generator of position). If I remember correctly, commutators arise naturally when we generate a Lie Algebra (not entirely sure about this). - That's not necessarily the case. For instance, besides position and momentum, there are also other observables like energy (and no, time is a parameter and not an operator). More strikingly, for spin, there is no natural pair. There is an angular momentum operator corresponding to the component in each spatial direction, but directions don't come naturally in pairs. - The generalized uncertainty principles relates any two observables (see eg the Schrödinger uncertainty relation). It probably can be further generalized to more than two observables, possibly by starting from $$\langle x,x\rangle\langle y,y\rangle\langle z,z\rangle \geq \frac13 \left(\langle x,x\rangle\|\langle y,z\rangle\|^2 + \langle y,y\rangle\|\langle x,z\rangle\|^2 + \langle z,z\rangle\|\langle x,y\rangle\|^2\right)$$ instead of straight Cauchy-Schwarz in case of three observables, but this is just a guess on my part - I did not check any literature. Related to this are canonically conjugate observables like $x,p_x$, for which the classical uncertainty principle holds because they are governed by the commutation relation $$[x,p_x]=i\hbar$$ corresponding to the fundamental Poisson bracket $$\{x,p_x\}=1$$ of classical mechanics. However, these observables are not ordinary: In particular, the observable $x$ corresponds to a coordinate of the underlying configuration space. In classical mechanics, the existence of a conjugate momentum follows from the symplectic structure of phase space, in quantum mechanics, it follows from the properties of the Fourier transform on our state space $L^2$. - It's because physical observables are operators in quantum mechanics. For two operators $F,G$, one may always compute the commutator $$[F,G] = FG-GF$$ which is the simplest quantity that should be zero in classical physics but it is nonzero in quantum mechanics. In general, the commutator is another observable (it may be a $c$-number, too). In the leading approximation, the commutator is equal to $i\hbar$ times the classical "Poisson bracket" of the classical observables with the same name. But if you define your classical initial state by the values of observables $Q_i$ (includes both positions and velocities or momenta in the familiar case), you may study configurations close to $Q_i$. In this way, you linearize the phase space and the commutator $$[Q_i,Q_j] = \omega_{ij}$$ may be approximated by a constant antisymmetric matrix that I called $\omega$. (It becomes a nonconstant function of $Q_i$ if you allow large deviations from the predetermined point.) It defines the so-called symplectic structure on the phase space. In this linearized approximation where the commutator is constant, the symplectic structure may always be diagonalized by choosing better combinations of $Q_i$, so that $\omega$ becomes a block diagonal matrix composed of $$\left ( \begin{array}{rr}0&+1\\-1&0\end{array} \right )$$ blocks, and these blocks guarantee that the observables defining the initial conditions - coordinates on the phase space - may be divided to pairs. In a simpler way, your question is really "because $\omega_{ij}$ has two indices". The rearrangement of a larger number of operators, e.g. $XYZT-YTZX$, may always be reduced to transpositions of the neighbors i.e. to commutators of two operators. That's why the ordinary commutators are always more important. In a group-theory context of mathematics, the commutators define the so-called "Lie algebra". Generalizations of commutators with more than 2 objects in the brackets also exist but they seem to be much less relevant physically. In particular, they're not naturally defined for operators - and physical observables become operators in QM. - [Peter: You asked this some time ago, but it's just too delightful of a question to pass over without adding my two conjugated cents... :) You already have several good answers (I liked David Zaslavsky's in particular), so this one will be concept-heavy and math-light.] One ways to visualize why quantum uncertainty occurs in pairs is to picture the universe as having two spaces that share a symmetric, two-way relationship. The first of these spaces is the one we call spacetime, consisting of xyz (space) plus t (time). For brevity I'll call this xyzt space "length space", and will refer to it even more briefly as just x. Length space is of course the space we know best, and is the space in which most classical physics phenomena take place. The second of spaces is technically an abstraction, but in terms of measurable effects it's about as real of an abstraction as you are likely to encounter. It's called momentum space, and it too has four dimensions: $p_{x}p_{y}p_{z}$, the momentum axes, plus $E$, the energy axis. I'll call $p_{x}p_{y}p_{z}E$ "momentum space," and will also refer to it as just p. Momentum and energy are of course very much part of classical mechanics. However, in classical mechanics, x dominates over p in much the same way that air dominates over water in a cloud by breaking up the water into many tiny and isolated droplets. This structural domination of x over p means that p shows up mostly in the dynamic properties of how large object move and interact in x, rather than as a space. However, for objects that are very small, very low in energy, or very low in momentum, that situation changes. For example, there are cases where ordinary matter can "condense" or form fluids that are most easily described as residing in momentum space. You might thing that such bizarre fluids would be rare and exotic, but that’s not the case at all. For example, if you look in a mirror or at a shiny piece of metal you are looking directly at the surface of a type of p space condensate called a Fermi sea. The Fermi sea is composed of conduction electrons, and only electrons at the very top of that sea are capable of reflecting light. It's often useful to think of p as approximately the place where the odd behaviors of quantum mechanics take place. Why that is I'll explain below, since it's directly related to your question about quantum uncertainty. Now to the meat of your question: For reasons I think is best described as the way our universe happens to works, there exists an extraordinarily deep and mostly symmetric relationship between x and p. I can describe that relationship this way: Each point in each space x and p behaves a tuned radio receiver for some frequency along the equivalent axis in the other space. For example, picture the $p_x$ axis of p as an old-style analog radio tuner, the kind with a little red dot indicating the frequency you have selected. Sliding the red dot up to the position labeled “103.5 megahertz” would create in a sinusoidal wave with just that frequency over on the $x$ axis of x. That infinitely long sinusoidal wave then becomes another way to "view" the red dot, since each precisely defines the other. That’s all well and good, but it’s also important to realize that this works both ways. That is, if you instead decide to select a “dial point” along the length space axis $x$, that too will select a pure frequency along the corresponding axis of the other space, that is, along the $p_x$ of p. That's a nice symmetry! It's a symmetry with some profound consequences. Remember, if you select a compact point in p you get an very long sinusoidal frequency over in x, and vice-versa. So let's say you don't want to have your particle scattered about so much in x, and you decide to pull it all in to one compact point in x. Oops! The moment you do that, the symmetric relationship causes the particle's representation in p to explode into an indefinitely long sinusoidal wave. Now, since x corresponds to length (and time) and p to momentum (and energy), what does this curious see-saw relationship remind you of? If you answered quantum uncertainty, you are exactly correct. In fact, that is what quantum uncertainty is, at its deepest levels: the impossibility of representing the same particle compactly in both x and p space at the same time, due to their symmetric "tuning" relationship to each other. The radio-dial like frequency tuning of these spaces has a more precise name: It's called a Fourier transform. So, if you have an old-style radio receiver and use it to move a dot back and forth along a linear scale, you can correctly say that you are doing a Fourier transform from frequency of the radio wave to the linear position on your radio dial. So, here at last is the answer to your original question quantum uncertainty always seems to express itself in pairs, rather than in triples or some other combination. The answer is that uncertainty is defined by pairs of spaces connected in both directions by Fourier transforms, which in turn just seems to be how our universe is constructed. To create a quantum uncertainty triplet, you would need to postulate a new universe in which there was a three-way Fourier-like relationship, presumably extending across three parallel spaces. Ouch! That’s not intuitive at all. But also, it’s not necessarily impossible. Some of the most interesting and important insights in both mathematics and physics have come about from people trying to do nominally “impossible” things that turned out to have interesting and non-trivial solutions. Dirac’s prediction of antimatter certainly comes to mind, since in that case he dug in and got stubborn about wanting only “linear” solutions for some equations that didn’t have linear solutions. By using complex matrix mathematics, he finally found a way to do it -- and simultaneously uncovered one of the deepest equations in physics. His instincts, and his stubborn insistence in following through on them, ended in a unique and spectacular success. (And warning to the wise: Very few people indeed have physics instincts like Dirac did, and even Dirac got stubborn in some quite wrong-headed ways in his later years. So if anyone wants to try exploring ideas like these for fun, please use a bit of caution on adopting the "stubborn” part of the Dirac insight recipe!) And finally: I said earlier that x and p are not exactly symmetric. What did I mean by that? The back-and-forth Fourier relationship is exactly symmetric, as best we can tell. So where does the breakdown in symmetry come in? The main break is this: While adding size or distance costs nothing x (length) space, adding size in p (momentum) space turns out to be very costly. This means that there is no problem with placing a particle very precisely in p, even if it results in an absolutely enormous (e.g. light years across) wave function in ordinary x space. Since adding more size to the x wave function costs nothing, it can just keep spreading until something disrupts it. If instead you try to stuff the particle into a very precise location in x, the wave function in p again become enormous. However, since momentum cost energy, a large wave function in p quickly becomes hugely expensive in terms of the mass-energy needed to sustain it. That’s one reason why particle accelerators have to pile so much energy into their particles. The extra energy expands their energy-intensive p space wave until they are large enough to make their sizes in ordinary x space very small indeed. Such additions of energy can for example be used to locate electrons precisely enough in space for them to "see" the motions of individual quarks inside protons and neutrons. So, massive overkill, but again: That was a good question. And who knows? Perhaps some clever mathematically inclined person will someday come up with an unexpected n-space generalization of the Fourier 2-space relationship that seems to guide our universe. As the old saying goes, the only sure way to fail is never to try. -	{"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": 10, "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": 0.9441651701927185, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/151665/integral-with-4-radicals-hat	# Integral with 4 radicals-hat I'd like to find out a simple way for calculating the value of: $$\int_{0}^{1}\sqrt{1+\sqrt{1 + {\sqrt{1+ \sqrt{x}}}}}\,dx .$$ Of course, I thought of some variable change, but it seems pretty complicated. On the other hand, I wonder if there can be made a generalization when having to deal with the expression with $k$ radicals, $k>1$. - as $k \to \infty$ it seems like the integral goes to $\phi$ – picakhu May 30 '12 at 17:36 1 surprisingly, this is Wolfram integrable. – user20266 May 30 '12 at 17:39 1 Nothing surprising - just substitute $t=\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}$. – Norbert May 30 '12 at 17:42 (with 3 ones) After that substitution, Maple gets $$\frac{16 t^{17}}{17} - \frac{112 t^{15}}{15} + \frac{288 t^{13}}{13} - \frac{320 t^{11}}{11} + \frac{112 t^{9}}{9} + \frac{48 t^{7}}{7} - \frac{32 t^{5}}{5}$$ – GEdgar May 30 '12 at 17:50 @ GEdgar: i suppose that things get worse when dealing with more radicals in place. – Chris's wise sister May 30 '12 at 17:52 show 7 more comments ## 2 Answers 1. Let $$\begin{eqnarray} u &=&\sqrt{1+\sqrt{1+\sqrt{x}}} \Leftrightarrow &x=\left( \left( u^{2}-1\right) ^{2}-1\right) ^{2}=u^{8}-4u^{6}+4u^{4}. \end{eqnarray}$$ Since $$\begin{equation} dx=\left( 8u^{7}-24u^{5}+16u^{3}\right) du \end{equation}$$ we have $$I :=\int_{0}^{1}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}dx\\=\int_{\sqrt{2}}^{\sqrt{1+\sqrt{2}}}\sqrt{1+u}\left(8u^{7}-24u^{5}+16u^{3}\right) du.\quad\textit{(computation below)}^†$$ Each term can be integrated using the substitution $t=\sqrt{1+u}$ $$\begin{equation} \int_{a}^{b}\sqrt{1+u}u^{n}du=2\int_{\sqrt{1+a}}^{\sqrt{1+b}}t^{2}\left( t^{2}-1\right) ^{n}\,dt,\quad a=\sqrt{2},b=\sqrt{1+\sqrt{2}}. \end{equation}$$ 2. Generalization to $k=5$ radicals $$\begin{equation} J:=\int_{0}^{1}\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}}dx. \end{equation}$$ Similarly to above the substitution is now $$\begin{eqnarray} v &=&\sqrt{1+\sqrt{1+\sqrt{1+\sqrt{x}}}}\Leftrightarrow x=\left( \left( \left( v^{2}-1\right) ^{2}-1\right) ^{2}-1\right) ^{2} \\ x &=& v^{16}-8v^{14}+24v^{12}-32v^{10}+14v^{8}+8v^{6}-8v^{4}-1, \end{eqnarray}$$ and $$\begin{equation} dx=\left( 16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv. \end{equation}$$ Hence $$\begin{eqnarray} J &=&\int_{\alpha }^{\beta }\sqrt{1+v}\left( 16v^{15}-112v^{13}+288v^{11}-320v^{9}+112v^{7}+48v^{5}-32v^{3}\right) dv \\ \alpha &=&a=\sqrt{1+\sqrt{2}},\beta =\sqrt{1+\sqrt{1+\sqrt{2}}}. \end{eqnarray}$$ -- †In SWP I obtained $$\begin{eqnarray} I &=&-\frac{26\,704}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{\sqrt{2}+1} \sqrt{2} \\&&+\frac{83\,584}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{\sqrt{2}+1} \\ &&+\frac{344\,096}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}} \\ &&+\frac{67\,328}{109\,395}\sqrt{\sqrt{2}+1} \\ &&-\frac{256}{3003}\sqrt{\sqrt{2}+1}\sqrt{2} \\ &&-\frac{17\,168}{765\,765}\sqrt{1+\sqrt{\sqrt{2}+1}}\sqrt{2} \\ &\approx &1.584\,9. \end{eqnarray}$$ - Hints: 1) First substitute $$\,t=\sqrt{1+\sqrt{x}}\Longrightarrow dt=\frac{dx}{4\sqrt{x}\sqrt{1+\sqrt{x}}}\Longrightarrow dx=4(t^2-1)tdt$$ , and now change the limits to $\,1\,,\,\sqrt{2}$ 2) Next, you have $$4\int_1^{\sqrt{2}}\,t(t^2-1)\sqrt{1+\sqrt{1+t}}\,dt$$ , and now substitute $$y=\sqrt{1+\sqrt{1+t}}$$and etc. You end up with a not-so-terrible polynomial function. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 8, "mathjax_display_tex": 14, "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": 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": 0.8987991213798523, "perplexity_flag": "middle"}
http://stats.stackexchange.com/questions/40859/combination-of-2-random-variables-that-is-perfectly-correlated-with-another-rand	# Combination of 2 random variables that is perfectly correlated with another random variable I'd appreciate some pointers for the following question: Given 3 random variables $X_i$, $i = 1, 2, 3$ such that none of them have a pairwise correlation of $+1$ or $−1$. Can we find a combination of $X_1$ and $X_2$ that has a correlation of $+1$ with $X_3$? - This is going to depend on the nature of the underlying relationships amongst these variables. BTW, is this homework? If so, please edit to add the homework tag. – gung Oct 21 '12 at 21:23 ## 1 Answer Yes, it's possible, but not likely to occur in real life. An artificial example ````x1 <- rnorm(100) x2 <- rnorm(100) x3 <- x1+x2 cor(x1, x2) cor(x1, x3) cor(x2, x3) ```` and, of course, since `x3 = x1 + x2`, the correlation between `(x1 + x2)` and `x3` is 1.0. If you had specified that the three variables are independent, I don't think it's possible, but I am not 100% sure. - 4 If $X_1$ and $X_2$ are uncorrelated with $X_3$, the linear combination $aX_1+bX_2$ is uncorrelated with $X_3$ since covariance is bilinear. – Douglas Zare Oct 21 '12 at 11:02 The question didn't say $X_1$ and $X_2$ were uncorrelated with $X_3$, just that the correlations weren't 1.0 or -1.0. – Peter Flom Oct 21 '12 at 11:13 Yes, I was referring to your last line, which talks about what might happen if the variables are pairwise independent, hence uncorrelated. – Douglas Zare Oct 21 '12 at 11:26 Oh, sorry. Cool, thanks for that. That was my intuition, but it's nice to know I was right. :-). – Peter Flom Oct 21 '12 at 11:30	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 16, "mathjax_display_tex": 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": 0.9611327052116394, "perplexity_flag": "middle"}
http://mathhelpforum.com/differential-geometry/162168-recursive-sequence-convergence-print.html	Recursive Sequence Convergence Printable View • November 5th 2010, 01:10 AM valerian Recursive Sequence Convergence $x_0 > x_1$ and r and s are positive numbers with p + q = 1. $x_n = px_{n-1} + qx_{n-2}$ for all $n \geq 2$. Prove that $(x_n)$ is convergent and find limits in terms of $x_0, x_1$, p and q. Okay I began by letting p = 1 - q so I could eliminate that variable. Then I began working it out up to $x_7$ but I couldn't find any pattern in terms of coefficients of the polynomial. It's a polynomial with respect to p. I am trying to get this to work out to something that I can put into summation notation, prove it is monotone, and then find an upper and/or lower bound but I am unsure how to go about putting it in summation notation - it's entirely possible this is not the way to do it but it seems most logical to me. The first term (in order of descending exponent on p) of the expansion is always $p^{n-1}a_1$. The last term is $a_0$ when n is even and $a_1$ when n is odd. These are the only real patterns I've been able to discern though. Any insight is much appreciated. Thanks. • November 5th 2010, 01:47 PM TheEmptySet Here is an idea for you The sequence can be written as follows $\begin{bmatrix}0 && 1 \\ (1-p) && p \end{bmatrix}\begin{bmatrix} x_{n-2} \\x_{n-1}\end{bmatrix}=\begin{bmatrix} x_{n-1} \\x_{n}\end{bmatrix}$ This matrix can be diagonalized and solved explicitly For the two eigenvalues I got $\lambda = p-1 \text{ or } \lambda =1$ I hope this helps. • November 5th 2010, 02:20 PM chisigma Quote: Originally Posted by valerian $x_0 > x_1$ and r and s are positive numbers with p + q = 1. $x_n = px_{n-1} + qx_{n-2}$ for all $n \geq 2$. Prove that $(x_n)$ is convergent and find limits in terms of $x_0, x_1$, p and q. Okay I began by letting p = 1 - q so I could eliminate that variable. Then I began working it out up to $x_7$ but I couldn't find any pattern in terms of coefficients of the polynomial. It's a polynomial with respect to p. I am trying to get this to work out to something that I can put into summation notation, prove it is monotone, and then find an upper and/or lower bound but I am unsure how to go about putting it in summation notation - it's entirely possible this is not the way to do it but it seems most logical to me. The first term (in order of descending exponent on p) of the expansion is always $p^{n-1}a_1$. The last term is $a_0$ when n is even and $a_1$ when n is odd. These are the only real patterns I've been able to discern though. Any insight is much appreciated. Thanks. The sequence is defined by the 'recursive relation'... $\displaystyle x_{n} - p\ x_{n-1} - q\ x_{n-2}=0$ (1) ... that is 'linear and homogeneous' and the solution of which is... $\displaystyle x_{n} = c_{1}\ \alpha_{1}^{n} + c_{2}\ \alpha_{2}^{n}$ (2) ... where $c_{1}$ and $c_{2}$ are 'arbitrary constants' that depend from $x_{0}$ and $x_{1}$ , $\alpha_{1}$ and $\alpha_{2}$ the solution of the 'characteristic equation'... $\displaystyle \alpha^{2}- p\ \alpha - q= 0 \implies \alpha= \frac{p \pm \sqrt{p^{2}+4\ p -4}}{2}$ (3) Now for $0< p= 1-q <1$ is $\|\alpha_{1}\|<1$ and $\|\alpha_{2}\|<1$ , so that in any case is... $\displaystyle \lim_{n \rightarrow \infty} x_{n} = 0$ (4) Kind regards $\chi$ $\sigma$ • November 5th 2010, 02:24 PM Also sprach Zarathustra Put x_n=m^n... and try to solve the quadratic equation... (hmmm...) [EDIT: 4 MINUTES LATE] All times are GMT -8. The time now is 03:11 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": 35, "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": 0.9620489478111267, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?t=588821	Physics Forums New Reply Page 1 of 2 1 2 > ## Schrodinger equation for one dimensional square well 1. The problem statement, all variables and given/known data the question as well as the hint is shown in the 3 attachments 2. Relevant equations 3. The attempt at a solution i know how to normalize an equation, however i do not understand what the hint is saying, or how to do these integrals, any guidance would be greatly appreciated Attached Thumbnails 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 hint is just telling you that each Sin(a pi x) is orthogonal to the other Sins I'll give you my hint - all you need to know is how to integrate Sin^2(x) Do you know how to do the other two problems? im not sure exactly what you mean by that, what is the integral i need to take? ## Schrodinger equation for one dimensional square well Well, do you know what it would mean for $\psi(x)$ to be normalized? normalizing ψ(x) means to determine if ∫ \|Ψ(x)\|^2 dx = 1 right? Quote by stigg normalizing ψ(x) means to determine if ∫ \|Ψ(x)\|^2 dx = 1 right? That is correct Do you know what it means if two functions are orthogonal? no i do not Okay, orthogonality is what the hint is describing two functions, f and g are orthogonal if $\int f*g\ dx = 0$ the hint is just restating that each of the Sins are orthogonal to the other ones, that is; $\int Sin(\pi x)Sin(2 \pi x) dx = 0$ etc Using this, how do you think you should proceed in determining A? do i expand the equation and cancel out the terms with Sin(πx)Sin(2πx) because they will integrate to 0? Yes, show me what you get does ∫Sin(πx)Sin(3πx)dx=0 and ∫Sin(2πx)Sin(3πx)dx=0 ? yes, if you work the integral out yourself you'll find that $\int Sin[n \pi x] Sin[m \pi x] dx = 0$ if $m \ne n$ alright so then in that case i will only be dealing with the sin2 functions and therefore \|Ψ(x)\|^2= (1/10a)sin2($\pi$x/a)+(aA2/a)sin2(2$\pi$x/a)+(9/5a))sin2(3$\pi$x/a) \|Ψ(x)\|^2= (1/10a)sin2(πx/a)+(aA2/a)sin2(2πx/a)+(9/5a))sin2(3πx/a) That is not exactly true, the terms aren't zero on their own, what IS true is that $\int \|Ψ(x)\|^2dx=\int (1/10a)sin2(πx/a)+(aA2/a)sin2(2πx/a)+(9/5a))sin2(3πx/a)dx$ yes yes youre right i jumped the gun on that, now will the limits of my integral be from -a to a or from -$\infty$ to$\infty$ Quote by stigg yes yes youre right i jumped the gun on that, now will the limits of my integral be from -a to a or from -$\infty$ to$\infty$ Since this is the particle in a box problem, the potential outside of the box is set to infinity and so we make ψ = 0 everywhere outside, so it doesn't matter where you set the limits of integration (as long as the box is contained in them of course) since we pick up exactly 0 from the outside region. If you look at the problem statement, you'll see that the box isn't -a<x<a, it's 0<x<a ah yes my mistake so using 0 to a as the limits of integration i get that it is equal to (1/10a)+(9/5a)+(2A2/a) New Reply Page 1 of 2 1 2 > Tags quantum, quantum well, schrodinger equation, wave equation Thread Tools \| \| \| \| \|---------------------------------------------------------------------------\|-------------------\|---------\| \| Similar Threads for: Schrodinger equation for one dimensional square well \| \| \| \| Thread \| Forum \| Replies \| \| \| Classical Physics \| 0 \| \| \| Quantum Physics \| 5 \| \| \| Classical Physics \| 9 \| \| \| Quantum Physics \| 8 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 13, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8989626169204712, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/99699-length-given-curve.html	# Thread: 1. ## length of the given curve one of my HW problems is to find the arc length of the given curve for $x=t^2, y=(3^{1/2}4/3)t^{3/2}, z=3t, 1 <= t <= 4$ now i have the definite integral that goes from 1 to 4, with the equation $(4t^2 + 12t + 9) dt$. but how do I solve this integral? it seems too hard 2. Originally Posted by Arez one of my HW problems is to find the arc length of the given curve for $x=t^2, y=(3^{1/2}4/3)t^{3/2}, z=3t, 1 <= t <= 4$ now i have the definite integral that goes from 1 to 4, with the equation $(4t^2 + 12t + 9) dt$. but how do I solve this integral? it seems too hard $<br /> l=\int_1^4 \sqrt{\left(\frac{\partial x}{\partial t}\right)^2+\left(\frac{\partial y}{\partial t}\right)^2+\left(\frac{\partial z}{\partial t}\right)^2} \ dt \ = \ \int_1^4 \sqrt{4t^2+12t+9} \ dt \ = \ \int_1^4 (2t+3) \ dt<br />$	{"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": 5, "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": 0.9443800449371338, "perplexity_flag": "head"}
http://mathoverflow.net/questions/102013/is-the-springer-resolution-a-blow-up	## Is the Springer resolution a blow-up? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let's consider the Springer resolution of the nilpotent cone $\mathcal{N}$ of a complex semisimple Lie algebra $\mathfrak{g}$, which is $$\widetilde{\mathcal{N}}=T^*\mathcal{B}\rightarrow \mathcal{N}.$$ It is a resolution of singularity of $\mathcal{N}$. ( For the construction of Springer resolution you can see section 6 of "Geometric Methods in Representation Theory of Hecke Algebras and Quantum Groups" .) My question is: is it a blow-up along some subscheme of $\mathcal{N}$? Or for the simplest case, where $\mathfrak{g}=sl(2,\mathbb{C})$. Now $$\mathcal{N}= {x^2=yz} \in \mathbb{C}^3$$ In this case is $\widetilde{\mathcal{N}}$ the same as the blow-up at $0$? - 1 In the simplest case of $\mathfrak{sl}_2$, the Springer resolution is easily seen to be the blow-up at the origin. In general, however, the Springer resolution is not a blow-up. – Mike Skirvin Jul 12 at 6:12 Why is it not? I think it is not too but I cannot think of a reason why... – Bugs Bunny Jul 12 at 10:11 5 A projective birational morphism $X\to Y$ (e.g., the Springer resolution) is the blow-up of $Y$ along some (non-unique) subscheme; see Hartshorne. But it can be hard to identify such a subscheme. The Springer resolution is an example where nature provides a resolution where it is hard to find an appropriate subscheme. Others are: the Hilbert scheme of $0$-dimensional subschemes of a smooth surface, resolving the corresponding symmetric product; determinantal varieties; theta divisors on Jacobians; vector bundles over homogeneous spaces $G/P$ collapsing to some cone in a representation of $G$. – inkspot Jul 12 at 12:46 1 The result from Hartshorne which inkspot is referencing is Theorem II.7.17. So $\widetilde{\mathcal{N}}$ is a blow up of $\mathcal{N}$, on very general grounds, but it is not yet clear to me whether we can take it to be the blow up of a subscheme supported on the singular locus. – David Speyer Jul 12 at 14:04 1 David: iirc another, maybe more familiar, example of the same phenomenon is the small resolution of the 3-fold ODP (as features in the Atiyah flop). Here there are non-Cartier Weil divisors through the singular point, and blowing up such a Weil divisor produces a small resolution. (Sorry for the off-topic comment.) – Artie Prendergast-Smith Jul 12 at 15:07 show 3 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": 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": 0.8781797289848328, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/229710/what-is-the-composition-of-relations-like-this-with-no-transitive-relations-bet	# What is the composition of relations like this? With no transitive relations between them? Given $R_1 = \{(1,2),(5,3)\}\quad\quad R_2 = \{(6,4),(5,7)\}$ What is $R_2 \circ R_1$? Because in my understanding, using the example $R_3 = \{(1,2),(3,4)\} \quad\quad R_4 = \{ (2,5),(6,7)\}$ Then $R_4\circ R_3 = \{(1,5)\}.\;$ Am I correct? So given that case, if $R_2 \circ R_1 = (a,b),\;a$ comes from $R_1$, right? While $b$ comes from $R_2$? - ## 1 Answer We have by definition (assembled from your example) $$R_2 \circ R_1 = \{(a,b) \mid \exists c : (a,c) \in R_1, (c,b) \in R_2\}$$ In the $R_4 \circ R_3$ example $(1,5) \in R_4 \circ R_3$ as we can "plug in" $2$ in the middle where $(1,2) \in R_3$, $(2,5) \in R_4$. If there is no "matching" pair from $R_1$ and $R_2$ as it is in your case, there is no $(a,b)$ fulfilling the condition for being a member of $R_2 \circ R_1$, we have by definition $R_2 \circ R_1 = \emptyset$. -	{"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": 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": 0.9691492915153503, "perplexity_flag": "head"}
http://mathhelpforum.com/math-topics/124251-expanding-brackets.html	# Thread: 1. ## multiplication hi, i need to expand these brackets , i know the answer will be 20 but i need to know what = 2. Originally Posted by andyboy179 hi, i need to expand these brackets , i know the answer will be 20 but i need to know what = $(2-4\sqrt{3})(2+4\sqrt{3}) \Rightarrow 2^2 - (4\sqrt{3})^2 \Rightarrow 4-48 = -44$ $-2\times4\sqrt{3} = -8\sqrt{3}$ Edit:apparently I didn't go down the right path with this, as per definition of "expansion" 3. at school we are told to work it out like this, (the blank spaces are the ones i don't know) 4. RULE: (x + y)(x - y) = x^2 - y^2 By the way, that's not "expanding brackets": it's a multiplication. 5. Originally Posted by andyboy179 at school we are told to work it out like this, (the blank spaces are the ones i don't know) BUT 4sqrt(3) times 4sqrt(3) = 48, not 24 6. Originally Posted by Wilmer BUT 4sqrt(3) times 4sqrt(3) = 48, not 24 oh ye, i know what i did wrong! so what would -2 x 4sqrt(3)= ? 7. Originally Posted by andyboy179 oh ye, i know what i did wrong! so what would -2 x 4sqrt(3)= ? Doesn't matter, because you also need to do +2 x 4sqrt(3), so the 2 will result in zero. 8. Originally Posted by andyboy179 at school we are told to work it out like this, (the blank spaces are the ones i don't know) The method that you use in school, although correct, is not the method used by most (and when I say most-I mean almost all) institutions. A really good explaination of how these problems are done can be found here. algebra.help -- Simplifying using the FOIL Method BTW: Your method is not an "expansion" in the normal sense of the word, as someone has already mentioned. THe term expansion implies that there will be more terms after the mutiplication is complete (in most cases). And that thes terms may have likenesses.	{"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": 0.9610018730163574, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/17517/how-do-formulas-get-figured-out	# How do formulas get figured out? Many of the formulas I learn in school are derived from more basic formulas — as long you your math is right and assuming the more basic formulas you used are correct, you are bound to get to a correct result... But how do people 'invent' the elementary formulas, like $F=ma$; how can you be sure that that's really the right equation for force? - Many of them are experimental relationships (fits). – Vladimir Kalitvianski Nov 27 '11 at 18:26 It started with $F \propto a$ and then $m$ was chosen so that this works. But the first is experimental. And as you see in relativity, formulas are only good to some point, not perfect. – queueoverflow Nov 27 '11 at 18:50 Both of those could be posted as answers... – David Zaslavsky♦ Nov 27 '11 at 23:34 – Ron Maimon Nov 28 '11 at 8:55 ## 2 Answers The way in which formulas are figured out is through a process which is difficult to describe, because it only happens once per formula, and it is different when the formula can be deduced in some way, vs when it is a purely new relation between quantities which have not been related before. The fundamental laws of a new regime of physics is never determined purely empirically, it always requires a principle of some sort, in addition to the empirical data, which serves to suggest empirical laws, and to weed out false theories. One of the best examples of a physical argument which produces a new equation is Einstein's argument for $E=mc^2$, described in my answer to this question: Did Einstein prove $E=mc^2$ correctly? The most important tools which physicists have are the symmetry principles, these can be used to deduce many classical relations. But these are not enough, and there is no general way to describe how to reason about physics to derive new laws--- it must be learned through examples. There are many illuminating examples, but perhaps the best is Heisenberg's 1925 derivation of quantum mechanics. I find this one to be most strange, the most insightful, and requiring the greatest leaps of imagination. It is described on the Wikipedia page for Matrix Mechanics. The equation you give is not the best example, because you can always say that it is the definition of force. This is not correct, because the force is implicitly assumed to be a physical property of the configuration of the particles, not of the particular particle trajectory, but in order to get rid of the circularity completely, let us assume you know statics, so that you have Archimedes definition of force. Archimedes demonstrates that you can think of gravity as producing a force on a lever proportional to the mass. Then you know from Galileo's experiments that the acceleration is constant in gravity, and you know from Galileo's arguments that there is a principle of Galilean relativity. You deduce that the force should be proportional to the acceleration, which is invariant under Galilean boosts, and because all the acceleration is at the same rate, that the Archimedes notion of gravitational mass is also the notion of inertial mass in F=ma. Something like this is probably what was going through the heads of the 17th century physicists. It is hard to tell, because F=ma was appreciated by everyone in Newton's time, not just by Newton. - 2 to add, dimensionality also plays an important role in determining that a formula is correct or not... – Vineet Menon Nov 28 '11 at 9:38 1 Dimensionality is only useful in regimes where you have at least one dimension, meaning away from quantum gravity regime. – Ron Maimon Nov 28 '11 at 17:44 They are figured out based on their ability to "economize thought," as Ernst Mach would say. To "economize thought" means to succinctly summarize the results of physical experiments or observations. Since there are many ways of "saving the phenomena" of experiments or observations, there are also many theories and thus many corresponding physics formulae. For example, consider the following three theories of gravity applied to planetary motion: 1. epicyclic theory 2. Newton's $F\propto1/r^2$ theory 3. Einstein's theory of General Relativity (GR) All three of these theories can explain, within certain limits, a given set of observations of the motions of the planets, but they all use completely different mathematical formulae: 1. The epicyclic theory basically uses a complex Fourier series (cf. this). 2. Newton's theory uses a simple algebraic equation. 3. GR uses tensors. Newton thought that his universal theory of gravitation, $F=Gm_1m_2r^{-2}$, was uniquely, exactly, and logically deduced from Kepler's observations, but this clearly is false because Kepler's observations showed perturbations from a perfect $1/r^2$ law due to the solar system being comprised of many masses. It is also false because, e.g., Einstein's GR theory superseded Newton's theory of gravitation. Thus, one physics theory (e.g., $p=mv$) is not more logically correct than another (e.g., $p=m+v$), although one might certainly be better at summarizing the results of experiments and observations than another. Physics formulae do not derive from mathematics like a geometric proof derives from Euclid's axioms. Physics formulae derive from observations and experiment; mathematics does not force a physics formula to be a certain way. For an excellent book on this whole topic, see The Aim & Structure of Physical Theory by the French physicist, historian, and philosopher of physics, Pierre Duhem. -	{"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": 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": 0.9502025246620178, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/122722/meaning-of-a-logical-operator?answertab=active	# Meaning of a Logical Operator Is it possible to know what those operator mean if they must be involved in this logicical condition? What is all the possible meaning of those two symbol if you don't know the symbol's meaning beforehand: (¬A) ⊕ A is always true, A ⊕ A is always false and with just those definition given, others unknown If it is not possible to get the outcome, how to prove it is insuffient? - 1 Insufficient data for meaningful answer. – Peter Tamaroff Mar 21 '12 at 0:00 How to prove this is insufficient? – Victor Mar 21 '12 at 0:01 1 – Peter Tamaroff Mar 21 '12 at 0:03 No, i think it could be of great useful example if you could prove it insufficient to get the outcome – Victor Mar 21 '12 at 0:04 @PeterT.off I don't think it is insufficient. "Given two rules specifying the operation of ⊕, find ⊕": that's a legitimate question. See my answer. – user2468 Mar 21 '12 at 0:10 ## 2 Answers Based on the description given in the question, we can build the following truth table: $$\begin{matrix} A & A & \| & ⊕ \\ \hline F & F & \| & F & \color{red}{\text{A ⊕ A}}\\ F & T & \| & T & \color{blue}{\text{¬A ⊕ A}}\\ T & F & \| & T & \color{blue}{\text{A ⊕ ¬A}}\\ T & T & \| & F & \color{red}{\text{A ⊕ A}} \\ \end{matrix}$$ Now, we can deduce $⊕$ is the exclusive or operation. Deduced directly from the rules stated in the question. Addednum: We can easily deduce the meaning of $\neg.$ Given the set $\mathbb{B} = \{ T, F \},$ any unary operator $\neg : \mathbb{B} \to \mathbb{B}$ will either operate as identity or as negation. Since the 2 equations in the given problem differ only by $\neg,$ $$A \oplus \neg A = T, A \oplus A = F,$$ we can deduce that $\neg$ can not be identity (assume $\neg A \equiv A$ and you'll get contradiction in the given system of formulas). Hence $\neg$ is negation. QED. Now we proceed as above. - i mean you don't even know the other symbol ¬ mean beforehand... – Victor Mar 21 '12 at 0:12 @Victor I can easily deduce the meaning of $\neg.$ Given the set $\mathbb{B} = \{ T, F \},$ any unary operator $\neg$ will either operate as identity or as negation. Since the 2 equations in the given problem differ only by $\neg,$ I can deduce that $\neg$ can not be identity, i.e., it is negation. – user2468 Mar 21 '12 at 0:19 @PeterT.off I've just posted a comment above this one answering this very same question. – user2468 Mar 21 '12 at 0:19 ## Did you find this question interesting? Try our newsletter The statement $A ⊕ B$ is true when either $A$ or $B$, but not both, are true. $A ⊻ B$ means the same. The statement $¬A$ is true if and only if $A$ is false.	{"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": 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": 0.9259545207023621, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/144256/sum-from-0-to-n-of-n-choose-i/144279	# Sum from 0 to n of $n \choose i$? [duplicate] Possible Duplicate: Algebraic Proof that $\sum\limits_{i=0}^n \binom{n}{i}=2^n$ Evaluation $\sum\limits_{k=0}^n \binom{n}{k}$ Is there a simple proof for this equality: $$\sum_0^n {n \choose i} = 2^n$$ thanks and sorry I forgot the basics - – Kirthi Raman May 12 '12 at 16:15 @J.M. Of course. This has been answered many times before. – Peter Tamaroff May 12 '12 at 16:22 MSE ought to incorporate some AI techniques to catch duplicate questions automatically. – Kirthi Raman May 12 '12 at 16:25 @KVRaman when you compose a question, it shows you some suggestions of (what it thinks) similar questions. But you're right. I can see some classification (machine learning) techniques applicable here. – user2468 May 12 '12 at 16:31 I don't usually answer a question that already has four answers including some very good ones, but I've posted an answer below that I think may be the simplest way of viewing the matter and is worth knowing about. – Michael Hardy May 12 '12 at 17:12 ## marked as duplicate by Peter Tamaroff, J. M., J.D., anon, Martin SleziakMay 12 '12 at 17:20 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. ## 5 Answers Recall the relation $\displaystyle{n+1\choose i}={n\choose i}+{n\choose i-1}$, valid for every $1\leqslant i\leqslant n$. Hence, $$\sum_{i=0}^{n+1}{n+1\choose i}=1+\sum_{i=1}^n\left[{n\choose i}+{n\choose i-1}\right]+1,$$ that is, $$\sum_{i=0}^{n+1}{n+1\choose i}=1+\sum_{i=1}^n{n\choose i}+\sum_{i=0}^{n-1}{n\choose i}+1=2\sum_{i=0}^n{n\choose i}.$$ The initial value $\displaystyle\sum\limits_{i=0}^0{0\choose i}=1$ completes the recursion. - for your efforts – ca11111 May 12 '12 at 16:28 $$(1+1)^n = \sum_{i=0}^n \begin{pmatrix} n \\ i \end{pmatrix}1^i 1^{n-i}.$$ But I do not know if this is as simple as you wish. - Thanks, added your suggestion! – Siminore May 12 '12 at 16:12 This is beautiful. – 000 May 12 '12 at 18:53 The standard combinatorial proof is that • The LHS counts the number of ways to choose $0$, $1$, $2, \ldots ,$ or $n$ things from a total of $n$ objects. • The RHS counts the number of ways to go through each of $n$ objects and mark them as "choose" or "don't choose". With a little thought, these are equal. An algebraic proof has been posted by Siminore. - There are many quite simple proofs. One of them is application of well known Newton binomial theorem: $$\sum_0^n {n \choose i} = \sum_0^n {n \choose i} 1^i 1^{n-i} = (1+1)^n = 2^n.$$ One can also prove this by combinatorial argument. Observe that ${n \choose i}$ is the number of subsets of cardinality $i$ of a set of cardinality $n$. Then $\sum_0^n {n \choose i}$ is the number of all subsets of cardinality $0, 1, 2, \dots, n$ of a set of cardinality $n$. Hence the sum counts all subsets of a $n$-set. But we know that thare are $2^n$ subsets of such set. - Here's a variation on the theme of Didier's answer. Each number in Pascal's triangle gets added twice to the row below it. The first $1$ below gets added to the next row to get the $1$ at the end, and also gets added to the next row to contribute to the $9$. Then the $8$ gets added to the next row to contribute to the $9$, and also gets added to the next row to contribute to the $36$. And so on. $$\begin{array}{ccccccccccccccccccc} & 1 & & 8 & & 28 & & 56 & & 70 & & 56 & & 28 & & 7 & & 1 \\ \\ 1 & & 9 & & 36 & & 84 & & 126 & & 126 & & 84 & & 36 & & 9 & & 1 \end{array}$$ Since each number is added twice to the next row, the sum of the numbers in the next row is twice as big. -	{"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": 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": 0.9499462246894836, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/273198/how-do-i-prove-if-f-n-is-a-sequence-of-extended-real-measurable-functions	# How do i prove if $\{f_n\}$ is a sequence of extended-real measurable functions, then $\sup_n f_n$ is measurable Rudin-RCA p.15 Let $X$ be a measurable space. Let $\{f_n\}$ be a sequence of extended-real measurable functions on $X$. How do i prove that $\sup_n f_n$ is measurable? Rudin uses a criterion to prove this, that is, if for every real $\alpha$, $f((\alpha,\infty])$ is measurable, then $f$ is measurable. I don't understand why this is sufficient to prove this.. Help me - 3 I just proved it, sorry for disturbing.. – Katlus Jan 9 at 1:06 2 You can answer your own question and even accept your own answer so the question doesn't go unanswered in the system. – Clayton Jan 9 at 1:18	{"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": 0.9331079125404358, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/62670-center-mass-problem.html	# Thread: 1. ## Center of Mass problem Let $y=9-x^2, y=0, x=0$ Find a number h such that y=h would cut the region into two equal areas. Now, I found out that the center of mass is (1,3.6), but how do I find the line? Do I just find the line that cut across (0,0) and (1,3.6)? The problem is the hint says that I need to integrate the area with respect to y, but I don't know how that is going to help. Thanks! 2. What you really seach for is the mean of y. The definition of the mean is $\frac{\int yf(y)dy}{\int f(y)dy}$* In your case $h = \frac{\int_0^9 y\sqrt{9-y}dy}{\int_0^9 \sqrt{9-y}dy}=\frac{324}{5}\frac{1}{18}=\frac{18}{5}$ You must do the top part using part integration. $\int y\sqrt{9-y}dy = -\frac{12}{5}(9-y)^{3/2}-\frac{2}{5}y(9-y)^{3/2}$ 3. Originally Posted by tttcomrader Let $y=9-x^2, y=0, x=0$ Find a number h such that y=h would cut the region into two equal areas. Now, I found out that the center of mass is (1,3.6), but how do I find the line? Do I just find the line that cut across (0,0) and (1,3.6)? The problem is the hint says that I need to integrate the area with respect to y, but I don't know how that is going to help. Thanks! The area between $y=0$ and $y=h \le 9$ is: $A(h)=\int_{y=0}^h \sqrt{9-y}\ dy=\frac{2}{3}[(9-h)^{3/2}-27]$ So the total area is $A(9)=18$ , so we look for $h$ such that $A(h)=9$ CB 4. Originally Posted by vincisonfire What you really seach for is the mean of y. The definition of the mean is $\frac{\int yf(y)dy}{\int f(y)dy}$* In your case $h = \frac{\int_0^9 y\sqrt{9-y}dy}{\int_0^9 \sqrt{9-y}dy}=\frac{324}{5}\frac{1}{18}=\frac{18}{5}$ You must do the top part using part integration. $\int y\sqrt{9-y}dy = -\frac{12}{5}(9-y)^{3/2}-\frac{2}{5}y(9-y)^{3/2}$ For this to be correct would require that median = mean (that is we seek the median not the mean and these are not in generaly equal) CB 5. I used to do these problems by simply finding the bounds of the area of integration - in this case you need to know the y min and max values of the region. Then you just say $\int_{a}^{h} f(x)dx = \int_{h}^{b} f(x)dx$, where a and b are your min and max. This is essentially the same thing that CB suggested, but to me it is easier this way.	{"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": 15, "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": 0.9562504291534424, "perplexity_flag": "head"}
http://mathforum.org/mathimages/index.php?title=Henon_Attractor&oldid=8310	Henon Attractor From Math Images Revision as of 15:13, 10 July 2009 by Mkelly1 (Talk \| contribs) Henon Attractor Fields: Dynamic Systems and Fractals Image Created By: Piecewise Affine Dynamics Website: Lozi Maps Henon Attractor This image is a Henon Attractor (named after astronomer and mathematician Michel Henon), which is a fractal in the division of the chaotic strange attractor. Basic Description The Henon Attractor is a special kind of fractal that belongs in a group called Strange Attractors, and can be modeled by two general equations. The Henon Attractor is created by applying this system of equations to a starting value over and over again and graphing each result. Making the Henon Attractor Say we took a single starting point (x,y) and plotted it on a graph. Then, we applied the two Henon Attractor equations to the initial point and emerged with a new point that we graphed. Next, we took this new point and again applied the two equations to it and graphed the next new point. If we continued to apply the two equations to each new point in a process called iteration and plotted every outcome from this iteration, we would create a Henon Attractor. Click here to learn more about iterated functions. Furthermore, if we plotted each outcome one at a time, we would observe that the points jump from one random location to another within the image. If you take a look at the animation, you can see the irregularity of the plotted points. Eventually, the individual points become so numerous that they appear to form lines and an image emerges. Magnification of the Henon Attractor 1X 8X 64X 512X If you magnify this image, you would find that the lines (really many, many points) that appear to be single lines on the larger image are actually sets or bundles of lines, that, if magnified closer, are bundles of lines and so on. This property is called self-similarity, which means that even as you look closer and closer into the image, it continues to look the same. In other words, the larger view of the image is similar to a magnified part of the image. History of the Henon Attractor Michel Henon was a French mathematician and astronomer who developed the Henon Attractor in the 1970s. At that time, Henon was interested in dynamical systems and especially the complicated orbits of celestial objects. The Henon Attractor emerged from Henon's attempt to model the chaotic orbits of celestial objects (like stars) in the mist of a gravitational force. A More Mathematical Explanation Note: understanding of this explanation requires: *Algebra Fractal Properties [Show] [Hide] Zooming in on the Henon Attractor The Henon Attractor is often described as being similar to the Cantor Set. Let us zoom into the Henon Attractor near the doubled-tip of the fractal (as seen in the animation). We can see that as we continue to magnify the lines that form the structure of the Henon Attractor, these lines become layers of increasingly deteriorating lines that appear to resemble the Canter Set. The Fractal Dimension of the Henon Attractor is not calculable using a single equation$D = \frac{log(n)}{log(e)}$, but it is estimated to be about 1.261. Chaotic System Original Henon Attractor, a = 1.4, b = 0.3 The Henon system can be described as chaotic and random. However, the system does have structure in that its points settle very close to an underlying pattern called a chaotic attractor. The basic Henon Attractor can be described by the equations, where $x_n$ is the x-value at the nth iteration. $x_{n+1} = y_n + 1 - ax^2_n$ $y_{n+1} = bx_n\,$ Astronomer Michel Henon created the original Henon Attractor using the values a = 1.4 and b = 0.3 and starting point (1,1). These are also the values used by the artist to create the featured image at the top of the page. However, by changing the values of a and b, we can obtain Henon Attractors that look slightly different. Changing "a" and "b" [Show] [Hide] Although the original Henon Attractor uses a = 1.4 and b = 0.3, we can alter those values slightly to produce various-looking Henon Attractors. However, the values of a and b are limited to a small range of values, outside of which the fractal ceases to resemble the Henon Attractor. Here are some more examples of Henon Attractors with different a and b values. a = 1; b = 0.542 a = 1.2; b = 0.3 a = 1.3; b = 0.3 (points need to be enlarged) a = 1.4; b = 0.3, original Henon Attractor a = 1.5; b = 0.2 a = 1.4; b = 0.1 Fixed Points [Show] [Hide] Original Henon Attractor with fixed points 1 and 2 Looking at the system of equations $x_{n+1} = y_n + 1 - ax^2_n$ and $y_{n+1} = bx_n\,$ that describe the fractal, the Henon Attractor uses only two variables (x and y) that are evaluated into themselves. This results in two equilibrium or fixed points for the attractor. Fixed points are such that if the two Henon Attractor equations are applied to the fixed points, the resulting points would be the same fixed points. In algebraic terms: $x_{n+1} = x_n\,$ and $y_{n+1} = y_n\,$ where $x_n$ is the x-value at the nth iteration and $x_{n+1}$ is the x-value at the next iteration. Therefore, if the system ever plotted onto the fixed points, the fractal would become stagnant. By solving the Henon Attractor's system of equations with a = 1.4 and b = 0.3, we can find that the fixed points for the original Henon Attractor are (0.6314 , 0.1894) and (-1.1314 , -0.3394). [Show Solving the System of Equations][Hide Solving the System of Equations] To solve the system of equations: $x_{n+1} = y_n + 1 - ax^2_n$ $y_{n+1} = bx_n\,$ Since $x_{n+1} = x_n\,$ and $y_{n+1} = y_n\,$, we can simplify the equations and refer to the variables as just $x$ and $y$, respectively $x = y + 1 - ax^2$ $y = bx\,$ By substituting the value of $y$ defined by the second equation into the $y$ in the first equation, we get $x = bx + 1 - ax^2$ Using the quadratic equation $x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $x = \frac{-(b-1) \pm \sqrt{(b-1)^2 - 4(-a)(1)}}{2(-a)}$ $x = \frac{-(b-1) \pm \sqrt{(b-1)^2 + 4a}}{-2a}$ Using a = 1.4, b = 0.3: $x = 0.6314, -1.1314 \,$ Using y = bx: $y = 0.1894, -0.3394 \,$ There are two types of fixed points, stable and unstable. The first fixed point (0.6314, 0.1894), labeled "1" on the image, is located within the attractor itself and is stable. This means that if a point is plotted close to the fixed point, the next iterated points will remain close to the fixed point. The second fixed point (-1.1314 , -0.3394), labeled "2", is considered unstable, and it is located outside of the bounds of the attractor. An unstable fixed point is such that if the system gets close to the fixed point, the next iterated points rapidly move away from the fixed point. Teaching Materials There are currently no teaching materials for this page. Add teaching materials. About the Creator of this Image Piecewise Affine Dynamics is a wiki site that was created by a group of French mathematicians that is dedicated to providing information about "dynamic systems defined by piecewise affine transformations". References Glenn Elert, The Chaos Hypertextbook Heinz-Otto Peitgen, Hartmut Jürgens, Dietmar Saupe, Chaos and fractals Bill Casselman, Simple Chaos-The Hénon Map www.ibiblio.org Henon Strange Attractors Michele Henon, Michele Henon Future Directions for this Page A better, less vague description of how sections of the Henon Attractor resembles the Cantor Set Also, the description of the Henon Attractor can be expanded to include a discussion about the fractal's "basin of attraction". For more information, click here. 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": 26, "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": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9212789535522461, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/208774/an-exercise-on-set-theory	# An exercise on set theory The type of exercise I'm about to speak of seems quite basic however I didn't have any exposure to it until recently so please provide me with some pointers on how to work it out. Show that if the sets A, B and C satisfy the following relations simultaneously: $A\cup B = C\\ (A\cup C) \cap B = C\\ (A \cap C) \cup B = A$ then they are the same. - 1 For example, it is possible to read $B\subseteq A\subseteq C\subseteq B$ from the given statements (how?), hence equality must hold. – Hagen von Eitzen Oct 7 '12 at 14:57 ## 2 Answers Recall that $X=Y$ if and only if $X\subseteq Y$ and $Y\subseteq X$. Now using this we have that: $B\subseteq A\cup B\subseteq C$ from the first line; and since $C=B\cap (\text{something})$ we have that $C\subseteq B$. From those we have that $B=C$. I am leaving it to you to show how $A=B$ as well, and to write a rigorous proof. - Thanks to Asaf Karagila's answer and my last Logics course I understood how to solve this type of exercise. I have attached my solution down below in a picture however I intend to rewrite this answer using latex equations when I'll get some free time. -	{"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": 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": 0.9675693511962891, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/231961/the-group-homomorphism-from-z-to-q	# The group homomorphism from Z to Q* Let F be non trivial group homomorphism F: Z -> Q. Want to prove that either Ker(F)={0} or Ker(F)= 2Z. Okay here is what i did; since i know that if there is homomorphism between two groups then there should be and isomorphism T such that, T: Z/Ker(F) -> Image(F). so i took Z/{0}=Z implies Z is isomorphic to Image(F). and i took Z/2Z={2Z, 1+2Z} is isomorphic to Image(F). and i take Z/3Z to be isomorphic to Image(F). Then since all three are isomorphic to Image(F) then they should be isomorphic to each other i.e. Z/3Z should be isomorphic to Z/2Z. (hence we arrive to a contradiction!). Is my method correct? If not then i need some directions. And once I'm done answering this question, can i deduce from it that (Q, x) is not cyclic? Because to show that we need to show isomorphism of Q* with Z.. - Your method is not correct because $F$ is different for all three homomorphisms, so the image of $F$ is different for all three. – Gerry Myerson Nov 7 '12 at 6:26 1 Note that the group of integers is generated by 1. Thus, if you know $F(1)$, you know $F$. Now think about how different values of $F(1)$ affect the kernel. – Gerry Myerson Nov 7 '12 at 6:28 ## 2 Answers Your method is not correct. It proves(?) for example that necessarily $\operatorname{Ker}F= \{0\}$. Here is a proof: Note that the only finite subgroups of $\mathbb{Q}^$ are $\{1\},\{1,-1\}$. Also the only subgroups of $\mathbb{Z}$ are $n \mathbb{Z}, \mathbb{Z}$ and $\frac{\mathbb{Z}}{n \mathbb{Z}}$ has $n$ elements. (Why ?) Now as you correctly said whatever $F$ is, it induce a group isomorphism $T:\frac{\mathbb{Z}}{\operatorname{Ker} F} \to \operatorname{Im}F$. Since $\operatorname{Im}F$ is a subgroup of $\mathbb{Q}^$, if $\operatorname{Im}F$ is finite we only have two possibilities for $\operatorname{Im}F$. But $F$ is non trivial so $\operatorname{Im}F \neq \{1\}$. So we left with only one possibility $\operatorname{Im}F = \{1,-1\}$. Therefore $\operatorname{Card}(\frac{\mathbb{Z}}{\operatorname{Ker} F})=2$. Now if $\operatorname{Im}F$ is infinite $\operatorname{Card}(\frac{\mathbb{Z}}{\operatorname{Ker} F})= \infty$. This mean $\operatorname{Ker}F= 2 \mathbb{Z}$ or $\operatorname{ker}F = \mathbb{Z}$. - thanku very much!! – d13 Nov 7 '12 at 7:15 1 @d13 You are welcome. – P.. Nov 7 '12 at 7:22 A homomorphism on $Z$ is completely determined by the image of 1. So if the homomorphism sends 1 to $q$ in $Q^*$, then it sends an arbitrary integer $n$ to $q^n$. The kernel consists of those $n$ for which $q^n=1$. If $q$ is not equal to plus or minus 1, then $q^n$ will never equal 1, except trivially at $n=0$. If $q=1$, then the homomorphism is trivial. If $q=-1$, then the kernel consists of those $n$ for which $(-1)^n=1$, i.e. the even integers. - thanx a lot 4 ur help. – d13 Nov 7 '12 at 7:15	{"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": 0.9222182631492615, "perplexity_flag": "head"}
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http://math.stackexchange.com/questions/36356/how-to-find-the-number-of-continued-fraction-from-a-periodic-representation	# How to find the number of continued fraction from a periodic representation? Problem Find the number that represented by $[2,2,2 \ldots]$ I know it wasn't difficult, but I was absent the last two classes. So I just want to make sure that I got it right. My attempt was, Consider $x = [2;\overline{2}] \implies [2;x]$ Hence, $x = 2 + \dfrac{1}{x} \Leftrightarrow x = \dfrac{2x + 1}{x} \Leftrightarrow x^2 - 2x - 1 = 0$. Solving the quadratic equation we have $x = \sqrt{2} + 1$ since $x > 0$. Therefore $[2,2,2, \ldots] = \sqrt{2} + 1$ Is it correct? Thank you, - 2 That looks right to me. – yunone May 2 '11 at 3:45 2 Yes that looks right, and is in fact a different proof that $\sqrt{2}$ is irrational. – Aryabhata May 2 '11 at 3:51 2 Yes - now try something a little harder, like $[2,3,4,3,4,3,4,3,4,\dots]$, with the $3,4$ repeating. – Gerry Myerson May 2 '11 at 4:12 @all: Thanks a lot. – Chan May 2 '11 at 4:24 If you put a number into Wolfram Alpha it'll give you back its continued fraction representation. This isn't the direction you're trying to go, but it'll help you check your work. – Michael Lugo May 2 '11 at 16:05	{"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": 0.9688932299613953, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/54742/gaussian-curvature-radius/54765	## Gaussian curvature radius ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In the paper Surface sampling and the intrinsic Voronoi diagram (2008), Ramsay Dyer defines the Gaussian curvature radius at a point $x$ of a surface $S$ to be $\rho_K(x) = 1/\sqrt{K(x)}$ where $K(x)=\kappa_1(x) \kappa_2(x)$ is the Gaussian curvature at $x$. Trying to track back the notion in Berger's A panoramic view of Riemannian geometry, and in Lee's Riemannian manifolds and in Chavel's Riemannian Geometry yielded nothing. My question is two-folded: 2. Is there a reason not to define it as $\rho_K(x) = 1/\|K(x)\|$? Otherwise, this definition is only valid for non-negatively curved surfaces. EDIT As pointed out by Deane Yang, there is no sense in the definition I suggested. Nevertheless, if one wants to relate the Gaussian curvature to a radius (for either negatively or positively curved surfaces) how about this alternative: $\rho_{K}(x)=1/\sqrt{\|K(x)\|}$? - ## 3 Answers As for question #2, why does your definition make sense for a negatively curved surface? For a positively curved surface it does not give the right answer for spheres, since presumably you would want a sphere of radius $r$ to have a Gauss curvature radius of $r$. In particular, the word "radius" reflects a linear measurement and therefore should scale linearly if you rescale the surface. The "radius of curvature" at a point on a curve is the radius of an osculating circle and turns out to be the reciprocal of the geodesic curvature. On a point of a surface in $R^3$, you get a radius of curvature for each tangent direction, corresponding to the osculating circle in that direction. In particular, there are the two principal radii of curvature corresponding to the principal directions. The Gauss curvature radius, as defined above, is the geometric average. Since it can be defined in terms of Gauss curvature only, it has the advantage of being intrinsic. You could also define the "mean radius" by taking the arithmetic average. I don't recall seeing this before, but it also seems reasonable to study. I recommend working out the example of $z = f(x,y)$ at the origin, where $f(0, 0) = \partial_xf(0,0) = \partial_yf(0,0) = 0$. - @Deane: By mean radius, do you refer to $1/H(p)$ where $H(p)=\kappa_1(p)+\kappa_2(p)/2$ is the mean curvature at the point $p$. – Dror Atariah Feb 8 2011 at 15:53 I defined "mean radius" to be the mean of the principal radii of curvature. That's not the same as the reciprocal of the mean of the principal curvatures. – Deane Yang Feb 8 2011 at 18:47 @Deane: Considering eq. (3) mathworld.wolfram.com/MeanCurvature.html what you actually defined is nothing but the quotient $H/K$. Is it correct? Is there some standard geometrical meaning of this quantity? – Dror Atariah Feb 10 2011 at 9:13 Dror, it's what I said it is, namely the average radius of curvature. It is indeed half of $H/K$. Any function of the principal radii of curvature can also be written as a function of $H$ and $K$. – Deane Yang Feb 15 2011 at 16:29 @Deane: I don't know what is the hidden message in your example. Here, the Gaussian curvature at $(0,0)$ is nothing but the determinant of the Hessian of $f$ and the mean curvature is the trace. Were you aiming at something more specific? – Dror Atariah Feb 21 2011 at 15:39 show 1 more comment ### 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. It so happens that $1/\sqrt{K}$, where $K$ is the Gaussian curvature, is, in a sense, the average of the arithmetic mean radius of curvature and the radius of harmonic mean curvature. The calculation is explained in the "Merged radius of curvature" subsection of the Wikipedia article on radius of curvature. It is called the arithmetic-harmonic mean radius of curvature. - 1 In short, and hopefully as correct, $\sqrt{\frac{1}{K}}$ is the arithmetic-harmonic mean of the two principal curvature radii $\frac{1}{\kappa_i}$. – Dror Atariah Feb 8 2011 at 15:15 Deane's answer is similar to what I would have tried to say if I'd got here on time. I don't recall seeing the "Gaussian curvature radius" defined before, so I can't point you to other references. The definition is natural. On the one hand the bound on the distance to a conjugate point (Morse-Schönberg lemma) is given in terms of a bound on the Gaussian curvature radius, and on the other hand the Gaussian curvature radius provides an upper bound to the "maximal curvature radius"(reciprocal of the maximum of the absolute values of the principal curvatures). As Deane pointed out, these two curvature radii coincide on the sphere. Since we are only using it as an upper bound, we just define it to be infinite if the curvature is non-positive. In flat or negatively curved spaces, conjugate points are not an issue; geodesics diverge. As to your alternative, $\rho_K(x) = 1/\sqrt{\|K(x)\|}$, I guess it depends on what you want to do. You are making a smaller sizing function, but why? The spirit of the Morse-Schönberg lemma is better captured without the absolute value signs. If the infinite values disturb you, you have not avoided them when the Gaussian curvature vanishes. - @Ramsay: Why is this smaller? Smaller then what? I agree that it is not very helpful definition w.r.t. Morse-Schoenberg lemma. But it is a sizing function which is well defined for non-flat surfaces. At the moment I'm not sure what I want to do with this definition; I found it a natural generalization to a negatively curved surface case and I was wondering if it was investigated in the literature. – Dror Atariah Feb 18 2011 at 10:22 By smaller I mean that $\rho_K(x) \leq \rho_G(x)$ for all $x$, where $\rho_G$ is the original definition. They agree everywhere except when the curvature is negative and $\rho_G$ provides no bound. I would be surprised if you found this to be a useful way to capture the geometry of negatively curved surfaces. Since it is purely intrinsic, you will never be able to control triangle normals this way, for example. It is not clear to me what is represented by the bound you're proposing to introduce. I don't recall seeing it anywhere previously. – Ramsay Feb 18 2011 at 17:04	{"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": 30, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9161391258239746, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=1858676	Physics Forums Thread Closed Page 1 of 2 1 2 > ## Integral of a delta function from -infinity to 0 or 0 to +infinity Hello everyone Today in my QM class, a discussion arose on the definition of the delta function using the Heaviside step function $\Theta(x)$ (= 0 for x < 0 and 1 for x > 0). Specifically, $$\Theta(x) = \int_{-\infty}^{x}\delta(t) dt$$ which of course gives $$\frac{d\Theta(x)}{dx} = \delta(x)$$ Some books (esp those on Communications and Signal analysis) define $\Theta(0) = 1/2$. However, if I set $x = 0$ in the above integral, I get $$\int_{-\infty}^{0}\delta(t) dt = \Theta(0) = \frac{1}{2}$$ To me, this is an ambiguous result, because even though this would follow if $\delta(x)$ were a "normal function" by virtue of its evenness, the point $x = 0$ is a singularity point of the integrand and besides, the normal Riemann integral would implicitly assume an open interval formed by the limits of integration: $(-\infty,0)$ and not a closed interval $(-\infty,0]$. Now, I have the following question: Is the expression $\int_{-\infty}^{0}\delta(t) dt = \frac{1}{2}$ correct? If I construct a sequence of well behaved functions (rectangular, gaussian, or something else) $\{\delta_{n}(x)\}$ which converge to $\delta(x)$, if the elements of this sequence are even then indeed $$\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}$$ But can one infer $$\int_{-\infty}^{0}\delta(t) dt = \lim_{n \rightarrow \infty}\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}$$ from this always? I think this should depend on the definition of the sequence, and that such a result in general does not make sense as there is an ambiguity when one writes 0 as the upper limit: does it mean 0- or 0+? (For 0-, the integral is zero and for 0+, the integral is 1). Can a rigorous justification and answer be given for this? Thanks in advance. 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 Quote by maverick280857 Hello everyone Today in my QM class, a discussion arose on the definition of the delta function using the Heaviside step function $\Theta(x)$ (= 0 for x < 0 and 1 for x > 0). Specifically, $$\Theta(x) = \int_{-\infty}^{x}\delta(t) dt$$ which of course gives $$\frac{d\Theta(x)}{dx} = \delta(x)$$ Some books (esp those on Communications and Signal analysis) define $\Theta(0) = 1/2$. However, if I set $x = 0$ in the above integral, I get $$\int_{-\infty}^{0}\delta(t) dt = \Theta(0) = \frac{1}{2}$$ To me, this is an ambiguous result, because even though this would follow if $\delta(x)$ were a "normal function" by virtue of its evenness, the point $x = 0$ is a singularity point of the integrand and besides, the normal Riemann integral would implicitly assume an open interval formed by the limits of integration: $(-\infty,0)$ and not a closed interval $(-\infty,0]$. Now, I have the following question: Is the expression $\int_{-\infty}^{0}\delta(t) dt = \frac{1}{2}$ correct? If I construct a sequence of well behaved functions (rectangular, gaussian, or something else) $\{\delta_{n}(x)\}$ which converge to $\delta(x)$, if the elements of this sequence are even then indeed $$\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}$$ But can one infer $$\int_{-\infty}^{0}\delta(t) dt = \lim_{n \rightarrow \infty}\int_{-\infty}^{0}\delta_{n}(t) dt = \frac{1}{2}$$ from this always? I think this should depend on the definition of the sequence, and that such a result in general does not make sense as there is an ambiguity when one writes 0 as the upper limit: does it mean 0- or 0+? (For 0-, the integral is zero and for 0+, the integral is 1). Can a rigorous justification and answer be given for this? Thanks in advance. I think you are right; defined in that way, without specifing the symmetry of the delta function, H(0) could have any value between 0 and 1. The ambiguity is also indicated here: http://en.wikipedia.org/wiki/Heaviside_step_function Quote by lightarrow I think you are right; defined in that way, without specifing the symmetry of the delta function, H(0) could have any value between 0 and 1. The ambiguity is also indicated here: http://en.wikipedia.org/wiki/Heaviside_step_function Thanks for your reply lightarrow, but can you point me to a source where this problem is discussed precisely with regard to the integration limits and different sequence definitions for the delta function, so that I could show it to my instructor. So far, since we've only worked with even functions converging to the Dirac Delta in the limit, it is not obvious to anyone that the integral does not "have" to be half but rather is ambiguous since I can always construct a delta function from say a rectangular function with height 1/A and width A extending from x = 0 to x = A, and not necessarily from x = -A/2 to x = +A/2. For such a definition, the integral would be 0 and not 1/2. ## Integral of a delta function from -infinity to 0 or 0 to +infinity Quote by maverick280857 Thanks for your reply lightarrow, but can you point me to a source where this problem is discussed precisely with regard to the integration limits and different sequence definitions for the delta function, so that I could show it to my instructor. So far, since we've only worked with even functions converging to the Dirac Delta in the limit, it is not obvious to anyone that the integral does not "have" to be half but rather is ambiguous since I can always construct a delta function from say a rectangular function with height 1/A and width A extending from x = 0 to x = A, and not necessarily from x = -A/2 to x = +A/2. For such a definition, the integral would be 0 and not 1/2. being an EE who works in signal processing, i have had many conversations (some disputed) with others regarding the meaning of the Dirac delta function (what we EEs like to call the "unit impulse function" - we also call the Heaviside function the "unit step function"). from a strict mathematical POV, the Dirac delta "function" is not really a function, but something they call a distribution and there is supposedly some whole theory behind this. but, as far as engineers are concerned, we treat it as a function that is the limit of those "nascent" delta functions that you call $\delta_n(t)$. the problem (or one of them) that the mathematicians have with this, is the integral of two functions, $f(t)$ and $g(t)$ that are equal almost everywhere (everywhere except a countable number of infinitely thin points on the t-axis), that those two integrals (over the same limits) must also be equal. if you set $f(t)$ to $\delta(t)$ and $g(t)$ to 0, you will see that they agree almost everywhere, yet the integral (from some negative t to some other positive t) of $\delta(t)$ is 1 yet the integral of $g(t)$ is 0. so there is, from a pure mathematical POV a problem. from my POV (not as anal-retentive as this distribution or generalized function theory is), i resolve the problem by simply letting $\delta(t)$ be one of those nascent $\delta_n(t)$, say the rectangular function, where the width of the delta function is one Planck Time in width. that's a legit function for the mathematicians, and it's close enough to the zero width $\delta(t)$ that it would make no physical difference in any physical situation. if the Planck Time is not narrow enough, make it a half or tenth of a Planck Time. one of the reasons that i prefer the symmetric definition of the Dirac delta is so that we can equate the integral of it to the step function: $$\int_{-\infty}^{t} \delta(u) du = H(t)$$ and also define the step function in terms of the sign or signum function: $$\frac{1}{2}(1 + \mathrm{sgn}(t)) = H(t)$$ these simple definitions sometimes makes our lives easier in signal processing. Quote by rbj being an EE who works in signal processing, i have had many conversations (some disputed) with others regarding the meaning of the Dirac delta function (what we EEs like to call the "unit impulse function" - we also call the Heaviside function the "unit step function"). from a strict mathematical POV, the Dirac delta "function" is not really a function, but something they call a distribution and there is supposedly some whole theory behind this. but, as far as engineers are concerned, we treat it as a function that is the limit of those "nascent" delta functions that you call $\delta_n(t)$. the problem (or one of them) that the mathematicians have with this, is the integral of two functions, $f(t)$ and $g(t)$ that are equal almost everywhere (everywhere except a countable number of infinitely thin points on the t-axis), that those two integrals (over the same limits) must also be equal. if you set $f(t)$ to $\delta(t)$ and $g(t)$ to 0, you will see that they agree almost everywhere, yet the integral (from some negative t to some other positive t) of $\delta(t)$ is 1 yet the integral of $g(t)$ is 0. so there is, from a pure mathematical POV a problem. from my POV (not as anal-retentive as this distribution or generalized function theory is), i resolve the problem by simply letting $\delta(t)$ be one of those nascent $\delta_n(t)$, say the rectangular function, where the width of the delta function is one Planck Time in width. that's a legit function for the mathematicians, and it's close enough to the zero width $\delta(t)$ that it would make no physical difference in any physical situation. if the Planck Time is not narrow enough, make it a half or tenth of a Planck Time. does anyone know why my $g(t)$ is rendered "g(i)" rather than g(t) by LaTeX? i clearly put a "t" in the equation. I agree with you rbj, but when you talk of $\delta(x)$, the Dirac delta distribution itself, the ambiguity at x = 0 cannot be avoided. Working with the sequence of functions $\delta_{n}(x)$ each member of which is well behaved, there is no such problem. Somehow, I've always had a problem reconciling the definitions...I've always believed in the distribution theory version and not the 'practical' way out...to define the integral to be half, just because the Dirac delta is even. As you rightly pointed out, its not a function anyway. PS--I am aware of the EE definitions, and it was in a course on signals and systems that I first came across this conceptual difficulty, when a discussion with some mathematics and EE professors led to the conclusion that the said integral has no meaning whatsoever. So while one can define the Heaviside step function to be equal to 1/2 at x = 0, the corresponding integral of delta(x) from x = -infinity to x = 0 cannot be unambiguously defined. I am looking for a rigorous reference for the same. Distribution theory textbooks that I have looked at, have proceeded step by step to derive the standard properties of such distributions, but have not discussed such weird situations. Quote by rbj one of the reasons that i prefer the symmetric definition of the Dirac delta is so that we can equate the integral of it to the step function: $$\int_{-\infty}^{t} \delta(u) du = H(t)$$ and also define the step function in terms of the sign or signum function: $$\frac{1}{2}(1 + \mathrm{sgn}(t)) = H(t)$$ these simple definitions sometimes makes our lives easier in signal processing. Agreed, but then again the extreme limits of a Riemann integral from a to b are really x = a+ to x = b-, irrespective of how you partition the set. So thats why I said it creates a problem...perhaps no computational issues will arise, but conceptually this doesn't seem rigorous enough to me. To cite an example, what would you compute $$\int_{-\infty}^{0}dx (x^2 -4) \delta(x)$$ as? I would write it as zero, because the point x = 0 is excluded in $(-\infty,0)$. But if you assume 0 here means 0+, then the answer would be -4. This is too simple an example and so perhaps I need to cook up something better :-P I believe the problem at hand deals with the fact that the engineers aren't always as rigorous as the physicists (who aren't always as rigorous as the mathematicians). At the fundamental level, I believe in the ambiguity of H(0). However, in engineering practice, it may be sometimes convenient to say H(0)=1/2 without questioning it further. I believe this "definition" comes from the Dirichlet theorem of Fourier analysis (which forms the basis of signals and systems). According to the Dirichlet theorem, the Fourier series of a signal converges to the midpoint at jump discontinuities. Therefore, if H'(x) is the Fourier expansion of H(x), then H'(0)=1/2. Conveniently (see: sloppily), the engineers just say that this implies H(0)=1/2. But hey, if it makes my cell phone work, who am I to complain? Note, a while back I ran across a journal article that dealt with some subtleties of the integral definition of the delta function. Sadly I didn't really read it in-depth, but it may be worth it to look into again. It was by David Griffiths in the American Journal of Physics. Quote by maverick280857 Agreed, but then again the extreme limits of a Riemann integral from a to b are really x = a+ to x = b-, irrespective of how you partition the set. So thats why I said it creates a problem...perhaps no computational issues will arise, but conceptually this doesn't seem rigorous enough to me. To cite an example, what would you compute $$\int_{-\infty}^{0}dx (x^2 - 4) \delta(x)$$ as? I would write it as zero, because the point x = 0 is excluded in $(-\infty,0)$. But if you assume 0 here means 0+, then the answer would be -4. and if we used the midpoint definition, then the answer is -2 and we can say that, in general $$\int_{-\infty}^{0} f(x) \delta(x) dx \ + \ \int_{0}^{+\infty} f(x) \delta(x) dx = \int_{-\infty}^{+\infty} f(x) \delta(x) dx = f(0)$$ at least in the electrical engineering and signal processing context (can't say diddley about QM), life is much easier with the (even) symmetrical $\delta(x)$ definition. Quote by cmos I believe the problem at hand deals with the fact that the engineers aren't always as rigorous as the physicists (who aren't always as rigorous as the mathematicians). it's true, regarding the Dirac delta function (and maybe the subtle differences between Riemann and Lebesque integration). dunno if it's true about other stuff. we try to be rigorous. At the fundamental level, I believe in the ambiguity of H(0). However, in engineering practice, it may be sometimes convenient to say H(0)=1/2 without questioning it further. I believe this "definition" comes from the Dirichlet theorem of Fourier analysis (which forms the basis of signals and systems). According to the Dirichlet theorem, the Fourier series of a signal converges to the midpoint at jump discontinuities. Therefore, if H'(x) is the Fourier expansion of H(x), then H'(0)=1/2. Conveniently (see: sloppily), the engineers just say that this implies H(0)=1/2. sure, you're right. but, at least this engineer says that real Dirac delta functions don't really exist in physical reality and these functions are useful to deal with impulsive-like physical events (like elastic collisions of really hard objects, or what happens when you connect an uncharged capacitor to a well-regulated voltage source). i, personally, have not found the strict definition and treatment of the Dirac delta to be useful. this has had practical implications. i have no trouble with certain expressions with the Dirac delta that lives outside of an integral, although i recognize that, eventually, it needs to find itself inside an integral in order to really do something with it. e.g., in the Nyquist-Shannon Sampling and Reconstruction Theorem: $$\sum_{k=-\infty}^{+\infty} \delta(t-k) = \sum_{n=-\infty}^{+\infty} e^{i 2 \pi n t}$$ i use that in our (perhaps sloppy) derivation of the results of the Sampling Theorem and i've had anal-retentive mathematicians tell me the above equation is meaningless and cannot be used in any derivation. i beg to differ. Quote by rbj and if we used the midpoint definition, then the answer is -2 and we can say that, in general $$\int_{-\infty}^{0} f(x) \delta(x) dx \ + \ \int_{0}^{+\infty} f(x) \delta(x) dx = \int_{-\infty}^{+\infty} f(x) \delta(x) dx = f(0)$$ I would write this as $$\int_{-\infty}^{0} f(x) \delta(x) dx \ + \int_{0-}^{0+} f(x)\delta(x) dx + \int_{0}^{+\infty} f(x) \delta(x) dx = \int_{-\infty}^{+\infty} f(x) \delta(x) dx = f(0)$$ The first and third terms in the left most expression would then be zero and the contribution would come only from the integral over (0-,0+) Quote by maverick280857 I would write this as $$\int_{-\infty}^{0} f(x) \delta(x) dx \ + \int_{0-}^{0+} f(x)\delta(x) dx + \int_{0}^{+\infty} f(x) \delta(x) dx = \int_{-\infty}^{+\infty} f(x) \delta(x) dx = f(0)$$ The first and third terms in the left most expression would then be zero and the contribution would come only from the integral over (0-,0+) yeah, but what does it gain you? you introduce another extraneous notation. i know we engineers see it in our first introduction to the Laplace Transform ("0-" which is another way to say $\lim_{\epsilon \to 0}-\epsilon^2$), and it's for this very same reason; so we make sure we include all of the Dirac impulse, no matter how it's defined. but it's not necessary if the Laplace Transform is defined as the double-sided L.T.: $$X(s) \ \equiv \ \mathcal{L}\{x(t)\} \ = \int_{-\infty}^{+\infty} x(t) e^{-st} dt$$ as is the Fourier Transform. not having to restrict the limits to 0- and 0+ can be convenient at times when setting up a problem. Ok, let me restate my question, since so far all our discussions have centered around convenience as a key idea in these definitions. What is the value of the following integrals [tex]\int_{0}^{\infty}\delta(x)dx[/itex] [tex]\int_{-\infty}^{0}\delta(x)dx[/itex] I am looking for a mathematically rigorous argument that can justify whether the integrals are 0 or 1/2. This should preferably be from distribution theory. If someone can point me to a source on the internet or a book where precisely these issues have been dealt with and such integrals are explicitly listed with a sufficiently rigorous and non-handwaived justification, I would be very grateful. Thanks! Quote by maverick280857 Ok, let me restate my question, since so far all our discussions have centered around convenience as a key idea in these definitions. What is the value of the following integrals $$\int_{0}^{\infty}\delta(x)dx$$ $$\int_{-\infty}^{0}\delta(x)dx$$ I am looking for a mathematically rigorous argument that can justify whether the integrals are 0 or 1/2. This should preferably be from distribution theory. If someone can point me to a source on the internet or a book where precisely these issues have been dealt with and such integrals are explicitly listed with a sufficiently rigorous and non-handwaived justification, I would be very grateful. Thanks! I could be totally wrong, but i don't see how is possible to justify that those integrals are 0 or 1/2. The definition of the delta function, at least for what I know, is: $$1.\ \int_{-\infty}^{\infty}\delta(x)dx\ =\ 1$$ $$2.\ \int_{-\infty}^{\infty}\delta(x) f(x)dx\ =\ f(0)$$ So, a delta function defined as in your previous example as a limit rectangle all in the +x, or all in the -x, or something else not symmetric, does satisfy that definition and so $$\int_{0}^{\infty}\delta(x)dx$$ or $$\int_{-\infty}^{0}\delta(x)dx$$ cannot have a unique value but depends on how you constructed the delta function. Recognitions: Gold Member Science Advisor Staff Emeritus Quote by maverick280857 Ok, let me restate my question, since so far all our discussions have centered around convenience as a key idea in these definitions. What is the value of the following integrals [tex]\int_{0}^{\infty}\delta(x)dx[/itex] [tex]\int_{-\infty}^{0}\delta(x)dx[/itex] I am looking for a mathematically rigorous argument that can justify whether the integrals are 0 or 1/2. This should preferably be from distribution theory. The proof goes like: We have previously made the definition $$\int_{-\infty}^{0}\delta(x)dx = 1/2$$ Therefore, the value of $$\int_{-\infty}^{0}\delta(x)dx$$ is 1/2. The clearest approach to this is probably purely algebraic. We have a linear functional $\int_{-\infty}^0$ which has already been defined on the set of test functions. We have simply built a new functional (which we denote by the same symbol) that extends this one to (some) distributions, by specifying it's value at a particular point (i.e. $\delta$). Really, the only thing there is to check is that this new functional has the properties we desire. Quote by Hurkyl The proof goes like:[indent]We have previously made the definition $$\int_{-\infty}^{0}\delta(x)dx = 1/2$$ So we can't prove that equality, if we have to define it, it's this you are saying? I don't think you can give a rigorous meaning to the integral. The reason is that in the rigorous approach you have to work with the delta functional defined as: $$\delta[f]= f(0)$$ where f is an arbitrary infinitely differentiable function that is equal to zero outside some compact set (if I remember correctly). The integral would correspond to applying the delta distribution to a test function which is equal to 1 for negative x and equal to zero for x>=0. But such a test function is not infinitely differentiable so it is not a legal test function. If remember correctly, the fact that you can define distributions, which are, sort of, very wildly behaved functions, is due to the fact that the set of test functions is so well behaved. There is a duality here, as you can view the test functions as functionals on the set of distributions. The more well behaved one set is the less well behaved the other can be. Thread Closed Page 1 of 2 1 2 > Thread Tools Similar Threads for: Integral of a delta function from -infinity to 0 or 0 to +infinity Thread Forum Replies Calculus & Beyond Homework 5 Calculus & Beyond Homework 5 Linear & Abstract Algebra 46 Calculus & Beyond Homework 4 Calculus 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": 51, "mathjax_display_tex": 32, "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": 0.9444341659545898, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/49742/list	## Return to Answer 4 added 148 characters in body The Turing degrees are an immensely intricate poset $\mathcal{D}$. Here are some of their remarkable properites: 1. Every countable poset is embeddable in $\mathcal{D}$. 2. $\mathcal{D}$ contains minimal degrees. (a non-zero degree $\mathbf{m}$ with no degree between $\mathbf{0}$ and $\mathbf{m}$) 3. For every non-zero degree $\mathbf{d}$, there is a degree that is incomparible with $\mathbf{d}$. 4. $\mathcal{D}$ contains an antichain of size $2^{\aleph_0}$. 5. No infinite strictly increasing chain in $\mathcal{D}$ has a least upper bound. 6. For every non-zero degree $\mathbf{d}$, there is a degree $\mathbf{c} < \mathbf{d}$ such that $\mathbf{c}'=\mathbf{d}$. (here Here $\mathbf{c}'$ denotes the jumpset of indices of oracle Turing machines that halt when using $\mathbf{c}$)\mathbf{c}\$ as an oracle. Note that one must check that this is well-defined on degrees.) 7. For any two recursively enumerable degrees, there is a recursively enumerable degree strictly between them. 8. Any finite distributive lattice can be embedded in the recursively enumerable degrees. 3 added 830 characters in body; added 10 characters in body The Turing degrees are probably the most an immensely intricate poset ever studied $\mathcal{D}$. Here are some of their remarkable properites: 1. Every countable poset is embeddable in $\mathcal{D}$. 2. $\mathcal{D}$ contains minimal degrees. (by humans at least)a non-zero degree $\mathbf{m}$ with no degree between $\mathbf{0}$ and $\mathbf{m}$) 3. For every degree $\mathbf{d}$, there is a degree that is incomparible with $\mathbf{d}$. 4. $\mathcal{D}$ contains an antichain of size $2^{\aleph_0}$. 5. No infinite strictly increasing chain in $\mathcal{D}$ has a least upper bound. 6. For every degree $\mathbf{d}$, there is a degree $\mathbf{c} < \mathbf{d}$ such that $\mathbf{c}'=\mathbf{d}$. (here $\mathbf{c}'$ denotes the jump of $\mathbf{c}$) 7. For any two recursively enumerable degrees, there is a recursively enumerable degree strictly between them. 8. Any finite distributive lattice can be embedded in the recursively enumerable degrees. 2 added 63 characters in body The poset of Turing degrees are probably the most intricate poset ever studied (by humans at least). 1 [made Community Wiki] The poset of Turing degrees.	{"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": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.903825581073761, "perplexity_flag": "middle"}
http://unapologetic.wordpress.com/2008/01/31/some-theorems-about-metric-spaces/?like=1&source=post_flair&_wpnonce=8149a33672	# The Unapologetic Mathematician ## Some theorems about metric spaces We need to get down a few facts about metric spaces before we can continue on our course. Firstly, as I alluded in an earlier comment, compact metric spaces are sequentially compact — every sequence has a convergent subsequence. To see this fact, we’ll use the fact that compact spaces are the next best thing to finite. Specifically, in a finite set any infinite sequence would have to hit one point infinitely often. Here instead, we’ll have an accumulation point $\xi$ in our compact metric space $X$ so that for any $\epsilon>0$ and point $x_m$ in our sequence there is some $n\geq m$ with $d_X(x_n,\xi)<\epsilon$. That is, though the sequence may move away from $\xi$, it always comes back within $\epsilon$ of it again. Once we have an accumulation point $\xi$, we can find a subsequence converging to $\xi$ just as we found a subnet converging to any accumulation point of a net. Let’s take our sequence and define $F_N=\mathrm{Cl}(\{x_n, n\geq N\})$ — the closure of the sequence from $x_N$ onwards. Then these closed sets are nested $F_1\supseteq F_2\supseteq...\supseteq F_N\supseteq...$, and the intersection of any finite number of them is the smallest one, which is clearly nonempty since it contains a tail of the sequence. Then by the compactness of $X$ we see that the intersection of all the $F_N$ is again nonempty. Since the points in this intersection are in the closure of any tail of the sequence, they must be accumulation points. Okay, that doesn’t quite work. See the comments for more details. Michael asks where I use the fact that we’re in a metric space, which was very astute. It turns out on reflection that I did use it, but it was hidden. We can still say we’re looking for an accumulation point first and foremost, because if the sequence has an accumulation point there must be some subsequence converging to that point. Why not a subnet in general? Because metric spaces must be normal Hausdorff (using metric neighborhoods to separate closed sets) and first-countable! And as long as we’re first-countable (or, weaker, “sequential”) we can find a sequence converging to any limit point of a net. What I didn’t say before is that once we find an accumulation point there will be a subsequence converging to that point. My counterexample is compact, and any sequence in it has accumulation points, but we will only be able to find subnets of our sequence converging to them, not subsequences. Unless we add something to assure that our space is sequential, and metric spaces do that. We should note in passing that the special case where $X$ is a compact subspace of $\mathbb{R}^n$ is referred to as the Bolzano-Weierstrass Theorem. Next is the Heine-Cantor theorem, which says that any continuous function $f:M\rightarrow N$ from a compact metric space $M$ to any metric space $N$ is uniformly continuous. In particular, we can use the interval $\left[a,b\right]$ as our compact metric space $M$ and the real numbers $\mathbb{R}$ as our metric space $N$ to see that any continuous function on a closed interval is uniformly continuous. So let’s assume that $f$ is continuous but not uniformly continuous. Then there is some $\epsilon>0$ so that for any $\delta>0$ there are points $x$ and $y$ in $M$ with $d_M(x,y)<\delta$ but $d_N(f(x),f(y))\geq\epsilon$. In particular, we can pick $\frac{1}{n}$ as our $\delta$ and get two sequences $x_n$ and $y_n$ with $d_M(x_n,y_n)<\frac{1}{n}$ but $d_N(f(x),f(y))\geq\epsilon$. By the above theorem we can find subsequences $x_{n_k}$ converging to $\bar{x}$ and $y_{n_k}$ converging to $\bar{y}$. Now $d_X(x_{n_k},y_{n_k})<\frac{1}{n_k}$, which converges to ${0}$, and so $\bar{x}=\bar{y}$. Therefore we must have $d_Y(f(x_{n_k}),f(y_{n_k})$ also converging to ${0}$ by the continuity of $f$. But this can’t happen, since each of these distances must be at least $\epsilon$! Thus $f$ must have been uniformly continuous to begin with. ### Like this: Posted by John Armstrong \| Point-Set Topology, Topology ## 10 Comments » 1. So, what kinds of spaces are compact, but not sequentially compact? I don’t see how the fact that the space has a metric enters into the proof of the Bolzano-Weierstrass Theorem. Comment by Michael Brazier \| February 1, 2008 \| Reply 2. Damn.. you’ve got a point there. All my references gave proof sketches for subspaces of $\mathbb{R}^n$ and then just asserted it worked for general metric spaces, so I tried to throw something together to work more generally, but it doesn’t seem to work. Thanks for catching that. As for an example where compactness doesn’t imply sequential compactness, the best one I know of is the space of all functions from the unit interval to the set $\{0,1\}$, using the topology of pointwise convergence which I haven’t really discussed yet, but which is equivalent to the product topology on a continuum’s worth of copies of $\{0,1\}$. This is compact by the Tychonoff theorem, but a diagonalization argument can show that there’s no convergent subsequence. Comment by \| February 1, 2008 \| Reply 3. Can you give a more or less explicit example of such a s sequence? Comment by \| February 6, 2008 \| Reply 4. In answer to Michael Livshits, I’d be fairly surprised if the following does not work (tweaking John’s example just a bit): consider all functions f: [0, 1] –> [0, 1] with the topology of pointwise convergence, and define a sequence f_n where f_n takes a real number r in [0, 1] to the fractional part of nr. If this had a convergent subsequence, indexed by an increasing set of integers a_n, then it would mean that the fractional parts {a_n r} would converge for every r in [0, 1]. My intuition says, to the contrary, that no matter what increasing sequence a_n you pick, there should be an r such that {a_n r} is equidistributed. If that intuition is correct, then you can apply similar reasoning to John’s example (round down to 0 if the fractional part is less than 1/2; otherwise round up to 1). Comment by Todd Trimble \| February 6, 2008 \| Reply 5. On further thought, the idea behind my example becomes even clearer if you consider instead the sequence of functions f_n(r) = {10^n r}. For any increasing sequence of natural numbers a_n, you can find an explicit r whose decimal expansion has, in its (a_{n}+1)th place after the decimal point, the digit given by the remainder of n mod 10 (take all the other digits of r to be 0, say). Quite clearly, the sequence of fractional parts {10^{a_n} r} cannot converge for that r. Neither does the sequence of 0′s and 1′s gotten by rounding up to 1 or rounding down to 0. Comment by Todd Trimble \| February 6, 2008 \| Reply 6. Nice example, Todd. Replacing $n$ by $n_k$ in your argument, we can see that no subsequence of your sequence convereges pointwise. Comment by \| February 6, 2008 \| Reply 7. It looks like you can use $f_n(r)=sin(2^nr)$ instead. Comment by \| February 6, 2008 \| Reply 8. [...] if . That is, if we have a sequence of points in (we only need to consider sequences because metric spaces are sequential) that converges to , then the image of this sequence converges to [...] Pingback by \| September 17, 2009 \| Reply 9. [...] It’s no loss of generality, then, to assume that is increasing. We also remember that the Heine-Cantor theorem tells us that since is compact, is actually uniformly [...] Pingback by \| January 12, 2010 \| Reply 10. [...] and thus compact. The collection must be finite or else it would have an accumulation point by Bolzano-Weierstrass, and thus wouldn’t be [...] Pingback by \| March 1, 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": 56, "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": 0.9429618120193481, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/178000/series-can-the-result-be-zero-for-a-continuous-interval-of-its-argument	series: can the result be zero for a continuous interval of its argument? I'm considering the series $$f_c(x) = \sum_{k=c}^\infty \left( c^{k-1} \binom{k}{c} \cdot \prod_{j=1}^{k-1} (x-1/j) \right)$$ where the parameter $c \in \mathbb N ,c \gt 0$ and fixed for a certain discussion. Let's denote the product-term $\prod_{j=1}^{k-1} (x-1/j)$ as $y_k(x)$ First, I see, that for $x =1/m, m\in \mathbb N$ the series is finite (the higher terms $y_{k \ge m}(1/m) = 0$) and gives an exact value. While this is obvious, it is not obvious to me, that, for all such x , and even for all c (!), it seems that it is also true that $f_{c \gt 0} (1/m) = 0$ which I can approximate faily well using Pari/GP. Q1: This looks somehow like a telescoping effect. Can it be shown, that for the so-constructed finite sums the result is always zero? Next, if $\|x\| \lt 1 \ne 1/m$ is a rational value with a non-simple fraction or is even irrational (the single terms do not vanish) then $f_c(x)$ becomes a series with infinitely many terms. Still it seems, that for any such x the series converges for c=1 and again it converges to zero. For example, using 400 terms, I got $f_1(0.8+\pi /1000,400) \lt 1e-38$ . Increasing the number of terms to 800 I got $f_1(0.8+\pi /1000,800) \lt 1e-76$ . Letting x approach 1 from below, I just needed more terms to always approximate zero (for negative x the terms have alternating signs, and their absolute values increase first before they begin to decrease). For c=2 it is similar, only that it must be $\|x\| \lt 1/2$ and in general it seems $f_c(x)$ approaches zero whenever we have $\|x\| \lt 1/c$ Q2: True? Is there an algebraic proof? If $x \lt -1/c$ the absolute value of the terms increase, but it seems, the series can always summed by Euler-summation, because the terms alternate in sign. Using that summation: again it seems, that the series always approaches zero. Q3: From the properties of power series I know, that when a continuous interval of the argument leads to the same function-value, then the function must be the constant function. But this seems to be different here. Is there possibly a trivial answer for this? Remark: This is an extension of an earlier discussion of mine where I considered the construction of the formal series $f_c(x)$ based on a matrix-problem. That older question was not yet answered, so I thought, to develop it into a question on series (here with a Newton-basis) might help to get more insight. Here are some references to older questions: The question as problem with an infinite matrix Learning the expression "Newton-basis" -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9510538578033447, "perplexity_flag": "head"}
http://mathoverflow.net/questions/82444?sort=oldest	## Elliptic pseudodifferential operator estimate [closed] ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If $P$ is an elliptic pseudodifferential operator of order 1 in the sense that its principal symbol is invertible, then we have the a priori estimate `$\\|u\\|_{H^1(U)} \le C (\\|Pu\\|_{L^2(W)} + \\|u\\|_{L^2(W)}), \quad (C>0 \text{ a constant})$` for any $\overline{U}\subset W\subset \mathbb{R}^n$ with $W$ bounded. Say I'm only interested in an upper bound of `$\\|u\\|_{L^2(U)}$`. Then the term $\\|u\\|_{L^2(W)}$ on the right hand side above feels a bit redundant. Can I obtain `$\\|u\\|_{L^2(U)} \le C\\|Pu\\|_{L^2(W)} \quad \text{?}$` - ## 1 Answer The answer to your question is no. Take any non-injective operator $P.$ - I forgot to mention I work semi-classically, so for a small parameter elliptic operators are injective. Anyway, I think I solved the problem. Thank you. – Alex A Dec 2 2011 at 12:41 2 If your operator is injective then you can even estimate the H^1 -norm of u in terms of the L^2-norm of u. This is true because any injective operator with closed image satisfies an injectivity estimate. – Orbicular Dec 2 2011 at 12:50	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 5, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8935949802398682, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/tagged/regression+return	# Tagged Questions 1answer 127 views ### Regression giving the return on a stock I have this regression equation: $$R_{stock} = 3,28\% + 1,65*R_{market}$$ Where $R_{stock}$ is the expected return on a stock and $R_{market}$ being the market risk premium. I have a one-year ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": false, "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": 0.9387872219085693, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/79621?sort=oldest	## Two rectangular parallelepiped ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Prove that if we have two rectangular parallelepiped (cuboids) such that one of them is placed inside the other then the sum of the three lengths of the inner parallelepiped is at most the sum of the three lengths of the exterior parallelepiped. In 2 dimensions the problem is trivial. Does this hold in higher dimensions? There is a way to prove it in higher Euclidean dimensions? - ## 3 Answers This version works for all parallelepipeds, not only rectangular ones: If you replace each parallelepiped by all points that have distance at most $\varepsilon$ to a point in the parallelepiped, you can still place the smaller inside the bigger one. In particular, the smaller object has a smaller volume. We divide up the extended parallepipeds by extending the planes corresponding to the six faces. This gives the volume of the original solid in the center, parallelepipeds of height $\varepsilon$ on top of each face, partial cylinders (with slanted parallel ends) of radius $\varepsilon$ and length the corresponding edge, and partial spheres of radius $\varepsilon$ around each vertex. The partial spheres add up to exactly one whole sphere simply by translation. The partial cylinders corresponding to parallel edges add up to one whole cylinder by translation. Now let $\varepsilon$ tend to infinity (yes, really). The term with $\varepsilon^3$ comes from the sphere of radius $\varepsilon$ and does not depend on the parallelepiped at all, . The coefficient of $\varepsilon^2$ comes from the cylinders and is clearly the sum of the edges times some constant. So, for the inequality to be valid the sum of the edges of the smaller box must be smaller than the sum of the edges of the larger box. I don't see any problem with the generalization to higher dimension. - I think this works, at least in three dimensions, but with some corrections: the term with $\epsilon^3$ is just the volume of a sphere of radius $\epsilon$, independent of the number of vertices: the spherical caps on each vertex form a dissection of a whole sphere. The $\epsilon^2$ term for an edge depends on its dihedral angle, but the twelve edges can be grouped into sets of four such that within each set the lengths are equal and the complementary dihedrals add to 2π, so for the parallelepiped (but not for other polyhedra) the total depends only on length. – David Eppstein Oct 31 2011 at 21:48 @David Eppstein: What you write is absolutely correct, but note that in the OP we had rectangular parallelepipeds, so I mentioned the number of vertices to indicate that it is not necessary to think more about this in this case. Sorry for being both too lazy to repeat the word "rectangular" and to give the more general argument. I don't like the English term "rectangular parallelepiped". – thei Oct 31 2011 at 21:56 3 This is an old chestnut. I first heard this problem from Isaac Kornfeld, told it to Bela Bollobas who included in his book of coffee time problems. Since airlines (and the post office) charge for packages on the basis of the linear dimension' (i.e. the sum of the dimensions), this question can also be phrased can you cheat and get a cheaper price by enclosing the smaller box in a larger one?'. – Anthony Quas Oct 31 2011 at 22:27 1 This is a really nice argument. It does work for all parallelepipeds and in all dimensions. To see that everything fits perfectly around edges and virtices it helps to look at $\epsilon$-neighborhoods of virtices and lines in $\mathbb R^n$ with the $\mathbb Z^n$ action generated by the translations along edges of the parallellepiped. – Vitali Kapovitch Nov 1 2011 at 1:54 @Vitali: Indeed this is a very nice proof. Thanks! – heartwork Nov 2 2011 at 13:01 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This is true for 3-dimensional parallelepipeds. Let the sides of the ambient parallelepiped be $A,B,C$ and of the inner one $a,b,c$. Then it's clear that $A^2+B^2+C^2\ge a^2+b^2+c^2$ since the diameter of the ambient body is obviously bigger. Also, it's easy to see that the there is a 1-Lipschitz map of $\mathbb R^3$ onto the smaller parallelepiped (just take the nearest point projection map). Being 1-Lipschitz this map does not increase 2-dimensional area and therefore, applying this to the surface of the bigger parallellepiped we get $2(AB+BC+AC)\ge 2(ab+bc+ac)$. Adding the above two inequalities we get $(A+B+C)^2\ge (a+b+c)^2$. I learned of this problem from Anton's Petrunin's list of exercises in orthodox geometry (which has a lot of other cool problems btw). I don't know if the same holds in higher dimensions. - Nice. I don't see what part fails for dimension $4$ and higher. – Noam D. Elkies Oct 31 2011 at 16:06 1 @Noam: you need to prove only that the projection of 2-skeleton is ONTO. But it seems to be true. – Ilya Bogdanov Oct 31 2011 at 16:17 1 @Noam D. Elkies. This argument doesn't work already in dimension 4 (at least I don't see how to make it work) because we need an inequality on 2-dimensional area of the 2-skeletons and the projection map doesn't preserve them. But the argument by thei below works in all dimensions and for all parallelepipeds, not just for rectangular ones. – Vitali Kapovitch Nov 1 2011 at 1:59 What if we ask the same problem with tetrahedron instead of parallelepiped. Consider two tetrahedrons one placed inside the other. Prove that the sum of all 6 sides of inner tetrahedron is at most the sum of the 6 sides of exterior tetrahedron. This looks much harder! - Please don't add comments as answers. If you want to ask a different question, ask it as a different question. – Yemon Choi Dec 17 2011 at 8:53 1 This is not an answer to the question. – S. Carnahan♦ Dec 17 2011 at 8:54	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 21, "mathjax_display_tex": 0, "mathjax_asciimath": 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": 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": 0.9235198497772217, "perplexity_flag": "head"}
http://mathoverflow.net/questions/34314?sort=newest	## When is an algebra of commuting matrices (contained in one) generated by a single matrix? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let C be an nxn matrix, then the polynomials in C (with appropriate coefficients) form an algebra of commuting matrices. I feel that I should know if the converse is true but I do not. So my first question (adding several conditions) is: Suppose A and B are symmetric integer matrices with AB=BA, must there be a matrix C such that both A and B are polynomials in C? This would imply that the algebra which A and B generate has dimension n or less. In general: Let A be an algebra of pairwise commuting nxn matrices over an integral domain D, Under what conditions (if any) on D and A can we conclude that there is a matrix C so that A is contained in the D-algebra generated by C? I am thinking of conditions such as D is an algebraically closed field, D has characteristic zero, the matrices are symmetric. etc. - This comment does not affect the answers given but I feel that it should be mentioned. In questions of when matrices are functions of a given one, the relevant concept is that of the bicommutant. Thus if $A$ and $B$ are two square matrices (of the same order), then $A$ is a function of $B$ if and only if $A$ is in the bicommutant of $B$. This is a standard result and can be proved using the Jordan normal form (compute the bicommutant of a Jordan block). (I am assuming the the underlying field is that of the complex numbers). – jbc Nov 3 at 10:10 ## 4 Answers Thanks for the answers. Just to wrap up a bit, here are a few examples. 1) Sometimes an ACM (algebra of commuting matrices) is sure to be generated by one of its members 2) Other times it has dimension too large to possibly be (embedded in) an ACM with a single generator. 3) An ACM might be generated by 2 matrices, not generated by any of its members, but embed in a larger ACM which does have a single generator. 4) An ACM might be generated by 2 matrices, not generated by any of its members, but not embed in a larger ACM which does have a single generator (even in the 3x3 case). 1) If the matrices are all normal then they can be simultaneously diagonalized. This reduces the problem to an algebra of diagonal matrices, which is easy to understand. Such an algebra is actually generated by one of its members. 2) The 5 dimensional algebra ${\cal{A}}_5$ mentioned by Mariano (4x4 matrices with 2x2 blocks $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ has dimension too large to be generated by a single matrix. Furthermore, each member M generates only the 2 dimensional algebra of matrices $jI+kM$ so no subalgebra of dimension 3 or 4 has a single generator. 3) Consider the subalgebra ${\cal{A}}_3$ of ${\cal{A}}_5$ generated by $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&B\\0&0\end{smallmatrix}\right)$ with Let A and B be 2x2 invertible matrices (neither a scalar multiple of the other). As mentioned, we can't embed ${\cal{A}}_3$ in a singly generated 4 dimensional subalgebra of ${\cal{A}}_5$ However it also embeds in other 4 dimensional algebras. For example a 4x4 matrix $\left(\begin{smallmatrix}BA^{-1}&C\\0&A^{-1} B\end{smallmatrix}\right)$ will generate an ACM which commutes with everything in ${\cal{A}}_3$. I guess in this case it would automatically contain ${\cal{A}}_3$. I certainly verified that randomly filling in the C does this in several cases. In many cases I tested one can get away with one or both of A and B having rank 1... but not always. The two 4x4 matrices made from matrices with $A=\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)$ and $B=\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$ give an example of that. One can shrink this to 3x3, so I will: 4) The two $3 \times 3$ matrices $\left(\begin{smallmatrix}0&1&0\\0&0&0\\0&0&0\end{smallmatrix}\right)$ and $\left(\begin{smallmatrix}0&0&1\\0&0&0\\0&0&0\end{smallmatrix}\right)$ generate an algebra A with general member $\left(\begin{smallmatrix}a&b&c\\0&a&0\\0&0&a\end{smallmatrix}\right)$. A is maximal but not generated by any of its members as each member generates a 2 dimensional subalgebra. - 4) is not quite correct: If $X,Y$ are the two matrices, the subalgebra generated by $P=a+bX+cY$ may have dimension $3$ and therefore equals $\langle X,Y \rangle$ since $P^2=a^2+2abX+2acY$. But over the base field $\mathbb{F}_2$ the dimension is $\leq 2$. – Martin Brandenburg Mar 10 at 13:14 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. The open problem that surprised me was on the lower bound of dimensions of maximal commutative subalgebras of $M_n(\mathbf C)$. A maximal commutative sublagebra of $M_n(\mathbf C)$ can have dimension strictly lass than $n$. See for example Courter's article, where he gives a 13 dimensional maximal commutative subalgebra of $M_{14}(\mathbf C)$. I think, In general the problem of finding lower bound for the dimension of maximal commutative subalgebras in $M_n(\mathbf C)$ is still open. - This question seems relevant <a=href"mathoverflow.net/questions/29087/…; – Bruce Westbury Aug 3 2010 at 6:52 Here is unadulterated link to the relevant answer to the question mentioned by Bruce: mathoverflow.net/questions/29087/… – Victor Protsak Aug 3 2010 at 7:59 Thanks for the link Bruce and Victor. – Pooja Aug 3 2010 at 13:29 For symmetric matrices over the reals, the answer is yes: if $A$ and $B$ commute, then they can be simultaneously diagonalized. That is to say, there exists an orthogonal matrix $U$ ($U^TU = I$) such that $U^TAU$ and $U^TBU$ are both diagonal. It follows that for any diagonal matrix $D$ with distinct diagonal entries, both $A$ and $B$ are polynomials in $C = UDU^T$. In general it is instructive, and loses no generality, to assume that $A$ is in a normal (Jordan, Smith, etc.) form. Over a field, for example, the dimension of the space of polynomials in $A$ is the same as the degree of the minimal polynomial of $A$, which is the sum, over all eigenvalues $\lambda$, of the size of the largest Jordan block for $\lambda$ in the Jordan normal form of $A$ over the algebraic closure of the field. It's not too hard to verify that every matrix commuting with $A$ is a polynomial in $A$ if and only if the minimal polynomial is the characteristic polynomial, that is if every eigenvalue has a single Jordan block (as happens automatically for example if there are $n$ distinct eigenvalues). I can't remember the precise statement of something that surprised me, some unsolved problem related to the dimension of a commuting set in terms of Jordan block sizes. Someone please comment if you are familiar with this open problem and can state it precisely. - I agree that normal form is the way to go, I'd thought about Jordon form but hadn't considered any other. I wondered what is known. It is not true that the algebra generated by three commuting matrices always embeds in one generated by one matrix ( shown by Mariano's example for 4x4 with a little work). I don't think that an algebra generated by two commuting matrices can have dimension greater than n (at least over an algebraically closed field). That does not mean that it embeds, but at least it blocks a dimension proof. – Aaron Meyerowitz Aug 3 2010 at 5:38 And see [this answer](mathoverflow.net/questions/29087/…) to another question. Notably it is indeed a theorem (Gerstenhaber 1961) that an algebra generated by two commuting matrices cannot have dimension greater than n (but it can be of dimension n while not being single-generated), and it is an open question whether an algebra generated by 3 matrices can. – Marc van Leeuwen 0 secs ago – Marc van Leeuwen Mar 11 at 5:08 Consider the subalgebra of $M_{2n}(k)$ spanned by the matrices of the form $\left(\begin{smallmatrix}0&A\\0&0\end{smallmatrix}\right)$ (all blocks are $n\times n$) together with the identity, which is commutative. Its dimension is larger than $2n$ when $n$ is sufficiently large, so it is not generated by a single matrix. - That is a good point! To slightly recover my dignity I'll ask: what about 3x3 matrices? Symmetric matrices? – Aaron Meyerowitz Aug 3 2010 at 2:22 2 Aaron, the point is that if matrices aren't semisimple, bad stuff happens. "Symmetric" doesn't help in general (because it doesn't imply s.s. except over $\mathbb{R}$), but 3x3 matrices are small enough that you can classify commutative subalgebras by hand (work with joint eigenvectors, etc). – Victor Protsak Aug 3 2010 at 4:32 It may be useful to look at Jacobson's paper ams.org/mathscinet/pdf/…. Where he proves that maximum dimension of commutative subalgebras in $M_n(\mathbf C)$ is $[n^2/4] +1$. Which essentially matches with Mariano's examples with an additional identity matrix. – Pooja Aug 3 2010 at 6:28	{"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": 63, "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": 0.9358352422714233, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4231587	Physics Forums ## Equation for Induction of Electric Current in Magnetic Field I am looking for something that should be relatively simple to find, but as such, I have not been successful. I know that an electric current can be created by moving through a magnetic field. I have found equations that indicate this can be based on the velocity and the length of said object moving through the field. However, I have not been able to identify a single equation to suit my needs. Basically - I am looking for an equation to calculate the electric current (and voltage? in order to calculate the power...) of a loop or coil of conducting material (wire) moving through a magnetic field (say, the Earth's?). Any help will be much appreciated. Thank you. 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 Depends on whether or not you are looking for understanding, or just for a formula to use for a particular case. For the latter: the e.m.f. ('voltage'), ε, generated in a straight wire of length L moving at speed v at right angles to its own length and to a magnetic field of flux density B is ε = BLv. If the wire is part of a circuit of resistance R, the current, I, will be I = BLv / R. But the rest of the circuit must stay stationary while the wire of length L moves. For example, the wire may slide along stationary rails, making contact with them at each end of the wire. The rails connect to the rest of the circuit. If the whole of the circuit, including the wire, moves all together, there will be no emf and no current. To understand why there is no emf in a circuit which moves without turning through a uniform field requires a proper understanding of electromagnetic induction. You need, in particular, to understand what is meant by flux linkage and cutting of flux. Alright, that makes sense. Is there any way to generate a continuous EMF by moving a loop/coil (for greater length, L) through the Earth's magnetic field, B, at some velocity, v? Recognitions: Gold Member ## Equation for Induction of Electric Current in Magnetic Field Quote by pbhuter Alright, that makes sense. Is there any way to generate a continuous EMF by moving a loop/coil (for greater length, L) through the Earth's magnetic field, B, at some velocity, v? It is possible, but completely unfeasible. The Earths magnetic field is very weak and on our everyday scale it is very uniform. Amplifying Drakkith's response, translating a whole circuit (that is moving it without turning it) through a uniform field will produce zero emf and zero current. The Earth's field, over a large scale, is not uniform, and if you moved your circuit fast enough over a large enough distance for the non-uniformity to show up (i.e. a distance comparable with the Earth's radius), then there would be an emf, but this is almost certainly impracticable. What IS feasible is to rotate a coil about its diameter in the Earth's field, making connections to an external circuit by means of slip rings. This is, of course, none other than an a.c. generator (with a very weak magnetic field!). Look up fluxgate magnetometer http://en.wikipedia.org/wiki/Magnetometer There have been several projects for amateur construction of these in electronics hobby magazines. Recognitions: Homework Help nice explanation, Philip. I hadn't really thought about this before. So the idea is that the loop is rigid, and is not being rotated, so the velocity at all parts of the loop is the same, and the magnetic field is uniform. And the equation for the emf due to the magnetic field is: $$\oint (\vec{v} \wedge \vec{B}) \ \cdot d \vec{L}$$ But since the integrand is a constant vector, then we can simply write it as the gradient of some scalar field, for example if: $$\vec{v} \wedge \vec{B} = q \hat{z}$$ (where q is just a number, not charge or anything) then we can write it as the gradient of the scalar field $\phi=qz$ And since we can write it in this form, the emf is: $$\oint \nabla (\phi) \ \cdot d \vec{L}$$ But luckily for us, this is the definition of: $$\oint d \phi$$ And this must equal zero because it is around a closed loop. Therefore the emf in this situation must equal zero. (Sorry if this is using too much vector calculus). I am interested in the explanation using cutting of flux and flux linkage. Because I never really got comfortable with those concepts. I love your vector calculus derivations. You certainly know how to wield these tools! In my earlier posts I used the principle that the emf is equal to the rate of change of flux linking the loop. [The nice thing about this it that it works for both the cases of a circuit moving in a magnetic field (when the charge carriers in the circuit experience a magnetic force q(vxB) and for a stationary circuit sitting in a changing B, when the charge carriers experience an electric force qE.] In the case of a circuit translating in a uniform field, the flux linked with the circuit never changes (if you like, new lines of flux replace old lines), so the emf is zero. It is, of course possible to show the equivalence of the emf line integrals in both cases (moving circuit and stationary circuit+changing B) to d$\Phi$/dt. I used to be able to do it without looking it up! The induced current & voltage will always be related per Ohm's law, i.e. emf = IR. But the induced voltage "emf" is related to the net flux* per emf = Blv, where B is the net flux density. This net flux density is the external field density Bext, combined with any flux density generated by the induced current I, which is Bint (internal). Lenz' law states that Bint has a polarity opposite to Bext. If the closed path resistance R is high, greater than XL for the loop, then the induced emf is Bextlv, & the Bint (internal flux density) can be ignored as it is way smaller than Bext. A small induced current generates a small B field which is negligible compared with Bext. But as R decreases, the internal B field starts increasing, cancelling the external B. The net result is that the induced emf is less than in the open loop case. But emf/I = R in all cases. I just wanted to convey that there is no inherent "voltage regulation" here. As the loop resistance R decreases, so does emf. Claude Tags induced current, magnetic field Thread Tools \| \| \| \| \|-----------------------------------------------------------------------------------\|-------------------------------\|---------\| \| Similar Threads for: Equation for Induction of Electric Current in Magnetic Field \| \| \| \| Thread \| Forum \| Replies \| \| \| Classical Physics \| 3 \| \| \| Classical Physics \| 0 \| \| \| General Physics \| 2 \| \| \| Introductory Physics Homework \| 1 \| \| \| Introductory Physics Homework \| 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": 2, "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": 0.9264042973518372, "perplexity_flag": "middle"}
http://nrich.maths.org/6787/solution	### 14 Divisors What is the smallest number with exactly 14 divisors? ### Summing Consecutive Numbers Many numbers can be expressed as the sum of two or more consecutive integers. For example, 15=7+8 and 10=1+2+3+4. Can you say which numbers can be expressed in this way? ### Rule of Three If it takes four men one day to build a wall, how long does it take 60,000 men to build a similar wall? # Weekly Problem 24 - 2010 ##### Stage: 3 Short Challenge Level: $\frac{4}{5} = \frac{12}{15}$ and $-\frac{2}{3} = -\frac{10}{15}$, so the number halfway between these is $\frac{1}{2} (-\frac{10}{15}+ \frac{12}{15})$ which is $\frac{1}{15}$. This problem is taken from the UKMT Mathematical Challenges. View the previous week's solution View the current weekly problem 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": 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": 0.9103310108184814, "perplexity_flag": "middle"}

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