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http://mathhelpforum.com/trigonometry/129448-double-angle-formulae.html	# Thread: 1. ## Double angle formulae Can someone please show me the process of answering this question? By writing sin3X as sin(2X+X), show that sin3X=3sinX−4sin^3X I know that sin2X=2sinXcosX and the other trig identities, but I cannot understand how to substitute it in, in particular the answer follows as: sin3X = sin2XcosX + cos2XsinX - it is this step that I am struggling with! Thanks 2. Originally Posted by CSG18 Can someone please show me the process of answering this question? By writing sin3X as sin(2X+X), show that sin3X=3sinX−4sin^3X I know that sin2X=2sinXcosX and the other trig identities, but I cannot understand how to substitute it in, in particular the answer follows as: sin3X = sin2XcosX + cos2XsinX - it is this step that I am struggling with! Thanks Use the hint given at the start of the question and recall that $\sin(A+B) = \sin A \cos B + \cos B \sin A$ $\sin(2X+X) = \sin (2x) \cos (x) + \cos (2x) \sin(x)$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 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": 0.8871083855628967, "perplexity_flag": "middle"}
http://en.m.wikipedia.org/wiki/Magnetic_damping	# Magnetic damping Magnetic Damping is a form of damping that occurs when a magnetic field moves through a conductor (or vice versa). ## Definition When a magnetic field (magnet) moves through a conductor an eddy current is induced in the conductor due to the magnetic field's movement. The flow of electrons in the conductor creates an opposing magnetic field to the magnet which results in damping of the magnet and causes heating inside of the conductor similar to heat buildup inside of power cords. The loss of energy used to heat up the conductor is equal to the loss of kinetic energy by the magnet. Eddy currents induced in conductors are much stronger as temperatures approach cryogenic temperatures. This allows for critical damping for cryogenic applications and testing in the aerospace industry. ↑Jump back a section ## Equation The differential equation of motion of a magnet dropped vertically through or near a conductor, where “M” is the mass of the magnet, “K” is the damping coefficient, “V” is the velocity, “g” is gravity and “a” is the acceleration of the magnet: $Ma = Mg - KV\,$ ↑Jump back a section ## Uses • Vehicle braking • Roller coaster braking • Elevators • Near critical damping at cryogenic temperatures ↑Jump back a section ## Observing Magnetic Dampening Magnetic damping can be observed easily due to the development of strong magnets made of neodymium and other rare earth metals. To observe magnetic damping with a minimum of materials and effort, use any length of copper tubing and a neodymium magnet with a smaller diameter than the tube. Ideally the neodymium magnet will have a diameter only slightly smaller than the tube to maximize the magnetic field. Hold the tube vertically and drop the magnet through it to see magnetic damping at work. Dropping a chained line of these magnets through the copper tube shows magnetic damping, however connecting the magnets in a ring before dropping them shows minimal magnetic damping. This is due to the collapse of the magnetic field as the north and south poles of the chained magnet are connected. ↑Jump back a section ## Dangers Neodymium magnets larger than 1.5 cm are very strong and should be handled with extreme care since they can shatter at high velocities, neodymium magnets of this size can also pinch and even break bones as they are attracted to other magnets or metals. ↑Jump back a section ## Availability Radially magnetized neodymium cylinders ranging from 0.1 to 2 cm diameter are readily available for purchase online. ↑Jump back a section	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9061949849128723, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/21521?sort=votes	## Murray-von Neumann classification of local algebras in Haag-Kastler QFT ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) The Haag-Kastler approach to quantum field theory (QFT) is one of the oldest approaches to rigorously define what a QFT is, it deals with nets of operator algebras: You start with a spacetime and assign von Neumann algebras (or $C^$-algebras, but my question is about the von Neumann algebra situation only) to certain subsets of the spacetime subject to certain axioms. I am interested in results about the Murray-von Neumann classification of these algebras, i.e. which kind of factors can occur in the central decomposition (the decomposition of the algebra as a direct integral of factors). To make this more precise, here is an example: One can define vacuum representations on Minkowski spacetime, for details please see Haag-Kastler vacuum representation on the nLab. A net of a vacuum representation is said to satisfy duality, or be a dual net, if one has $\mathcal{M}(\mathcal{O}') = \mathcal{M}'(\mathcal{O})$, put in words: the algebra of the causal complement of a bounded open set $\mathcal{O}$ is the commutant of the algebra associated with $\mathcal{O}$. Then it is a theorem that algebras associated to diamonds can only have factors of type $III_1$ in their central decomposition. 1. Is the assumption of duality necessary or is causality enough? Is Haag duality enough? (Haag duality means that the duality condition does not have to hold for all algebras associated to bounded open regions, but for diamonds only). 2. What are the necessary assumptions to deduce that algebras associated to diamonds are a factor of type $III_1$, i.e. have trivial center? What are the necessary assumptions to get that these algebras are hyperfinite? 3. Are there similar results about the factor decomposition of algebras associated to more general subsets than diamonds of the Minkowski space, like open bounded subsets? 4. Are there similar results about more general spacetimes, like globally hyberbolic ones? - ## 1 Answer There is a nice overview about algebraic quantum field theory by Halverson and Müger, which covers some of the stuff I mention below and can be found at http://arxiv.org/abs/math-ph/0602036 Concerning your question(s): Having a factor of type III means that every (non-zero) projection in it is Murray-von Neumann equivalent to the identity. In the Haag-Kastler approach Borchers came up with a slightly weaker notion of "type III-ness", which is sometimes called 'property B'. Definition: Let $\mathcal{O} \rightarrow \mathcal{A}(\mathcal{O})$ be a net of von Neumann algebras on some common Hilbert space $H$. $\mathcal{A}$ has property B if for any two double cones $\mathcal{O}_1$ and $\mathcal{O}_2$ such that $\overline{\mathcal{O}_1}\subset \mathcal{O}_2$ and for any non-zero projection $E \in \mathcal{A}(\mathcal{O}_1)$ there is an isometry $V \in \mathcal{A}(\mathcal{O}_2)$, such that $VV^ = E$ and $V^V= 1$ (in other words: $E$ is equivalent to $1$ in the algebra $\mathcal{A}(\mathcal{O}_2)$). The point is the following theorem proven by Borchers in A remark on a theorem of B. Misra, H.J. Borchers, Communications in Mathematical Physics, Volume 4 (5), page 315-323. Theorem: Let $\mathcal{O} \rightarrow \mathcal{A}(\mathcal{O})$ be a net of von Neumann algebras satisfying microcausality, the spectrum condition, and weak additivity. Then the net $\mathcal{A}$ satisfies property B. So, what are these notions? • microcausality means that the algebras associated to spacelike separated double cones commute • additivity means that for any double cone $\mathcal{O}$ the set of all $\mathcal{A}(\mathcal{O} + x)$ for $x \in T$, where $T$ denotes the translation group generates the associated (universal) $C^$-algebra (I am not quite sure, what weak means in the statement). • the spectrum condition means there is a subset $T_+ \subset T$ with $T_+ \cap (-T_+) = {0}$ and the spectrum of the unitary representation of $T$ is contained in $T_+$. Since these are all motivated by physics, they are all kind of natural to demand for nets of von Neumann algebras in AQFT. Anyway there are also some results concerning the type of local algebras. For example, if you take wedge shaped regions $W$ in Minkowski spacetime, then it is shown in On the Net of von Neumann algebras associated with a Wedge and Wedge-causal Manifolds, H.J. Borchers, http://www.lqp.uni-goettingen.de/papers/09/12/09120802.html that the local algebra associated to $W$ is actually of type $III_1$ for nets satisfying similar assumptions motivated by physics. Furthermore there is the paper The Universal Structure of Local Algebras, Buchholz, D., D'Antoni and Fredenhagen, K., Communications in Mathematical Physics 111 (1), page 123-135 in which it is shown that if you assume that your net is derived from a Wightman QFT and satisfies asymptotic scale invariance and nuclearity, then the local algebras associated to double cones are of the form $\mathfrak{R} \otimes Z(\mathcal{O})$, where $\mathfrak{R}$ is the unique hyperfinite type $III_1$-factor and $Z(\mathcal{O})$ is the center of the local algebra $\mathcal{A}(\mathcal{O})$. So, this and the fact that in many examples of nets you can check directly that the local algebras are of type $III_1$ is the justification for physicists to assume this to be the case in everything physically relevant. - Thanks, that is the kind of answer I was looking for. I knew that in Minkowski spacetime the algebra of the wedge is also of type $III_1$, but did not know the paper of Borchers about "wedge-local manifolds". I'm also interested in the consequences that physicists think this fact implies, but did not ask explicitly because I considered it not to be appropriate here (it's not about mathematics). I did not mark this as the accepted answer because I hope to get further input :-) – Tim van Beek Apr 16 2010 at 12:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9235857129096985, "perplexity_flag": "head"}
http://mathhelpforum.com/geometry/91914-locus.html	# Thread: 1. ## Locus What is the locus of all points from which a given segment a is seen at a given angle A? It has a name? In spanish it is the "arco capaz" and in french "arc capable". What about english? 2. Originally Posted by mercedes What is the locus of all points from which a given segment a is seen at a given angle A? It has a name? In spanish it is the "arco capaz" and in french "arc capable". What about english? Hi mercedes. I found the article Arco capaz on the Spanish Wikipedia site, and from what I understand, an arco capaz is the locus of all points P from two distinct fixed points A and B such that the angle APB is a fixed angle. I don’t think there is a special name for this arc in English. Clearly the locus is an arc of a circle, so we just call it that in English. If the arc is smaller than a semicircle (i.e. if $\angle\,\mbox{APB}$ is greater than 90°) it is called a minor arc; if it is larger than a semicircle (i.e. the fixed angle is acute) it is called a major arc. Espero que esto te ayude.	{"extraction_info": {"found_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": 0.3, "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": 0.9245472550392151, "perplexity_flag": "middle"}
http://www.physicsforums.com/showthread.php?p=3869605	Physics Forums ## Functional integral theorem I've been watching Sidney Coleman's QFT lectures (http://www.physics.harvard.edu/about/Phys253.html). I've gotten up to his discussion of functional integration, and I have some questions. He starts out by discussing a finite-dimensional integral of a Gaussian function: $\int{\frac{d^n x}{(2\pi)^{n/2}}e^{-\frac{1}{2}xAx}} = (det A)^{-1/2}$, where $x$ is an n-dimensional vector, and $A$ an n-dimensional symmetric matrix. So far, that makes sense--if you diagonalize $A$, it just turns into the product of $n$ Gaussian integrals. He then goes on to discuss the integral of a polynomial times a Gaussian, $\int{\frac{d^n x}{(2\pi)^{n/2}}P(x)e^{-\frac{1}{2}xAx}}$, where $P(x)$ is a polynomial. Seemingly out of nowhere, he gives the result of this integral as $P(-\frac{\partial}{\partial b})(det A)^{-1/2}$. I have absolutely no idea where this comes from. Google is turning up bits and pieces of information on this, but nothing I can make a complete picture out of. The best I've been able to work out is that it's in some way related to differentiating under the integration sign, but I can't quite put the pieces together. This is clearly going to become important in the subsequent sections, where we're going to go on to develop the path integral formulation of QFT, so I'd really like to figure this out. Can anybody shed any light on how this works? PhysOrg.com physics news on PhysOrg.com >> Promising doped zirconia>> New X-ray method shows how frog embryos could help thwart disease>> Bringing life into focus Recognitions: Science Advisor Check out http://en.wikipedia.org/wiki/Gaussian_integral especially the section titled "Integrals of similar form". Zee's QFT textbook also covers it a bit, iirc. Recognitions: Homework Help Science Advisor Near as I can tell, there are some missing pieces to the formula you wrote. Perhaps these were not copied correctly in the notes? The general expressions is $$f(A,b) = \int \frac{d^n x}{(2\pi)^{n/2}} \exp{\left(-\frac{x A x}{2} + b x\right)} = \left(\det{A}\right)^{-1/2} \exp{\left(\frac{b A^{-1} b}{2}\right)}.$$ The final equality follows by completing the square under the integral and then shifting the integration variable. Now assuming that you can exchange integration and differentiation you can write $$\int \frac{d^n x}{(2\pi)^{n/2}} x^k \exp{\left(-\frac{x A x}{2}\right)} = \int \frac{d^n x}{(2\pi)^{n/2}} \left[\frac{\partial^k}{\partial b^k} \exp{\left(-\frac{x A x}{2}+bx\right)}\right]_{b=0} = \left[\frac{\partial^k}{\partial b^k} f(A,b) \right]_{b=0}.$$ The expression for general P then follows from the Taylor expansion of P. There is some indexology that I have left schematic, but let me know if you have trouble working it out. Also, note that if P has an infinite Taylor expansion then the exchange of summation and integration may not be valid i.e. one could use P(d/db) inside the integral but not outside it. ## Functional integral theorem Ahh, that's where the $b$ comes from. Yeah, he added a linear term earlier on in the lecture when discussing the integral of a generalized Gaussian, but by the time we got to this point, he'd absorbed everything into a general quadratic form $e^{-Q(x)}$, so I didn't realize that was the $b$ we were talking about. I didn't include it in my initial description because I didn't realize it was relevant, but now I see what's going on. Quote by Physics Monkey Also, note that if P has an infinite Taylor expansion then the exchange of summation and integration may not be valid i.e. one could use P(d/db) inside the integral but not outside it. $P(x)$ is given as being a finite-order polynomial, so I can say with reasonable certainty that its Taylor expansion terminates. Thank you very much--this makes perfect sense now. Thread Tools \| \| \| \| \|--------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Functional integral theorem \| \| \| \| Thread \| Forum \| Replies \| \| \| Quantum Physics \| 0 \| \| \| Calculus \| 3 \| \| \| Quantum Physics \| 2 \| \| \| Calculus & Beyond Homework \| 0 \| \| \| Advanced Physics Homework \| 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": 12, "mathjax_display_tex": 2, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9432618021965027, "perplexity_flag": "head"}
http://stats.stackexchange.com/questions/2/what-is-normality/357	# What is normality? In many different statistical methods there is an "assumption of normality". What is "normality" and how do I know if there is normality? - 1 – robin girard Jul 25 '10 at 20:13 ## 7 Answers The assumption of normality is just the supposition that the underlying random variable of interest is distributed normally, or approximately so. Intuitively, normality may be understood as the result of the sum of a large number of independent random events. More specifically, normal distributions are defined by the following function: where $\mu$ and $\sigma^2$ are the mean and the variance, respectively, and which appears as follows: This can be checked in multiple ways, that may be more or less suited to your problem by its features, such as the size of n. Basically, they all test for features expected if the distribution were normal (e.g. expected quantile distribution). - 4 Do you get bonus points for the first graph in the stats exchange? – csgillespie Jul 19 '10 at 20:53 One note: The assumption of normality is often NOT about your variables, but about the error, which is estimated by the residuals. For example, in linear regression $Y = a + bx + e$; there is no assumption that $Y$ is normally distributed, only that $e$ is. - 8 +1. Finally someone has pointed out what perhaps is the most important aspect of this question: in most situations, "normality" is important in regard to residuals or to sampling distributions of statistics, not in regard to the distributions of the populations! – whuber♦ Oct 21 '11 at 20:27 3 I would add that if $e$ is normally distributed, then Y is at least conditionally normal as well. I think its this that gets missed - people think that Y is marginally normal but its actually conditional normality that is needed. The simplest example of this is a one way ANOVA. – probabilityislogic Oct 6 '12 at 1:34 A related question can be found here about the normal assumption of the error (or more generally of the data if we do not have prior knowledge about the data). Basically, 1. It is mathematically convenient to use normal distribution. (It's related to Least Squares fitting and easy to solve with pseudoinverse) 2. Due to Central Limit Theorem, we may assume that there are lots of underlying facts affecting the process and the sum of these individual effects will tend to behave like normal distribution. In practice, it seems to be so. An important note from there is that, as Terence Tao states here, "Roughly speaking, this theorem asserts that if one takes a statistic that is a combination of many independent and randomly fluctuating components, with no one component having a decisive influence on the whole, then that statistic will be approximately distributed according to a law called the normal distribution". To make this clear, let me write a Python code snippet ````# -- coding: utf-8 -- """ Illustration of the central limit theorem @author: İsmail Arı, http://ismailari.com @date: 31.03.2011 """ import scipy, scipy.stats import numpy as np import pylab #=============================================================== # Uncomment one of the distributions below and observe the result #=============================================================== x = scipy.linspace(0,10,11) #y = scipy.stats.binom.pmf(x,10,0.2) # binom #y = scipy.stats.expon.pdf(x,scale=4) # exp #y = scipy.stats.gamma.pdf(x,2) # gamma #y = np.ones(np.size(x)) # uniform y = scipy.random.random(np.size(x)) # random y = y / sum(y); N = 3 ax = pylab.subplot(N+1,1,1) pylab.plot(x,y) # Plotting details ax.set_xticks([10]) ax.axis([0, 2*N 10, 0, np.max(y)1.1]) ax.set_yticks([round(np.max(y),2)]) #=============================================================== # Plots #=============================================================== for i in np.arange(N)+1: y = np.convolve(y,y) y = y / sum(y); x = np.linspace(2np.min(x), 2np.max(x), len(y)) ax = pylab.subplot(N+1,1,i+1) pylab.plot(x,y) ax.axis([0, 2N 10, 0, np.max(y)1.1]) ax.set_xticks([2i 10]) ax.set_yticks([round(np.max(y),3)]) pylab.show() ```` As can be seen from the figures, the resulting distribution (sum) tends towards a normal distribution regardless of the individual distribution types. So, if we do not have enough information about the underlying effects in the data, normality assumption is reasonable. - You can't know whether there normality and that's why you have to make an assumption that's there. You can only prove the absence of normality with statistic tests. Even worse, when you work with real world data it's almost certain that there isn't true normality in your data. That means that your statistical test is always a bit biased. The question is whether you can live with it's bias. To do that you have to understand your data and the kind of normality that your statistical tool assumes. It's the reason why Frequentist tools are as subjective as Bayesian tools. You can't determine based on the data that it's normally distributed. You have to assume normality. - 3 You cannot prove anything using statistics. A proof is meant to be exact. Statistics is about probabilities. Even a p=0.99 result of a Chi squared does not "prove" that the underlying distribution is not normal. It's just damn unlikely that it's normal. – xmjx Jun 16 '11 at 7:53 @xmjx: You can't even say that a given distribution is probably normal distributed. If you have a distribution with where 99.99% of your values are 1 but 0.01% of your values are 1000000 a statistic test that samples 100 values has a good chance to tell you wrongly that your distribution is normally distributed. – Christian Oct 21 '11 at 19:44 1 I'm not much of a statistical expert, so this may seem like a silly question... doesn't "true normality" exist in the underlying process that generates the variable rather than the data? It may seem like a silly distinction, but perhaps it could save some soul-searching. If the gathered data is not exactly normal, but the underlying random process works in a basically normal way, is that a situation where you could decide to "live with the bias"? – Jonathan Oct 21 '11 at 21:28 @Christian - your comment that "...100 values has a good chance..." isn't borne out at all by my hacking: x=c(rep(1,99),rep(1000000,1)); ks.test(x,pnorm) > The assumption of normality is still "rejected" by the K-S Test. – rolando2 Oct 5 '12 at 22:27 I like this answer (+1) but it is a bit pessimistic about what can be done with the assumption of normality. It is usually a good starting point for any modelling, and you can generalise to a very wide class of distributions by taking either mixtures or functions of normally distributed random variables. – probabilityislogic Oct 6 '12 at 1:46 The assumption of normality assumes your data is normally distributed (the bell curve, or gaussian distribution). You can check this by plotting the data or checking the measures for kurtosis (how sharp the peak is) and skewdness (?) (if more than half the data is on one side of the peak). - 2 What levels of kurtosis and skewdness are acceptable to meet the assumption of normality? – A Lion Jul 19 '10 at 19:38 5 Most statistical methods assume normality, not of the data, but rather of an assumed random variable, e.g. the error term in a linear regression. Checking involves looking at the residuals, not the original data! – Statprof Jul 19 '10 at 22:24 Other answers have covered what is normality and suggested normality test methods. Christian highlighted that in practice perfect normality barely exists. I highlight that observed deviation from normality does not necessarily mean that methods assuming normality may not be used, and normality test may not be very useful. 1. Deviation from normality may be caused by outliers that are due to errors in data collection. In many cases checking the data collection logs you can correct these figures and normality often improves. 2. For large samples a normality test will be able to detect a negligible deviation from normality. 3. Methods assuming normality may be robust to non-normality and give results of acceptable accuracy. The t-test is known to be robust in this sense, while the F test is not source(permalink). Concerning a specific method it's best to check the literature about robustness. - I think the reason why normality is a good assumption is because of its relative lack of use of the data - only the first two moments are used in estimation with the normal distribution. This makes diagnostic checking of a least squares model very easy - basically you just look for outliers which could influence the sufficient statistics. – probabilityislogic Oct 6 '12 at 1:41 To add to the answers above: The "normality assumption" is that, in a model $Y=\mu+X\beta +\epsilon$, the residuak term $\epsilon$ is normally distributed. This assumption (as i ANOVA) often goes with some other: 2) The variance $\sigma^2$ of $\epsilon$ is constant, 3) independence of the observations. Of this three assumptions, 2) and 3) are mostly vasly more important than 1)! So you should preoccupy yourself more with them. George Box said something in the line of ""To make a preliminary test on variances is rather like putting to sea in a row boat to find out whether conditions are sufficiently calm for an ocean liner to leave port!" - [Box, "Non-normality and tests on variances", 1953, Biometrika 40, pp. 318-335]" This means that, unequal variances are of great concern, but actually testing for them is very difficult, because the tests are influenced by non-normality so small that it is of no importance for tests of means. Today, there are non-parametric tests for unequal variances that DEFINITELY should be used. In short, preoccupy yourself FIRST about unequal variances, then about normality. When you have made yourself an opinion about them, you can think about normality! Here is a lot of good advice: http://rfd.uoregon.edu/files/rfd/StatisticalResources/glm10_homog_var.txt - I am quite sure my interpretation is right. Box has also written at length about this in Box, Hunter & Hunter: Statistics for Experimenters which I have read thoroughly. But now I see that what I wrote about whas not what I meant, it should say ...then about normality! unequal variances are much more important than normality. Of course , independence is the mother of all assumptions. – kjetil b halvorsen Oct 5 '12 at 22:06 1 will edit, I saw the error in my answer while writing. – kjetil b halvorsen Oct 5 '12 at 22:11	{"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": 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": 0.916331946849823, "perplexity_flag": "middle"}
http://nrich.maths.org/7531&part=	nrich enriching mathematicsSkip over navigation ### Modular Fractions We only need 7 numbers for modulus (or clock) arithmetic mod 7 including working with fractions. Explore how to divide numbers and write fractions in modulus arithemtic. ### Readme Decipher a simple code based on the rule C=7P+17 (mod 26) where C is the code for the letter P from the alphabet. Rearrange the formula and use the inverse to decipher automatically. ### Double Time Crack this code which depends on taking pairs of letters and using two simultaneous relations and modulus arithmetic to encode the message. # Function Pyramids ##### Stage: 5 Challenge Level: This problem explores structures like those found in Number Pyramids and More Number Pyramids. A function pyramid is a structure where each entry in the pyramid is determined by the two entries below it, together with some function: The function used in the problems Number Pyramids and More Number Pyramids could be expressed as: $$f(a,b) = a+b$$ Here is a function pyramid for you to explore. Type some numbers into the three spaces on the bottom layer. Can you figure out how the rest of the pyramid is generated? Here are some questions you might like to consider: Can you choose numbers for the bottom layer so that the number 1 appears in the top cell? Can you choose numbers for the bottom layer so that the number 5 appears in the top cell? Can you identify the function used to determine the next layer, given the bottom layer? Can you choose numbers for the bottom layer so that the number in the top cell is negative? Why not make up some function pyramids of your own, and ask yourself some similar questions? 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": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.8637193441390991, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/65202/how-to-prove-log-n-n/65205	# How to prove $\log n < n$? Sorry if this is a silly question but most books claim $\log n < n$ for $n \geq 1$ without providing any proof, saying it's too obvious. Could someone give me a rigorous proof? Is there some trick or method I forgot about? - 1 In fact, $\log x \leq x-1$ for any $x \geq 0$. This is equivalent to the inequality $1+t \leq e^t$ for all $t \in \mathbb R$. – Srivatsan Sep 17 '11 at 3:06 Hint: $e^x = 1 + x + x^2/2 + x^3/6 ... > 1 + x$ when $x>0$. – Thomas Andrews Sep 17 '11 at 4:19 1 – Arturo Magidin Sep 17 '11 at 4:57 ## 6 Answers If we let $f(n) = \log n$ and $g(n) = n$, then $f'(n) = \frac{1}{n}$ and $g'(n)=1$. Since $f(1) < g(1)$ and $f'(n) \leq g'(n)$ for $n \geq 1$, we must have $f(n) < g(n)$ for all $n \geq 1$. The idea in the last sentence is sometimes called the racetrack principle: If $f(n)$ and $g(n)$ denote the positions of horses $f$ and $g$ at time $n$, horse $f$ starts behind horse $g$ (i.e, $f(1) < g(1)$), and at any given time horse $f$ is never faster than horse $g$ (i.e, $f'(n) \leq g'(n)$) then horse $f$ will always be behind horse $g$ (i.e, $f(n) < g(n)$ for $n \geq 1$). OP asks for a proof of the racetrack principle. The racetrack principle: If $f(a) < g(a)$ and $f'(n) \leq g'(n)$ for $n \geq a$, then $f(n) < g(n)$ for $n \geq a$. Proof: Let $h(n) = f(n) - g(n)$. Then $h'(n) = f'(n) - g'(n) \leq 0$. The mean value theorem tells us that there exists some point $x \in [a,n]$ such that $$h'(x) = \frac{h(n) - h(a)}{n-a}.$$ Since we know the claim is true for $n=a$, take $n > a$. We already know $h'(x) \leq 0$, so we get $h(n) - h(a) \leq 0$, which means $f(n) - g(n) \leq f(a) - g(a) < 0$, and so $f(n) < g(n)$ for $n \geq 1$. - Oops - cross-posted the same thing. Sorry! – Billy Sep 17 '11 at 3:06 How do I prove this general theorem? Is it possible to do using the Big-O definition? – Mark Sep 17 '11 at 3:08 @Mark: I'll add a proof. – Mike Spivey Sep 17 '11 at 3:12 On the account of your proof,I guess we can call it a variation of racetrack principle :-) – Quixotic Sep 17 '11 at 3:35 How rigorous do you need? If it's rigorous enough in your context, prove that log(1) < 1 (shouldn't be hard!), and then show that the derivative of log(x) is less than or equal to the derivative of x for all $x \geq 1$. - Could you prove this using Big-O definition? – Mark Sep 17 '11 at 3:16 2 @Mark: No, this isn't what big O notation is about. Big O describes limiting behaviour (e.g. as two functions go to infinity), and says nothing about what they do elsewhere. – Billy Sep 17 '11 at 3:18 You are right, but I wonder how one should prove this without using calculus. Are there more primitive tools? – Mark Sep 17 '11 at 3:21 2 @Mark: That depends on how you define the natural logarithm in the first place. The most direct definitions I am aware of, either as $\log x = \int_1^x dt/t$, or as the inverse of the exponential function defined as $d/dx \exp x = \exp x$ and $\exp 0 = 1$, are definitions which themselves require calculus. – Rahul Narain Sep 17 '11 at 3:57 just simply look at the function $f(t)=t-log(t)$. You can show that this function is always increasing and that $f(n)\ge f(1)=1$ for every $n$. - I will assume that by $\log n$ we mean the natural logarithm of $n$. If $b > 1$, then $\log b$ is the area under the curve $y=1/x$, above the $x$-axis, from $x=1$ to $x=b$. This area is clearly less than the area of the rectangle with base $b-1$ and height $1$, so $\log b <b-1$. Comment: It is not uncommon in calculus courses to define $\log b$ as above, and then introduce the exponential function. If we want to make the argument rigorous, we would probably introduce the definite integral before introducing the derivative, and define $\log b$ as $\int_1^b \frac{1}{t} dt$. The inequality we need is an easy consequence of the definition of Riemann integral. - For $n\geq 1$, $\log n < n$ iff $n < e^n = 1 + n + n^2/2! + n^3/3! + \cdots$ - In your comments you seem ask about this in the context of the big O notation -- e.g., the concept frequently used in computer science that when analyzing an algorithm for time (or resource consumption like RAM). In big O notation, the following order appears (with constant time as best, and exponential time being worst): • $O(1)$ - constant time • $O(\log N)$ - logarithmic • $O(N)$ - linear • $O(N \log N)$ - loglinear • $O(N^2)$ - quadratic (followed by other polynomial times e.g., $O(N^3)$ - cubic, etc.) • $O(2^N)$ - exponential time You aren't going to come up with a mathematical proof for this ordering that shows for every $N$ that a $O(\log N)$ function will be smaller than every $O(N)$ function, because it simply isn't true. To demonstrate with a counterexample, let $f(N) = 10^{100} \log N$ (an $O(\log N)$ algorithm; you ignore the constant multiplier), and let $g(N) = N$ ($O(N)$ algorithm). While $N \lt 10^{98}$, $f$ the logarithmic function will be larger (and hence slower; less optimal) than $g$ the linear-time function, opposite to what you usually expect. The point of big O notation is that the scaling what usually matters most is how the functional will scale for large N. Comparing any logarithmic and linear function, the logarithmic function will always be smaller than the linear function for all values of $N$ larger than some finite number. You would say that a $O(\log N)$ function grows asymptotically slower than a $O(N)$ function. Note in many cases like comparing $f(N) = N$ and $g(N) = \log N$ it will be true over the entire domain of $N$ (equivalent to $N \gt 0$). The power of big-O notation is that is if you know you have an $O(N)$ and an $O(\log N)$ function and both take 2 seconds to do when $N = 100$, when you need to process $N = 100000$ cases, the O(N) function will worst-case take roughly 2000 seconds, but the $O(log N)$ function will take only about 5 seconds. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 101, "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": 0.9094063639640808, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/356/how-can-i-measure-the-mass-of-the-earth-at-home?answertab=active	# How can I measure the mass of the earth at home? How can I measure the mass of the earth at home? How was the mass of the earth first measured? - 1 Archimedes could have done it with a lever: "give me a place to stand on, and I will move the earth" (not a place at home however). – Gerard Nov 8 '10 at 10:22 2 – sigoldberg1 Nov 9 '10 at 22:53 I added two new tags of "experimental-physics" and "education" to your question. I hope you don't mind. For me, a complete success with this question is for a couple of kids to get together and actually do the experiment, maybe asking questions along the way. Then I would add the tag "experiment" to their question, to indicate that it is dealing with a real, live, experiment. – sigoldberg1 Nov 10 '10 at 21:25 ## 4 Answers Yes we/you can. I recall seeing a famous video of a homemade version of the Cavendish torsion balance experiment from the early 1960's, made I think for the PSSC high school course. Basically, the physicist hung a torsion balance from a high ceiling by a long (>10 m?) piece of computer data tape (chosen because it would not stretch). He carefully minimized air currents. The torsion masses were two .5 kg bottles of water on a wooden bar (no magnetic interference). Mass, in the form of boxes of sand, say 20kg was piled around on the floor as static mass and then reversed in position with respect to the suspended masses. There was a clear plastic box around the balance (with a hole in its top for the suspending tape to pass through) also to minimize the effect of air currents, since the lateral force on each bottle is about Gm1m2/r^2 = (6.7e-11)0.5kg20kg/(0.1m)^2 N ~ 6.7e-8 N, i.e. a lateral force on each bottle equivalent to that generated by a weight of about 7 micrograms, about that of a 1 mm^3 grain of sand. This is visible to us because the long arms of the torsion balance convert this small force into a torque on the suspending filament, and the restoring torque is itself very small. I found an Italian dubbed version of the video on Youtube. See http://www.youtube.com/watch?v=uUGpF3h3RaM&feature=related and a slightly longer version at http://www.youtube.com/watch?v=V4hWMLjfe_M&feature=related. I believe the demonstrator was Prof. Jerrold Zacharias from MIT and the PSSC staff. If anyone can point me to the original undubbed black and white film loop, I'd appreciate it. It looked really crude but qualitatively it worked. The mirror moved upon reversal of the mass positions. Yeah, experimental physics!! Calculate it out. Use your laser pointer. Glue mirrors. Calibrate. Give it as an experiment in class. Make a (music?) video. Put it on Youtube and embed it here. Social physics. I also found some other do it your self experimenters with crude equipment, experimental tips (try fishing line) and different masses. See http://funcall.blogspot.com/2009/04/lets-do-twist.html http://www.hep.fsu.edu/~wahl/phy3802/expinfo/cavendish/cenco_grav.pdf and http://www.fourmilab.ch/gravitation/foobar/, which uses a ladder, some cobblestones, monofilament fishing line and has videos. For the experiment in this last reference, you don't need mirrors, since you can see the balance masses move directly because their excursion is so large. See also http://www.youtube.com/watch?v=euvWU-4_B5Y For all these experiments there is no calibration of the restoring force of the twisted filament (which Cavendish did from the free torsion period of the balance), the balance beam of one appears to be styrofoam, (so I would worry about subtle charge effects), and the beam hits the support of the fixed masses so that it bounces and we do not see the harmonic angular acceleration we might expect. This last problem is apparently well known to amateur experimenters in this field. Another exposition and video is at http://www.juliantrubin.com/bigten/cavendishg.html The best summary and historical exposition I found is at http://wapedia.mobi/en/Torsion_bar_experiment . I did not realize that the experiment was originally designed by John Michell, a contemporary, whose designs and apparatus passed to Cavendish upon his death. See http://wapedia.mobi/en/John_Michell. Newton had considered the deviation from vertical that a stationary pendulum would have near a terrestrial mountain in the Principia (1686). Although he considered the deviation too small to measure, it was measured 52 years later at Chimborazo, Ecuador in 1738, which was the first experiment showing that the Earth was not hollow, apparently a live hypothesis at the time. The same experiment was repeated in Scotland in 1774. See http://wapedia.mobi/en/Schiehallion_experiment . Mitchell devised the torsion balance experiment in 1783, and started construction of a torsion balance. Cavendish did his experiment in 1797-1798. To me this is all quite inspiring. Editorial (I'll move this positive rant to meta soon) - given the obviously widely varied audience on this site, I would very much like to see more questions like this one relating to amateur or home experiments. The analysis of the data and possible sources of errors in these experiments is often subtle, and is very instructive. To have real physicists and other clever students publicly criticize some aspect of an experiment provides something that many students may never get otherwise. The social network framework will help many newcomers from different countries learn what real science is in a way that yet another dose of imperfectly understood theory never will. And it's fun too. - I thought of this recently as a "seeding" question, but I wasn't really expecting to see it. – Mark C Nov 9 '10 at 18:09 I would love to see this video too. – j.c. Nov 9 '10 at 21:38 1 I found a Spanish dubbed version, see above. – sigoldberg1 Nov 11 '10 at 12:38 You can make estimates of the earth's mass $M_e$ by estimating it's average density $\rho$ and using the formula $M_e = \rho \cdot V$, where of course $V= \frac{4}{3} \pi R^3$, so you have to know the earth's radius $R$. This method is rather a gamble because you don't know the earth's average density (suppose the earth would be hollow?). Newton made such an estimate himself. Newton's law of gravitation $F_{grav} = G \cdot \frac{M_e \cdot m}{R^2} = m \cdot g$ describes the attraction between a mass $m$ on the earth's surface and the earth. While $m$, $g$, so $F_{grav}$ and $R$ can be determined, there are two unknowns: $M_e$ and $G$. So one can only determine the mass of the earth together with this constant $G$: the gravitational constant. You cannot measure the mass of the earth at home, unless you have access to a Cavendish torsion balance to measure the gravitational constant. The Cavendish experiment is described in this Wikipedia article. Approx. 100 years after Newton the first experiment was done to measure $G$. The value of $G \cdot M_e$ can be measured very accurately (3,9860042·1014 m³/s²). The value of $G$ can be measured less accurately (6,674·10-11 m³/kg·s²) so that's the bottleneck for the determination of the earth's mass. The earth's mass is not constant, it is estimated that every day 40 tons of dust from meteorites hit it. This is totally neglectible for a mass of 5,9736 ∙ 10^24 kg. - As Gerard stated you can't directly measure the mass of the Earth. You can measure the acceleration of gravity at the Earth's surface using a simple pendulum. You need to know two things: The length of the pendulum, which can be accurately and easily measured with a ruler. The period of the oscillation, which can be measured by taking a stop-watch and timing a number of small swings. Then use the formula: $g = \frac{4 \pi ^2 L}{T^2}$ Where L is the length and T is the period. This is then related to the mass by: $M = \frac{g R^2}{G}$ Where R is the radius of the earth and G is Newtons gravitational constant. You still need to know these other quantities, as Gerard said. I don't know how you would find R or G at home. Edit: Formatted the Formulae using Latex. - 1 just start using latex and it will parse automatically – Mark Eichenlaub Nov 9 '10 at 11:18 Thank you, originally I tried it, but I didn't realise you had to wait and click outside the edit box so that the parser has time to change it in the preview. – Dom Nov 10 '10 at 15:25 You can't measure the mass of Earth directly, as others have stated. You can calculate it knowing: • The value of $g$, the gravitational acceleration (approximately $9.8 m/s$) • The value of $R_e$, the radius of the Earth (approximately $6378.1 km$) • The value of $G$, the gravitational constant (approximately $6.67×10^{-11} N m^2/kg^2$) and solving the following equation: $mg = \frac{GM_em}{R_e^2}$ or $M_e = \frac{g R_e^2}{G}$ Now: • to measure $g$ you can use a pendulum - this can be done at home. • to measure $R_e$ the simplest experiment is Eratosthenes' experiment - this cannot be done at home • to measure $G$ you need to use a Cavendish balance - which cannot be done at home because it's a notoriously difficult experiment (the constant is really small, requires custom apparatus, a very long time, etc.). - 2 Eratosthenes's experiment can be done, however, by two kids at two homes via Skype. In 300BC it involved two big technologies of his day, i.e. the calendar and the mail. His experiment apparently started when he, as librarian at Alexandria got a letter from someone saying that on a certain day, he could see the sun reflected from the water at the bottom of a deep vertical well in his home town at noon. Eratosthenes did the experiment and the math. Today we could use Skype and kids in different places, vertical plumb lines, protractors (Greek?!), and do the math. Don't give up so easily. – sigoldberg1 Jan 27 '11 at 4:16 @sig stress, but you stil need a copule of things which disqualify the experiment from the "home" category: a large vertical structure of known height, and a friend living at a significantly different latitude. – Sklivvz♦ Jan 27 '11 at 7:27	{"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": 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": 0.9422292113304138, "perplexity_flag": "middle"}
http://en.wikisource.org/wiki/Page%3AA_Treatise_on_Electricity_and_Magnetism_-_Volume_2.djvu/71	# Page:A Treatise on Electricity and Magnetism - Volume 2.djvu/71 From Wikisource Jump to: navigation, search This page has been proofread, but needs to be validated. 420.] 39 GENERATION OF A SOLID ANGLE. During this motion the element QQ' will generate an area in the form of a parallelogram whose sides are parallel and equal to QQ' and PP'. If we construct a pyramid on this parallelogram as base with its vertex at P, the solid angle of this pyramid will be the increment dω which we are in search of. To determine the value of this solid angle, let θ and θ' be the angles which ds and dσ make with PQ respect ively, and let φ be the angle between the planes of these two angles, then the area of the projection of the parallelogram ds . dσ on a plane perpendicular to PQ or r will be $ds d\sigma \sin \theta \sin \theta' \sin\phi,\,$ and since this is equal to r2dω, we find $d\omega = \Pi ds d\sigma = \frac{1}{r^2} \sin\theta \sin\theta' ds d\phi.$ (2) Hence $\Pi = \frac{1}{r^2} \sin\theta \sin\theta' \sin\phi.$ (3) 420.] We may express the angles θ, θ', and φ in terms of r, and its differential coefficients with respect to s and σ, for $\cos\theta = \frac{dr}{ds}, \quad \cos\theta = \frac{dr}{ds\sigma}, \quad \text{and } \sin\theta \sin\theta'\cos\phi = r \frac{d^2r}{dsd\sigma}.$ (4) We thus find the following value for Π2, $\Pi^2 = \frac{1}{r^4} \left[1 - \left(\frac{dr}{ds}\right)^2 \right] \left[1 - \left(\frac{dr}{d\sigma}\right)^2 \right] - \frac{1}{r^2} \left(\frac{d^2}{dsd\sigma}\right)^2.$ (5) A third expression for Π in terms of rectangular coordinates may be deduced from the consideration that the volume of the pyramid whose solid angle is dω and whose axis is r is $\frac{1}{3} r^3 domega = \frac{1}{3} r^3 \Pi ds d\sigma.$ But the volume of this pyramid may also be expressed in terms of the projections of r, ds, and dσ on the axis of x, y and z, as a determinant formed by these nine projections, of which we must take the third part. We thus find as the value of Π, $\Pi = \frac{1}{r^3} \begin{vmatrix} \xi - x, & \eta - y, & \zeta - z, \\ \frac{d\xi}{d\sigma}, & \frac{d\eta}{d\sigma}, & \frac{d\zeta}{d\sigma}, \\ \frac{dx}{ds}, & \frac{dy}{ds}, & \frac{dz}{ds}. \end{vmatrix}$ (6)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 7, "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": 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": 0.9111486673355103, "perplexity_flag": "middle"}
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http://physics.stackexchange.com/questions/11939/what-is-the-lagrangian-for-a-relativistic-charge-that-includes-the-self-force/11946	# What is the Lagrangian for a relativistic charge that includes the self-force? The usual Lagrangian for a relativistically moving charge, as found in most text books, doesn't take into account the self force from it radiating EM energy. So what is the Lagrangian for a relativistic charge that includes the self-force? - 3 – Arun Nanduri Jul 6 '11 at 15:34 To properly address self-energy issues, one has to abandon a classical description and consider full-fledge quantum-electro-dynamics (QED) and renormalization. – Qmechanic♦ Jul 6 '11 at 15:50 2 I wouldn't go as far as what Qmechanic says, but it is clear that one must abandon classical Maxwellian description of electromagnetism. It is actually quite an active research topic whether there can be non-linear theories of classical electrodynamics that can treat self-force. For an example, see M. Kiessling, J. Stat. Phys. (2004) vol 116 p1057. Kiessling has a second paper in the same issue of the journal starting on p1123 dealing with the quantum version of the procedure. – Willie Wong Jul 6 '11 at 16:38 @Willie is the problem that an acceleration is modelled as the cause of the proper self force? Usually, proper force is the cause of acceleration, and independent of the acceleration it causes. – Larry Harson Jul 7 '11 at 10:40 ## 5 Answers classical electrodynamics mainly deals with two kinds of proplems: a) The action of a field on a charged particle and b) the fields arising from the motion of such a field. Of course, this can only be approximative but it turns out that a lot of phenomena can be described in this way. However, you are right, an entire treatment would include a) and b) simultaniously - including the whole dynamics of such a system with radiative reaction (or, the Abraham-Lorentz-Force and its relativistic counterpart). But as Qmechanic pointed out in a comment, there may be no fully consistent way to do so within the framework of classical electrodynamics. Jackson (Chapter 16) states: The difficulties presented by this problem touch one of the most fundamental aspects of physics, the nature of an elementary particle. Although partial solutions, workable within limiting areas, can be given, the basic problem remains unsolved. Thus, you will have to search for a really satisfactory answer within the description of Quantum Electrodynamics. Otherwise, you may include the effect phenomenologically. This was done e.g. by Barone and Mendes in Lagrangian description of the radiation damping (PRA, 2007) and they give a Lagrangian of the form $$\mathcal{L}=\frac{m}{2}\dot{\mathbf{r}}_1\cdot\dot{\mathbf{r}}_2 - \frac{\gamma}{2}\epsilon\left(\dot{\mathbf{r}}_1,\ddot{\mathbf{r}}_2\right) -V\left(\mathbf{r}_1,\mathbf{r}_2\right),$$ with $\gamma := 2e^2/3c^3$, with $\epsilon = \epsilon_{ij}dx^idx^j$ beeing the Levi-Cevita tensor, and the $\mathbf{r}_i$ arise from the special treatment of the problem used in the paper employing some kind of image phase-space representation of the system. Furthermore, $V\,$ is the potential related to the Abraham–Lorentz–Dirac-Force which is given explicitely in the paper. - Solving both, Lorentz force and Maxwell's equation, does include radiation reaction. Radiation reaction emerges from the interaction of the charge with the field it has emitted itself, the self-field. If you do not want to solve Maxwell's equations and treat the fields as external, you can use Lorentz-Abraham-Dirac equation with some known inconsistencies for unphysical initial conditions, or the Landau-Lifshitz equation as approximation. Some authors claim that Landau-Lifshitz equation is not an approximation but agrees with solutions Lorentz-Abraham-Dirac equation for physical initial conditions. Lorentz-Abraham-Dirac and Landau-Lifshitz equations cannot be derived from Hamilton's principle because radiation emission is a dissipative process. For a pedagogical review see http://arxiv.org/abs/gr-qc/9912045. - It is the usual CED Lagrangian for a charge and electromagnetic field; it includes the free field Lagrangian, a free particle Lagrangian and the interaction Lagrangian $jA$. Self-action is encoded in it when you consider the system as a whole, as coupled unknown variables. Of course, it yields rubbish: an infinite "electromagnetic mass" that you need to discard. I say "rubbish" because, on one hand, we define the charge as a stable, fixed entity; on the other hand we introduce a self-force to get some dynamics due to it. Our definition contradicts our intentions. - I would say that it is already in the Lagrangian, problem being: in the classical ED lagrangian you describe point-like particles, and the self interaction is classically described as the classical lorentz force a charge creates on itself if it is a sphere-like particle, considering retarded potentials. A classical solution would be to include the shape in the lagrangian, but the lagrangian is local, so it deals with points. If someone talks to you about QED, it is because classically you reach the conclusion that the reaction creates a mass like term opposing acceleration, and this mass can become infinite, in QED you also have some infinites, but you can remove them. Nonetheless I would say you remove it because what actually interests QED is the measure you make, and you never really reach the size of a point while measuring, but it might be an over interpretation. BUT I have a good news for you there exists a classical formulation that includes the retroaction, to the expense of retrocausality, or waves flowing backward in time, or a kind of action at a distance: it is the lagrangian you can find in the Feynman-Wheeler absorber theory. very interesting but Feynman abandoned it. - If the charge is point-like then its classical self-energy is infinite. As the self-force comes from the gradient of potentials, the result is undetermined. Consider a theory with a local interaction. Then we must take into account some field, e.g. the electromagnetic field. Now for a massive relativistic particle moving at constant velocity there is always an inertial frame of reference where the particle is not moving. Therefore the magnetic field vanishes and the electric field is that given by Coulomb's law, i.e. $$\mathbf E(\mathbf r)=\frac q{4\pi\epsilon}\frac{\mathbf r}{\Vert \mathbf r\Vert^3}.$$ Since the particle stays at $\mathbf r=0$, the force is undefined. So far the only massless particle known is the photon, which is neutral. - -1 The OP question is about Lagrangian. – Vladimir Kalitvianski Jul 6 '11 at 22:55 1 Potential energy enters into the Lagrangian... – Phoenix87 Jul 7 '11 at 8:25 1 Yes, the potential energy is a part of the Lagrangian but it depends on not-yet-determined variables there. One does not use it for calculations because the Lagrangian serves first of all to get the equations. In particular, from CED Lagrangian and the Noether theorem it follows that the total energy is conserved, and this energy includes a self-action contribution. The force is a term from equations of motion, not from the Lagrangian. – Vladimir Kalitvianski Jul 7 '11 at 14:04 – Ben Crowell Aug 6 '11 at 17:46	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 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": 0.9294325113296509, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/43269/violation-of-newtons-3rd-law-and-momentum-conservation	# Violation of Newton's 3rd law and momentum conservation Why and when does newtons 3rd law violate in relativistic mechanics? Check this link http://www.animations.physics.unsw.edu.au/jw/Newton.htm. - 2 It would help a lot if you include in this question the reason why you would expect Newton's 3rd law to be violated in relativistic mechanics. Don't rely on people being able to click the link. – David Zaslavsky♦ Nov 2 '12 at 16:48 – Qmechanic♦ Nov 2 '12 at 17:32 ## 1 Answer Newton's third law is naively violated in relativistic mechanics when there is field potential momentum. This happens in basically any magnetic field situation where there are also charged objects. Feynman gives a simple example, two charge particles, one moving directly towards the other and the other one moving in some other random direction (not towards the first). The electric forces are nearly equal and opposite (up to relativistic corrections that are lower order) and the magnetic force (the first relativistic correction to the Newton's third law consistent Coulomb repulsion) is only nonzero on one of the particles, since there is no magnetic field along the line of motion by symmetry. Newton's third law always holds in special relativity if it is expressed as conservation of particle momentum plus field momentum, since it is a consequence of translational invariance plus a Lagrangian formulation, or of translational invariance plus a Hamiltonian formulation, or of translational invariance plus quantum mechanics. This implies conservation of momentum, which implies that any forces between two bodies must be balanced (since the force is the flow of momentum). For direct 3-body forces, Newton's 3rd law, as stated by Newton, fails even in the nonrelativistic limit, as explained in this answer: Deriving Newton's Third Law from homogeneity of Space . There is a stronger formulation of momentum conservation that holds in relativistic field theories. Energy momentum is not just conserved, it is locally conserved, which means that there is a stress energy tensor $T^{\mu\nu}$ which obeys $$\partial_\mu T^{\mu\nu} = 0$$ This equality tells you that the flow of momentum density ($T^{0i}$ for $i=1,2,3$) across any surface is conservative, with a current equal to the stress (the force per unit area across an infinitesimal surface). This is the local form of Newton's third law "The force is equal and opposite". If you also add "The force between distant objects is collinear" (which is implied by conservation of momentum), the relativistic version is that there is a stress tensor choice which obeys $$T^{\mu\nu} = T^{\nu\mu}$$ This says that the flow of the j-component of momentum in the i-th direction is equal to the flow of the i-th component of momentum in the j-th direction, which implies collinearity of force for distant objects transferring momentum through a field. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 3, "mathjax_display_tex": 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": 0.9388008117675781, "perplexity_flag": "head"}
http://mathoverflow.net/questions/63109?sort=votes	Approximation in $L^2$ by piecewise constant functions Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Dear all, in order to prove the validity of my Galerkin approach of a certain variational problem, I need to check the so-called approximability property. In my case, it boils down to showing that for all $w\in L^2(\Omega)$, $\lim_{h\rightarrow 0}\inf_{w^h\in V^h}\|\|w-w^h\|\|=0$, where $\Omega=[0; 1]^d$, and $V^h$ is the space of piecewise constant functions on a regular (orthogonal) grid, with step $h$. It is probably a classical result. However, I've browsed quickly the finite element literature, and the regularity requirements on $w$ are usually stronger ($H^1$ for example). So, does the approximability property of $L^2$ functions by piecewise constant function hold? If yes, what theorem/author can I refer to? Thanks in advance. Best regards, Sebastien - 3 Answers A few word about the setting of $L^2$ approximation scheme by simple functions. If $(\Omega,\mathcal{A},\mu)$ is a measure space and $\mathcal{B}$ is a finite sub-algebra of $\mathcal{A}$, generated by the partition $\mathcal{E}$, the class $L^2(\Omega,\mathcal{B},\mu)$ is a finite dimensional subspace of $L^2(\Omega,\mathcal{A},\mu)$ consisting of simple functions which are constant on each element of $\mathcal{E}$. The best approximation of $w\in L^2(\Omega,\mathcal{A},\mu)$ is the simple function that takes the value $\frac{1}{\mu(E)}\int_E vd\mu$ on each set $E\in\mathcal{E}$. Further, any sequence of finite algebras $(\mathcal{B_n})_{n\in \mathbb{N}}$ that generate $\mathcal{A}$ as a $\sigma$ algebra, produce this way an approximation of the identity (the corresponding orthogonal projectors converge strongly to the identity). In general, you may be interested in the notion of conditional expectation relative to a $\sigma$-algebra (the orthogonal projector). - 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's obvious. Continuous functions are dense in $L^2$, and piecewise constant functions approximate a continuous, even in the uniform metric. Sorry, for my English. - Dmitry's answer addresses your question. If you're looking for the rate of convergence in h, then -you need to say a bit more about the regularity of w. I'd recommend the book by Brenner and Scott. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": false, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.92374187707901, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/7152/simple-question-concerning-jump-process-levy-process-model-for-a-risky-actif-p	# Simple question concerning Jump process (Lévy process) model for a risky actif price process [closed] Consider $X= \left( X_t \right)_{t\geq 0}$ is a Lévy process whose characteristic triplet is $\left( \gamma, \sigma ^2, \nu \right)$ and where its Lévy measure is $$\nu \left( dx\right) = A \sum_{n=1} ^{\infty} p^n \delta_{-n}\left( dx \right) + Bx^{\beta-1}\left( 1+x \right)^{-\alpha -\beta}e^{-\lambda x } \mathbf{1}_{\left ]0,+\infty \right[}\left( x\right)dx.$$ I'd like to know how to show that a price process $S_t = S_0 \exp\left( r t + X_t \right)$ of a risky actif under interest rate $r >0$ is well defined and admits first and second order moments. Someone could help on it, please? - This question was cross-posted here and here, where it has been answered. I'm going to close the question on Quant.SE. – chrisaycock♦ Feb 24 at 23:50 ## closed as not a real question by chrisaycock♦Feb 24 at 23:51 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.	{"extraction_info": {"found_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": 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": 0.9763640761375427, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/108945?sort=oldest	## Defining isogenies over smaller fields ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I'm having some issues with abelian varieties and fields of definition. This already became clear in my previous question on Jacobians. Here's another question. If somebody can explain some nice facts on fields on definition this would help me a lot (because these aren't the only questions I have concerning fields of definition). Let $A$ be an abelian variety over a number field $K$. Let $L/K$ be a finite field extension. Suppose that there exists an isogeny $A_L\to B$ defined over $L$. Is this isogeny defined over $K$ if $B$ can be defined over $K$? I think the answer is negative, but I have to admit that my set of examples is very scarce and, therefore, I can't find an easy counterexample. - ## 2 Answers No: Consider the elliptic curve $E: y^2 = x^3 + x$ defined over $\mathbb{Q}$. Then the isogeny $y \mapsto iy$ and $x \mapsto -x$ is defined over $\mathbb{Q}(i)$ but obviously not over $\mathbb{Q}$. In general if $K \subset L$ is Galois, then a morphism defined over $L$ comes from one over $K$ if and only if it is invariant under the action of $Gal(L/K)$. This follows from the so-called "theory of descent". See for example Bjorn Poonen's notes Rational points on varieties. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There is another obvious obstruction: by definition two twists are isomorphics above an extension, but are not isomorphic above their field of definition (and also not isogenous). I believe that these are the only two obstructions, meaning that if (1) there exist a K-rational isogeny $f:A \to B$ and (2) $End^0_L(A)=End^0_K(A)$ then every $L$-isogeny $A \to B$ is in fact $K$-rational. Indeed, since there exist a $K$-isogeny, the $\mathbb{Q}$-rank $r$ of $End^0_K(A)$ is the same as $End^0_K(B)$, and this is also the same as the $\mathbb{Z}$-rank of $Hom_K(A,B)$ (the module of $K$-isogenies), since the endomorphism ring is an order in its endomorphism algebra and $Hom_K(A,B)$ is not empty. Now by hypothesis (2), we have also the the $\mathbb{Z}$-rank of $Hom_L(A,B)$ is of rank $r$. Since $Hom_K(A,B) \subset Hom_L(A,B)$ are of same rank, for every $L$-isogeny $g:A \to B$ there exist $m$ such that $mg$ is $K$-rational. Since $mg=gm$ is $K$-rational and $m$ is $K$-rational, there exist a $K$-rational isogeny $g_2$ such that $gm=g_2 m$ by the universal property of isogenies. But then $m(g-g_2)=0$ which mean that $g=g_2$ and our isogeny $g$ was in fact $K$-rational. -	{"extraction_info": {"found_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": 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": 0.9558736085891724, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/64852/list	Return to Question 2 deleted 5 characters in body Let $k$ be a field, let $G$ be an algebraic group scheme flat over $k$ and let $T = \textrm{Spec } k[x, x^{-1}]$ be a one-dimensional torus. Does there exist • a scheme $X$ over $k$, • an algebraic $T$-action on $X$, • and a principal $G$-bundle on $X$ which cannot be $T$-equivariantly trivalized in the fppf topology? In other words, a principal $G$-bundle $P \to X$ such that there does not exist a scheme $U$ over $k$ with a $T$-action and a $T$-equivariant fppf morphism $U \to X$ such that the pull-back bundle $P \times_X U \to U$ is trivial? Here "fppf" means faithfully flat and finite presentation. 1 T-Equivariant trivialization of a principal G-bundle Let $k$ be a field, let $G$ be an algebraic group scheme flat over $k$ and let $T = \textrm{Spec } k[x, x^{-1}]$ be a one-dimensional torus. Does there exist • a scheme $X$ over $k$, • an algebraic $T$-action on $X$, • and a principal $G$-bundle on $X$ which cannot be $T$-equivariantly trivalized in the fppf topology? In other words, a principal $G$-bundle $P \to X$ such that there does not exist a scheme $U$ over $k$ with a $T$-action and a $T$-equivariant fppf morphism $U \to X$ such that the pull-back bundle $P \times_X U \to U$ is trivial? Here "fppf" means faithfully flat and finite presentation.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "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": 0.8550617694854736, "perplexity_flag": "head"}
http://mathoverflow.net/questions/16691/geometrically-interpreting-the-answer-to-a-vector-calculus-question-involving-tan	## Geometrically interpreting the answer to a vector calculus question involving tangent line segments to ellipses. ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let E be an ellipse centered at the origin on the x, y plane with major radius b and minor radius a. The length of the shortest line segment tangent to E that begins on the x-axis and ends on the y-axis is a+b. This can be shown using Lagrange multipliers. This answer is very simple and leads us to ask the following question: Can you give a geometric reason for why the length is a+b? This was originally asked to me by Frank Jones a few years ago. - ## 2 Answers There is a geometric way to show that $n$-gon circumscribed around an ellipse has minimal perimeter if it is inscribed in a confocal ellipse. From Poncelet porism (and generalization of optical property) it follows that we have continuous family of "minimal" polygons. If we know it, then it is easy to understand that the circumscribed rhomb (from your question) and the circumscribed rectangular (with perimeter $4(a+b)$) are minimal polygons. So, side of the rhomb equals $a+b$. - That is neat. akopyan, are you the coauthor of Geometry of conics: books.google.com/…? (I deleted a question asking for a reference, because I see that it can be found in that book.) – Jonas Meyer Feb 28 2010 at 22:57 1 That is a beautiful answer. Thank you! – Khalid Bou-Rabee Feb 28 2010 at 23:18 1 Yes, I'm coauthor of this book. Also, look for the famous book "Geometry" by Marcel Berger. In volume 2 you can find many interesting statements similar to the fact in the initial question. – akopyan 53 secs ago – akopyan Mar 1 2010 at 3:28 1 Poncelet's porism is quite elementary to see in this case. Consider the case when $a=b=1$ of a circle. Then the two rectangles are the same, and are inscribed in a circle of radius $\sqrt{2}$. Poncelet's porism just says that there is a 1-parameter family of circumscribing/inscribing squares, in this case related by rotation. The affine transformation $(ax, by)$ gives a projective transformation taking this picture to the ellipse with major and minor axes $a,b$. Then the rotations are sent projectively to a 1-parameter family of projective transformations fixing both ellipses. – Agol Mar 1 2010 at 17:51 3 2 Agol, it is not. In this case you obtain two similar ellipses with different foci. Reason is, under a affine transformation length of segments doesn't preserve. – akopyan Mar 1 2010 at 19:11 show 2 more comments ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. By working simultaneously in the 4 quadrants, this becomes a question of minimizing the perimeter of enclosing rhombi with diagonals on the coordinate axes. Proving the inequality was Problem of the Week No. 13 in Spring 2005 at Purdue. Here is Steven Landy's solution to the problem. The proof is geometric in the sense of Descartes rather than Euclid, and shows that the minimum is at least a+b. There's no calculus, so maybe this is close to what you're looking for. (Edit: I entertained the delusion that this might be close to answering your question only before akopyan's answer was posted.) -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9336944222450256, "perplexity_flag": "head"}
http://en.m.wikipedia.org/wiki/Section_(category_theory)	# Section (category theory) In category theory, a branch of mathematics, a section is a right inverse of a morphism. Dually, a retraction is a left inverse. In other words, if $f\colon X\to Y$ and $g\colon Y\to X$ are morphisms whose composition $f\circ g\colon Y\to Y$ is the identity morphism on $Y$, then $g$ is a section of $f$, and $f$ is a retraction of $g$. Every section is a monomorphism, and every retraction is an epimorphism; in algebra the sections are also called split monomorphisms and the retractions split epimorphisms. In an abelian category, if f:X→Y is a split epimorphism with section g:Y→X, then X is isomorphic to the direct sum of Y and the kernel of f. ## Examples In the category of sets, every monomorphism (injective function) with a non-empty domain is a section and every epimorphism (surjective function) is a retraction; the latter statement is equivalent to the axiom of choice. In the category of vector spaces over a field K, it is also true that every monomorphism and every epimorphism splits; this follows from the fact that linear maps can be uniquely defined by specifying their values on a basis. In the category of abelian groups, the epimorphism Z→Z/2Z which sends every integer to its image modulo 2 does not split; in fact the only morphism Z/2Z→Z is the 0 map. Similarly, the natural monomorphism Z/2Z→Z/4Z doesn't split because there is no non-trivial homomorphism Z/4Z→Z/2Z. The categorical concept of a section is important in homological algebra, and is also closely related to the notion of a section of a fiber bundle in topology: in the latter case, a section of a fiber bundle is a section of the bundle projection map of the fiber bundle. Given a quotient space $\bar X$ with quotient map $\pi\colon X \to \bar X$, a section of $\pi$ is called a transversal. ↑Jump back a section ## See also This category theory-related article is a stub. You can help Wikipedia by expanding it. ↑Jump back a section	{"extraction_info": {"found_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": 11, "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": 0.9053182005882263, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/239524/asymptotic-behavior-of-1-2-2-3-3-4-4-5-cdots-n-1-n-times-n	# Asymptotic behavior of $(1/2 + 2/3 + 3/4 + 4/5 + \cdots + (n-1)/n ) \times n$ I am interested in the following questions: given: $$G(n) = \left(\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n\right)n$$ 1. what is a $F(n)$ which could be an upper bound (clearly as tight as possible) for $G(n)$ for $n$ arbitrarily large ? 2. Does the series: $$\frac12 + \frac23 + \frac34 + \frac45 + \cdots + \frac{n-1}n$$ have a "name" and a sum (any reference)? - This question most likely does not belong here, but should be migrated to math.SE. This site is specifically about software Mathematica, even though Mathematica can do symbolic mathematics. If you have access to Mathematica, you can just type your query in and it will provide you with the answer. If you do not have access to Mathematica, you could try using Wolfram\|Alpha which is built using Mathematica and is quite capable of evaluating such a simple sum as yours. – Sasha Nov 17 '12 at 23:16 ## 4 Answers Let $$F(n) = \left(\dfrac{2-1}{2} + \dfrac{3-1}{3} + \dfrac{4-1}{4} + \cdots + \dfrac{n-1}{n} \right)$$ we then have that $$F(n) = \left(1 - \dfrac12 + 1 - \dfrac13 + 1 - \dfrac14 + \cdots + 1 - \dfrac1n \right) = (n-1) - \left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right)$$ Now note that $$-\left( \dfrac12 + \dfrac13 + \cdots + \dfrac1n \right) = 1 - H_n = 1 - \left(\log(n) + \gamma + \dfrac1{2n} - \dfrac1{12n^3} + \mathcal{O}(1/n^5) \right)$$ Hence, $$F(n) = n - \left(\log(n) + \gamma + \dfrac1{2n} - \dfrac1{12n^3} + \mathcal{O}(1/n^5) \right)$$ $$nF(n) = n^2 - n \log n -\gamma n - \dfrac12 + \dfrac1{12n^2} + \mathcal{O}(1/n^4)$$ You could make use of the fact that $$H_n = \log (n) + \gamma - \dfrac{\zeta(0)}n + \sum_{k=1}^{\infty} \dfrac{\zeta(-k)}{n^{k+1}}$$ to get better approximations/bounds. - Clearly $$\begin{align} \frac12+\frac23+\frac34+\frac45+\ldots+\frac{n-1}n&=n-1-\sum_{k=2}^n\frac1k\\ &=n-\sum_{k=1}^n\frac1k\\ &=n-H_n\;, \end{align}$$ where $H_n$ is the $n$-th harmonic number, so $$G(n)=n^2-nH_n\;.$$ There are good approximations of $H_n$ by relatively nice functions: $$H_n\sim\ln n+\gamma+\frac1{2n}-\frac1{12n^2}\;,$$ for instance, and the approximation can be improved by taking more terms of the series $$H_n\sim\ln n+\gamma+\frac1{2n}-\sum_{k\ge 1}\frac{B_{2k}}{2kn^{2k}}\;,$$ where the numbers $B_k$ are the Bernoulli numbers and $\gamma$ is the Euler-Mascheroni constant. - Your expression in the parentheses is $(n-1)-\left(\frac12+\frac13+\cdots+\frac{1}{n}\right)$ which is asymptotically related to $n-\ln n$. So in total you have something like $n^2-n \ln n.$Does that help? - Regarding your second question: Just type "sum 1/2 + 2/3 + 3/4 + 4/5 + 5/6 + ... + (n - 1)/n" or something similar in WolframAlpha. According to WolframAlpha this sum does not have a name but is equal to $n - {H_n}$ where ${H_n}$ is the $n$-th harmonic number. Source: WolframAlpha. - 1 Great idea, it also gives 2nd interpretation as `n-polygamma(0, n+1)-gamma` – Vitaliy Kaurov Nov 17 '12 at 23:48 – Vitaliy Kaurov Nov 18 '12 at 4:42	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 15, "mathjax_display_tex": 12, "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": 0.929256796836853, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/271927/why-historically-do-we-multiply-matrices-as-we-do/271935	# Why, historically, do we multiply matrices as we do? Multiplication of matrices — taking the dot product of the $i$th row of the first matrix and the $j$th column of the second to yield the $ij$th entry of the product — is not a very intuitive operation: if you were to ask someone how to mutliply two matrices, he probably would not think of that method. Of course, it turns out to be very useful: matrix multiplication is precisely the operation that represents composition of transformations. But it's not intuitive. So my question is where it came from. Who thought of multiplying matrices in that way, and why? (Was it perhaps multiplication of a matrix and a vector first? If so, who thought of multiplying them in that way, and why?) My question is intact no matter whether matrix multiplication was done this way only after it was used as representation of composition of transformations, or whether, on the contrary, matrix multiplication came first. (Again, I'm not asking about the utility of multiplying matrices as we do: this is clear to me. I'm asking a question about history.) - As to the last, matrix multiplication definitely came first (centuries first), and I'm reasonably certain from a compact representation of systems of linear equations. Leibniz already had a determinant formula. As I have no historic sources for first use, this doesn't answer your question though. – gnometorule Jan 7 at 4:19 Matrices are linear operators and have meaning only when it acts on vectors. Given matrices $A$ and $B$, what would we want the operator/matrix $BA$ to mean? Ideally, we would want $BA$ to mean the following. For all vectors $x$, we want $(BA)x = B(Ax)$ Once we have this i.e. $(BA)x = B(Ax)$ for all $x$, then we are forced to live with the way we currently multiply matrices. And as to why matrix-vector product is defined in the way it is, the primary reason for introducing matrices was to handle linear transformation in a notationally convenient way. – user17762 Jan 7 at 4:21 @HarryStern, explicitly, I thought. Anyway, yes, that's what I want. – msh210 Jan 7 at 4:44 I actually missed your explicit question in the middle of the body. Perhaps you should change that to the title. – Harry Stern Jan 7 at 4:46 There is another question, of course, which is not so much why matrix multiplication was defined like this, but why it stuck - why this apparently curious definition took off, and didn't die the death of so many putative definitions. And that was because it proved mathematically fruitful. – Mark Bennet Jan 7 at 8:47 ## 3 Answers Matrix multiplication is a symbolic way of substituting one linear change of variables into another one. If $x' = ax + by$ and $y' = cx+dy$, and $x'' = a'x' + b'y'$ and $y'' = c'x' + d'y'$ then we can plug the first pair of formulas into the second to express $x''$ and $y''$ in terms of $x$ and $y$: $$x'' = a'x' + b'y' = a'(ax + by) + b'(cx+dy) = (a'a + b'c)x + (a'b + b'd)y$$ and $$y'' = c'x' + d'y' = c'(ax+by) + d'(cx+dy) = (c'a+d'c)x + (c'b+d'd)y.$$ It can be tedious to keep writing the variables, so we use arrays to track the coefficients, with the formulas for $x'$ and $x''$ on the first row and for $y'$ and $y''$ on the second row. The above two linear substitutions coincide with the matrix product $$\left( \begin{array}{cc} a'&b'\\c'&d' \end{array} \right) \left( \begin{array}{cc} a&b\\c&d \end{array} \right) = \left( \begin{array}{cc} a'a+b'c&a'b+b'd\\c'a+d'c&c'b+d'd \end{array} \right).$$ So matrix multiplication is just a bookkeeping device for systems of linear substitutions plugged into one another (order matters). The formulas are not intuitive, but it's nothing other than the simple idea of combining two linear changes of variables in succession. Matrix multiplication was first defined explicitly in print by Cayley in 1858, in order to reflect the effect of composition of linear transformations. See paragraph 3 at http://darkwing.uoregon.edu/~vitulli/441.sp04/LinAlgHistory.html. However, the idea of tracking what happens to coefficients when one linear change of variables is substituted into another (which we view as matrix multiplication) goes back further. For instance, the work of number theorists in the early 19th century on binary quadratic forms $ax^2 + bxy + cy^2$ was full of linear changes of variables plugged into each other (especially linear changes of variable that we would recognize as coming from ${\rm SL}_2({\mathbf Z})$). For more on the background, see the paper by Thomas Hawkins on matrix theory in the 1974 ICM. Google "ICM 1974 Thomas Hawkins" and you'll find his paper among the top 3 hits. - 1 This is interesting, but does not really answer the question, which is mainly historical. – MJD Jan 7 at 4:33 3 Huh? The question is, historically, why do we multiply matrices the way we do. I answered that question: it is a shorthand for substituting one linear change of variables into another. – KCd Jan 7 at 4:36 2 Who thought of multiplying matrices in that way? When? How do you know that this was the first reason found for defining matrix multiplication in this way? – MJD Jan 7 at 4:38 1 What @MJD said. – msh210 Jan 7 at 4:45 1 @MJD: Cayley thought of multiplying matrices that way when he defined matrix multiplication. I won't say that it was the reason for defining matrix multiplication, but just that it was a way to think about the total process. If you read about the way linear algebra developed, you'll find that the subject came from several directions; adding one system of linear equations (with the same variables) led to addition of matrices and the algebraic calculations that arise when dealing with substitutions of linear changes of variables led to matrix multiplication. – KCd Jan 7 at 5:01 show 4 more comments ## Did you find this question interesting? Try our newsletter \begin{align} u & = 3x + 7y \\ v & = -2x + 11y \\ \\ \\ \\ p & =13u-20v \\ q & = 2u+6v \end{align} Given $x$ and $y$, how do you find $p$ and $q$? How do you wrote \begin{align} p & = \bullet x + \bullet y \\ q & = \bullet x+\bullet y\quad\text{?} \end{align} What numbers go where the four $\bullet$s are?	{"extraction_info": {"found_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": 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": 0.9499601721763611, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/184663-finding-values-b-c.html	Thread: 1. Finding the values of a, b and c. The tangent to the curve of y=ax^2 + bx +c at the point where x=2 is parallel to the line y=4x. There is a stationary point at (1, -3). Find the value of a, b and c. I was really unsure of how to work out this question, so first I differentiated y. dy/dx=2ax+b The point (1, -3) is on the curve and therefore: -3=a(1)^2+b(1)+c -3=a^2+b+c But I don't know how to find another equation, and I honestly don't even know if what I've done so far is on the right track. 2. Re: Finding the values of a, b and c. Hello, grooverandshaker! You were off to a good start . . . $\text{The tangent to the curve of }y\:=\:ax^2 + bx +c\text{ at the point where }x=2$ . . $\text{is parallel to the line }y\,=\,4x.$ $\text{There is a stationary point at }(1,-3).$ $\text{Find the values of }a, b\text{ and }c.$ $\text{The slope of the line }y\,=\,4x\text{ is }4.$ The slope of a tangent to the curve is: . $y' \:=\:2ax + b$ When $x = 2$, the slope of the tangent is $4.$ . . $\text{We have: }\:4a + b \:=\:4$ There is a stationary point at $(1,-3).$ That is, when $x = 1$, the derivative equals zero. . . We have: . $2a + b \:=\:0$ The point $(1,-3)$ is on the curve. That is, when $x = 1,\:y = -3$ . . We have: . $a + b + c \:=\:-3$ We have a system of equations: . $\begin{Bmatrix}4a + b\qquad &=& 4 & [1] \\ 2a + b \qquad &=& 0 & [2] \\ a + b + c &=& \text{-}3 & [3] \end{Bmatrix}$ Subtract [1] - [2]: . $2a \,=\,4 \quad\Rightarrow\quad \boxed{a \,=\,2}$ Substitute into [2]: . $4 + b \:=\:0 \quad\Rightarrow\quad \boxed{b \:=\:-4}$ Substitute into [3]: . $2 - 4 + c \:=\:-3 \quad\Rightarrow\quad \boxed{c \:=\:-1}$ Therefore, the function is: . $y \:=\:2x^2 - 4x - 1$ 3. Re: Finding the values of a, b and c. Thank you so much mate! That makes perfect sense. =)	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 20, "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": 0.8753382563591003, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/23649/limit-approaches-infinity-on-one-side-and-negative-infinity-on-other-side	# Limit approaches infinity on one side and negative infinity on other side I know this is a simple question for most of you, but I am currently studying for a Calculus exam and was just wondering why an online calculator I am using to double-check my work was disagreeing with me on this question: $$\lim_{x\to 0} \cot(x)\sec(x)$$ I reduce this down to $\frac{1}{\sin(x)}$, and in that case $x\to 0^-$ the limit is equal to negative infinity; and if $x\to 0^+$, the limit is equal to positive infinity. Doesn't this mean that the limit as $x\to 0$ does not exist? I use the calculator (linked below), and while it verifies that the two sides approach opposite infinity, it solves the entire limit as approaching "infinity". What does this mean? http://www.numberempire.com/limitcalculator.php - It doesn't mean anything special. The limit doesn't exist. The one sided limits (if you allow these symbols) are $+\infty$ and $-\infty$. So the 'calculator' is actually wrong. I wouldn't rely on it. – Weltschmerz Feb 25 '11 at 3:58 ## 1 Answer Your analysis is correct. Alternatively, $\sec(x)\to 1$ as $x\to 0$, and you can deal with $\cot(x)$, which goes to $\infty$ as $x\to 0^+$ and to $-\infty$ as $x\to 0^-$. Note, though, the fact that each one-sided limit does not exist is already enough to tell you the limit does not exist. (Saying that the limit equals $\infty$ or $-\infty$ is not saying that the limit exists, it is saying that the limit does not exist and explaining why: because the values of the function grow without bound, either in the positive direction or in the negative direction, respectively). Even though we write things like $$\lim_{x\to 0}\frac{1}{x^2} = \infty$$ this limit does not exist. As to the limit calculator at your link, I don't know what it means when it says as two-sided limit is $\infty$, since it says the same thing for $\lim\limits_{x\to 0}\frac{1}{x}$. In other words, it means that the on-line calculator is either not giving the correct answer, or else it means something other than what we think it means. -	{"extraction_info": {"found_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": 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": 0.9601044058799744, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/5030-practice-test-problems-can-you-help.html	# Thread: 1. ## Practice Test Problems- Can you help? I have a few practice problems that I would like solved so I can see how to do them. If you can help I will be so grateful!! 1. The sum of three consecutive integers is 201. Find the integers. 2. The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52cm find the dimensions of the rectangle. 3. Anna has 12 bills in her wallet, some \$5 and some \$10. The total value of the bills is \$100. How many of each bill does Anna have? These are just practice and I need to see how they are done so that I understand how to do them. My test is tommorow so any help before then will be appreciated. I thank you all dearly. Hope to hear from some of you soon!!! 2. Originally Posted by foofergutierrez I have a few practice problems that I would like solved so I can see how to do them. If you can help I will be so grateful!! 1. The sum of three consecutive integers is 201. Find the integers. Let the middle integer be $x$ then the sum of the three integers is: $<br /> (x-1)+x+(x+1)=201<br />$ rearranging: $<br /> 3x=201<br />$, so $x=67$. RonL 3. Originally Posted by foofergutierrez 2. The length of a rectangle is 2 cm more than twice its width. If the perimeter of the rectangle is 52cm find the dimensions of the rectangle. Let the width of the rectangle be $w$, then the length is $l=2w+2$. The perimiter is: $<br /> 2w+2l=2w+4w+4=52<br />$. So we have the following equation for $w$ to solve: $<br /> 6w+4=52<br />$ RpnL 4. Originally Posted by foofergutierrez 3. Anna has 12 bills in her wallet, some \$5 and some \$10. The total value of the bills is \$100. How many of each bill does Anna have? Let the number of tens be $t$ and the number of fives be $f$. Then the first condition in the problem tells us that $t+f=12$ (total of 12 bills). The second condition tells us that: $<br /> 5f+10t=100<br />$ (total value of the bills), now divide through by 2 to get: $<br /> f+2t=20<br />$ which leaves you with a pair of simultaneous equations to solve: $<br /> {f+t=12 \atop f+2t=20}<br />$ RonL	{"extraction_info": {"found_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": 0.9325788617134094, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/27191/chsh-violation-and-entanglement-of-quantum-states	# CHSH violation and entanglement of quantum states How is the violation of the usual CHSH inequality by a quantum state related to the entanglement of that quantum state? Say we know that exist Hermitian and unitary operators $A_{0}$, $A_{1}$, $B_{0}$ and $B_{1}$ such that $$\mathrm{tr} ( \rho ( A_{0}\otimes B_{0} + A_{0} \otimes B_{1} + A_{1}\otimes B_{0} - A_{1} \otimes B_{1} )) = 2+ c > 2,$$ then we know that the state $\rho$ must be entangled. But what else do we know? If we know the form of the operators $A_{j}$ and $B_{j}$, then there is certainly more to be said (see e.g. http://prl.aps.org/abstract/PRL/v87/i23/e230402 ). However, what if I do not want to assume anything about the measurements performed? Can the value of $c$ be used to give a rigourous lower bound on any of the familar entanglement measures, such as log-negativity or relative entropy of entanglement? Clearly, one could argue in a slightly circular fashion and define an entanglement measure as the maximal possible CHSH violation over all possible measurements. But is there anything else one can say? - 2 Your questions is answered here: arXiv:0907.2170. BTW, device-independent is the key phrase to search for. – Piotr Migdal Nov 18 '11 at 15:42 @PiotrMigdal Thanks for your comment. I had not thought of goggling for "device-independent" and was not aware of that paper. It seems to answer my question, though I'm still going through some of the details. – Earl Nov 21 '11 at 10:37 @PiotrMigdal: Perhaps you should post that as an answer. – Joe Fitzsimons Nov 21 '11 at 15:33 ## 1 Answer In a paper C.-E. Bardyn et al., PRA 80(6): 062327 (2009), arXiv:0907.2170, they discuss constrains on the state, given how much the CHSH equality is violated ($S=2+c$), but without putting any assumptions on the operator used. In general people consider schemes, when operators (for a Bell-type measurement) are random or one or more parties cannot be trusted. One of the key phrases is device-independent and maybe also loophole-free (as even a slight misalignment of operators may change the results dramatically). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 9, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9280750751495361, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/47957/alternate-proof-for-one-one-function/47976	# Alternate Proof for one-one function I need an alternate proof for this problem. Show that the function is one-one, provide a proof. $f:x \rightarrow x^3 + x : x \in \mathbb{R}$ I needed to show that the function is a one-one function. I tried doing $f(x) = f(y) \Rightarrow x = y$, It ended up with $x(x^2 + 1) = y(y^2 + 1)$. I couldn't figure out how to solve this further. Instead I had good idea of what the graph looked like, so I tried to show that it is a strictly increasing function. $f(x_2) - f(x_1) = (x_2^3 - x_1^3) + (x_2 - x_1)$ Thus, $f(x_2) - f(x_1) > 0$ Hence, f(x) is strictly increasing. And hence is one-one. Unfortunately this solution isn't acceptable. :( Can you guys help me work out how to do it the right way. Thanks again for your help. - 5 Why isn't your proof acceptable? It is a nice proof... btw, not sure what "right way" would mean... – Aryabhata Jun 27 '11 at 11:44 Did you study about derivatives? – Dennis Gulko Jun 27 '11 at 11:54 I have been doing some extra reading(monotonic functions, etc), I am supposed to solve it within covered course material. – mathguy80 Jun 27 '11 at 12:09 @math: What has been covered? Has this been covered: If $a \gt b$ and $c \gt 0$ then $ac \gt bc$ and $-ac \lt -bc$? – Aryabhata Jun 27 '11 at 12:27 1 However there is a little piece missing: you need to say if $x_2 > x_1$ then $f(x_2) - f(x_1) > 0$ That would make the comment above a proof, but without the "if" part it doesn't work. – Mark Bennet Jun 27 '11 at 14:58 show 2 more comments ## 2 Answers Note that $f'(x) = 3x^{2} + 1 > 0$ for $x \in \mathbb{R}$. So $f$ is monotone. Try doing this: $$x^{3} + x =y^{3}+y \Longrightarrow (x^{3}-y^{3})= -(x-y)$$ Then use $x^{3}-y^{3} = (x-y) \cdot (x^{2}+xy+y^{2})$. Using this you have $$(x-y) \cdot \Bigl[ x^{2}+xy+y^{2} +1\Bigr] =0$$ - Umm, $f'(x)=3x^2+1$. – Thomas Andrews Jun 27 '11 at 12:00 2 Question is tagged [algebra-precalculus] so talking of derivatives in the very first sentence might not be appropriate. – Aryabhata Jun 27 '11 at 12:05 1 Thanks, I had forgotten the a^3 - b^3 expansion, seems to have come in handy! I get that you then imply that either (x-y) = 0 or $(x^2 + xy + y^2 - 1) = 0$ But how do you show that $(x^2 + xy + y^2 - 1) \ne 0$.? – mathguy80 Jun 27 '11 at 12:16 2 btw, you have $x^2 + xy + y^2 -1$ instead of $x^2 + xy + y^2 + 1$. As stated, you can put $x=0, y=1$. – Aryabhata Jun 27 '11 at 12:22 1 @Chandru: The answer, though accepted, should be corrected. Note that in the first displayed line, you want something like $x^3-y^3=-(x-y)$, which, if $x \ne y$, leads to $x^2+xy+y^2=-1$. Then, for completeness, one shows that $x^2+xy+y^2 \ge 0$ always, maybe by completing the square. – André Nicolas Jun 27 '11 at 13:16 show 5 more comments You need to show that if $x \neq y$ then $x^3 + x \neq y^3 + y$. If $x < y$, then also $x^3 < y^3$, hence $x+x^3 < y+y^3$; if $x > y$, then $x^3 > y^3$, hence $x+x^3 > y+y^3$. Hence $x \neq y$ implies $x^3 + x \neq y^3 + y$. - Isn't this what OP has and does not want? I guess OP is missing some details... – Aryabhata Jun 27 '11 at 11:55 @Aryabhata: Let's wait to OP's response... – Shai Covo Jun 27 '11 at 12:01 I'm sorry I am not entirely sure I follow. To prove it's one-one, i need to show $f(x) = f(y) \Rightarrow x = y$, ie:- $x^3 + x = y^3 + y \Rightarrow x = y$. What you are showing is $x \ne y \Rightarrow x^3 + x \neq y^3 + y$. This then implies that $x = y \Rightarrow x^3 + x = y^3 + y$. Hence one-one. Do I interpret this correctly? – mathguy80 Jun 27 '11 at 12:29 1 @mathguy, no, it implies that if $x^3+x=y^3+y$, then $x=y$, and that's exactly what you need. – Gerry Myerson Jun 27 '11 at 12:46 I really don't see how this is different from proving that the function is increasing. btw, your $x \gt y$ part of the proof is redundant, I suppose. – Aryabhata Jun 27 '11 at 13:10	{"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": 42, "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": 0.9383388757705688, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/49170/what-is-an-example-of-a-regular-realization-of-c-5-over-mathbbqx	## What is an example of a regular realization of $C_5$ over $\mathbb{Q}(x)$? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) It's known that all abelian groups are regularly realizable over $\mathbb{Q}(x)$, but it occurred to me that I don't even have an example of a cyclic regular extension of $\mathbb{Q}(x)$ handy. So: what is an example of a regular realization of $C_5$ over $\mathbb{Q}(x)$? - 3 Dear James: Geometrically conn'd finite abelian coverings of geom. conn'd smooth proj. curves over a perfect field $k$ (such as $\mathbf{Q}$, $\mathbf{C}$, $\mathbf{F}_q$, etc.) can be understood nicely via generalized Jacobians. One constructs such covers of the curve by pullback of isogenies to an appropriate generalized Jacobian (depending on ramification). This provides insight into covers much as class field theory does for global fields (with its classical "modulus"), not giving "explicit" eqns for the abelian ext'n. Would you be happy with a construction along such lines? – BCnrd Dec 12 2010 at 21:46 2 Dear James: Jensen, Ledet and Yui in their book Generic polynomials: constructive aspects of the inverse Galois problem describe both generic $C_5$ and $D_5$-extensions. The google book reference is books.google.co.uk/books?isbn=0521819989 and their family for $C_5$ is on page 44 – Tim Dokchitser Dec 12 2010 at 22:56 1 At the start of sect. 3 of the paper math.ucalgary.ca/~aksilves/papers/lecacheux.pdf a messy quintic polynomial f_t(X) is described with coeff. in Z[t] (Lecacheux's quintics). Theorem 2 in the paper says that for rational t, if f_t(X) is irred. over Q then its Galois group over Q is Z/5Z. If instead we let t be an indeterminate and you can show f_t(X) to be irred. in Q(t)[X], then calculations in section 3 of the paper show the Galois group of f_t(X) over Q(t) is either Z/5 or D_5 and probably the calculations that eliminate D_5 if t is in Q also work if t is indeterminate. – KConrad Dec 12 2010 at 23:17 2 A simpler example than the one in my previous comment is Emma Lehmer's quintic, which is given on the first page of Darmon's article jstor.org/pss/2008408. Darmon gives a reference at the bottom of the first page for a proof that Lehmer's quintic defines a regular Z/5Z Galois extension of Q(t). – KConrad Dec 12 2010 at 23:22 4 James, it's not bizarre that an example is complicated. Over Q if you want to prove the existence of a cyclic extension of degree n, the easiest way I know is to first see that (Z/p)* is a Galois group over Q using the explicit and non-complicated cyclotomic polynomial (x^p-1)/(x-1), and then for each n you find a prime p which is 1 mod n and pick out the degree-n subextension of the p-th cyclotomic field. This argument does not give you an explicit polynomial defining the degree-n extension of Q and you can't expect the explicit realization of such a polynomial to be a simple polynomial. – KConrad Dec 12 2010 at 23:38 show 7 more comments ## 2 Answers Emma Lehmer's quintic ```\[\begin{align} y^5 +& x^2y^4 - (2x^3 + 6x^2 + 10x + 10)y^3 +\\ &(x^4 + 5x^3 + 11x^2 + 15x + 5)y^2 + (x^3 + 4x^2 +10x + 10)y + 1 \end{align}\]``` has Galois group $C_5$ over ${\mathbf Q}(x)$ and the splitting field over ${\mathbf Q}(x)$ is a regular extension. Her paper is "Connection between Gaussian periods and cyclic units" Math. Comp. 50 (1988), 535--541. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I wanted to make a certain comment, but it got too long, so here it is as an "answer" that is constructive in principle (and applicable over any field of characteristic 0, suitable to make cyclic regular covers ramified in a controlled locus). Let $k$ be a field of characteristic 0 and let $X$ be the projective line over $k$, so its usual Jacobian vanishes. Thus, its generalized Jacobians are affine. Being commutative and connected, they're products of a unipotent $k$-group and a $k$-torus, so isogenies to them from smooth connected commutative $k$-groups must arise from the torus part (as we're in char. 0). Hence, in language of geometric class field theory (see sec. 2 in Ch. I and sec. 3 in Ch. V of Serre's book "Algebraic groups and class fields"), we can stick with reduced moduli; i.e., reduced divisors $D$ on $X$ and their generalized Jacobians $J_D$, which are $k$-tori. The generalized Jacobian is isomorphic to the Picard variety ${\rm{Pic}}^0_{X_D/k}$ where $X_D$ is the geometrically integral curve over $k$ obtained by "crushing $D$ to a $k$-point" (defined via Galois descent of an analogous procedure over a splitting field of $D$). By 9.2/10 in "Neron Models", this is described rather explicitly as a $k$-torus. It is the quotient of a product of Weil restrictions: $$J_D = (\prod_{x \in D} {\rm{R}}_{k(x)/k}(\mathbf{G}_m))/\mathbf{G}_m.$$ We seek connected commutative extensions of this $k$-torus by (cyclic) finite constant $k$-groups. A cofinal system of tori equipped with isogenies onto $J_D$ over $k$ is given by the multiplication maps $[n]:J_D \rightarrow J_D$, whose kernels are (by the snake lemma with commutative $k$-groups) $$J_D[n] = (\prod_{x \in D} {\rm{R}}_{k(x)/k}(\mu_n))/\mu_n.$$ as $k$-groups. Let $C_{D,n} = J_D[n]/H_{D,n}$ denote the maximal $k$-group quotient of $J_D[n]$ that is constant (i.e., trivial as a Galois module). Using $\infty$ as a $k$-rational base point and assuming $\infty\not\in D$, there is a canonical $k$-morphism `$$f_D:X - D \rightarrow J_D,$$` and geometrc CFT says that the $f_D$-pullbacks of the $C_{D,n}$-coverings $J_D/H_{D,n} \rightarrow J_D$ are a cofinal system of geometrically connected finite abelian coverings of $X$ (i.e., regular finite abelian extensions of $k(X)$) unramified outside of $D$. So if $f_D$ could be made explicit, then we'd have a pretty "explicit" way of making finite abelian regular extensions of $X$ unramified outside of $D$, provided we can understand the structure of the Galois module of geometric points of $J_D[n]$. For $k = \mathbf{Q}$ this Galois module seems like something one can understand pretty systematically, provided that one understands how $\mathbf{Q}(\zeta_n)$ interacts with each of the fields $k(x)$ for $x \in D$. Example: Suppose $n = p$ is an odd prime and $k = \mathbf{Q}$ and $D$ is the single closed point $x \in \mathbf{A}^1_k$ with degree $p-1 > 1$ corresponding to the primitive $p$th roots of unity. Then ${\rm{R}}_{k(x)/k}(\mu_n)$ has the $p$-torsion Galois module of geometric points given by $\oplus_{0 \le i < p} \omega^i$ where $\omega$ is the mod-$p$ cyclotomic character, with $\mu_p = \omega$ embedded in the evident manner (the factor for $i = 1$). Hence, there's exactly one cyclic quotient of order $p$ with trivial Galois action (projection to the factor for $i = 0$) and so we conclude that $\mathbf{P}^1_{\mathbf{Q}}$ admits a unique degree-$p$ geometrically connected cyclic covering whose ramification is supported at precisely the primitive $p$th roots of unity. This corresponds to a particularly natural degree-$p$ regular abelian extension of $\mathbf{Q}(t)$. What is it? (I have no idea, even for $p = 3$ or $p = 5$.) In general, here is how I think $f_D$ can be made explicit. Change coordinates so $0 \not\in D$ (no problem, since $X(k)$ is infinite), and increase $D$ (harmless) so that $\infty \in D$. Then `$D' := D - \{\infty\}$` is the zero locus of a unique separable $h \in k[t]$ such that $h(0) = 1$, and $J_D = \prod_{x \in D'} {\rm{R}}_{k(x)/k}(\mathbf{G}_m)$. For $x \in D'$, let $h_x\|h$ be the unique irreducible factor over $k$ with $h_x(0) = 1$ and $h_x(x) = 0$, and let $y_x = y \bmod h_x(y) \in k[y]/(h_x) = k(x)$ be the "canonical" zero of $h_x$ in $k(x)$. The $x$th component of $f_D$ corresponds to a unit in $k(x)[t][1/h]$. This unit is probably $1 - (t/y_x)$ or its inverse (but I have not checked rigorously). -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 103, "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": 0.9248316287994385, "perplexity_flag": "head"}
http://mathoverflow.net/questions/10569/smooth-proper-schemes-over-rings-of-integers-with-points-everywhere-locally/11356	Smooth proper schemes over rings of integers with points everywhere locally Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) [Edit: Question 1 has been moved elsewhere so that an answer to Question 2 can be accepted.] Question 2. Is there a number field $K$, and a smooth proper scheme $X\to\operatorname{Spec}(\mathfrak{o})$ over its ring of integers, such that $X(K_v)\neq\emptyset$ for every place $v$ of $K$, and yet $X(K)=\emptyset$ ? I believe the answer is Yes. Remark. Let $K$ be a real quadratic field, $\mathfrak{o}$ the ring of integers of $K$, and $A$ the quaternion algebra over $K$ which is ramified exactly at the two real places. Then the conic $C$ corresponding to $A$ is a smooth projective $\mathfrak{o}$-scheme such that $C(\mathfrak{o})=\emptyset$ (because $C(K_v)=\emptyset$ for each of the real places $v$). But if we insist that $C(K_v)\neq\emptyset$ at these two real places $v$, then $A$ would have to split at these $v$ (in addition to all the finite places), and we would have $C=\mathbb{P}_{1,\mathfrak{o}}$. More generally, let $K$ be a number field, $\mathfrak{o}$ its ring of integers, and let $C$ be a smooth proper $\mathfrak{o}$-scheme whose generic fibre $C_{K}$ is a twisted $K$-form of the projective space of some dimension $n>0$. If $C$ has points everywhere locally, then $C=\mathbb{P}_{n,\mathfrak{o}}$. This remark shows that $X$ cannot be a twisted form of a projective space. - I cannot for the life of me find a definition of a proper scheme online. Just so I understand why an ordinary variety violating the Hasse principle isn't an answer to this question, could someone provide such a definition? – Qiaochu Yuan Jan 6 2010 at 0:00 1 "Proper" is a property of a map, not a scheme. It's the map to Spec(Z) that's proper, not the scheme. It morally means "all fibres compact". The reason an ordinary (projective) variety violating the Hasse principle may not be a counterexample is because the OP wants the map to be smooth too, and smooth over Spec(Z) means good reduction at all primes. – Kevin Buzzard Jan 6 2010 at 15:56 Part of what is so annoying about this question is that I know very few examples of schemes which are proper and smooth over Spec Z. So I basically keep trying to modify Kevin Buzzard's example, and failing. – David Speyer Jan 7 2010 at 12:48 4 Answers Chandan asked Vladimir and me for an example of an elliptic curve over a real quadratic field that has everywhere good reduction and non-trivial sha, with an explicit genus 1 curve representing some element of sha. Here's one we found: The elliptic curve y^2+xy+y = x^3+x^2-23x-44 over Q (Cremona's reference 4225m1) has reduction type III at 5 and 13. These become I0* over K=Q(sqrt(65)), and I0* can be killed by a quadratic twist. Specifically, the original curve can also be written as y^2 = x^3+5x^2-360x-2800 over Q, and its quadratic twist over K E: sqrt(65)Uy^2 = x^3+5x^2-360x-2800 has everywhere good reduction over K; here U = 8+sqrt(65) is the fundamental unit of K of norm -1. Now 2-descent in Magma says that the 2-Selmer group of E/K is (Z/2Z)^4, of which (Z/2Z)^2 is accounted by torsion. So it has either has rank over K or non-trivial Sha[2], and according to BSD its rank is 0 as L(E/K,1)<>0 (again in Magma). Actually, because K is totally real, I think results like those of Bertolini and Darmon might prove that E has Mordell-Weil rank 0 over K unconditionally. So it has non-trivial Sha[2]. After some slightly painful minimisation, one of its non-trivial elements corresponds to a homogeneous space C: y^2 = (23562U+1462)x^4 + (4960U+240)x^3 + (1124U-291)x^2 + (141U-833)x + (50U-733) with U as above. So here is a curve such that J(C) has everywhere good reduction and the Hasse principle fails for C. Hope this helps! Tim - That's very nice. Thank you. – Chandan Singh Dalawat Jan 11 2010 at 4:06 You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. There is no doubt that such examples as in David Speyer's response exist: indeed, they exist in great abundance in the following sense: Let $k_1$ be any number field, and let $E_{/k_1}$ be any elliptic curve with integral $j$-invariant. Then it has potentially good reduction, meaning that there is a finite extension $k_2/k_1$ such that $E_{/k_2}$ is the generic fiber of an abelian scheme over `$\mathbb{Z}_{k_2}$`. Furthermore, let $N$ be your favorite integer which is greater than $1$. Then there exists a degree $N$ field extension $k_3/k_2$ such that the Shafarevich-Tate group of $E_{/k_3}$ has an element of order $N$ (in fact, one can arrange to have at least $M$ elements of order $N$ for your favorite positive integer $M$): see Theorem 3 of http://math.uga.edu/~pete/ClarkSharif2009.pdf Since good reduction is preserved by base extension, the genus one curve $C_{/k_3}$ corresponding to the locally trivial principal homogeneous space of $E_{/k_3}$ of period $N$ gives an affirmative answer to Question 2. Specific examples of elliptic curves over quadratic fields with everywhere good reduction are known: see e.g. the survey paper http://mathnet.kaist.ac.kr/pub/trend/shkwon.pdf where the following example appears and is attributed to Tate: $E: y^2 + xy + \epsilon^2 y = x^3, \ \epsilon = \frac{5+\sqrt{29}}{2}$, has everywhere good reduction over `$k = \mathbb{Q}(\sqrt{29})$`. Indeed, the given equation is smooth over $\mathbb{Z}_k$, since the discriminant is $-\epsilon^{10}$ and $\epsilon$ is a unit in $\mathbb{Z}_k$. If this elliptic curve happens itself to have nontrivial Sha, great. If not, the theoretical results above imply that a quadratic extension of it will have a nontrivial $2$-torsion element of Sha, i.e., there will exist some hyperelliptic quartic equation $y^2 + p(x)y + q(x) = 0$ with $p(x), q(x)$ in the ring of integers of some quadratic extension $K$ of $\mathbb{Q}(\sqrt{29})$, which is smooth over $\mathbb{Z}_K$ and violates the local-global principle. If someone is interested in actually computing the equation, I would say a better strategy is searching for elliptic curves defined over quadratic fields with everywhere good reduction until you find one which already has a 2-torsion element in its Shafarevich-Tate group. (I don't see how to guarantee this theoretically, but I would be surprised if it were not possible.) Then it is easy to write down the defining equation. - My choice of Tate's example was made independently of Chandan's comment above: it's just a coincidence. Anyway, there are plenty of other examples, probably including some which have nontrivial Sha over the ground field. – Pete L. Clark Jan 8 2010 at 9:03 That's very nice, Pete; the question seems to have been tailor-made for you! I'm tempted to "accept" your answer, but perhaps we should wait for someone to write down an explicit equation of a genus-$1$ curve $C$ over a quadratic field $K$ whose jacobian $J$ has good reduction everywhere and such that $[C]$ is an order-$2$ element of $\operatorname{Sha}(J,K)$. It would have the same appeal as Selmer's example ($3x^3+4y^3+5z^3=0$) or Tate's example ($y^2+xy+\varepsilon^2y=x^3$). – Chandan Singh Dalawat Jan 8 2010 at 10:42 Regarding question 2, does the following work? Let $E$ be a rational elliptic curve with integer $j$-invariant. Then there is a number field $K$ so that `$E \times_{\mathbb{Q}} K$` has a smooth model over `$\mathcal{O}_K$`. Roughly, $\mathbb{Q}(j^{1/6})$ should work, but there might be some subtleties at 2 and 3. If Sha of `$E \times_{\mathbb{Q}} K$` is nontrivial, then I think an element of Sha should correspond to a torsor for $E \times_{\mathbb{Q}} K$ with the required property. I don't understand the elliptic curve tables well enough to know how to search them for an example like this, but presumably one of our readers does. - You are right; this is the reason why I said I believe the answer is Yes. There are examples in the literature of abelian varieties $A$ over number fields $K$ which have everywhere good reduction, and let's hope someone can find an explicit example where moreover $\operatorname{Sha}(A,K)\neq0$, as you suggest, and write down all the details. – Chandan Singh Dalawat Jan 8 2010 at 4:46 Tate's example : over $K=\mathbb{Q}(\sqrt{29})$ , the elliptic curve $E:y^2+xy+\varepsilon^2y=x^3$ , where $\varepsilon=(5+\sqrt{29})/2$, has good reduction everywhere because its discriminant is a unit. I don't know whether there is a finite extension $L\|K$ with $\operatorname{Sha}(E,L)\neq0$. Another possibility is to start with a genus-$1$ curve $C$ over some number field $K$ which has points everywhere locally but no $K$-points, and look for a finite extension $L\|K$ over which the jacobian of $C$ acquires good reduction but $C$ does not acquire an $L$-point. – Chandan Singh Dalawat Jan 8 2010 at 8:22 @David: By the way, I don't think $\mathbb{Q}(j^{\frac{1}{6}})$ is even roughly correct -- for instance, nothing stops $j$ from being a perfect $6$th power! I know that it is sufficient to trivialize the $3$ and $4$ torsion, but this leads to an extension of quite large degree. Frustratingly, I think I used to know more about this but I am having trouble remembering it now. – Pete L. Clark Jan 8 2010 at 9:01 If the fibres of the morphism $f: X\rightarrow\mathrm{Spec}(\mathbb{Z})$ have dimension $\leq 1$ the following facts are interesting regarding question 1: 1. By a theorem of Minkowski the field $\mathbb{Q}$ has no unramified extensions. If the fibres of $f$ have dimension $0$ smoothness is the same as being etale thus leading to an unramified extension of $\mathbb{Q}$. 2. A theorem of Fontaine proved in 1985 says that there exist no proper smooth curves over $\mathbb{Q}$ of genus $g\geq 1$ with good reduction everywhere. Thus the case of a non-rational curve of genus $0$ remains, that is a curve with $g=0$ and without a rational point. - That doesn't work. If f:C -> Z is a curve of genus zero, then omega_{C/Z}^{-1} has degree 2 and f^*(omega_{C/Z}^{-1}) is a free Z module of rank 3. (Locally free over a general base, but Z is a PID.) So C can be written as a conic in P^2_{Z}. The Hasse principle applies to conics. – David Speyer Jan 8 2010 at 11:44 Put another way, the only smooth proper curve over $\mathbb{Q}$ which has good reduction everywhere is $\mathbb{P}_1$. – Chandan Singh Dalawat Jan 8 2010 at 12:18 1 In fact this was already explained by Poonen in a "motivational comment" to his question linked to above. – Pete L. Clark Jan 8 2010 at 13:54	{"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": 107, "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": 0.9433180689811707, "perplexity_flag": "head"}
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I was wondering if it would be possible to make a small tokamak; not one ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9313870668411255, "perplexity_flag": "middle"}
http://cs.stackexchange.com/questions/9706/non-deterministic-turing-machine-that-halts-on-at-least-one-branches-of-computat	# Non-deterministic Turing machine that halts on at least one branches of computation I'm looking at my textbook here from Michael Sipser and he says that a nondeterministic Turing machine is a decider if all its computation branches halt on all inputs. I think I recall seeing somewhere what you'd call a nondeterministic Turing machine that halts on at least one branch for all inputs, but may loop on others. Is there a name for such a thing? I see later in this chapter the word verifier, but that doesn't seem to fit... I think that refers to an algorithm. A verifier for a language $A$ is an algorithm $V$, where $$A=\{w\mid V\text{ accepts }\langle w,c\rangle\text{ for some string c}\}.$$ We measure the time of a verifier only in terms of the length of $w$, so a polynomial time verifier runs in polynomial time in the length of $w$. A language $A$ is polynomially verifiable if it has a polynomial time verifier. - Perhaps just in the definition of the language recognized by a NTM? An NTM accepts a string $w$ if there exists at least one computation path that ends in the accepting state ... but not necessarily this happens for all input strings (otherwise L(NTM) = \Sigma^* ) – Vor Feb 12 at 16:46 I believe that you would say that the machine "accepts" the language. – Philip Feb 12 at 19:03 ## 1 Answer The idea is that a deterministic TM will always answer Yes/No in finite time (else the whole idea makes no sense). And to do that, the deterministic simulation of the NTM can't just go off into lala-land on some branches, i.e., every branch must end in yes/no at a finite depth. It decides (gives you a definite answer). If not all branches halt, it can verify (i.e., given a word in the language it is guaranteed to answer Yes; if not, perhaps it anwers No, perhaps it loops). - Well, clearly an NTM can indeed have branches that are infinitely deep just by virtue of trying to solve an undecidable problem. What do you mean then to say that it can't? Or am I misunderstanding something? – agent154 Feb 20 at 20:39	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9336438775062561, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/38304/do-we-say-that-phonon-has-effective-mass-through-its-dispersion-relation/52600	# Do we say that phonon has effective mass through its dispersion relation? The effective mass is proportional to the second derivative of the dispersion relation d2k/dE2. Do we say that phonon have effective mass through it ? Spin wave have. - ## 4 Answers Phonon dispersions are generally indicated by a linear spectrum, so the second derivative is 0. Thus phonons are effectively massless, and have a velocity given by the first derivative. - 1 – user12445 Sep 25 '12 at 18:14 1 @user12445: Optical modes are massive. – Ron Maimon Sep 26 '12 at 7:24 My opinion: Phonons are collective excitations of the crystal lattice vibration. They are massless Goldstone bosons resulting from the violation of the continuous translational symmetry of free space, by the crystal lattice. Phonons must carry momentum because they interact with electrons and change the momentum of the latter. An example is the electronic excitations in indirect gap semiconductors (I think this has been mentioned previously). Also, experiments of neutron scattering by a crystal indicate that phonons must have momentum. However, although phonons are massless, they have momentum by virtue of their wavelength. The effective mass equation 1/m = 1/h2d2E/dk2 arises from a semi-classical treatment of the motion of electrons and holes in the crystal lattice, and represents the effective mass of electrons and holes. I don’t think it applies to phonons. Similarly, it would be incorrect to conclude that phonons must travel at the speed of light because they are massless. This is because, unlike photons, phonons are not relativistically invariant objects - they don’t obey Dirac’s or Maxwell’s equations. They are excitations of a classical acoustic wave. I hope this contributes a little bit to this interesting discussion. - This figure shows the phonon dispersions of ZnO. It is clear that while some phonons have very roughly linear dispersions many do not (especially close to zone center). The second derivative would be non zero in these regions. I hope this has added to the conversation. - The effective mass of modes that are linear near the centre of the Brillouin zone will be zero since $\frac{d^{2}k}{d\omega^{2}}$ vanishes, but near extremes of the BZ this often isn't true and they will of course gain an effective mass. Furthermore, optical modes are often quadratic near the BZ centre, which gives them an effective mass, and the same is generally true of surface acoustic modes. -	{"extraction_info": {"found_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": 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": 0.9494411945343018, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/5252-simple-de-some.html	# Thread: 1. ## A simple DE...for some. If $y=e^{ax}cos(bx)$ and $y'''+2y'+3y=0$ what is a and b? That is the problem to solve and I don't even know what to do. Can anyone help me out a little and give me a suggestion or two for the solution of this one? 2. Originally Posted by a4swe If $y=e^{ax}$ and $y'''+2y'+3y=0$ what is a and b? That is the problem to solve and I don't even know what to do. Can anyone help me out a little and give me a suggestion or two for the solution of this one? You don't appear to have a "b". But never mind. What you do is calculate the derivatives from your given form for y(x), and substitute them into the DE. That will give you an algebraic equation (after dividing the common factor $e^{ax}$ which will occur on the left hand side of the equation), this will allow you to determine what a (and b?) have to be. RonL 3. Excuse me, that should be $<br /> y=e^{ax}cos(bx)<br />$ 4. Originally Posted by a4swe Excuse me, that should be $<br /> y=e^{ax}cos(bx)<br />$ The basic idea is to keep differentiating, and at each stage substitute for complicated expression simpler ones base on what you already know. Finally you should arrive at an equation of the required form, then you can equate coefficients between the given DE and what you have found from repeated differentiateion to get some algebraic equations for a and b. (You will need to check the algebra in what follows) If: $<br /> y=e^{ax}\cos(bx)<br />$ Then: $<br /> y'=ae^{ax}\cos(bx)- be^{ax}\sin{bx}$ $=a y - be^{ax}\sin{bx}<br />$, and so: $<br /> y''=a y' - bae^{ax}\sin{bx} -b^2e^{ax}\cos(bx)$ $<br /> =ay'-b^2y- bae^{ax}\sin{bx}=ay'-b^2y+ay'-a^2y<br />$, hence: $<br /> y''=2ay'-(a^2+b^2)y<br />$ so: $<br /> y'''=2ay''-(a^2+b^2)y'=4a^2y'-2a(a^2+b^2)y-(a^2+b^2)y<br />$ which you should now rearrange into the form $y'''+Ay'+By=0$ then equate coefficients with the given form to get the algebraic equations for $a$ and $b$. RonL Thank you very much, this helps a lot. 6. Originally Posted by a4swe If $y=e^{ax}cos(bx)$ and $y'''+2y'+3y=0$ what is a and b? That is the problem to solve and I don't even know what to do. Can anyone help me out a little and give me a suggestion or two for the solution of this one? Here is another way. Since this is a homogenous linear differencial equation of order three with constand coefficients look at the charachachteriistic polynomial. --- $k^3+2k+3=0$ Trivially, $k=-1$ is a solution. Synthetic division, $k^3+2k+3\div k+1=k^2-k+3$ The solution of, $k^2-k+3=0$ are, $k=\frac{1}{2}\pm i\frac{\sqrt{11}}{2}$ Therefore the general solution of this differential equation is, $C_1e^{-x}+C_2e^{1/2 x}\sin \left( \frac{\sqrt{11}}{2} x\right)+C_3e^{1/2 x}\cos \left( \frac{\sqrt{11}}{2} x\right)$ The particular solution you have is when $C_1=C_3=0,C_2=1$ Thus, $e^{1/2 x}\sin \left( \frac{\sqrt{11}}{2} x \right)=e^a \sin (bx)$ Thus, $a=1/2 \mbox{ and }b=\frac{\sqrt{11}}{2}$	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 28, "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": 0.9526243209838867, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/109108/is-it-true-that-any-matrix-can-be-decomposed-into-product-of-rotation-reflection/281087	# Is it true that any matrix can be decomposed into product of rotation,reflection,shear,scaling and projection matrices? It seems to me that any linear transformation in $R^{n\times m}$ is just a series of applications of rotation(actually i think any rotation can be achieved by applying two reflections, but not sure), reflection, shear, scaling and projection transformations. One or more of each kind in some order. This is how I have been imagining it to myself, but I was unable to find proof of this on the internet. Is this true? And if this is true, is there a way to find such a decomposition? EDIT: to make it clear, I am asking whether it is true that $\forall A\in R^{n \times m}$,$$A=\prod_{i=1}^{k}P_i$$ Where $P_i$ is rotation, reflection, shear, scaling, or projection matrix in $R^{n_i\times m_i}$. Also $n,m,k\in N$,and $n_i,m_i\in N$ for all i. And if it is true then how can we decompose it into that product. - The generality in which "linear transformation" makes sense is not the generality in which the other terms you're using make sense. – Qiaochu Yuan Feb 14 '12 at 1:41 @QiaochuYuan Can you elaborate? – Sunny88 Feb 14 '12 at 2:01 "Linear transformation" makes sense for vector spaces over any field, as does "shear" and "scaling." However, to define a notion of "rotation" requires an inner product (so we should ideally be working over $\mathbb{R}$ or $\mathbb{C}$), and depending on what you want out of "projection" or "reflection" those notions may also require an inner product. – Qiaochu Yuan Feb 14 '12 at 2:05 @QiaochuYuan Sorry, I had in mind matrices of real numbers when I asked this question. I modified my question to reflect that. – Sunny88 Feb 14 '12 at 2:20 – Inquest Feb 14 '12 at 10:30 show 3 more comments ## 3 Answers The question is not posed completely clearly, but I think that something close to what the questioner wants should follow quickly from the singular value decomposition, which states that any real matrix $A$ can be written in the form $$A=UDV,$$ where $U$ and $V$ are square real orthogonal matrices and $D$ is a (possibly rectangular) diagonal matrix with nonnegative entries on the diagonal. Since $U$ and $V$ are orthogonal they are products of rotations and reflections, while $D$ can be thought of as a product of projections and scalings. For example, if $$A=\left(\begin{array}{cc}1&2x\\0&1\end{array}\right),$$ then $$A= \left(\begin{array}{cc}\cos \phi&-\sin\phi\\\sin\phi&\cos\phi\end{array}\right) \left(\begin{array}{cc}\sqrt{x^2+1}-x&0\\0&\sqrt{x^2+1}+x\end{array}\right) \left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right),$$ where $$\phi=-\frac{\pi}{4}-\frac{1}{2}\arctan x, \qquad \theta=\frac{\pi}{4}-\frac{1}{2}\arctan x.$$ In reply to the comments below: Interpreting a diagonal matrix with positive entries along the diagonal as a scaling relies on allowing the scaling to be non-uniform, i.e., allowing it to scale different axes by different amounts. If the scaling matrices are restricted to be uniform, then, by using examples like the one above, you can write a square diagonal matrix with positive entries as a product of orthogonal matrices, shears, and a uniform scaling. - So it means that shear is also a product of reflections, projections and scaling? Can that be true? – Sunny88 Jan 18 at 16:34 I added an example of the SVD of a shear above. – David Moews Jan 18 at 22:22 How can a diagonal matrix be thought of as a product of projections and scalings? [Also, the middle matrix in your decomposition of the shear isn't diagonal.] – user108903 Jan 18 at 22:27 I made a typographical error; thank you for spotting that. A diagonal matrix with nonnegative entries projects onto the set of axes given by its non-zero entries, and then scales these axes. The scaling may be non-uniform as the diagonal matrix may use a different scale factor for each axis. – David Moews Jan 18 at 22:35 So are you saying that every positive semidefinite diagonalizable matrix is a scaling? I had assumed that "scaling" means a scalar multiple of the identity, and certainly no invertible diagonal matrix is a product of (orthogonal) projections and scalar multiples of $I$. – user108903 Jan 18 at 22:39 show 1 more comment The claim "any linear transformation is a series of applications projection transformations" is not correct. J. A. Erdos showed that every noninvertible $n\times n$ matrix is a finite product of projection matrices. An elmentary proof can be found here [An Elementary Proof That Every Singular Matrix Is a Product of Idempotent Matrices by J. Araújo and J. D. Mitchell, The American Mathematical Monthly, Vol. 112, No. 7 (Aug. - Sep., 2005), pp. 641-645] And it is not the case for invertible matrices generally. - If "any linear transformation is a series of applications projection transformations" is not true, I don't see how it says anything about "any linear transformation is a series of applications of projection, shear,reflection,rotation and scaling transformations". Now if it was true, then that is different matter :) – Sunny88 Feb 14 '12 at 1:27 There are several things that are true and known for a long time, related to the question. First, we can consider non-singular matrices without much loss of generality, since any singular matrix can be written (in many ways) as non-singular composed with (orthogonal) projection. As in another answer, the "singular value decomposition" (an example of a Cartan decomposition, valid in many other scenarios) expresses an arbitrary non-singular (real, for specificness) matrix as a product of rotation, diagonal matrix, and another rotation. Yes, this does express "shearing" as such a composition. Another decomposition of non-singular real matrices is as a product of shear (upper triangular with 1's on the diagonal), diagonal, and then rotation. Among its other names, this is a special case of "Iwasawa decomposition". Yet another is as product of upper triangular, permutation matrix (=exactly one non-zero entry, a "1", in each row and column), then another shear. This is a special case of "Bruhat decomposition". Yes, each of these applies to the components of the others. -	{"extraction_info": {"found_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": 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": 0.9280107021331787, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=4175309	Physics Forums ## Prove that Operators are Hermitian 1. The problem statement, all variables and given/known data Prove that i d/dx and d^2/dx^2 are Hermitian operators 2. Relevant equations I have been using page three of this document http://www.phys.spbu.ru/content/File...pe/qm07-03.pdf and the formula there. 3. The attempt at a solution I have taken a photo and attached it as I'm not good with Latex and it will just look messy, hope that isn't a problem. I get stuck as I have done the integration for both sides of the equation, I get answers very similar, apart from the complex conjugates. I'm confused as to what do with them or if I've even done it right. Thank You 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 Recognitions: Gold Member Homework Help Quote by chris_avfc 1. The problem statement, all variables and given/known data Prove that i d/dx and d^2/dx^2 are Hermitian operators 2. Relevant equations I have been using page three of this document http://www.phys.spbu.ru/content/File...pe/qm07-03.pdf and the formula there. 3. The attempt at a solution I have taken a photo and attached it as I'm not good with Latex and it will just look messy, hope that isn't a problem. I get stuck as I have done the integration for both sides of the equation, I get answers very similar, apart from the complex conjugates. I'm confused as to what do with them or if I've even done it right. Thank You Sorry, but I wasn't able to follow your work all the way through. It sort of looked to me though that you might have accidentally dropped a $d \psi$ or $dx$ maybe, in addition to maybe some other stuff that doesn't look quite right. Since this is a two part problem, I'll start with the first part, proving that i d/dx is Hermitian. $$\int \left( i \frac{d}{dx} \psi \right)^* \psi \ dx,$$ and finagle that until you can show that it equals $$\int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx$$ And you are correct that you want to use integration by parts to do this. Example (ii) on page 3 of your attachment pretty much does it for you. However, this example takes a few shortcuts. I'll try to give you some hints about how to do the math without taking the shortcuts. Hint 1: If A and B are complex variables, then (AB)* = AB. Hint 2: $$\left( \frac{d}{dx} U \right) V \ dx = \left( d U \right) V = V \ dU$$ The important part of that hint is to note that the "dx"s cancel. The example (ii) in your attachment doesn't follow through with this cancellation, but it could have. The reason it doesn't follow through with the cancellation is so that it doesn't have to add the "dx"s back in at the end of the calculations (or concern itself with the limits of integration, although those wouldn't change anyway for this particular problem). That's one of the shortcuts the example is doing. Hint 3: You don't really have to make the substitutions using U's, V's dU's and dV's if you don't want to. The equation you start with is so close to the integration by parts definition to begin with, it might just be easier to stick with $\psi$'s, $\psi^$'s, $d \psi$'s and $d \psi^$'s. (However if you'd like to make the substitutions to U's, V's dU's and dV's, you are certainly free to do so.) Hint 4: The example in your attachment leaves out the limits of integration so as not to clutter up the equations. But they are still there. The limits of integration are x = -∞ to ∞. You can't forget about those for this problem. Restating the general formula for integration by parts, without forgetting limits, we have $$\int_a^b U \ dV = UV \|_a^b - \int_a^b V \ dU$$. I think you missed part of that in your attempted solution. You need to evaluate the UV term using the limits (the actual limits in this case are at x = -∞ to ∞). Evaluating that term is the crux of solving this part of the problem. After all that, refer to example (iii) for the d2/dx2 operator. That example uses some shortcuts (and substitutions) too, but they are no less straightforward than the previous example. Quote by collinsmark Hint 1: Hint 2: Hint 3: Hint 4: . Thank you very much for that mate, I see I have made a whole number of stupid mistakes, especially forgetting all about limits. I still can't get it right though, I'm taking the constant $i$ or $i^$ out of the differential, cancelling the $dx$, then doing integration by parts with what is left which is: $$i \int \psi^ d\psi$$ on the right hand side, and $$i^* \int \psi d\psi^$$ on the left hand side. I'm ending up with: $$i(\psi \psi^ - \infty^2)$$ on the right hand side. and $$i^(\psi \psi^ - \infty^2)$$ on the left hand side. The whole infinite thing makes me think I have done it wrong again. Recognitions: Gold Member Homework Help ## Prove that Operators are Hermitian Quote by chris_avfc Thank you very much for that mate, I see I have made a whole number of stupid mistakes, especially forgetting all about limits. I still can't get it right though, I'm taking the constant $i$ or $i^$ out of the differential, cancelling the $dx$, then doing integration by parts with what is left which is: $$i \int \psi^ d\psi$$ on the right hand side, Okay, that's fine, but there's really no need to simplify anything on the right side of the equation. The right side of the equation is your end goal. The task at hand is to manipulate the left side of the equation until it looks like $$\int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx.$$ Once it looks like that you're finished. Of course there's nothing wrong with simplifying it $$\int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx = i \int \psi^* d \psi$$ and then "meeting in the middle." I'm just sayin', I think your instructor probably wants you to start with $$\int \left(i \frac{d}{dx} \psi \right)^* \psi \ dx$$ and end with $$\int \psi^* \left( i \frac{d}{dx} \right) \psi \ dx.$$ (Note, I left the limits of integration off too, to avoid clutter. You can add them in if you wish.) But what you did is fine, so let's just continue. and $$i^* \int \psi d\psi^$$ on the left hand side. Okay, so far so good. Hint 5: The complex conjugate of $i$ is $-i$. $$i^ = -i$$ I'm ending up with: $$i(\psi \psi^* - \infty^2)$$ on the right hand side. and $$i^(\psi \psi^ - \infty^2)$$ on the left hand side. The whole infinite thing makes me think I have done it wrong again. Now you kind of lost me. Although we are doing integration by parts, there's no way to fully evaluate the integral. There's no way to evaluate $\int \psi^* d \psi$ without knowing more about the function $\psi(x)$. So you'll just need to leave that in integral form. Let me help out a little more, setting up the integration by parts. $$-i \left[ \int \psi d \psi^* \right] = -i \left[ \psi \psi^* \|_{x=-\infty}^{\infty} - \int \psi^* d \psi \right]$$ I mentioned before, you can't evaluate $\int \psi^* d \psi$. But you can evaluate $\psi \psi^* \|_{x=-\infty}^{\infty}$. In order for $\psi (x)$ to be a valid wavefunction, it must be square integrable. And in order for it to be square integrable, $\| \psi (x) \|^2 = \psi \psi^(x)$ must taper off to zero as x approaches infinity or negative infinity. I think you were assuming that $\psi$ approaches infinity as x approaches infinity. That's not true. (If it was, $\psi$ would not be a valid wavefunction.) Quote by collinsmark Okay, so far so good. Hint 5: The complex conjugate of $i$ is $-i$. $$i^ = -i$$ And in order for it to be square integrable, $\| \psi (x) \|^2 = \psi \psi^(x)$ must taper off to zero as x approaches infinity or negative infinity. I think you were assuming that $\psi$ approaches infinity as x approaches infinity. That's not true. (If it was, $\psi$ would not be a valid wavefunction.) Ah I see, I didn't realise you couldn't evaluate the integral and I was assuming that it went to infinite as well. But how do you get the $dx$ back in at the end, do you just insert them back? God knows how I am going to manage the rest of this sheet now, thank you though. By the way the link in your sig is such a great help, I've bookmarked it. Recognitions: Gold Member Homework Help Quote by chris_avfc But how do you get the $dx$ back in at the end, do you just insert them back? You could note that $d \psi = \left( \frac{d}{dx}\psi \right) dx$ and substitute. Quote by collinsmark You could note that $d \psi = \left( \frac{d}{dx}\psi \right) dx$ and substitute. Ah of course! So looking at example iii on the sheet I linked, how do they manage to move the $p$ out of the conjugate like that? Recognitions: Gold Member Homework Help Quote by chris_avfc Ah of course! So looking at example iii on the sheet I linked, how do they manage to move the $p$ out of the conjugate like that? Again, the example skipped a lot of steps. To solve that example without skipping steps, you'll need to integrate by parts again, but this time you'll be doing it twice. Note that $$\hat p = -i \hbar \frac{d}{dx}$$ The example didn't explicitly show that. Instead example (iii) builds off of the previous example (which is almost the same as what we just did above). Building off of the previous example, you can use $$\left[ \left( \frac{d}{dx} \right) g(x) \right]^ =g^(x) \left(\frac{d}{dx} \right)$$ twice. If you don't want to build off of the previous example, you can do example (iii) by noting that $$\hat T = \frac{\hat p ^2}{2m} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$ and do integration by parts twice, the long way. Recognitions: Gold Member Homework Help 'Just to avoid any possible misinterpretation, when I claimed, Quote by collinsmark $$\left[ \left( \frac{d}{dx} \right) g(x) \right]^ =g^(x) \left(\frac{d}{dx} \right)$$ I should have made a note that that only applies if $g(x)$ is square integrable. Of course, we're assuming that it is here. Good luck! Quote by collinsmark 'Just to avoid any possible misinterpretation, when I claimed, I should have made a note that that only applies if $g(x)$ is square integrable. Of course, we're assuming that it is here. Good luck! So if I were to apply that to the second part of my question, I could take the $d^2/dx^2$ out of the conjugate? Recognitions: Gold Member Homework Help I'm sorry, but I made a big mistake. The last thing I intended to do was confuse you, but I think I might have done so. In my two most recent posts, I mistakenly wrote, $$\left[ \left( \frac{d}{dx} \right) g(x) \right]^ =g^(x) \left(\frac{d}{dx} \right) \mathrm{\ \ \ \ \ [sic]}$$ That's wrong! I forgot an $i$ in a couple places. What I should have wrote was $$\left[ \left(i \frac{d}{dx} \right) g(x) \right]^ =g^(x) \left(i \frac{d}{dx} \right)$$ It turns out that $i$ is important. Why? because $\frac{d}{dx}$ by itself is not a Hermitian operator. That $i$ in front of it is needed to make it Hermitian. The reason it is allowed to pull the $\left( i \frac{d}{dx} \right)$ operator out from under the conjugate, and to the right (instead of to the left), is because it is Hermitian. If the operator was not Hermitian, we wouldn't be allowed to do that. Quote by chris_avfc So if I were to apply that to the second part of my question, I could take the $d^2/dx^2$ out of the conjugate? Yes and no. 'Yes' because $\frac{d^2}{dx^2}$ is Hermitian. 'No' because you haven't proven that yet. The whole point of the second part of the problem statement is to prove that $\frac{d^2}{dx^2}$ is Hermitian. Until you do that, you're not allowed to pull it out (to the right) directly. Since I confused you once (by forgetting the $i$), let me try to make up for it by helping out a little more. As I mentioned before, there are a couple ways to approach the second part of the problem statement. You could prove it by integrating by parts, twice. Or alternately, you could prove it by building off the first part of the problem statement. Following the path taken by page 3, (iii) if your attachment, I'll help you with the latter approach. Although you're not allowed to pull out a $\frac{d^2}{dx^2}$ from underneath the conjugate, and to the right, in this particular problem (until you prove it is Hermitian), you can pull out a $\left( i \frac{d}{dx} \right)$, because you've already proven that that operator is Hermitian. $$\int \left( \frac{d^2}{dx^2} \psi \right)^ \psi \ dx.$$ The first thing we can do is split up the $\frac{d^2}{dx^2}$ into two $\frac{d}{dx}$ operations. $$\int \left[ \frac{d}{dx} \left( \frac{d}{dx} \psi \right) \right]^* \psi \ dx.$$ Next we note that $-i^2 = 1$, and multiply everything in the square brackets by $-i^2$. It should be obvious now that that doesn't change anything, because we're just multiplying everything within the brackets by 1. $$\int \left[ -i^2 \frac{d}{dx} \left( \frac{d}{dx} \psi \right) \right]^* \psi \ dx.$$ Let's redistribute the $i$'s. $$\int - \left[ i \frac{d}{dx} \left(i \frac{d}{dx} \psi \right) \right]^* \psi \ dx.$$ Now we can start pulling out $\left( i \frac{d}{dx} \right)$ terms, and to the right. I can't do the whole problem for you but I'll show you how to do the first one (you'll have to do it twice, total). [tex] \int - \left[ \color{red}{i \frac{d}{dx}} \left(i \frac{d}{dx} \psi \right) \right]^* \psi \ dx = \int - \left[ \left( i \frac{d}{dx} \psi \right)^* \color{red}{\left(i \frac{d}{dx} \right)} \right] \psi \ dx.[/tex] Now let me just get rid of a set of brackets. $$\int - \left( i \frac{d}{dx} \psi \right)^* \left(i \frac{d}{dx} \right) \psi \ dx.$$ Now you take it from here. Start by pulling out a $\left( i \frac{d}{dx} \right)$ term from under the conjugate, and to the right, which is now in red: $$\int - \left( \color{red}{i \frac{d}{dx}} \psi \right)^* \left(i \frac{d}{dx} \right) \psi \ dx.$$ Simplify after that. Once you get the answer, I suggest doing the problem again from the beginning, without using this thread as a guide, just so you know you can do it on your own. Good luck! Tags hermitian Thread Tools \| \| \| \| \|---------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Prove that Operators are Hermitian \| \| \| \| Thread \| Forum \| Replies \| \| \| Quantum Physics \| 1 \| \| \| Advanced Physics Homework \| 9 \| \| \| Calculus & Beyond Homework \| 13 \| \| \| Advanced Physics Homework \| 3 \| \| \| Calculus & Beyond Homework \| 3 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 54, "mathjax_display_tex": 31, "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": 0.9566395282745361, "perplexity_flag": "head"}
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http://mathhelpforum.com/statistics/1944-help-probability.html	# Thread: 1. ## help-probability Don't know how to do these questions. Understand how to use Venn diagrams but to many terms for my mind. Can anybody tell me how to figure this out? Attached Thumbnails 2. Originally Posted by Arbitur Don't know how to do these questions. Understand how to use Venn diagrams but to many terms for my mind. Can anybody tell me how to figure this out? Since $P(A \mbox{ and }B)=P(A)P(B)$ where $A,B$ are disjoint. Thus, the answer to the first question is $(.2)(.3)=.06$. Since $P(A\mbox{ or }B)=P(A)+P(B)$ where $A,B$ are disjoint. Thus, the answer to the second question is $(.2)+(.3)=(.5)$. Since $P(A)+P(B)+P(C)=1$ where $A,B,C$ are disjoint. Thus, the answer to the third question is $1-.5=.5$. Q.E.D. #### Search Tags Copyright © 2005-2013 Math Help Forum. All rights reserved.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 9, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9529594779014587, "perplexity_flag": "head"}
http://en.wikibooks.org/wiki/Waves/Plane_Superposition	# Waves/Plane Superposition Waves : 2 and 3 Dimension Waves 1 - 2 - 3 - 4 - 5 - 6 - 7 Problems ## Contents ### Superposition of Plane Waves We now study of wave packets in two dimensions by asking what the superposition of two plane sine waves looks like. If the two waves have different wavenumbers, but their wave vectors point in the same direction, the results are identical to those presented in the previous chapter, except that the wave packets are indefinitely elongated without change in form in the direction perpendicular to the wave vector. The wave packets produced in this case march along in the direction of the wave vectors and thus appear to a stationary observer like a series of passing pulses with broad lateral extent. Superimposing two plane waves which have the same frequency results in a stationary wave packet through which the individual wave fronts pass. This wave packet is also elongated indefinitely in some direction, but the direction of elongation depends on the dispersion relation for the waves being considered. One can think of such wave packets as steady beams, which guide the individual phase waves in some direction, but don't themselves change with time. By superimposing multiple plane waves, all with the same frequency, one can actually produce a single stationary beam, just as one can produce an isolated pulse by superimposing multiple waves with wave vectors pointing in the same direction. #### Two Waves of Identical Wavelength If the frequency of a wave depends on the magnitude of the wave vector, but not on its direction, the wave's dispersion relation is called isotropic. In the isotropic case two waves have the same frequency only if the lengths of their wave vectors, and hence their wavelengths, are the same. The first two examples in figure 2.6 satisfy this condition. In this section we investigate the beams produced by superimposed isotropic waves. We superimpose two plane waves with wave vectors $\mbox{k}_1 = (- \Delta k_x , k_y )$ and $\mbox{k}_2 = ( \Delta k_x , k_y )$. The lengths of the wave vectors in both cases are $k_1 = k_2 = ( \Delta k_x^2 + k_y^2 )^{1/2}$: $A = \sin ( -\Delta k_x x + k_y y - \omega t ) + \sin (\Delta k_x x + k_y y - \omega t ) .$ (3.15) If $\Delta k_x << k_y$, then both waves are moving approximately in the $y$ direction. An example of such waves would be two light waves with the same frequencies moving in slightly different directions. Figure 2.7: Wave fronts and wave vectors of two plane waves with the same wavelength but oriented in different directions. The vertical bands show regions of constructive interference where wave fronts coincide. The vertical regions in between the bars have destructive interference, and hence define the lateral boundaries of the beams produced by the superposition. The components $\Delta k_x$ and $k_y$ of one of the wave vectors are shown. Applying the trigonometric identity for the sine of the sum of two angles (as we have done previously), equation (3.15) can be reduced to $A = 2 \sin (k_y y - \omega t) \cos (\Delta k_x x) .$ (3.16) This is in the form of a sine wave moving in the $y$ direction with phase speed $c_{phase} = \omega / k_y$ and wavenumber $k_y$, modulated in the $x$ direction by a cosine function. The distance $w$ between regions of destructive interference in the $x$ direction tells us the width of the resulting beams, and is given by $\Delta k_x w = \pi$, so that $w = \pi / \Delta k_x .$ (3.17) Thus, the smaller $\Delta k_x$, the greater is the beam diameter. This behavior is illustrated in figure 2.7. Figure 2.8: Example of beams produced by two plane waves with the same wavelength moving in different directions. The wave vectors of the two waves are $\mbox{k} = (\pm 0.1 , 1.0)$. Regions of positive displacement are illustrated by vertical hatching, while negative displacement has horizontal hatching. Figure 2.8 shows an example of the beams produced by superposition of two plane waves of equal wavelength oriented as in figure 2.7. It is easy to show that the transverse width of the resulting wave packet satisfies equation (2.17). #### Two Waves of Differing Wavelength In the third example of figure 2.6, the frequency of the wave depends only on the direction of the wave vector, independent of its magnitude, which is just the reverse of the case for an isotropic dispersion relation. In this case different plane waves with the same frequency have wave vectors which point in the same direction, but have different lengths. More generally, one might have waves for which the frequency depends on both the direction and magnitude of the wave vector. In this case, two different plane waves with the same frequency would typically have wave vectors which differed both in direction and magnitude. Figure 2.9: Wave fronts and wave vectors of two plane waves with different wavelengths oriented in different directions. The slanted bands show regions of constructive interference where wave fronts coincide. The slanted regions in between the bars have destructive interference, and as previously, define the lateral limits of the beams produced by the superposition. Mathematically, we can represent the superposition of these two waves as a generalization of equation (2.15): $A = \sin [ - \Delta k_x x + (k_y + \Delta k_y ) y - \omega t ] + \sin [ \Delta k_x x + (k_y - \Delta k_y ) y - \omega t ] .$ (3.18) In this equation we have given the first wave vector a $y$ component $k_y + \Delta k_y$ while the second wave vector has $k_y - \Delta k_y$. As a result, the first wave has overall wavenumber $k_1 = [ \Delta k_x^2 + (k_y + \Delta k_y )^2 ]^{1/2}$ while the second has $k_2 = [ \Delta k_x^2 + (k_y - \Delta k_y )^2 ]^{1/2}$, so that $k_1 \ne k_2$. Using the usual trigonometric identity, we write equation (2.18) as $A = 2 \sin (k_y y - \omega t ) \cos ( - \Delta k_x x + \Delta k_y y ) .$ (3.19) To see what this equation implies, notice that constructive interference between the two waves occurs when $- \Delta k_x x + \Delta k_y y = m \pi$, where $m$ is an integer. Solving this equation for $y$ yields $y = ( \Delta k_x / \Delta k_y )x + m \pi / \Delta k_y$, which corresponds to lines with slope $\Delta k_x / \Delta k_y$. These lines turn out to be perpendicular to the vector difference between the two wave vectors, $\mbox{k}_2 - \mbox{k}_1 = ( 2 \Delta k_x , -2 \Delta k_y )$. The easiest way to show this is to note that this difference vector is oriented so that it has a slope $- \Delta k_y / \Delta k_x$. Comparison with the $\Delta k_x / \Delta k_y$ slope of the lines of constructive interference indicates that this is so. Figure 2.10: Example of beams produced by two plane waves with wave vectors differing in both direction and magnitude. The wave vectors of the two waves are $\mbox{k}_1 = ( -0.1 , 1.0)$ and $\mbox{k}_2 = ( 0.1, 0.9)$. Regions of positive displacement are illustrated by vertical hatching, while negative displacement has horizontal hatching. An example of the production of beams by the superposition of two waves with different directions and wavelengths is shown in figure 2.10. Notice that the wavefronts are still horizontal, as in figure 2.8, but that the beams are not vertical, but slant to the right. Figure 2.11: Illustration of factors entering the addition of two plane waves with the same frequency. The wave fronts are perpendicular the the vector average of the two wave vectors, $\mbox{k}_0 = ( \mbox{k}_1 + \mbox{k}_2 )/2$, while the lines of constructive interference, which define the beam orientation, are oriented perpendicular to the difference between these two vectors, $\mbox{k}_2 - \mbox{k}_1$. Figure 2.11 summarizes what we have learned about adding plane waves with the same frequency. In general, the beam orientation and the lines of constructive interference are not perpendicular to the wave fronts. This only occurs when the wave frequency is independent of wave vector direction. #### Many Waves with the Same Wavelength As with wave packets in one dimension, we can add together more than two waves to produce an isolated wave packet. We will confine our attention here to the case of an isotropic dispersion relation in which all the wave vectors for a given frequency are of the same length. Figure 2.12: Illustration of wave vectors of plane waves which might be added together. Figure 2.12 shows an example of this in which wave vectors of the same wavelength but different directions are added together. Defining $\alpha_i$ as the angle of the $i$th wave vector clockwise from the vertical, as illustrated in figure 2.12, we could write the superposition of these waves at time $t = 0$ as $A = \sum_i A_i \sin ( k_{xi} x + k_{yi} y )$ $= \sum_i A_i \sin [kx \sin ( \alpha_i ) + ky \cos ( \alpha_i ) ]$ (3.20) where we have assumed that $k_{xi} = k \sin ( \alpha_i )$ and $k_{yi} = k \cos ( \alpha_i )$. The parameter $k = \vert \mbox{k} \vert$ is the magnitude of the wave vector and is the same for all the waves. Let us also assume in this example that the amplitude of each wave component decreases with increasing $\vert \alpha_i \vert$: $A_i = \exp [-( \alpha_i / \alpha_{max} )^2 ] .$ (3.21) The exponential function decreases rapidly as its argument becomes more negative, and for practical purposes, only wave vectors with $\vert \alpha_i \vert \le \alpha_{max}$ contribute significantly to the sum. We call $\alpha_{max}$ the spreading angle. Figure 2.13: Plot of the displacement field $A(x,y)$ from equation (2.20) for $\alpha _{max} = 0.8$ and $k = 1$. Figure 2.13 shows what $A(x,y)$ looks like when $\alpha_{max} = 0.8 \mbox{ radians}$ and $k = 1$. Notice that for $y = 0$ the wave amplitude is only large for a small region in the range $-4 < x < 4$. However, for $y > 0$ the wave spreads into a broad semicircular pattern. Figure 2.14: Plot of the displacement field $A(x,y)$ from equation (2.20) for $\alpha _{max} = 0.2$ and $k = 1$. Figure 2.14 shows the computed pattern of $A(x,y)$ when the spreading angle $\alpha_{max} = 0.2 \mbox{ radians}$. The wave amplitude is large for a much broader range of $x$ at $y = 0$ in this case, roughly $-12 < x < 12$. On the other hand, the subsequent spread of the wave is much smaller than in the case of figure 2.13. We conclude that a superposition of plane waves with wave vectors spread narrowly about a central wave vector which points in the $y$ direction (as in figure 2.14) produces a beam which is initially broad in $x$ but for which the breadth increases only slightly with increasing $y$. However, a superposition of plane waves with wave vectors spread more broadly (as in figure 2.13) produces a beam which is initially narrow in $x$ but which rapidly increases in width as $y$ increases. The relationship between the spreading angle $\alpha_{max}$ and the initial breadth of the beam is made more understandable by comparison with the results for the two-wave superposition discussed at the beginning of this section. As indicated by equation (2.17), large values of $k_x$, and hence $\alpha$, are associated with small wave packet dimensions in the $x$ direction and vice versa. The superposition of two waves doesn't capture the subsequent spread of the beam which occurs when many waves are superimposed, but it does lead to a rough quantitative relationship between $\alpha_{max}$ (which is just $\tan^{-1} (k_x / k_y )$ in the two wave case) and the initial breadth of the beam. If we invoke the small angle approximation for $\alpha = \alpha_{max}$ so that $\alpha_{max} = \tan^{-1} (k_x / k_y ) \approx k_x / k_y \approx k_x / k$, then $k_x \approx k \alpha_{max}$ and equation (2.17) can be written $w = \pi / k_x \approx \pi / (k \alpha_{max} ) = \lambda / (2 \alpha_{max} )$. Thus, we can find the approximate spreading angle from the wavelength of the wave $\lambda$ and the initial breadth of the beam $w$: $\alpha_{max} \approx \lambda / (2d) \quad \mbox{(single slit spreading angle)} .$ (3.22)	{"extraction_info": {"found_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": 86, "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": 0.9377202391624451, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/265666/what-does-it-mean-for-a-domain-to-lie-left-to-a-path	# What does it mean for a domain to lie left to a path? I am studying the book "Complex variables and applications" by James Ward Brown, Ruel Vance Churchill and I don't understand what they mean by the domain to lie left to a path. How can we tell that the domain is left to the path from the drawing ? - I think you should have met them in Green's Theorem. It's just the orientation of surface with respect to path. It simply means that when you move along the curve, in direction pointed by arrow, the domain will be towards the lefthand side. – 007resu Dec 27 '12 at 0:37 1 Imagine walking along the path in the direction in which it’s oriented. Put your arms out; the left arm points into the region. – Brian M. Scott Dec 27 '12 at 0:38 1 By the way, it doesn't say 'lie left to the path' as you have it in your question, but 'lie to the left of the path'. – Tara B Dec 27 '12 at 0:40 @BrianM.Scott - thanks a lot! Since this was the explanasion I understood I suggest you make it an answer so I can accept it – Belgi Dec 27 '12 at 0:41 2 Of course, all of this assumes the (natural) convention that when we walk around the curve, we are standing so that our head is pointed upward from the printed page (or outward from the monitor). :) – Ted Dec 27 '12 at 0:51 show 1 more comment ## 4 Answers Imagine walking along the path in the direction in which it’s oriented. Put your arms out; the left arm points into the region. - Oh good, I was going to suggest you write your comment as an answer, since I thought you explained it better than either of us. – Tara B Dec 27 '12 at 1:49 When you walk in one direction around a simple closed curve, the "inside" of the curve is either always on your left or always on your right. This simply means that the integration along the curves follow the path around the curve so that the interiors are on the left of the direction in which you traverse the curve. - It means that if you are walking around the path in the direction shown by the arrow, then the domain will be on your left. - The accepted answer is good enough from the perspective of a mathematician visualizing the complex plane containing a path, spread out before her. But that is a human situation. A mathematical definition of "left" in the complex plane is slightly tricky. In going from the negative to the positive part of the real line, the half plane to the left of our path is the one containing $\mathrm i$. Unfortunately, there is no way of telling which of $\mathrm i$ and$-\mathrm i$ is which; it is an arbitrary choice. As long as we place $\mathrm i$ in what from our own perspective is the "upper" half of the plane, the orientive conventions of the plane will accord with our native ones. -	{"extraction_info": {"found_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": 0.9571688771247864, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/105934?sort=votes	## Defining the integral of a function using the product measure ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Imagine that we're trying to define the expression $$\int_U f(x)dx$$ in a rigorous way. Assume that $f:X \rightarrow \mathbb{R}^{\geq 0}$ where $(X,\mu)$ is a measure space, and suppose that $U$ is a measurable subset of $X$. That most typical approach to making this integral rigorous is the method of Lebesgue, whereby we partition the range of $f$ into increasingly small horizontal strips. This seems very elaborate to me - why not just define the integral in the obvious way as the "(product) measure of the set of all points under the curve"? (if its defined; our integrable functions would then be precisely those for which the product measure is indeed defined). We can make this idea precise by writing $$\int_U f(x)dx := (\mu \times \lambda)(\lbrace (x,y) : x \in U \wedge 0 \leq y \leq f(x)\rbrace)$$ where $\mu$ is the measure on $X$ and $\lambda$ is the standard measure on $\mathbb{R}$. My question is, why isn't this the "standard" definition of the integral? - ## 2 Answers Linearity of this integral is very mysterious. Moreover, the definition of the product measure using integration, i.e. $\mu \otimes \lambda (M) =\int \int I_M(x,y) d\mu(x) d\lambda(y)$, is very easy (up to a technical problem concerning measurability) and can be understood without knowing Caratheodory's construction of measures. - ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. I'm not sure I buy Jochen's comment that product measure can be so easily defined using integration --- it seems like you're going to have to do some work to show that his double integral is well-defined for every set $M$ in the $\sigma$-algebra generated by the measurable rectangles. The real problem may be that you actually "need" integration theory to define product measures via the standard Caratheodory construction, when you show that $(\mu\times\nu)(A \times B) = \mu(A)\nu(B)$ defines a premeasure on the algebra generated by the measurable rectangles. That is, if $A \times B$ can be expressed as a disjoint union $\bigcup A_i \times B_i$, we need $\mu(A)\nu(B) = \sum \mu(A_i)\nu(B_i)$. And as far as I can see you pretty much have to use the monotone convergence theorem to prove that. - Yes, it is not obvious that the iterated integral is well defined. But this is a technical problem which (for $\sigma$-finite measures) can be solved e.g. using Dynkin systems and, as you said, the monotone convergence theorem. However, imagine an average student in a course on probability theory: He could disregard all existence and measurability problems and nevertheless understand the proofs of almost all important results including the construction of product measures. – Jochen Wengenroth Aug 31 at 6:34	{"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": 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": 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": 0.9553738832473755, "perplexity_flag": "head"}
http://en.wikipedia.org/wiki/Cartesian_product	Cartesian product "Cartesian square" redirects here. For Cartesian squares in category theory, see Cartesian square (category theory). In mathematics, a Cartesian product is a mathematical operation which returns a set (or product set) from multiple sets. That is, for sets A and B, the Cartesian product A × B is the set of all ordered pairs (a,b) where a ∈ A and b ∈ B. The simplest case of a Cartesian product is the Cartesian square, which returns a set from two sets. A table can be created by taking the Cartesian product of a set of rows and a set of columns. If the Cartesian product rows × columns is taken, the cells of the table contain ordered pairs of the form (row value,column value). If the Cartesian product is columns × rows is taken, the cells of the table contain the ordered pairs of the form (column value,row value). A Cartesian product of n sets can be represented by an array of n dimensions, where each element is an n-tuple. An ordered pair is a 2-tuple. The Cartesian product is named after René Descartes,[1] whose formulation of analytic geometry gave rise to the concept. Examples A deck of cards An illustrative example is the Standard 52-card deck. The standard playing card ranks {Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, 2} form a 13 element-set. The card suits {♠, ♥, ♦, ♣} form a 4-element set. The Cartesian product of these sets returns a 52-element set consisting of 52 ordered pairs which correspond to all 52 possible playing cards. ranks × suits returns a set of the form {(Ace, ♠), (King, ♠), ..., (2, ♠), (Ace, ♥), ..., (3, ♣), (2, ♣)}. suits × ranks returns a set of the form {(♠, Ace), (♠, King), ..., (♠, 2 ), (♥, Ace), ..., (♣, 3), (♣, 2)}. A two-dimensional coordinate system An example in analytic geometry is the Cartesian plane. The Cartesian plane is the result of the Cartesian product of two sets X and Y, which refer to points on the x-axis and points on the y-axis, respectively. This Cartesian product can be denoted as X × Y. This produces the set of all possible ordered pairs whose first component is a member of X and whose second component is a member of Y (e.g., the whole of the x–y plane). Alternatively, the Cartesian product can be denoted as Y × X, in which case the first component of the order pair is a member of Y and the second component of the ordered pair is a member of X. The Cartesian product is therefore not commutative. $X\times Y = \{\,(x,y)\mid x\in X \ \and \ y\in Y\,\}.$ [2] $Y\times X = \{\,(y,x)\mid y\in Y \ \and \ x\in X\,\}.$ $X\times Y \neq Y\times X$ Most common implementation (set theory) Main article: Implementation of mathematics in set theory A formal definition of the Cartesian product from set-theoretical principles follows from a definition of ordered pair. The most common definition of ordered pairs, the Kuratowski definition, is $(x, y) = \{\{x\},\{x, y\}\}$. Note that, under this definition, $X\times Y \subseteq \mathcal{P}(\mathcal{P}(X \cup Y))$, where $\mathcal{P}$ represents the power set. Therefore, the existence of the Cartesian product of any two sets in ZFC follows from the axioms of pairing, union, power set, and specification. Since functions are usually defined as a special case of relations, and relations are usually defined as subsets of the Cartesian product, the definition of the two-set Cartesian product is necessarily prior to most other definitions. Non-commutativity and non-associativity Let A, B, C, and D be sets. The Cartesian product A × B is not commutative, $A \times B \neq B \times A,$ because the ordered pairs are reversed except if at least one condition is satisfied:[3] • A is equal to B, or • A or B is an empty set. For example: A = {1,2}; B = {3,4} A × B = {1,2} × {3,4} = {(1,3), (1,4), (2,3), (2,4)} B × A = {3,4} × {1,2} = {(3,1), (3,2), (4,1), (4,2)} A = B = {1,2} A × B = B × A = {1,2} × {1,2} = {(1,1), (1,2), (2,1), (2,2)} A = {1,2}; B = ∅ A × B = {1,2} × ∅ = ∅ B × A = ∅ × {1,2} = ∅ Strictly speaking, the Cartesian product is not associative (unless the above condition occurs). $(A\times B)\times C \neq A \times (B \times C)$ Intersections, unions, and subsets The Cartesian product acts nicely with respect to intersections. $(A \cap B) \times (C \cap D) = (A \times C) \cap (B \times D)$[4] Notice that in most cases the above statement is not true if we replace intersection with union. $(A \cup B) \times (C \cup D) \neq (A \times C) \cup (B \times D)$ Here are some rules demonstrating distributivity with other operators:[3] $A \times (B \cap C) = (A \times B) \cap (A \times C),$ $A \times (B \cup C) = (A \times B) \cup (A \times C),$ $A \times (B \setminus C) = (A \times B) \setminus (A \times C),$ $(A \times B)^c = (A^c \times B^c) \cup (A^c \times B) \cup (A \times B^c).$[4] Other properties related with subsets are: $\text{if } A \subseteq B \text{ then } A \times C \subseteq B \times C,$ $\text{if both } A,B \neq \emptyset \text{ then } A \times B \subseteq C \times D \iff A \subseteq C \and B \subseteq D.$[5] Cardinality The cardinality of a set is similar to the number of elements of the set. For a simpler example, defining two sets: A = {a, b} and B = {5, 6}. Both set A and set B consist of two elements each. Their Cartesian product, written as A × B, results in a new set which has the following elements: A × B = {(a,5), (a,6), (b,5), (b,6)}. Each element of A is combined with each element of B. Each pair makes up one element of the output set. The number of values in each pair is equal to the number of sets whose cartesian product is being taken; hence 2 in this case. The cardinality of the result set is equal to the product of the cardinalities of all the input sets. That is, \|A × B\| = \|A\| · \|B\| and similarly \|A × B × C\| = \|A\| · \|B\| · \|C\| and so on. The cardinality of A × B is also infinity if A or B is an infinite set.[6] Cartesian square and Cartesian power The Cartesian square (or binary Cartesian product) of a set X is the Cartesian product X2 = X × X. An example is the 2-dimensional plane R2 = R × R where R is the set of real numbers - all points (x,y) where x and y are real numbers (see the Cartesian coordinate system). Higher powers of a set The cartesian power of a set X can be defined as: $X^n = \underbrace{ X \times X \times \cdots \times X }_{n}= \{ (x_1,\ldots,x_n) \ \| \ x_i \in X \ \text{for all} \ 1 \le i \le n \}.$ An example of this is R3 = R × R × R, with R again the set of real numbers, and more generally Rn. Generalized powers from different sets n-ary product The Cartesian product can be generalized to the n-ary Cartesian product over n sets X1, ..., Xn: $X_1\times\cdots\times X_n = \{(x_1, \ldots, x_n) : x_i \in X_i \}.$ It is a set of n-tuples. If tuples are defined as nested ordered pairs, it can be identified to (X1 × ... × Xn-1) × Xn. The n-ary cartesian power of a set X is isomorphic to the space of functions from an n-element set to X. As a special case, the 0-ary cartesian power of X may be taken to be a singleton set, corresponding to the empty function with codomain X. Infinite products It is possible to define the Cartesian product of an arbitrary (possibly infinite) indexed family of sets. If I is any index set, and $\left\{X_i\,\|\,i\in I\right\}$ is a collection of sets indexed by I, then the Cartesian product of the sets in X is defined to be $\prod_{i \in I} X_i = \left\{ f : I \to \bigcup_{i \in I} X_i\ \Big\|\ (\forall i)(f(i) \in X_i)\right\},$ that is, the set of all functions defined on the index set such that the value of the function at a particular index i is an element of Xi . Even if each of the $X_i$ is nonempty, the Cartesian product may be empty in general. The axiom of choice postulates that the product is nonempty. For each j in I, the function $\pi_{j} : \prod_{i \in I} X_i \to X_{j},$ defined by $\pi_{j}(f) = f(j)$ is called the j -th projection map. An important case is when the index set is $\mathbb{N}$, the natural numbers: this Cartesian product is the set of all infinite sequences with the i -th term in its corresponding set Xi . For example, each element of $\prod_{n = 1}^\infty \mathbb R = \mathbb R \times \mathbb R \times \cdots$ can be visualized as a vector with countably infinite real-number components. This set is frequently denoted $\mathbb{R}^\omega$, or $\mathbb{R}^{\mathbb{N}}$ . The special case Cartesian exponentiation occurs when all the factors Xi involved in the product are the same set X. In this case, $\prod_{i \in I} X_i = \prod_{i \in I} X$ is the set of all functions from I to X, and is frequently denoted ${X^I}$. This case is important in the study of cardinal exponentiation. The definition of finite Cartesian products can be seen as a special case of the definition for infinite products. In this interpretation, an n-tuple can be viewed as a function on {1, 2, ..., n} that takes its value at i to be the i-th element of the tuple (in some settings, this is taken as the very definition of an n-tuple). Nothing in the definition of an infinite Cartesian product implies that the Cartesian product of nonempty sets must itself be nonempty. This assertion is equivalent to the axiom of choice. Other forms Abbreviated form If several sets are being multiplied together, e.g. X1, X2, X3, …, then some authors[7] choose to abbreviate the Cartesian product as simply ×Xi. Cartesian product of functions If f is a function from A to B and g is a function from X to Y, their cartesian product f×g is a function from A×X to B×Y with $(f\times g)(a, b) = (f(a), g(b)).$ This can be extended to tuples and infinite collections of functions. Note that this is different from the standard cartesian product of functions considered as sets. Definitions outside of Set theory Category theory Although the Cartesian product is traditionally applied to sets, category theory provides a more general interpretation of the product of mathematical structures. This is distinct from, although related to, the notion of a Cartesian square in category theory, which is a generalization of the fiber product. Exponentiation is the right adjoint of the Cartesian product; thus any category with a Cartesian product (and a final object) is a Cartesian closed category. Graph theory In graph theory the Cartesian product of two graphs G and H is the graph denoted by G×H whose vertex set is the (ordinary) Cartesian product V(G)×V(H) and such that two vertices (u,v) and (u′,v′) are adjacent in G×H if and only if u = v and u' is adjacent with v' in H, or u' = v' and u is adjacent with v in G. The Cartesian product of graphs is not a product in the sense of category theory. Instead, the categorical product is known as the tensor product of graphs. References 1. Warner, S: Modern Algebra, page 6. Dover Press, 1990. 2. ^ a b Singh, S. (2009, August 27). Cartesian product. Retrieved from the Connexions Web site: http://cnx.org/content/m15207/1.5/ 3. ^ a b 4. Cartesian Product of Subsets. (2011, February 15). ProofWiki. Retrieved 05:06, August 1, 2011 from http://www.proofwiki.org/w/index.php?title=Cartesian_Product_of_Subsets&oldid=45868 5. Peter S. (1998). A Crash Course in the Mathematics Of Infinite Sets. St. John's Review, 44(2), 35–59. Retrieved August 1, 2011, from http://www.mathpath.org/concepts/infinity.htm 6. Osborne, M., and Rubinstein, A., 1994. A Course in Game Theory. MIT Press.	{"extraction_info": {"found_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": 30, "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": 0.8912218809127808, "perplexity_flag": "head"}
http://simple.wikipedia.org/wiki/Multiset	# Multiset The bags that can be used for shopping are at bag. A multiset (sometimes called a bag) is a concept from mathematics. In many ways, multisets are like sets. Certain items are either elements of that multiset, or they are not. However, multisets are different from sets: The same type of item can be in the multiset more than once. For this reason, mathematicians have defined a relation (function) that tells, how many copies of a certain type of item there are in a certain multiset. They call this multiplicity. For example, in the multiset { a, a, b, b, b, c }, the multiplicities of the members a, b, and c are respectively 2, 3, and 1. A multiset is illustrated by means of a histogram. A multiset can also be considered an unordered tuple: • The tuples (a,b) and (b,a) are not equal, and the tuples (a,a) and (a) are not equal either. • The multisets {a,b} and {b,a} are equal, but the multisets {a,a} and {a} are not equal. • The sets {a,b} and {b,a} are equal, and the sets {a,a} and {a} are equal too. ## Examples One of the simplest examples is the multiset of prime factors of a number n. Here the underlying set of elements is the set of prime divisors of n. For example the number 120 has the prime factorisation $120 = 2^3 3^1 5^1$ which gives the multiset {2, 2, 2, 3, 5}. Another is the multiset of solutions of an algebraic equation. A quadratic equation, for example, has two solutions. However, in some cases they are both the same number. Thus the multiset of solutions of the equation could be { 3, 5 }, or it could be { 4, 4 }. In the latter case it has a solution of multiplicity 2. ## References • Stanley, Richard P. (1997, 1999). Enumerative Combinatorics, Volumes 1 and 2. Cambridge University Press. ISBN 0-521-55309-1, 0-521-56069-1.. This short article about mathematics can be made longer. You can help Wikipedia by .	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 1, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.9061154723167419, "perplexity_flag": "middle"}
http://physics.aps.org/articles/print/v3/40	# Viewpoint: Blurry vision belongs to history and , Experimental Biomolecular Physics, Department of Applied Physics, Royal Institute of Technology, SE-10691 Stockholm, Sweden Published May 10, 2010 \| Physics 3, 40 (2010) \| DOI: 10.1103/Physics.3.40 Making simple modifications to laser-scanning microscopes—like those found in many laboratories—can beat the classical diffraction limit by a factor of $2$. The classical view that lens-based optical microscopes cannot resolve details separated by less than half the wavelength of light is increasingly becoming a dated one. Several microscopy techniques, used either alone or in combination, can now beat the diffraction limit by at least a factor of $2$. Now, in a paper appearing in Physical Review Letters, Claus Müller and Jörg Enderlein at Georg August University in Göttingen, Germany, add to this arsenal of techniques a method that essentially involves modifying an existing bench-top confocal-laser scanning microscope [1]. This type of microscope—called a CLSM for short—is already a fundamental tool for research, particularly in the biological sciences, and Müller and Enderlein’s proposal could influence a broad community. Müller and Enderlein’s technique is built on the method of structured illumination microscopy [2]. When two fine patterns—like that of two combs with different teeth spacing—superimpose multiplicatively, they create moiré fringes (Fig. 1, left). In the case of structured illumination, one pattern is the fluorescence of an unknown sample, while the other pattern comes from the “structured illumination” source, which has a known spatial dependence. The product of the sample and the illumination patterns generates an image that has spatial frequencies that are both sum and difference frequencies between those contained in the original patterns. The difference frequencies, which make up the moiré fringes, have a lower spatial frequency than either of the original patterns and it is these fringes that can often be resolved by a microscope, even if the spatial frequencies of the original patterns themselves are too high to be resolved (see the sequence of panels on the right of Fig. 1). To extract the unknown sample pattern, one needs to do the Fourier back-transformation from frequency space to real space, including a “filter function” that takes the illumination pattern into account. Structured illumination can resolve objects a factor of $2$ better than expected from the classical optical resolution limit and, experimentally, the approach is relatively straightforward. The main challenge lies in the Fourier analysis of the image. When a wide area of the sample is illuminated, the structured illumination approach is conceptually clear and well established [2]. Müller and Enderlein’s insight was to realize that a diffraction-limited laser focus is itself a special type of structured illumination, containing all possible Fourier modes allowed by the confocal microscope. In order to image the entire spatial pattern of the fluorescence generated by such a laser excitation, that is, the convoluted image of the laser focus and the sample, Müller and Enderlein use a CCD camera in place of the point detector found in conventional confocal-laser microscopes. At each point in the laser’s scan across the sample, an image is captured on the CCD camera, which, like the moiré fringes in Fig. 1, contains the information of the spatial distribution of the illumination (the laser focus) and the local sample pattern. Now, if one knows the spatial distribution of the laser focus, it is possible to filter out the unknown sample pattern from the CCD image, improving the resolution by up to a factor of $2$. Although this gain in resolution is moderate, it is enough to enter dimensions where important biological questions await answers. Müller and Enderlein’s technique is also, in contrast to some others, simple. In principle, most confocal-laser scanning microscopes could be adapted to take advantage of the intrinsic structured illumination in the laser focus. In a more general perspective, structured illumination microscopy—either in the wide-field approach or that proposed by Müller and Enderlein—is just one out of several tricks that have proven extremely useful for enhancing far-field optical resolution. An important example is the resolution enhancement that comes from combining controlled photoswitching of fluorescent molecules with the nonlinear fluorescent response of molecules to high-intensity light. This is the principle behind stimulated emission by depletion (STED) microscopy [3]. In STED microscopy, an additional irradiance field is applied in the peripheral parts of the excitation field to drive fluorescent molecules into a dark ground state by stimulated emission (Fig. 2, left). Similar effects can be accomplished by using other photo-induced states [4]. Another way to improve the resolution takes advantage of the fact that the intensity profiles of single fluorescent molecules can be determined with high precision. Optical resolution limits the width of these profiles, but their peak intensities can be determined with a precision far beyond the classical resolution limit [5]. By switching the fluorescent molecules on and off with light, the location of a few, spatially well-separated, individual molecules can be determined at a time (Fig. 2, middle). Repeating this process and overlapping images of many individual molecules can generate an image of even densely labeled samples [6, 7]. Finally, there is a way to enhance the resolution of structured illumination (at least theoretically) by an unlimited amount. The idea takes advantage of the nonlinear photophysical response of fluorescent molecules [8, 9]. At low intensities, the pattern of fluorescence in the sample will follow that of the illumination, but as the illumination intensity increases, molecules experiencing the most intense illumination will saturate (Fig. 2, right). For example, a sinusoidal illumination pattern will produce a squarelike pattern of fluorescence in the sample, which basically contains all Fourier frequencies, from the most fundamental frequencies to frequencies far beyond those that are observable. Although these latter frequencies cannot be detected, their contribution is known from the saturation behavior of the fluorophores and it is possible to reconstruct images over a wider spatial frequency range and with a correspondingly increased spatial resolution. Müller and Enderlein point out that one could also take advantage of saturation effects in their approach. Moreover, it should be possible to adapt any existing confocal-laser scanning microscope to follow their technique, even incorporating standard features on the microscope, such as multicolor imaging and a three-dimensional sectioning. Müller and Enderlein’s method is one out of several important recent contributions to the very active development of ultrahigh resolution far-field optical imaging. This development, based on the interplay between optics and photophysical spectroscopy, has still not come to an end. Simultaneously with this work, the use of laser scanning microscopy for structured illumination has also been suggested for high-resolution two-photon fluorescence microscopy [10]. Indeed, spatially varying illumination patterns of a laser focus for resolution enhancement could possibly also be based on other nonlinear spectroscopic transitions, not necessarily involving fluorescence generation, for instance, stimulated Raman spectroscopy or photothermal spectroscopy. Overall, the ongoing development in high-resolution optical microscopy will virtually open, or rather sharpen, the eyes of the observers in many research fields, most notably in biology, where a wealth of relevant structures and interactions fall beyond the classical resolution limits of light microscopy. ### References 1. C. B. Müller and J. Enderlein, Phys. Rev. Lett. 104, 198101 (2010). 2. M. G. L. Gustafsson, J. Microsc. 198, 82 (2000). 3. S. W. Hell and J. Wichmann, Opt. Lett. 19, 780 (1994). 4. S. W Hell, Science 316, 1153 (2007). 5. W. Heisenberg, The physical principles of the quantum theory (Dover Publications, New York, 1930)[Amazon][WorldCat]. 6. M. J. Rust et al., Nat. Methods 3, 793 (2006). 7. E. Betzig et al., Science 313, 1642 (2006). 8. R. Heintzmann et al., J. Opt. Soc. Am. A 19, 1599 (2002). 9. M. G. L. Gustafsson, Proc. Natl. Acad. Sci. USA 102, 13081 (2005). 10. J. Lu et al., Nano Lett. 9, 3883 (2009). ### Highlighted article #### Image Scanning Microscopy Claus B. Müller and Jörg Enderlein Published May 10, 2010 \| PDF (free) ### Figures ISSN 1943-2879. Use of the American Physical Society websites and journals implies that the user has read and agrees to our Terms and Conditions and any applicable Subscription Agreement.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 4, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.8878688812255859, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/41784/convert-any-number-to-positive-how?answertab=active	# Convert any number to positive. How? How to convert any number (negative or positive) into a positive number.. For example, my input can be 4545 or -4545, I need the out come to be positive value 4545. - ## 2 Answers You just have to simply multiply by $-1$. For example if you have a number $-a$ then multiply by $-1$ to get $-a \times -1 =a$. If the number is positive then multiply by $1$. Or if you are aware of the Absolute Value, then take the absolute value of the number. - This is the problem. I am getting the first value from an equation. I do not know it would be positive or negative. But I want it in positive. So I need to apply some equation on it if I am right. – Muneer May 28 '11 at 7:04 cool. Absolute Value is working. – Muneer May 28 '11 at 7:11 With a calculator, you could make a number positive in one of two (simple) ways: • $\text{abs}(x)$ • $\sqrt{(x^2)}$ The first one is recommended, but the second one will work as well as the square root function on most calculators returns the positive root. Squaring a real number always makes it positive, so taking the square root of a number squared returns the positive number. - 2 Yes, by all means use $\operatorname{abs}(x)$! You don't want to waste time squaring and taking roots (you'll also have to cope with imprecisions)... – t.b. May 28 '11 at 7:10 1 Precisely. The second option is a work-around only (like if your calculator doesn't have an absolute-value function). – El'endia Starman May 28 '11 at 7:12 2 In textbooks, the first one is often written $\lvert x\rvert$. – FUZxxl May 28 '11 at 8:57	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 0.9145206809043884, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/23129/why-is-euclidean-geometry-scale-invariant/23140	# Why is Euclidean geometry scale-invariant? In Euclidean geometry, I frequently use concepts related to invariance under scaling. For example, I know that if two squares have different side lengths, the ratio of their side lengths is the square root of the ratio of their areas. I became curious about the justification of this principle recently when I was asked to explain pi, the ratio of circumference to diameter of a circle, to a young student. It occurred to me half way through the explanation that I do not know a reason why pi exists; I'm not sure why all circles have the same ratio. I haven't studied non-Euclidean geometry, but I presume this scale-invariance is a special feature of Euclidean geometry. On the surface of a sphere, for example, the sum of the angles of a triangle is related to the ratio of the area of the triangle to the surface area of the sphere, whereas Euclidean triangles have the same sum of angles regardless of size. I know that Euclidean geometry is distinguished from other geometries by the parallel postulate, the statement that for any line and any point not on the line, there exists a unique line parallel to the original and through the point. My question is, "How can I go from this parallel postulate and other relevant features of geometry to understanding why Euclidean geometry has no innate length scale involved?" - Do you want an explanation entirely in terms of Euclid's axioms (which sounds rather tedious) or would you be okay with a more modern treatment (using inner products, etc.)? – Qiaochu Yuan Feb 21 '11 at 20:17 1 In hyperbolic geometry for instance, the sum of the angles of a triangle is less than 180°. Interestingly, the difference between the sum and 180° is what is called the defect and the area of the triangle is proportional to the inversse of the curvature of the hyperbolic space times the defect of the triangle. – Raskolnikov Feb 21 '11 at 20:19 @Qiaochu I know some linear algebra and a little abstract algebra from studying physics, so yes a more modern treatment would be interesting, but depending on what tools it uses I might need a reference to understand the background. – Mark Eichenlaub Feb 21 '11 at 20:20 3 A super-simple answer is that curvature is not scale-invariant. A curved object, when you look at it at smaller and smaller scales tends to look flat. So in some sense Euclidean geometry is designed to be flat. – Ryan Budney Feb 22 '11 at 0:38 1 @Alexei: the flat torus of course refers to $\mathbb{R}^2 / \mathbb{Z}^2$. – Willie Wong♦ Feb 28 '11 at 14:06 show 2 more comments ## 3 Answers Similarity is a special feature of Euclidean geometry. It follows from the Euclidean parallel postulate: For each line and each point not on the line, there is a unique second line, parallel to the first, that contains the point. As a matter of fact, the following claim, called Wallis's postulate, is equivalent to the parallel postulate: Given any triangle $\triangle ABC$ and any scale factor, r, there exists a triangle $\triangle DEF$ similar to $\triangle ABC$ where r is the ratio of the lengths of corresponding sides. There is a list of 15 claims all logically equivalent to the Euclidean parallel postulate on the Wikipedia Parallel Postulate article. One of which, not surprisingly, is the Pythagorean theorem. Perhaps the following will provide some insight. Assume AAA Similarity Theorem: If two triangles have all three pairs of angles congruent, then the triangles are similar. the proof of which does not require the parallel postulate but is rather involved. Nonetheless, with it, we can show: Similar Triangle Construction Theorem: If $\triangle ABC$ is triangle and $\overline{DE}$ any segment, then there is a point $F$ such that $\triangle ABC$ is similar to $\triangle DEF$. Proof: We can construct a unique ray $\overrightarrow{DP}$ so that $\angle EDP \cong \angle A$. We can similarly construct a unique ray $\overrightarrow{EQ}$ on the same side of $\overleftrightarrow{DE}$ so that $\angle DEQ \cong \angle B$. Since the angle measures of the constructed angles, $\angle EDP$ and $\angle DEQ$, sum to less than $180$, by Euclid's fifth postulate (equivalent to Euclidean parallel postulate), the two rays must intersect at some point $F$ forming a triangle $\triangle DEF$. By the angle-sum theorem (also logically equivalent to the Euclidean parallel postulate), $\angle C \cong \angle F$. Now by the AAA similarity theorem, the two triangles are similar. $\square$ Then given any scale factor $r$, We can construct a segment $\overline {DE}$ so that $$r=\frac{DE}{AB}.$$ From this, Wallis's postulate follows straight forwardly. Conversely, let us assume Wallis's postulate and from it derive the Euclidean parallel postulate. Proof of EPP from Wallis's Postulate: Suppose we have a line $\ell$ and a point $P$ not on the line. We can drop a perpendicular from $P$ to line $\ell$ and let the perpendicular intersect $\ell$ at point $F$. Then construct a new line $m$ at $P$ at $90$ degrees to our perpendicular. This new line will be parallel $\ell$. If it were not, $m$ and $\ell$ would intersect forming a triangle which would violate the exterior angle theorem. It remains to show the line $m$ is unique. Suppose there were a different line $n$ also through $P$. We must show $n$ is not parallel, that is, line $n$ intersects $\ell$. If $n$ is our perpendicular $\overleftrightarrow{PF}$ obviously it intersects $\ell$. So if $n$ is not $\overleftrightarrow{PF}$, lets choose a point $Q$ of $n$ between lines $\ell$ and $m$ but not on $\overleftrightarrow{PF}$. Let $R$ be the point of intersection of a perpendicular dropped from $Q$ onto line $\overleftrightarrow{PF}$. Let $r=\frac{PF}{PR}$. Wallis's theorem assures us that there is somewhere a triangle $\triangle ABC$ that is similar to $\triangle PQR$ with scale factor $r$. We can construct a new triangle $\triangle PSF$ on the segment $\overline {PF}$ congruent to $\triangle ABC$ by placing $S$ on $\ell$ at distance $BC$ from $F$ and on the same side of $\overleftrightarrow{PF}$ as $Q$. This triangle is congruent to $\triangle ABC$ by the side-angle-side congruence theorem and so is similar to $\triangle PQR$. So the angles $\angle FPS$ and $\angle RPQ$ are the same angle. Hence the lines $n$, $\overleftrightarrow{PQ}$, and $\overleftrightarrow{PS}$ are the same line, and $n$ does indeed intersect $\ell$. $\square$ - You give a statement of the AAA Similarity Theorem but usually people take as the definition of similarity the statement: Two triangles whose corresponding angles are congruent are called similar. In the hyperbolic plane (multiple parallels) one can prove that if two triangles are similar then they are congruent. – Joseph Malkevitch Feb 22 '11 at 5:09 2 @Joseph: The definition that I've used is: Two triangles △ABC and △DEF are similar under a correspondence A↔D, B↔E, and C↔F iff $\angle A \cong \angle D$,$\quad \angle B \cong \angle E$, $\: \angle C \cong \angle F$ and there is a positive number r such that $$r=\frac{AB}{DE}=\frac{BC}{EF}=\frac{CA}{FD}.$$ Using the reals with geometry for distance, angle measure and proportion, following G. D. Birkhoff, seems more natural than Hilbert's numberless approach. I suspect if the foundations of the reals was in better shape when he did his geometry work, Hilbert might have taken the same approach. – Eric Nitardy Feb 22 '11 at 7:05 @Eric Yes, there are other approaches to axiomatics other than Hilbert's approach. The version of Birkhoff's that I am most familiar with appears in R. Millman and G. Parker's book, Geometry: A Metric Approach with Models. Loosely speaking Hilbert wanted to preserve a "synthetic" view. – Joseph Malkevitch Feb 22 '11 at 14:53 @Eric (continued) On page 219 of (first edition, 1981) Millman and Parker write: "We shall define similar triangles to be triangles with corresponding angles congruent." Take a look at Millman and Parker's Theorem 9.1.9. A neutral geometry satisfies EPP (parallel postulate) if and only if there are non-congruent triangles ABC and DEF with angles A and D congruent, B and E congruent and C and F congruent. – Joseph Malkevitch Feb 22 '11 at 14:54 @Joseph: The definitions and development I know best is from an unpublished manuscript by John M. Lee titled "Axiomatic Geometry". It is based in part on Birkhoff and in part on the School Mathematic Study Group postulates. For the past few weeks, I've been reading "Euclidean and Non-Euclidean Geometries" by Marvin J. Greenberg. I prefer John Lee's approach so far. I'll put Millman and Parker on my reading list. – Eric Nitardy Feb 22 '11 at 16:28 show 2 more comments Maybe this reason isn't "deep" enough, but it's because the Euclidean distance formula (the Pythagorean identity) is scale invariant: $$a^2 + b^2 = c^2$$ $$(Xa)^2 + (Xb)^2 = X^2(a^2+b^2) = (Xc)^2$$ whereas the distance formulas in spherical geometry: $$\cos a \cos b = \cos c$$ and hyperbolic geometry: $$\cosh a \cosh b = \cosh c$$ lack this property. - 1 Interesting point. Then all I need to do is understand how why the Pythagorean theorem is special to Euclidean geometry. The most famous proofs that pop to mind for me involve things like similar triangles, though, and so using them would be circular reasoning. – Mark Eichenlaub Feb 21 '11 at 20:24 I am currently studying Perspective and Projective geometry in my math history class and I have learned a bit about the differences in geometries from it... Perspective and Projective geometry differ from Euclidean geometry in that they use descriptive properties and Euclidean geometry follows metric properties. This means that, in Euclidean geometry, you must maintain distance/length, ratios of distance/length, angles, etc. Therefore, scaling an object (similar triangles for example) would be invariant. Also, in Perspective/Projective geometry, distances and ratios of distance and angles are not maintained like in Euclidean geometry which is why parallel lines meet at a point! Unlike Euclidean geometry where parallel lines NEVER MEET. I'm not sure if this is making any sense, but this is what I've learned so far from studying Projective geometry. Hope it helps... -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 80, "mathjax_display_tex": 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": 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": 0.9315260648727417, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/tagged/weight	# Tagged Questions The weight tag has no wiki summary. 1answer 47 views ### Floating Objects and Weight The Situation: A ball is placed in a beaker filled with water and floats. It is also attached to the bottom of the beaker via a string. The Question: The ball is attached to the beaker, thus ... 1answer 280 views ### Archimedes principle and specific gravity A physical balance measures the gravitational mass of a body. I conducted an experiment to find out the specific gravity of a bob. I first measured the mass of the bob in air, and then in water. The ... 2answers 132 views ### Place each foot on a scale: can you add the two to find your weight? I frequent a blog from a British psychologist, and every Friday he likes to pose an interesting puzzle or riddle. The Monday after that he posts the answer. They're good fun, and IANAP but this week's ... 0answers 69 views ### Weight distribution of rod [closed] A line with mass - a metal rod of mass $X$, with a scale at each end. As the rod is vertical the scale at the bottom will read the full mass, as the rod reclines to horizontal each scale will read ... 2answers 180 views ### Forces on an aircraft - thrust, lift, drag, weight I'm extremely sceptical about the wikipedia page on aircraft flight mechanics. When describing 'straight climbing flight', it says: lift is unable to alter the aircraft's potential energy or ... 1answer 47 views ### If a cart hits a wall, does the weight of it affect how it moves, when the center of gravity is constant? I have a model that represents a bicycle (a wood block with wheels), and I'm balancing the center of gravity so it's the same as a real bike. However, when the center of mass is kept constant, does ... 4answers 458 views ### Why do we weigh less when falling? I don't want to go to science world to find out because it would be a long round-trip. I understand that acceleration/deceleration would effect the weight and I can also imagine that someone at ... 2answers 240 views ### What is the weight equation through general relativity? The gravitational force on your body, called your weight, pushes you down onto the floor. $$W=mg$$ So, what is the weight equation through general relativity? 1answer 173 views ### How much does a proton weigh when it is going around the LHC at CERN? Considering that speed increases weight and the proton is going at almost the speed of light, I would like to know how much a speeding proton would weigh in the LHC. 3answers 302 views ### If something weighs 25 kg, how do I find the mass of the object? An object is falling and it weighs 25 kg (on a scale, presumably). What is its mass? I know that weight is measured in Newtons and mass in kilograms, but what if a problem states that something ... 1answer 170 views ### Question about finding $k$ in Hooke's Law My textbook (Advanced Engineering Mathematics by Dennis Zill) offers the following explanation of Hooke's Law: By Hooke's Law, the spring itself exerts a restoring force $F$ opposite to the ... 1answer 56 views ### Is $kg_f$ the same everywhere in the universe? A textbook question says that a vehicle weighs $25kg_f$ on Earth, and asks us to consider certain issues related to its behaviour on the moon. My question is, does the unit kilogram-force $kg_f$ ... 9answers 1k views ### What is the difference between weight and mass? My science teacher is always saying the words "weight of an object" and "mass of an object," but then my physics book (that I read on my own) tells me completely different definitions from the 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": 6, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9498574733734131, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-statistics/177020-probability-distribution-function-help.html	# Thread: 1. ## Probability Distribution Function Help Given the probability distribution function: Determine the: 1. Probability density function. 2. The mean. 3. The median. Hello, I am really struggling with this subject area, this is an example I have found, would someone be able to go through a solution so I can begin to understand it a bit more. Appreciate any help in advance. Thanks. 2. Originally Posted by jscfc Given the probability distribution function: Determine the: 1. Probability density function. 2. The mean. 3. The median. Hello, I am really struggling with this subject area, this is an example I have found, would someone be able to go through a solution so I can begin to understand it a bit more. Appreciate any help in advance. Thanks. First you have a typo in your problem. Second we are not mind readers so please tell us that the function is the Cumulative distribution function Cumulative distribution function - Wikipedia, the free encyclopedia Your book or the above wiki will tell you that the CDF is the integral of the PDF $\displaystyle F(X)=\int_{-\infty}^{x}f(t)dt$ So to undo an integral we will use the fundamental theorem of calculus to take the derivative of $F(x)$ This will give $f(x)$ the PDF This will answer the rest of you question. Exponential distribution - Wikipedia, the free encyclopedia If you get stuck please post some work so we can see where.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.8763407468795776, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/43459/qr-factorization-how-to-get-decreasing-r-ii	QR factorization: How to get decreasing r_ii Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Hi, I'm attempting to implement a QR factorization with column pivoting so that the returned R matrix has decreasing diagonal elements (that is, $r_{i,i} \leq r_{i-1,i-1}$ for all $i\geq 2$). Mathematically, it would involve finding the matrix $P$ so that $AP=QR$, or $A=QRP^T$. I'm using Gram-Schmidt to compute QR. One source I found was this: http://www.mathworks.de/matlabcentral/newsreader/view_thread/250632 Quote from the answer: "During the iteration #k, it is easy to show that if we pick among the set of remaining vectors (i.e., not yet included in the span) the vector that has the largest orthogonal component to the current subspace (generated by k first vectors), then the diagonal of R must decrease." Sounds simple enough, but HOW do I determine which vector of the remaining ones that have the largest orthogonal component to the subspace that's already been found? Thanks a lot in advance for any help on this matter! Best regards Hallgeir - 3 Is there any reason why you prefer Gram-Schmidt (you're using the modified version I hope, classical Gram-schmidt has piss-poor numerical stability) over using Householder reflections? – J. M. Oct 25 2010 at 4:28 2 Seconded. Also, check if you can link to the relevant LAPACK subroutine (netlib.org/lapack/lug/node42.html) instead of coding it from scratch. – Federico Poloni Oct 25 2010 at 9:19 Hi, these are reasonable points of course, but the whole point was to learn how to actually implement it. And I do indeed use the modified version. :) I learn better by doing, and as such I find that implementing the algorithm is a useful way of achieving that. – Hallgeir Oct 26 2010 at 5:31 1 Answer At each step $k$, choose the column of the "reduced" working matrix $A(k:n,k:n)$ with largest Euclidean norm and bring it in front with a permutation. Notice that $r_{11}$ is the Euclidean norm of the first column... - A little late reply from me (seems the email notification came quite late), but thank you very much! :) Seems to work perfectly! – Hallgeir Oct 26 2010 at 5:34	{"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": 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": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9279897809028625, "perplexity_flag": "middle"}
http://en.wikipedia.org/wiki/Ring_of_periods	# Ring of periods For a more frequently used sense of the word "period" in mathematics, see Periodic function. In mathematics, a period is a number that can be expressed as an integral of an algebraic function over an algebraic domain. Sums and products of periods remain periods, so the periods form a ring. Maxim Kontsevich and Don Zagier (2001) gave a survey of periods and introduced some conjectures about them. ## Definition A real number is called a period if it is the difference of volumes of regions of Euclidean space given by polynomial inequalities with rational coefficients. More generally a complex number is called a period if its real and imaginary parts are periods. The values of absolutely convergent integrals of rational functions with algebraic coefficients, over domains in $\mathbb{R}^n$ given by polynomial inequalities with algebraic coefficients are also periods, since integrals and irrational algebraic numbers are expressible in terms of areas of suitable domains. ## Examples Besides the algebraic numbers, the following numbers are known to be periods: • The natural logarithm of any algebraic number • π • Elliptic integrals with rational arguments • All zeta constants (the Riemann zeta function of an integer) and multiple zeta values • Special values of hypergeometric functions at algebraic arguments • Γ(p/q)q for natural numbers p and q. An example of real number that is not a period is given by Chaitin's constant Ω. Currently there are no natural examples of computable numbers that have been proved not to be periods, though it is easy to construct artificial examples using Cantor's diagonal argument. Plausible candidates for numbers that are not periods include e, 1/π, and Euler–Mascheroni constant γ. ## Purpose of the classification The periods are intended to bridge the gap between the algebraic numbers and the transcendental numbers. The class of algebraic numbers is too narrow to include many common mathematical constants, while the set of transcendental numbers is not countable, and its members are not generally computable. The set of all periods is countable, and all periods are computable, and in particular definable. ## Conjectures Many of the constants known to be periods are also given by integrals of transcendental functions. Kontsevich and Zagier note that there "seems to be no universal rule explaining why certain infinite sums or integrals of transcendental functions are periods". Kontsevich and Zagier conjectured that, if a period is given by two different integrals, then each integral can be transformed into the other using only the linearity of integrals, changes of variables, and the Newton–Leibniz formula $\int_a^b f'(x) \, dx = f(b) - f(a).$ A useful property of algebraic numbers is that equality between two algebraic expressions can be determined algorithmically. The conjecture of Kontsevich and Zagier implies that this is also possible for periods. It is not expected that Euler's number e and Euler–Mascheroni constant γ are periods. The periods can be extended to exponential periods by permitting the product of an algebraic function and the exponential function of an algebraic function as an integrand. This extension includes all algebraic powers of e, the gamma function of rational arguments, and values of Bessel functions. If, further, Euler's constant is added as a new period, then according to Kontsevich and Zagier "all classical constants are periods in the appropriate sense". ## References • Belkale, Prakash; Brosnan, Patrick (2003), "Periods and Igusa local zeta functions", International Mathematics Research Notices (49): 2655–2670, doi:10.1155/S107379280313142X, ISSN 1073-7928, MR 2012522 • Kontsevich, Maxim; Zagier, Don (2001), "Periods", in Engquist, Björn; Schmid, Wilfried, Mathematics unlimited—2001 and beyond, Berlin, New York: Springer-Verlag, pp. 771–808, ISBN 978-3-540-66913-5, MR 1852188 • Waldschmidt, Michel (2006), "Transcendence of periods: the state of the art", Pure and Applied Mathematics Quarterly 2 (2): 435–463, ISSN 1558-8599, MR 2251476	{"extraction_info": {"found_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": 11, "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": 0.8569900393486023, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/tagged/groupoids+algebraic-topology	# Tagged Questions 1answer 67 views ### Properties of the Category of topological spaces with $n$ basepoints. I've recently encounted a problem in my reading which would seem to be more naturally phrased if the category we work in shifted from the category $\textbf{Top}^*$ of pointed topological spaces, to ... 0answers 58 views ### 2-morphisms from spans of spans I have a question about the construction of 2-morphisms from spans of spans in the paper "2-vector spaces and groupoid" by Jeffrey Morton . Suppose we have a span of span of groupoids as follows and ... 1answer 388 views ### Does May's version of groupoid Seifert-van Kampen need path connectivity as a hypothesis? May's A Concise Course in Algebraic Topology gives the following statement of the Seifert-van Kampen theorem for fundamental groupoids $\Pi(X)$ (section 2.7): Theorem (van Kampen). Let ... 3answers 153 views ### can the statement “a simplicial set is the nerve of a category if and only if it satisfies a horn-filling condition” be tweaked for groupoids? For some reason I convinced myself that a simplicial set (or maybe I mean directly Kan complex) is homotopy equivalent to the nerve of a groupoid if and only if it has no higher homotopy groups. Is ... 1answer 236 views ### covering spaces and the fundamental groupoid Briefly, my question is whether there is a basepoint-free statement of the basic theorem on covering spaces. For a nice space $X$, I would hope that there is an equivalence of categories between ... 1answer 215 views ### Topological Space as an $(\infty,0)$-category Given a topological space $X$, we may wish to consider it as an $(\infty,0)$-category, where the objects are the points of the space, the 1-morphisms are continuous paths between points, the ... 4answers 496 views ### definition of a groupoid Notation: Underlining $\underline{G}$ denotes a category and $\underline{G}(x,y)$ the class of morphisms from $x$ to $y$. On the Wiki page about groupoids, it is written (I write here my own more ... 4answers 286 views ### What are the ramifications of the fact that the first homotopy group can be non-commutative, whilst the higher homotopy groups can't be? Does this mean that the first homotopy group in some sense contains more information than the higher homotopy groups? Is there another generalization of the fundamental group that can give rise 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": 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": 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": 0.9157530069351196, "perplexity_flag": "head"}
http://mathhelpforum.com/discrete-math/43282-proving-supremum-exists.html	# Thread: 1. ## Proving a supremum exists. The question is: Let A = {p/q : p,q are positive integers and p < 25q } Prove, using the definition of the supremum, that supA = 25. I have proved that $\alpha$ = 25, which is an upper bound. What remains to prove is that it is the least upper bound and this is done on the answer sheet by reductio ad absurdum. The answer sheet reads "Suppose that $\gamma$ is an upper bound of A and $\gamma$ < $\alpha$. By the Archimedian property of the rational numbers, there is an x= p/q $\in$ Q such that $\gamma < x < 25 = \alpha$ Since we can assume both p and q are positive integers we see that x < 25 implies that p < 25q. Therefore x $\in$ A and x > $\gamma$. Therefore $\gamma$ is not an upper bound. Therefore, if $\gamma$ is an upper bound of A, then $\gamma \ge \alpha$. Thus $\alpha$ = 25 is the supremum." I draw your attention to the inequality on a line on its own. (I couldn't seem to make the symbols bold.) Surely if $\gamma$ is an upper bound to A, it is by definition greater than all x in A? A very confused mathematician. 2. That proof is correct. Here is what it shows. $\alpha$ is an upper bound and no number less than $\alpha$ is an upper bound, therefore $\alpha$ is the least upper bound, supremum. 3. Ah, thank you for clarifying things. I think I was approaching it from the wrong angle. Is it saying that no x can exist that will fulfil the condition $\gamma < x < 25 = \alpha$ whilst the condition that $\gamma$ is an upper bound, so $\gamma$ cannot exist? 4. Originally Posted by iamahorse Ah, thank you for clarifying things. I think I was approaching it from the wrong angle. Is it saying that no x can exist that will fulfil the condition $\gamma < x < 25 = \alpha$ whilst the condition that $\gamma$ is an upper bound, so $\gamma$ cannot exist? Yes, that is what I said.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 22, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9628360271453857, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/tagged/number-theory?sort=faq&pagesize=15	# Tagged Questions Number theory is the study of the properties and construction of numbers, particularly integers. Prime numbers are of particular interest to number theorists and consequently cryptographers as they are considered the "building blocks" of numbers and produce many interesting results which are useful ... 3answers 429 views ### Would the ability to efficiently find Discrete Logs have any impact on the security of RSA? This answer makes the claim that the Discrete Log problem and RSA are independent from a security perspective. RSA labs makes a similar statement: The discrete logarithm problem bears the same ... 4answers 6k views ### How can I generate large prime numbers for RSA? What is the currently industry-standard algorithm used to generate large prime numbers to be used in RSA encryption? I'm aware that I can find any number of articles on the Internet that explain how ... 3answers 2k views ### What is the relation between RSA & Fermat's little theorem? I came across this while refreshing my cryptography brain cells. From the RSA algorithm I understand that it somehow depends on the fact that, given a large number (A) it is computationally ... 3answers 1k views ### For Diffie-Hellman, must g be a generator? Due to a number of recently asked questions about Diffie-Hellman, I was thinking this morning: must $g$ in Diffie-Hellman be a generator? Recall the mathematics of Diffie-Hellman: Given public ... 3answers 500 views ### Modern integer factorization software What are the modern software packages that can be used to factoring large numbers into primes. By modern I mean developed and made public within the last 5 years. I'm interested in things that are ... 2answers 480 views ### Selecting a large random prime Say I want a random 1024-bit prime $p$. The obviously-correct way to do this is select a random 1024-bit number and test its primality with the usual well-known tests. But suppose instead that I do ... 5answers 817 views ### Galois fields in cryptography I don't really understand Galois fields, but I've noticed they're used a lot in crypto. I tried to read into them, but quickly got lost in the mess of heiroglyphs and alien terms. I understand they're ... 3answers 372 views ### Is it reasonable to assure that p-1 and q-1 aren't smooth? I came across the requirement that, in RSA, $p-1$ and $q-1$ shouldn't be smooth, shouldn't consist of lots of small factors. Therefore my question: How complicated is it to check whether $p-1$ is ... 1answer 178 views ### A discrete-log-like problem, with matrices: given $A^k x$, find $k$ Let $p$ be a large prime; we will work in $GF(p)$. Let $A$ be a $n\times n$ matrix. Also, let $x$ be a $n$-vector and $k$ a positive integer. Suppose we are given $p$, $A$, $x$, and $y$. The goal ... 0answers 124 views ### Finding where I am in a linear recurrence relation Suppose I have a linear recurrence relation $$a(n) = c_1 a(n-1) + \dots + c_k a(n-k) + d,$$ where the constants $c_1,\dots,c_k,d$ are given and the initial values $a(0),\dots,a(k-1)$ are given as ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 20, "mathjax_display_tex": 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": 0.9425811171531677, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/tagged/forcing	## Tagged Questions 2answers 209 views ### collapsing successor of singular Let $\lambda$ be a singular cardinal. Is it consistent that there is a forcing of size $\lambda^+$ that collapses $\lambda^+$ while preserving all cardinals below $\lambda$? (Not … 1answer 161 views ### Question about Shelah’s version of “Shooting a club” found in PIF Suppose $S \subset \omega_{1}$ is stationary co-stationary. Then there is a forcing notion $P_{S}$ which shoots a closed unbounded $C \subset S$ without collapsing cardinals (or … 1answer 305 views ### Forcing mildly over a worldly cardinal. A cardinal $\theta$ is worldly if $V_{\theta}$ is a model of ZFC. We could force to collapse $\theta$ to a successor cardinal, for example, and destroy the worldliness of $\theta$ … 1answer 103 views ### Which $\omega_1$-trees are proper? Consider a tree $(T, <_T)$ of height $\omega_1$, with countable levels. One can view $T$ as a forcing poset by calling a condition $s\in T$ stronger than $t\in T$ if \$t <_T … 2answers 127 views ### Examples of stationary set preserving forcings that are not semiproper? A notion of forcing $P$ is called stationary set preserving iff each stationary subset of $\omega_1$ remains stationary in $V^P$. It is standard to show that semiproper (and of cou … 1answer 207 views ### The transcendence degree of $\mathbb R$ after adding a Cohen Let $V\models\sf ZFC$, and let $V[r]$ be a generic extension obtained by adding one Cohen real, or equivalently $\omega$ Cohen reals. It is clear that $\Bbb R^{V[r]}$ and \$\Bbb R^ … 1answer 122 views ### Hereditarily Countable Names and Proper Forcing The 'hereditarily countable names' are as defined in Shelah's Proper and Improper Forcing, Chapter 3 Definition 4.1. Let $\mathbb{P}$ be a proper forcing notion and $\dot{Q}$ a \$\m … 1answer 320 views ### Forcing Diamond It is well known that adding a subset of a regular cardinal $\kappa$ with partial functions of size $< \kappa$ forces $\Diamond_\kappa$. One can also see that if $S \in V$ is a … 2answers 882 views ### Similarities between Post’s Problem and Cohen’s Forcing Remark: I have since learned that G.H. Moore addresses this question in the third reference listed at the end of this post, beginning on p. 157 in which he cites a letter from Krei … 1answer 138 views ### Stationary many subsets of $\kappa^+$ whose order type is a cardinal and whose intersection with $\kappa$ is an inaccessible cardinal Is anything known about the consistency strength of the following statement? $\kappa$ is a Mahlo cardinal and there is a stationary set of $a \in \mathcal{P}_\kappa(\kappa^+)$ su … 2answers 201 views ### Set forcing and ultrapowers The following is a result of Woodin (the proof is found after Theorem 5 of "Generalizations of the Kunen Inconsistency" by Hamkins, Kirkmayer and Perlmutter): (Woodin) Let $V[G]$ … 0answers 132 views ### What are the enforceable models of local artinian rings? I was reading Hodges' "Model Theory" Chapter 8 a propos existentially closed models of $\forall_2$ theories in a countable first order language $L$. He extends the proof of the om … 0answers 193 views ### Random reals and strongly meager sets Adding a single Cohen real makes the set of reals from the ground model strong measure zero (see this question). The notion of strong measure zero sets has its dual concept in the … 1answer 292 views ### subalgebra of a simple forcing Let $\alpha > 0$ be any ordinal. Consider the forcing $Add(\omega,\alpha) * Add(\omega_1,1)$. Let $B$ be its boolean completion. Let $\dot{X}$ be the canonical $B$-name for the … 2answers 321 views ### From the product lemma to to a result about powersets Recall the product lemma from Easton's famous paper, which tells us something about when we have a forcing notion (which may be a proper class) that splits as a product with one fa … 15 30 50 per page	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 43, "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": 0.9211784601211548, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/274837/confusion-regarding-the-convexity-of-a-function	# Confusion regarding the convexity of a function I want to know how come the function f(y)=1/y is convex? - How do you understand the word 'convex'? Aka what is your definition of a convex function? – Calvin Lin Jan 9 at 23:58 @CalvinLin The function is convex only if y > 0 if y < 0 it is concave. So how can we say it is convex – user34790 Jan 10 at 0:01 @user34790 The function is not convex over the entire $\mathbb{R}$. However, probably in your setting, you are studying the function for $y>0$ only and hence it is convex. – user17762 Jan 10 at 0:11 @user34790 Great! (I think you mean $y''>0$) I'm glad you pointed that out the 2 different regions. When people define functions, they also talk about the domain of the function. In this case, you will have to look at the domain, which as Marvis pointed out. – Calvin Lin Jan 10 at 15:32 ## 1 Answer One way to check convexity is to check if the second derivative, if it exists, is positive. It is easy to check that the second derivative is positive for $y>0$ in your case. We have $f(y) = \dfrac1y$. Hence, $f'(y) = -\dfrac1{y^2}$, $f''(y) = \dfrac2{y^3} > 0$, for $y>0$. Hence, the function is convex for $y>0$ and concave for $y<0$. -	{"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": 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": 0.9187220335006714, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/tagged/computational-geometry?page=1&sort=votes&pagesize=50	# Tagged Questions The study of computer algorithms which admit geometric descriptions, and geometric problems arising in association with such algorithms. The two major classes of problems are (a) efficient design of algorithms and data classes using geometric concepts and (b) representation and modelling of curves ... 18answers 5k views ### How to check if a point is inside a rectangle? There is a point (x,y), and a rectangle a(x1,y1),b(x2,y2),c(x3,y3),d(x4,y4), how can one check if the point inside the ... 1answer 546 views ### Geometry of nose in and nose out parking in parking lots I would like some computational evidence in favor of my observation that one can park a car in tighter (parking lot) spaces by backing in rather than nose in. I have been doing this successfully for ... 5answers 945 views ### Is it possible to solve any Euclidean geometry problem using a computer? 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Background : We know that the famous theorem of Mordell and Weil says that " Group of rational points on an ...	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 77, "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": 0.9263885021209717, "perplexity_flag": "head"}
http://quant.stackexchange.com/questions/7194/what-is-the-relationship-between-forward-measures-and-lmm	# What is the relationship between Forward measures and LMM? I am reading the book "Term-structure Models: A Graduate Course" by D. Filipovic. In chapter 7 they define the $T$-Forward measure through the density process $$Z_t:=E_Q[\frac{P(T,T)}{P(0,T)B(T)}\|\mathcal{F}_t]=\frac{P(t,T)}{P(0,T)B(t)}$$ or simply on $\mathcal{F}_T$ $$\frac{dQ^T}{dQ}=\frac{1}{P(0,T)B(T)}\tag{1}$$ This is called the $T$-forward measure. However, introducing in chapter 11 the LIBOR Market Model (LMM), we define equivalent measures through $$\frac{dQ^{T_{M-1}}}{dQ^{T_M}}=\mathcal{E}_{T_{M-1}}(\sigma_{T_{M-1},T_M}\bullet W^{T_M}) \tag{2}$$ where $\sigma_{T_{M-1},T_M}(t)=\frac{\delta L(t,T_{M-1})}{\delta L(t,T_{M-1)}+1}\lambda(t,T_{M-1})$ which is a bounded and progressive process (hence $(2)$ really defines a equivalent measure). Continue by backward induction to obtain a whole finite family of such measures. They call any such $Q^{T_m}$ a forward measure. I do not see why the call them forward measures, since the radon nikodym derivative in $(2)$ is obviously different from the one in $(1)$. So why do the call this measure forward measure? - ## 3 Answers Another definition of the $T_{i+1}$-forward measure: Under a $T_{i+1}$-forward measure the forward (LIBOR) rate with (natural) payment in $T_{i+1}$ is a martingale. Since there is a famility of forward (LIBOR) rates, there is a familiy of such measures. I do not have the book at hand, but as far as I understand the situation is as follows. ${Q}_{T_M}$ is the $T$ forward measure for $T=T_{M}$. This is the induction start. Assumed that ${Q}_{T_M}$ is the $T=T_{M}$-forward measure, then equation (2) will indeed give that ${Q}_{T_{M-1}}$ is the $T$ forward measure for $T=T_{M-1}$. Without this initial assumption (which is given at the start of the Section, afaik) it would indeed not make sense to call that family $T$-forward measures. What is the relation between T-forward measure and LMM With respect to the question in the title I would take a didactically different approach than the author (and actually, the equation (2) is used in the reversed direction $M-1 \rightarrow M$ in the following): If you look at a single forward rate $L_{T_{i}}$ in the LMM, then its SDE is a Black model. Looking at this is enough if you just consider simple options in that rate, like a caplet. This SDE becomes very simple under ${Q}_{T_{i+1}}$ - the corresponding forward measure, since the drift is zero under ${Q}_{T_{i+1}}$. Hence we know the drift of each single rate of the LMM, when we consider each forward rate under its respective forward measure. However, to have a consistent model, we need to write down the SDE-system under one measure. For example, the terminal measure ${Q}_{T_{M}}$, that is the forward measure of the last rate. Now equation (2) provides the step vise change of measure moving from ${Q}_{T_{i+1}}$ towards ${Q}_{T_{M}}$ getting all SDEs under one single measure. - Thanks for your answer. Just to be sure I understand your argument: If we assume that $Q^{T_M}$ is the $T_M$ forward measure, and the radon nikodym derivative of $dQ^{T_{M-1}}/Q^{T_M}$ is given as a stochastic exponential, hence we call this again a forward measure? I think this is indeed the motivation of the author. However, for me it seems from a mathematical viewpoint a little bit ambiguous. – hulik Feb 5 at 9:42 Yes. If if you have that assumption as an induction start, then (2) will give you that you have a family of forward measures. I update my answer a bit.... – Christian Fries Feb 5 at 13:50 If you look in Chapter 7.1 where you find equation (1), you will see just below that: $$\frac{d\mathbb{Q}^T}{d\mathbb{Q}} \mid_\mathcal{F_t}=\mathcal{E}_t(v(\cdot,T) \bullet W^)$$ where $W^$ is a $\mathbb{Q}$-Brownian motion. Besides, you'll notice on the book that your equation (2) is described as a way to induce the probability measure $\mathbb{Q}_{T_M} \sim \mathbb{Q}_{T_{M-1}}$ on $\mathcal{F}_{T_{M-1}}$. I think the "$\mid_{\mathcal{F}_{T_{M-1}}}$" is implied here. So there you can see your relationship and why it is called a forward measure. - 1 Thank you for you answer. I know this particular form of the density process as a stochastic exponential. However this is more an assumption on HJM framework. But I do not see how this answer the question. Maybe I do not understand you conclusion. If you say that a Forward measure has a density process of a particular form, i.e. a stochastic exponential, then I do not agree. They define the forward measure as $\frac{dQ^T}{dQ}=\frac{1}{P(0,T)B(T)}$. Otherwise assuming the filtration is generated by Brownian Motion, every martingale would introduce a forward measure – hulik Feb 3 at 12:20 @hulik I thought you were saying that the two formulas didn't look the same at all in your question, so I tried to show that actually they look very much the same. – SRKX♦ Feb 3 at 23:16 thanks a lot for your comment. So you would define a forward measure as a measure, which has a density process of a particular form? Sorry for bothering you and thanks for your patience! – hulik Feb 4 at 8:58 A Forward measure is simply a measure such that the $T$-discounted forward bond starting at time $t$ is a martingale. This allows you to work with reference underlying $S(t) = P(t,T_1)/P(t,T_2)$ with $T_1$ and $T_2$ fixed. Flipovic says that $\frac{dQ^T}{dQ} = \frac{1}{P(0,T)B(T)}$ is "called" a forward measure and then proceeds to show that the discounted $t$-forward bond is a martingale under that measure. In my opinion it would have been better, in order to avoid confusion, to define the forward measure as the one that makes the discounted forward bond to be a martingale. - Thanks for your comment! It seems that it is didactically probably not the best way how the author introduced forward measure. – hulik Feb 6 at 8:01 By $T$-discounted you mean just dividing by $B(T)$. However, this is not enough. You also have to divide by $P(0,T)$ to obtain martingales. Or what exactly do you mean by "A Forward measure is simply a measure such that the T-discounted forward bond starting at time t is a martingale."? Is it: $P(t,T)/B(T)$ should be a martingale under this measure? – hulik Feb 6 at 8:41 I would stick to the other definition: The forward measure is the measure under which the forward is a martingle. The advantage is that this definition is also valid in a more general context (like multi-curve / OIS-discounting). – Christian Fries Feb 6 at 10:27 @hulk: I believe the "$T$-discounted" in this answer is misleading. The forward bond (without disocunting!) is a martingale. This is the case if you choose $P(t,T_{2})$ as numeraire (not B(t)!). – Christian Fries Feb 6 at 10:29 @ChristianFries You are right about the numeraire. That was a mistake in my previous comment. By forward rate, do you mean instantaneous forward rate, i.e. $f(t,T)$? If so, why should this be a martingale under $Q^T$? For that you need further assumption on $\sigma(\cdot,T)$ – hulik Feb 6 at 12:22 show 4 more comments	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 52, "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": 0.9609942436218262, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/184533/two-variable-functions-with-positive-laplacian?answertab=active	# two variable functions with positive laplacian The function $f(x,y)=x^3+y-1$ in $\omega = (1,2)^2$ is such that $f\times \Delta f \ge 0$ on $\omega.$ I am wondering about the existence of a $C^2-$extension $F$ of $f$ in $\Omega = (0,2)^2$ such that $f\times \Delta f \ge 0$ on $\Omega$. Thanks - 6 questions, 0 accepted answers, last seen August 20. Typical ask-and-run. – user31373 Sep 20 '12 at 0:09 ## 1 Answer This particular function can be written as $f(x,y)=(x^3-1/2)+(y-1/2)$ where both summands are 1-variable functions that are convex and strictly positive on $(1,2)$. It is not hard to extend such functions to $(0,2)$ (or the entire line $\mathbb R$, if you wish) while keeping both convexity and positivity. (And of course, convexity implies subharmonicity). Here is a concrete extension to $\Omega$, using the Iverson bracket: $$f(x,y)=\left((x^3-1/2)+10(1-x)^3[x<1]\right)+\left((y-1/2)+(1-y)^3[y<1]\right)$$ You can check directly that both expressions in big parentheses are positive on $(0,1)$. Convexity is clear, as is $C^2$ smoothness. -	{"extraction_info": {"found_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": 0.9258918166160583, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3969826	Physics Forums ## Plotting series in Matlab I am going to plot two series in Matlab. They are as following. please help me to write the appropriate code for them: 1-f(t)=(2/pi)+(4/pi)Ʃ(1/(1-4n^2))cos(2npit) -1≤t≤1 n=1 forN=200 2-f(t)=(4/0.25^2)Ʃ(sin(0.25npi)/(npi)^2)sin(npit) -2≤t≤2 n=1,3,5 forN=200 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug Recognitions: Homework Help Welcome to PF: You want to construct a vector for n, the use that and your formula to compute a vector of each element in the series. The complication appears to be that you have two variables: n and t. So you'll probably have to construct vectors for the different frequencies as well as the coefficients ... with the aim of ending up with a 2D matrix whose columns are the elements of the sum (corresponding to each n) and the rows are the functions at discrete samples (corresponding to values of t). Summing the columns provides a row vector whose entries are f(t) sampled for discrete values of t. Then you can plot(t,f). Could you please confirm the parentheses are correct in the two expressions? As written, I interpret the summation to be over n, ie {sum [1/(1-4n^2)cos(2npit)] over n=1..200} but the expression 1 appears to be written as cos(2npit){sum[1/(1-4*n^2)] over n=1..200} (et sim for 2). Do you want f as a function of t and variable n or just as a function of t with the specified range for n? Recognitions: Homework Help ## Plotting series in Matlab I ust figured everything after the sigma was included in the sum. for 1.$$f(t)=\frac{2}{\pi} + \frac{4}{\pi} \sum_{n=1}^{200} \frac{\cos(2n\pi t)}{1-4n^2}$$... and for 2.$$f(t)=64\sum_{n=1}^{200} \frac{\sin(\frac{n\pi}{4})}{(n\pi)^2}\sin(n\pi t)$$ I still think the strategy I've outlined is the way to go here - we should avoid actually providing the code though. Quote by Simon Bridge Welcome to PF: You want to construct a vector for n, the use that and your formula to compute a vector of each element in the series. The complication appears to be that you have two variables: n and t. So you'll probably have to construct vectors for the different frequencies as well as the coefficients ... with the aim of ending up with a 2D matrix whose columns are the elements of the sum (corresponding to each n) and the rows are the functions at discrete samples (corresponding to values of t). Summing the columns provides a row vector whose entries are f(t) sampled for discrete values of t. Then you can plot(t,f). That would be a possible solution.What if we do not consider sum command? Quote by Simon Bridge I ust figured everything after the sigma was included in the sum. for 1.$$f(t)=\frac{2}{\pi} + \frac{4}{\pi} \sum_{n=1}^{200} \frac{\cos(2n\pi t)}{1-4n^2}$$... and for 2.$$f(t)=64\sum_{n=1}^{200} \frac{\sin(\frac{n\pi}{4})}{(n\pi)^2}\sin(n\pi t)$$ I still think the strategy I've outlined is the way to go here - we should avoid actually providing the code though. In the second one n=1,3,5. everything else is right. Well, here's how it might be done in Mathcad. I have gut feeling (a colon sensation, one might say) that there is sum way of doing a similar thing in Matlab. I have a forboding that loops may occur to somebody, but they should probably be avoided. Attached Thumbnails Attached Files phys - 12 06 24 summation 01.mcd (10.0 KB, 1 views) Recognitions: Homework Help Quote by Mphil1984 That would be a possible solution.What if we do not consider sum command? What exact steps you use are up to you - I have not suggested any exact code, just the mathematical structures you'd need to employ. Technically you don't even need a 2D matrix to store all the terms, you could use a different 1D vector for each of the 200 terms instead. The approach outlines is a "brute force" method - there will be more elegant approaches. Note: for the second one where n=1,3,5...200 (or is it n=1,3,5...399 for N=200 terms?) Anyway = in matlab you need only do the step, so all the odd numbers are n=1:2:199; for all the odd numbers from 1 to 200. Writing it out in sigma notation is trickier than that... I need to sum over integer m from 0 to 99 and n=2m+1 and it looks like:$$f(t)=64\sum_{n=1}^{200} \sin\left ( \frac{(2n-1)\pi}{4}\right ) \frac{\sin((2n-1)\pi t)}{((2n-1)\pi)^2}$$... for the first 200 odd numbers. [One way matlab/octave is easier than pencil and paper :) ] Note - the mathcad notes look like what I was suggesting only using jargon. Thread Tools \| \| \| \| \|------------------------------------------------\|----------------------------------------------\|---------\| \| Similar Threads for: Plotting series in Matlab \| \| \| \| Thread \| Forum \| Replies \| \| \| Programming & Comp Sci \| 0 \| \| \| Math & Science Software \| 8 \| \| \| Engineering, Comp Sci, & Technology Homework \| 0 \| \| \| Engineering, Comp Sci, & Technology Homework \| 3 \| \| \| Math & Science Software \| 0 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 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": 0.9214593768119812, "perplexity_flag": "middle"}
http://mathhelpforum.com/advanced-algebra/176392-orthogonal-matrices-print.html	# orthogonal matrices Printable View • March 30th 2011, 09:23 PM Jskid orthogonal matrices Show that if A and B are orthogonal matrices, then AB is an orthogonal matrix. I think I need to show that the columns of A are orthonormal to each other, but I don't know how. • March 30th 2011, 09:25 PM Drexel28 Quote: Originally Posted by Jskid Show that if A and B are orthogonal matrices, then AB is an orthogonal matrix. I think I need to show that the columns of A are orthonormal to each other, but I don't know how. You often pick the hardest characterization to use. Try picking another one--solution below--use at your own risk Spoiler: You sure you want to look so soon? Spoiler: Fine, do it Spoiler: Isn't it true that $A$ is real orthogonal if and only if $A^{-1}=A^\top$? So that $(AB)^{-1}=B^{-1}A^{-1}=B^\top A^\top=(AB)^\top$? • March 30th 2011, 09:55 PM FernandoRevilla A "tiny" alternative: $M\in\mathbb{R}^{n\times n}$ is orthogonal iff $MM^{t}=I$ So, $(AB)(AB)^t=\ldots =I$ All times are GMT -8. The time now is 10:35 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": 6, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "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": 0.9089231491088867, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/114796-finding-absolute-maximum-absolute-minimum.html	# Thread: 1. ## finding absolute maximum and absolute minimum I need to find the absolute maximum and minimum of the function below on the interval [0,4] g(x) = x / x^2 + 4 I know that i need to find g' which i think is g'(x) = (x^2 + 4)1 - x(2x) (x^2 + 4)^2 g'(x) = 4 - x^2 (x^2 - 4)^2 Am I correct so far? Do I need to do g'' or should I factor out and find the critical values? 2. I get $g(x) = \frac{x}{x^2+4}$ $g'(x) = \frac{(x^2+4)\times 1 - 2x \times x}{(x^2+4)^2}$ $g'(x) = \frac{x^2+4 - 2x^2}{(x^2+4)^2}$ $g'(x) = \frac{4 - x^2}{(x^2+4)^2}$ making g'(x)=0 $0 = \frac{4 - x^2}{(x^2+4)^2}$ $0 = 4 - x^2$ $x=\pm 2$ You can sub these values into g''(x) if you want to know if they are mins or maxs. Or if you know the general shape of the orignal function it shouldn't be required.	{"extraction_info": {"found_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": 7, "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": 0.82354736328125, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/tagged/complete-intersection	## Tagged Questions 0answers 129 views ### Bezout’s theorem for non-proper intersections? Is there a version of Bézout's theorem for non-proper intersections? For my specific problem, the setup is as follows: Let $P_1,P_2,P_3,P_4\in\mathbb{C}[z_1,z_2,z_3,z_4]$, and su … 2answers 239 views ### Parameter space for complete intersections and their discriminant Consider globally complete intersections in $\mathbb{P}^n$, of codimension $k$, of some fixed multi-degree $(d_1,\dots,d_k)$. Is there some nice (i.e. "explicit") parameter spac … 1answer 182 views ### Irreducible components of reduced complete intersection Let $Z$ be an irreducible and reduced scheme. Does there exist a reduced complete intersection $Y$ such that $Z$ is an irreducible component of $Y$? 1answer 275 views ### what can be reached by flat degeneration of (globally) complete intersection? Let $X\subset\mathbb{P}^n$ be a (globally) complete intersection, let $(X_t)_{t\in\mathbb{C}^1}$ be a flat family, with $X_1=X$. Which types of schemes can we get as $X_0$? Or, co … 0answers 205 views ### Factoriality of complete intersections Let $X\subset\mathbb{P} _{\mathbb{C}}^N$ be a normal complete intersection. $X$ is called factorial if every Weil divisor on it is Cartier; equivalently if all local rings \$\mathc … 1answer 263 views ### If $X$ is an affine variety, is $X$ one component of a complete intersection with two? This is an idle question, but I give the example that motivated me below. Say $X \subseteq {\mathbb A}^n_k$ is irreducible and $k$ is infinite. Then by picking a regular point of … 1answer 356 views ### Looking for an inequality between Chern and Todd classes (something in style of Bogomolov-Miyaoka-Yau) Consider a smooth projective surface $S\subset\Bbb P^N_{\Bbb C}$ which is a complete intersection of hypersurfaces of degrees $(d_1,..,d_{k\ge2})$ with $d_i\ge2$ for all i. Is it … 1answer 250 views ### Sections of a fibration in intersections of quadrics Suppose that we have a smooth variety $X$ of dimension $n$ that fibers (a flat morphism) over a curve $Y$, and s.t. the fibers of $X \to Y$ are all complete interesections of two q …	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 25, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": 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": 0.881303071975708, "perplexity_flag": "head"}
http://mathoverflow.net/questions/21651/cohomology-of-flag-varieties/21659	Cohomology of Flag Varieties Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) For `$K$` a compact Lie-group with maximal torus `$T$`, I'd like to know the cohomology `$\text{H}^{\ast}(K/T)$` of the flag variety `$K/T$`. If I'm not mistaken, this should be isomorphic to the algebra of coinvariants of the associated root system, according to the "classical Borel picture", as it is often called in the literature (sadly, often without a reference). Unfortunately, Borels original paper is quite long and so I have the following Question: Does anybody know a short proof for the theorem? For example, can it be proved using a direct computation of the Lie-algebra cohomology `$\text{H}^{\ast}({\mathfrak k},{\mathfrak t})$`? As a side-question: Is it correct that for a complex semisimple Lie-group `$G$` with Borel `$B$`, compact real form `$K$` and maximal torus `$T$` the map `$K/T\to\ G/B$` is a homotopy equivalence? What is a good reference for these things? I'm sorry if this is too elementary for MO, but apart from Borel's original paper I couldn't find good sources. - 3 Answers Borel's lengthy 1953 Annals paper is essentially his 1952 Paris thesis. It was followed by work of Bott, Samelson, Kostant, and others, which eventually answers your side question affirmatively. For a readable modern account in the setting of complex algebraic groups rather than compact groups, try to locate a copy of the lecture notes: MR649068 (83h:14045) 14M15 (14D25 20F38 57N99 57T15) Hiller, Howard, Geometry of Coxeter groups. Research Notes in Mathematics, 54. Pitman (Advanced Publishing Program), Boston, Mass.-London, 1982. iv+213 pp. ISBN 0-273-08517-4. (This was based on his 1980 course at Yale. Eventually he left mathematics to work for Citibank.) The identification of the cohomology ring with the coinvariant algebra of the Weyl group has continued to be important for algebraic and geometric questions, for instance in the work of Beilinson-Ginzburg-Soergel. While Hiller's notes are not entirely self-contained, they are helpfully written. (But note that his short treatment of Coxeter groups has a major logical gap.) ADDED: In Hiller's notes, Chapter III (Geometry of Grassmannians) is most relevant. For connections with Lie algebra cohomology, the classical paper is: MR0142697 (26 #266) 22.60 (17.30) Kostant, Bertram, Lie algebra cohomology and generalized Schubert cells. Ann. of Math. (2) 77 1963 72–144. Nothing in this rich circle of ideas can be made quick and easy; a lot depends on what you already know. P.S. Keep in mind that Hiller tends to give explicit details just for the general linear group and grassmannians, but he also points out how the main results work in general, with references. I don't know a more modern textbook reference for this relatively old work. But the intuitive connection between the Borel picture and the Bott/Kostant cohomology picture is roughly this: The Lie subalgebra spanned by negative root vectors plays the role of tangent space to the flag manifold/variety. In the Lie algebra cohomology approach you get an explicit graded picture for each degree in terms of number of elements in the Weyl group of a fixed length, whereas the Borel description in terms of Weyl group coinvariants makes the algebra structure of cohomology more transparent. (What I don't know is whether a simpler proof of Borel's theorem can be derived using Lie algebra cohomology.) Concerning the relationship between `$K/T$` and `$G/B$`, this goes back to the work around 1950 on topology of Lie groups (Iwasawa, Bott, Samelson): all the topology of a connected, simply connected Lie group comes from a maximal compact subgroup. So the two versions of the flag manifold are homeomorphic. In later times, emphasis has often shifted to treating `$G$` as a complex algebraic group, so that `$G/B$` is a projective variety. For me the literature is hard to compactify. One more reference, which treats the Borel theorem in a semi-expository style: MR1365844 (96j:57051) 57T10 Reeder, Mark (1-OK), On the cohomology of compact Lie groups. Enseign. Math. (2) 41 (1995), no. 3-4, 181–200. - Thank you, these links were very helpful to me. – B. Bischof Apr 17 2010 at 19:23 1 Another important paper in the direction of Kostant's paper is that by Bernstein, Gel'fand, Gel'fand, "Schubert cells and Cohomology of the spaces G/P" – B. Bischof Apr 19 2010 at 14:37 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. Note that your statement is only true in rational cohomology. For example, $H^\ast(SO(5)/T)$ is not generated in degree $2$ (though it is rationally). The easiest proof I know starts from equivariant cohomology: $H^\ast_T(K/T) = H^\ast_{T\times T}(K) = H^\ast_{T\times K\times T}(K\times K) = H^\ast_K(K/T \times K/T)$ So far this uses $H^\ast_F(X) = H^\ast(X/F)$ for free actions. Now use the equivariant Künneth formula: $... = H^\ast_K(K/T) \otimes_{H^\ast_K} H_K(K/T) = H^\ast_T \otimes_{H^\ast_K} H^\ast_T$ Rationally, the base ring $H^\ast_K$ is $(H^\ast_T)^W$, the invariants. Since you didn't want equivariant but ordinary cohomology, kill the left factor, leaving ${\mathbb Q} \otimes_{(H^\ast_T)^W} H^\ast_T$, which is your desired ring of coinvariants. (I'm having a bunch of trouble with $H$ vs. $H^\ast$ in typesetting here, sorry!) - Having had the same trouble myself, I fixed the typesetting by using \ast instead of * for the asterisks. (Backticks also work.) – Tim Perutz Apr 17 2010 at 16:59 3 I am not sure I agree that equivariant cohomology is supreme as a candidate for the quickest way to go if one wants an algebro-topological proof. One can use the Eilenberg-Moore spectral sequence which has the form (with a field $k$ as coefficient ring say) `$$ \mathrm{Tor}^_{H^(K)}(H^(T)),k) \implies H^(K/T)$$` and as $H^(T)$ is free as $H^(K)$-module (when $k$ has characteristic $0$) this degenerates to give what one wants. In any case if either of these proofs qualifies as "short" depends on whether your definition of length is recursive or not.... – Torsten Ekedahl Apr 17 2010 at 18:10 Comment to my own comment: When I wrote $H^(K)$ and $H^(T)$ I meant $H^(BK)$ and $H^(BT)$. – Torsten Ekedahl Apr 19 2010 at 5:26 Maybe even faster: $H_K(K/T)=H_T$, so $H(K/T)=\mathbb Q \otimes_{H_K} H_T=\mathbb Q \otimes_{H_T^W} H_T$ the ring of coinvariants. – Jan Weidner Oct 25 2011 at 12:06 Concerning the side question, the map K/T → G/B is actually an isomorphism of real manifolds, not just a homotopy equivalence. Not sure about the references, but this is essentially textbook material (unfortunately I forgot where I learned about this). Considering the case of U(n) ⊂ GL(n) acting on the flags in Cn is instructive. - Oddly, in the infinite-dimensional case the most relevant map is the other way, and is only a homotopy equivalence, as one can read about in [Pressley-Segal] "Loop Groups". – Allen Knutson Apr 19 2010 at 1:11 1 One way of getting this isomorphism is that it is a special case of the Iwasawa decomposition for a general semi-simple (real) Lie group. This allows us to get many specific references, Helgason for instance. – Torsten Ekedahl Apr 19 2010 at 5:34 Oh, so this is called the Iwasawa decomposition. I used to know this, but it was a long time ago. – Leonid Positselski Apr 19 2010 at 20:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 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": 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": 0.9161049127578735, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/105157?sort=newest	## Triangular grid with 4 edges per vertex ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I am trying to create a triangular grid/mesh for a rectangular domain in $\mathbb{R}^2$ with the property that each vertex is shared by (at most) four edges. Is this possible to accomplish? - Finite or infinite? – Brendan McKay Aug 21 at 14:04 As long as you do not need more than 8 regions. Gerhard "Otherwise You Need Different Geometry" Paseman, 2012.08.21 – Gerhard Paseman Aug 21 at 14:59 1 I can only see how to do $6$ regions. An octahedron is a closed manifold. – Will Sawin Aug 21 at 17:46 ## 2 Answers Since there is no requirement that the outer region must be a triangle, the question is not quite as trivial as indicated in earlier comments. The rectangle $[0,n] \times [0,1]$ can be triangulated by dividing it into unit squares and then inserting the SW-NE diagonal in each square. Still, this might not be the kind of grid/mesh one wants. To see the problem, it might be easier to think in terms of angles than to use Euler's polyhedron formula: If there are interior points in the triangulation, then the angles at those points have to be at least $90^\circ$ on average, while the average angle in a triangle is only $60^\circ$. It follows that most vertices of the triangulation have to be on the boundary of the region. - How is your suggestion a triangulation? Is it for only some pairs (a,b)? When I try it I get quadrilaterals. Gerhard "Maybe This Is Different Geometry" Paseman, 2012.08.21 – Gerhard Paseman Aug 21 at 19:04 Actually I try it and I see more triangles, but also degree 6 vertices. I must be doing something wrong. Gerhard "A Picture Would Really Help" Paseman, 2012.08.21 – Gerhard Paseman Aug 21 at 19:05 Gerhard, maybe I was a bit sloppy, answer edited. – Johan Wästlund Aug 21 at 19:20 OK,now I understand your edited version. Have you any triangulations with more than 3 vertices in the interior? My guess is that even one vertex in the interior seriously limits the possibilities. Gerhard "Ask Me About System Design" Paseman, 2012.08.21 – Gerhard Paseman Aug 21 at 19:24 ### You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you. This question is too elementary for MO, but here's a hint. (See the FAQ for alternative sites to ask your question.) 1. Familiarize yourself with the notion of Euler characteristic 2. Convince yourself that the Euler characteristic of your triangulation of a rectangle is 1. 3. Deduce some inequalities on the numbers of vertices, edges and faces, assuming all faces are triangles and all vertex valences are less than or equal to 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": 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": 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": 0.9175876379013062, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/35488?sort=oldest	## Values of the multiplicative group over a ring spectrum ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In his notes on elliptic cohomology, Lurie defines the multiplicative group $\mathbb{G}_m$ over a ring spectrum $A$ as $\operatorname{Spec} A[\mathbb{Z}]$. What is the value $\mathbb{G}_m(B)$ of the represented functor at an $A$-algebra $B$? If this is too hard to say in general: Are there at least any specific examples, other than Eilenberg-MacLane spectra, where one does know the answer? - ## 1 Answer It is slightly complicated. One has a number of adjunctions: ```$$ \begin{eqnarray} \mathbb{G}_m(B) &=& Hom_{A-alg}(A[\mathbb{Z}],B) \\ &\simeq& Hom_{E_\infty-rings}(\mathbb{S}[\mathbb{Z}],B)\\ &\simeq& Hom_{E_\infty-spaces}(\mathbb{Z},GL_1(B))\\ &\simeq& Hom_{spectra}(H\mathbb{Z},gl_1(B)). \end{eqnarray} $$``` (Note these adjunctions are weak equivalences of spaces, and the last two adjunctions require a fair amount of theory to make rigorous.) The problem is that it is usually quite difficult to compute the maps out of the Eilenberg-Mac Lane spectrum $H\mathbb{Z}$ unless the target is also an Eilenberg-Mac Lane space. In the case where the algebra $B$ comes from a simplicial commutative ring, this is true and so one at least knows that the set of homotopy classes of maps `$[H\mathbb{Z}, gl_1(B)]$` surjects onto `$\pi_0(B)^\times$`. Even for complex K-theory, the calculation is somewhat involved (but doable), but the only method that I can immediately think of involves the Bousfield-Kuhn functor. - I forgot an important special case of this. If `$\pi_0(B)$` is a $\mathbb{Q}$-algebra, then `$\mathbb{G}_m(B)$` is isomorphic to `$\pi_0(B)^\times$` after passing to homotopy classes of maps. – Tyler Lawson Aug 13 2010 at 14:24 1 Tyler, can't we say that if $\pi_0(B)$ is a $\mathbb{Q}$-algebra, then $G_m(B)\simeq GL_1(B)$? – Charles Rezk Aug 13 2010 at 17:52 @Charles: Well, yes. I guess I should have decided once and for all whether I was talking about spaces of maps or homotopy classes of maps – Tyler Lawson Aug 13 2010 at 18:05	{"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": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9229905605316162, "perplexity_flag": "head"}
http://mathoverflow.net/questions/52270?sort=votes	## Is H^2(W_p,C^times) well-known? ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) Let $W_p$ be the Weil group of $\mathbf{Q}_p$. What is the Galois cohomology group $H^2(W_p,\mathbf{C}^{\times} )$ (with trivial action)? Is it zero, or something huge and complicated? (This group comes up, at least for me, when you want to compare two Weil group representations whose projectivizations agree.) - Just to nitpick a bit: the Weil group $W_p$ is not a profinite group, so "the Galois cohomology group $H^2(W_p,\mathbb{C}^{\times})$" is not well-defined. I believe though that this group can be made sense of in some reasonable way, and Marty's answer addresses this. – Pete L. Clark Jan 18 2011 at 9:20 ## 1 Answer It is known that $H^2(W, C^\times)$ is trivial, when $W$ is the Weil group of a global or local field, with the trivial action on $C^\times$, and the cohomology is taken in the sense of Moore (measurable cochains). This is the main result of C.S. Rajan, "On the vanishing of the measurable cohomology groups of Weil groups", Compositio Math. 140 (2004) 84-98 (also easy to find online). Is this well-known? I don't know. I only found this paper recently, and hopefully now it will become better-known! - Beautiful, thanks! – David Hansen Jan 17 2011 at 2:57 As is explained in the introduction to Rajan's paper, Tate had proved the analogous result for Galois groups instead of Weil groups. – Chandan Singh Dalawat Jan 18 2011 at 4:56	{"extraction_info": {"found_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": 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": 0.9167326092720032, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/70391/convex-bodies-with-constant-maximal-section-function-in-odd-dimensions	## Convex bodies with constant maximal section function in odd dimensions ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) In 1970 or so, Klee asked if a convex body in $\mathbb R^n$ ($n\ge 3$) whose maximal sections by hyperplanes in all directions have the same volume must be a ball. The counterexample in $\mathbb R^4$ is trivial and can be described as follows: Let $f:[-1,1]\to\mathbb R$ be continuous, strictly concave and satisfy $f(-1)=f(1)=0$. For every such function, let $Q_f$ be the body of revolution given by $y^2+z^2+t^2\le f(x)^2$. Then $Q_f$ and $Q_g$ have the same maximal sections in every direction if (and, actually, only if) $f$ and $g$ are equimeasurable, i.e., $\|{f>t}\|=\|{g>t}\|$ for all $t>0$. (Of course, there are plenty of concave functions equimeasurable with $\sqrt{1-x^2}$). With some extra work, one can construct something like this in $\mathbb R^n$ when $n$ is even though I do not know any similarly nice geometric description of such bodies for $n\ge 6$. What I (and my co-authors) are currently stuck with is the case of odd $n$ (say, the usual space $n=3$). In view of such simple example in $\mathbb R^4$, I suspect that we are just having a mental block. Can anybody help us out? - what about $n=2$? – Will Jagy Jul 15 2011 at 2:14 Planar domains of constant width have been well-known for a long time (normally one thinks of them as having constant projections but all their projections are realized as sections as well). – fedja Jul 15 2011 at 2:42 1 en.wikipedia.org/wiki/Curve_of_constant_width – Mark Sapir Jul 15 2011 at 2:43 That's right. I knew about curves of constant width, I did not connect that properly with your question. – Will Jagy Jul 15 2011 at 2:46 1 Not really :(. The bodies of constant width (=constant maximal 1-dimensional sections) are not hard to construct because basically you just have to relate pairs of points. When you raise the section dimension, you have to deal with integral transforms. I suspect that this simple R^4 construction went unnoticed because the customary way is to represent convex bodies by their radial or support functions even for the bodies of revolution and the equimeasurability condition is a total mess in such terms. So, it seems like we need a fresh look with some little twist... – fedja Jul 15 2011 at 3:29 show 1 more comment ## 1 Answer To those who are still interested: we've finally made it but it's so ugly that a nice alternative approach will be always welcome :) We are still stuck with Bonnesen's question about the possibility to recover a convex body from the volumes of its maximal sections and projections in odd dimensions, so some help will be appreciated. The even-dimensional case can be found here. I feel a bit like a student asking for help with his homework, of course, but why not? We all get stuck now and then :). This should really be a comment but it's too long to fit the number of characters restriction. - More affirmation of Klee's prowess as a problem-poser! May I ask: Is the essence of your proof in odd dimensions concentrated (in the intellectual sense) in $\mathbb{R}^3$? – Joseph O'Rourke Jan 4 2012 at 2:24 It doesn't matter for us whether it is 3,5,7, or any other odd number: there is no simplification or complication when going from one odd dimension $d>1$ to another, if that's what you meant. However, as I said, the odd dimension case is harder than the even one due to the square root popping up everywhere and the non-locality of the Radon transform. – fedja Jan 4 2012 at 12:23	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9360281229019165, "perplexity_flag": "head"}
http://medlibrary.org/medwiki/Cladding_(fiber_optics)	# Cladding (fiber optics) 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: Cladding is one or more layers of materials of lower refractive index, in intimate contact with a core material of higher refractive index. The cladding causes light to be confined to the core of the fiber by total internal reflection at the boundary between the two.[1] Light propagation in the cladding is suppressed in typical fiber. Some fibers can support cladding modes in which light propagates in the cladding as well as the core. (From Federal Standard 1037C and from MIL-STD-188) The numerical aperture of a fiber is a function of the indices of refraction of the cladding and the core by: $\mathrm{NA}=\sqrt{n_\mathrm{core}^2-n_\mathrm{clad}^2}$[2] ## References 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 "Cladding (fiber optics)", available in its original form here: http://en.wikipedia.org/w/index.php?title=Cladding_(fiber_optics) • ## 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": 1, "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": 0.8857016563415527, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/18060/list	2 Fixed title 1 # Correspondeces in Topology I had one or two little fights with correspondences in the context of algebraic geometry where an elementary correspondence $C:X\to Y$ of connected smooth $k$-Schemes seems to be defined as an irreducible closed (reduced) subscheme $C\hookrightarrow Y\times_k X$, such that the projection to $X$ is finite an surjective. Further, I have vaguely heard of correspondences in topology (invented by Lefschetz?) where it seems to me, that such a thing is a cohomology class in $H^(Y\times X)$ for compact, oriented manifolds `$X,Y$`. Using Poincare duality and the cohomology pushforward functor `$(-)_!$` I can associate a cohomology class `$(\Delta_f)_!(1)$` in `$H^(Y\times X)$` to a map $X\to Y$. My questions are: 1. I can not really see the analogy of the concepts, except that in booth cases one can associate a corresponces to a morphism by its graph. So, what (or how deep) is the analogy? 2. I know (only a very few) applications of correspondences in algebraic geometric, but of none in topology. What are they good for? Where can I find applications?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9589897394180298, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/44018/what-is-voltage-strict-but-intuitive-definition-from-accumulators-perspective?answertab=votes	# What is voltage: strict but intuitive definition from accumulator's perspective I know, that voltage is analogous to pressure for charge, but analogies lie. I don't see charge pressing anything and I don't understand definition of $U=A/q$ (voltage = work/charge), cause I can't see what is that work. Voltage in a circuit is determined by an accumulator. I suppose, that accumulator is characterized with voltage output and current output. I don't get how they are related, but can imaging some tweaks to the accumulator: 1) If I double the area of both metal plates, contacting with respective solutions, would it double the voltage output? Intuitively it will cause twice the amount of Ions bump into plates and give their charges to them, so it would double the current and won't change any "charge pressure", I can think of. 2) If I double the concentration of each solution, would it double the current or charge. My intuition that it is the same as if I doubled the area of plates - it would just double the amount of ions transferring their charge per moment of time. Nernst equation suggests a change of voltage in this case, but I can't see what it is. What I want to understand, is what is voltage in terms of ball-like ions, forces and such intuitive entities. Thanks. - If you short-circuit the electrodes like that, there can't be a potential difference (voltage). – RedGrittyBrick Nov 12 '12 at 11:47 All right, there is something in between. I don't want to say, that there is a resistance in between, cause my next question is going to be what is resistance. – Bob Nov 12 '12 at 11:57 ## 3 Answers 1) If I double the area of both metal plates, contacting with respective solutions, would it double the voltage output? No, the voltage is dependent on the chemistry. "D" and "AAA" sized zinc-carbon cells both produce about 1.5 volts. Bigger batteries can produce larger currents. This is presumably because a larger electrode surface area allows for a greater number of ions to be reacted with. If I double the concentration of each solution, would it double the current or charge. It would increase the charge. Empirically you measure the charge in a lead-acid battery by measuring the specific gravity of the electrolyte. There are limits to how concentrated a solution can be. A fully-charged lead-acid battery contains sulphuric acid, a fully discharged lead-acid battery contains water. - 1) If I double the area of both metal plates, contacting with respective solutions, would it double the voltage output? Intuitively it will cause twice the amount of Ions bump into plates and give their charges to them, so it would double the current and won't change any "charge pressure", I can think of. Remember that the potential is energy per charge. So it doesn't what the absolute number of ions bumb into the surface, the thing that matters is only the energy given per electron. Therefore if you make the area twice as large, you don't get twice as much energy given to each electron, you just get twice as many electrons with the same energy. So you get twice the current, but the same potential. 2) If I double the concentration of each solution, would it double the current or charge. My intuition that it is the same as if I doubled the area of plates - it would just double the amount of ions transferring their charge per moment of time. Nernst equation suggests a change of voltage in this case, but I can't see what it is. You need to be careful with your intuition when considering this. The potential is dependent on what goes on at the surface which isn't just linearly linked to the bulk concentration. If you for instance have a surface with a fixed low number of catalytic sites, once you move beyond a concentration that allows all of them to be occupied, you might not see that much change in current as you increase concentration. Also you need to remember that a solution with double the concentration isn't just something with double the amount of ions, you have a change in the solvent/ion ratio, which may cause other things to take place, perhaps changing how ions are organised on the surface. - I agree about the first point. I disagree about the second. Nernst equation holds even for low concentrations of solutions (e.g. micromole/litre is a typical concentration of a substance in living cell and Nernst equation is applied to it), so that interactions between ions can be neglected and doubling/halving its concentration won't have any spatial effect. – Bob Nov 12 '12 at 20:06 @Bob You seem to be confused. Are you trying to because there does exist a special case where the concentration differences will obey Nernst equation that my assessment that you cannot assume a general linearity between potential and concentration is wrong? If that is the case, please remember that Nernst equation is not a linear relationship firstly, secondly, even if it was, you could not argue against a trend not being general true with one special case where it is. eg. arguing against; All humans are not male, with: yes they are, this one is! – Vincent Nov 13 '12 at 11:30 What I want to understand, is what is voltage in terms of ball-like ions, forces and such intuitive entities. First you'd need to clarify the situation in your drawing: where exactly does the voltage appear? Answer: it's all in the microscopic gap between each metal surface and the electrolyte. In other words, if you dip a chunk of copper into a bowl of salt solution, a constant voltage appears, with the metal charged negative and the liquid charged positive. In other words, all batteries actually contain two batteries: one on the surface of one electrode, and the other on the other electrode surface. Google "half-cell" to find info on this. Simplified intuitive explanation: when metal touches water, the polar water atoms start rapidly dissolving the metal. The water atoms surround metal atoms and yank them out of the surface. The metal starts dissolving rapidly, as rapidly as salt or sugar. But something else happens too, otherwise metals in water would vanish within minutes. When water pulls a metal atom out of the metal surface, it ends up with a positive metal ion. Metal atoms have a 'loose' outer electron which becomes part of the bulk electron-sea of the metal and does not adhere to each atom. So, as metal atoms (positive metal ions) leave the surface and move into the water, the water acquires an enormous positive charge. The metal surface is full of left-behind electrons and develops an enormous negative charge. In other words, when a blob of metal touches a blob of water, we end up with a self-charging capacitor! Now obviously this capacitor cannot charge up to infinite voltage. The "dielectric" is a thin layer of water molecules separating the positively-charged water from the negatively charged metal. A few volts across a nanometers-wide dielectric will give us an e-field of megavolts/cm. This strong field pushes backwards on the positive metal ions and slows the corrosion process. Fewer ions leave the metal, the voltage rises more slowly, and finally we have equilibrium where a few ions are randomly falling back to the metal, and an equal number are still being dissolved. So, a hunk of metal in a bowl of water doesn't dissolve in minutes, but instead develops a significant constant voltage where the metal is negative and the water is positive. (And, if you could somehow reduce this voltage, the corrosion process would run wild, and the metal would rot away as you watched. Or if you INCREASED the voltage instead, you could force any ions already in the water to plate back onto the surface, and the hunk of metal would start growing larger.) What then is a battery? If you put two hunks of copper in a bowl of water, both hunks charge up to the same negative volts (a few volts wrt the water,) so if you touched them together, nothing would happen. Ah, but different types of metal will spontaneously charge up to different voltage. Touching them together then decreases the water-metal voltage at one surface, and increases it for the other. Try copper and zinc. WOW! They get hot, and one hunk starts dissolving furiously, while the other starts growing larger. This would be very magical if noticed in centuries past, since the metal-eating reaction only happens when the two metal objects bump against each other ...or when both are touched against a 3rd piece of metal. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 5, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9451118111610413, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/34252/eigenvalues-of-ab-where-a-is-symmetric-positive-definite-and-b-is-diagonal	## Eigenvalues of A+B where A is symmetric positive definite and B is diagonal ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) If I have a symmetric positive definite matrix A and a diagonal matrix B, and I know the eigenvalues of both A and B (by iterative numerical computation in A's case and trivially for B), is there any way I can rapidly find the eigenvalues of the matrix M=A+B? (I would be surprised if it helps, but I actually have the stronger condition that A is Laplacian. Unfortunately, the entries of the matrix B are large, and so B cannot be considered a small "perturbation" to A. Finally, I only really have the extremal eigenvalues of A, though I am only hoping to find the extremal eigenvalues of A+B.) - What do you mean by "the Laplacian"? Should one understand it as the discretization of the continuous Laplacian in a box? If yes all eigenvalues and eigenvectors can be computed by hand. The answer to your question is "no" in some generic sense. There are obstructions to what you want to do in the limit of infinite matrix size. Maybe the best known one would be that for "B random" the matrix "A + B" exhibits localization, whereas "A" has extended states. So the eigenvectors differ (the eigenvalues also do, since one set obeys clock statistics and the other poisson). – Helge Aug 2 2010 at 14:42 Is B also positive? – Loick Aug 2 2010 at 14:45 "$A$ is Laplacian" --- do you mean it is the Laplacian matrix of some simple graph? – Laurent Lessard Aug 2 2010 at 17:07 Helge: Laplacian means it is the Laplacian matrix of a graph Loick: None of the elements of B are negative, but they are often zero. Laurent: Right, that's what I mean. – Fumiyo Eda Aug 3 2010 at 11:59 ## 5 Answers I doubt it. At least it shouldn't be easier than the case where you have the sum of two arbitrary positive definite matrices A',B' with known eigenvalues and eigenvectors. Then you could use an orthogonal basis of eigenvectors for B' and set $A=PA'P^{-1}$ and $B=PB'P^{-1}$. B would be diagonal and AB would have the same eigenvalues as A'B'. Couldn't one even make B=I by choosing an orthonormal basis? - 3 If you only know the eigenvalues, there are a bunch of linear inequalities constraining the answer. People usually think about the (nonobviously) equivalent problem where the matrices are Hermitian; see front.math.ucdavis.edu/9908.5012 – Allen Knutson Aug 5 2010 at 20:36 ### 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 ambiguity in your question is the word 'rapidly'. If you want to have an information on the eigenvalues of $A+B$, without any extra information besides those given in the question, then this is the problem raised By H. Weyl in 1912. The answer was conjectured in 1962 by A. Horn, and this conjecture was proved by A. Knutson and T. Tao in 1999. It is one of the works for which Tao received a Fields medal. So, the answer is that the spectrum may be any vector in a polytope in ${\mathbb R}^n$ whose definition is given recursively in terms of the size $n$ of the matrix. A nice expository paper is R. Bhatia, Linear algebra to quantum cohomology: the story of A. Horn’s inequalities. Amer. Math. Monthly, 108 (2001), pp 289–318. Historically, the interest in this question came from Quantum Mechanics. - The work that you have put into finding the eigenvalues of $A$ is not going to save you time, except that it does give you a bound (together with the maximum eigenvalue of $B$) for how large you have to make $\alpha$ to find the maximum eigenvalue of $\alpha I - A -B$ iteratively--and the row sums would give you a bound on that anyway. In any case finding the extremal eigenvalues of $A+B$ shouldn't be any harder than for $A$, since they are both the same size and equivalently sparse, but if you are repeating this many times with the same $A$ and different $B$ I don't see any shortcuts to iterating each one. - This previous MO question may be relevant. - If $A=\pmatrix{1&0\cr0&9\cr}$ and $B$ is diagonal with eigenvalues 4 and $-4$ then $A+B$ could have eigenvalues 5 and 5 or it could have eigenvalues 13 and $-3$. Doesn't really look like you can get the eigenvalues of $A+B$ from the given information. - 2 I think I have confused you - or maybe I am confused myself! I know not only the eigenvalues of the matrices A and B but the matrices themselves. So, for example, I know that (e.g.) A=diag(1,9) and B=diag(4,-4). Which imply that the eigenvalues of A+B must be 5 and 5. Does that make sense? I'm not trying to solve the more general problem of "I know I have two matrices with the following eigenvalues, now what are the eigenvalues of their sum?" I'm just trying to compute the eigenvalues of the sum A+B rapidly, because there are fast ways to get the eigenvalues of Laplacian and diagonal matrices. – Fumiyo Eda Aug 2 2010 at 13:05	{"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": 20, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9410133957862854, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-math-topics/13241-unique-geodesic.html	# Thread: 1. ## Unique Geodesic My book defines, the Riemannian Metric (below) for hyperbolic geometry. Given a set of points say in R^2 defined by the Riemannian Metric, is there always a geodesic between them? Furthermore, is it unique. (My book never mentions that and I am curios to know). C is some smooth curve joining two points in the set. Attached Thumbnails 2. There's no need to restrict this conversation to H^2. If a manifold with a metric (a.k.a Riemannian manifold) is connected, then any two of its points can be joined by a curve - this won't necessarily be a geodesic. Geodesics are defined as "curves of least length", say like a great circle on a sphere. This is not so precise, as there are great circles that do not minimize length. What actually goes, is that on a geodesic, the tangent vector can be transported parrallel to itself - and this condition is given by a system of differential equations. This explains the local nature of geodesy - Using standard ODE arguments. For given initial conditions, such as a point on the manifold and a direction on the tangent space, the system can be solved locally (i.e. there is always a geodesic between that point and the ones sufficiently close to it) and the solution is unique. Hope this clarifies things a bit. 3. Ah! But what about the sphere? There is more than one geodesic between the poles. 4. Yes, but only one along a specified direction on the tangent plane ps. Anyhow, the solution to a first order system of ODEs is not necessarily unique, if you just specify two points the solution must cross. 5. Originally Posted by Rebesques Yes, but only one along a specified direction on the tangent plane Okay, whatever you say, I do not know differencial geometry. 6. Hm, maybe some more explanations are in order The equations for a geodesic lead to a system of the form ${\bf y}'={\bf f}({ \bf y}(t),t)$,with f differentiable - thankfuly. For a unique solution, we must specify ${ \bf y}(0)={\bf p}$ (a point on the manifold) and ${\bf y}'(0)={\bf v_p}$ (a tangent vector at this point). ODE theory grants us a unique solution for t sufficiently small. This is our geodesic!	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 3, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9227933287620544, "perplexity_flag": "head"}
http://mathhelpforum.com/calculus/32839-help-needed-algebra.html	# Thread: 1. ## Help needed with algebra Hi I've been trying out some exam questions but have become stuck on one and was hoping someone out there might be able to help me. The question is: Evaluate using the method of residues $\int_{0}^{2\pi}\frac{\cos2\theta \: d\theta}{1-2acos\theta + a^2}$ So far I have that $\int_{0}^{2\pi}\frac{\cos2\theta\:d\theta}{1-2acos\theta+a^2}$ = $\frac{1}{2i}\int\frac{z^4+1}{z^3(a^2-az-az^-1+1)}$ as $cos2\theta= \frac{(z^2+z^-2)}{2}$and $d\theta=\frac{dz}{iz}$ I've tried to multiply all of this out and tidy up but the algebra is causing me problems from this point on and I was hoping you might be able to tell me where I'm going wrong. Any suggestions would be gratefully appreciated Thanks 2. What is $a$? What are the restictions on it? Is it $\|a\|<1$? 3. Originally Posted by michaela-donnelly $\int_{0}^{2\pi}\frac{\cos2\theta\:d\theta}{1-2acos\theta+a^2}$ = $\frac{1}{2i}\oint\frac{z^4+1}{z^3(a^2-az-az^{-1}+1)}dz$ The denominator factorises as $z^2(a-z)(az-1)$. The double pole at z=0 is obviously inside the contour. The other poles are at z=a and z=a^{-1}. One of these will be inside the contour and the other one outside (unless \|a\|=1, in which case you are in trouble). 4. ## Help needed with algebra thank you for your reply. I don't know if there are any restrictions on a as the question doesn't say. I presume it is that $\|a\|<1$ but I'm not sure....	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 11, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9747102856636047, "perplexity_flag": "head"}
http://wiki.chemprime.chemeddl.org/index.php/CoreChem:Kinetic_Theory_of_Gases:_The_Total_Molecular_Kinetic_Energy	CoreChem:Kinetic Theory of Gases: The Total Molecular Kinetic Energy - ChemPRIME # CoreChem:Kinetic Theory of Gases: The Total Molecular Kinetic Energy ### From ChemPRIME Equation (1) from Postulates of the Kinetic Theory can tell us a lot more than this about gases, however. If both sides are multiplied by V, we have $PV=\tfrac{\text{1}}{\text{3}}Nm(u^{\text{2}})_{\text{ave}}$ (1) The kinetic energy of an individual molecule is ½ m (u2)ave, and so the average kinetic energy (Ek)ave of a collection of molecules, all of the same mass m is (Ek)ave = (½ m u2)ave = ½ m (u2)ave The total kinetic energy Ek is just the number of molecules times this average: Ek = N × (Ek)ave = N × ½ m (u2)ave or, multiplying both sides by 3/3 (i.e., by 1) $E_{k}=\tfrac{\text{3}}{\text{3}}\text{ }\times \text{ }\tfrac{\text{1}}{\text{2}}Nm(u^{\text{2}})_{\text{ave}}=\tfrac{\text{3}}{\text{2}}\text{ }\times \text{ }\tfrac{\text{1}}{\text{3}}Nm(u^{\text{2}})_{\text{ave}}$ Substituting from Eq. $E_{k}=\tfrac{\text{3}}{\text{2}}PV$ (2) or $PV=\tfrac{\text{2}}{\text{3}}E_{k}$ (3) The product of the pressure and the volume of a gas is two-thirds the total kinetic energy of the molecules of the gas. Now we can understand why PV comes out in joules—it is indeed energy. According to postulate 4 of the kinetic theory, gas molecules have constant total kinetic energy. This is reflected on the macroscopic scale by the constancy of PV, or, in other words, by Boyle's law. The kinetic theory also gives an important insight into what the temperature of gas means on a microscopic level. We know from the [[CoreChem:The Ideal Gas Equation\|ideal gas law]] that PV = nRT. Substituting this into Eq. (3), nRT = ⅔ Ek (4) If we divide both sides of Eq. (4) by n and multiply by $\tfrac{3}{2}$, $\frac{E_{k}}{n}=\tfrac{\text{3}}{2}RT$ The term Ek/n is the total kinetic energy divided by the amount of substance, that is, the molar kinetic energy. Representing molar kinetic energy Em by we have $E_{\text{m}}=\tfrac{\text{3}}{2}RT$ (5) The molar kinetic energy of a gas is proportional to its temperature, and the proportionality constant is $\tfrac{3}{2}$ times the gas constant 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": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "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": 0.8774439096450806, "perplexity_flag": "middle"}
http://quant.stackexchange.com/questions/1594/what-are-some-simple-algorithms-for-hedging-vanilla-bonds/1596	# What are some simple algorithms for hedging vanilla bonds? My team will soon be implementing an auto hedger for our bond trading desk which will be integrated tightly with our risk application and I am interested in researching how this may work. Any advice or information or general thoughts would be appreciated, especially on algorithms used to suggest hedges with bond futures. I am guessing at the simplest level comparing the DV01 and instrument maturity to find the closest matching bond future would work but I am keen to know what other factors could be taken into account. Many thanks. - ## 1 Answer The top reference for this topic is Risk Management: Approaches for Fixed Income Markets by Golub and Tilman. The main measures you will want to calculate for hedging the yield curve risks of a bond portfolio are the key rate durations. The wikipedia article gives a brief overview. If you have access to Lehman/Barclays data, they calculate key rate durations daily for every bond in their indices. You can also calculate it yourself as $krd_i=-\frac 1P \frac {P_{i,up}-P_{i,down}} {2\Delta r_i}$ if you have a pricing model for the bonds in which you can vary a set of interest rates $r_i$. See chapter 2 of the book for details. - KR01 hedging is useful depending on your basis functions. Linear functions are easy but can be inaccurate. For KR01 hedging, this is where time should be spent. You could also try bucket hedging which hedges certain chunks of the term structure. Mr. Smith, is your team doing rate modeling or do you get the modeled structure from elsewhere? – strimp099 Aug 6 '11 at 12:03	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9287937879562378, "perplexity_flag": "middle"}
http://mathoverflow.net/revisions/53511/list	## Return to Question 2 added 70 characters in body Let $(A,\mathfrak{m}_A)$ be a local Artinian $k$-algebra with residue field $k$. Then the scheme $\mathrm{Spec}(A)$ can be loosely seen as a "fat point", or an "infinitesimal neighbourhood" of a point $\mathrm{Spec}(k)$. If the maximal ideal satisfies the condition $\mathfrak{m}^k=0$, then I think the spec of $A$ can be seen as a "$(k-1)$-th order infinitesimal neighbourhood" of the closed point. Recall: a small extension of $A$ is an extension of local Artinian $k$-algebras $0\to I \to B\to A \to 0$ such that $\mathfrak{m}_B\cdot I =0$. Does the concept of a small extension have any geometric interpretation? Does it mean that the closed embedding $\mathrm{Spec}(A)\hookrightarrow \mathrm{Spec}(B)$ is such that it only "fattens" already existent "directions" of $\mathrm{Spec}(A)$, making it into a "higher order" fattening of the closed point but without adding new "directions" of fattening? Is my suggestion totally misleading? Edit: as comments point out, my interpretation was not correct. 1 # Geometric meaning of small extensions ? Let $(A,\mathfrak{m}_A)$ be a local Artinian $k$-algebra with residue field $k$. Then the scheme $\mathrm{Spec}(A)$ can be loosely seen as a "fat point", or an "infinitesimal neighbourhood" of a point $\mathrm{Spec}(k)$. If the maximal ideal satisfies the condition $\mathfrak{m}^k=0$, then I think the spec of $A$ can be seen as a "$(k-1)$-th order infinitesimal neighbourhood" of the closed point. Recall: a small extension of $A$ is an extension of local Artinian $k$-algebras $0\to I \to B\to A \to 0$ such that $\mathfrak{m}_B\cdot I =0$. Does the concept of a small extension have any geometric interpretation? Does it mean that the closed embedding $\mathrm{Spec}(A)\hookrightarrow \mathrm{Spec}(B)$ is such that it only "fattens" already existent "directions" of $\mathrm{Spec}(A)$, making it into a "higher order" fattening of the closed point but without adding new "directions" of fattening? Is my suggestion totally misleading?	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9320105910301208, "perplexity_flag": "head"}
http://unapologetic.wordpress.com/2009/09/30/	# The Unapologetic Mathematician ## Differentiability Implies Continuity In the course of showing that the differential of a function at a point — if it exists at all — is unique (and thus we can say “the” differential), we showed that given an orthonormal basis we have all partial derivatives. We even have all directional derivatives, with pretty much the same proof. We replace $e_k$ with an arbitrary vector $u$, and pick the scalar $\tau$ so that $\frac{\delta}{\lVert u\rVert}>\lvert\tau\rvert>0$. We find that $\left[D_uf\right](x)=df(x;u)=df(x)u$. So when a function is differentiable not only do all directional derivatives exist, they’re all given by a single linear functional applied to the direction vector. Notice that this does not hold for the pathological example we used to show that having all directional derivatives didn’t imply continuity. Okay, so now does having a differential at a point imply that a function is continuous there? Remember that this was the major reason we rejected both partial and directional derivatives as insufficient as generalizations of differentiation in one variable. But, happily, it does. Firstly, we’re going to pick a basis and show that the function satisfies a Lipschitz condition (five minutes of furtive laughter in any advanced calculus class starts…. now) (it’s worse than doing quantum mechanics with bras in front of high schoolers). That is to say, there is a positive number $M$ and some neighborhood $N$ of ${0}$ so that if $t\in N$ but $t\neq0$, then $\lvert f(x+t)-f(x)\rvert<M\lVert t\rVert$. Or, in more conceptual terms, any displacement near enough to ${0}$ can only be made $M$ times bigger after running it through $f$. This gives us some control on what the function does as we move our input point around. So, first we take $\epsilon=1$ in the definition of the differential, to find $\displaystyle\lvert\left[f(x+t)-f(x)\right]-df(x;t)\rvert<\lVert t\rVert$ we can add $\lvert df(x;t)\rvert$ to both sides and use the triangle inequality to find $\displaystyle\lvert f(x+t)-f(x)\rvert\leq\lvert\left[f(x+t)-f(x)\right]-df(x;t)\rvert+\lvert df(x;t)\rvert<\lvert df(x;t)\rvert+\lVert t\rVert$ But once we pick a basis we can write out the differential as $\displaystyle\lvert df(x;t)\rvert=\left\lvert\sum\limits_{i=1}^n\left[D_{e_i}f\right](x)t^i\right\rvert\leq\sum\limits_{i=1}^n\left\lvert\left[D_{e_i}f\right](x)\right\rvert\lvert t^i\rvert\leq\left(\sum\limits_{i=1}^n\left\lvert\left[D_{e_i}f\right](x)\right\rvert\right)\lVert t\rVert$ I’ve written out the sum explicitly here because it’s necessary in the last term. So if we pick $\displaystyle M=1+\sum\limits_{i=1}^n\left\lvert\left[D_{e_i}f\right](x)\right\rvert$ then we have the Lipschitz condition we want. And then it just so happens that a Lipschitz condition implies continuity. Indeed, given an $\epsilon>0$ pick a $\delta>0$ small enough that both $\delta<\frac{\epsilon}{M}$, and also the ball of radius $\delta$ fits inside the neighborhood $N$ from the Lipschitz condition. Then for $\delta>\lVert t\rVert>0$ we find $\displaystyle\lvert f(x+t)-f(x)\rvert<M\lVert t\rVert<M\delta<M\frac{\epsilon}{M}=\epsilon$ and we have continuity. Now to really understand this, go back and walk it through with a function of one variable. See if you can find where the old proof that a having a derivative implies continuity is sitting inside this Lipschitz condition proof. Posted by John Armstrong \| Analysis, Calculus \| 5 Comments ## About this weblog This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”). I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 27, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 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": 0.9282953143119812, "perplexity_flag": "head"}
http://mathhelpforum.com/advanced-math-topics/3475-solutions.html	# Thread: 1. ## solutions Dear sir/Madam, I am praveen here, please find SCANNED Attachments.These contains some problems. i need solution for these problems. I request to send solutions. thanking you Regards, Praveen Attached Thumbnails 2. 3b) $x\Gamma(x)=\lim_{n\to\infty}\frac{n^xn!}{(1+x)(2+x )....(n+x)}$ 3c) Therefore the poles are where the denominator is zero, $(1+x)(2+x)....=0$ Thus, $x=-1,-2,-3,...$ 3. 1a)Maybe? Since this infinite series can be diffrenciated it must be countinous. Since diffrenciablility implies couninutitity. 1b) The solution(s) to $x^2+\alpha x+\beta=0$ are, $x=\frac{-\alpha\pm \sqrt{\alpha-4\beta}}{2}$ Since there are pure imaginaries, $-\alpha =0$ thus, $\alpha =0$ Then you are looking at, $x^2+\beta=0$ Thus, $x^2=-\beta$ Has pure imaginaries solutions only when, $-\beta<0$ Thus, $\beta>0$ --- Solution, $\alpha=0,\beta>0$ 4. 1a) If the space is finite, then it is certainly compact. Conversely, suppose the space X is compact. Consider the (open) covering $\cup_{x\in X} B(x,1/2)$. Since X is compact, there exists a finite subcovering $X\subset B(x_1,1/2)\cup\ldots\cup B(x_n,1/2).$ As our metric is the discreet metric, $B(x_1,1/2)=\{x_1\},\ldots, B(x_n,1/2)=\{x_n\}.$ So $X\subset \{x_1,\ldots, x_n\}\Rightarrow$ X is finite. Now for C[a,b]. We should show that there exists some numerable and dense subset of C[a,b]. Surely, this cannot be done off the top of our heads. Weierstrass' Approximation Theorem, sais that the set of polynomials P[a,b] $\subset$C[a,b] is dense. If we could create a numerable and dense subset of P[a,b], we would be done. Mumble, mumble... What about polynomials with rational coefficients? Worth a try. 1b) Remember $\|\|T\|\|={\rm sup}_{x\neq 0}\frac{\|\|Tx\|\|_Y}{\|\|x\|\|_X}$, so $\|\|T\|\|={\rm sup}_{x\neq 0}\frac{\|\|Tx\|\|_Y}{\|\|x\|\|_X}\geq \frac{\|\|Tx\|\|_Y}{\|\|x\|\|_X}, \ \forall x\in X$ so again $\|\|T\|\|\geq \frac{\|\|Tx\|\|_Y}{\|\|x\|\|_X}, \ \forall x\in X.$ 5. Bored to do any work (again), so lets try 3a. -Note that sinz=-z+z^3/3!-... so sinz/z^7=(1/z^6)(-1+...) This means z=0 is a pole of order 6. -Now e^{1/z}=1+1/z+1/(2!z^2)+... and so there are infinite terms in the series to diverge at z=0; this means z=0 is an essential singularity. -Also (1-cosz)/z=(-z^2/2!-z^3/3!-...)/z=-z/2!-z^2/3!-... so z=0 is a simple root. 6. ...And just thought of an answer to 2)a). There is a set of unit vectors $S=\{e_1,\ldots,e_n,\dots\}$ to form a basis () for X. Consider any linear mapping $T:X\rightarrow Y$ not to vanish on S (we can have this as Y is nonempty) and define $\overline{T}(e_n)=\frac{n}{\|\|T(e_n)\|\|}T(e_n), \ \forall n\in \mathbb{N}.$ This mapping is well defined as S is a basis. We then have $\|\|\overline{T}\|\|={\rm sup}_{\|\|x\|\|=1}\frac{\|\|\overline{T}(x)\|\|}{\|\|x\|\|} \geq {\rm sup}_{n} \frac{\|\|\overline{T}(e_n)\|\|}{\|\|e_n\|\|}=+\infty$ by the way $\overline{T}$ is defined. (*) a correction: we can extract such a set S from the actual basis set B -which I admit, need not be countable. Define the operator to be zero on B-S and everything works fine. 7. Lunch time... No lunch for those not paid, so let's try 2b), which actually I could have figured out much sooner if I was not wasting braincells on myspace. i) $\ell^p, \ 0<p<1$ is reflexive. Just show that for every $f\in \ell^p'$ there exists a sequence $(c_n)\in\ell^q, \ (1/q)+(1/p)=1$ with $\|\|f\|\|=\|\|(c_n)\|\|_{\ell^q}$. For this, consider a (continuous linear) functional $f$ on $\ell^p$, and let $e_{n}=(0,0,\ldots,0,1,0,\ldots),$ the unit occupying the n-th place. This sequence is a basis for $\ell^p$, as they are linearly independent and $(a_n)=\sum a_n(e_n)$. We exploit linearity to obtain $f((a_n))=\sum a_n f((e_n))$, so all functionals determine (and are completely determined by) the sequence $(f((e_n)))=<img src=$c_n)" alt="(f((e_n)))=c_n)" />. To show this belongs to $\ell^q$, take $(a_n)=(\|c_1\|^{q-1},\ldots,\|c_k\|^{q-1},0,\ldots)$ for any natural k, and using continuity $\|f((a_n))\|\leq \|\|f\|\| \cdot\|\|(a_n)\|\|_{\ell^p}$ or $\bigg\{\sum_{i=1}^k \|c_i\|^q\bigg\}^{1/q}\leq \|\|f\|\|$ and since k was arbitrary, $\|\|f\|\|\geq \|\|(c_n)\|\|_{\ell^q}$. On the other hand, Holder's inequality gives $\|f((b_n))\|=\|\sum b_n c_n\|\leq \|\|b_n\|\|_{\ell^p}\|\|c_n\|\|_{\ell^q}, \ \forall (b_n)\in \ell^p$ so also $\|\|f\|\|\leq \|\|(c_n)\|\|_{\ell^q}$, so these are equal. ii) Direct application of the Open Mapping Theorem: There is $\epsilon>0$ such that corresponding balls satisfy $B_Y(0,\epsilon)\subset T(B_X(0,1))$, so for $y\in Y, \ \|\|y\|\|_Y=1$ we have $\|\|T^{-1}(\epsilon y)\|\|_X\leq 1$ or $\|\|T^{-1}(y)\|\|_X\leq 1/\epsilon$. So $T^{-1}$ is continuous.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 49, "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": 0.8968682885169983, "perplexity_flag": "middle"}
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I'm trying to figure out a way to ... 0answers 27 views ### Topology for a metric $R^2$ + graph using mathlab For selfstudy I am taking a topology course and learning about metric spaces, the teacher talk about topology for a metric, this is: $B_{1}((0,0))$ for $p=1,2,..$ in $R_p^{1}$ and $R_p^{2}$, where ... 4answers 64 views ### Sketching the graph of a trig function How would I sketch the following function. $f(x)=\cos(x)+\sin(x)$ on $[0,2\pi]$ for my first derivative I got $f'(x)=-\sin(x)+\cos(x)=0$ but How would I find the critical points I mean I know ... 1answer 60 views ### Burger's Equation 'Shocks' Not matching the characteristics Album to view all the images that are described below. I have am using Burger's Equation with the initial condition of a Gaussian. The blue curve is the initial function before any time has passed, ... 2answers 40 views ### Limit aproaching infinity $f(x) = -\infty$ ?!? On one of my homework problems it's asking for me to sketch a graph of a function which fulfills five requirements. All of the requirements make sense to me other than the following: ... 0answers 21 views ### random graph's #edges based on probability If probability of occurrence of edges is 1/2, then will there be n c 2 * (1/2) total edges. Does this applies irrespective of where edges do occur.	{"extraction_info": {"found_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": 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": 0.9318116903305054, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/83197/how-to-build-a-covering-space?answertab=active	# how to build a covering space? I just started to learn for an exam but I am stuck in this exercise: Let Y be a topological space with $\pi_1(X)=\mathbb{Z}/_{19}$ . Is there a covering space (order 4) ? How do you construct a covering space in this case? Thank you very much! - 4 You're given information about the fundamental group of $Y$, and want to get information about the covering spaces of $Y$. You should review what you've learned about the relationship between fundamental groups and covering spaces. – Chris Eagle Nov 17 '11 at 22:43 ## 1 Answer Hint: A connected covering space of $Y$ of order $4$ would correspond to a subgroup of $\pi_1(Y)$ of index $4$. If you don't require connectedness then it is easy to find a covering space of any order. - 1 @kobloau: Do you know about Lagrange's theorem in finite group theory? – Zhen Lin Nov 20 '11 at 0:11	{"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": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9126715660095215, "perplexity_flag": "head"}
http://math.stackexchange.com/questions/35449/history-of-dot-product-and-cosine/136914	# History of dot product and cosine The fact that the dot product and the cosine of the angle between two vectors are mutually computable is easy to show (see the two sides in the two answers at Dot product in coordinates). But looking at the dot product, I would never have thought that it somehow captures something about the angle (and vice versa). How did the connection get discovered? Who were the major players? Did it just fall out of the development of matrix operations for linear algebra (or did the dot product come first) or are these only related by hindsight or what? - Very nice question! I'm eager to learn the answer. Hope, somebody can come up with an idea. – Torbjoern Apr 27 '11 at 16:10 3 – J. M. Apr 27 '11 at 16:12 6 I don't know the answer, but addressing the "I would never have thought..." part: If you look at the length of $\vec v-\vec w$ in terms of the dot product and distribute, you get $\\|\vec v-\vec w\\|^2 = \\|\vec v\\|^2 + \\|\vec w\\|^2 - 2\vec v\cdot\vec w$. This is reminiscent of the Law of Cosines. If you look at the triangle with lengths $\\|v\\|$, $\\|w\\|$, and $\\|v-w\\|$ and apply the Law of Cosines with the angle between $v$ and $w$, all the squared lengths cancel right away to give the result. To me it seems natural enough that you or I could have "discovered" it (after the fact). – Jonas Meyer Apr 27 '11 at 18:36 @Jonas: that sounds 'answerish' (actually, the easiest derivation of the dot-product/cosine connection is through the law of cosines. – Mitch Apr 27 '11 at 18:54 1 i think you need the cauchy-schwarz-bunyakowsky inequality $\|u.v\| \le \|u\|\|v\|.$ so that $u.v$ can be written as $\|u\|\|v\|\cos (\theta)$ for some $\theta.$ that this $\theta$ is the angle between the vectors $u,v$ in $R^n$ is fortunate? – abel Apr 27 '11 at 20:19 show 1 more comment ## 2 Answers I suspect that J.M. is right: historically, quaternions came before vectors. But before the quaternions you had complex numbers. For two complex numbers $z$ and $w$, the fact that $\text{Re}(\overline{z} w) = \|z\| \|w\| \cos \theta$, where $\theta$ is the angle between them measured at the origin, is clear from the polar representation. Write this out in terms of the real and imaginary parts of $z$ and $w$ and you have the two-dimensional version of your relationship. Hamilton, of course, knew all of this, and a main motivation behind his development of the algebra of quaternions was to get a way to study three-dimensional space analogous to the use of complex numbers to study two-dimensional space. If $P = a i + b j + c k$ and $Q = d i + e j + f k$ are quaternions representing what we would call "vectors", the real part of the quaternion $\overline{P} Q = (-ai-bj-ck)(di+ej+fk)$ is $ad + be + cf$, and this should be (and is) $\|P\| \|Q\| \cos \theta$. Gibbs et al dispensed with the quaternion framework and wrote this as the "dot product" of the two vectors $(a,b,c)$ and $(d,e,f)$. - The dot product as an explicit algebraic operation is not so necessary to the 'dischord of appearances', it is the $a d + b e + c f$ a decidedly open calculation in comparison to the obscurantist length of a vector and angle formula (requiring much more machinery). Any idea if this (what Hamilton knew) was popular before Gibbs? – Mitch Apr 25 '12 at 19:25 Looking at mathword under D you'll find: DOT PRODUCT is found in 1901 in Vector Analysis by J. Willard Gibbs and Edwin Bidwell Wilson: The direct product is denoted by writing the two vectors with a dot between them as A·B This is read A dot B and therefore may often be called the dot product instead of the direct product. [This citation was provided by Joanne M. Despres of Merriam-Webster Inc.] If you get a hold of a copy I'm sure they'll have a good discussion on the matter. Gibbs was the big proponent of vectors and I'm willing to bet he'd have discussed the connection between the two. That'd probably be a good starting point to tracking down an answer to your question. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 28, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.956513524055481, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/3390/can-anybody-provide-a-simple-example-of-a-quantum-computer-algorithm/3407	# Can anybody provide a simple example of a quantum computer algorithm? Does anybody give a good textbook description of a quantum computer algorithm and how its different from an ordinary algorithm? - 1 +1. Will the real @[Peter Shor] please stand up :) – user346 Jan 20 '11 at 5:52 ## 2 Answers I'll avoid talking about Shor's algorithm and leave it for you to read on your own -- Shor's algorithm is the quantum algorithm, and is the reason quantum computing has become such a hot topic, and is not only a must read, but Scott Aaronson has provided a better introduction than I could ever manage, http://www.scottaaronson.com/blog/?p=208 Instead, I will guide you through Deutsch's algorithm, which solves a fantastically useless problem exponentially faster than any classical algorithm: Imagine I have a function $f$ on a configuration of $n$ bits, $C=\{0,1\}^n$ that takes each configuration to a single bit. I promise you beforehand that this function is either: 1. Constant - all configurations are mapped to the same value. 2. Balanced - exactly half of the configurations map to $1$, the other half to $0$. So classically, in the worst case you must evaluate the function for $2^{n-1}+1$ configurations ($O(2^n)$) to verify which category $f$ falls into. Now, a quantum computer isn't just a parallel processor, where you can give it a superposition of the configurations and get back $f$ evaluated on all of them. At the end of the day, you have to make a measurement which destroys our carefully crafted superposition -- so we have to be clever! The fundamental feature of a quantum algorithm is that we use unitary operations to transform our state and use interference between states so that when we measure the state at the end, it tells us something unambiguous. So without further ado, the Deutsch algorithm for $n=2$ qubits (you can do this for many more qubits, of course, but you wanted a simple example). The "quantum version" of our function is a unitary operation (a "quantum gate") that takes me from $\|a\rangle\|b\rangle$ to $\|a\rangle\|b+f(a)\rangle$ (where the addition is modulo 2). Also at my disposal is a gate that takes any single bit and puts it in a particular superposition: $$\|1\rangle\rightarrow(\|0\rangle-\|1\rangle)/\sqrt{2}$$ $$\|0\rangle\rightarrow(\|0\rangle+\|1\rangle)/\sqrt{2}$$ The first step in the algorithm is to prepare my two qubits as $\|0\rangle\|1\rangle$ and then apply this transformation, giving me the state $(\|0\rangle+\|1\rangle)(\|0\rangle-\|1\rangle)/2$. Now I evaluate the function by applying the gate I described earlier, taking my state to $$[\|0\rangle(\|0+f(0)\rangle-\|1+f(0)\rangle)+\|1\rangle(\|0+f(1)\rangle-\|1+f(1)\rangle)]/2$$ If we stare at this carefully, and think about arithmetic mod 2, we see that if $f(0)=1$ the first term picks up a minus sign relative to its state before, and if $f(0)=0$ nothing happens. So we can rewrite the state as $$[(-1)^{f(0)}\|0\rangle(\|0\rangle-\|1\rangle)+(-1)^{f(1)}\|1\rangle(\|0\rangle-\|1\rangle)]/2$$ or regrouping $$(\|0\rangle+(-1)^{f(0)+f(1)}\|1\rangle)(\|0\rangle-\|1\rangle)/2$$ Now we shed the second qubit (we don't need it anymore, and it's not entangled with the first bit), and apply the second transformation I listed once more (and regrouping, the algebra is simple but tedious) -- at last, our final state is $$[(1+(-1)^{f(0)+f(1)})\|0\rangle+((1-(-1)^{f(0)+f(1)})\|1\rangle]/2$$ And lastly, we measure! It should be clear from the state above that we've solved the problem... If $f$ is constant, then $f(0)=f(1)$ and we will always measure $\|0\rangle$, and if $f$ is balanced then $f(0) \neq f(1)$ and we always measure $\|1\rangle$. Some final comments: hopefully this gives you some taste for the structure of a quantum algorithm, even if it seems like kind of a useless thing -- I strongly recommend going and reading the article I linked to at the beginning of this answer. - nice job.+1. Its good to see the increasing number of well-formed questions and answers on this site! – user346 Jan 20 '11 at 5:53 2 "Now, a quantum computer isn't just a parallel processor, where you can give it a superposition of the configurations and get back f evaluated on all of them." Actually, it isn't -even- a parallel processor. Quantum parallelism is not the same as having an exponentially parallel classical computer, although an exponentially parallel classical computer can simulate a quantum computer efficiently. – Joe Fitzsimons Jan 20 '11 at 10:02 Thanks Joe, that's really what I meant to point out, although my word choice was rather poor. – wsc Jan 20 '11 at 22:17 Ok, since wsc gave you an explanation of Deutsch's algorithm, let me talk you through another, and more useful, algorithm: Grover's search algorithm. This is a search algorithm for an unsorted database, or for searching a black box for the input which results in a specific output (a very useful primitive in algorithms). For a set of $N$ entries, assuming there is only entry satisfying the search query, the algorithm takes only $O(\sqrt{N})$ queries, where as the best possible classical algorithm takes $O(N)$ queries of the database. Further, Grover's algorithm (with a small modification I won't go into here) is provably optimal. So, how does it work? Let's assume there is a database of $2^n$ entries. First we prepare a superposition of $2^n$ classical states with positive phase: $\mid S \rangle = 2^{-\frac{n}{2}} \sum_{x=1}^{2^n} \| x \rangle$. This is can be done simply by applying a single quantum gate (in this case a Hadamard gate) to each of $n$ qubits initially prepared in the zero state. I'll represent the states encountered by a diagram, where the length of each line represents it's amplitude in the superposition, and whether it is direction from the center line to indicate its phase (in this case it will always be either positive or negative). The state we obtain after this initialisation step is then depicted as below. The algorithm itself consists of $r$ rounds which each consist of two steps: 1. The phase of a classical state in the superposition is flipped if (and only if) the entry in the database at that location matches the search query, as illustrated below. Here we assume entry $m$ is the entry matching the search query. Note that this is only a single query to the database, which is having different effects in each branch of the wavefunction. This is achieved by applying an operator $U_m = I - 2 \| m \rangle \langle m \|$ 1. The 'inversion about the mean' operator is applied. This is simply a unitary operator given by $U_S = 2 \| S \rangle \langle S \| - I$, where as before $\| S \rangle = 2^{-\frac{n}{2}} \sum_{x=1}^{2^n} \| x \rangle$. The effect of this operator is very simple: the amplitude $A_i$ of each entry $i$ is taken to $2\mu - A_i$, where $mu$ is the average amplitude. This is equivalent to inverting the distance above or below the mean. The first diagram below illustrates where the mean now lies, while the second shows the effect of applying this operator. Clearly the effect of this operation is to amplify the amplitude of the entry matching the search query, and this will continue until the mean lies below the zero line (after $r$ rounds). This occurs after $O(\sqrt{N})$ queries. To see this, we first note that the state of the system after each round $i$ can always be written in the form $\|\psi_i\rangle = \cos \theta_i \|s\rangle + \sin \theta_i \|m\mid$, where $\|s\rangle = (2^n - 1)^{-\frac{1}{2}} \sum_{x\neq m} \| x \rangle$. Clearly $\|s\rangle$ and $\|m\rangle$ are orthogonal, so we can represent any state encountered as a unit vector in the 2D plane spanned by $\|s\rangle$ and $\|m\rangle$. Let's consider the effect of a single round. First $U_m$ is applied and then $U_S$ is applied. Note that $U_S = 2 \|S\rangle\langle S\| - I = (\frac{2(2^n - 1)}{2^n}-1)\|s\rangle\langle s \| + \frac{2\sqrt{2^n - 1}}{2^n}\|s\rangle\langle m \| + \frac{2\sqrt{2^n - 1}}{2^n}\|m\rangle\langle s \| + (\frac{2}{2^n} - 1)\|m\rangle\langle m \|$. When we multiply this by $U_m$ to get the total operation applied in a round $U_S U_m = (\frac{2(2^n - 1)}{2^n}-1)\|s\rangle\langle s \| - \frac{2\sqrt{2^n - 1}}{2^n}\|s\rangle\langle m \| + \frac{2\sqrt{2^n - 1}}{2^n}\|m\rangle\langle s \| + (\frac{2(2^n - 1)}{2^n})\|m\rangle\langle m \|$. This is simply a rotation through a constant angle $\phi=\sin^{-1}(\frac{2\sqrt{2^n - 1}}{2^n}) \approx \frac{2}{\sqrt{2^n}}$ for large $n$. Since we initially start very close to the state $\|s\rangle$, we thus require $r \approx \frac{2}{\sqrt{2^n}}$, meaning that we achieve the maximum amplitude for $\|m\rangle$ in just ~$2\sqrt{N}$ queries of the database. It is easy to see that this is impossible classically, since if you make $m$ queries to the database, there are still $N-m$ entries which have not been checked which may possibly match the search term, meaning that a linear number of entries need to be checked to insure even a constant probability of finding the correct entry. Hope this is useful - It is really hard when one is forced to choose one good answer over another +1 – Humble Jan 20 '11 at 10:47	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 60, "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": 0.9351893663406372, "perplexity_flag": "head"}
http://crypto.stackexchange.com/questions/6637/understanding-feldmans-vss-with-a-simple-example/6650	# Understanding Feldman's VSS with a simple example I'm trying to understand Feldman's VSS Scheme. The basic idea of that scheme is that one uses Shamir secret sharing to share a secret and commitments of the coefficients of the polynomial to allow the other party to verify that the share they received is valid. I implemented it in code following the wikipedia page, but my verification function doesn't work. Here is a simple example of the failure: We will work in the field $\mathbb{Z}_{11}$ with generator $g=3$. Let $f(x)=5+3x+8x^2$ be our Shamir polynomial. Thus $f(1)=5$. So let $s_1=5$ be the share that party $p_1$ gets. The dealer also must commit to the coefficients. So the commitments are $3^5=1, 3^3=5, 3^8=5$. (These are computed as $g^c$ for each coefficient $c$ of the polynomial). For verification that $s_1$ is correct, we compute $g^{s_1}=3^5=1$ and compare this to $155=3$ (the exponents are all $1$ since $i=1$, otherwise this step is done as $k^{i^j}$ where $k$ is the commitment, $j$ is the index of the commitment). Since $1\neq3$, the verification fails. But why? It should pass assuming I have done things correctly. Update So working the math a little in more detail: $g^{s_1}=1$ and $g^5 g^{3^{1^1}} g^{8^{1^2}}= g^{5+3^{1^1}+8^{1^2}}= g^{5+3\cdot 1^1 + 8\cdot 1^2} = g^{16} = g^6 = 3$ and $1\neq 3$. Note that the exponent is $f(1)$. Wikipedia offers a small hint: (Typically, one takes a subgroup of $\mathbb{Z}_q^$, where $q$ is a prime such that $q$ divides $p-1$. Is my problem that I am not working in such a subgroup? If so, why does it not work in general for $\mathbb{Z}_p$ (Wikipedia only says that typically one works in this subgroup)? Is there a standard way to set up such a subgroup? - ## 1 Answer It has to do with which modulus you use. You did all your arithmetic modulo 11. However, when using Feldman's VSS, you gotta use two different moduli (using each one in the appropriate spot). In your example, you shouldn't do all arithmetic modulo 11. Instead, you should be doing some arithmetic modulo 11, and some arithmetic modulo 5 (the order of $g$ in this case). If you do that, everything will work out. In general, you need to pick primes $p,q$ such that $q \| p-1$. Then, some arithmetic is done modulo $p$, and some arithmetic is done modulo $q$. In particular: • The polynomial $f(x)$ is treated modulo $q$, so when you compute the shares, you need do arithmetic modulo $q$. • The commitments and all computations with the commitments are treated modulo $p$, so when you verify that you got the correct share, you work modulo $p$. The reason we do it this way is that when doing a computation like $3^{15}$ (modulo $11$), we can reduce the base modulo 11, but we have to reduce the exponent modulo 10. Roughly speaking, commitments are in the base, whereas the value of the polynomial (the shares; the coefficient of the polynomial) are in the exponent -- so you gotta use different moduli for these two different kinds of values. We can modify your example to take this into account. We could take $p=11$, $q=5$, and generator $g=3$ of the subgroup of order $q$ of $(\mathbb{Z}/p\mathbb{Z})^$. However, you can no longer have the polynomial $f(x)=5+3x+8x^2$: the polynomial is interpreted modulo $q$, so all coefficients have to be in the range $0..4$. This means we'll need to change things a little bit. So, here's a corrected example. You could use the polynomial $f(x)=0+3x+3x^2$. Since $f(1)=1$, you'll get the share $s_1=1$. The commitments are $3^0=1$, $3^3=5$, and $3^3=5$. For verification of the correctness of the share, we first compute $3^{s_1}=3^1=3$. Next, we compute the check value $1 \times 5^1 \times 5^{1^2} = 3$. You can see that $3^{s_1}$ is equal to the check value, so everything verifies, and the share is correct. - Awesome, that helped a lot. Is there a proper way to generate secure values for $p,q$? Basically we would one $p-1$ to have a large prime factor I'm assuming. But we'd also need to know that factor. Or even better, is there a semi-standard $p,q$ that will work? – mikeazo♦ Mar 11 at 16:42 1 All you need is that $p,q$ are both prime and that the discrete logarithm in the size-$q$ subgroup of $(\mathbb{Z}/p\mathbb{Z})^$ is hard. This is the same as the requirement for, e.g., ElGamal or DSA. So, use any standard algorithm for generating such $p,q$, or any such $p,q$. Or, randomly pick a random 160-bit value $q$, test if it is prime, pick a random 1888-bit value $k$, and test if $p=kq+1$ is prime; repeat until both $p,q$ are prime (there are ways to optimize this so it runs faster, but this will work). – D.W. Mar 11 at 20:59 Choosing $p,q$ in this manner ensures that the Legendre symbol of the secret $s$ is not leaked by the commitment $g^s$, correct? Specifically, the article on Wikipedia says the description of Feldman's VSS as written there is not secure as $g^s$ leaks information about $s$ (which I'm assuming is the Legendre symbol). Does choosing $p,q$ as you specify fix this problem? – mikeazo♦ 2 days ago 1 @mikeazo, that's correct. With those parameters, the Legendre symbol doesn't leak anything about $s$: the Legendre symbol $(g^s\|p)$ will always be $1$, regardless of $s$, so no leakage. Actually, with those parameters, I don't see any way that $g^s$ leaks anything about $s$, so I don't know what the Wikipedia comment is referring to. (Maybe it's what happens if you don't choose $p,q$ that way? Maybe it's the fact that you can verify a guess at $s$ using $g^s$? Or something else entirely? The statement is unsourced, so I can't tell.) – D.W. 2 days ago I think wikipedia is referring to the fact that if $g$ is a generator of $\mathbb{Z}_p^$, then $(g^s \mid p)$ leaks the least significant bit of $s$ (HAC 3.9.1). Wikipedia only suggests choosing $g$ with prime order $q$ where $q\mid p-1$ in a parenthetical statement as kind of an afterthought. I'm assuming that is why they say $g^s$ leaks information. – mikeazo♦ 2 days ago	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 91, "mathjax_display_tex": 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": 0.9267535209655762, "perplexity_flag": "head"}
http://mathhelpforum.com/differential-geometry/126823-providing-conditions.html	# Thread: 1. ## Providing Conditions I need to provide a condition on \|x-1\| that will assure that \|x^2 - 1\| < 1/n. Any ideas? 2. Originally Posted by frenchguy87 I need to provide a condition on \|x-1\| that will assure that \|x^2 - 1\| < 1/n. Any ideas? $\left\|x^2-1\right\|=\left\|x-1\right\|\|x+1\|$...so... come on. 3. Originally Posted by frenchguy87 I need to provide a condition on \|x-1\| that will assure that \|x^2 - 1\| < 1/n. Any ideas? $\|x^2- 1\|= \|x-1\|\|x+1\|$ to make that less than $\frac{1}{n}$, you need to have $\|x- 1\|< \frac{1}{n\|x+1\|}$. But we need a constant, not a function of x on the right. That is, we need $\|x- 1\|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\frac{1}{x+1}$. If \|x-1\|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as \|x-1\|< 1, \|x+1\|> 2 so [tex]\frac{1}{\|x+1\|}< \frac{1}{2}. Then $\|x-1\|< \frac{1}{n\|x+1\|}< \frac{1}{2n}$. In order that both of those be true, you must have \|x-1\|< the smaller of 1 and $\frac{1}{2n}$. Assuming that n is a positive integer, of course, $\frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\|x-1\|< \frac{1}{2n}$. 4. Thanks a lot! Very helpful 5. You actually kind of lose me here, why do you say Originally Posted by HallsofIvy If \|x-1\|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as \|x-1\|< 1, \|x+1\|> 2 so [tex]\frac{1}{\|x+1\|}< \frac{1}{2}. 6. Originally Posted by HallsofIvy $\|x^2- 1\|= \|x-1\|\|x+1\|$ to make that less than $\frac{1}{n}$, you need to have $\|x- 1\|< \frac{1}{n\|x+1\|}$. But we need a constant, not a function of x on the right. That is, we need $\|x- 1\|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\frac{1}{x+1}$. If \|x-1\|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as \|x-1\|< 1, \|x+1\|> 2 so [tex]\frac{1}{\|x+1\|}< \frac{1}{2}. Then $\|x-1\|< \frac{1}{n\|x+1\|}< \frac{1}{2n}$. In order that both of those be true, you must have \|x-1\|< the smaller of 1 and $\frac{1}{2n}$. Assuming that n is a positive integer, of course, $\frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\|x-1\|< \frac{1}{2n}$. Shouldn't we choose \|x-1\|= min{ 1, $\frac{1}{3n}$ ??? ,by adding 2 to all sides of -1< x-1<1 give you 1<x+1< 3 7. Originally Posted by frenchguy87 I need to provide a condition on \|x-1\| that will assure that \|x^2 - 1\| < 1/n. Any ideas? . $\|x-1\|<\frac{1}{3n}$=> \|x-1\|<1 => \|x\|-1<1 => \|x\|+1<3 => \|x+1\|< 3 => $\|x-1\|\|x+1\|\leq 3\frac{1}{3n}=\frac{1}{n}$ 8. Originally Posted by xalk . $\|x-1\|<\frac{1}{3n}$=> \|x-1\|<1 => \|x\|-1<1 => \|x\|+1<3 => \|x+1\|< 3 => $\|x-1\|\|x+1\|\leq 3\frac{1}{3n}=\frac{1}{n}$ Why do you guys all say $\|x-1\|<1$, I can't figure out where that comes from... 9. Originally Posted by CrazyCat87 Why do you guys all say $\|x-1\|<1$, I can't figure out where that comes from... The point is to get some kind of bound on x so we can bound x+1. Since we want \|x-1\| small, \|x-1\|< 1 is handy. You could use any positive number other than 1 on the right, but 1 is easiest. 10. Originally Posted by HallsofIvy $\|x^2- 1\|= \|x-1\|\|x+1\|$ to make that less than $\frac{1}{n}$, you need to have $\|x- 1\|< \frac{1}{n\|x+1\|}$. But we need a constant, not a function of x on the right. That is, we need $\|x- 1\|< \frac{1}{n(x+1}< \delta$ which means we need an lower bound on $\frac{1}{x+1}$. If \|x-1\|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). As long as \|x-1\|< 1, \|x+1\|> 2 so [tex]\frac{1}{\|x+1\|}< \frac{1}{2}. Then $\|x-1\|< \frac{1}{n\|x+1\|}< \frac{1}{2n}$. In order that both of those be true, you must have \|x-1\|< the smaller of 1 and $\frac{1}{2n}$. Assuming that n is a positive integer, of course, $\frac{1}{2n}$ is less than 1 and so is the smaller. The condition you want is $\|x-1\|< \frac{1}{2n}$. I can almost follow this except for this part, If \|x-1\|< 1 so that -1< x-1< 1, then 2< x+1< 3 (add 2 to each part). Shouldn't we have 1 < x+1 < 3, which would then give \|x+1\|> 1 so $\frac{1}{\|x+1\|}< 1$. Then $\|x-1\|< \frac{1}{n\|x+1\|}< \frac{1}{n}$ ? 11. This has gone on too long. Do you agree that $\|x-1\|<1$ implies that $\|x+1\|<3$? If so, then let $\|x-1\|<\frac{1}{3n}<1$. That gives $\|x^2-1\|={\color{red}\|x-1\|}{\color{blue}\|x+1\|}<{\color{red}\left(\frac{1}{ 3n}\right)}{\color{blue}(3)}=\frac{1}{n}$. Now you have your condition: $\frac{1}{3n}$.	{"extraction_info": {"found_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": 42, "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": 0.9437566995620728, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/146148/probability-of-random-shuffling-of-cards	# Probability of random shuffling of cards I have a pack of cards and use the following method to shuffle them • Pick a random card from the deck and replace the first card with it • Put the first card back in the deck • Move to the second card and repeat the process till the final card So I select any one of 51 cards to replace the first card, 50 for the second, 49 for the third and so on till the 51st card. Cards once selected and replaced aren't to be put back in the deck Given the above method • Does all cards have equal probability of ending up in any position? • If the answer is yes or no, how to calculate and prove it? - When you put the first card back in the deck, where in the deck do you put it? Do you mean it switches places with the random card chosen to replace it? And what do you mean, "cards once selected and replaced aren't to be put back in the deck"? If a card is replaced and not put back in the deck, then the deck has fewer than 52 cards, which is not what's usually meant by shuffling. – Gerry Myerson May 17 '12 at 6:50 @GerryMyerson Cards once replaced aren't put back. Its a biased shuffle (shuffle may be the wrong term). So for the first card I pick from the next 51 cards, for the second card, I leave the first card and select from the remaining 50 cards and so on. So there would be no selection for the last card since it is automatically select. And just 2 options for the 51 st card. – Ubermensch May 17 '12 at 9:16 I'm still not with you. You write that you replace the first card, and that you put the first card back in the deck, but now you write that cards once replaced aren't put back. So, is the first card put back in the deck, or isn't it? If it isn't, you only have 51 cards in the deck (and as you continue, you'll have fewer and fewer), and the chance of the first card ending up in any position is zero, because, once replaced, it wasn't put back. So what is going on here? – Gerry Myerson May 17 '12 at 11:31 @GerryMyerson For the first card (7C), I pick any card from the remaining 51 cards (6S). Now I am going to put the original card (7C) back in the deck. And for the next card (6D), I am going to select from the remaining 50 cards excluding 6S and 6D and so on. So what is the probability that 6S,6D,7C or any other card would end up in the list and is it same for all the cards in the pack? – Ubermensch May 17 '12 at 11:49 OK, so where in the deck do you put that 7C when you put it back in the deck? Might you put it on top? Do you always put it on the bottom? Where? – Gerry Myerson May 17 '12 at 12:53 show 4 more comments ## 2 Answers Hint: Can the first choice be the first card, which then stays put? Given your answer, what is the probability that a given card ends up first? What is the probability that a given card ends up second? - @RossMilikan The first card can be selected in 1/51 ways, the second in 1/50 ways and so on. But I can't think beyond it as to what's the chance of each card being anywhere in the shuffled deck – Ubermensch May 17 '12 at 4:46 @Ubermensch: If the first card can't be the one that starts out on top, the shuffle isn't totally random and you are done. Looking at the starting pack there are many arrangements that are forbidden (probability 0). To have a random shuffle each card has to have 1/52 chance of being in each place and beyond this there cannot be any correlation between which cards end up in each position. More interesting is the case where the first selection includes the top card. – Ross Millikan May 17 '12 at 5:07 As Ross Millikan points out, if you must swap the first card with some card strictly below it, then the card that starts out on top has probability zero of winding up on top, so the answer to your (first) question is, no. As Ross also suggests, things are more interesting when you allow the first card (and each subsequent card) to stay put; make the top card swap with the $k$th card with probability $1/52$ for each $k$, $1\le k\le52$, then make the second card swap with the $k$th card with probability $1/51$ for each $k$, $2\le k\le52$, etc. etc. Then indeed all cards have equal probability of ending up in any given position; in fact, more is true, as all 52-factorial possible permutations of the deck are equally likely. This is most easily seen by proving that it's true not just for a 52-card deck, but for an $n$-card deck, for any positive integer $n$, using induction on $n$. The base case is easy. Then assume it's true for $n$-card decks, and assume you have an $n+1$-card deck. Make the first swap; any card could now be on top, all with equal probability, and the card that's now on top never gets moved after this, so you are now dealing with just the remaining $n$ cards. Now apply the induction hypothesis! - Thanks. The second case(the interesting one) is the one I am working on. Got the intuition. Would work towards the proof. – Ubermensch May 18 '12 at 7:34	{"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": 14, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 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": 0.9564672708511353, "perplexity_flag": "middle"}
http://mathoverflow.net/questions/75888?sort=oldest	## Line bundles on Ind Schemes ### Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points) I've learned when you have a integral smooth scheme line bundles are the same as Cartier divisors are the same as Weil divisors. My question is to what extent does this continue to hold (if at all) when you are talking about ind objects. Since things become more subtle in positive characteristic lets say we are working over an algebraically closed field of characteristic 0. Background I think the situation can be quite different because smoothness, or algebraic smoothness is more complicated. In the non ind scheme case you can define smoothness of a point $p \in X$ as $S^q(m/m^2) \to m^q/m^{q+1}$ being an isomorphism for all $q\ge 0$, where $m$ is the maximal ideal of the local ring at $p$. In the ind case you have varieties $(X_n)_{n \ge 0}$ and $p \in X_n$ for all sufficiently large $n$ so you have $S_n^q(m_n/m_n^2) \to m^q_n/m^{q+1}_n$ for all sufficiently large $n$. Algebraic smoothness at $p$ means that the map $\varprojlim S_n^q(m_n/m_n^2) \to \varprojlim m^q_n/m^{q+1}_n$ is an isomorphism for all $q\ge 0$ Thoughts It is true that if $X$ is the union of $X_n$ and each $X_n$ is smooth then $X$ is algebraically smooth. However algebraic smoothness is not inductive in the sense that there are examples of $p \in X$ being algebraically smooth even when $p \in X_n$ is a singular point for all $X_n$ containing $p$. However line bundles are inductive in the sense that a line bundles on $X$ determines line bundles on each $X_n$ compatible under pull back and vice versa. This suggests to me that you could have a collection of compatible Weil divisors in each $X_n$ which are not the zero section of any section of any line bundle. But I don't have an example. - ## 1 Answer I don't know the general story, but I think the correspondence between Weil divisors and line bundles breaks down already for $\mathbf{A}^\infty$, which is pretty much the smoothest ind-variety out there. Using the standard ind-structure on $\mathbf{A}^\infty$, the example I have in mind is the union of all hyperplanes $x_n-x_1=0$. This is a codimension one closed ind-subscheme, but it's not the zero locus of any polynomial in $\mathcal{O}_{\mathbf{A}^\infty}$. - 1 Variant: fix a sequence $(a_n)_{n\in\mathbb{N}}$ in $k^\mathbb{N}$; for each $n$, put $X_n=\mathbb{A}^1=\mathrm{Spec}\,k[x]$ (with $\mathrm{Id}$ as transition map), and let $D_n$ be the Cartier divisor on $X_n$ defined by $\prod_{0\leq j\leq n} (x-a_n)$. – Laurent Moret-Bailly Sep 20 2011 at 7:37 That's a nice example of compatible inclusions that don't define a divisor on $X$ (I think) -- the union $D$ of all the $D_n$'s is not even a closed subset of $X$, since $D \cap X_n$ is not closed. – Dave Anderson Sep 20 2011 at 17:15 @Dave: right, but the same holds for your example: both your "ind-subscheme" (whatever this means) and mine are Zariski-dense. In fact, my example can be obtained from yours by intersecting with a suitable line. – Laurent Moret-Bailly Sep 21 2011 at 8:16 @Laurent, I may be making a silly mistake, but I think the intersection of my divisor with $X_n$ is the hyperplane arrangement with $x_i=x_1$ for $i=1,...,n$, plus $x_1=0$. (I'm using the standard ind-structure, where $X_n$ has $x_i=0$ for all $i>n$.) I agree that both are Zariski-dense, but I'm using the ind-topology on $\mathbf{A}^\infty$. (A subset is closed iff its intersection with each $X_n$ is.) – Dave Anderson Sep 21 2011 at 14:34 OK, then what I don't understand is what we should mean by "ind-subscheme" or "ind-divisor". – Laurent Moret-Bailly Sep 21 2011 at 15:43	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 50, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 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": 0.9408252239227295, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/8144/at-what-scale-does-uv-ir-mixing-happen-in-string-theory	# At what scale does UV/IR mixing happen in string theory? Imagine, if you will, a string background with an extremely tiny value for the string coupling. The Planck scale is many orders of magnitude smaller in size than the string scale. Is the UV/IR mixing scale the Planck scale, or the string scale? What happens at intermediate scales? Gravitons are string modes, and as such, their spatial extent is the string scale. They can never resolve anything smaller than that. On the other hand, D-branes are more sensitive. - ## 1 Answer In perturbative string theory, for $g\ll 1$, it's always the string scale where the UV/IR mixing happens first. It's the lightest mass scale associated with the lightest (possibly extended) objects - and in perturbative string theory regimes, it's always the fundamental strings. For example, the low-energy spectrum at masses below $M_{\rm string}$ is always related to the high-energy spectrum at masses above $M_{\rm string}$ by the modular invariance (of the toroidal world sheet) - and similarly for open strings and cylindrical world sheets. Different effects may see different forms of UV/IR mixing for which different scales are relevant, however. The term UV/IR mixing is a somewhat vague, umbrella term that covers many effects in string theory or quantum gravity. D-branes may resolve shorter distances than the string scales but there is actually always some sense in which even e.g. D0-branes are linked to the string scale. If they resolve distances $\Delta x$ and times $\Delta t$, then $\Delta x \cdot \Delta t > L_{\rm string}^2$. Yoneya would play with insights like this. D0-branes may resolve sub-stringy distances but only if their velocity is very small, in which case they can only measure the time with much-worse-than-stringy resolution. There are also - less understood - aspects of the UV/IR mixing that arise at the Planck scale. But some of them are known so incompletely that the arguments are only valid up to the assumption that $g$ is considered to be of order one. The relevant scale where geometry - with degrees of freedom at each "layer" of scales being independent of each other - fails to be applicable is always given by the scale associated with the first higher-order corrections to the Einstein-Hilbert action etc. and it's always the string scale i.e. the scale associated with the tension of the lightest fundamental objects. - 2 Can you elaborate upon the less understood aspects of UV/IR mixing at the Planck scale? Thanks. – Shankar Apr 6 '11 at 11:22 If UV/IR mixing happens at the string scale, what happens if we have S-duality between two different string theories? e.g. type IIB, or type I with heterotic SO(32)? There are now two different string scales. Surely UV/IR mixings can't happen at both scales simultaneously? The only scale invariant under S-duality is the Planck scale. – Shankar Apr 7 '11 at 14:12	{"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": 7, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9297624826431274, "perplexity_flag": "middle"}
http://psychology.wikia.com/wiki/Bayesian_statistics	# Bayesian inference Talk0 31,725pages on this wiki ## Redirected from Bayesian statistics Assessment \| Biopsychology \| Comparative \| Cognitive \| Developmental \| Language \| Individual differences \| Personality \| Philosophy \| Social \| Methods \| Statistics \| Clinical \| Educational \| Industrial \| Professional items \| World psychology \| Statistics: Scientific method · Research methods · Experimental design · Undergraduate statistics courses · Statistical tests · Game theory · Decision theory Bayesian inference is a statistical inference in which evidence or observations are used to update or to newly infer the probability that a hypothesis may be true. The name "Bayesian" comes from the frequent use of Bayes' theorem in the inference process. Bayes' theorem was derived from the work of the Reverend Thomas Bayes.[1] ## Evidence and changing beliefs Bayesian inference uses aspects of the scientific method, which involves collecting evidence that is meant to be consistent or inconsistent with a given hypothesis. As evidence accumulates, the degree of belief in a hypothesis ought to change. With enough evidence, it should become very high or very low. Thus, proponents of Bayesian inference say that it can be used to discriminate between conflicting hypotheses: hypotheses with very high support should be accepted as true and those with very low support should be rejected as false. However, detractors say that this inference method may be biased due to initial beliefs that one holds before any evidence is ever collected. (This is a form of inductive bias). Bayesian inference uses a numerical estimate of the degree of belief in a hypothesis before evidence has been observed and calculates a numerical estimate of the degree of belief in the hypothesis after evidence has been observed. (This process is repeated when additional evidence is obtained.) Bayesian inference usually relies on degrees of belief, or subjective probabilities, in the induction process and does not necessarily claim to provide an objective method of induction. Nonetheless, some Bayesian statisticians believe probabilities can have an objective value and therefore Bayesian inference can provide an objective method of induction. See scientific method. Bayes' theorem adjusts probabilities given new evidence in the following way: $P(H\|E) = \frac{P(E\|H)\;P(H)}{P(E)}$ where • $H$ represents a specific hypothesis, which may or may not be some null hypothesis. • $P(H)$ is called the prior probability of $H$ that was inferred before new evidence, $E$, became available. • $P(E\|H)$ is called the conditional probability of seeing the evidence $E$ if the hypothesis $H$ happens to be true. It is also called a likelihood function when it is considered as a function of $H$ for fixed $E$. • $P(E)$ is called the marginal probability of $E$: the a priori probability of witnessing the new evidence $E$ under all possible hypotheses. It can be calculated as the sum of the product of all probabilities of any complete set of mutually exclusive hypotheses and corresponding conditional probabilities: $P(E) = \sum P(E\|H_i)P(H_i)$. • $P(H\|E)$ is called the posterior probability of $H$ given $E$. The factor $P(E\|H) / P(E)$ represents the impact that the evidence has on the belief in the hypothesis. If it is likely that the evidence $E$ would be observed when the hypothesis under consideration is true, but unlikely that $E$ would have been the outcome of the observation, then this factor will be large. Multiplying the prior probability of the hypothesis by this factor would result in a larger posterior probability of the hypothesis given the evidence. Conversely, if it is unlikely that the evidence $E$ would be observed if the hypothesis under consideration is true, but a priori likely that $E$ would be observed, then the factor would reduce the posterior probability for $H$. Under Bayesian inference, Bayes' theorem therefore measures how much new evidence should alter a belief in a hypothesis. Bayesian statisticians argue that even when people have very different prior subjective probabilities, new evidence from repeated observations will tend to bring their posterior subjective probabilities closer together. However, others argue that when people hold widely different prior subjective probabilities their posterior subjective probabilities may never converge even with repeated collection of evidence. These critics argue that worldviews which are completely different initially can remain completely different over time despite a large accumulation of evidence.[How to reference and link to summary or text] Multiplying the prior probability $P(H)$ by the factor $P(E\|H) / P(E)$ will never yield a probability that is greater than 1, since $P(E)$ is at least as great as $P(E \cap H)$ (where $\cap$ denotes "and"), which equals $P(E\|H)\,P(H)$ (see joint probability). The probability of $E$ given $H$, $P(E\|H)$, can be represented as a function of its second argument with its first argument held fixed. Such a function is called a likelihood function; it is a function of $H$ alone, with $E$ treated as a parameter. A ratio of two likelihood functions is called a likelihood ratio, $\Lambda$. For example, $\Lambda_E = \frac{L(H\|E)}{L(\mathrm{not}\,H\|E)} = \frac{P(E\|H)}{P(E\|\mathrm{not}\,H)}$, where the dependence of $\Lambda_E$ on $H$ is suppressed for simplicity (as $E$ might have been, except we will need to use that parameter below). Since $H$ and not-$H$ are mutually exclusive and span all possibilities, the sum previously given for the marginal probability reduces to $P(E) = P(E\|H)\,P(H)+P(E\|\mathrm{not}\,H)\,P(\mathrm{not}\,H)$ As a result, we can rewrite Bayes' theorem as $P(H\|E) = \frac{P(E\|H)\,P(H)}{P(E\|H)\,P(H)+ P(E\|\mathrm{not}\,H)\,P(\mathrm{not}\,H)} = \frac{\Lambda_E P(H)}{\Lambda_E P(H) +P(\mathrm{not}\,H)}$. We could then exploit the identity $P(\mathrm{not}\,H) = 1 - P(H)$ to exhibit $P(H\|E)$ as a function of just $P(H)$ (and $\Lambda_E$, which is computed directly from the evidence). With two pieces of evidence $E_1$ and $E_2$, that are marginally and conditionally independent of each other given the hypotheses, Bayesian inference can be applied iteratively. We could use the first piece of evidence to calculate an initial posterior probability, and then use that posterior probability as a new prior probability to calculate a second posterior probability given the second piece of evidence. Bayes' theorem applied iteratively yields $P(H\|E_1 \cap E_2) = \frac{P(E_2\|H)\;P(E_1\|H)\,P(H)}{P(E_2)\;P(E_1)}$ Using likelihood ratios, we find that $P(H\|E_1 \cap E_2) = \frac{\Lambda_1 \Lambda_2 P(H)}{[\Lambda_1 P(H) + P(\mathrm{not}\,H)]\;[\Lambda_2 P(H) + P(\mathrm{not}\,H)]}$, This iteration of Bayesian inference could be extended with more independent pieces of evidence. Bayesian inference is used to calculate probabilities for decision making under uncertainty. Besides the probabilities, a loss function should be evaluated to take into account the relative impact of the alternatives. ## Simple examples of Bayesian inference ### From which bowl is the cookie? To illustrate, suppose there are two full bowls of cookies. Bowl #1 has 10 chocolate chip and 30 plain cookies, while bowl #2 has 20 of each. Our friend Fred picks a bowl at random, and then picks a cookie at random. We may assume there is no reason to believe Fred treats one bowl differently from another, likewise for the cookies. The cookie turns out to be a plain one. How probable is it that Fred picked it out of bowl #1? Intuitively, it seems clear that the answer should be more than a half, since there are more plain cookies in bowl #1. The precise answer is given by Bayes' theorem. Let $H_1$ correspond to bowl #1, and $H_2$ to bowl #2. It is given that the bowls are identical from Fred's point of view, thus $P(H_1)=P(H_2)$, and the two must add up to 1, so both are equal to 0.5. The event $E$ is the observation of a plain cookie. From the contents of the bowls, we know that $P(E\|H_1) = 30/40 = 0.75$ and $P(E\|H_2) = 20/40 = 0.5$. Bayes' formula then yields $\begin{matrix} P(H_1\|E) &=& \frac{P(E\|H_1)\,P(H_1)}{P(E\|H_1)\,P(H_1)\;+\;P(E\|H_2)\,P(H_2)} \\ \\ \ & =& \frac{0.75 \times 0.5}{0.75 \times 0.5 + 0.5 \times 0.5} \\ \\ \ & =& 0.6 \end{matrix}$ Before we observed the cookie, the probability we assigned for Fred having chosen bowl #1 was the prior probability, $P(H_1)$, which was 0.5. After observing the cookie, we must revise the probability to $P(H_1\|E)$, which is 0.6. ### False positives in a medical test False positives result when a test falsely or incorrectly reports a positive result. For example, a medical test for a disease may return a positive result indicating that patient has a disease even if the patient does not have the disease. We can use Bayes' theorem to determine the probability that a positive result is in fact a false positive. We find that if a disease is rare, then the majority of positive results may be false positives, even if the test is accurate. Suppose that a test for a disease generates the following results: • If a tested patient has the disease, the test returns a positive result 99% of the time, or with probability 0.99 • If a tested patient does not have the disease, the test returns a positive result 5% of the time, or with probability 0.05. Naively, one might think that only 5% of positive test results are false, but that is quite wrong, as we shall see. Suppose that only 0.1% of the population has that disease, so that a randomly selected patient has a 0.001 prior probability of having the disease. We can use Bayes' theorem to calculate the probability that a positive test result is a false positive. Let A represent the condition in which the patient has the disease, and B represent the evidence of a positive test result. Then, probability that the patient actually has the disease given the positive test result is $\begin{matrix} P(A \| B) &=& \frac{P(B \| A) P(A)}{P(B \| A)P(A) + P(B \|\mathrm{not}\,A)P(\mathrm{not}\,A)} \\ \\ &= &\frac{0.99\times 0.001}{0.99 \times 0.001 + 0.05\times 0.999} \\ ~\\ &\approx &0.019 .\end{matrix}$ and hence the probability that a positive result is a false positive is about $1-0.019 = 0.98$, or 98%. Despite the apparent high accuracy of the test, the incidence of the disease is so low that the vast majority of patients who test positive do not have the disease. Nonetheless, the fraction of patients who test positive who do have the disease (.019) is 19 times the fraction of people who have not yet taken the test who have the disease (.001). Thus the test is not useless, and re-testing may improve the reliability of the result. In order to reduce the problem of false positives, a test should be very accurate in reporting a negative result when the patient does not have the disease. If the test reported a negative result in patients without the disease with probability 0.999, then $P(A\|B) = \frac{0.99\times 0.001}{0.99 \times 0.001 + 0.001\times 0.999} \approx 0.5$, so that $1-0.5 = 0.5$ now is the probability of a false positive. On the other hand, false negatives result when a test falsely or incorrectly reports a negative result. For example, a medical test for a disease may return a negative result indicating that patient does not have a disease even though the patient actually has the disease. We can also use Bayes' theorem to calculate the probability of a false negative. In the first example above, $\begin{matrix} P(A \|\mathrm{not}\,B) &=& \frac{P(\mathrm{not}\,B \| A) P(A)}{P(\mathrm{not}\,B \| A)P(A) + P(\mathrm{not}\,B \|\mathrm{not}\,A)P(\mathrm{not}\,A)} \\ \\ &= &\frac{0.01\times 0.001}{0.01 \times 0.001 + 0.95\times 0.999}\, ,\\ ~\\ &\approx &0.0000105\, .\end{matrix}$ The probability that a negative result is a false negative is about 0.0000105 or 0.00105%. When a disease is rare, false negatives will not be a major problem with the test. But if 60% of the population had the disease, then the probability of a false negative would be greater. With the above test, the probability of a false negative would be $\begin{matrix} P(A \|\mathrm{not}\,B) &=& \frac{P(\mathrm{not}\,B \| A) P(A)}{P(\mathrm{not}\,B \| A)P(A) + P(\mathrm{not}\,B \|\mathrm{not}\,A)P(\mathrm{not}\,A)} \\ \\ &= &\frac{0.01\times 0.6}{0.01 \times 0.6 + 0.95\times 0.4}\, ,\\ ~\\ &\approx &0.0155\, .\end{matrix}$ The probability that a negative result is a false negative rises to 0.0155 or 1.55%. ### In the courtroom Bayesian inference can be used in a court setting by an individual juror to coherently accumulate the evidence for and against the guilt of the defendant, and to see whether, in totality, it meets their personal threshold for 'beyond a reasonable doubt'. • Let $G$ denote the event that the defendant is guilty. • Let $E$ denote the event that the defendant's DNA matches DNA found at the crime scene. • Let $P(E\|G)$ denote the probability of seeing event $E$ if the defendant actually is guilty. (Usually this would be taken to be near unity.) • Let $P(G\|E)$ denote the probability that the defendant is guilty assuming the DNA match (event $E$). • Let $P(G)$ denote the juror's personal estimate of the probability that the defendant is guilty, based on the evidence other than the DNA match. This could be based on his responses under questioning, or previously presented evidence. Bayesian inference tells us that if we can assign a probability p(G) to the defendant's guilt before we take the DNA evidence into account, then we can revise this probability to the conditional probability $P(G \| E)$, since $P(G \| E) = \frac{P(G) P(E \| G)}{P(E)}.$ Suppose, on the basis of other evidence, a juror decides that there is a 30% chance that the defendant is guilty. Suppose also that the forensic testimony was that the probability that a person chosen at random would have DNA that matched that at the crime scene is 1 in a million, or 10−6. The event E can occur in two ways. Either the defendant is guilty (with prior probability 0.3) and thus his DNA is present with probability 1, or he is innocent (with prior probability 0.7) and he is unlucky enough to be one of the 1 in a million matching people. Thus the juror could coherently revise his opinion to take into account the DNA evidence as follows: $P(G \| E) = (0.3 \times 1.0) /(0.3 \times 1.0 + 0.7 \times 10^{-6}) = 0.99999766667.$ The benefit of adopting a Bayesian approach is that it gives the juror a formal mechanism for combining the evidence presented. The approach can be applied successively to all the pieces of evidence presented in court, with the posterior from one stage becoming the prior for the next. The juror would still have to have a prior estimate for the guilt probability before the first piece of evidence is considered. It has been suggested that this could reasonably be the guilt probability of a random person taken from the qualifying population. Thus, for a crime known to have been committed by an adult male living in a town containing 50,000 adult males, the appropriate initial prior probability might be 1/50,000. For the purpose of explaining Bayes' theorem to jurors, it will usually be appropriate to give it in the form of betting odds rather than probabilities, as these are more widely understood. In this form Bayes' theorem states that Posterior odds = prior odds x Bayes factor In the example above, the juror who has a prior probability of 0.3 for the defendant being guilty would now express that in the form of odds of 3:7 in favour of the defendant being guilty, the Bayes factor is one million, and the resulting posterior odds are 3 million to 7 or about 429,000 to one in favour of guilt. A logarithmic approach which replaces multiplication with addition and reduces the range of the numbers involved might be easier for a jury to handle. This approach, developed by Alan Turing during World War II and later promoted by I. J. Good and E. T. Jaynes among others, amounts to the use of information entropy. In the United Kingdom, Bayes' theorem was explained to the jury in the odds form by a statistician expert witness in the rape case of Regina versus Denis John Adams. A conviction was secured but the case went to Appeal, as no means of accumulating evidence had been provided for those jurors who did not want to use Bayes' theorem. The Court of Appeal upheld the conviction, but also gave their opinion that "To introduce Bayes' Theorem, or any similar method, into a criminal trial plunges the Jury into inappropriate and unnecessary realms of theory and complexity, deflecting them from their proper task." No further appeal was allowed and the issue of Bayesian assessment of forensic DNA data remains controversial. Gardner-Medwin argues that the criterion on which a verdict in a criminal trial should be based is not the probability of guilt, but rather the probability of the evidence, given that the defendant is innocent (akin to a frequentist p-value). He argues that if the posterior probability of guilt is to be computed by Bayes' theorem, the prior probability of guilt must be known. This will depend on the incidence of the crime, which is an unusual piece of evidence to consider in a criminal trial. Consider the following three propositions: A: The known facts and testimony could have arisen if the defendant is guilty, B: The known facts and testimony could have arisen if the defendant is innocent, C: The defendant is guilty. Gardner-Medwin argues that the jury should believe both A and not-B in order to convict. A and not-B implies the truth of C, but the reverse is not true. It is possible that B and C are both true, but in this case he argues that a jury should acquit, even though they know that they will be letting some guilty people go free. See also Lindley's paradox. Other court cases in which probabilistic arguments played some role were the Howland will forgery trial, the Sally Clark case, and the Lucia de Berk case. ### Search theory Main article: Bayesian search theory In May 1968 the US nuclear submarine Scorpion (SSN-589) failed to arrive as expected at her home port of Norfolk, Virginia. The US Navy was convinced that the vessel had been lost off the Eastern seaboard but an extensive search failed to discover the wreck. The US Navy's deep water expert, John Craven USN, believed that it was elsewhere and he organised a search south west of the Azores based on a controversial approximate triangulation by hydrophones. He was allocated only a single ship, the Mizar, and he took advice from a firm of consultant mathematicians in order to maximise his resources. A Bayesian search methodology was adopted. Experienced submarine commanders were interviewed to construct hypotheses about what could have caused the loss of the Scorpion. The sea area was divided up into grid squares and a probability assigned to each square, under each of the hypotheses, to give a number of probability grids, one for each hypothesis. These were then added together to produce an overall probability grid. The probability attached to each square was then the probability that the wreck was in that square. A second grid was constructed with probabilities that represented the probability of successfully finding the wreck if that square were to be searched and the wreck were to be actually there. This was a known function of water depth. The result of combining this grid with the previous grid is a grid which gives the probability of finding the wreck in each grid square of the sea if it were to be searched. This sea grid was systematically searched in a manner which started with the high probability regions first and worked down to the low probability regions last. Each time a grid square was searched and found to be empty its probability was reassessed using Bayes' theorem. This then forced the probabilities of all the other grid squares to be reassessed (upwards), also by Bayes' theorem. The use of this approach was a major computational challenge for the time but it was eventually successful and the Scorpion was found about 740 kilometers southwest of the Azores in October of that year. Suppose a grid square has a probability p of containing the wreck and that the probability of successfully detecting the wreck if it is there is q. If the square is searched and no wreck is found, then, by Bayes' theorem, the revised probability of the wreck being in the square is given by $p' = \frac{p(1-q)}{(1-p)+p(1-q)}.$ ## More mathematical examples ### Naive Bayes classifier See naive Bayes classifier. ### Posterior distribution of the binomial parameter In this example we consider the computation of the posterior distribution for the binomial parameter. This is the same problem considered by Bayes in Proposition 9 of his essay. We are given m observed successes and n observed failures in a binomial experiment. The experiment may be tossing a coin, drawing a ball from an urn, or asking someone their opinion, among many other possibilities. What we know about the parameter (let's call it a) is stated as the prior distribution, p(a). For a given value of a, the probability of m successes in m+n trials is $p(m,n\|a) = \begin{pmatrix} n+m \\ m \end{pmatrix} a^m (1-a)^n.$ Since m and n are fixed, and a is unknown, this is a likelihood function for a. From the continuous form of the law of total probability we have $p(a\|m,n) = \frac{p(m,n\|a)\,p(a)}{\int_0^1 p(m,n\|a)\,p(a)\,da} = \frac{\begin{pmatrix} n+m \\ m \end{pmatrix} a^m (1-a)^n\,p(a)} {\int_0^1 \begin{pmatrix} n+m \\ m \end{pmatrix} a^m (1-a)^n\,p(a)\,da}.$ For some special choices of the prior distribution p(a), the integral can be solved and the posterior takes a convenient form. In particular, if p(a) is a beta distribution with parameters m0 and n0, then the posterior is also a beta distribution with parameters m+m0 and n+n0. A conjugate prior is a prior distribution, such as the beta distribution in the above example, which has the property that the posterior is the same type of distribution. What is "Bayesian" about Proposition 9 is that Bayes presented it as a probability for the parameter a. That is, not only can one compute probabilities for experimental outcomes, but also for the parameter which governs them, and the same algebra is used to make inferences of either kind. Interestingly, Bayes actually states his question in a way that might make the idea of assigning a probability distribution to a parameter palatable to a frequentist. He supposes that a billiard ball is thrown at random onto a billiard table, and that the probabilities p and q are the probabilities that subsequent billiard balls will fall above or below the first ball. By making the binomial parameter a depend on a random event, he cleverly escapes a philosophical quagmire that was an issue he most likely was not even aware of. ### Computer applications Bayesian inference has applications in artificial intelligence and expert systems. Bayesian inference techniques have been a fundamental part of computerized pattern recognition techniques since the late 1950s. There is also an ever growing connection between Bayesian methods and simulation-based Monte Carlo techniques since complex models cannot be processed in closed form by a Bayesian analysis, while the graphical model structure inherent to statistical models, may allow for efficient simulation algorithms like the Gibbs sampling and other Metropolis-Hastings algorithm schemes. Recently Bayesian inference has gained popularity amongst the phylogenetics community for these reasons; applications such as BEAST, MrBayes and P4 allow many demographic and evolutionary parameters to be estimated simultaneously. As applied to statistical classification, Bayesian inference has been used in recent years to develop algorithms for identifying unsolicited bulk e-mail spam. Applications which make use of Bayesian inference for spam filtering include DSPAM, Bogofilter, SpamAssassin, SpamBayes, and Mozilla. Spam classification is treated in more detail in the article on the naive Bayes classifier. In some applications fuzzy logic is an alternative to Bayesian inference. Fuzzy logic and Bayesian inference, however, are mathematically and semantically not compatible: You cannot, in general, understand the degree of truth in fuzzy logic as probability and vice versa.[How to reference and link to summary or text] ## References 1. ↑ Douglas Hubbard "How to Measure Anything: Finding the Value of Intangibles in Business" pg. 46, John Wiley & Sons, 2007 • On-line textbook: Information Theory, Inference, and Learning Algorithms, by David MacKay, has chapters on Bayesian methods, including examples; arguments in favour of Bayesian methods (in the style of Edwin Jaynes); modern Monte Carlo methods, message-passing methods, and variational methods; and examples illustrating the connections between Bayesian inference and data compression. • Berger, J.O. (1999) Statistical Decision Theory and Bayesian Analysis. Second Edition. Springer Verlag, New York. ISBN 0-387-96098-8 and also ISBN 3-540-96098-8. • Bolstad, William M. (2004) Introduction to Bayesian Statistics, John Wiley ISBN 0-471-27020-2 • Bretthorst, G. Larry, 1988, Bayesian Spectrum Analysis and Parameter Estimation in Lecture Notes in Statistics, 48, Springer-Verlag, New York, New York • Carlin, B.P. and Louis, T.A. (2008) Bayesian Methods for Data Analysis, Third Edition. Chapman & Hall/CRC, Boca Raton, Florida. [1] ISBN 1-58488-697-8. • Dawid, A.P. and Mortera, J. (1996) Coherent analysis of forensic identification evidence. Journal of the Royal Statistical Society, Series B, 58,425-443. • Foreman, L.A; Smith, A.F.M. and Evett, I.W. (1997). Bayesian analysis of deoxyribonucleic acid profiling data in forensic identification applications (with discussion). Journal of the Royal Statistical Society, Series A, 160, 429-469. • Gardner-Medwin, A. What probability should the jury address?. Significance. Volume 2, Issue 1, March 2005 • Gelman, A., Carlin, J., Stern, H., and Rubin, D.B. (2003). Bayesian Data Analysis. Second Edition. Chapman & Hall/CRC, Boca Raton, Florida. [2] ISBN 1-58488-388-X. • Gelman, A. and Meng, X.L. (2004). Applied Bayesian Modeling and Causal Inference from Incomplete-Data Perspectives: an essential journey with Donald Rubin's statistical family. John Wiley & Sons, Chichester, UK. ISBN 0-470-09043-X • Giffin, A. and Caticha, A. (2007) Updating Probabilities with Data and Moments • Jaynes, E.T. (1998) Probability Theory: The Logic of Science. • Lee, Peter M. Bayesian Statistics: An Introduction. Second Edition. (1997). ISBN 0-340-67785-6. • Loredo, Thomas J. (1992) "Promise of Bayesian Inference in Astrophysics" in Statistical Challenges in Modern Astronomy, ed. Feigelson & Babu. • O'Hagan, A. and Forster, J. (2003) Kendall's Advanced Theory of Statistics, Volume 2B: Bayesian Inference. Arnold, New York. ISBN 0-340-52922-9. • Pearl, J. (1988) Probabilistic Reasoning in Intelligent Systems, San Mateo, CA: Morgan Kaufmann. • Robert, C.P. (2001) The Bayesian Choice. Springer Verlag, New York. • Robertson, B. and Vignaux, G.A. (1995) Interpreting Evidence: Evaluating Forensic Science in the Courtroom. John Wiley and Sons. Chichester. • Winkler, Robert L, Introduction to Bayesian Inference and Decision, 2nd Edition (2003) Probabilistic. ISBN 0-9647938-4-9 ## See also Template:Statistics portal # Photos Add a Photo 6,465photos on this wiki • by Dr9855 2013-05-14T02:10:22Z • by PARANOiA 12 2013-05-11T19:25:04Z Posted in more... • by Addyrocker 2013-04-04T18:59:14Z • by Psymba 2013-03-24T20:27:47Z Posted in Mike Abrams • by Omaspiter 2013-03-14T09:55:55Z • by Omaspiter 2013-03-14T09:28:22Z • by Bigkellyna 2013-03-14T04:00:48Z Posted in User talk:Bigkellyna • by Preggo 2013-02-15T05:10:37Z • by Preggo 2013-02-15T05:10:17Z • by Preggo 2013-02-15T05:09:48Z • by Preggo 2013-02-15T05:09:35Z • See all photos See all photos >	{"extraction_info": {"found_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": 79, "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": 0.9329801201820374, "perplexity_flag": "middle"}
http://physics.stackexchange.com/questions/tagged/newtonian-gravity+potential-energy	# Tagged Questions 4answers 96 views ### Potential energy sign conventions Almost every book on physics that I read have some weird and non-clear explanations regarding the potential energy. Ok, I do understand that if we integrate a force over some path, we'll get a ... 2answers 58 views ### Why is potential energy negative when orbiting in a gravitational field? I had to do a problem, and part of it was to find the mechanical energy of satellite orbiting around mars, and I had all of the information I needed. I thought the total mechanical energy would be the ... 2answers 49 views ### Escape velocity to intersection of two gravitational fields Find the minimal velocity needed for a meteorite of mass $m$ to get to earth from the moon. Hint: the distance between the center of earth and the center of moon is $\approx 60 R_E$, and the ... 1answer 76 views ### Gravitational potential energy of mass between two planets Suppose I want to launch a rocket from earth to some point $O$ between the center of earth and the center of moon (on a straight line connecting their centers), where the gravitational force of the ... 2answers 80 views ### Gravitational potential outside Lagrangian points or Lagrange points The diagram in Why are L4 and L5 lagrangian points stable? shows that the gravitational potential decreases outside the ring of Lagrange points — this image shows it even more clearly: If I ... 4answers 417 views ### what is the 2D gravity potential? In 3D, I can calculate the total force due to gravity acting on a point on the surface of the unit sphere of constant density, where I choose units so that all physical constants (as well as the ... 3answers 540 views ### Still trying to understand gravitational potential and Poisson's equation? A week or so back I asked a question about the gravitational potential field $$\phi=\frac{-Gm}{r}, \qquad r\neq 0,$$ and how to show the Laplacian of $\phi$ equals zero for $r\neq 0$? Eventually, ... 2answers 160 views ### Why no basis vector in Newtonian gravitational vector field? In my textbook, the gravitational field is given by$$\mathbf{g}\left(\mathbf{r}\right)=-G\frac{M}{\left\|\mathbf{r}\right\|^{2}}e_{r}$$ which is a vector field. On the same page, it is also given as a ... 1answer 406 views ### Trying to understand Laplace's equation I'm struggling here so please excuse if I'm writing nonsense. I understand that the gravitational potential field, a scalar field, is given by $$\phi=\frac{-Gm}{r}$$ where $\phi$ is 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": 6, "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": 0.9319077134132385, "perplexity_flag": "head"}
http://mathhelpforum.com/algebra/68991-geometric-progression.html	# Thread: 1. ## Geometric Progression A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive. Prove that r^5 = 1/2 (sqrt3 - 1). I manage to get up to 2(r^5 - 1) = r^15 - r^5. 2. Originally Posted by azuki A geometric series has first term 1 and the common ratio r is positive. The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive. Prove that r^5 = 1/2 (sqrt3 - 1). I manage to get up to 2(r^5 - 1) = r^15 - r^5. Sum of a geometric series: $S_n=\frac{a(r^n-1)}{r-1}$ $S_5=\frac{a(r^5-1)}{r-1}$ $S_{15}=\frac{a(r^{15}-1)}{r-1}$ The sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive implies: $S_5 = 2S_{15} - 2S_5$ $3S_5 = 2S_{15}$ $\frac{3a(r^5-1)}{r-1}=\frac{2a(r^{15}-1)}{r-1}$ $3a(r^5-1)=2a(r^{15}-1)$ $3(r^5-1)=2(r^{15}-1)$ $3(r^5-1)=2[(r^5)^3-1]$ (note: the expression in the brackets is now a difference of two CUBES, so we can use: $a^3-b^3=(a-b)(a^2+ab+b^2)$ So $(r^5)^3-1=(r^5-1)(r^{10}+r^5+1)$ $3(r^5-1)=2(r^5-1)(r^{10}+r^5+1)$ $3=2(r^{10}+r^5+1)$ $\frac{3}{2}=r^{10}+r^5+1$ $0=r^{10}+r^5-\frac{1}{2}$ $0=2r^{10}+2r^5-1$ $0=2(r^5)^2+2r^5-1$ Let $X=r^5$ $0=2X^2+2X-1$ Using the quadratic formula ( $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$), $X=\frac{-2\pm\sqrt{4-4\times2\times-1}}{4}$ $X=\frac{-2\pm\sqrt{12}}{4}$ $X=\frac{-2\pm2\sqrt{3}}{4}$ $X=\frac{-1\pm\sqrt{3}}{2}$ Putting $X=r^5$ back in: $r^5=\frac{-1\pm\sqrt{3}}{2}$ But the question said that $r$ is a positive number, so $r^5$ must ALSO be positive, therefore $r^5=\frac{-1+\sqrt{3}}{2}$ Proving that $r^5=\frac{1}{2}(\sqrt{3}-1)$ 3. Why sum of the first 5 terms is twice the sum of the terms from the 6th to the 15th inclusive implies: S_5 = 2S_15 – 2S_5	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 30, "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": 0.9409189820289612, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/285585/find-the-general-solution-of-the-pde	# Find the general solution of the PDE Find the general solution of the PDE $xu_x-xyu_y-y=0$ for all $u(x,y)$ and find the parametric form of the solution of the PDE which follows the side condition $u(s^2,s)=s^3$ I got part (a) of the solution. The general solution is $u(x,y)=-xf(ye^x)$ I have the solution that the parametric form of the PDE is $x(s,t)=s^2e^t$ but i am not sure on how to solve it. - ## 1 Answer You found wrongly about the general solution. $xu_x-xyu_y-y=0$ $xu_x-xyu_y=y$ $u_x-yu_y=\dfrac{y}{x}$ $\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$ $\dfrac{dy}{dt}=-y$ , letting $y(0)=y_0$ , we have $y=y_0e^{-t}=y_0e^{-x}$ $\dfrac{du}{dt}=\dfrac{y}{x}=\dfrac{y_0e^{-t}}{t}$ , we have $u(x,y)=y_0\int^t\dfrac{e^{-t}}{t}dt+f(y_0)=ye^x\int^x\dfrac{e^{-t}}{t}dt+f(ye^x)$ $u(s^2,s)=s^3$ : $se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt+f(se^{s^2})=s^3$ $f(se^{s^2})=s^3-se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt$ $\therefore u(x,y)=ye^x\int^x\dfrac{e^{-t}}{t}dt+f(ye^x)$ , where $f(s)$ satisfy $f(se^{s^2})=s^3-se^{s^2}\int^{s^2}\dfrac{e^{-t}}{t}dt$ -	{"extraction_info": {"found_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": 0.9199928045272827, "perplexity_flag": "head"}
http://www.physicsforums.com/showthread.php?p=3785999	Physics Forums ## Complex Variables Limit Problem(s) 1. The problem statement, all variables and given/known data a) $$\lim_{z\to 3i}\frac{z^2 + 9}{z - 3i}$$ b) $$\lim_{z\to i}\frac{z^2 + i}{z^4 - 1}$$ 2. Relevant equations ???? 3. The attempt at a solution I'm assuming both of these are very, very similar, but I'm not quite sure how to solve them. I would like a method other than using ε and $\delta$. If you simply plug in the limit, it's obviously indeterminate. Is there an easy method to solve these limits or is the only option to use ε and $\delta$? I'm not sure how to start, any suggestions would be helpful. Thanks. 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 Try factoring the numerator and/or denominators. It's quite simple from there. Wow, can't believe I didn't realize that. It helped me solve a), which I ended up getting to be 6i, but b) cannot be factored (I don't think?). If it were $$z^4 + 1$$ in the denominator then I could, but I'm pretty sure I cannot factor anything in that problem? ## Complex Variables Limit Problem(s) Quote by wtmore Wow, can't believe I didn't realize that. It helped me solve a), which I ended up getting to be 6i, but b) cannot be factored (I don't think?). If it were $$z^4 + 1$$ in the denominator then I could, but I'm pretty sure I cannot factor anything in that problem? The denominator is a difference of squares. Then one of the factors has the same type of factorization as (a), namely the trick that a sum of squares can be factored with the use of an imaginary number. Quote by SteveL27 The denominator is a difference of squares. Then one of the factors has the same type of factorization as (a), namely the trick that a sum of squares can be factored with the use of an imaginary number. So I have: $$\frac{z^2+i}{z^4-1}=\frac{z^2+i}{(z^2-1)(z^2+1)}=\frac{z^2+i}{(z-1)(z+1)(z-i)(z+i)}$$ Am I missing something in the numerator? EDIT: Would multiplying by the numerators conjugate be beneficial? Quote by wtmore So I have: $$\frac{z^2+i}{z^4-1}=\frac{z^2+i}{(z^2-1)(z^2+1)}=\frac{z^2+i}{(z-1)(z+1)(z-i)(z+i)}$$ Am I missing something in the numerator? EDIT: Would multiplying by the numerators conjugate be beneficial? Hmmm ... that didn't help. I'm stuck too now. Recognitions: Homework Help Science Advisor Quote by SteveL27 Hmmm ... that didn't help. I'm stuck too now. The second one isn't indeterminant. Quote by Dick The second one isn't indeterminant. Would the answer be $\pm∞$? Quote by wtmore Would the answer be $\pm∞$? Just $∞$. There's only one point at infinity in the extended complex numbers. If you think of the plane, you go to infinity when you go toward the edge of the plane in any direction. There's only one complex infinity, way out there beyond the edge of the plane. A really nice visualization is to add a single point at infinity, and identify it with the "circumference" of the plane ... take the entire plane and fold it into a sphere, with the point at infinity at the north pole. It's called the Riemann sphere. http://en.wikipedia.org/wiki/Riemann_sphere Recognitions: Homework Help Science Advisor Quote by wtmore Would the answer be $\pm∞$? Not in the complex numbers. It's a pole. Saying "does not exist" is probably safe. Thread Tools \| \| \| \| \|---------------------------------------------------------\|----------------------------\|---------\| \| Similar Threads for: Complex Variables Limit Problem(s) \| \| \| \| Thread \| Forum \| Replies \| \| \| Calculus & Beyond Homework \| 3 \| \| \| Linear & Abstract Algebra \| 0 \| \| \| Calculus & Beyond Homework \| 7 \| \| \| Calculus & Beyond Homework \| 2 \| \| \| Calculus & Beyond Homework \| 2 \|	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 6, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "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": 0.9413327574729919, "perplexity_flag": "middle"}
http://nrich.maths.org/7964/index?nomenu=1	## 'Polynomial Interpolation' printed from http://nrich.maths.org/ ### Show menu Steve wishes to draw a quadratic polynomial $y(x) = ax^2+bx+c$ through the three points $(x, y)=(1,4), (2, 5), (3, 7)$. He writes down this expression: $$y(x) = 4\frac{(x-2)(x-3)}{(1-2)(1-3)}+5\frac{(x-1)(x-3)}{(2-1)(2-3)}+7\frac{(x-1)(x-2)}{(3-1)(3-2)}$$ Does this solve the problem that Steve was trying to address? Using Steve's example as a guide, can you construct a quadratic polynomial which passes through the three points $(1,2), (2, 4), (4, -1)$? Is this the only such quadratic polynomial? Can you construct a cubic polynomial which passes through the four points $(1,2), (2, 4), (3, 7),(4, -1)$. Is this the only such cubic polynomial? By this stage, the particular examples you have constructed should give you ideas about how to construct a 'general' case. Use your insights to answer these questions: Can you write down an expression for a line passing through the two points $(2, 7)$ and $(8,-6)$ using this method? Can you always fit a quadratic polynomial through three points $(x_1, y_1), (x_2, y_2), (x_3, y_3)$? Can you always fit a quartic polynomial through five points $(x_i, y_i)$ ($i=1\dots 5$) where exactly two of $x_i$ are zero? How many different polynomials can you construct which would pass through the points $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)$? Explore sets of points through which it is not possible to fit a polynomial. Extension: For various numbers of points and degrees of polynomial you might wish to consider when the fitting is unique, when it is possible with multiple polynomials and when it is impossible.	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 11, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": 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": 0.9322267174720764, "perplexity_flag": "head"}
http://mathematica.stackexchange.com/questions/16198/matrix-plot-little-exercise	# Matrix Plot – Little Exercise Hi there, mathematicians. I'm not very good at coding plots in Mathematica, so I was hoping that one of you could help me solve a problem I'm having. I have the following matrix plot: ````Z = {Subscript[x, 0], Subscript[x, 1], Subscript[x, 2]} X = {{0, 8, 12}, {.1, 0, 0}, {0, .2, 0}} ```` Then I put it in a function as follows: ````P[x_] := X^x.Z ```` I would like Mathematica to display plots such that, whenever x iterates from 1 to 10 in the function, the x1, x2, and x3 from the matrix are respectively plotted in three plots. Can anyone help me? EDIT: Trying to be more clear I have the matrix (mat) and the vector (v): ````mat = {{0, 8, 12}, {.1, 0, 0}, {0, .2, 0}} v = {x0, x1, x2} ```` Then later I define the function for population development, `pD`, taking `t` (for time in days) as its only parameter. ````pD[t] = mat^t.v ```` Now I would like to show the development of the variables $x_1$,$x_2$, and $x_3$ assuming the values 30, 60, 30, respectively. I would do so by making plots of the three functions over the range 1 to 10 days. For each the first plot should display the values of $\{t,pD[t_{x_1}]\}$, the second the values of $\{t,pD[t_{x_2}]\}$, and the third the values of $\{t,pD[t_{x_3}]\}$. I hope this clarifies my question. Sorry for being so unconventional in the first place. However, having worked my eyes blind for quite some time, I presumed that it was anything but difficult to understand. Thanks for your patience! Best regards, Brinck10 - please do not start your variables and symbols and function names with UpperCase letter. start everything with lowerCase. UpperCase first letter is meant to be used by Mathematica only. – Nasser Dec 12 '12 at 18:51 1 You do not have numerical values for your `z` vector. So how can one plot anything here? i.e. your `{Subscript[x, 0], Subscript[x, 1], Subscript[x, 2]}` is just symbolic. Can you please explain more the problem using normal math? i.e. what are the values of $x_0$ , $x_1$ and $x_2$? thanks – Nasser Dec 12 '12 at 19:00 Let us assume that `x_0`, `x_1`, and `x_2` are respectively 30, 60, 30. Then I would like to plot three graphs, one graph representing each `x`, for each iteration of x in `P[x]`. Thanks in advance. – Frederik Brinck Jensen Dec 12 '12 at 19:27 You are really overloading $x$ in so many places so I am confused. You have matrix which you called $X$ and then you have $x_i$ and using $x$ inside `P`. But aside from this. $X^n . Z$ gives one vector. So as $n$ changes from $1$ to $10$, one vector is generated. So you want to plot just a line for each $n$? btw, I wish you name your variables to mean more what they are. i.e. rename $X$ to be `mat` and $Z$ to become `vec` and rename your function `P[x_]:= X^x.Z` to be `p[n_]:=(mat^n).vec` so things become more clear. Using good names for things makes the code more understandable. – Nasser Dec 12 '12 at 19:42 ## 2 Answers Assuming you want mat^t to mean the matrix power, one way to plot the three elements of the output is to make a table of the values and then ListPlot them: ```` pD[t_] := MatrixPower[mat, t].{30, 60, 30}; ListLinePlot[Transpose[Table[pD[t], {t, 1, 10, 1}]]] ```` - I am posting this, just to have something to use to try to converge to an answer. Since I am not sure I still understand the question. This just plots a vector in 3D. i.e a line from $(0,0,0)$ to the another point in 3D space. The other point is the result of doing $mat^n.vec$. where `mat` matrix and `vec` is vector. And the question asked to plot this for each $n$ I have a feeling I am missing something here. But at least now we have something to change to try to find out what is actually needed :) ````Manipulate[ Module[{vec, mat, pt}, vec = {30, 60, 30}; mat = {{0, 8, 12}, {.1, 0, 0}, {0, .2, 0}}; pt = p[n, mat, vec]; Graphics3D[ {Thick, Line[{{0, 0, 0}, p[n, mat, vec]}]}, Axes -> True ] ], {{n, 2, "n="}, 1, 10, 1}, Initialization :> ( p[n_, mat_, vec_] := mat^n.vec ) ] ```` - @Nasser-m-abbasi see my edit. Thanks for your trouble :) – Frederik Brinck Jensen Dec 12 '12 at 20:28 lang-mma	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 23, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 20, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9143878817558289, "perplexity_flag": "middle"}
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I am okay with understanding the math, but i don't quite know how to convert that stuff to code. basically i need the integral 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": 119, "mathjax_display_tex": 11, "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": 0.892035722732544, "perplexity_flag": "head"}
http://physics.stackexchange.com/questions/39542/why-would-klein-gordon-describe-spin-0-scalar-field-while-dirac-describe-spin-1/39601	Why would Klein-Gordon describe spin-0 scalar field while Dirac describe spin-1/2? The derivation of both Klein-Gordon equation and Dirac equation is due the need of quantum mechanics (or to say more correctly, quantum field theory) to adhere to special relativity. However, excpet that Klein-Gordon has negative probability issue, I do not see difference between these two. What makes Klein-Gordon describe scalar field while Dirac describe spin-1/2 field? Edit: oops. Klein-Gordon does not have non-locality issue. Sorry for writing wrongly. Edit: Can anyone tell me in detail why $\psi$ field is scalar in Klein-Gordon while $\psi$ in Dirac is spin-1/2? I mean, if solution to Dirac is solution to Klein-Gordon, how does this make sense? - 4 Answers Let's review how the KG equation is recovered from the Dirac: $$(i \gamma \delta - m)\Psi = 0$$ $$(-i \gamma \delta - m)(i \gamma \delta - m) = 0$$ $$(\gamma^\nu \gamma^\mu \delta_\nu \delta_\mu + m^2) \Psi = 0$$ $$(\delta^2+m^2)\Psi = 0$$ In order for us to recover KG, we had to assume $\gamma^\nu \gamma^\mu = \eta^{\mu\nu}$. In other words, you can think of gamma's as doing the dot product of delta's. To take a lousy example, it's like we had an equation describing a velocity "spinor" and then squared it, so now it describes the speed "scalar" which has one less degree of freedom. This doesn't explain much more than how an equation describing a spinor can reduce into an equation describing a scalar. The reason why the Dirac equation requires spinors and not scalars is because of special relativity. If it weren't for the pesky minus sign in $\eta^{\mu\nu}$, the algebra of the gamma's would be much simpler and we wouldn't need them to be 4x4 matrices. Then $\Psi$ could describe a scalar field. - Spin is a property of the representation of the rotation group $SO(3)$ that describes how a field transforms under a rotation. This can be worked out for each kind of field or field equation. The Klein-Gordon field gives a spin 0 representation, while the Dirac equation gives two spin 1/2 representations (which merge to a single representation if one also accounts for discrete symmetries). The components of every free field satistfy the Klein-Gordon equation, irrespective of their spin. In particular, every component of the Dirac equations solves the Klein-Gordon equation. Indeed, the Klein-Gordon equation only expresses the mass shell constraint and nothing else. Spin comes in when one looks at what happens to the components. A rotation (and more generally a Lorentz transformation) mixes the components of the Dirac field (or any other field not composed of spin 0 fields only), while on a $k$-component spin 0 field, it will transform each component separately. In general, a Lorentz transformation given as a $4\times 4$ matrix $\Lambda$ changes a $k$-component field $F(x)$ into $F_\Lambda(\Lambda x)$, where $F_\Lambda=D(\Lambda)F$ with a $k\times k$ matrix $D(\Lambda)$ that depends on the representation. The components are spin 0 fields if and only if $D(\Lambda)$ is always the identity. - If we say: "A field has a spin 0, spin 1/2 or spin 1 representation" then we in fact say something about how the field parameters transform if we go from one reference frame to another. spin 0: The values of the field do not change if we go from one reference frame to another spin 1: We have to apply the Lorentz transform matrix $\Lambda$ on the field parameters. spin 1/2: We have to apply $\Lambda^{1/2}$ on the field parameters. Remark: The use of an expression like $\Lambda^{1/2}$ should be interpreted in a somewhat symbolic way because vectors and bispinors are different objects. There is an extra factor 1/2 though in the exponent of the $\Lambda^{1/2}$ matrix. The spin (associated with rotation) gets in here because the transformation matrix $\Lambda$ handles both boosts as well as rotations. The peculiar factor 1/2 however arises also in the 1 dimensional version of the Dirac equation where there is no such thing as spin (or rotation) and the corresponding 1 space + 1 time dimension version of $\Lambda$ only describes boosts. The deeper reason for the factor 1/2 is that the Dirac equation relates two field components $\psi_R$ and $\psi_L$ which are equal to each other in the rest frame. In the 1 dimensional case these are the right-moving and left-moving components. The ratio of the two transforms as follows $(\psi_R:\psi_L)\longrightarrow\Lambda~(\psi_R:\psi_L)$ In the normalization of the plane wave eigen functions this then ends up like $\psi_R\longrightarrow\Lambda^{+1/2}\psi_R$ $\psi_L\longrightarrow\Lambda^{-1/2}\psi_L$ If we now go back to 3 spatial dimensions then $\Lambda$ includes both boosts and rotations and the factor 1/2 as an exponent on the rotation generation matrices leads two what we call spin 1/2 particles. Hans. - The spin 1/2 description is wrong. One has to apply $\Lambda$ and not its square root, but as it is applied to a spinor rather than a vector, the action of $\Lambda$ is different. – Arnold Neumaier Oct 11 '12 at 10:47 Each and every text book uses the expression $\Lambda^{1/2}$ as applied to the spinor components. Spinors are for the same reason also normalized to the square of the mass $\sqrt{m}$ – Hans de Vries Oct 11 '12 at 11:06 You are confusing observables such as the current $\bar{\psi}\gamma^\mu\psi$ which transforms like a vector with the field $\psi$ itself. – Hans de Vries Oct 11 '12 at 11:10 Rather than looking at the specific form of $\Lambda$, Look at the general scaling of a general boost: Vectors: $\cosh(\vartheta)+\Gamma\sinh(\vartheta)$ versus Spinors: $\cosh(\vartheta/2)+\Gamma\sinh(\vartheta/2)$ – Hans de Vries Oct 11 '12 at 11:35 I was not talking about currents. - ''Each and every text book uses the expression $Λ^{1/2}$''?? Then please show me where it is used in this form in Weinberg's Volume 1. – Arnold Neumaier Oct 11 '12 at 14:16 show 5 more comments Spin is part of what a field IS. The data for two fields of different spins are very different. The KG equation doesn't even make sense for a spin 1/2 field and likewise for the Dirac equation and spin 0 fields. -	{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 38, "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": 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": 0.9182396531105042, "perplexity_flag": "head"}
http://terrytao.wordpress.com/tag/topological-vector-spaces/	What’s new Updates on my research and expository papers, discussion of open problems, and other maths-related topics. By Terence Tao # Tag Archive You are currently browsing the tag archive for the ‘topological vector spaces’ tag. ## Locally compact topological vector spaces 24 May, 2011 in expository, math.FA, math.GN \| Tags: Hilbert's fifth problem, local compactness, topological vector spaces \| by Terence Tao \| 11 comments Recall that a (real) topological vector space is a real vector space ${V = (V, 0, +, \cdot)}$ equipped with a topology ${{\mathcal F}}$ that makes the vector space operations ${+: V \times V \rightarrow V}$ and ${\cdot: {\bf R} \times V \rightarrow V}$ continuous. One often restricts attention to Hausdorff topological vector spaces; in practice, this is not a severe restriction because it turns out that any topological vector space can be made Hausdorff by quotienting out the closure ${\overline{\{0\}}}$ of the origin ${\{0\}}$. One can also discuss complex topological vector spaces, and the theory is not significantly different; but for sake of exposition we shall restrict attention here to the real case. An obvious example of a topological vector space is a finite-dimensional vector space such as ${{\bf R}^n}$ with the usual topology. Of course, there are plenty of infinite-dimensional topological vector spaces also, such as infinite-dimensional normed vector spaces (with the strong, weak, or weak-* topologies) or Frechet spaces. One way to distinguish the finite and infinite dimensional topological vector spaces is via local compactness. Recall that a topological space is locally compact if every point in that space has a compact neighbourhood. From the Heine-Borel theorem, all finite-dimensional vector spaces (with the usual topology) are locally compact. In infinite dimensions, one can trivially make a vector space locally compact by giving it a trivial topology, but once one restricts to the Hausdorff case, it seems impossible to make a space locally compact. For instance, in an infinite-dimensional normed vector space ${V}$ with the strong topology, an iteration of the Riesz lemma shows that the closed unit ball ${B}$ in that space contains an infinite sequence with no convergent subsequence, which (by the Heine-Borel theorem) implies that ${V}$ cannot be locally compact. If one gives ${V}$ the weak-* topology instead, then ${B}$ is now compact by the Banach-Alaoglu theorem, but is no longer a neighbourhood of the identity in this topology. In fact, we have the following result: Theorem 1 Every locally compact Hausdorff topological vector space is finite-dimensional. The first proof of this theorem that I am aware of is by André Weil. There is also a related result: Theorem 2 Every finite-dimensional Hausdorff topological vector space has the usual topology. As a corollary, every locally compact Hausdorff topological vector space is in fact isomorphic to ${{\bf R}^n}$ with the usual topology for some ${n}$. This can be viewed as a very special case of the theorem of Gleason, which is a key component of the solution to Hilbert’s fifth problem, that a locally compact group ${G}$ with no small subgroups (in the sense that there is a neighbourhood of the identity that contains no non-trivial subgroups) is necessarily isomorphic to a Lie group. Indeed, Theorem 1 is in fact used in the proof of Gleason’s theorem (the rough idea being to first locate a “tangent space” to ${G}$ at the origin, with the tangent vectors described by “one-parameter subgroups” of ${G}$, and show that this space is a locally compact Hausdorff topological space, and hence finite dimensional by Theorem 1). Theorem 2 may seem devoid of content, but it does contain some subtleties, as it hinges crucially on the joint continuity of the vector space operations ${+: V \times V \rightarrow V}$ and ${\cdot: {\bf R} \times V \rightarrow V}$, and not just on the separate continuity in each coordinate. Consider for instance the one-dimensional vector space ${{\bf R}}$ with the co-compact topology (a non-empty set is open iff its complement is compact in the usual topology). In this topology, the space is ${T_1}$ (though not Hausdorff), the scalar multiplication map ${\cdot: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is jointly continuous as long as the scalar is not zero, and the addition map ${+: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is continuous in each coordinate (i.e. translations are continuous), but not jointly continuous; for instance, the set ${\{ (x,y) \in {\bf R}: x+y \not \in [0,1]\}}$ does not contain a non-trivial Cartesian product of two sets that are open in the co-compact topology. So this is not a counterexample to Theorem 2. Similarly for the cocountable or cofinite topologies on ${{\bf R}}$ (the latter topology, incidentally, is the same as the Zariski topology on ${{\bf R}}$). Another near-counterexample comes from the topology of ${{\bf R}}$ inherited by pulling back the usual topology on the unit circle ${{\bf R}/{\bf Z}}$. Admittedly, this pullback topology is not quite Hausdorff, but the addition map ${+: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is jointly continuous. On the other hand, the scalar multiplication map ${\cdot: {\bf R} \times {\bf R} \rightarrow {\bf R}}$ is not continuous at all. A slight variant of this topology comes from pulling back the usual topology on the torus ${({\bf R}/{\bf Z})^2}$ under the map ${x \mapsto (x,\alpha x)}$ for some irrational ${\alpha}$; this restores the Hausdorff property, and addition is still jointly continuous, but multiplication remains discontinuous. As some final examples, consider ${{\bf R}}$ with the discrete topology; here, the topology is Hausdorff, addition is jointly continuous, and every dilation is continuous, but multiplication is not jointly continuous. If one instead gives ${{\bf R}}$ the half-open topology, then again the topology is Hausdorff and addition is jointly continuous, but scalar multiplication is only jointly continuous once one restricts the scalar to be non-negative. Below the fold, I record the textbook proof of Theorem 2 and Theorem 1. There is nothing particularly original in this presentation, but I wanted to record it here for my own future reference, and perhaps these results will also be of interest to some other readers. 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http://math.stackexchange.com/questions/tagged/diophantine-equations+geometry	Tagged Questions 4answers 137 views Integral solutions of hyperboloid $x^2+y^2-z^2=1$ Are there integral solutions to the equation $x^2+y^2-z^2=1$? 0answers 66 views Finding integer coordinates on a sphere's surface [duplicate] Possible Duplicate: Integer coordinate set of points that is a member of sphere surface Assume $C$ is a sphere with radius $r$ and center in the origin (0,0,0). How can we find the set of ... 2answers 169 views rational triangles and cosines I've recently started to try working on exercises from this book on Diophantine equations before I need to return it to the library. This one has me slightly stumped. It asks to show that the cosine ... 1answer 194 views Heronian triangles How to prove that all Heronian triangles can be found using formulas described here? I understand that the described substitution will give Heronian triangle, but how to prove that using the ... 2answers 251 views Heronian triangle Generator I'm trouble shooting my code I wrote to generate all Heronian Triangles (triangle with integer sides and integer area). I'm using the following algorithm $$a=n(m^{2}+k^{2})$$ $$b=m(n^{2}+k^{2})$$ ... 1answer 458 views Integer coordinate set of points that is a member of sphere surface I have a graphic application to develop which involve many spheres. I should determine then on run time. Supposing that I have a sphere of radius r, how can I determine the sub set of the sphere ... 5answers 397 views Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$? Does $a^2+b^2=1$ have infinitely many solutions for $a,b\in\mathbb{Q}$? I'm fairly certain it does, but I'm hoping to see a rigorous proof of this statement. Thanks. Here is my motivation. I'm ... 3answers 423 views Primitive integer triangles Consider the triangles with integer sides a, b and c with a ≤ b ≤ c. An integer sided triangle (a,b,c) is called primitive if gcd(a,b,c)=1. How many primitive integer sided triangles exist with a ...	{"extraction_info": {"found_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": 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": 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": 0.9266940951347351, "perplexity_flag": "middle"}
http://cs.anu.edu.au/publications/toc/articles/v008a008/index.html	http://theoryofcomputing.org ISSN 1557-2862 Endorsed by ACM SIGACT Volume 8 (2012) Article 8 pp. 197-208 The Communication Complexity of Gap Hamming Distance by In the gap Hamming distance problem, two parties must determine whether their respective strings $x,y\in\{0,1\}^n$ are at Hamming distance less than $n/2-\sqrt n$ or greater than $n/2+\sqrt n.$ In a recent tour de force, Chakrabarti and Regev (2010) proved the long-conjectured $\Omega(n)$ bound on the randomized communication complexity of this problem. In follow-up work, Vidick (2010) discovered a simpler proof. We contribute a new proof, which is simpler yet and a page-and-a-half long.	{"extraction_info": {"found_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": 0.8214447498321533, "perplexity_flag": "middle"}
http://mathhelpforum.com/calculus/131737-improper-integral-sequences-series.html	Thread: 1. Improper Integral, Sequences, and Series Hey guys, I have a few problems that if possible, I would love to have someone check my solutions to them. Problem 1. Determine whether the improper integral $\int^{\infty}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx$ converges, or diverges. In the case of convergence, give its value. My Solution $= \lim_{t \to \infty}\int^{t}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx = \lim_{t \to \infty}[lnx - ln(x^2 + 1)]^{t}_{1} = \lim_{t \to \infty}[(lnt - ln(t^2 + 1) - (ln1 - ln2)] =$ $\lim_{t \to \infty}(lnt - ln(t^2 + 1) + ln2) = \infty - \infty + ln2 = ln2$ ... Converges to ln2 Problem 2. Determine whether the sequence with the given general term converges, or diverges. In the case of convergence, give its value. a.) $a_{n} = \frac{2n - 1}{3 - 5n}$ My Solution $<br /> \lim_{n \to \infty}\frac{2n - 1}{3 - 5n} = \frac{\infty}{\infty}$, Using L'Hopital's Rule. $= \lim_{n \to \infty}\frac{-2}{5} = \frac{-2}{5}$ ... Converges to $\frac{-2}{5}$ b.) $a_{n} = (-1)^{n}\sin{\frac{n\pi}{2}}$ My Solution $\lim_{n \to \infty}\|(-1)^{n}\sin{\frac{n\pi}{2}}\| = \lim_{n \to \infty}\|(-1)^{n}\|\|\sin{\frac{n\pi}{2}}\| = \lim_{n \to \infty}1 \sin{\frac{n\pi}{2}} =$ No Limit, Diverges c.) $a_{n} = (1 + \frac{1}{n})^n$ My Solution $1^{\infty}$ ... $y = (1 + \frac{1}{n})^{n}, lny = nln(1 + \frac{1}{n})$ $\lim_{n \to \infty} lny = \lim_{n \to \infty}\frac{ln(1 + \frac{1}{n})}{\frac{1}{n}} = \frac{0}{0},$ Using L'Hopital's Rule. $= \lim_{n \to \infty}\frac{\frac{-1}{n^2 + n}}{\frac{-1}{n^2}} = \lim_{n \to \infty}\frac{n}{n + 1} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{1}{1} = 1$ ... $lny \to 1,$ $y \to e$ ... Converges to e Problem 3. Calculate the sum of the given convergent geometric series. a.) $\frac{9}{4} - \frac{3}{2} + 1$ ... My Solution $a = \frac{9}{4}$, $r = \frac{-2}{3}$ $\frac{\frac{9}{4}}{1 + \frac{2}{3}} = \frac{27}{20}$ b.) $\Sigma^{\infty}_{n = 1}(-1)^{n}(\frac{3}{7})^n$ My Solution $a = \frac{-3}{7}$, $r = \frac{-3}{7}$ $\frac{\frac{-3}{7}}{1 + \frac{3}{7}} = \frac{-3}{10}$ Problem 4. Write the first three terms of the sequence of partial sums of the series $\Sigma^{\infty}_{k = 1}\frac{k}{2k - 1}$. My Solution $S_{1} = \frac{1}{1} = 1$ $S_{2} = 1 + \frac{2}{3} = \frac{5}{3}$ $S_{3} = \frac{5}{3} + \frac{3}{5} = \frac{34}{15}$ $1$, $\frac{5}{3}$, $\frac{34}{15}$ Thank you in advance for any help! 2. Originally Posted by mturner07 Hey guys, I have a few problems that if possible, I would love to have someone check my solutions to them. Problem 1. Determine whether the improper integral $\int^{\infty}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx$ converges, or diverges. In the case of convergence, give its value. My Solution $= \lim_{t \to \infty}\int^{t}_{1}(\frac{1}{x} - \frac{2x}{x^2 + 1}) dx = \lim_{t \to \infty}[lnx - ln(x^2 + 1)]^{t}_{1} = \lim_{t \to \infty}[(lnt - ln(t^2 + 1) - (ln1 - ln2)] =$ $\lim_{t \to \infty}(lnt - ln(t^2 + 1) + ln2) = \infty - \infty + ln2 = ln2$ ... Converges to ln2 Nop: $\ln t-\ln(t^2+1)=\ln\left(\frac{t}{t^2+1}\right)\xrighta rrow[t\to\infty]{}-\infty$ and thus the integral diverges. Of course, you can't write $\infty-\infty=0$ ... Problem 2. Determine whether the sequence with the given general term converges, or diverges. In the case of convergence, give its value. a.) $a_{n} = \frac{2n - 1}{3 - 5n}$ My Solution $<br /> \lim_{n \to \infty}\frac{2n - 1}{3 - 5n} = \frac{\infty}{\infty}$, Using L'Hopital's Rule. $= \lim_{n \to \infty}\frac{-2}{5} = \frac{-2}{5}$ ... Converges to $\frac{-2}{5}$ Nop. You can't use DIRECTLY L'Hospital with a discrete variable since L'H implies the use of derivative which use limits which need a continuous variable. Of course, you can use L'H with $\frac{2x-1}{3-5x}$ and then use that the limit stays the same no matter how $x\rightarrow \infty$ , and thus this is so if you choose to go to the limit along the naturals. Another way, perhaps more natural , to divide both numerator and denominator by the highest power of n and use arithmetic of limits: $\frac{2n-1}{3-5n}\cdot\frac{1\slash n}{1\slash n} = \frac{2-\frac{1}{n}}{\frac{3}{n}-5}\xrightarrow [n\to\infty]{}\frac{2-0}{0-5}=-\frac{2}{5}$ Tonio b.) $a_{n} = (-1)^{n}\sin{\frac{n\pi}{2}}$ My Solution $\lim_{n \to \infty}\|(-1)^{n}\sin{\frac{n\pi}{2}}\| = \lim_{n \to \infty}\|(-1)^{n}\|\|\sin{\frac{n\pi}{2}}\| = \lim_{n \to \infty}1 \sin{\frac{n\pi}{2}} =$ No Limit, Diverges c.) $a_{n} = (1 + \frac{1}{n})^n$ My Solution $1^{\infty}$ ... $y = (1 + \frac{1}{n})^{n}, lny = nln(1 + \frac{1}{n})$ $\lim_{n \to \infty} lny = \lim_{n \to \infty}\frac{ln(1 + \frac{1}{n})}{\frac{1}{n}} = \frac{0}{0},$ Using L'Hopital's Rule. $= \lim_{n \to \infty}\frac{\frac{-1}{n^2 + n}}{\frac{-1}{n^2}} = \lim_{n \to \infty}\frac{n}{n + 1} = \frac{\infty}{\infty} = \lim_{n \to \infty}\frac{1}{1} = 1$ ... $lny \to 1,$ $y \to e$ ... Converges to e Problem 3. Calculate the sum of the given convergent geometric series. a.) $\frac{9}{4} - \frac{3}{2} + 1$ ... My Solution $a = \frac{9}{4}$, $r = \frac{-2}{3}$ $\frac{\frac{9}{4}}{1 + \frac{2}{3}} = \frac{27}{20}$ b.) $\Sigma^{\infty}_{n = 1}(-1)^{n}(\frac{3}{7})^n$ My Solution $a = \frac{-3}{7}$, $r = \frac{-3}{7}$ $\frac{\frac{-3}{7}}{1 + \frac{3}{7}} = \frac{-3}{10}$ Problem 4. Write the first three terms of the sequence of partial sums of the series $\Sigma^{\infty}_{k = 1}\frac{k}{2k - 1}$. My Solution $S_{1} = \frac{1}{1} = 1$ $S_{2} = 1 + \frac{2}{3} = \frac{5}{3}$ $S_{3} = \frac{5}{3} + \frac{3}{5} = \frac{34}{15}$ $1$, $\frac{5}{3}$, $\frac{34}{15}$ Thank you in advance for any help! .	{"extraction_info": {"found_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": 67, "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": 0.8061180114746094, "perplexity_flag": "middle"}
http://math.stackexchange.com/questions/202990/why-are-there-5-groups-that-contain-the-2-enemies/202993	# Why are there 5 groups that contain the 2 enemies? Suppose I must form a committee of 3 people chosen out of 7. 2 of the 7 are enemies of each other. These 2 people would wreak havoc if they are both chosen to be on the committee. How may ways are there to form committees containing both of them? Why is the answer $\binom{2}{2}\binom{5}{1} = 5$? Why am I picking 2 out of 2 people and then 1 out of 5 people? - ## 3 Answers Think of the pool as consisting of two groups, the awkward two and the other five. To form one of these useless committees you must pick both of the members of the awkward group, which you can do in $\binom22$ ways, and then you can combine them with any one member of the other group, whom you can choose in $\binom51$ ways. You’re really just observing that in order to get a useless committee, you have to include both of the awkward folks and one of the others, so the only choice involved is in which one of the $5$ others you pick. - You wish to know how you can choose $3$ people out of $7$, when $2$ of them have to be chosen. This means that you have to choose $2$ out of the $2$ enemies, and another from the remaining five, that would be $\binom{5}{1}$. - You're picking three members. First you take the two enemies (since that's one requirement), for the first two members (there is one combination of two possible from two enemies, a.k.a $\binom{2}{2}$) Then there are 5 neutral people to choose from for the third member (there are five combinations of one possible from five neutral people $\binom{5}{1}$). -	{"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": 12, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9763592481613159, "perplexity_flag": "head"}
http://mathoverflow.net/revisions/16545/list	## Return to Answer Post Made Community Wiki by Scott Morrison♦ 2 added 277 characters in body To expand on Pete's comment where we have additional structure, in any context where duality makes sense, to show that $A=B$, it suffices to show that they both have the same duals. Of course this does not answer the question, since we still need a way of showing that their duals are equal, but sometimes this may be easier. Examples of duals are when $A$ and $B$ are both subsets of a common universe $U$ and the dual is just the complement. Or, if $A$ and $B$ are both subspaces of $R^n$ and the dual is the orthogonal complement. Edit: Although I'm not an expert, I'm guessing there are examples coming from topological structure as well. For example, to show that a subset $A$ of a topological space $X$ is equal to $X$, it suffices to show that $A$ is closed and contains a dense subset of $X$. 1 To expand on Pete's comment where we have additional structure, in any context where duality makes sense, to show that $A=B$, it suffices to show that they both have the same duals. Of course this does not answer the question, since we still need a way of showing that their duals are equal, but sometimes this may be easier. Examples of duals are when $A$ and $B$ are both subsets of a common universe $U$ and the dual is just the complement. Or, if $A$ and $B$ are both subspaces of $R^n$ and the dual is the orthogonal complement.	{"extraction_info": {"found_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": 0.968490719795227, "perplexity_flag": "head"}

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