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H: Using a weighted table? I'm writing a simple computer program that has to use the following weighted table to pick a random "payment" where the relative probabilities are given by "weight". How does one interpret the table above and use it to determine how much to "pay"? EDIT: After speaking with some folks it seems this table is simple enough that I can just create a table with 41 entries where each pay amount appears ( Weight times ) and select a random number between 0 and 40 and use that to index into the table. AI: Assuming that the weight vector defines a finite probability distribution, you can sample payments as follows. First pick a random integer $x \in [0, 41]$. Then the payment is as follows: $0 \le x \le 20$ : $200 + 50 \lfloor x/3 \rfloor$ $21 \le x \le 38$ : $550 + 50 \lfloor (x-21)/2 \rfloor$ $39 \le x \le 41$ : $1000 + 500 (x - 39)$
H: A puzzle with powers and tetration mod n A friend recently asked me if I could solve these three problems: (a) Prove that the sequence $ 1^1, 2^2, 3^3, \dots \pmod{3}$ in other words $\{n^n \pmod{3} \}$ is periodic, and find the length of the period. (b) Prove that the sequence $1, 2^2, 3^{3^3},\dots \pmod{4}$ i.e. $\{\ ^nn \pmod{4}\}$ is periodic and find the length of the period. (c) Prove that the sequence $1, 2^2, 3^{3^3},\dots \pmod{5}$ i.e. $\{\ ^nn \pmod{5}\}$ is periodic and find the length of the period. The first two were not terribly difficult (but might be useful exercises in Fermat's little theorm), but the third one is causing me problems, since my methods are leading to rather a lot of individual cases, and I would be interested to see if anyone here can find a neater way to solve it. (In (c), I have evaluated the first 15 terms, and not found a period yet - unless I have made a mistake.) AI: We will need the following slight generalization of the totient theorem: $a^{k + \varphi(n)} \equiv a^k \bmod n$ for all $a$ provided that $k$ is at least as large as the largest exponent in the prime factorization of $n$. It follows that $$a^b \equiv a^{b \bmod 4} \bmod 5$$ provided that $b \ge 1$, and $$b^c \equiv b^{c \bmod 2} \bmod 4$$ provided that $c \ge 2$ (where we need to take the remainder in $\{ 2, 3\}$). Finally, $$c^d \equiv c \bmod 2$$ provided that $d \ge 1$. In summary, $$a^{b^{c^d}} \equiv a^{b^c} \bmod 5$$ provided that $b \ge 1, c \ge 2, d \ge 1$. But this expression only depends on the value of $a \bmod 5, b \bmod 4, c \bmod 2$, so the sequence in question has eventual period dividing $20$, and computations of the first few terms suffices to show that the eventual period is an actual period. A similar argument works for $5$ replaced by any positive integer $m$ (but there is no guarantee that the eventual period is an actual period in this generality and I expect this to be false); we get that the eventual period divides $\text{lcm}(m, \varphi(m), \varphi(\varphi(m)), ...)$. And of course we can replace $\varphi(m)$ by the Carmichael function for larger $m$ to get better bounds.
H: Fibonacci numbers and proof by induction Consider the Fibonacci numbers $F(0) = 0; F(1)=1; F(n) = F(n-1) + F(n-2)$. Prove by induction that for all $n>0$, $$F(n-1)\cdot F(n+1)- F(n)^2 = (-1)^n$$ I assume $P(n)$ is true and try to show $P(n+1)$ is true, but I got stuck with the algebra. How do we reach $P(n+1)$ from $P(n)$? Also, strong induction may be used instead. AI: Here is a pretty alternative proof (though ultimately the same), suggested by the determinant-like form of the claim. Let $$M_n = \left(\begin{array}{cc} F(n+1) & F(n)\\ F(n) & F(n-1)\end{array}\right),$$ and note that $$M_1 = \left(\begin{array}{cc} 1 & 1\\ 1 & 0\end{array}\right),$$ and $$M_{n+1} = \left(\begin{array}{cc} 1 & 1\\ 1 & 0\end{array}\right) M_n.$$ It follows by induction that $$M_n = \left(\begin{array}{cc} 1 & 1\\ 1 & 0\end{array}\right)^n.$$ Taking determinants (and using $\det(A^n) = \det(A)^n$) now gives the result.
H: Proving pointwise convergence to a "Dirichlet-like" function Question: Let $\{r_1, r_2, \dots\}$ be the set of rationals in the interval $[0,1]$. For $x \in [0,1]$ and $n \in \Bbb N$, let $f_n(x)$ and $f(x)$ be given by the following: $$ f_n(x) = \begin{cases} 1 & \text{ if } x= r_1, \dots, r_n \\ 0 & \text{ otherwise } \end{cases} \qquad f(x) = \begin{cases} 1 & \text{ if } x \text{ rational}\\ 0 & \text{ if } x \text{ irrational} \end{cases} $$ Prove that $f_n \to f$ pointwise, but not uniformly. My Thoughts: I'm not sure how to show either convergence result. For pointwise convergence, $\| f_n(x) - f(x) \|$ becomes $0$ at $x$ irrational or $x \in \{r_1,\dots, r_n\}$, and $1$ at all the rationals not yet enumerated. How can I work this into my proof? Edit: Let's suppose to the contrary that there is a sufficiently large $N_0$ so that for some $\epsilon_0$ for all $x \in [0,1]\Rightarrow\|f_{N_0}(x) - f(x)\| \ge \epsilon_0$. Here is where I am stuck now Edit 2: I have found it! I went through my textbook, and I found the following key sentence: In pointwise convergence, one might have to choose a different $N$ for each different $x$. In uniform convergence, there is an $N$ which works for all $x$ in the set $E$. So the proofs follow: Proof of pointwise convergence: Let $\epsilon > 0$ be given. Then let $x_0$ be the $N$th rational number in $[0,1]$. Taking $n = N$, we have $\|f_n(x) - f(x)\| \le \epsilon$ for all $x \le x_0$, and we can successively take larger and larger $n$ to always guarantee that $\|f_n(x) - f(x)\| \le \epsilon$. Proof of lack of uniform convergence: Let $\epsilon > 0$ be given. Then if $f_n \to f$ uniformly, there exists an $M$ so that $n \ge M$ implies $\|f_n(x) - f(x)\| \le \epsilon$ for all $x$. Taking $x$ to be the $(n+1)$th rational number, we have that for sufficiently small $\epsilon$, $\|f_n(x) - f(x)\| \ge \epsilon$, so $f_n \not \! \to f$ uniformly. AI: Pick an $x\in[0,1]$. Then can you say that $f_n(x)=f(x)$ when $n$ is sufficiently large? Take any $n$. Then can you say that there is $x\in[0,1]$ such that $f_n(x)=0$ but $f(x)=1$?
H: Symbol for Euclidean norm (Euclidean distance) Which symbol is more commonly used to denote the Euclidean norm: $ \left \\| \textbf a \right \\| $ or $ \left \| \textbf b \right \|$? AI: As mentioned above, I don't know what is most common (statistically). However, ff you have a vector $V$ space over say the real numbers $\mathbb{R}$, then you can have a norm $\\|\cdot\\|$ on the vector field (so you get a normed space). One thing that you would like is: $$ \\| \alpha v\\| = \lvert\alpha\lvert\cdot\\| v\\|. $$ for $\alpha \in \mathbb{R}$, and $v\in V$. Here the single vertical lines is the norm on the real numbers and the double lines is the norm on the vector space. If you consider for example the real numbers as a vector space over itself, then you can use the absolute value as a norm. If you have the vector space $V = \mathbb{R}^n$ as a vector space over the real numbers, then I do believe that the standard notation is the doube lines $\\|\cdot \\|$. Again, this is because you want to have the single lines for the real numbers. Note that even though the absolute value and the norm seem like the same thing, they are different because the absolute value is evaluated at real numbers, the norm of the vectors. Indeed the Euclidean norm is defined from the absolute value. So for $v = (v_1, \dots, v_n)$, you have $$ \\| v \\|^2 = \lvert v_1\lvert^2 + \dots + \lvert v_n\lvert^2. $$ Note what happens when $n = 1$, then you have $\\|\cdot \\| = \lvert\cdot\lvert$.
H: What is my mistake when evaluating this limit? $\lim_{x\to 0^+} \frac{1}{x}=\infty$ But with l'Hospital's Rule $\lim_{x\to 0^+} \frac{1}{x}=\lim_{x\to 0^+} \frac{0}{1}=0$ So where's my naive mistake? AI: I believe l'Hopital's rule works only for indeterminate forms.
H: What is the advantage and disadvantage of Hilbert System? I have an exam in the course about higher order logic. I was looking for answer of the question "Explain the advantage and disadvantage of using Hilbert system". The disadvantage in the meaning of why it is hard to apply. Thanks. AI: From a practical point of view, the disadvantages of a Hilbert System are It is very cumbersome to use directly for deriving some formula. You need to prove at least some metatheorems before you can use such a system without too much overhead. It doesn't mirror the natural way to do deduction, which can work by proving subtheorems relying on additional assumption, or by doing case-by-case analysis, or by using proof by contradiction. It needs many axioms and axiom schema's to work. From a theoretical point of view, the advantages of a Hilbert System are It works. It is conceptually very simple. It has very few deduction rules, often only "modus ponens" and "generalization", which makes it easier to prove metatheorems, or to implement the scheme in a computer program (see metamath proof explorer for a practical example) At least in theory, it should allow to explore the consequences of different axiom systems easily. The relation between a Hilbert system and a natural deduction system is similar to the relation between machine language and a high level programming language. Of course you can build a high level programming language on top of machine language, and similarly you can prove metatheorems about Hilbert systems which allow you to use some of the more convenient proof techniques from natural deduction systems also with Hilbert systems.
H: Application of Hahn-Banach Theorem Let $E$ be Banach space over $\mathbb{R}$. Let $u$ and $v$ be such that $\|\|u\|\|=\|\|v\|\|=1$ and $\|\|2u+v\|\|=\|\|u-2v\|\|=3$. How do we show that there exists a linear functional $f$ defined on all of $E$ such that $\|\|f\|\|=1$, $f(u)=1$ and $f(v)=1$? AI: Note first that $u$ and $v$ are linearly independent. Now consider the linear functional $g$ on the linear span $V$ of $u$ and $v$ such that $g(u) = 1$ and $g(v) = 1$. Thus $g(au + bv) = a+b$. What is the norm of this? Draw a picture of the unit ball of $V$, using the fact that it must be symmetric and convex...
H: Is there such a thing as discrete Riemannian geometry? General relativity expresses gravity as a curvature in space time, created by the stress energy tensor. $$T_{\mu\nu} \approx R_{\mu\nu} - \frac{R}{2} g_{\mu\nu}$$ Given I put the fact that energy is quantizied and can only change within discrete levels into the stress energy tensor, then the Ricci tensor and as a result, the Christoffel symbols, as well as the metric would have to change into some finite difference form. For large space, this finite difference would approximate the equation above, but for small areas, it would behave differently. Is there any kind of discrete Riemannian geometry that satisfies this criteria? Any sources? AI: The first thing to realize is that nobody has a working theory of quantum gravity, so nobody can really answer your question. As timur has pointed out, quantization doesn't necessarily imply discretization. It also doesn't work the other way around: discretization doesn't imply quantization. You can certainly use finite difference approximations to solve the Einstein field equations, and people do indeed do this, but all this will give you is an approximation to classical (i.e., non-quantum-mechanical) GR. This is the kind of thing relativists do, for example, when they simulate violent classical processes like the mergers of black holes and the subsequent emission of gravitational waves. Having said all that, there are a couple of leading candidates for a theory of quantum gravity, which may or may not be equivalent if you work out their detailed implications (which nobody has been able to do). These are string theory and loop quantum gravity (LQG). LQG does in some sense quantize spacetime, but you shouldn't take that too literal-mindedly. It's not distances that are quantized but areas and volumes. A naive way of seeing that you probably can't quantize distance by using some kind of grid is that under a Lorentz transformation, the grid spacing would undergo length contraction and time dilation. (Area and volume are preserved by a Lorentz transformation.) A theory similar to LQG is causal dynamical triangulation (CDT). Scientific American has published a couple of popular-level articles about LQG and CDT: Smolin, "Atoms of Space and Time," Scientific American, Jan 2004 Jerzy Jurkiewicz, Renate Loll and Jan Ambjorn, "Using Causality to Solve the Puzzle of Quantum Spacetime ," Scientific American, July 2008, https://www.scientificamerican.com/article.cfm?id=the-self-organizing-quantum-universe Smolin has also written a nice popular-level book called Three Roads to Quantum Gravity, which is unfortunately getting to be out of date at this point.
H: Fast variance calculation Suppose to have a sequence $X$ of $m$ samples and for each $i^{th}$ sample you want to calculate a local mean $\mu_{X}(i)$ and a local variance $\sigma^2_{X}(i)$ estimation over $n \ll m$ samples of $X$. Excluding the "boundary" samples of $X$, one can incrementally calculate the local mean estimation as follows: \begin{equation} S(i+1) = S(i) + X(i+1+\frac{n}{2}) - X(i-\frac{n}{2}) \\ \mu_{X}(i+1) = \frac{1}{n} S(i+1) \end{equation} where we suppose to have the previous value of $S(i)$ which is \begin{equation} S(i) = \sum_{k=i-n/2}^{i+n/2}{X(k)} \end{equation} and therefore without needing to compute the whole $S(i)$ for each $i$. I'm looking for a similar "fast" way to calculate a local variance $\sigma^2_{X}(i)$ estimation. Notice I could accept an approximate extimation of this measure too, if precise enough. Can you help me? Thank you very much in advance for your attention. EDIT: now my doubt is the following: the local variance at $(i+1)^{th}$ sample should be: \begin{align} \sigma^2_{X}(i+1) &= \frac{1}{n}\sum_{k=i+1-n/2}^{i+1+n/2}{(X(k) - \mu_{X}(i+1))^2} \\ &= \frac{1}{n}\sum_{k=i+1-n/2}^{i+1+n/2}{ \left((X(k))^2 + (\mu_{X}(i+1))^2 - 2\mu_{X}(i+1)X(k) \right)} \\ &= \frac{1}{n}\sum_{k=i+1-n/2}^{i+1+n/2}{(X(k))^2} + \frac{1}{n}\sum_{k=i+1-n/2}^{i+1+n/2}{(\mu_{X}(i+1))^2} - \frac{1}{n}\sum_{k=i+1-n/2}^{i+1+n/2}{2\mu_{X}(i+1)X(k)} \\ &= \mu_{X^2}(i+1) + (\mu_{X}(i+1))^2 - 2\mu_{X}(i+1)\frac{1}{n}\sum_{k=i+1-n/2}^{i+1+n/2}{X(k)} \\ &= \mu_{X^2}(i+1) + (\mu_{X}(i+1))^2 - 2(\mu_{X}(i+1))^2 \\ &= \mu_{X^2}(i+1) - (\mu_{X}(i+1))^2 \end{align} where $\mu_{X^2}(i+1)$ can be calculated by doing \begin{equation} \mu_{X^2}(i+1) = \frac{1}{n}K(i+1) \end{equation} as proposed by Kartik Audhkhasi, and $(\mu_{X}(i+1))^2$ can be computed by using the already computed $\mu_{X}(i+1)$. Therefore, given \begin{equation} \mu_{X^2}(i+1) = \frac{1}{n}K(i+1) \quad,\quad \mu_{X}(i+1) = \frac{1}{n}S(i+1) \end{equation} the final result is: \begin{align} \sigma^2_{X}(i+1) &= \mu_{X^2}(i+1) - (\mu_{X}(i+1))^2 \\ &= \frac{1}{n}K(i+1) - \frac{1}{n^2}(S(i+1))^2 \\ &= \frac{1}{n}\left(K(i+1) - \frac{1}{n}S(i+1)S(i+1)\right) \end{align} Could this method be correct and what's the difference with the one proposed below by Kartik Audhkhasi. In other words, why it has been proposed an update formula which requires to store the previous mean value too for variance calculation? Thank you again for your attention. AI: I think one can adopt a similar approach to the mean computation. Specifically, consider the following sum of squares: $K(i) = \sum_{j=i-n/2}^{i+n/2} X(j)^2$ Hence, we can write the following update formula for $K$: $K(i+1) = K(i) + X(i+1+n/2)^2 - X(i-n/2)^2$ The relation between $K$ and the variance ($V$) is simple: $V(i) = \frac{1}{n}\sum_{j=i-n/2}^{i+n/2}[X(j) - \mu(i)]^2 = \frac{1}{n}K(i) - \mu(i)^2$ Using the above two formulae, you can get the following update for the variance: $V(i+1) = V(i) + \mu(i)^2 - \mu(i+1)^2 + \frac{1}{n}[X(i+1+n/2)^2 - X(i-n/2)^2]$ So you can use this in conjunction with the mean update formula. Note: I am assuming biased estimator for variance. You can use the unbiased one by having $n-1$ in the denominator and deriving the corresponding update formula.
H: Evaluation of $\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)} \, d\theta$ For a test I had to evaluate $\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta$. First I used substitution, with $z=e^{i\theta}$ and $d\theta=\frac{1}{iz}dz$, as shown: $$\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta=\int_{\|z\|=1}\frac{1}{1+\frac{3}{4}(z+z^{-1})(z+z^{-1})}\frac{1}{iz}dz=\frac{4}{i}\int_{\|z\|=1}\frac{z}{3z^4+10z^2+3}dz$$ Then I used the residue theorem: $$\int_{\|z\|=1}\underbrace{\frac{z}{3z^4+10z^2+3}}_{f(z)}dz=\int_{\|z\|=1}\frac{z}{(z^2+3)(z^2+\frac{1}{3})}dz=\int_{\|z\|=1}\frac{z}{(z-i\sqrt{3})(z+i\sqrt{3})(z-i\frac{i\sqrt{3}}{3})(z+\frac{i\sqrt{3}}{3})}dz$$ As only $\frac{i\sqrt{3}}{3}$ and $-\frac{i\sqrt{3}}{3}$ lie in the integration domain, I don't need to compute the other residues: $$\operatorname{Res}(f;\frac{i\sqrt{3}}{3})=\frac{3}{16}$$ and $$\operatorname{Res}(f;-\frac{i\sqrt{3}}{3})=\frac{3}{16}$$ Finally: $$\int_0^{2\pi} \frac{1}{1+3\cos^2(\theta)}d\theta=\frac{4}{i}2\pi i \frac{6}{16}=3\pi $$ However MATLAB doesn't agree with that: EDU>> f=@(z) 1./(1+3cos(z).^2); quad(f,0,2pi) ans = 3.1416 Where did I go wrong ? I don't see what's wrong in using theorems this way and I did the computations 3 times. Ask for more steps in calculation if needed. AI: Right after the first appearance of $f(z)$, when you factored the denominator, you forgot a factor $3$. Note: Complex analysis might not be the simplest approach here, since the change of variables $t=\tan(θ)$ yields directly that, for every $a\gt-1$, $$ \int_0^{2\pi}\frac{\mathrm d\theta}{1+a\cos^2(\theta)}=\frac{2\pi}{\sqrt{a+1}}. $$
H: An example of neither open nor closed set I need a very simple example of a set of real numbers (if there is any) that is neither closed nor open, along with an explanation of why it is so. AI: $[0,1)$ It is not open because there is no $\epsilon > 0$ such that $(0-\epsilon,0+\epsilon) \subseteq [0,1)$. It is not closed because $1$ is a limit point of the set which is not contained in it.
H: another topology multiple choice Let $S^1 = \{(x, y) \in \Bbb R^2 : x^2 + y^2 = 1 \}.$ Let $D = \{(x, y) \in\Bbb R^2 : x^2 + y^2 \le 1 \}$ and $E = \{(x, y) \in\Bbb R^2 : 2x^2 + 3y^2 \le 1\}$ be also considered as subspaces of $\Bbb R^2.$ Which of the following statements are true? a. If $f : D \to S^1$ is a continuous mapping, then there exists $x \in S^1$ such that $f(x) = x$. b. If $f : S^1 \to S^1$ is a continuous mapping, then there exists $x \in S^1$ such that $f(x) = x$. c. If $f : E \to E$ is a continuous mapping, then there exists $x \in E$ such that $f(x) = x$. AI: HINTS: (a) and (c) The Brouwer fixed-point theorem. For (a) don’t forget that $S^1\subseteq D$. (b) $S^1$ is just a circle; what happens if you rotate it a half turn?
H: Real, imaginary parts of $G(w)=\frac{1}{2}(\sqrt{2}-\sqrt{2}e^{iw})$ For the function $G(w)=\frac{1}{2}(\sqrt{2}-\sqrt{2}e^{iw})$ , show that; $\qquad\mathrm{Re}\,G(w)=\sqrt2\sin^2(w/2)\quad$ and $\quad\operatorname{Im}\,G(w)=-1/\sqrt2\sin w$. I really need help with understanding this. Anything you could do to help would be greatly appreciated. AI: Assumption: $w\in \Bbb R$ $G(w)=\frac{1}{\sqrt 2} (1-\cos w-i\sin w)=\frac{1}{\sqrt 2}(1-\cos w)+i(-\frac{1}{\sqrt 2}\sin w)=\sqrt 2 \sin^2(w/2)+i(-\frac{1}{\sqrt 2}\sin w)$ Thus, Re$(G(w))=\sqrt 2 \sin^2(w/2)$ and Im$(G(w))=-\frac{1}{\sqrt 2}\sin w$
H: is there unique name of inequality $ \leq $ and $ \geq $ I'm tutoring a 8th grader. Once that kid asked me if there is a unique name for $ \leq $ and $ \geq $. Question goes like this: "Since it holds both equality $ = $ and inequality $<, >$ why is it still named inequality?" How do I answer it? AI: The detail of $\leq$ is that it means $<$ or $=$. You might help him/her note that that $$1\leq 1$$ is a true statement, because $1$ is in fact equal or grater than $1$. It seems strange for many students to write $1\leq 1$ when it seems $1=1$ is "more true" or "better" than the former. The "problem" is that order relations (see below) are in fact defined and are analogous to the behaviour of $\leq$ and not $<$. As Halmos puts it: This part is for you: In general, given a relation on a set $S$, we call it a (partial) order, and use the symbol $\leq$ or any similar variant such as $\leqslant$, $\preceq$ if it has the following properties: NOTE: If $x\leq y$ we usually may say that $x$ is smaller or precedes $y$. $(1)$ Transitivity If $x\leq y$ and $y \leq z$ then $x\leq z$. In words: "If a number $n$ is smaller or equal than another number $m$, and this last one is smaller or equal than another number $p$ , then first number is smaller or equal than the last one." $(2)$ Antisymmetry If $x\leq y$ and $y\leq x$ then $y=x$. In easy words: "If a number $n$ is smaller or equal than another number $m$, and this number $m$ is smaller or equal than the first number $n$, then they must be equal. This is maybe a picky thing to explain since it has the weight of the logical operator "or": if we say $A$ or $B$ is true, then it can happen $A$ is true, $B$ is true, or both are true. $(3)$ Reflexivity For any $x$, $x\leq x $ is always true. Remember $\leq$ means "smaller or equal". The statement $1\leq 2$ is perfectly valid. So is $1\leq 1$. Students usually find this odd, and say "But $1$ is not smaller than $1$: true, but $\leq$ means smaller or equal. Since equality is true, the statement in question is true. You can check that given a set $X$, usual improper set inclusion is an order in the powerset $2^X$. The reason it is called a partial order is that sometimes we cannot relate two elements. For example, ${1,2}$ and ${3,4}$ are both in $2^{\{1,2,3,4\}}$, but neither $\{1,2\} \subseteq \{3,4\}$ nor $\{3,4\} \subseteq \{1,2\}$ hold. An order is called a total order if for any $x,y$ in the set in question, either $x\geq y$ or $y\geq x$ holds - intuitively, we can compare every pair of elements. The usual inequality of numbers is a strict order. Now, associated to each (partial) order $\leq$ is the relation $<$: we say that $x<y$ iff $x\leq y$ and $x\neq y$. This new order relation arising from the old one is transitive, and for no elements $x,y$ do both $x<y$ and $y<x$ hold simultanesouly. This is usually called the strict order relation corresponding to $\leq$. All that technicality is intended for you to have a good idea about what order relations are in general. Hope it helps,
H: Proof of Bayesian Prediction Formula I was wondering how to proof this formula, commonly used in Bayesian Prediction: $$ \mathrm{P}(x\|\alpha) = \int_\theta \mathrm{P}(x\|\theta)\mathrm{P}(\theta\|\alpha) \, \mathrm{d}\theta$$ The left hand side can be expressed as the following, through marginalizing: $$ \mathrm{P}(x\|\alpha) = \int_\theta \mathrm{P}(x, \theta \| \alpha) \, \mathrm{d}\theta \quad \quad \ldots \text{(1)}$$ Expanding the right hand side, $$ \int_\theta \mathrm{P}(x\|\theta) \mathrm{P}(\theta\|\alpha) \, \mathrm{d}\theta = \int_\theta \frac{\mathrm{P}(x,\theta)}{\mathrm{P}(\theta)} \frac{\mathrm{P}(\theta,\alpha)}{\mathrm{P}(\alpha)} \, \mathrm{d} \theta \quad \quad \ldots \text{(2)}$$ Note that in equation (1), there will be a $\mathrm{P}(x,\theta,\alpha)$ term, but in equation (2), I can't see how that term will appear. Thanks. AI: We have, in general $$ \mathrm{P}(x) = \int_\theta \mathrm{P}(x\|\theta)\mathrm{P}(\theta) \, \mathrm{d}\theta$$ and conditioning everything on $\alpha$ : $$ \mathrm{P}(x \| \alpha) = \int_\theta \mathrm{P}(x\|\theta \alpha) \mathrm{P}(\theta \| \alpha) \, \mathrm{d}\theta$$ In the Bayesian setting, $\mathrm{P}(x\|\theta \alpha)=\mathrm{P}(x\|\theta)$ because, if we are given the parameter $\theta$ we know the density of $x$, and the values of $\alpha$ adds nothing ($\alpha$ only gives us information about $\theta$ - once we know it, they contribute nothing).
H: Cauchy-Schwarz-type inequality Let $f\in C^1[0,\infty)$ such that $f(x)\to 0$ as $x\to\infty$. Prove that $$\int_0^\infty f(x)^2\, dx\le \Biggl(\int_0^\infty x^2f(x)^2\, dx\Biggr)^{1/2}\Biggl(\int_0^\infty f'(x)^2\, dx\Biggr)^{1/2}\,.$$ Hint: write $f(x)^2=-\int_x^\infty \Bigl(f(t)^2\Bigr)'\, dt$. So I tried doing this without the hint, as follows: For ease, all integrals are taken from $0$ to $\infty$. Consider $xf(x)f'(x)$ and integrate by parts with $u=xf(x)$ and $dv=f'(x)dx$. This gives $$\int xf(x)f'(x)=\underbrace{xf(x)^2\Bigr\|_0^\infty}_A-\int f(x)^2-\int xf(x)f'(x)\,.$$ But the term $A$ does not necessarily go to zero, which is a total buzzkill. I've putzed around with the hint a bit, but it's been fruitless. Any help is appreciated, thanks AI: Cauchy-Schwarz tells us that the right side of the inequality is greater than $$\int_0^\infty xf(x)f'(x)\ dx.$$ Using the hint, note $$\int_0^\infty f(x)^2 \ dx= \int_0^\infty -\int_x^\infty \left(f(t)^2\right)'\, dt\ dx= \int_0^\infty \int_x^\infty -2f(t)f'(t)\, dt\ dx. $$ We can use Fubini's theorem to interchange the order of integration. It is helpful to draw out the region of integration on paper when doing this. You get the sector bounded by the line $x=t$ and the $t$ axis. $$\int_0^\infty \int_x^\infty -2f(t)f'(t)\, dt\ dx = -\int_0^\infty \int_0^t 2f(t)f'(t)\ dx \ dt= -2\int tf(t)f'(t) \ dt.$$ At this point we are off by a sign and a constant, but hopefully the suggestion to switch the integrals was helpful. I will edit this if I find my mistake.
H: Is a right inverse of a surjective linear map linear? On a finite dimensional vector space, the answer is yes (because surjective linear map must be an isomorphism). Does this extend to infinite dimensional vector space? In other words, for any linear surjection $T:V\rightarrow V$, AC guarantees the existence of right inverse $R:V\rightarrow V$. Must $R$ be linear? How about $T:V\rightarrow W$ linear surjection in general? AI: No. Let $V = \text{span}(e_1, e_2, ...)$ and let $T : V \to V$ be given by $T e_1 = 0, T e_i = e_{i-1}$. A right inverse $S$ for $T$ necessarily sends $v = \sum c_i e_i$ to $\sum c_i e_{i+1} + c_v e_1$ but $c_v$ may be an arbitrary function of $v$.
H: Formulate mathematical equation I have a function: $$f(n) = 1 - \left(\frac{1}{n}\right)$$ I want to add another variable $k$ to this equation such that: if $k=n$ , then $f(n,k)=1$ if $n=1$, then $f(n,k)=0$ How can we do that ? AI: Consider the function $g(n,k) = 1 - \frac{1}{n} + \frac{1}{k}$ Then if $k = n$, you have $g(n,k) = 1$. (Note that $n$ and $k$ can not be equal to $0$.) As Ross Millikan suggested, for your new question, you can consider the function: $$h(n,k) = \begin{cases} g(n,k) & \quad n \neq 1 \\ 0 & \quad n = 1 \end{cases}$$ $h$ is a perfectly good function and satisfy the property that $n \neq 1$ and $k = n$, then $h(n,k) = 1$; and if $n = 1$, then for all $k$, $h(n,k) = 0$.
H: Counting necklaces with a fixed number of each bead I want to count the number of necklaces, with $n$ beads in total, where the alphabet of beads is $\{1,\ldots,k\}$, and where the number of beads with color $i$ is $n_i$. For example, if $n=4$, and $n_1 = n_2 = 2$, the following necklaces are the only possible $$1122$$ $$1212$$ I have been able to find formulas for different variations of the problem, but not this one. If there is no nice formula, a generating function is always a nice compromise. AI: There is a reasonably nice generating function coming from the Pólya enumeration theorem. It is given explicitly in, for example, this blog post: the number you want is the coefficient of $\prod r_i^{n_i}$ in $$\frac{1}{n} \sum_{d \| n} (r_1^{n/d} +\cdots + r_k^{n/d})^d \varphi \left( \frac{n}{d} \right).$$
H: Infinite intersection and union, less than or equal to and less than Can anyone show me why these sets are equal? $$\{f \leq a \} = \bigcap_{k=1}^{\infty}\{f<a + \frac{1}{k}\}$$ $$\{f < a \} = \bigcup_{k=1}^{\infty}\{f \leq a - \frac{1}{k}\}$$ $f(x)$ is a function that can take on values over the extended real line, and $x$ is a point in n-dimensional Euclidean space. $a$ is a member of the extended real line. AI: If $x \in \{f \leq a\}$, you have that $f(x) \leq a$. So $f(x) \leq a + \frac{1}{k}$ for all $k$. It has been shown that $\{f \leq a\} \subset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$. If $x \in \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$. Then $f(x) < a + \frac{1}{k}$ for all $k$. This can only happen if $f(x) \leq a$. So $\{f \leq a\} \supset \bigcap_{k = 1}^\infty \{f < a + \frac{1}{k}\}$ The first equality of sets has been shown. The second is similar. If $x \in \{f < a\}$, then $a - f(x) > 0$. Choose a $k$ such that $\frac{1}{k} \leq a - f(x)$. Then $f(x) \leq a - \frac{1}{k}$. Hence $x \in \{f \leq a - \frac{1}{k} \}$. Thus, $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$. Suppose $x \in \bigcup_{k = 1}^\infty \{f \leq a - \frac{1}{k}\}$. This means that $f(x) \leq a - \frac{1}{k}$ for some $k$. So $f(x) < a$. $x \in \{f < a\}$. Equality of the two sets has been shown.
H: Roots of unity in non-Abelian groups: when do they form subgroups? I haven't studied group theory in earnest beyond first courses, so my notation may be nonstandard and my question may be a 'standard fact', so bear with me: Consider a group $G$, and for each natural number $n \in \mathbb{N}$ define $$G_n := \{g \in G : g^n = 1\}$$ to be the set of $n^\text{th}$ roots of unity in $G$. If $G$ is Abelian, then each $G_n$ is a subgroup of $G$, but in general this may not be the case: for example, taking $$G = D_3 = \langle \sigma, \tau : \sigma^2 = \tau^3 = 1, \tau \sigma = \sigma \tau^2 \rangle,$$ we can compute that $$ G_2 = \{1,\sigma,\sigma \tau, \sigma \tau^2\}$$ which is not a subgroup of $D_3$. This leads to the first question, which I've already answered: Are there any non-Abelian groups $G$ such that every $G_n$ is a subgroup of $G$? The smallest such $G$ is the quaternion group, assuming my calculations are correct. So we refine this question: Which non-Abelian groups $G$ are such that every $G_n$ is a subgroup of $G$? I haven't made any calculations for groups larger than the quaternion group, but I thought I'd throw this question out into the open. However, the calculation above for $D_3$ shows that no dihedral groups $D_n$ with $n \geq 3$ satisfy this property, since $$(D_n)_2 = \{1,\sigma,\sigma \tau, \ldots, \sigma \tau^{n-1}\}$$ is not a subgroup of $D_n$. Have at it! AI: I am going to use the notation $\sqrt[n]{G}$ instead. In this answer all groups will be finite. If $G$ has a Sylow $p$-subgroup $P$ which is not normal, then the elements of order dividing $\|P\|$ can't form a subgroup (since they consist precisely of the conjugates of $P$). Hence in order for $\sqrt[n]{G}$ to always be a subgroup, every Sylow subgroup needs to be normal. Lemma: Let $G$ be a finite group all of whose Sylow subgroups are normal. Then $G$ is the direct product of its Sylow subgroups. Proof. Let $g$ be an element of a Sylow $p$-subgroup and $h$ be an element of a Sylow $q$-subgroup, $p \neq q$. Then $ghg^{-1} h^{-1}$ lies in a Sylow $q$-subgroup (since $ghg^{-1}$ is in a Sylow $q$-subgroup) but also in a Sylow $p$-subgroup (since $hg^{-1} h^{-1}$ lies in a Sylow $p$-subgroup), and the intersection of any two such subgroups is trivial. Hence $gh = hg$. It follows that the obvious map from the direct product of the Sylow subgroups of $G$ to $G$ is an injective homomorphism between two groups of the same size, hence an isomorphism. $\Box$ Write $G = P_1 \times ... \times P_k$ where $P_i$ is the Sylow $p_i$-subgroup ($p_i$ the primes in increasing order) and let $n = \prod p_i^{e_i}$. Then $$\sqrt[n]{G} = \prod \sqrt[p_i^{e_i}]{P_i}.$$ Consequently, $\sqrt[n]{G}$ is always a subgroup of $G$ if and only if $\sqrt[p_i^{e_i}]{P_i}$ is always a subgroup of $P_i$. So the problem reduces to the case of non-abelian $p$-groups, and here I have no idea what can happen in general, but here are infinitely many examples for every $p \ge 3$: the discrete Heisenberg group $H_3(\mathbb{F}_{p^n})$ is non-abelian of order $p^{3n}$, and every element can be written in the form $I + N$ where $N$ is a $3 \times 3$ nilpotent matrix over $\mathbb{F}_{p^n}$. Hence $$(I + N)^p = I + p N + {p \choose 2} N^2 = I$$ and we conclude that every element has order dividing $p$.
H: Integral: $\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t$ How to prove the following: $$\int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t=\frac{\pi}{2}\left(\ln(2)+\frac{\pi}{2} i\right)$$ AI: Deform the contour to pick up the residue at $t=i$, $$\begin{eqnarray} \int^\infty_{-\infty}\frac{\ln(t+1)}{t^2+1}\mathrm{d}t &=& 2\pi i\, \mathrm{Res}_{t=i} \frac{\ln(t+1)}{t^2+1} \\ &=& 2\pi i \frac{\ln(1+i)}{2i} \\ &=& \frac{\pi}{2}\left(\ln 2+\frac{\pi}{2} i\right). \end{eqnarray}$$ Here we have assumed the cut starting at $t=-1$ is slightly below the original contour. This appears to be the intended assumption. (If we instead assume the cut is slightly above the original contour we pick up the residue at $t=-i$ and get the complex conjugate of the quoted result.) Note that the contribution at infinity vanishes since it goes like $\ln(R)/R\,(R\to\infty)$.
H: Properties of ellipse x-y form $$5x^2+8xy+5y^2=1$$ $$1\left(\frac{x-y}{\sqrt{2}}\right)^2+9\left(\frac{x+y}{\sqrt{2}}\right)^2=1$$ I know that these two forms are equal, showing that the equation is an ellipse. I do know what happens when the ellipse is in the form of $ax^2 + by^2 = 1$ but not sure what happens in the combination of x-y cases. Can anyone help me with this? (e.g., what would be two axis?) AI: In the familiar case $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the axes of the ellipse are $x=0$ and $y=0$. In this case, they are $x-y=0$ and $x+y=0$. What we have here is a "standard" ellipse, rotated through $45^\circ$.
H: Computing an integral. How to compute $\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS$, where $\Sigma=\Big\{(x,y,z)\Big\| \cfrac{x^2}{2}+\cfrac{y^2}{2}+z^2=1,z\ge 0\Big\}$? AI: Since in $\Sigma$, $z>0$ so our domain is $D:\sqrt{1-(1/2)x^2-(1/2)y^2}>0$ or $D: x^2+y^2<1$. Clearly $S: f=x^2+y^2+2z^2=2$ which is an ellipsoid which we regard just the upper part of it (top of the $x-y$ plane in $\mathbb R^3$). Now $$\displaystyle\iint\limits_{\Sigma} z\sqrt{1+z^2}dS=\displaystyle\iint\limits_{D}z\sqrt{1+z^2}d\sigma$$ where $d\sigma=\frac{\|\|\nabla f\|\|}{\|\partial f/\partial z\|}dxdy=\sqrt{1+z_x^2+z_y^2}dxdy$. I think you can evaluate the second double integral over $D$.
H: question relating to analyticity and C-R equations and relationship between real differentiability, complex differentiability and C-R equations So if function is analytic( ~ holomorphic) in Ω⊂C, then it satisfies C-R equations. And if f satisfies C-R equations and the functions u(x,y) and v(x,y) have first partial derivatives which are continuous, then f is analytic(~holomorphic). Am I right? Then what about relationship between real differentiability and complex differentiability? Does they need C-R equations to distinguish them from each other? I mean that the relationship between those properties in case of complex function are defined by C-R equations . AI: It's not quite clear exactly what the quesion is. A real differentiable function satsifying Cauchy-Riemann's equations on an open set is holomorphic (or complex differentiable if you prefer), even if it's not a priori $C^1$. If it doesn't satisfy C-R, it's certainly not complex differentiable. (If you only assume existence of partial derivatives, it's more tricky, look up Looman-Menchoff's theorem.)
H: Getting wrong result calculating sums I try to solve the following sum: $$\sum_{k=0}^{n}\frac{2}{3^k} = \frac{2}{1} + \frac{2}{3} + \frac{2}{9} + \frac{2}{81} + ...$$ The recurring factor of all numbers is $\frac{1}{3}$, for example $\frac{2}{1} \times \frac{1}{3} = \frac{2}{3}$ and $\frac{2}{3} \times \frac{1}{3} = \frac{2}{9}$. According to the known formula for geometric sums: $$a + aq + aq^2 + ... + aq^n = \frac{a(q^{n+1} - 1)}{q - 1}$$ where $q$ is the factor and $n+1$ is the number of terms. Therefore: $$\sum_{k=0}^{n}\frac{2}{3^k} = \frac{\frac{2}{1}((\frac{1}{3})^{n+1} - 1)}{\frac{1}{3} - 1} = \frac{2((\frac{1}{3})^{n+1} - 1)}{-\frac{2}{3}} = $$ $$= \frac{6((\frac{1}{3})^{n+1} - 1)}{-2} = -3((\frac{1}{3})^{n+1} - 1) =$$ $$= -3((\frac{1}{3})^{n} \times (\frac{1}{3})^{1} - 1) = -3(\frac{1}{3^{n}} \times \frac{1}{3} - 1) =$$ $$= \frac{-3}{3^{n}} \times \frac{-3}{3} + 3 = \frac{-3}{3^{n}} \times (-1) + 3 =$$ $$= \frac{3}{3^{n}} + 3$$ This is clearly not the answer. For example when $n = 0$, this equation shows a term of $4$, which doesn't exist. Where am I going wrong? AI: $$-3\left(\frac{1}{3^{n}} \times \frac{1}{3} - 1\right) \neq \frac{-3}{3^{n}} \times \frac{-3}{3} + 3$$ There is no a distributive property with $a×(b×c) = (a×b)×(a×c)$! Be careful! $$-3\left(\frac{1}{3^{n}} \times \frac{1}{3} - 1\right) = \frac{-3}{3^{n}} \times \frac{1}{3} + 3$$ $$= \frac{-1}{3^{n}} + 3$$
H: Complete Lattice with unique negation Suppose i have a complete lattice (meet and join exist for any two subsets of the lattice) and for every element of the lattice, there is a unique element that is its complement (hence a Boolean negation). If such a lattice is also modular, then it is Boolean. Negation is a strong requirement.. If i don't know it is modular, then what kind of lattice is this? What if the lattice is also atomic? AI: There are lattices with unique complements that are not Boolean: In his article "Lattices with unique complements", R.P. Dilworth proves that every lattice can be embedded into a lattice with unique complements. So take any non-modular lattice (e.g. $N_5$). This may be embdedded into a lattice $L$ with unique complements. Then $L$ cannot be modular since it contains a non-modular sublattice. In this book review of "Lattices with Unique Complements" by V. N. Salii the author mentions at the end that the question whether there exist complete non-distributive lattices with unique complements is still unsolved (by 1990). However it is true that every atomic complete lattice with unique complements is Boolean (and isomorphic to the power set lattice of its atoms). For a proof, see Theorem 18 in §6 of G. Birkhoff's book "Lattice theory".
H: If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well. I'm having trouble understanding the statement: If $\kappa$ is a cardinal, $\aleph$ is any $\aleph$-number, and if $\kappa\leq\aleph$ then $\kappa$ can be well ordered as well. I understand the concept of well-ordering a set of cardinal numbers (any set of cardinal numbers is in fact well-ordered by the relation $\le$) but I'm not sure if I have a proper understanding of what exactly it means to well-order an individual cardinal number. I know that to each cardinal number there is associated a unique ordinal number and an ordinal number is set consisting of ordinal numbers strictly less than it so it's well-ordered and so to say $\kappa$ is well-ordered, is all that we mean is that the ordinal number that this cardinal numbers is associated to is well-ordered? If what I'm saying is correct, aren't all cardinal well-ordered? What do we need this condition for then: $\kappa \le \aleph$ (where $\aleph$ is a well-orderable cardinal)? Sorry if I'm confusing you too. AI: Assuming the axiom of choice cardinals are exactly initial ordinals, that is an ordinal which is not bijectible with any of its members. However without the axiom of choice there might be non well-orderable sets. For non well-orderable sets we define cardinality in a slightly different way. As it turns out cardinals are partially ordered in ZF, and the fact that they are totally ordered is equivalent to AC. The partial order is defined by injections between sets, and it is not hard to see that if cardinality is invariant under bijections then the partial order over the sets translates well to a partial ordering of the cardinals. When we say that $\kappa$ is a well-ordered cardinal we mean that $\kappa$ is an ordinal. When we say that $\kappa$ is a general cardinal we mean that it represents the cardinality of an arbitrary set, which may not be well-orderable. To say that $\kappa\leq\aleph_\alpha$ for some $\alpha$, is to say that every set $K$ such that $\|K\|=\kappa$ has an injection into the $\alpha$-th initial ordinal. If a set $K$ has an injection into an ordinal it inherits the well-ordering of the ordinal, namely: If $f\colon K\to\alpha$ is an injective function into an ordinal then $x\prec y\iff f(x)\in f(y)$ is a well-ordering of $K$. To summarize, to say that $\kappa$ is an arbitrary cardinal and $\kappa\leq\aleph$ for some $\aleph$ number, is to say that any set of cardinality $\kappa$ can in fact be well-ordered, therefore by the definition of cardinals $\kappa$ itself is an ordinal. To read more: Non-aleph infinite cardinals Defining cardinality in the absence of choice There's non-Aleph transfinite cardinals without the axiom of choice?
H: topology - analysis Book I need some notion about topology(I'm very interested in boundary points, open sets) and few examples of solved exercises about limits of functions($f:\mathbb{R}^{n}\rightarrow \mathbb{R}^m$) using $\epsilon, \delta$ and also some theory for continous functions. Please give me some links or name of the books which can help me. Thanks :) AI: Here is a fast free internet resource to get you started, while your text is in the mail.
H: Radius of convergence (root, ratio) Let $\sum c_n z^n$ be a power series. ($c_n,z_n \in \mathbb{C}$) Let $\alpha = \lim \sup {\|c_n\|}^{1/n}$ Then this series is convergent if $\|z\|<1/{\alpha}$. Let $\beta = \lim \sup \|c_{n+1}/{c_n}\|$ Here, series is convergent if $\|z\|<1/{\beta}$. Since $\alpha≦\beta$, $\|z\|<1/{\alpha}$ gives more choice of $\|z\|$ which makes the series convergent. It seems like the second one is easier to apply to problems. Is any constraint that makes $\alpha=\beta$? AI: The second is indeed easier to apply, however, you should be careful, since that second limit is only defined if for every $n$, $c_n\neq 0$. HINT: If $ \lim \|c_{n+1}/c_n\|$ exists then $\alpha = \beta$.
H: A criterion for convergence in the operator norm Let $L:H\rightarrow H$ be a continuous linear operator and $R_n:H \rightarrow H$ a sequence of continuous linear operators, where $H$ is a Hilbert space. If the $\sum_{n=1}^{k} R_n$ converge pointwise to $L$, meaning for every $x\in H$ we have $$Lx=\sum_{k=1}^{\infty} R_k x=\lim_{k\rightarrow \infty}\sum_{n=1}^{k} R_nx,$$ is then do we know then, that the operator $T$ defined by $$ Tx=\sum_{k=1}^{\infty} R_kx,$$ 1) equals L ? 2) is continuous ? 3) convergences in the operator norm to $L$ ? ( For 1) and 2) with the help of Asaf Karagila, I think the answer is "yes", since for 1) $L$ and $T$ agree for every value and for 2) we know that $L$ was continuous, so using 1), also $T$ has to be continuous, right? ) EDIT: Everything has been cleared except that I need a counterexample for 3) AI: Here is a simple counterexample for 3) : Suppose your Hilbert space $H$ is infinite dimensional and separable, meaning it has an infinite countable orthonormal complete system. Let denote such a system by $(e_n)_{n \ge 0}$. We know, from Parseval's identity, that every $x \in H$ is equal to $\displaystyle \sum_{n \ge 0} \langle x, e_n \rangle e_n$. Let $R_n$ defined by $R_n x = \langle x, e_n \rangle e_n$. Then $\sum_{k=0}^n R_k$ converge pointwise to the identity operator. But if the convergence occurs for the operator norm, since every operator $\sum_{k=0}^n R_k$ is of finite rank, the limit would be a compact operator. But the identity is not compact by Riesz theorem, since $H$ is not finite dimensional. The special case $H = \ell^2(\mathbb{N})$ with the complete orthonormal system given by $e_n(k) = \delta_{n,k}$ doesn't even require the use of Parseval's identity.
H: Let both $a$ and $b$ belong to the set $\{ 1,2,3,4 \}$. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots? Let both $a$ and $b$ belong to the set $\{ 1,2,3,4 \}$. What is the number of equations of the form $ax^2+bx+1=0$ which have real roots? for real roots, $a \gt 0$, $b^2-4{a}{c} \ge 0$ Here we have $c=1$, and $a \ge 0$ Now we need to have $b^2-4a \ge 0$ i.e. $(-2\sqrt{a} \ge b) \cup (b \ge 2\sqrt{a})$ AI: You have basically answered the question yourself. You just need $a>0$ which is automatic and $b^2\geq 4ac=4a$. So for $a=1$, $b$ can be $2,3,4$, for $a=2$, $b$ can be $3,4$ and for $a=3$ or $4$, $b$ can be only $4$. Thus there are $7$ possible polynomials.
H: Why are epsilon-delta proofs difficult? What conceptual difficulties do students learning epsilon-delta proofs have, or, why are the proofs difficult? Motivation I have to teach people and never found the epsilonistics particularly difficult, so I have trouble relating to the students and understanding their difficulties. Desirable answers Personal experiences References to essays or academic literature What students have told you I am not interested in Advice to use a particular method of teaching Advice to use a particular book in teaching Arguments about the desirability of teaching epsilonistics AI: The epsilon-delta definition of a limit is often a student's first exposure to universal and existential quantifiers in a formal setting. It's important to understand each one and how "for all, there exists" differs from "there exists, for all". It was not immediately apparent to me and others in my class that these two statements are wildly different. Getting the quantifiers straight is also important for helping students understand which variable to "fix" and which to "choose". When the definition says "there exists a $\delta$", students can confuse that as something that is given and that they somehow need to prove that this mysterious $\delta$ works for all $\epsilon$.
H: Finitely but not countably additive set function Let X be any countable! set and and let F be the cofinite set, i.e., $A \in F $ if A or $A^{c}$ is finite (this is an algebra). Then show that the set function $\mu: F \rightarrow [0,\infty)$ defined as $\mu(A)=0$ if A is finite $\mu(A)=1$ if $A^{c}$ is finite is finitely additive. I have started the argument by letting $A=\sqcup_i A_i$. If all $A_i$ are finite, then $\mu(A)=0=\sum_i\mu(A_i)$ and finite additivity follows. If there is at least one $A_i$ not finite then $\sqcup A_i$ is not finite. But $\sqcup A_i \in F$, which implies $(\sqcup_i A_i)^{c}$ is finite. But then $\mu(\sqcup_i A_i)=1 \neq \sum_i\mu(A_i)$. Could anyone let me know where am I going wrong with the second part of the argument and how to finish this off? I can imagine that finite additivity of $\mu$ relies on the fact that X is countable. AI: If there is at least one infinite $A_i$ then it is the only one: Let $j \neq i$, then since $A_i \cap A_j = \emptyset$ we have $A_j \subseteq A_i^C$, so $A_j$ is finite. Hence in the sum $\sum_i \mu(A_i)$ there is exactly one $1$, giving $$\sum_i \mu(A_i) = \mu\left(\bigcup_i A_i\right) = 1.$$
H: Existence of smooth function $f(x)$ satisfying partial summation We know that $\sum_{i=1}^{n}i=\frac{n(n+1)}{2}$, $\sum_{i=1}^{n}\frac{1}{i}=\psi_{0} (n+1)-\psi_{0} (1)$, where $\psi_{0}(x)$ is the digamma function. My problem is, (1).Is there a transformation such that it maps $x \to \frac{x(x+1)}{2}$ and $\frac{1}{x} \to \psi_{0}(x+1)$, and map a smooth $f(x)$ into another smooth function $g(x)$, such that $g(x)-g(x-1)=f(x)$ ? When I mention transformation, I mean an operator or algorithm for me to get $g(x)$ from $f(x)$. (2). Surely $g(x)$ if exists, it is not unique because $g(x)+C$ also satisfy the condition. Let's take $g(x)+C$ and $g(x)$ as the same case. Is there another smooth $h(x)\not = g(x)+C$ satisfying this condition? The problem came when I tried to evaluate $\sum_{i=1}^{n} \sqrt{i}$, I'd like to represent it by integral form. Thanks for attention! AI: You can use the zeta function and the Hurwitz zeta function to write your sum. As a more general case of your summation, we can have the following representation $$\sum_{i=1}^{n} i^s = \sum_{i=1}^{\infty}i^s - \sum_{i=0}^{\infty}(i+n+1)^s = \zeta(-s) - \zeta(-s, n+1)\,.$$ Now, substituting $s= -\frac{1}{2}$ in the above identity yields $$\sum_{1}^{n} i^{\frac{1}{2}} = \zeta\left(-\frac{1}{2}\right) - \zeta\left(-\frac{1}{2}, n+1\right) \,.$$ See convergence issues of the zeta function and the Hurwitz zeta function.
H: Non-abelian group $G$ of order $p^3$ I just need some hints to prove this: Let $\|G\|=p^3$ be a a non-abelian group. If every subgroup of $G$ is normal, then $p=2$ and $G=Q_8$. I know the following facts about a non-abelian group $G$ of order $p^3$: $Z(G)=G'=Z_p$ If $p$ is an odd number, then the function $\phi:G\longrightarrow Z(G)$ given by $ \phi(g)=g^p$ is a homomorphism and $\|\ker(\phi)\|=p^2$ or $\|\ker(\phi)\|=p^3$. AI: Suppose $p$ is odd. Consider the kernel of $\phi$ in the second fact. It consists exactly of the elements of order dividing $p$, and so there are either $p^2$ or $p^3$ of these; always more than $p$. However, by the first fact, there are exactly $p$ central elements of order dividing $p$. In particular, for every odd prime $p$ and non-abelian $p$-group $G$, there is a non-central element of order $p$, and the subgroup it generates is not normal (since it is order $p$ and not central by assumption). Suppose $p=2$. Then there are two very explicit cases, $D_8$ which doesn't work, and $Q_8$ which does. Groups like this, in which every element of order $p$ are central, have been studied by JG Thompson and others. Maps like $\phi$ always exist, and serve to build the upper exponent-$p$ series of the group. In particular, if $Z(G)$ is cyclic, $p$ is odd, and every element of order $p$ is central, then $G$ itself is cyclic. If $Z(G)$ has rank 2, then the “socle series” of $G$ has factors of rank at most 2.
H: Continuous solutions of $f(x) = \lambda \int_0^{\pi/2} \cos({x-y)}f(y)\,dy$ Possible Duplicate: Eigenvalues of an operator Find all the functions $f \in C([0,\frac{\pi}{2}])$ which are solutions of $$ f(x) = \lambda \int_0^{\pi/2} \cos({x-y)}f(y)\,dy, \qquad \lambda \in \mathbb R. $$ I just tried some algebraic manipulations and calculations, but I do not know ho to face this problem. Any help or hint is welcomed. Thanks a lot. AI: Expand $\cos(x-y)$ into $\cos(x)\cos(y)+\sin(x)\sin(y)$ to show that $f(x)=a\cos(x)+b\sin(x)$ with $$ a=\lambda\int_0^{\pi/2}\cos(y)f(y)dy,\qquad b=\lambda\int_0^{\pi/2}\sin(y)f(y)dy. $$ Since $$ \int_0^{\pi/2}\cos^2(y)dy=\int_0^{\pi/2}\sin^2(y)dy=\frac{\pi}4,\qquad \int_0^{\pi/2}\cos(y)f(y)dy=\frac12, $$ this yields $$ 4a=\lambda(a\pi+2b),\qquad 4b=\lambda(2a+\pi b). $$ If $\lambda\pi-4\ne\pm2\lambda$, the only solution is $a=b=0$, hence $f=0$. If $\lambda\pi-4=\pm2\lambda$, then $\lambda\ne0$ hence every $a=\mp b$ is a solution. For $\lambda=4/(\pi\pm2)$, $f(x)=a(\cos x\pm\sin x)$, or, equivalently, $f(x)=c\cos(x\mp\pi/4)$.
H: Geometry of N-dimensional hypercubes and their N-M surfaces. I'm working on a computer code to mesh a N-dimensional space with N-dimensional hypercubes and do some physics in it. I am wondering if I can produce a generic code (with C++ templates, but that is not the subject here) that is ok for N = 2, 3, 4... My question is about the extraction of the N-M dimensionnal faces of the cubes. These faces are N-M dimensionnal hypercubes in a N dimensionnal space (for exemple the faces of a 3-dimensional hypercubes (a cube) are 2-dimensional hypercubes (squares) in a 3-dimensional space). Is there a generic way to define the data I need to characterize these N-M faces (different from storing all their vertexes). To illustrate that here is a example in N=3 dimensions : For the N=3 cube I need at least : a center (3 coordinates) + the side length (1 number) For its N-1=2 dimensional faces (faces) I need at least: a center (3 coordinates) + the side length (1 number) + the plane normal vector (3 coordinates) For its N-2=1 dimensional faces (edges) I need at least : the coordinates of the two points (6 coordinates) For its N-3 = 0 dimensional faces (vertexes) I need a least : 3 coordinates This way to define the faces is very heterogeneous and not generic at all. What is the most optimal (from a memory point of view) generic way to define a N-M hypercube face in a N dimensional space ? Thank you very much. AI: The $n-m$ dimensional faces of the $n$-dimensional unit cube are obtained by fixing $m$ of the coordinates each to a fixed value that is either $0$ or $1$. So you need to record an $m$-subset of $[n]$ (your favourite $n$-set), together with a bitvector of size $m$, holding the chosen coordinates (so there are $\binom nm2^m$ such faces). I'll leave it to you to figure out how to parameterise size and position of a scaled, translated copy of the unit cube. If you just want to record the face without associating it to a particular parent $n$-cube, then you must record the $m$-set of coordinates you want to fix, the $m$ coordinate values you want to fix them to, and for the $n-m$ coordinates that vary their minimal value (the maximal value is $1$ more than that, or $\textit{size}$ more if the size is variable).
H: Universal covers and homotopy From Wikipedia (about homotopy of spheres): $\pi_2 (S^1) = 0$: This is because $S^1$ has the real line as its universal cover which is contractible (it has the homotopy type of a point). In addition, because $S^2$ is simply connected, by the lifting criterion, any map from $S^2$ to $S^1$ can be lifted to a map into the real line and the nullhomotopy descends to the downstairs space. Question 1: Am I right in assuming that I don't need a universal cover? (just any covering space is good enough to apply the lifting criterion) The "lifting criterion" tells me that if $p: (\tilde{X},a) \to (X,b)$ is a covering map and $f: (Y,c) \to (X,b)$ is any continuous map then $f$ lifts to $\tilde{f}: (Y,c) \to (\tilde{X}, a)$ if and only if $f^\ast \pi_1 (Y) \subset p^\ast \pi_1 (\tilde{X})$. Hence For $\tilde{X} = \mathbb R, X = S^1, Y = S^2$, $p^\ast \pi_1 (\tilde{X}) = \{0\}$ and hence since $f : S^2 \to S^1$ lifts to $\tilde{f} : S^2 \to \mathbb R$, $f^\ast \pi_1 (Y)$ must be the trivial group, too. Question 2: But how do I get from there to $\pi_{\color{red}{2}} = 0$? The lifting criterion only tells me something about $\pi_1$. Thanks for your help. AI: You're right, the claim made by wikipedia is valid for any covering map. And the proof is also valid. Let $p : \widetilde X \to X$ a covering map. As you said, the lifting criterion clearly implies that any map $f$ from a simply connected space to $X$ lifts to $\widetilde X$. So, for example, any $f : S^n \to X$ ($n \geq 2$) lifts to $\widetilde f : S^n \to \widetilde X$. In your case, $X$ is contractible, so there is an homotopy $(\widetilde f_t)$ from $\widetilde f = \widetilde f_0$ to a constant map $\widetilde f_1$. The homotopy $f_t = p \circ \widetilde f_t$ then goes from $f_0 = f$ to a constant map. We have proved that $\pi_n X = 0$ as soon as $\pi_n \widetilde X = 0$. In fact, one can see along the same lines that every covering map $p : \widetilde X \to X$ induces isomorphisms $p_* : \pi_n \widetilde X \to \pi_n X$ for all $n \geq 2$ (doesn't work for $n=1$, of course: $S^1$ isn't simply connected).
H: a problem about boundedness implying integrability Let $X_k,k\in N$ be independent identically distributed random variables, and suppose that for some $p\in (0;2)$ the sequence $\{\frac{1}{n^{1/p}}\sum_{k\leq n} X_k\}_{n\ge 1}$ is a.s. bounded. Prove that $X_1$ is in $L^p$. The question cannot be that hard, but I got lost at the moment :( AI: Hints: Let $S_n=\sum\limits_{k\leqslant n}X_k$. (1.) The event $[n^{-1/p}S_n\ \text{bounded}]$ is included in the event $[n^{-1/p}X_n\ \text{bounded}]$. (2.) For each $x\gt0$, the series $\sum\limits_n\mathrm P(\|X_1\|\gt xn^{1/p})$ converges if and only if $X_1$ is in $L^p$. (3.) By Borel-Cantelli, if the series $\sum\limits_n\mathrm P(\|X_n\|\gt xn^{1/p})$ diverges, then $\limsup\limits_nn^{-1/p}\|X_n\|\geqslant x$ almost surely. (4.) Assume that $X_1$ is not in $L^p$, then use the contrapositive of (2.), then (3.), and finally (1.) to conclude that $n^{-1/p}S_n$ is not almost surely bounded.
H: necessary conditions to be relatively compact Let $X$ be a Banach space and $A\subset X$ be bounded. Suppose that, for any $\varepsilon>0$, $\exists\; F_\varepsilon$ a subspace of $X$ of finite dimension such that $$\text{dist}(x,F_\varepsilon)\leq\varepsilon,\quad\forall x\in A.$$ Then prove that $A$ is relatively compact in $X$, i.e. its closure is compact in $X$. Can anybody give me an hint? AI: I would reason as follows. Fix $\varepsilon>0$ and accordingly the finite dimensional subspace $F_\varepsilon$ granted by the hypotheses of the problem. Then $F_\varepsilon$ is closed. In particular $\forall x\in A$ exists $v_x\in F_\varepsilon$ such that $\\|x-v_x\\|=\text{dist}(x,F_\varepsilon)\leq \varepsilon$. Since $A$ is bounded, the set of all such $v_x$ is bounded, because $\\|v_x\\|\leq \\|x-v_x\\|+\\|x\\|\leq \varepsilon+M$. Then it istotally bounded because bounded in a finite dimensional space. Suppose $v_p$ is an $\varepsilon$ net in this set. Then we may choose, for any $p$, $x_p$ as (one of) the element(s) of $A$ such that $\\|x_p-v_p\\|=\text{dist}(x_p,F_\varepsilon).$ Notice that $x_p$ exists by construction of the set of the $v_x's$. Then the $\{x_p\}$ are a $3\varepsilon$ net in $A$ because if we choose $x\in A$, let $v_p$ be such that $\\|v_x-v_p\\|\leq \varepsilon$ by definition of $\varepsilon$ net. Then $\\|x-x_p\\|\leq \\|x-v_x\\|+\\|v_x-v_p\\|+\\|v_p-x_p\\|\leq 3\varepsilon.$ Hope this is correct.
H: Calculation of residues Let be $a\in\mathbb C\setminus\mathbb Z$ a fixed complex number, and define the following function: $$f(z)=\frac{a\pi\cot\pi z}{z(a-z)}$$ It has simple poles in $z\in\mathbb Z\setminus\{0\}$ and in $z=a$, and a double pole in $z=0$. my textbook says that it is straightforward that: $\operatorname{Res}(f,n)=\frac{a}{n(a-n)}\;\;$ for $n\in\mathbb Z\setminus\{0\}$ $\operatorname{Res}(f,a)=-\pi\cot\pi a$ $\operatorname{Res}(f,0)=\frac{1}{a}$ but I don't understand why these fact are true. AI: For simple poles, it's easy: $Res(f,n) = \lim_{z\to n} (z-n)f(z) = \frac{a}{n(a-n)}$ for $n \in \mathbb Z$ Because $\lim_{z\to n} (z-n) \cot(\pi z) = \frac{1}{\pi}$. We also have: $Res(f,a) = \lim_{z\to a} (z-a)f(z) = -\pi \cot(\pi a)$ For the double pole in $b$, you can use the formula: $Res(f,b) = \lim_{z\to b} \frac{d}{dz} \left( (z-b)^2f(z) \right)$ $\frac{d}{dz}\left( (z-b)^2f(z) \right) = \frac{-\pi\,a\,z\,{\csc\left( z\right) }^{2}}{a-z}+\frac{\pi\,a\,z\,\cot\left( z\right) }{{\left( a-z\right) }^{2}}+\frac{\pi\,a\,\cot\left( z\right) }{a-z}$ So the limit seems to be $\frac{\pi}{a}$ and not $\frac1a$
H: Prove that $J^T J+\lambda I$ is a singular matrix If $J$ is a singular matrix, then $(J^T J)^{-1}$ is singular too. I'm trying to prove that $J^T J+\lambda I$ is a singular matrix, where $I$ denotes identity matrix. Any suggestions please? Thanks AI: Let $x$ be nonzero such that $Jx=0$. Then $J^\mathrm{T} Jx=0$, so $J^\mathrm{T} J$ is singular, and $(J^\mathrm{T} J)^{-1}$ does not even exist. As for the second sentence in your question, $J$=the zero matrix and its transpose are certainly singular, but $J^\mathrm{T} J+\lambda I$ will certainly not be if $\lambda\neq 0$.
H: Axiom of Union? I'm reading Comprehensive Mathematics for Computer Scientists 1. On the second chapter: Axiomatic Set Theory. He first states the axiom of the empty set, the axiom of equality and then he proceeds to the axiom of union: Axiom 3 (Axiom of Union) If $a$ is a set, then there is a set: $\{$$x$ \| there exists an element $b\in a$ such that $x\in b$$\}$. This set is denoted by $\bigcup a$ and is called the union of $a$. Notation 2 If a = {b,c}. or a = {b,c,d}, respectively, one also writes b $\cup$ c, or b $\cup$ c $\cup$ d, respectively, instead of $\cup$a I've learned the definition of Union while I was in school, but it wasn't with axioms, they just gave an intuitive example: $a=\{1,2,3\}$ $b=\{4,5\}$ $a\bigcup b=\{1,2,3,4,5\}$ I can't see how the notion of this intuitive example happens on the axiom of union. In my example, it's easy to understand because there's a mention to another set, where's the mention in this axiom? AI: The connection between your example and the more general definition is that $\bigcup\{a,b\}=a\cup b$. Written out in all its gory details, this is $$\bigcup\Big\{\{1,2,3\},\{4,5\}\Big\}=\{1,2,3\}\cup\{4,5\}=\{1,2,3,4,5\}\;.$$ Let’s check that against the definition: $$\begin{align} &\bigcup\Big\{\{1,2,3\},\{4,5\}\Big\}\\ &\qquad=\left\{x:\text{there exists an element }y\in\Big\{\{1,2,3\},\{4,5\}\Big\}\text{ such that }x\in y\right\}\\ &\qquad=\Big\{x:x\in\{1,2,3\}\text{ or }x\in\{4,5\}\Big\}\\ &\qquad=\{1,2,3\}\cup\{4,5\}\\ &\qquad=\{1,2,3,4,5\}\;. \end{align}$$ Take a slightly bigger example. Let $a,b$, and $c$ be any sets; then $$\begin{align} \bigcup\{a,b,c\}&=\Big\{x:\text{there exists an element }y\in\{a,b,c\}\text{ such that }x\in y\Big\}\\ &=\{x:x\in a\text{ or }x\in b\text{ or }x\in c\}\\ &=a\cup b\cup c\;. \end{align}$$ One more, even bigger: for $n\in\Bbb N$ let $A_n$ be a set, and let $\mathscr{A}=\{A_n:n\in\Bbb N\}$. Then $$\begin{align} \bigcup\mathscr{A}&=\Big\{x:\text{there exists an }n\in\Bbb N\text{ such that }x\in A_n\Big\}\\ &=\{x:x\in A_0\text{ or }x\in A_1\text{ or }x\in A_2\text{ or }\dots\}\\ &=A_0\cup A_1\cup A_2\cup\dots\\ &=\bigcup_{n\in\Bbb N}A_n\;. \end{align}$$
H: Cauchy's functional equation for $\mathbb R^n$ Suppose $f(x+y)=f(x)+f(y)$. If $f:\mathbb R\to \mathbb R$ and is measurable, then $f(x)=cx$. This is referred to as Cauchy's functional equation. Suppose $f:\mathbb R^n\to \mathbb R^n$ instead. Does it still hold that $f$ is linear? Wikipedia says that Hilbert's fifth problem is a generalization of this functional equation, but I can't parse that page well enough to understand how it relates. AI: I think we can deduce the $n$-dimensional result from the $1$-dimensional result. We know $$f(x+y)=f(x) + f(y)$$ for all $x,y\in\mathbf{R}^n$ and we want to prove that in fact $$f(\lambda x + \mu y) = \lambda f(x) + \mu f(y)$$ for all $x,y\in\mathbf{R}^n$ and $\lambda,\mu\in\mathbf{R}$. It clearly suffices to prove that $$f(\lambda x) = \lambda f(x)$$ for all $x\in\mathbf{R}^n$ and $\lambda\in\mathbf{R}$. But, for fixed $x,e\in\mathbf{R}^n$, $g:\lambda\mapsto \langle f(\lambda x), e\rangle$ is a measurable function $\mathbf{R}\to\mathbf{R}$ satisfying $g(\lambda + \mu) = g(\lambda) + g(\mu)$, from which it follows by the $1$-dimensional result that $g(\lambda) = \lambda g(1)$. In other words, for all $x,e\in\mathbf{R}^n$ we have $\langle f(\lambda x),e\rangle = \langle \lambda f(x),e\rangle$. Since $e$ is arbitrary, this implies $f(\lambda x) = \lambda f(x)$ for all $x\in\mathbf{R}^n$ and $\lambda\in\mathbf{R}$, so $$f(\lambda x + \mu y) = f(\lambda x) + f(\mu y) = \lambda f(x) + \mu f(y).$$
H: Is it true that if function $f$ is analytic( ~ holomorphic) in $\Omega \subset\Bbb C$, then it satisfies C-R equations? Is it true that if function $f$ is analytic ($\sim$ holomorphic) in $\Omega \subset \Bbb C$, then it satisfies the Cauchy-Riemann (C-R) equations? And is it t true that if $f$ satisfies C-R equations and the functions $u(x,y)$ and $v(x,y)$ (the real and imaginary parts of $f(x,y)$ respectively) have first partial derivatives which are continuous, then $f$ is analytic ($\sim$ holomorphic)? AI: Analytic and holomorphic are a priori different. A complex-analytic function is one that can be written as series with terms of the form $a_nz^n$. A holomorphic function is one that satisfies the Cauchy-Riemann equations. Your question is if they are equivalent, and the answer is yes. Analytic functions are holomorphic, because you can differentiate the series term by term, and each term satisfies the Cauchy-Riemann equations. It is important here to distinguish complex-analyticity from real-analyticity, because for real-analytic functions this result does not hold. A real-analytic function in 2D is one that can be written as series with terms of the form $a_{n,m}x^ny^m$. The result that says that holomorphic functions are analytic is called the Cauchy-Taylor theorem. One way to explain why it is true is that functions satisfying Cauchy-Riemann equations are very smooth, and their derivatives does not grow fast as you differentiate many times.
H: How can I prove the properties of equality? Say I have an arbitrary group, and a relation inside this group, e.g. $aba^{-1} = b^{-1}$. How can I prove that left-multiplying $a$ on both sides will also give a valid equation? Thanks AI: No justification is needed. In the formal predicate calculus with equality, the following is a basic axiom. For any function symbol $f$ (say in two variables, but this doesn't matter), we have the axiom $$\forall x\forall x'\forall y\forall y'(((x=x')\land (y=y'))\implies (f(x',y')=f(x,y))).\tag{$1$}$$ (There is a similar axiom for relation symbols $R$.) These axioms enable us to prove formally the fact that you are asking about. However, these properties are so basic that when we are doing ordinary mathematics, we use them without comment, just as you have in the past used without worry the fact that if $x=2$ and $y=3$ then $x+y=2+3$.
H: Finding the last digit of largest mersenne prime I was wondering what the best way was to find the last digit of the largest known Mersenne prime $2^{6972593} - 1$? Is there any logical way to do this or will I have to just some how compute the answer and then find the last digit? AI: The answer is yes, there is in fact a "logical" (i.e. simple) way to do this! When computing the last digit of a number, we only need to consider the number mod 10 (that is, we only need to compute its remainder when divided by 10). So let's do that! We see the powers of $2 \pmod{10}$ cycle nicely: $2^1 \equiv 2 \pmod{10}$ $2^2 \equiv 4 \pmod{10}$ $2^3 \equiv 8 \pmod{10}$ $2^4 \equiv 6 \pmod{10}$ $2^5 \equiv 2 \pmod{10}$ $2^6 \equiv 4 \pmod{10}$ ...etc. And now the pattern appears obvious. Since the powers of $2$ mod 10 repeat every 4 numbers, we only need to compute $6972593 \pmod 4$. This is simply $1$, and so $2^{6972593} \equiv 2^1 \pmod{10}$ and so it ends in a $2$. Thus, $2^{6972593}-1$ ends in $\boxed{1}$, as desired. Ask me if you don't understand anything :)
H: The different of a Galois extension of an algebraic number field Since the ramification indices of a prime are the same in a Galois extension, the following proposition is likely to be true. If it is true, how do we prove it? Proposition Let $K$ be an algebraic number field. Let $L$ be a finite Galois extension of $K$. Let $G$ be the Galois group of $L/K$. Let $\mathfrak{D}_{L/K}$ be the different. Then $\sigma(\mathfrak{D}_{L/K}) = \mathfrak{D}_{L/K}$ for every $\sigma \in G$. AI: Isn't it better to use the defn of inverse different as the "dual module" to the integers $\mathfrak o_L$ under trace to $K$? That is, $$ \hbox{inverse different}\;=\; \{\alpha\in L\;:\; \hbox{tr}^L_K(\alpha\cdot \mathfrak o_L)\subset \mathfrak o_K\} $$ Trace is certainly Galois-invariant, and taking inverse of a fractional ideal is, also.
H: Writing 1/3 as a sum of other numbers Is it possible to write $0.3333(3)=\frac{1}{3}$ as sum of $\frac{1}{4} + \cdots + \cdots\frac{1}{512} + \cdots$ so that denominator is a power of $2$ and always different? As far as I can prove, it should be an infinite series, but I can be wrong. In case if it can't be written using pluses only, minuses are allowed as well. For example, $\frac{1}{2}$ is $\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\cdots$ So, what about $\frac{1}{3}$? AI: There is no way to write it as a finite sum. For if you bring such a sum to a common denominator, that denominator will be a power of $2$. Minus signs won't help. It can be expressed as the infinite "sum" $$\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^3}+\frac{1}{4^4}+\cdots.$$ For note that if $\|r\|\lt 1$ the geometric series $a+ar+ar^2+ar^3+\cdots$ has sum $\frac{a}{1-r}$. Put $a=\frac{1}{4}$ and $r=\frac{1}{4}$, and simplify. Another interesting representation of $\frac{1}{3}$ is $\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\cdots$.
H: Solving an augumented matrix (A\|B) with same matrix coeff I am taking a linear algebra course at my university. We just have the first day of class but I am reading by myself, so I decided to solve some problems. I do not understand how to solve this problem: "The two systems: $a)$$2x + y = 3\tag{1}$ $4x + 3y= 5 \tag{2}$ $b)$ $2x + y = -1\tag{3}$ $4x + 3y= 1 \tag{4}$ have the same coefficient matrix but different right-hand sides. Solve both systems simultaneously by eliminating the $(2,1)$ entry of the augmented matrix $$\Bigg(\begin{array}{cc\|cc} 2 & 1 & 3 &-1\\ 4 & 4 & 5 &-1\\ \end{array}\Bigg) $$ and then performing back substitutions for each columns corresponding to the right hand sides." What does the$(2,1)$ entry mean? Can someone explain me how to solve this? I know that the problem must be easy, but I haven't seen an example yet. AI: The $(i,j)$ entry is the entry in row $i$, column $j$, so the $(2,1)$ entry is the entry in row $2$, column $1$. Starting with the matrix $$\left[\begin{array}{rr\|rr} 2&1&3&-1\\ 4&4&5&-1 \end{array}\right]\;,$$ subtract twice the first row, which is $\begin{bmatrix}4&2&6&-2\end{bmatrix}$, from the second row to get $$\left[\begin{array}{rr\|rr} 2&1&3&-1\\ 0&2&-1&0 \end{array}\right]\;.\tag{0}$$ This actually corresponds to two matrices, one for each of the linear systems: $$\left[\begin{array}{rr\|rr} 2&1&3\\ 0&2&-1 \end{array}\right]\tag{1}$$ for the first system, and $$\left[\begin{array}{rr\|rr} 2&1&-1\\ 0&2&0 \end{array}\right]\tag{2}$$ for the second. From the second row of $(1)$ I see that $2y=-1$, so $y=-\frac12$. The first row of $(1)$ corresponds to the equation $2x+y=3$; I now know that $y=-\frac12$, so this is $2x-\frac12=3$, $2x=\frac72$, and $x=\frac74$. Similarly, the second row of $(2)$ corresponds to the equation $2y=0$, so $y=0$. The first row corresponds to $2x+y=-1$, and $y=0$, so $2x=-1$, and $x=-\frac12$. I now have the solutions to both of the original systems. With a little practice you won’t have to split $(0)$ into $(1)$ and $(2)$; you’ll be able to do the same work directly from $(0)$.
H: Solving a Linear Equation over $2^X$ I was reading Vinberg's "A Course in Algebra" and in the opening line of chapter 2 he says, "Fix a field $\mathbb K$. We are going to abuse the language slightly and call elements of $\mathbb K$ numbers." Anyway, linear algebra and solving systems of equations is present (it seems) in tons of abstract books for various reasons (analysis, algebra, geometry, ...) but the most "abstract" I've seen the discussion become is the use of the field $\mathbb Q$ or $\mathbb C$. I'm really wanting to make solving systems of equations a more abstract thing, where we don't even deal with equations of "numbers". For example, if $X$ is a set and $2^X$ is the set of all its subsets, it can be shown that $2^X$ is a ring with respect to the operations of symmetric difference: $M \Delta N = (M - N) \cup (N- M)$ and intersection, taken for addition and multiplication, respectively. (The ring is commutative and associative). Well, what does solving a system of equations of this sort look like? Are there other "abstract" systems of equations that you can think of, which are interesting. Further, are there any connections of this idea with other things we are familiar with? Positive definiteness? Eigenvalue problem? Uniqueness and Existence of solution, etc... AI: If $\Bbb K$ is a field, then it is equipped with operators $+$ and $\cdot$, and those operators are well-defined and the field axioms are satisfied. Numbers in $\Bbb Q$ are just elements of $\Bbb Q$ where we have a familiar interpretation for the operators $+$ and $\cdot$. But there's nothing preventing us from making a field of letters, and defining $A+B = C$, $A+C = D$, and so on... The solution of a linear system of $n$ equations in $\Bbb Q$ yields some tuple in $\Bbb Q^n$ such that each element of the tuple can be written in terms of combinations of elements in $\Bbb Q$ operated on in some order by $+$ and $\cdot$. So solutions in a linear system in $\mathbf{2^X}$ would look like a combination of elements of $\mathbf{2^X}$ with the operations $\Delta$ and $\cap$ respectively. Since $\mathbf{2^X}$ is a ring, it is closed under these operations, and the resulting element will just be another element in $\mathbf{2^X}$. In other words, it looks exactly like a solution in $\Bbb Q$, but with different operations and different elements. As far as what we can define in this system, we need to define a vector space over $\mathbf{2^X}$. Once this vector space is defined, we can define matrix-vector multiplication in all the same ways, and so we could have things like eigenvalues, etc. Things get a little unfamiliar when $\mathbf{X}$ is finite; however, to find an analog, you could look how linear systems work in a field of prime characteristic, such as $\Bbb Z_p.$ Let's first consider a simple case, say $\Bbb Z_5$. Define a 2-dimensional vector space, $\Bbb Z_5^2$. Let's set $x = 3$ and $y = 2$ (but pretend that we don't know that!). Then, we can write the following two equations $$\begin{align} 2x + 3y & = & 2 \\ 4x + 2y & = & 1 \end{align}$$ The matrix representation is exactly what we expect: $$\begin{pmatrix} 2 & 3 \\ 4 & 2 \end{pmatrix}\begin{pmatrix} x \\ y\end{pmatrix} = \begin{pmatrix} 2 \\ 1 \end{pmatrix}.$$ We can perform Gaussian elimination on this matrix, where we must simply remember that we're working in a field of characteristic 5. $$\begin{pmatrix}2 & 3 \\ 0 & 2-3\cdot 2\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 & 3 \\ 0 & 1\end{pmatrix}\begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 \\ 1-2\cdot 2\end{pmatrix} = \begin{pmatrix} 2 \\ 2\end{pmatrix}.$$ This gives us $y = 2$ and $2x+1=2$, the latter of which is solved as $2x = 1 \Longrightarrow x = 3$.
H: How to divide using addition or subtraction We can multiply $a$ and $n$ by adding $a$ a total of $n$ times. $$ n \times a = a + a + a + \cdots +a$$ Can we define division similarly using only addition or subtraction? AI: To divide $60$ by $12$ using subtraction: $$\begin{align} &60-12=48\qquad\text{count }1\\ &48-12=36\qquad\text{count }2\\ &36-12=24\qquad\text{count }3\\ &24-12=12\qquad\text{count }4\\ &12-12=0\qquad\;\text{ count }5\;. \end{align}$$ Thus, $60\div 12=5$. You can even handle remainders: $$\begin{align} &64-12=52\qquad\text{count }1\\ &52-12=40\qquad\text{count }2\\ &40-12=28\qquad\text{count }3\\ &28-12=16\qquad\text{count }4\\ &16-12=4\qquad\;\text{ count }5\;. \end{align}$$ $4<12$, so $64\div 12$ is $5$ with a remainder of $4$.
H: Ways to compare a set of values against a base value I have a statistics doubt. Consider I have the following values. Base Model Model 1 Model 2 5 7 4 10 12 8 15 17 12 20 23 21 25 30 27 30 31 31 35 34 37 40 43 43 45 45 44 50 51 50 I want a better way to represent Model1 and Model2 by comparing against the Base Model. I used an error metric: $$\frac{\|\text{My_Model} - \text{Base_Model}\|}{\text{Base_Model}} \times 100$$ What other methods can I use to represent these. I want to plot these as a graph and see how good Model 1 and Model 2 are w.r.t Base Model. Can I use any form of distributions? AI: In the chi square goodness of fit test the measure is ∑(observed-expected)$^2/expected. The change to your formula would be to square the numerator and not express it as a percentage.
H: 2D basis functions orthogonal under exponential kernel In one dimension, the Laguerre polynomials are orthogonal under exponential weighting: $$ \int_0^\infty L_n(x) L_m(x) e^{-x} \, dx = 0, n \ne m $$ Does anyone know what the corresponding basis functions would be in 2 dimensions? $$ \int_{-\infty}^\infty \int_{-\infty}^\infty F_n(x,y) F_m(x,y) e^{-r} \, dx \, dy = 0, r= \sqrt{x^2+y^2}, n \ne m $$ The Zernike polynomials are orthogonal, but with uniform weight and over the unit disk. The underlying problem is to compute an estimator for a missing pixel. A series of orthogonal functions are helpful since you can then incrementally compute a 1-st order estimator, then a second order, then a third order, and so on. The exponential arises since (in natural scenes anyway) presumably pixels further away have less influence. AI: Supposing $F_n$'s are functions of $r$, your orthogonality relation is $$\int_0^\infty F_m F_n\ r e^{-r} dr = 0$$ so $F_m$ is the generalized Laguerre polynomial $L^{(1)}_n$, see here and here. To get an orthogonal basis of $L^2(\mathbb{R}^2,e^{-r}dx\,dy)$, you need to use all the functions $L^{(1)}_n(r)\, e^{ik\phi}$.
H: Pre-Calculus - Solving for $x^3$ In my Pre-Calculus class we were discussing $i$, and all of it's mysterious imaginary wonders, but I had a moment of confusion during the lecture. This was the problem: Solve for $x$. $$ x^3 - 1 = 0 $$ Not too bad; what he said was to factor it: $$ (x - 1)(x^2 + x + 1) = 0 $$ Then use the quadratic formula for the trinomial, ending up with: $$ (x - 1)(\frac{-1 \pm i \sqrt 3} 2) $$ It's pretty straight forward, but then, if you're solving for $x$ in: $$ x^2 - 1 = 48 $$ You'd move the 1 over and Square Root both sides making $x = \pm 7$. My question is, what are you actually doing if you move the 1 over in the original problem and Cube Root both sides? AI: Not really. If you "square root" both sides, you get $\|x\|=7$, which means $x=7$ or $x=-7$. Alternatively, you can subtract $48$ to get $$x^2-49=0$$ $$x^2-7^2=0$$ $$(x-7)(x+7)=0$$ $$x-7=0\text{ or } x+7=0$$ The cube root has the "nice" property is is a one one correspondence, from where there is "no harm" if you "cube root" you equation. This means $$x^3=y^3 \iff x=y$$ That is, two real numbers have the same third power if and only if they are the same number in the first place. The problem with the square root (and in general even powers of $x,y$) is that $$x^n=y^{n} \iff x=y$$ is false since the additive inverse also works, for example $(-7)^2=7^2$ yet $7=-7$ is manifestly false. Your equation is $$x^3-1=0$$ If you're working over $\Bbb R$, that is, considering real numbers as solutions, then there is no harm on doing what you suggest $$x^3-1=0$$ $$x^3=1\iff x=1$$ Note $-1$ is not a solution since $(-1)^3=-1$. Now, the problem arises when we extend our "workplace" to the complex number system. Then our useful assertion that $$\tag 1 x^3=y^3\iff x=y$$ becomes false. In such case, you must factor the polynomial, as you did, and find complex solutions, since using $(1)$ would be wrong, for it is a true statement for real numbers only. You have found the counterexamples just now $$\left(\frac{-1+i\sqrt 3}{2}\right)^3=\left(\frac{-1-i\sqrt 3}{2}\right)^3=1^3=1$$ but $$\frac{{ - 1 + i\sqrt 3 }}{2} \ne \frac{{ - 1 - i\sqrt 3 }}{2} \ne 1$$
H: Optimization problem involving step function I've got to optimize the following function with respect to $\phi$: $q(\phi, x) = \frac{1}{n} \sum_{i=1}^{n}{H(y_i)}$ where $y_i = k - \phi l - x_i$ and $H(.)$ denotes the Heaviside function. $k$ and $l$ are constants, and $x$ follows either (1) a continuous uniform distribution or (2) a normal distribution. This is part of a quite standard programming problem but I'm a little stuck with finding the optimal $\phi$ I'm sure this is a totally simple question but I can't quite figure it out... any help is greatly appreciated... AI: The step functions all add up in the same direction, since $\phi$ has the same sign in all $y_i$ – thus $q$ is minimal for all $\phi$ such that all step functions are $0$, which occurs for $\phi\lessgtr(k-x)/l$, where the inequality is $\lt$ or $\gt$ and $x$ is the greatest or least of the $x_i$, depending on whether $l$ is negative or positive, respectively.
H: is the converse true: in a simply connected domain every harmonic function has its conjugate The question is. Is the converse true: In a simply connected domain every harmonic function has its conjugate? I am not able to get an example to disprove the statement. AI: No, to the question as stated by OP in the comments. Take the real part of a holomorphic function on an annulus. On the other hand, look at the comment by Jonas.
H: Clarification on pairs of sets that make up a power set From the Wikipedia article on power sets, ... one can select any element x of a set, S, and gather up the subsets of S itself into pairs (A-, A+) that are identical excepting that A- does not include x and A+ does. So if S = {1,2,3}, and we choose x to be 3, I get pairs {1},{1,3} {2},{2,3} {1,2},{1,2,3} But there should be 23 = 8 sets. Which am I missing? AI: Let us write the power set of $S=\{1,2,3\}$: $$\mathcal P(S)=\Bigg\{\varnothing,\underbrace{\{1\},\{2\},\{3\}}_{\text{singletons}},\overbrace{\{1,2\},\{1,3\},\{2,3\}}^{\text{pairs}},\{1,2,3\} \Bigg\}$$ You have chosen $3$, so we essentially going to write $S$ as $S\setminus\{3\}\cup\{3\}$. Note that this is to write a set of three elements as $2+1$, taking a power set, in terms of size it to raise by a power of $2$, so we have $2^3=2^{2+1}=2^2\times 2$. This means that indeed we can write $\mathcal P(S)$ as four pairs of the form: $\langle A,B\rangle$ where $A\subseteq S\setminus\{3\}$ and $B=A\cup\{3\}$. Of course we can replace $S$ by an arbitrary set (even infinite!) and $3$ by an arbitrary element of this set. The calculation pans out the same way: First we write a table of all the subsets of $\{1,2\}$, and fill the $B$ coloumn: $$ \begin{array}{lcr} \begin{array}{\|c\|c\|} \hline A & B\\\hline \varnothing & \\\hline \{1\} & \\\hline \{2\} & \\\hline \{1,2\} & \\ \hline \end{array} & \text{We add the }3: \begin{array}{\|c\|c\|} \hline A & B\\\hline \varnothing & \varnothing\cup\{3\}\\\hline \{1\} & \{1\}\cup\{3\}\\\hline \{2\} & \{2\}\cup\{3\}\\\hline \{1,2\} & \{1,2\}\cup\{3\}\\ \hline \end{array} & \text{and finally:} \begin{array}{\|c\|c\|} \hline A & B\\\hline \varnothing & \{3\}\\\hline \{1\} & \{1,3\}\\\hline \{2\} & \{2,3\}\\\hline \{1,2\} & \{1,2,3\}\\ \hline \end{array} \end{array}$$ Indeed four rows, eight elements. We wrote all $\mathcal P(S)$ as a list of pairs as wanted.
H: How to find $k$'s quotient of $\frac{-18}{(-2)^k}$ I try to solve a geometric sum, but I'm stuck. The original terms are: $$s = 9 - \frac{9}{2} + \frac{9}{4} - \frac{9}{8} + \cdots - \frac{9}{128}$$ After some work I found that the sum would be: $$\sum_{k=1}^8 \frac{-18}{(-2)^k}$$ My next step towards finding the sum of all terms is to use the general formula. However, this includes getting the quotient, and I don't understand how to do so in this type of scenario. AI: I expect that you know that $$a+ar+ar^2+\cdots +ar^n=a\frac{r^{n+1}-1}{r-1}=a\frac{1-r^{n+1}}{1-r}$$ (if $r\ne 1$). It remains to identify $a$, $r$, and $n$. We have $a=9$. And always we obtain the "next" term by multiplying the previous term by $-\frac{1}{2}$, so now we know that $r=-\frac{1}{2}$. Since $128=2^7$, we know that $n=7$. Substitute in the general formula, either version, but for this case I prefer the second form, it makes the denominator positive. We have $1-r=1-\left(-\frac{1}{2}\right)=\frac{3}{2}$. For the top, we will need to calculate $1-\left(-\frac{1}{2}\right)^8$. This is $1-\frac{1}{256}$, which you may want to simplify to $\frac{255}{256}$. Now it is just a question of putting the pieces together, and maybe simplifying somewhat.
H: Lift of homeomorphism is homeomorphism Suppose $p:Y \rightarrow X$ is the universal covering map of $X$. Given a contiunuous $f: X \rightarrow X$ then a well known theorem for existence of lifts states that there exist a continuous lift $$\tilde f : Y \rightarrow Y \text{ with } p \circ \tilde f = f \circ p.$$ If we additionally suppose f is a homeomorphism. Then I think $\tilde f$ is a homeomorphism, too. I tried to prove that but failed. My idea was to take a lift of $f^{-1}$ and compose it with $\tilde f$ in order to obtain a lift of the identity. But that´s not really helping here. Anybody knows how to prove or disprove that? Thanks in advance AI: If we fix $x_0 \in X$, $y_0\in p^{-1}(x_0)$ and let $y_1 = \tilde f(y_0)$, then there is a unique lifting of $f$ such that $\tilde f(y_0) = y_1$ (for a reference and sketch of proof, see here). By lifting $f^{-1}$ to $Y$ in such a way that $\widetilde {f^{-1}}(y_1) = y_0$, we then obtain the following commuting diagram (of pointed spaces): The $\mathrm{id}_Y$-arrow follows from the fact that $\mathrm{id}_Y$ is a lifting of $\mathrm{id}_X$ and therefore is the unique such lifting. It follows that $\widetilde {f^{-1}} \circ \tilde f = \mathrm{id}_Y$ and with a similar diagram we also see that $\tilde f \circ \widetilde {f^{-1}}= \mathrm{id}_Y$. So $\tilde f$ is indeed a homeomorphism.
H: Dual of a vector space Let $S$ be a subspace of a vector space $V$ so that $V/S$ is finite dimensional. How do we show that $S$ is the intersection of kernels of a finite number of elements in the dual of $V$? If $S$ is finite dimensional, what can we conclude about the dimension of $V$? AI: Put $\,S=Span\{v_i\;\;\|\;\;i\in I\}\,$ and $\,W:= V\setminus S\Longrightarrow W=Span\{w_1,...,w_n\}\,$ , and we get at once that $$\{w_1,...,w_n,w_k\;\;\|\;\;k\in K:=\text{ a set of indexes}\,\,,\,K\cap\{1,2,...,n\}=\emptyset\}$$ is a basis of $\,V\,$ . Define (with $\,\Bbb F:=\,$ the definition field) $$\forall\,\,i=1,2,...,n\,\,,\,f_i: V\to \Bbb F\,\,,\,f_iw_j=\delta_{ij}=\left\{\begin{array} {}1 &\text{, if}\,\,1\leq i=j\leq n\\0&\text{, if}\,\,1\leq i\neq j\leq n\,\,\text{ or}\,\,j\in K\end{array}\right.$$ and extend this definition by linearity. Clearly $\,f_i\in V^*\,$ , and letting $$v=\sum_{r=1}a_rw_r+\sum_{k\in K}a_kv_k\,\,,\,\,a_k=0\,\,\text{but for a finite number of indexes in} \,K$$ we get that $$\ker f_i:=\{v\in V\;\;\|\;\;f_iv=0\}=\{v\in V\;\;\|\;\;a_i=0\,,\,1\leq i\leq n\,,\,\,\text{in the above expression for}\,\,v\}$$ And thus $$\bigcap_{i=1}^n\ker f_i=\{v\in V\;\;\|\;\;a_1=...=a_n=0\,\,\text{in the above expression for}\,\,v\}$$ Well, now end the argument.
H: Consistency of Peano axioms (Hilbert's second problem)? (Putting aside for the moment that Wikipedia might not be the best source of knowledge.) I just came across this Wikipedia paragraph on the Peano-Axioms: The vast majority of contemporary mathematicians believe that Peano's axioms are consistent, relying either on intuition or the acceptance of a consistency proof such as Gentzen's proof. And then went back to the article on mathematical induction: The principle of mathematical induction is usually stated as an axiom of the natural numbers; see Peano axioms. I'm not a math person, but I'm curious - how can something which sounds like it is believed to be consistent, be the base of such widely used proof technique? (The Wikipedia article on Peano-Axioms also states: "When the Peano axioms were first proposed, Bertrand Russell and others agreed that these axioms implicitly defined what we mean by a "natural number".) AI: The answer is relatively simple, but complicated. We cannot prove that Peano axioms (PA) is a consistent theory from the axioms of PA. We can prove the consistency from stronger theories, e.g. the Zermelo-Fraenkel (ZF) set theory. Well, we could prove that PA is consistent from PA itself if it was inconsistent to begin with, but that's hardly helpful. This leads us to a point discussed on this site before. There is a certain point in mathematical research that you stop asking yourself whether foundational theory is consistent, and you just assume that they are. If you accept ZF as your foundation you can prove that PA is consistent, but you cannot prove that ZF itself is consistent (unless, again, it is inconsistent to begin with); if you want a stronger theory for foundation, (e.g., ZF+Inaccessible cardinal), then you can prove ZF is consistent, but you cannot prove that the stronger theory is consistent (unless... inconsistent bla bla bla). However what guides us is an informal notion: we have a good idea what are the natural numbers (or what properties sets should have), and we mostly agree that a PA describes the natural numbers well -- and even if we cannot prove it is consistent, we choose to use it as a basis for other work. Of course, you can ask yourself, why is it not inconsistent? Well, we don't know. We haven't found the inconsistency and the contradiction yet. Some people claim that they found it, from time to time anyway, but they are often wrong and misunderstand subtle point which they intend to exploit in their proof. This works in our favor, so to speak, because it shows that we cannot find the contradiction in a theory: it might actually be consistent after all. Alas, much like many of the mysteries of life: this one will remain open for us to believe whether what we hear is true or false, whether the theory is consistent or not. Some reading material: How is a system of axioms different from a system of beliefs?
H: How do you calculate "excess returns"? (Although this is perhaps a very simple question, math-wise, it still has me puzzled so it's okay to ask for help here I guess. I really need to understand this, so any help is appreciated) First, assume the S&P100 market index has performed like this on the first 3 days of the month: day open close return (abs) return (%) ---------------------------------------------------- 1 574.19 562.51 -11.68 -2.03 2 564.56 570.72 6.16 1.09 3 571.99 572.93 0.94 0.16 In other words, "the market" went down $11.68 or -2.03% on day 1. Now asssume that I have bought and sold the following stocks on each of these 3 days: day stock buy sell return (abs) return (%) ------------------------------------------------------------ 1 HPQ 40.81 40.14 -0.67 -1.64 2 FDX 89.79 87.89 -1.90 -2.12 3 PEP 63.49 63.24 -0.25 -0.39 So on day 1, I bought and sold HPQ for a return of -$0.67 or -1.64%. I now want to calculate, in %, how my trades performed against the market: day market return (%) my return (%) excess return (%) ------------------------------------------------------------------- 1 -2.03 -1.64 0.39 (-1.64 - -2.03) 2 1.09 -2.12 3.21 3 0.16 -0.39 -0.55 ------------------------------------------------------------------- total -0.78 -4.15 -3.37 (-4.15 - -0.78) For instance, on day 1 the market went down 1.64% but the stock I bought only went down 1.64%. In other words, my trade "outperformed" the market by 0.39%. When totalling, the market lost 0.78% during the three days, but my trades lost -4.15%. The way I interpret this, my trades underperformed 3.37% against the market. Alternatively, if I take the first (574.19) and last (572.93) price of the market indexfund, the market only declined ((572.93 - 574.19) / 574.19) * 100 = -0.22%. Based on this number, my trades underperformed the market by -3.37 - -0.22 = -3.15% Question 1 How do the 2 market performances that I calculated (-0.78% and -0.22%) relate to each other? Question 2 How do I correctly culculate the relative performance of my trades against the market? Is is -3.37% or -3.15%? And if it's neither, how do I calculate with how many % my trading outperformed (or underperformed) the market? [edit after thinking about incorporating overnight returns] day stock close@yesterday open@today close@today return (abs) return (%) ------------------------------------------------------------------------------ 1 HPQ 40.00 40.81 40.14 0.14 0.35 2 FDX 90.00 89.79 87.89 -2.11 -2.34 3 PEP 64.00 63.49 63.24 -0.76 -1.19 ---------------------------------------------------------------------------- + -3.18 (today's opening price is just there for reference. I've calculated return (abs) as close@today - close@yesterday and return (%) as (return (abs) / close@yesterday x 100)) AI: Normally the market return of a given day is calculated from the previous day's close, not from that day's open, so the return on day 2 is $570.72-562.51=8.21$ or When you add the returns on the three days you miss the rises in the index that happened overnight. This is why you get different returns by adding the return of the three days from taking the whole span. If you want to cover day 1 open to day 3 close, the $-0.22\%$ is correct. Similarly, you should calculate your return from purchase (presumably at market open day 1) to sale (market close day 3), but it should be based on the whole portfolio. On the assumption you bought the same dollar value of each stock, the portfolio performance is the average of the performances of the three stocks, not the sum. If you didn't buy the same amount, you need to weight the change by how much of each stock you bought. If you bought the same amount of each stock, your return is $-1.38\%$. Your excess return is then $-1.16\%$
H: definition of winding number, have doubt in definition. could any one tell me why in the definition of index number or winding number of a curve $\gamma(t)$ around some point $a$ we take this integral : $$\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z-a} $$ why not $$\frac{1}{2\pi i}\int_{\gamma}\frac{1}{z^2+2z+1+e^z-a} $$? AI: It is a non-trivial fact from topology that given any curve $\gamma:[0, 1]\rightarrow\mathbb{C}\setminus\{a\}$ we can find a continuous polar coordinate expression about $a$ such that $$\gamma(t) = a + r(t)\cdot e^{i\theta(t)}$$ where $\theta$ and $r$ are continuous themselves (if the curve is differentiable, so are $r$ and $\theta$). Moreover $\theta$ and $r$ are uniquely defined up to multiples of $2\pi$ which is a freedom in the choice of the starting angle. The main consequence of the above is that the natural definition of the winding number should be $$\mathrm{Wnd}(a,\ \gamma) = \frac{\theta(1) - \theta(0)}{2\pi}$$ i.e. the total continuous angular change of the curve divided by $2\pi$. It should not be hard to convince yourself that this is always an integer for closed curves and represents the number of times $\gamma$ "winds" about $a$. It remains to relate this intuitive definition to our integral definition and the key component is the logarithm $$\log(z) = \ln\|z\| + i\theta(z)$$ where the imaginary component of the logarithm is essentially an angle tracking function, provided that we can make it vary continuous. (The logarithm is inherently discontinuous along some branch cut, so one remedy is the above continuous polar expression. Another is to piece together multiple integrals to make the argument vary continuously.) So we have \begin{align}\oint_\gamma \frac{1}{z-a} dz &= \int_0^1 \frac{r'(t)e^{i\theta(t)} + i\theta'(t)r(t)e^{i\theta(t)}}{r(t)e^{i\theta(t)}} dt\\ &=\int_0^1 \frac{r'(t)}{r(t)} + i\theta'(t) dt\\ &=\log(r(1))-\log(r(0)) + i\left[\theta(1) - \theta(0)\right]\end{align} Since the curve is closed, it follows that $r(1) = r(0)$ so the expression evaluates to $$i\left[\theta(1) - \theta(0)\right]= 2\pi i\cdot\mathrm{Wnd}(a,\ \gamma)$$ which is the justification for the integral definition.
H: natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ Please help me find the natural solutions for $9m+9n=mn$ and $9m+9n=2m^2n^2$ where m and n are relatively prime. I tried solving the first equation in the following way: $9m+9n=mn \rightarrow (9-n)m+9n=0 $ $\rightarrow m=-\frac{9n}{9-n}$ Thanks in advance. AI: $$mn=9n+9m \Rightarrow (m-9)(n-9)=81$$ This equation is very easy to solve, just keep in mind that even if $m,n$ are positive, $m-9,n-9$ could be negative. But there are only 6 ways of writing 81 as the product of two integers. The second one is trickier, but if $mn >9$ then it is easy to prove that $$2m^2n^2> 18mn > 9m+9n $$ Added Also, since $9\|2m^2n^2$ it follows that $3\|mn$. Combining this with $mn \leq 9$ and $m\|9n, n\|9m$ solves immediately the equation. P.S. Your approach also works, if you do Polynomial long division you will get $\frac{9n}{n-9}=9 +\frac{81}{n-9}$. Thus $n-9$ is a divisor of $81$. P.P.S. Alternately, for the second equation, if you use $2\sqrt{mn} \leq m+n$ you get $$18 \sqrt{mn} \leq 9(m+n)=2m^2n^2$$ Thus $$(mn)^3 \geq 81$$ which implies $mn=0 \text{ or } mn \geq 5$.
H: Seemingly similar but different probability games Burger King is currently running its "family food" game in which each piece can be modeled as a scratch off game where exactly one of three slots is a winner and you may only scratch one slot as your guess. As I was standing in line the other day I realized that their advertisement of "make it a large to double your chances of winning" (where large drinks/fries have two pieces on them) was actually not exactly true. The real probability of having at least one winning with two tickets is $\frac{5}{9}$ rather than $\frac{2}{3}$ which would really be double the chance. This is because the probability of both losing is $\frac{2}{3}\times\frac{2}{3}$ so the probability of at least one winning is $1-\frac{2}{3}\times\frac{2}{3}=\frac{5}{9}$. Now this was all very clear to me but while I sipped my strawberry banana smoothie, I wondered why this game of two three-slot tickets where each one has one winner and you can scratch one from each is different from the game of a single six-slot ticket where there are two winning slots and you get two scratches. The games must be different because the probabilities of winning are different. The six-slot game has $1-\frac{4}{6}\times\frac{3}{5}=\frac{3}{5}$ chance of getting at least one win. The two games seem the same to me intuitively. Can anyone explain how they are different? EDIT: I had noticed that in the two tickets, by making one guess, you actually eliminate 2 other possibilities with it so maybe this is the core of it, but I am still trying to see it more intuitively. AI: Your probability of winning on the first scratch is the same in each instance (this is the intuitive sense in which the two games are the same), so let's look at what happens if you don't win on the first scratch and want to win on the second scratch. This is where the games have to start differing. In the six-slot case, you know there are $2$ winners left out of a possible $5$ slots, so your probability of winning on the second scratch is $\frac{2}{5}$. In the pair of three-slot case, you still know that there are $2$ winners left out of a possible $5$ slots, except that the $5$ slots are divided into two groups: $2$ slots that you aren't allowed to scratch, and that contain $1$ winner, $3$ slots that you are allowed to scratch, and that contain $1$ winner. In other words, you aren't allowed to take advantage of the fact that there are $2$ winners left because you can't scratch the slots corresponding to one of them, and those slots are more likely to contain a winner than the ones you are allowed to scratch. This is where the games differ.
H: Learning to differentiate, with respect to $f_{\overline{z}}$ Let $R,\phi$ be real valued $C^1$ functions over a complex variable $z$. Consider the following function: $f=Re^{i\phi}$ . Where we define $f_{\overline{z}}= \frac{1}{2}(f_x+if_y)$ I want to differentiate this function with "respect to $\overline{z}$". I proved that this rule of derivation also has the property on the multiplication. $(fg)_\overline{z}=fg_{\overline{z}}+gf_{\overline{z}}$ So: $f_{\overline{z}}=R_\overline{z}e^{i\phi}+R(e^{i\phi})_{\overline{z}}$ But I don't know how to differentiate: $(e^{i\phi})_{\overline{z}}=\frac{1}{2}((e^{i\phi(z)})_{{x}}+(e^{i\phi(z)})_{{y}})$ I don't know what can I do in this step, Even if I view the function as a function over two real variables $(x,y)=z$ :/ Please help me AI: Hint - try using: $$e^{i\phi} = \cos(\phi) + i\sin(\phi)$$
H: A "Matrix Trigonometry" $e^X$ for matrix $X$ is defined as an always-converging taylor series (provided that $X$ is a $n \times n $ complex matrix): $$e^X:=\sum_{k=0}^{\infty}\frac{X^k}{k!} $$ A thought occurred to me that we might as well define $\cos(X):=\frac12(e^{iX}+e^{-iX})$ and $\sin(X):=\frac1{2i}(e^{iX}-e^{-iX})$. Now some obvious questions arise: Is there a generalization for $2\pi$, the period of sine and cosine? Perhaps the best way to do so is to generalize the Euler's Identity $e^{2i\pi}=1$; Is there matrix $T$ such that $e^T=1$? This implies that $\cos (X+T)=\cos (X), \sin(X+T)=\sin (X)$. A simple calculation shows that $\cos^2(X)+\sin^2(X)=I$. Can we generalize other trigonometric identities any further? Can this concept be used further to derive some useful results? My senses tell me this should find its place in applied mathematics. If there's any previous reference (which I think is likely) please inform me. AI: Yes, any diagonalizable matrix $T$ whose eigenvalues are integer multiples of $2 \pi i$ has this property, but this does not imply the identity you want in general unless $X$ and $T$ commute. So in particular $T$ can be scalar. Yes, any trigonometric identity which is a consequence of a polynomial identity between expressions of the form $e^{ix}$ holds also for commuting matrices. Really the useful operation is the matrix exponential (solving differential equations, relating Lie algebras and Lie groups, etc.) and everything else is just derived from it. But a keyword you might be interested in is "functional calculus."
H: Can the dimension of the image of a linear map "increase"? Suppose we have a linear transformation $f: V \to V$. How is it possible that $\dim(\operatorname{im}(f \circ f))$ is larger than $\dim(\operatorname{im}(f))+\dim(\operatorname{im}(f)) - \dim(V)$? It's a question on a past exam, so there should be an example that proves this. AI: For example, take $f$ on ${\mathbb R}^2$ given by the matrix $\pmatrix{1 & 0\cr 0 & 0\cr}$. Then $im(f \circ f) = im(f) = {\mathbb R} \times \{0\}$, and $1 > 1 + 1 - 2$. I don't know what you mean by "increase" in the title of this question, there's nothing increasing here.
H: Prove that $2^{pq} - 1 = (2^p - 1)(\sum^{q-1}_{i=0} 2^{pi})$ for two natural numbers $p, q$ How can I prove the below? $$2^{pq} - 1 = (2^p - 1)\left(\sum^{q-1}_{i=0} 2^{pi}\right)$$ for two natural numbers $p, q$ It looks like I need proof by induction? But how? There's two variables? AI: Setting $a=2^p$ then $$ 2^{pq}-1=a^q-1=(a-1)\sum_{i=0}^{q-1}a^i=(2^p-1)\sum_{i=0}^{q-1}(2^p)^i=(2^p-1)\sum_{i=0}^{q-1}2^{pi}. $$
H: Can the completion of a domain be a non-domain? Suppose $R$ is a domain, finitely generated over an algebraically closed field, and $\mathfrak{m}\subset R$ is a maximal ideal. Is $\underleftarrow{\lim} R/\mathfrak{m}^n$ necessarily a domain? AI: This phenomenon of posing a question in the title and the posing its negation in the body is very curious to me. Anyway, it is possible for the completion of a domain to be a non-domain. Let $R = k[x, y]/(y^2 - x^3 - x^2)$ where $k$ is a field of characteristic not equal to $2$ and let $\mathfrak{m} = (x, y)$. In the completion $\sqrt{1 + x}$ exists (e.g. by Hensel's lemma, or more explicitly by the binomial theorem) and consequently the defining polynomial admits a factorization $$(y - x \sqrt{1 + x})(y + x \sqrt{1 + x})$$ so the completion has a zero divisor. This corresponds geometrically to the fact that $y^2 = x^3 + x^2$ has a singularity at the origin (the curve looks locally like an intersection of a pair of lines and this can be detected by completing with respect to $\mathfrak{m}$). For a number-theoretic example let $R = \mathbb{Z}[x]/(x^2 + 3)$ and let $\mathfrak{m} = (2, x + 1)$. In the completion $\sqrt{1 - 4}$ exists in $\mathbb{Z}_2$ (again by either Hensel's lemma or the binomial theorem), so the same argument as above produces a zero divisor. Morally speaking this is for the same reason as above: $\text{Spec } R$ is singular at $\mathfrak{m}$.
H: What preposition to use when fitting data? Please could some real experts give an opinion on this question on English.SE: What preposition to use when fitting data? Do we fit data with, by, or as a linear function? Forgive me as I do not know how best to highlight this and I'm happy for this question be be deleted. AI: One does no such thing. One fits a linear function to data, not the other way around. Unless of course one is inventing "data" that will nearly fit a preconceived pattern, a practice not entirely unknown in Science.
H: Test the convergence of $\sum_{n=0}^{\infty} \frac{n^{k+1}}{n^k + k}$ Problem: Test the convergence of $\sum_{n=0}^{\infty} \frac{n^{k+1}}{n^k + k}$, where $k$ is a positive constant. I'm stumped. I've tried to apply several different convergence tests, but still can't figure this one out. AI: Hint $$ \frac{n^{k+1}}{n^k +k} =n \frac{1}{1+\frac{k}{n^k}}$$ What happens when $n \to \infty$?
H: Find out the dimensions of the rectangular parallelepiped The volume of a rectangular parallelepiped is given as 144 cc. Its surface area is given as 192 sq cm. And its corner to corner diagonal is given as 13 cm. How do I find out the three sides. I have assumed the sides to be $a,b,c$. Now, $a^2+b^2+c^2=169, 2(ab+bc+ca)=192$, and $abc=144$. How do I solve the three equations without forming a cubic equation? (This is a class 9 problem so i am not suppoesd to use solution to cubic equations.) AI: The quickest way is just to reacall pythagorean triples $(3,4,5)$ and $(5,12,13)$ due to 5 being common to both triples and 13 being the diagonal length thus establishing that 3, 4,and 12 satisfy $3^2+4^2+12^2=13^2$ as desired for the correct diagonal length. Notice that the product of these three numbers in the triples is 144 and they give the correct surface area.
H: Can the coefficients of a given term in this family of power series have a common divisor? Let $g_m(b)$ be the coefficient of $x^m$ the power series $$\dfrac{1}{1-x-bx^2}$$ (When $b=1$, this is just the generating function of Fibonacci numbers.) Of course, $g_m(b)$ depends on both $m$ and $b$. Now fix $m$, and let $b$ run from 1 to $\infty$. Can the $g_m(b)$'s have a non-trivial common divisor for some $m$? It seems reasonable to believe this cannot happen, but I don't see how to prove it. The relation among the $g_m(b)$'s for a fixed $b$ is simple. But for a fixed $m$, I don't see how to relate the $g_m(b)$'s. AI: No. Note that $g_m$ has constant term $1$. If all values of $g_m(b)$ have a common divisor $d$, then $d \| g_m(d)$, but $g_m(d) \equiv g_m(0) = 1 \bmod d$. Initially I wanted to resolve this problem using the following lemma, which is unnecessary. But I thought I would keep it here in case anyone was interested. Lemma: Let $f(x)$ be a polynomial which attains integer values when $x$ is an integer. If $d$ is an integer such that $d \| f(x)$ for all positive integers $x$, then $d \| f(x)$ for all integers $x$. Proof. We need to observe the following facts about the finite difference operator $\Delta f(x) = f(x + 1) - f(x)$. $d \| f(x)$ for all positive integers $x$ if and only if $d \| f(1)$ and $d \| \Delta f(x)$ for all positive integers $x$. $d \| f(x)$ for all integers $x$ if and only if $d \| f(1)$ and $d \| \Delta f(x)$ for all integers $x$. If $f$ is a polynomial, then $\Delta f$ is a polynomial of degree $\deg f - 1$. It follows that $d \| f(x)$ for all positive integers $x$ if and only if $$d \| f(1), \Delta f(1), \Delta^2 f(1), ... \Delta^{\deg f} f(1)$$ and $d \| \Delta^{\deg f+1} f(x)$ for all positive integers $x$. But $\Delta^{\deg f+1} f$ is identically zero, hence $d \| \Delta^{\deg f+1} f(x)$ for all integers $x$, hence $d \| f(x)$ for all integers $x$. Incidentally, these polynomials are closely related to the Chebyshev polynomials of the second kind.
H: What is the maximum number of consecutive composite numbers possible? Can anyone tell me what is the maximum number of consecutive composite numbers possible? I mean can I get say 1000 consecutive natural numbers. Is there any general theorem that when I have a n-digit number there will always be p consecutive composite numbers? AI: Let $p_1,p_2, p_3,\dots,p_n$ be the first $n$ primes, and let $P_n$ be their product. Then the $p_{n+1}-2$ consecutive integers $P_n+2,P_n+3, \cdots, P_n+(p_{n+1}-1)$ are all composite. For let $P_n+x$ be one of these numbers. Since $2\le x\lt p_{n+1}$, $x$ is divisible by some prime $p\le p_n$ ($x$ could itself be prime). But $P_n$ is also divisible by $p$, so $P_n+x$ is divisible by $p$. Clearly $P_n+x\gt p$, so $P_n+x$ is composite. We can in general get a very slightly cheaper string by starting at $P_n-2$ and going backwards. These procedures get us arbitrarily long strings of consecutive composites, since there are infinitely many primes. But one can do a lot better than $P_n$ in general. The subject of Prime Gaps has been extensively studied. You will find detailed information in this Wikipedia article.
H: On the number of possible solutions for a quadratic equation. Solving a quadratic equation will yield two roots: $$\frac{-\sqrt{b^2-4 a c}-b} {2 a}$$ and: $$\frac{\sqrt{b^2-4a c}-b}{2 a}$$ And I've been taught to answer it like: $$\frac{\pm\sqrt{b^2-4a c}- b}{2 a}$$ Why does it yields only two solutions? Aren't there infinite solutions for that? Is there a proof on the number of possible solutions for a quadratic equation? AI: One can see that there are only two solutions from the way it is solved. If $ax^2+bx+c=0$ with $a\neq 0$, it follows that $ax^2+bx=-c$, so that $x^2+\frac{b}{a}x=-\frac{c}{a}$. Trying to complete the square, one obtains $(x+\frac{b}{2a})^2=x^2+\frac{b}{a}x+(\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}=\frac{b^2-4ac}{4a^2}$. Taking square roots, the result follows.
H: Proof of $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$ For positive real numbers $a$, $b$ and $c$, how do we prove that: $$(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)>\frac{(ab+bc+ca)^3}{3}$$ AI: By Cauchy-Schwarz: $(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2) \ge (a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2$ By the power mean inequality: $(a^{1.5}b^{1.5}+b^{1.5}c^{1.5}+c^{1.5}a^{1.5})^2 \ge \frac{(ab+bc+ca)^3}{3}.$ (Note: The > sign in your post is false; let $a=b=c.$)
H: Distance Formula in Three Dimensions The distance formula in 3-D space is defined as: $$\|P_1\, P_2\| = \sqrt{(x_2- x_1)^2 + (y_2 -y_1)^2 + (z_2- z_1)^2}$$ My question is that if I have 2 points that have negative coordinates, do I have to use the absolute value on all the points? For example my two points are $\,P(3, -2, -3)\,\,,\,\, Q(7,0,1)$ When I plug in the y and z values of P do I have to use positive 2 and positive 3? Thanks in advance. AI: Do it exactly like it says: $7-3=4$, $0-(-2)=2$, $1-(-3)=4$, so the distance is $$\sqrt{4^2+2^2+4^2}.$$ This turns out to be $\sqrt{36}$, which is $6$. Another example: Suppose that $(x_1,y_1,z_1)=(-11, 9, -4)$ and $(x_2,y_2,z_2)=(-3, 5,6)$. Then $x_2-x_1=(-3)-(-11)=8$, $y_2-y_1=5-9=-4$, and $z_2-z_1=6-(-4)=10$, so the distance is $\sqrt{8^2+(-4)^2+10^2}$, which is $\sqrt{180}$. As usual, one has to take care with the minus signs, since slips are all too easy to make.
H: If $x$ and $y$ are prime , which of the following cannot be their sum The question is If $x$ and $y$ are prime numbers , which of the following can not be their sum? $5$,$9$,$13$,$16$ or $23$. The answer is $23$. How did they get this? As far as I can tell is that when prime numbers are added I am suppose to get an even value for example $5+7 = 12 $ or $7+7 = 14 $ or $7+11 = 18 $. Could anyone please tell me what I am missing here? AI: All of the odd answers must have 2 as one of things you are adding. Otherwise, adding two primes always ends up as an even number. You can easily get 2 and 3, 2 and 7, 2 and 11. Now when you do 2 and 23, you see that the other number to add is 21. 21 is not prime, so 23 is your answer . Basically, the methodology is to to take all the odd numbers and subtract 2 from it. If the result is not prime, then that will be your answer. If all odd numbers can be produced by 2 + a prime, then you'll have to guess and check for the even numbers.
H: How to determine if a 3D triangle given by points is a right triangle? How do I figure out if a triangle is a right triangle in 3-D space if you are given three points: $P = (3, -2, -3)$, $Q = (7, 0, 1)$, $R = (1, 2, 1)$? I know that it is an isosceles triangle (two sides are the same $6$ units). But how would I now calculate this to see if it is a right triangle? AI: Compute the length of the three sides and then use Pythagoras Theorem to check if there is a right angle.
H: Does an orientable subbundle of an orientable vector bundle always have a orientable complement? If I have an orientable vector bundle $E$ and a subbundle $F$ on a manifold $M$, where both the bundles are orientable, does $F$ have a complement in $E$ which is also orientable? Does it have a complement bundle at all? That is, a subbundle of $E$ that is pointwise a complement of $F$. What if $F$ is of codimension $1$? Is the complement always trivial in this case? Does anything change if $E$ is specifically the tangent bundle $TM$? AI: If $E=F\oplus G$, then $\Lambda^\det E=\Lambda^\det F\otimes\Lambda^\det G$. If both $\Lambda^\det E$ and $\Lambda^\det G$ are trivial, then so is $\Lambda^\det F$, because the three being line bundles, you can «divide by $\Lambda^\det G$» in that equality.
H: Help understanding a step in the proof of Jordan's Lemma I don't understand this step on the Wikipedia article on Jordan's Lemma: $$ \begin{align} I_R:=\biggl\|\int_{C_R} f(z)\, dz\biggr\| &\le R\int_0^\pi\bigl\|g(Re^{i\theta})\,e^{aR(i\cos\theta-\sin\theta)}\,ie^{i\theta} \bigr\|\,d\theta\\ &=R\int_0^\pi \bigl\|g(Re^{i\theta})\bigr\|\,e^{-aR\sin\theta}\,d\theta\,. \end{align} $$ AI: The very last step follows from $$\|i\|=\|e^{ix}\|=1\,\,,\,\,x\in\Bbb R$$
H: The intersection of two conics - matrix solution I recently had to compute the intersection of two conics, and found it to be a long and complicated procedure. Looking at Wikipedia, the advice to find the points of intersection of two conics in general, which is echoed in the answer to this question, is to write the conics in matrix form, say $C_1$ and $C_2$, and then solve $\det( \lambda C_1+\mu C_2)=0$ for values of $\lambda$ and $\mu$, then consider the degenerate conic $C_0=\lambda C_1+\mu C_2$ and the intersection of this with your original conics will be easier to compute (and coincides with the intersection of the two conics originally). My question is: originally we were looking for the solutions of two degree 2 polynomials in two variables. How is solving $\det(\lambda C_1 + \mu C_2)=0$, a degree three polynomial in two variables, any easier (in general)? Certainly there are specific matrices $C_1$ and $C_2$ for which this is quite obvious, but in general it does not seem to me as though we have made our problem simpler. AI: It's not "really" a polynomial in two variables: it's homogeneous in $\lambda$ and $\mu$, so really it behaves like a polynomial in one variable $\frac{\lambda}{\mu}$. The use of the homogeneous polynomial is that we don't have to artificially exclude the case that $\mu = 0$. The essential point is the following: you want to compute the set of points satisfying $x^T C_1 x = x^T C_2 x = 0$. This is the same as computing the set of points satisfying $$x^T (a C_1 + b C_2) x = x^T (c C_1 + d C_2) x$$ for any $a, b, c, d$ such that $ad - bc \neq 0$. Consequently it is potentially possible to simplify the problem a lot by choosing $a, b, c, d$ carefully and this determinant calculation is just one way to do that. (Also it generalizes to conics in higher dimensions.)
H: Simplify the complex equation For the function $$G(w) = \frac{\sqrt2}{2}-\frac{\sqrt2}{2}e^{iw},$$ show that $$G(w) = -\sqrt2ie^{iw/2} \sin(w/2).$$ Ive been told to use the equation below by use of the complex sine definition, but i would like to know the exact steps to get there and then how to simplify it to get my result. $-\sqrt{2}ie^{i\frac{w}{2}}\sin\frac{w}{2} =-\sqrt{2}ie^{i\frac{w}{2}}(\frac{1}{2i}(e^{i\frac{w}{2}} - e^{-i\frac{w}{2}})) $ AI: $$\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}e^{iw} = \frac{\sqrt{2}}{2}(1 - e^{iw})$$ then pull out $e^{\frac{iw}{2}}$ and you get $$ = \frac{\sqrt{2}}{2}e^{\frac{iw}{2}}\left(e^{-\frac{iw}{2}} - e^{\frac{iw}{2}}\right) $$ then pull out one $(-i)$ and push the $\frac{1}{2}$ inside. "Pull out one $(-i)$" might seem slightly weird since there was no real factor of $-1$ or $i$ multiplied in the entire expression, so what were really doing is multiplying by $1$ in a very convenient way. The reasoning behind multiplying by $1$ in this way is that we know the formula for complex sin ($\sin(x) = \frac{e^{ix} -e^{-ix}}{2i}$) and this clever multiplication allows us to get it. So then we have $$ = -\sqrt{2}ie^{\frac{iw}{2}}\left(\frac{e^{\frac{iw}{2}} - e^{-\frac{iw}{2}}}{2i}\right) $$ so now since in our last step we set up our equation to have the formula for complex sine, we'll now substitute the complex sine and we have $$= -\sqrt{2}ie^{\frac{iw}{2}}\sin\left(\frac{w}{2}\right) $$
H: Finding a paper by John von Neumann written in 1951 There's a 1951 article by John von Neumann, Various techniques used in connection with random digits, which I would really like to read. It is widely cited, but I can't seem to find an actual copy of the paper, be it free or paying. Is there a general strategy to find copies of relatively old papers like this one? EDIT: I've searched quite a lot before posting this question and fond the following reference: Journal of Research of the National Bureau of Standards, Appl. Math. Series (1951), 3, 36-38 Unfortunately, my library doesn't have it, and it is not in NIST's online archive (neither at http://www.nist.gov/nvl/journal-of-research-past-issues.cfm nor at http://nistdigitalarchives.contentdm.oclc.org/cdm/nistjournalofresearchbyvolume/collection/p13011coll6) AI: One of the citations gives the bibliographic info, von Neumann J, Various Techniques Used in Connection with Random Digits, Notes by G E Forsythe, National Bureau of Standards Applied Math Series, 12 (1951) pp 36-38. Reprinted in von Neumann's Collected Works, 5 (1963), Pergamon Press pp 768-770. That should be enough information for any librarian to find you a copy.
H: need help understanding Cauchy-Riemann criteria I don't understand why both limits have to be equal in both directions. $$\lim_{ \Delta x \rightarrow 0} {\Delta u+i\Delta v \over \Delta x} \|_{\Delta y = 0} = \lim_{\Delta y \rightarrow 0}{\Delta u+i\Delta v \over i\Delta y}\|_{\Delta x = 0} $$ Shouldn't the limit shift in which direction $x ,y$ moves on? AI: Think of an analogy with the existence of limit of a function of one real variable. You need both the left and right hand limit to exist and be equal to the value of the function at that point. This is because there are two directions of approaching the particular point. Now, in case of a function of one complex variable, you have directions of approach in $x$ and $y$ directions in the complex plane. So you need to consider both $x$ and $y$ limits.
H: Why is the last step of proof by induction necessary? Why do we assume the condition holds up to certain number $n$ and prove it holds for $n+1$? Is there any example where something holds up to $n$ but fails for $n+1$? AI: If I understand you correctly this is a confusion about how the variable n is quantified over. The induction step is "for all n (if P(n) then P(n+1))". The induction step is NOT "if (for all n (P(n)) ) then (for all n (P(n+1)))." You are right that the second statement is trivially true for any property P.
H: inequality fraction I saw this problem in an Australian maths olympiad: $6/10 < a/b < 10/15$ The problem asked for the lowest possible value of $b\in \mathbb{Z}$. I tried manipulating but couldnt derive one of their answers. AI: That's how I did it. First of all, note that \begin{equation} \frac{11}{15} - \frac{7}{10}= \frac{1}{30} \end{equation} thus b=31 certainly works, i.e. is a higher bound to the answer. Then notice that $b$ cannot be any divisor of 10 or 15, as $\frac{11}{15} < \frac{8}{10}$ and $\frac{10}{15}< \frac{7}{10}$. It remains to check a few values. As $\frac{7}{10}=0.70$ you are looking for a fraction which is 'slightly more' than $0.70$. $b=4$ does not work as $\frac{3}{4} > \frac{11}{15}$ and $b=6$ does not work either because we know $\frac{4}{6} < 0.70$ and $\frac{5}{6}$ is clearly too big. Now check $b=7$ and you find $\frac{5}{7}$ is in the required range, so $b=7$ is your answer.
H: how do we know that integral is non-elementary? Possible Duplicate: How can you prove that a function has no closed form integral? Is there a condition that states that the indefinite integration is non-elementary? AI: There is a decision procedure called the Risch algorithm that will either tell you that the intergral is non-elementary, or produce an elementary anti-derivative. It is not an easy algorithm to execute, or even implement in a computer algebra system (although the latter has been done), so there is no hope of finding an easy condition for the existence of an anti-derivative.
H: expected value properties I'm reading an algorithm proof in which there is a passage I can't understand. Substantially we have 2 statistically independent random arrays $X$ and $Z$, and a third random array $Y$. It's known that $Y=XZ$ and $E[Z] = 1$. What the author provides are textually the following passages: \begin{align} VAR[Y] &= E[(Y -E[Y])^2]\\ &= E[(X(Z-1) + (X-E[X]))^2]\\ &= (VAR[X]+E[X^2])VAR[Z]+VAR[X] \end{align} where $E[X]$ and $VAR[X]$ are the expected value (i.e. mean) and the variance of $X$, respectively. Trying to understand that proof, I'm getting a different result, as it is: \begin{align} VAR[Y] &= E[(Y -E[Y])^2]\\ &= E[(XZ - E[XZ])^2]\\ &= E[(XZ -X +X -E[X])^2]\\ &= E[(X(Z-1) + (X-E[X]))^2]\\ &= E[X^2(Z-1)^2 + (X - E[X])^2 + 2X(Z-1)(X - E[X])]\\ &= E[X^2]E[(Z-1)^2] + VAR[X] + 2E[X(Z-1)(X - E[X])]\quad \text{(then, as $E[Z] = 1$)}\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2Z - ZXE[X] - X^2 + XE[X]]\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2Z] - 2E[ZXE[X]] - 2E[X^2] + 2E[XE[X]]]\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2]E[Z] - 2E[XE[X]]E[Z] - 2E[X^2] + 2E[XE[X]]]\\ &= E[X^2]VAR[Z] + VAR[X] + 2E[X^2] - 2E[XE[X]] - 2E[X^2] + 2E[XE[X]]]\\ &= E[X^2]VAR[Z] + VAR[X] \neq (VAR[X]+E[X^2])VAR[Z]+VAR[X] \end{align} Can you help me to understand where am I wrong? Thank you very much in advance for your attention. EDIT: This is the original author proof AI: I don't see in what sense $X$, $Y$ and $Z$ are arrays, especially given $E[Z]=1$; I'll assume that they're just random variables. (Also where you write "pass" it seems you mean "passage".) Your calculation is correct. You can check this using a simple example, for instance $X$ and $Z$ both uniformly drawn from $\{0,2\}$. You could considerably shorten your calculation by noting that the expected value of any term of the form $f(X)(Z-1)$ vanishes; that gets rid of the mixed term that the last few lines focus on.
H: Can a transformation matrix be expressed in terms of the vector to be transformed? I'm currently learning linear algebra with my friend via an online course, and we have a disagreement that we would like settled. Upon learning that vectors can be projected onto lines by a simple function, and that this function is a linear transformation, I recommended that we find a way to calculate the transformation matrix for the function. This function was characterized by a vector $\vec{v}$ pointing along the line to be projected onto. The non-matrix form of the function we were given was this: $$proj_{\vec{v}}(\vec{x}) = (\vec{x} \cdot \hat{u})\hat{u}$$ where $\hat{u}$ is the normalized form of $\vec{v}$ calculated by ($\frac{1}{\|\|\vec{v}\|\|}\vec{v}$). When trying to construct this as a matrix, my friend came up with this (this is just for $\mathbb{R}^{3}$, but you get the idea): $$proj_{\vec{v}}(\vec{x}) = \left[\begin{array}{ccc} \vec{x} \cdot \hat{u} & 0 & 0 \\ 0 & \vec{x} \cdot \hat{u} & 0 \\ 0 & 0 & \vec{x} \cdot \hat{u} \end{array}\right] \cdot \hat{u}$$ I see multiple problems with this. First of all, transformation matrices cannot be expressed in terms of the vector to be transformed, can they? To my understanding, they are supposed to contain constant values that are the same no matter what vector is being transformed. Second of all, linear transformations multiply $\vec{x}$ by the transformation matrix, not any other vector, right? I'm not particularly interested in the actual transformation matrix for this problem - I plan to get some practice with what i've learned by calculating that :) - I'd just like confirmation as to whether or not the things my friend has done are valid. Thank you! AI: The short answer is: You are right. The representing matrix has to be independent of $x$. Moreover a matrix $A$ represents a linear function $f$ if and only if $Ax=f(x)$ for all $x$, so $A$ has indeed to be multiplied with the vector in question. Even though you didn't ask for it you can write: $$(x\cdot u)u=u(x\cdot u)=u(u\cdot x)=u(u^tx)=(uu^t)x.$$ So $uu^t$ is the representing matrix. (I write the representing matrix, since the basis is understood to be fixed. In genereal there is more than one representing matrix.)
H: How many students turned up for renting the rooms? Sara has a house which she wants to convert to a hostel and rent it out to students of a nearby women’s college. The house is a two story building and each floor has eight rooms. When one looks from the outside, three rooms are found facing North, three found facing East, three found facing West and three found facing South. Expecting a certain number of students, Sara wanted to follow certain rules while giving the sixteen rooms on rent: All sixteen rooms must be occupied. No room can be occupied by more than three students. Six rooms facing north is called north wing. Similarly six rooms facing east, west and south are called as east wing, west wing and south wing. Each corner room would be in more than one wing. Each of the wings must have exactly 11 students. The first floor must have twice as many students as the ground floor. However Sara found that three fewer students have come to rent the rooms. Still, Sara could manage to allocate the rooms according to the rules. How many students turned up for renting the rooms? 1) 24 2) None of these 3) 27 4) 30 5) 33 AI: The answer is c) $27$; this can be shown by obtaining bounds on the number of students. There is at least one student in each room, and thus at least $8$ on the ground floor, $16$ on the first floor, and $6$ per wing. $8$ students on each floor are spread out to occupy all rooms, but on the first floor there are at least another $8$ students to be distributed. If we want to have as few students as possible while fulfilling the requirement that each wing has $11$, we should put the extra students on the first floor in the corner rooms, where they count more towards the wing requirement. There's space for exactly $8$ students left, so we can fill the corner rooms, but that still only makes $10$ students per wing. Thus we have to add at least one more student in a ground-floor corner room and two more students in first-floor edge rooms. That fulfills all the requirements, so the least number of students is $27$. If we want to have as many students as possible while fulfilling the requirement that each wing has $11$, we should put the extra students on the first floor in the edge rooms, where they count less towards the wing requirement. There's space for exactly $8$ students left, so we can fill the edge rooms, which gets us up to $8$ students per wing. That leaves a total of $4\cdot3=12$ of the wing requirement. If we add more students, one can go in an edge room on the ground floor, but two more have to go on the first floor, and these can now only go in corner rooms. Thus each group of $3$ students added uses up another $5$ of the wing requirement, and since there was only $12$ left, we can add at most two groups of $3$ students, which only gets us up to $30$ students. Thus the minimum number of students is $27$ and the maximum is $30$.
H: about the differentiability : the general case Let $U$ be an open set in $\mathbb{R}^{n}$ and $f :U \subset \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ be a given function. We say that $f$ is differentiable at $x_{0}\in U$ if the partial derivatives of $f$ existi at $x_{0}$ and if $$\displaystyle \lim_{x \rightarrow x_{0}} \frac{\\|f(x)-f(x_{0})-T(x-x_{0})\\|}{\\|x-x_{0}\\|}=0$$ where $T=Df(x_{0})$ is the $ m \times n$ matrix with elements $\displaystyle \frac{\partial f_{i}}{\partial x_{j}}$ evaluated at $x_{0}$ and the $T(x-x_{0})$ means the product of $T$ with $x-x_{0}$ (regarded as a column matrix). We call $T$ the derivative of $f$ at $x_{0}$. Now I consider a particular case($m=n=2$) $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ Following the definition I obtain : $$\displaystyle f(a,b)-f(a_{0}, b_{0})- \begin{pmatrix}\frac{\partial f_{1}}{\partial a} & \frac{\partial f_{1}}{\partial b}\\ \frac{\partial f_{2}}{\partial a} & \frac{\partial f_{2}}{\partial b} \end{pmatrix} \begin{pmatrix}a-a_{0} \\ b-b_{0} \end{pmatrix} = f(a,b)-f(a_{0},b_{0})- \begin{pmatrix}\frac{\partial f_{1}}{\partial a} \cdot(a-a_{0})+\frac{\partial f_{1}}{\partial b}\cdot(b-b_{0})\\ \frac{\partial f_{2}}{\partial a}\cdot(a-a_{0})+\frac{\partial f_{2}}{\partial b}\cdot(b-b_{0}) \end{pmatrix}$$ where $f(a,b)=(f_{1}(a,b),f_{2}(a,b))$. My question is : How can I compute this limit because the last element is a matrix and first two aren't. And why $f(x)-f(x_{0})-T(x-x_{0})$ I have to put into $\\| \\|$ ? I have an idea why I have to put into norm but I'm not sure and can you give an concrete example how I compute the limit - for example $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2, f(a,b)=(a^2+b^2, a^2+b^2)$ when $ (a,b) \rightarrow (1,1)$. Thanks :) AI: Let me compute for the example, $f(x,y)=(x^2+y^2, x+y)$. We write $f_1(x,y)=x^2+y^2$ and $f_2(x,y)=x+y$. Then $$ \frac{\partial f_1}{\partial x}(x,y) = 2x,\quad \frac{\partial f_1}{\partial y}(x,y) = 2y,\quad \frac{\partial f_2}{\partial x}(x,y) = 1,\quad \frac{\partial f_2}{\partial y}(x,y) = 1. $$ Since these four functions are continuous then $f$ is differentiable. THEN $T$ is already the Jacobian matrix: $$ T=\begin{pmatrix} 2x& 2y\\ 1&1\end{pmatrix}. $$ Finally $$ f(x,y)-f(x_0,y_0)-T\begin{pmatrix}x-x_0\\y-y_0\end{pmatrix}= \begin{pmatrix}x^2+y^2-x_0^2-y_0^2-2x_0(x-x_0)-2y_0(y-y_0) \\ x+y-x_0-y_0 -(x-x_0+y-y_0)\end{pmatrix}, $$ taking limits $$ \lim_{(x,y)\to(x_0,y_0)} \frac{(x-x_0)^2+(y-y_0)^2}{\sqrt{(x-x_0)^2+(y-y_0)^2}}= \lim_{(x,y)\to(x_0,y_0)} \sqrt{(x-x_0)^2+(y-y_0)^2}=0. $$
H: Continuity of function $\lim_{t \rightarrow x} f(t)$ Let a function $f:A \rightarrow \mathbb{R}$, defined on a set $A\subset \mathbb{R}$ without isolated points, has a finite limit $g(x)$ in each point $x\in A$. I wish to show that $g: A \rightarrow \mathbb{R}$, $g(x)=\lim_{t \rightarrow x} f(t)$, is continuous. AI: Let $x$ in $A$ and $\varepsilon\gt0$. By definition of $g(x)$ as a limit, there exists $\eta\gt0$ such that every $z$ in $A$ such that $\|z-x\|\lt2\eta$ and $z\ne x$ is such that $\|f(z)-g(x)\|\leqslant\varepsilon$. Let $x'$ in $A$ such that $\|x'-x\|\lt\eta$ and $x'\ne x$. Then every $z$ in $A$ such that $\|z-x'\|\lt\min\{\|x'-x\|,\eta\}$ and $z\ne x'$ is such that $z\ne x$ and $\|z-x\|\lt2\eta$, hence $\|f(z)-g(x)\|\leqslant\varepsilon$. In particular, the limit $g(x')$ of $f(z)$ when $z\to x'$ with $z$ in $A$ and $z\ne x'$, is such that $\|g(x')-g(x)\|\leqslant\varepsilon$. This is valid for every $x'$ in $A$ such that $\|x'-x\|\lt\eta$ and $x'\ne x$, hence $g$ is continuous at $x$.
H: If $\sim$ is an equivalence relation on $X$, and there is a strict total order on $X/\sim$, what kind of ordering does $X$ have? I would like to know if there's a special name for this kind of ordering. When I say there is a strict total order on $X/\sim$, what I mean is that two distinct elements in the same equivalence classes are considered equal, but an element of one equivalence class is either less than or greater than an element of another equivalence class. In other words, what kind of ordering does the set of poker hands have? For example, one equivalence class is the set of straight flushes with a high card of 10 (i.e., 6 through 10 of the same suit). Any two from these four hands are considered equal. However, a hand from this set is greater than any poker hand that's less than a straight flush (e.g. four of a kind, straight flush, etc.) or a straight flush with a high card less than 10, and less than any straight flush with high card greater than 10. AI: We say that $R$ is a preorder of $X$ if $R$ is reflexive and transitive. Then equivalence relation $x\sim y\iff xRy\land yRx$ has the property that $X/\sim$ is partially ordered. If the partial order is a total order, then $R$ is a total preorder. The inverse holds too, if $X/\sim$ is totally ordered by $\leq$ we define $xRy\iff x/\sim\leq y/\sim$. It is a nice exercise to see that this is a total preorder of $X$.
H: Convergence of a power series when radius of convergence is 1 Suppose radius of convergence of $\sum c_n z^n$ is 1. ($c_n, z \in \mathbb{C}$) (i)Here, if $\{c_n\}$ is monotonically decreasing and $\lim c_n = 0$, then $\sum c_n z^n$ converges at every point on the circle $\|z\|$, except possibly at $z=1$. (Of course, $c_n$ is assumed to be non-negative real here) (ii) $\{\|c_n\|\}$ is monotonically decreasing and $\lim c_n = 0$, then $\sum c_n z^n$ converges at every point on the circle $\|z\|$, except possibly at $z=1$. ($c_n \in \mathbb{R}$) (iii) Same as (ii), but $c_n \in \mathbb{C}$ I know the statement (i) is true. However, are (ii) and (iii) false? I think (ii) is at least true. Please give me a proof if it true, otherwise give me a counterexample. Help! AI: That (ii) is false is implied by the exception in (i). For any series according to (i) that doesn't converge at $z=1$, e.g. $c_n=1/n$, change the signs of the odd coefficients to obtain a series according to (ii) that doesn't converge at $z=-1$.
H: uniform sampling of the sphere w.r.t. inner products other than the usual one I will have a small (usually 2 or 3, probably no bigger than 6) positive integer $n$ and symmetric positive-definite real $n$-by-$n$ matrices $A$, and will want to sample an $\:\mathbf{R}^n$-valued$\:$ random variable $\mathbf{x}$ such that for any/all linear map(s) $\: f : \mathbf{R}^n \to \mathbf{R}^n \:$ such that $\Big[$for all vectors $v$, $\;\;v^{\hspace{0.015 in}T} Av\:=\:\left(\|\|f(v)\|\|_2\right)^2\;\;\Big]$, $\big[\|\|f(\mathbf{x})\|\|_2 = 1 \:$ almost surely$\big]$ and $\big[f(\mathbf{x})$ is distributed uniformly on the unit sphere$\big]$ . The way I thought of to do this is to Gram-Schmidt orthonormalize the standard basis with respect to the inner product induced by $A$, and then take the linear combination with coefficients given as the entries of a uniform sample from the unit sphere. $\:$ However, since I will be sampling from $\mathbf{x}$ only once for each $A$, I'm hoping for a rather more efficient method. Is there any faster way to do the sampling I want? (The matrices will be changing somewhat slowly, which might help.) AI: I think we can dispense with the "all linear maps" part, since all such linear maps are given by one such map followed by an arbitrary rotation of the unit sphere, which doesn't affect the uniform distribution; thus we can focus on a particular such map, $f(x)=A^{1/2}x$. The problem is that you can easily apply $A$, but not $A^{-1/2}$. If the eigenvalues of $A$ are sufficiently similar and you have an efficient way to bound the least eigenvalue $\def\lmin{\lambda_{\text{min}}}\lmin\gt\kappa$ of $A$, you can get around this using rejection sampling: sample $x$ uniformly from the unit ball, reject the sample if $x^\top Ax\gt\kappa$, else divide it by $\sqrt{x^\top Ax}$ and use it. (Ideally, if you know $\lmin$, then $\kappa=\lmin$ yields the lowest rejection rate.)

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