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H: If $M$ is finitely generated then $M/N$ is finitely generated. Let $N$ be a submodule of $R$-module $M$. Prove that if $M$ is finitely generated then $M/N$ is finitely generated. Help me some hints. AI: Suppose $M$ is generated by $n$ elements, say $\{a_i\}_{i=1}^n$. Claim : $M/N = (a_i + N \ \|\ i \in [1,n])$. $(\supseteq)$ It is clear. $(\subseteq)$ Suppose $x + N \in M/N$ with $x = \sum r_i a_i$. Then $x + N = \sum r_i (a_i + N)$, as desired. Thus $\{a_i + N\}_{i=1}^n$ is a generating set for the $R$-module $M/N$.
H: Ring made from a Bijection of Positive Integers and Rationals to Reals (or Complex Numbers) I've been working on this problem in my head and I'm not quite how to phrase this in a way that makes sense since I haven't been doing pure math in some time so I hope this makes sense. What I'm looking for is something like this: $f: \mathbb{Q} \times \mathbb{Z}^+ \to S$ where $f$ is a bijection, $S$ is a ring, and $S \subset \mathbb{C}$ (preferably $S \subset \mathbb{R}$) If anyone can get me a very simple example of a function like this, I would greatly appreciate it. I apologize for the open-endedness of this question. UPDATE I wanted to mention another constraint I neglected to say when I originally posted this. It must necessarily be the case that the following is true: $f(x, 1) = x$ I believe that is the only constraint I neglected to mention. Again, I appreciate any simple answers to this question. AI: Any bijection to either the rationals or the integers in $\mathbb{R}$ works. They're not particularly interesting though as you lose any structure of $\mathbb{Q}$ and $\mathbb{Z}^+$. With the edited question we need to be a little more clever, but not much more. We just pick a countable subring of $\mathbb{R}$ which has $\mathbb{Q}$ as a further proper subring. Let's use $\mathbb{Q}[\sqrt{2}]$ for convenience, although any irrational would do just as well. Now, map $(x,1)\mapsto x$ and then with the remaining elements $\mathbb{Q}\times (\mathbb{Z}\setminus 1)$ we map this set bijectively to the remaining part of the domain $\mathbb{Q}[\sqrt{2}]\setminus\mathbb{Q}$ which is still countably infinite and so admits such a bijection.
H: Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring. Prove that if $R$ is a principal ideal ring and $S$ is a multiplicatively closed subset of $R$ then $S^{-1}R$ is also a principal ideal ring. Thanks for any insight. AI: $S^{-1}R$ is a subring of a field (the field of fractions of $R$), and thus is an integral domain. Let $I \subseteq S^{-1}R$ be an ideal. Let $D = \{a \in R \ \|\ \frac{a}{b} \in I \}$. claim that $D$ is an ideal of $R$. note that $\frac{0}{s} \in I$, so that $0 \in D$. Now let $\frac{a}{d}$, $\frac{b}{e} \in I$; then $\frac{a}{d} - \frac{e}{d}\frac{b}{e} = \frac{a-b}{d}$, so that $a-b \in D$. let $r \in R$ and $a \in D$, with $d \in S$. Then $\frac{r}{d} \frac{a}{d} = \frac{ra}{d^2}$, so that $ra \in D$. Thus $D$ is an ideal of $R$, and thus we have $D = (\alpha)$. Let $d \in S$ such that $\frac{\alpha}{d} \in I$; then $(\frac{\alpha}{d}) \subseteq I$. Now let $\frac{a}{b} \in I $with $a = c\alpha$; we have $\frac{a}{b} = \frac{cd}{b} \frac{\alpha}{d}$, so that $I = (\frac{\alpha}{d})$ is principal. Thus,$S^{-1}R$ is a principal ideal domain.
H: Vectors Dependent or not? Am I doin it right? $3-2x + x^2,6-4x+2x^2 $ in $P2$ Check if they are dependent or not? I am taking their determinant as written $\| 3-2x\space\space\space\space\space\space\space\space\space 6-4x \|$ $\|x^2\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space2x^2\space\space\space\|$ the determinant is non-zero so they are Linearly Dependent The question is am I taking the determinant right as it is a polynomial and what if I want to use $AX=0$ to check for trivial or non-trivial solutions how will I make A by the given statement i.e $3-2x + x^2,6-4x+2x^2 $ in $P2$ Thanks in advance and I really need to workout on my basic concepts of Linear Algebra :( AI: Non-zero determinant means that the columns/rows are Linearly independent. But taking a determinant is not in general a good way to determine whether vectors are linearly independent (since it only works for $n$ vectors at a time, and since it is computation intensive). The best way to do it is to take the two vectors you're dealing with and row eliminate. In this case the matrix looks like $$\left[ \begin{matrix} 3 & -2 & 1 \\ 6 & -4 & 2 \end{matrix} \right].$$ When you row eliminate, you can find that this matrix is of rank one (the second row goes away right away). Comment: a good way to remember that row elimination preserves the nullspace (and so can tell you the rank) is that row elimination is left-multiplication by invertible matrices, and so $Mx = 0 \iff E_1 \dotsb E_m Mx = 0$ (since $E_1, \dotsc, E_m$ all have trivial nullspace).
H: linearly independence of functions over $\mathbb{R}$ I have been asked to prove that $\{\tan(ax)\|a\in \mathbb{R}^+\}$ is linearly independent. I was wondering if there is a generic method/idea for proving linear independence of functions over $\mathbb{R}$. AI: Hint: Since $\tan \in C^{\infty}(\mathbf{R})$. Pick any finite number of $a_i$'s from $\mathbf{R}$. Show the Wronskian of $\tan{a_1x},\dots,\tan{a_nx}$ is nonzero.
H: Why complex conjugate roots of polynomials with real coefficients have the same multiplicity? Let $$f(x)=(x-z_1)^{k_1}(x-\overline{z_1})^{l_1} \cdot ... \cdot (x-z_n)^{k_n} (x-\overline{z_n})^{l_n}g(x),$$ where $z_i$ are different complex numbers with nonzero imaginaris parts, $k_i, l_i $ are natural numbers, $g$ is a real polynom and $g(z_i)\neq 0$ for $i=1,...,n$. Why, if $f$ is a real polynom, then $k_i=l_i$ for all $i=1,...,n$? AI: If $z$ is a zero of $f(x)$, let $m$ be the multiplicity of $z$ as a root of $f$. Then $(x-z)^m$ divides $f(x)$. Taking conjugates, and using that $f(x)$ has real coeffs., we see that $(x -\overline{z})^m$ also divides $f(x)$. Thus $\overline{z}$ has multiplicity at least $m$. Applying the same argument, with $\overline{z}$ in place of $z$, and using that $\overline{\overline{z}} = z$, we see that in fact the multiplicities of $z$ and $\overline{z}$ as roots of $f$ coincide.
H: Simplify $1 - \cos x + \sin x - \tan x$. I somehow can't find a way to do this, the expression already seems simplified to me. I've tried factoring, I got $$(\cos x - \sin x)(1 - \cos x) \over \cos x$$ but obviously that's not simplified and I can't see how I could continue. AI: You're not far off. If you distribute the $\cos x$ into the first factor on the numerator you get $$1-\cos x+\sin x-\tan x = (1-\tan x)(1-\cos x)$$ I think this is about as 'simplified' as you can expect it to be.
H: Prove the following $(A \cap B') \cup (A\cap C) = A \cap( B' \cup C)$ I need help to prove: $(A \cap B') \cup (A\cap C) = A \cap( B' \cup C)$ Thanks AI: Double inclusion. For example: $$x\in (A\cap B')\cup (A\cap C)\implies (x\in A\;\wedge\;x\notin B)\;\vee\;(x\in A\;\wedge\;x\in C)\implies$$ $$x\in A\;\wedge (x\notin B\;\vee\;x\in C)\implies x\in A\cap(B'\cup C)$$
H: Is the following convergence of the series of functions uniform? For the series f(x) = x + [sum from j=1 to j=infinity, x((1-x)^j)], I already proved that this series converges pointwise for x on [0, 1]: f(x) converges pointwise to 1 if x is on (0, 1]; f(x) converges pointwise to 0 if x=0. Note that I found that for x on (0, 1], the partial sum of this series is Sn(x) = 1 - ((1-x)^(n+1)) However, is such convergence uniform? How can I formally prove it? Thanks a lot! AI: If the convergence were uniform then by the uniform convergence theorem the limit function would also be a continuous function, but the limit function you calculated is not continuous. Hence the convergence is not uniform.
H: Brownian Motion conditional distribution Let $\{X(u),u\geq0\}$ be a standard Brownian motion. What is the conditional distribution of $X(t)$ given $\{X(t_{1}),\dots,X(t_{n})\}$, where $0<t_{1}<\cdots<t_{n}<t_{n+1}=t$? --So far, I have derived the joint pdf of $X(t_{n+1})$ and $X(t_{1}),\dots,X(t_{n})$ using the fact that each increment $X(t_{i})-X(t_{i-1})$ is independent and normally distributed. The pdf if (I think) given by $$ \begin{align} f(x_{1},x_{2},\dots,x_{n},x_{n+1})&=f_{t_{1}}(x_{1})f_{t_{2}-t_{1}}(x_{2}-x_{1})\cdots f_{t_{n}-t_{n-1}}(x_{n}-x_{n-1})f_{t_{n+1}-t_{n}}(x_{n+1}-x_{n})\\ &=\frac{\exp\left\{-\frac{1}{2}\left[\frac{x_{1}^{2}}{t_{1}}+\frac{(x_{2}-x_{1})^{2}}{t_{2}-t_{1}}+\cdots+\frac{(x_{n}-x_{n-1})^{2}}{s-t_{n-1}}+\frac{(x_{n+1}-x_{n})^{2}}{t_{n+1}-t_{n}}\right]\right\}}{(2\pi)^{(n+1)/2}[t_{1}(t_{2}-t_{1})\cdots(s-t_{n-1})(t-s)]^{1/2}}. \end{align} $$ So the $X(t_{i})$s are jointly normal. Now to find the conditional distribution, I believe we need to use multivariate distribution theory. I.e., if $\mathbf{X}$ and $\mathbf{Y}$ have a joint normal distribution, then the conditional distribution of $\mathbf{Y}$ given $\mathbf{X}$ is also normally distributed such that $$\mathbf{Y}\|\mathbf{X}\sim\operatorname{Multivariate\ Normal}(\boldsymbol\mu^{},\boldsymbol\Sigma),$$ where $\boldsymbol\mu^{}=\boldsymbol\mu_{y}+\boldsymbol\Sigma_{yx}\boldsymbol\Sigma_{xx}^{-1}(\mathbf{x}-\boldsymbol\mu_{x})$ and $\boldsymbol\Sigma^{*}=\boldsymbol\Sigma_{yy}-\boldsymbol\Sigma_{yx}\boldsymbol\Sigma_{xx}^{-1}\boldsymbol\Sigma_{xy}$. How would I go about finding each of these components for $X(t)\|X(t_{1}),\dots,X(t_{n})$? Thank you. AI: One knows that the marginal distributions of Brownian motion are normal and that $X(t)-X(t_n)$ is independent of $\sigma(X(s);s\leqslant t_n)$. Hence, the conditional distribution of $X(t)$ conditionally on $\sigma(X(t_k);1\leqslant k\leqslant t_n)$, for every $t_k\leqslant t_n$ (or, conditionally on $X(t_n)$ only) is normal with mean $X(t_n)$ and variance $t-t_n$.
H: How do I find this $\begin{vmatrix} A&B\\ B&A \end{vmatrix}$? Let $A, B$ be matrices such that $A=(a_{ij})_{n\times n}, B=(b_{ij})_{n\times n}$, $$b_{ii}=2a_{ii},$$ and $$b_{ij}=\begin{cases} a_{ij}&i>j\\ -a_{ij}&i<j \end{cases}$$ for $(i,j=1, 2, 3, \cdots , n)$. Find $$\begin{vmatrix} A&B\\ B&A \end{vmatrix}.$$ My try: $$\begin{vmatrix} A&B\\ B&A \end{vmatrix}=\begin{vmatrix} a_{11}&a_{12}&\cdots&a_{1n}&a_{11}&-a_{12}&-a_{13}&\cdots&-a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}&a_{21}&a_{22}&-a_{23}&\cdots&-a_{2n}\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\cdots&\\ a_{n1}&a_{n2}&\cdots&a_{nn}&a_{n1}&a_{n2}&a_{n3}&\cdots&a_{nn} \end{vmatrix}$$ then I can't carry on. Maybe there's some other nice method. Thank you. AI: Proof of Albert's formula: $$\left\|\begin{smallmatrix} A&B\\B&A\end{smallmatrix}\right\|=\left\|\begin{smallmatrix} A-B&B-A\\B&A\end{smallmatrix}\right\|=\left\|\begin{smallmatrix} A-B&0\\B&A+B\end{smallmatrix}\right\|=\|A-B\|\|A+B\|$$ Step 1: Subtract 2nd row of blocks from 1st row. Step 2: Add 1st column of blocks to 2nd column. Step 3: Matrix is block triangular.
H: Why is the total number of binary relations $2^{n^2}$ Why is the formula for finding the total number of binary relations in a set A equal to $2^{A^2}$ ? I have looked on the web (including here) and find the answer is because the relation can be in or out. This is the exact thing my professor told me but did not explain meaning in or out. I asked what that and he looked at me like I was stupid and did not explain further. My understanding is if the set is in you have for example A = {a,b} so a relation could be (a,a) which would be in But what if the relation is out would that not be the empty set? So for an example I had A is a set of 2 elements. So that would mean the number of binary relations would be $2^4$ which is 16 but this is what I got -only 14 elements. (a), (b), (a,b), (a,a), (b,b), (a,b,a), (a,b,b), (b,b,b), (a,a,a,), (b,a,b,b), (b,a,b,a), (b,a,a,a), (b,b,b,b), (a,a,a,a) What are the only two elements? Is the empty set a part of this? What does it mean to be in or out? AI: A binary relation $R$ on a set $A$ looks like a bunch of statements of the form $a\ R\ b$ for $a, b \in A$. So for each pair $(a,b)$ you have two choices: does $a\ R\ b$ or not? Supposing $A$ has size $n$, there are $n^2$ possible pairs $(a,b)$, and for each of these pairs you have two choices of whether or not they appear in your relation. So the number of possible relations is $$\underbrace{2 \times 2 \times \cdots \times 2}_{\text{one for each of the}\ n^2\ \text{pairs}} = 2^{n^2}$$ Comment: In your list: (a), (b), (a,b), (a,a), (b,b), (a,b,a), (a,b,b), (b,b,b), (a,a,a,), (b,a,b,b), (b,a,b,a), (b,a,a,a), (b,b,b,b), (a,a,a,a) I'm not sure what these are meant to represent... they're certainly not binary relations.
H: Is $([0, \sqrt 2] \cap \mathbb Q) \subset \mathbb Q$ closed, bounded, compact? As far as I can tell it is bounded, as it's within $[0, \sqrt 2]$, and is closed as there cannot be an open neighbourhood about 0, and as it's closed and bounded it is therefore compact. However I'm not sure if closed and bounded imply compact in this situation, as I've only ever used this property in $\mathbb R$. Am I wrong? Edit: the question specifies this is $\mathbb Q$ with the Euclidean metric, so no need to allow for different ones. AI: It is not compact. Note that $\left[0,\sqrt2\right]\cap\Bbb Q=\left[0,\sqrt2\right)\cap\Bbb Q$, and it can be covered by the open sets $U_n=\left[0,\sqrt2-\frac1n\right)\cap\Bbb Q$, but this cover has no finite subcover. It closed, however, as a subset of the rationals. This shows that a closed and bounded subset of the rationals need not be compact.
H: How to show field extension equality I've seen similar field extension questions on SE, but nothing with a third root, and I'm having trouble adapting any of those solutions to this problem. So I'm trying to prove that $\mathbb{Q}(\sqrt{2}+{5}^{1/3})=\mathbb{Q}(\sqrt{2},{5}^{1/3})$. Now, $\mathbb{Q}(\sqrt{2},{5}^{1/3})$ contains both $\sqrt{2}$ and ${5}^{1/3}$, so $\mathbb{Q}(\sqrt{2}+{5}^{1/3})\subseteq\mathbb{Q}(\sqrt{2},{5}^{1/3})$, right? How then do I go about proving the reverse, that $\mathbb{Q}(\sqrt{2},{5}^{1/3})\subseteq\mathbb{Q}(\sqrt{2}+{5}^{1/3})$? I not even sure where to start. Any advice, tips? AI: An idea $$r=\sqrt2+\sqrt[3]5\implies r^3-3\sqrt2\,r^2+6r-2\sqrt2=5\implies$$ $$(r^3+6r-5)^2=\left[(3r^2+2)\sqrt2\right]^2$$ The above shows that $\;\dim_{\Bbb Q}\Bbb Q(\sqrt 2+\sqrt[3]5)= 6\;$ (why? Show the polynomial $\;(r^3+6r-5)^2-\left[(3r^2+2)\sqrt2\right]^2\in\Bbb Q[x]\;$ is irreducible ), but since $\;\dim_{\Bbb Q}\Bbb Q(\sqrt2\,,\,\sqrt[3]5)=6\;$ (why?) and we already know $\;\Bbb Q(\sqrt2+\sqrt[3]5)\le\Bbb Q(\sqrt2\,,\,\sqrt[3]5)\;$ we get equality.
H: Radius of convergence of a Taylor series. I am looking for the shortest possible way to find out the radius of convergence of the Taylor series expansion about $x = a \in \mathbb{R}$ of the function $$f(x) = \frac{1}{1 + x^2}$$ Taylor series expansion of the function $f(x)$ about $a$ will be $f(x) = \sum_{n = 0}^{\infty} a_n (x - a)^n$ where $a_n = \frac{f^n(a)}{n!}$. So one possible way is to calculate $f^n(a)$ and consider different values of $a_n$. Then apply root test, ratio test etc. to find out the radius of convergence of the power series. I want to reduce calculation. So the above process will not be applicable. Thank you for your help. AI: The function $$f(z) = \frac{1}{1+z^2}$$ is meromorphic in the entire plane. Therefore, the Taylor series about any point $a$ will converge in the largest disk with centre $a$ that does not contain a pole of $f$. Since $f$ has only two poles, in $i$ and $-i$, the radius of convergence of the Taylor series is $\min \{ \lvert a-i\rvert, \lvert a+i\rvert\}$. For real $a$, the two distances are equal, and the radius of convergence is $\sqrt{1+a^2}$.
H: Does the following question regarding functions and relations make sense? I have a function $f: P(\mathbb{N} \times \mathbb{N}) \to P(\mathbb{N} \times \mathbb{N})$ defined: $f(R)=R \cup R^{-1}$ The question I'm struggling with is "Find $f(P(\mathbb{N} \times \mathbb{N}))$". My main problem is that I can't make sense out of this question. Function $f$ takes a subset of $P(\mathbb{N} \times \mathbb{N})$, which is a relation, and returns a sum of this relation and the opposite relation. Makes sense. But $P(\mathbb{N} \times \mathbb{N})$ is not a relation, it's an infinite set of relations, isn't it? Perhaps they meant $\bigcup P(\mathbb{N} \times \mathbb{N})$? Sorry if I'm missing something obvious here, but I'm just completely lost and confused... Thanks for any hints! AI: A function $f:X\to Y$ induces a function $F:\wp(X)$ to $\wp(Y)$ by $F(A)=\{f(x):x\in A\}$ for $A\subseteq X$. Instead of explicitly introducing $F$, we often just write $f[A]$ for $\{f(x):x\in A\}$. Many — perhaps most — people further confuse the issue by writing $f(A)$ instead of $f[A]$, so that it looks as if $A$ is an element of the domain of $f$ instead of a subset of it. Assuming that there is no typo in your source, I suspect that that is what’s intended here, and that you’re supposed to find what I would write $$\begin{align} f[\wp(\Bbb N\times\Bbb N)]&=\{f(R):R\in\wp(\Bbb N\times N)\}\\ &=\{f(R):R\subseteq\Bbb N\times\Bbb N\}\\ &=\{R\cup R^{-1}:R\subseteq\Bbb N\times\Bbb N\}\;. \end{align}$$
H: Matrix of orthogonal projection It was required to find the orthogonal projection of the vector $u =(0,1,0,2)$ onto $W=${$(x,y,z,t) \in \mathbb{R}^4 : x+y-t = 0$} and the matrix of the projection. First, I've found a basis for $W$ and, use Gram-Schimidt process, an orthonormal basis. With it, I could do the first part of the problem. Having an orthonormal basis for my space, can this make the job of finding the matrix of the projection easier? If it does, how can I found this matrix? Thanks in advance! AI: Having an orthonormal basis for my space, can this make the job of finding the matrix of the projection easier? If it does, how can I found this matrix? Yes! If $\{\mathbf{q}_i\}$ forms an orthonormal basis for the subspace $W$ that you want to orthogonally project onto, then the matrix of the orthogonal projection is is just $QQ^T$, where the columns of $Q$ are the $\mathbf{q}_i$. More generally, if $\{\mathbf{p}_i\}$ only forms an orthogonal (but not necessarily orthonormal) basis for $W$, then the matrix of the orthogonal projection on to $W$ is $P(P^T P)^{-1}P^T$, where $\mathbf{p}_i$ form the columns of $P$.
H: Equation for this graph I am trying to find an equation for the graph that crosses the Y axis at (0,100), X axis at (100, 0), is a curve with adjustable degree of "bending" and has an axis of symmetry y(x) = x. Here are few examples of such graphs: I've tried quadratic, cubic, quartic equations but I can't adjust the degree at which the curve is bent. Any hints/tips appreciated. AI: I know a function like the second one. Have a look at it to find if it works or not. That is $$x^{\frac{2}3}+y^{\frac{2}3}=100^{\frac{2}3}$$
H: Rationalizing a denominator. The question instructs to rationalize the denominator in the following fraction: My solution is as follows: The book's solution is which is exactly the numerator in my solution. Can someone confirm my solution or point what what I'm doing wrong? Thank you. AI: Going from the first to the second line of your working, you seem to have said $$(\sqrt{6} -2)(\sqrt{6}+2) = 6+4$$ on the denominator of the fraction. This isn't true: in general $(a-b)(a+b)=a^2-b^2$, not $a^2+b^2$.
H: Isometry of a metric space with proper subset In Irving Kaplansky's "Set Theory and Metric Spaces", exercise 17 on page 71 asks for an example of a metric space which is isometric to a proper subset of itself. Any infinite discrete space and any $\ell^p$ space are such spaces, for different reasons. I want some kind of middle ground, that is, some example which is more intriguing than a discrete space and doesn't require knowledge of functional analysis or manifolds. Thanks in advance! AI: Let $\langle Y,d\rangle$ be any metric space, and fix a point $p\in Y$. Let $A$ be any infinite set, let $P=\{p\}\times A$, and let $X_A=(Y\times A)/P$, the set resulting from $Y\times A$ by identifying $P$ to a point. By abuse of notation call this point $p$. Define a metric $\rho_A$ on $X_A$ as follows: $$\rho_A(\langle y,\alpha\rangle,\langle z,\beta\rangle)=\begin{cases} d(y,z),&\text{if }\alpha=\beta\\ d(y,p)+d(p,z),&\text{if }\alpha\ne\beta\;, \end{cases}$$ and $\rho_A(p,\langle y,\alpha\rangle)=d(p,y)$. $\langle X_A,\rho_A\rangle$ is a kind of ‘hedgehog’ with $\|A\|$ spines all radiating from the centre $p$. If $B$ is any subset of $A$ such that $\|B\|=\|A\|$, then $\langle X_B,\rho_B\rangle$ is isometric with $\langle X_A,\rho_A\rangle$.
H: Integral involving logarithm and cosine For $a,b>0$ I would like to compute the integral $$I=\int_0^{2\pi} -\log{\sqrt{a^2+b^2-2ab\cos{t}}}~dt.$$ Numerical computations suggest that $$I=\min\{-\log{a},-\log{b}\}.$$ How can I prove this? I tried to find the antiderivative using Mathematica, but the result looks awfull. Have you seen such an integral? Some reference would be great! AI: The term under the square root can be written as either $(ae^{it}-b)(ae^{-it}-b)$ or $(a-be^{it})(a-be^{-it})$. You want to choose the one where neither factor crosses the negative real axis. That is, you want to write $$ a^2+b^2-2ab\cos t=(c-de^{it})(c-de^{-it}), $$ where $c=\max(a,b)$ and $d=\min(a,b)$. Then the integral becomes $$ \int_{0}^{2\pi}-\log\sqrt{(c-de^{it})(c-de^{-it})}dt=-\frac{1}{2}\int_{0}^{2\pi}\log(c-de^{it})dt-\frac{1}{2}\int_{0}^{2\pi}\log(c-de^{-it})dt. $$ Each term can be evaluated as a contour integral. Letting $z=e^{it}$ in the first term, and $z=e^{-it}$ in the second, we have $$ -\frac{1}{i}\int_{\Gamma}\frac{\log(c-dz)}{z}dz, $$ where $\Gamma$ is the unit circle. Because $c\ge d>0$, the contour avoids the branch cut, so the only contribution is from the pole at $z=0$: $$ I=-{2\pi}\log c=-2\pi\log\max(a,b). $$ This agrees with your result except for the factor of $2\pi$.
H: Real Analysis: Use Intermediate value Thm to prove the functions Let $f \colon [0,2] \to \Bbb R$ be a continuous function such that $f(0) = f(2)$. Use the intermediate value theorem to prove that there exist numbers $x, y \in [0, 2]$ such that $f (x) = f (y)$ and $\|x − y\| = 1$. Hint: Introduce the auxiliary function $g \colon [0, 1] \to \Bbb R$ defined by $g(x) = f (x + 1) − f (x)$. I still do not know how to prove it. Could anyone help? AI: The given function $\;g\;$ is continous in $\;[0,1]\;$ and $$\begin{cases}g(0)=f(1)-f(0)\\{}\\g(1)=f(2)-f(1)\end{cases}\implies g(0)g(1)<0\;\;\text{, unless}\;f(1)=f(0)\;\;\text{(why?)}$$ and so by the IVM there exists $\;c\in (0,1)\;$ s.t. $\;g(c)=0\;\ldots$ ...end the exercise now,
H: Does closed imply bounded? Definitions:1. A set $S$ in $\mathbb{R}^m$ is bounded if there exists a number $B$ such that $\mathbf{\|\|x\|\|}\leq B$ for all $\mathbf{x}\in S$, that is , if $S$ is contained in some ball in $\mathbb{R}^m$.2. A set in $\mathbb{R}^m$ is closed if, whenever $\{\mathbf{x}_n\}_{n=1}^{\infty}$ is convergent sequence completely contained in $S$, its limit is also contained in $S.$3. A set $S$ in $\mathbb{R}^m$ is compact if and only if it is both closed and bounded. Does closedness not imply boundedness in general? If so, why does a compact set need to be both closed as well as bounded? AI: The set $\{\,(x,y)\in\mathbb R^2\mid xy=1\,\}$ is closed but not bounded. Even simpler, $\mathbb R^n$ itself is closed (but not bounded).
H: Why $R=\{(a,b)\mid a=b \mbox{ or } a=-b\}$ is not anti symmetric? Why $R=\{(a,b)\mid a=b \mbox{ or } a=-b\}$ is not anti symmetric? I read because this is symmetric so it is not anti symmetric, but $R=\{(a,b) \mid a=b \}$ is both symmetric and anti symmetric. AI: It’s not antisymmetric because $\langle 1,-1\rangle\in R$ and $\langle -1,1\rangle\in R$, but $1\ne -1$. That’s a clear violation of the definition of antisymmetry. Remember, antisymmetry means that if $\langle a,b\rangle\in R$ and $\langle b,a\rangle\in R$, then $a=b$. Note also that it is entirely possible for a relation to be both symmetric and antisymmetric. The relation of equality on any set is both symmetric and antisymmetic.
H: How to compress data dimension I have n points on the plane $(x_1, y_1) \ldots (x_n, y_n)$ - it's the points of one stroke. I want to reduce number of points without significant information loss. I read some information about PCA, but I don't sure if it's what I need. Can anyone recommend methods for doing what I want. AI: You can model it as a function $y=f(x)$, trying various functions to find one you like. The simplest is to fit the best straight line through your points. You can use many functional forms, usually being guided by looking at your data. Any numerical analysis text can help you.
H: Limit of $n/\ln(n)$ without L'Hôpital's rule I am trying to calculate the following limit without L'Hôpital's rule: $$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$ I tried every trick I know but nothing works. You don't have to prove it by definition. AI: Every time you make $n$ twice and big, you increase $\ln n$ by less than $1$ (since $e$ is less than $2$). So imagine repeatedly doubling the numerator and each time you do it, adding just $1$ to the denominator, and think about what that will approach. That does it. PS inspired by a comment: $$ \frac{n}{\ln n} > \frac{n}{\log_2 n}. $$ A comparison test then says that if the latter goes to $\infty$, then so does the former. Each value of $n$ is located between two powers of $2$: we have $2^m\le n\le 2^{m+1}$. That number $m$ is $m=\lfloor \log_2 n\rfloor$. So $m\le\log_2 n<m+1$. $$ \frac{2^m}{m+1}\le\frac{n}{m+1}<\frac{n}{\log_2 n}\le\frac n m. $$ So it is enough to show that $\dfrac{2^m}{m+1}\to\infty$. Show by induction that for some constant $c$, we have $$ \frac{2^m}{m+1} \ge \left(\frac 3 2\right)^m $$ and then do another comparison.
H: Why does $\operatorname{Spec}(\prod_1^\infty \Bbb F_2)$ have connected components that are not open? To be honest, I don't even know how to describe all prime ideals in $\prod_1^\infty \Bbb F_2$. I know we get one for each $n \in \Bbb N$ corresponding to the set of elements that are zero in the $n$-th coordinate, and at least an extra one corresponding to the ultrafilter $F := \{x\|x_i\ne 0 \,\text{for at most finitely many }i\}$. AI: What you write down is a filter, not an ultrafilter. The ideals in a product of fields correspond to the filters on the index set. If $\mathcal{F}$ is a filter, the corresponding ideal is $\{a : \{i : a_i = 0 \} \in \mathcal{F}\}$. It follows that the prime ideals (which equal the maximal ideals in that case) correspond to the ultrafilters. The spectrum identifies with the Stone-Čech compactification of the index set. So your question is really a topological one: Why does $\beta \mathbb{N}$ have a connected component which is not open? Well, it is totally disconnected and not discrete. So take any $p \in \beta \mathbb{N} \setminus \mathbb{N}$, then $\{p\}$ is a connected component which is not open.
H: Find number of triangles. This is an extremely basic question posed to my brother in fifth standard and i was only able to find $20$ triangles in this figure however the answer according to the answer key was 24. $Side -1: 12 $ $Side -2: 6 $ $Side -3: 2 $ Is there any way to conclusively prove that there are only 20 triangles? AI: There are $12$ "simple" triangles. Each peak belongs to one of six size-two triangles and there are two size-three triangles. So the $20$ seems to be correct. An alternative approach: Each triangle is bounded by a horizontal straight line, an ascending $60^\circ$ line and a descending $60^\circ$ line. Since there are three lines of each type, we'd arrive at $3\cdot3\cdot 3$ triangles. However, the seven points of intersection (center and vertices of hexagon) fail to produce a triangle. $27-7=20$.
H: Prove or refute contingent: If A implies B is contingent, then B is too The question is: If $A, A \to B$ are contingent, then so is $B$ $A, A \to B$ (implies) is a contingent, but how exactly to show «so is $B$»? If I'm using a truth table, how should I show that $B$ is also contingent? AI: This is not necessarily true. We explore each of the alternatives to contingency: Suppose $B$ is a tautology. Suppose $B$ is a contradiction. (If both of these lead to contradicting our premises, then $B$ would necessarily be contingent. Otherwise, it is not. Suppose $B$ is a tautology (hence always evaluating to true), then $A\rightarrow B$ would necessarily be true always...hence a tautology. This contradicts one of our premises. $\therefore$ $B$ cannot be a tautology. Suppose $B$ is a contradiction (always false). This is entirely possible, given the contingency of $A$ and of $A\rightarrow B$. That is, $B$ could always be false, and then the truth value of $A\rightarrow B$ is entirely determined by (contingent upon) the truth-value of $A$, which is, in turn, contingent. $\therefore\; B$ could be a logical contradiction. Hence, by definition then, $B$ is not necessarily contingent given only the constraints of the premises. $B$ could very well be a logical contradiction, and in that event, it is not contingent.
H: If $X$ is a noetherian topological space, then any union connected components of $X$ is clopen. If $X$ is a noetherian topological space, then any union connected components of $X$ is clopen. This is exercise 3.6P of Vakil. I can see that a union of connected components is closed. This is so because connected components are closed, and, in a Noetherian topological space, an arbitrary union of closed sets is closed. My guess is that in order to show that a single connected component is open (which is clearly a sufficient and necessary condition) one should use the fact that, in a Noetherian space, a connected component is a finite union of irreducible components. If we could extend this union to a larger union of irreducible components that covers all of $X$, we'd be done, but I'm not sure if we can do this. AI: HINT: Show that $X$ has only finitely many connected components. Let $\mathscr{C}$ be the family of closed subsets of $X$ that have infinitely many connected components. If $\mathscr{C}\ne\varnothing$, $\mathscr{C}$ must have a minimal element $C$ (minimal with respect to inclusion). Use $C$ to get a contradiction, thereby showing that $\mathscr{C}=\varnothing$.
H: How to find dim(S∩T) I have S and T in the subspace of $R^4$ and defined by $S={[s, t, 0, 0]\|s, t ∈ R}$ $T={[0, s, t, 0]\|s, t ∈ R}$ Then I am asked to find $dim(S)$, $dim(T)$, and $dim(S∩T)$. From what I have gathered, the dimensions of the first two are 4 each, but I'm lost on how to find the dimension of an intersection. Thanks! AI: Roughly, the dimension is the number of degrees of freedom. So for both $S,T$, how many parameters can be varied independently? $x \in S$ iff $x \in \operatorname{sp} \{ e_1, e_2 \}$, $x \in T$ iff $x \in \operatorname{sp} \{ e_2, e_3 \}$, and the vectors $e_k$ are linearly independent. From this you should be able to read off $\dim S, \dim T$. Now note that $x \in S \cap T$ iff $x \in S$ and $x \in T$ iff $x \in \operatorname{sp} \{e_2 \}$. Now you should be able to read off the dimension of $T \cap S$.
H: Solving for upper limit of integration, given a lower limit and integrand Consider $f$ a real integrable function, usually we want to evaluate the integral $\int_a^bf(x)dx$ for some $a<b$ given. Now, suppose we know $a$ but we don't know $b$, further, we know the value of this integral, let $\int_a^bf(x)dx=\lambda$. My question is, in what conditions we can find $b$, and how? A simple case is when $f$ has a known primitive $F$ and $F$ has inverse $F^{-1}$. $$ \int_a^bf(x)dx=\lambda\implies F(b)-F(a)=\lambda\implies F(b)=F(a)+\lambda\implies b=F^{-1}(F(a)+\lambda)$$ In this case we can evaluate $b$, but if this is not the case, what can be done? Thanks. PS: More approaches are welcome, approaches not relying on $f$ primitive. AI: I'm not sure how satisfying it will be, but the work you've shown in your Question proves that talk about which $b$ attains which $\lambda$ is equivalent to talk about a specific antiderivative $F$ attaining values $\lambda$, namely the antiderivative of $f$ such that $F(a) = 0$. Since all antiderivatives of $f$ are obtained by adding the famous "+C" to any one of them, the latter constraint can always be arranged (provided the function $f$ is integrable on a suitable interval containing $a$). The issue is whether solutions exist, and how to find them: $$ F(b) = \lambda $$ Conceivably $f$ will be given in a form that we can easily evaluate, but expressions for $F$ are not available or are not easily evaluated. In such cases we are apt to fall back on numerical integration schemes to try and show solutions exists and narrow down the approximation of them. The best "quadrature rule" depends on the smoothness of $f$ and on what precision is needed in locating $b$. Knowing that $F$ is increasing where $f$ is positive and decreasing where $f$ is negative, we can use knowledge of the sign of $f$ to help bracket intervals that might contain solutions $b$. Indeed, since the derivative of $F$ is simply $f$, using root finding methods that exploit derivatives is apt to make refining precision of approximate solutions go that much faster.
H: $\Bbb{Z}/p^k \Bbb{Z} \otimes_{\Bbb{Z}} A $ is isomorphic to the Sylow $p$-subgroup of $A$ Let $A$ be a finite abelian group of order $n$ and let $p^k$ be the largest power of the prime $p$ dividing $n$. Then $\Bbb{Z}/p^k \Bbb{Z} \otimes_{\Bbb{Z}} A $ is isomorphic to the Sylow $p$-subgroup of $A$. Hints on proving this? AI: Hint: $\Bbb{Z}/p^k \Bbb{Z} \otimes_{\Bbb{Z}} A $ is isomorphic to $A/p^k A $.
H: Is binary-relation $\left\{\left(a,b\right)\mid a,b\in\mathbb{N}\wedge a,b \text{ are even numbers}\right\}$ reflexive? I'm a novice in set theory and I'm not clear about reflexive relation. My question is the title. Is binary-relation $R:=\left\{\left(a,b\right)\mid a,b\in\mathbb{N}\wedge a,b \text{ are even numbers}\right\}$ reflexive? The relation set $R$ is the set of all positive ordered pair of even numbers, and it satisfies $\alpha R\alpha$. But $\alpha$ does not demonstrates all elements in $\mathbb{N}$ where the definition of reflexive is $\left(\forall a\in \mathrm{A}\right)\left(aRa\right)$. Is this means for all $a$ in $\mathrm{A}$ or in $R$? Added OK. I think I've understood. For checking -> In the case above, $R$ is not reflexive binary-relation on $\mathbb{N}$. But it is reflexive on $R$. Is this correct? AI: Recall that $R$ is reflexive on the set $A$ if and only if for every $a\in A$, the pair $(a,a)$ is in $R$. This means that the same relation can be reflexive on one set, but not on another. For example $\{(0,0)\}$ is reflexive on $\{0\}$ but not on $\{0,1\}$. So the question depends on who is your $A$. If $A$ is the set of even numbers, then the answer is yes. If it's not then the answer is no, because there will be some $x\in A$ which is not even, and therefore $(x,x)\notin R$.
H: Why is this set a Borel set on $R^2$? Consider a measurable space $([0,1]\times [0,1], \mathcal{B}([0,1]) \times \mathcal{B}([0,1]))$, and a subset $A:=\{(x,y):x=y\}$ (the diagonal). According to the text book, $A \in \mathcal{B}([0,1]) \times \mathcal{B}([0,1])$. Can any body tell me the reason and in general, how do we check if a given set is a Borel set? Thanks:) PS: what I cannot see is, since $\mathcal{[0,1]} \times \mathcal{[0,1]}$ by definition is generated by the class $\mathcal{C}=\{I_1 \times I_2\}$, where $I_1, I_2$ are intervals on $[0,1]$, I cannot express $A$ as any countable union/intersection, complement of any $I_1 \times I_2$. AI: By the definition of the product $\sigma$-algebra, $\mathcal{B}([0,1]) \times \mathcal{B}([0,1])$ is generated by sets of the form $X \times Y$ where $X,Y \in \mathcal{B}([0,1])$. If $x$ and $y$ are distinct real numbers in the interval $[0,1]$ then we can find disjoint subintervals $X$ and $Y$ of $[0,1]$ containing $x$ and $y$ respectively, with rational endpoints. We have $X,Y \in \mathcal{B}([0,1])$ and the rectangle $X \times Y$ is disjoint from the diagonal. Therefore the complement of the diagonal is the union of countably many elements of the product $\sigma$-algebra $\mathcal{B}([0,1]) \times \mathcal{B}([0,1])$, so the diagonal itself is in this $\sigma$-algebra.
H: How do i find the lapalace transorm of this intergral using the convolution theorem? $$\int_0^{t} e^{-x}\cos x \, dx$$ In the book, the $x$ is written as the greek letter "tau". Anyway, I'm confused about how to deal with this problem because the $f(t)$ is clearly $\cos t$, but $g(t)$ is not clear to me. Please help. AI: $f(t) = \int_0^t e^{-\tau} \cos \tau d \tau = e^{-t }\int_0^t e^{t-\tau} \cos \tau d \tau = (e^{-t })\left( (x \mapsto u(x)e^{x}) * (x \mapsto u(x)\cos x) \right)(t)$. $({\cal L}f )(s) = ({\cal L}(x \mapsto u(x)e^{x}) )(s+1) ({\cal L}(x \mapsto u(x) \cos x) )(s+1) $. Note the $s+1$ parameter to deal with the multiplication by $e^{-t}$.
H: a question on : why sin(x) = sin(180-x) I was trying to follow a proof from another question in this forum: Why is $\sin(x) = \sin(180^{\circ}-x)$ I was following the geometric proof given by forum poster : egreg. I could follow his proof for the cosine. But I think the proof for the sine contains an error. Egreg stated that from the triangle in the circle you get the following: sin(alpha)/a = 2R BUT when I analyzed the small triangle with angle alpha in it, I get: sin(alpha)/a = 1/2R It really doesn't affect the proof much at all. But I am wondering if there was an error, or there is something wrong with my own trig analysis. Hope that the forum guy Egreg can reply since he is the originator of this proof. Regards, P AI: You are correct. Recall the Law of Sines, which states $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R$. This means $\dfrac{\sin A}{a} = \dfrac{1}{2R}$. This is probably a typo on his part; it doesn't affect the proof at all. The section of the proof you are referring to is just simply a clever proof of the Law of Sines.
H: If $L\cdot\{\epsilon,0\}$ regular language, is $L$ regular? I've encountered a question during my studies: If $L\cdot\{\epsilon,0\}$ regular language, is $L$ regular? I thought to disprove it by using $A\subseteq 2\mathbb{N}, L=\{w\in\{0\}^:\|w\|\notin A\}$ but I need to prove that L is irregular and using the pumping lemma in my proof is forbidden. Any help? AI: If you’re not allowed to use basic tools like the pumping lemma and the Myhill-Nerode theorem, the only thing that I can suggest is a cardinality argument. There are only countably many regular expressions over the alphabet $\{0\}$, so there are only countably many regular languages that are subsets of $\{0\}^$. However, there are uncountably many subsets $A\subseteq 2\Bbb N$, and each gives rise to a different language $L_A=\{w\in\{0\}^*:\|w\|\notin A\}$, so there must be an $A\subseteq 2\Bbb N$ such that $L_A$ is not regular.
H: Application of the Dominated Convergence Theorem $$lim_{n \to \infty} \int_0^1(1 - e^{\frac{-x^2}{n}})x^{-1/2}dx$$ I want to use the Dominated Convergence Theorem to solve this. Let $f_n = (1 - e^{\frac{-x^2}{n}})x^{-1/2}$. Step 1: Determining convergence of $f_n$ Fix $x$ to be some constant number. Now, bringing the limit inside the integral, we have $lim_{n \to \infty}(1 - \frac{1}{e^{\frac{k}{n}}})t$ where $k, t$ are constants. As $n \to \infty, \ \frac{1}{e^{\frac{k}{n}}} \to 1$ So we get $(1 - 1)t = 0$ So $f_n$ converges to the zero function $f_0$. Note that it doesn't converge at $x = 0$ as we have a division by zero but on $(0, 1]$ it converges to $f_0$ and the measure of $\{0\}$ is $0$ so we have convergence almost everywhere. Step 2: Determining a dominating function $g$ Now, when $x \in (0, 1)$, for all $n$ we have, $$e^{\frac{1}{n}} > 1$$ $$\implies e^{\frac{x^2}{n}} \ge 1$$ $$\implies 0 < \frac{1}{e^{\frac{x^2}{n}}} \le 1$$ $$\implies -1 \le - \frac{1}{e^{\frac{x^2}{n}}} < 0$$ $$\implies 0 \le 1 - \frac{1}{e^{\frac{x^2}{n}}} < 1$$ $$\implies \mid 1 - \frac{1}{e^{\frac{x^2}{n}}}\mid < 1$$ $$\implies \mid (1 - \frac{1}{e^{\frac{x^2}{n}}})x^{-1/2}\mid < x^{-1/2}$$ Let $g = x^{-1/2}$ $g$ is integrable as $\int_0^\pi x^{-1/2} = 2\sqrt{x} \mid_0^\pi = 2\sqrt\pi$ So we have a sequence of integrable functions $f_n$ that converges a.e. to the function $f_0$. Let $g = \frac{1}{\sqrt{x}}$. By the dominated convergence theorem as $\mid f_n \mid \le g$ on $(0, 1)$ for all $n$ we have that $$\int_0^1 f_0 = lim_{n \to \infty} \int_0^1 f_n $$ And the integral of the zero function $f_0$ is $0$ so we have that $$lim_{n \to \infty} \int_0^1(1 - e^{\frac{-x^2}{n}})x^{-1/2}dx = 0$$ Have I made any mistakes here? In particular, I have an issue with the way I brought the limit inside the integral in step 1. I couldn't apply the Monotone Convergence Theorem as we have decreasing sequence, so is there some other way to justify bringing it inside the integral? AI: Your argument is ok. In step 1 there are no integrals; you are just considering the limit of $f_n$.
H: Find the vector so that it is parallel Find a vector with length 7 and is parallel to the line $y=\frac{12}{13}x-1$ Is there some more advanced mathematics behind this than elementary maths or? Could use a nudge. AI: The line $y=12/13 x$ is parallel to the line $y=12/13x -1$. Consider a vector in it, for example $(13, 12)$. Divide it by its length and multiply by 7.
H: Arbitrary meaning two different things in two different settings The arbitrary in induction: Induction step: We assume that P(k) is true and then we need to show that P(k+1) is true as well. k is arbitrary and means "any one" Consider this example of arbitrary used in the rule of Universal Instantiation Rule: (FOR ALL x) P(x) => P(c) where c is some arbitrary element of the universe. c is arbitrary and essentially means "all". My issue here is, how can the value c represent all values at once? The argument is probably because c is arbitrary, that's how it can do it. But if c is arbitrary, notice that k in the induction type of problem is also arbitrary and it clearly cannot represent "all". Are there two different definitions of arbitrary? What is wrong here? Am I misunderstanding something? AI: You have $\forall x P(x)$ which means that $P(x)$ is true for every element in your space. This implies, of course, that $P(c)$ is true for any $c$. I do not see the problem in interpretation. When you use induction you want to prove $P(k) \Rightarrow P(k+1)$ for every $k$. Arbitrary means you can pick whatever element you want from your space. If you talk about $k \in \Bbb{N}$ arbitrary then $k$ does not represent all natural numbers but one natural number. At least that's how I understand it.
H: Why $\sum_{k=1}^{\infty}\frac{\ln(k)}{k}$ diverges? So my solution manual says this diverges since it is a p-series and $p=1 \le 1$ I understand why that p-series converges, but I don't understand conceptually how they got there using the comparison test (as the instructions advise). Again, here is my series: $$\sum_{k=1}^{\infty}\frac{\ln(k)}{k}$$ The instructions used the following comparison: $$\frac{\ln(k)}{k} \gt \frac{1}{k}$$ This is what confuses me about this problem and the solution. Why was the above comparison used? I understand the concept that if $a_n \lt b_n$ and $b_n$ converges then $a_n$ converges. This makes sense to me because I can see a graph in my head of a larger series converging so the smaller similar series must also converge. Is the above comparison checking to see if the smaller series diverges and if so the larger must also diverge? Please help me to understand this. AI: It is also true that if $a_n > 0, \quad b_n \gt 0$, $\sum a_n$ diverges and $a_n < b_n $, then $\sum b_n$ also diverges. Here, we are comparing $$a_n = \dfrac{1}{n}\;\;\text{ and }\;\;b_n = \dfrac {\ln n}n$$ and we have: $a_n >0, b_n>0$ $a_n$ diverges; (it's the harmonic series, which diverges; you should make sure you know this - and if you haven't already explored the proof of its divergence, do so!), $b_n > a_n$. So our series with the term $b_n$ diverges.
H: Radical extension over $\mathbb{Q}$ Let $K=\mathbb{Q}(\sqrt[n]{a})$, where $a\in\mathbb{Q}$, $a>0$ and suppose $[K:\mathbb{Q}]=n$. Let $E$ be any subfield of $K$ and let $[E:\mathbb{Q}]=d$. Prove that $E=\mathbb{Q}(\sqrt[d]{a})$. Hint: Consider $N_{K/E}(\sqrt[n]{a})\in E$. Attempt: I proved that the norm is in $E$ and if $E\supset\mathbb{Q}(\sqrt[d]{a})$ I could prove that $E=\mathbb{Q}(\sqrt[d]{a})$, but I can't prove that $E\supset\mathbb{Q}(\sqrt[d]{a})$. AI: Let us use multiplicativity of the norm. Write $z=N_{K/E}(\root n\of a)\in E$. Then $$z^n=N_{K/E}((\root n\of a)^n)=N_{K/E}(a)=a^{n/d},$$ because $N_{K/E}(y)=y^{n/d}$ for all $y\in E$. But $E$ is a subfield of the reals, and in the reals the only solutions of $z^n=a^{n/d}$ are $(\pm) \root d\of a,$ where the minus sign is a live possibility only when $n$ is even. So $z=\pm \root d\of a$. This implies that $\Bbb{Q}(\root d\of a)\subseteq E$. It sounds like you know how to complete the argument.
H: Undetermined coefficients guessing particular solution's shape I try to find general solution for: $$y'' - y = 8 t e^t$$ I find comp. as $(D-1)(D+1)=0$ and general solution of it as $y_g= C_1 e^t + C_2 e^-t$ at this point. I know particular solution shape is: $$y= A t e^t+ B t^2 e^t$$ But what is the way to find it? In other words how can I guess that? AI: We are using the Method of Undetermined Coefficients. The homogeneous solution is: $$y_h(t) = c_1 e^t + c_2 e^{-t}$$ Since we have an $e^t$ in the homogeneous, the method has us multiply by an extra $t$ term, so we try: $$y_p(t) = t(a e^t + b t e^t)$$ You can see a good discussion of the reasons and more examples on the linked web site, particularly example $10$. It shows that you try several approaches until you find one that works. After a little experience, you see a better approach for a choice as there is a pattern.
H: Finding the Maximum Likelihood Estimation (Self learning) I'm trying to learn the subject of Maximum Likelihood Estimation by myself. I'm facing one of the first questions in the textbook which is: `The waiting time (in minutes) on a queue to the dentist is the random variable $X$ with the following pdf: $$f(x) = \left\{\begin{matrix} 2\theta xe^{-\theta x^2} & x > 0\\ 0 & x \leq 0 \end{matrix}\right.$$ Find a maximum likelihood estimator for $\theta$ based on the waiting times of $n$ people that are waiting for the dentist, $X_1, X_2, ... X_n$. (The formula) Find the maximum likelihood estimation in a model of 3 people that were waiting 20, 50 and 30 minutes. (The exact number). Again, I am not interested in the exact answers but a way of looking and thinking of questions like this. I tried sketching the graph of $f(x)$ as $f(\theta, x)$ but I found it difficult. AI: In general, the likelihood is $$ L(\theta; x_1, x_2, \dots, x_n)=f(x_1, x_2, \dots, x_n ;\theta)=\prod_{i=1}^n f(x_i; \theta) $$ where I assumed independence (which is not explicitly stated here). What you want to sketch then is the likelihood. Or the log likelihood, which is very common to work with (due to invariance property of MLEs). In this case, that is $$ \log L(\theta; x_1, x_2, \dots, x_n)=\log\left(\prod_{i=1}^n 2\theta x_ie^{-\theta x_i^2}\right)=n\log2+n\log\theta+\sum_{i=1}^n\log x_i-\theta\sum_{i=1}^nx_i^2. $$ To find the maximum likelihood estimator, you differentiate this with respect to $\theta$, set it to 0 and solve for $\theta$. If it's a global maximum, it's your MLE. Once you get used to this method, it is very formulaic. To find the when you have data, you simply set $n=3$ and plug in your $x_i$'s in your estimator.
H: A self-convolution formula that counts bracket expressions Problem: Consider an alphabet of size $m+2$, consisting of the two bracket symbols $\ [ \ ] \ $ plus $m$ non-bracket symbols ($m \ge 0$). Define $f_m(n)$ to be the number of length-$n$ strings on this alphabet, in which brackets may occur only in the usual "balanced" manner of possibly-nested parentheses. We want a formula for $f_m(n)$. How to prove the following formula? (Is there some combinatorial derivation?) Conjecture: If $m,n$ are nonnegative integers, then $f_m(n) = a(n+2)$, where $a()$ is the sequence defined by the following recursion: $$\begin{array}{lcl} a(1) & = & m/2 \\ a(2) & = & 1 \\ a(i) & = & a(1)a(i-1) + a(2)a(i-2) + \cdots + a(i-1)a(1), \ \ \ i \ge 3 \end{array}$$ (The $a(i)$ should be written $a_m(i)$ to show they depend on $m$, but this is suppressed for simplicity.) In other words, for each $m$ the infinite sequence $(f_m(0), \ f_m(1) , \ f_m(2), \ \dots)$ is generated by starting with the initial segment $(m/2, \ f_m(0))$ in an auxilliary sequence, then computing each additional element as the self-convolution of its prefix. Example: $f_1(4) = 9$. There are 9 admissible length-4 strings on an alphabet with 1 non-bracket symbol (say $x$); namely, $\ xxxx, \ xx[], \ x[x], \ [xx], \ x[]x, \ [x]x, \ []xx, \ [[]], \ [][]$. The formula gives $f_1(4) = a(4+2) = 9$ in the sequence that starts with ($1/2, 1$): $$\begin{array}{lcl} 1/2 \\ 1 \\ (1/2)(1)+(1)(1/2)=1 \\ (1/2)(1)+(1)(1)+(1)(1/2)=2 \\ (1/2)(2)+(1)(1)+(1)(1)+(2)(1/2)=4 \\ (1/2)(4)+(1)(2)+(1)(1)+(2)(1)+(4)(1/2)=9 \end{array}$$ Motivation: In certain primitive computer languages (e.g., P'' and its brainfuck variations), programs are just the kind of strings we're considering, and I wanted to count the length-n programs. (For P'', there are $m = 3$ non-bracket symbols, and for brainfuck $m = 6$.) Here's a table of some $f_m(n)$ values computed by brute-force enumeration: Some f_m(n) values computed by enumeration m OEIS n = 0 1 2 3 4 5 6 7 8 9 10 -- ------- -------------------------------------------------------------------- 0 A126120 1 0 1 0 2 0 5 0 14 0 42 1 A001006 1 1 2 4 9 21 51 127 323 835 2188 2 A000108 1 2 5 14 42 132 429 1430 4862 16796 58786 3 A002212 1 3 10 36 137 543 2219 9285 39587 171369 4 A005572 1 4 17 76 354 1704 8421 42508 218318 5 A182401 1 5 26 140 777 4425 25755 152675 6 A025230 1 6 37 234 1514 9996 67181 458562 Every row of this table corresponds to a sequence found at OEIS, but I've found no reference connecting all of them to a single combinatorial problem (e.g., our "bracket string" counting problem) nor generating all of them from a single parametric formula (e.g., our conjecture). Aside: It may seem peculiar that the formula uses half-integers to generate integer sequences (for odd $m$). I think this is part of a general pattern in which an initial segment of some number of rationals may be used to parameterize a family of rational sequences generated from the initial segment by self-convolution. If the initial segment is a half-integer followed by some integers, the result will always be an integer sequence. AI: Your formula is correct, but perhaps not expressed in the most enlightening way. Write it instead as $\displaystyle f_m(n) = mf_m(n-1) + \sum_{k=0}^{n-2} f_m(k)f_m(n-2-k)$. The first term corresponds to those strings which begin with a non-bracket character. Otherwise, the first character is an left bracket, followed by some number of characters (a string of this type of some length $k$) followed by a right bracket, followed by a string of this type of length $n-2-k$ for a total length of $n$. (This is the same idea as in the proof of the Catalan recurrence.) Alternatively, one has $\displaystyle f_m(n) = \sum_{k=0}^{\lfloor \frac{n}{2}\rfloor} \binom{n}{2k}m^{n-2k}C_k$, where $C_k$ are the Catalan numbers.
H: Why is this binary-relation symmetric? From the example of binary-symmetric-relation demonstrated in Wikipedia, how can they say the relation "$x$ and $y$ are odd numbers" is symmetric without stating any set of $x$, $y$? If such set is $\mathbb{N}$, then the relation is not symmetric because relation set does not contain even numbers. Am I thinking correctly or just missing something? AI: The relation is defined as $x\sim y$ if $x,y$ are both odd numbers. They do not specify the set $X$, on the other hand, this $X$ can be whatever you want. It could be the set of naturals, or the set of all real numbers, or even the set of all topologies on some set. To check that this is indeed symmetric, let $x\sim y$. Then $x$ and $y$ are odd numbers. But then we can also say "$y$ and $x$ are odd numbers", hence $y\sim x$. You can visualize $x\sim y\iff x,y\in A$ as a subset of $X\times X$ by thinking of the Cartesian square $(A\cap X)\times(A\cap X)$ within $X×X$. This way one sees easily that it is symmetric and transitive, and it's reflexive only if $A\supseteq X$. That is why normally we add to this relation the diagonal in $X×X$. We then get the equivalence relation $x∼y\iff x=y\vee x,y\in A$.
H: Finding the velocity vector Am I finding the equation of the slope of the tangent line at c(t)? $\frac{dy/dt}{dx/dt}$ = $\frac{2t}{3t^2-8}$ AI: The position of the particle at any particular time will be given by the vector r(t) = (x(t),y(t)) You want to differentiate r(t), i.e, v(t) = dr(t)/dt = (dx(t)/dt,dy(t)/dt) Note that the speed will be given by s(t) = d/dt of sqrt(x(t)^2+y(t)^2), but that is not the velocity or a vector
H: Prove that the following sequence has a finite partial limit Let $ a_n $ be a sequence. Prove that if the sequence $\|a_n\| $ does not converge to $\infty$, $a_n$ has a finite partial limit. AI: As $\|a_n\|$ doesn't converge there must be a subsequence $b_i$ such that $b_i \in [-K,K]$ for some $K \in \mathbb{R}$ (If such a subsequence doesn't exist $\text{lim}_{n \rightarrow \infty}\|a_n\|=\infty$). Now take any convergent subsequence of $b_i$ (exists due to Bolzano Weierstrass).
H: Prove that [0,1] is equivalent to (0,1) and give an explicit description of a 1-1 function from [0,1] onto (0,1) The problem is stated as follows: Show that there is a one-to-one correspondence between the points of the closed interval $[0,1]$ and the points of the open interval $(0,1)$. Give an explicit description of such a correspondence. Now, I think I can prove the first part of the problem by demonstrating the following: Define $f: (0,1) \to \mathbb{R}$ as follows. For $n \in \mathbb{N}$, $n \geq 2$, $\space{ }f(\frac{1}{n}) = \frac{1}{n-1}$ and for all other $x \in (0,1)$, $\space{}f(x) = x$ Prove that $f$ is a $1-1$ function from $(0,1)$ onto $(0,1]$ Slightly modify the above function to prove that $[0,1)$ is equivalent to $[0,1]$ Prove that $[0,1)$ is equivalent to $(0,1]$ Since the "equivalent to" relation is both symmetric and transitive, it should follow that $[0,1]$ is equivalent to $(0,1)$. Hence, there does exist a one-to-one correspondence between $[0,1]$ and $(0,1)$. I have no trouble with the above. My problem is in "finding an explicit description of such a correspondence." Can I modify the above function, or will that not suffice? AI: Steps 2 and 3 are not necessary. The function $g:(0,1] \to [0,1]$ defined by $g(1) = 0$ and $g(x) = f(x)$ if $x \neq 1$ is a bijection. This shows that $(0,1]$ is equivalent to $[0,1]$ and, by transitivity, that $(0,1)$ is equivalent to $[0,1]$. Furthermore, the function $g \circ f$ is a one-to-one correspondence between $(0,1)$ and $[0,1]$ that you can describe explicitly.
H: Quotient groups and dihedral groups I came across a question asking for me to find all the possible quotient groups for the dihedral group $D_6$. How must I go about this? AI: I assume that your definition of $D_6$ is $$D_6=\{1,r,r^2,s,sr,sr^2\}$$ where $r^3=s^2=rsrs=1$. A quotient set $D_6/N$ is a group if and only if $N\lhd D_6$. Hence, we want to find all normal subgroups of $D_6$. We always have the trivial ones $$D_6/D_6\cong\{1\}$$ and $$D_6/\{1\}\cong D_6.$$ The subgroup $\langle r\rangle = \langle r^2\rangle= \{1,r,r^2\}$ is of index $2$ and thus normal. So $$D_6/\langle r\rangle$$ is a quotient group also. If $N\lhd D_6$ is a different normal subgroup, then $\|N\|=2$, so either $N=\langle s \rangle$, $N=\langle sr\rangle$ or $N=\langle sr^2\rangle$. But none of them are normal. For example $\langle s \rangle$ is not normal since $(sr)s(sr)^{-1}=sr^2\notin\langle s \rangle$. Hence, the only non-trivial quotient group is $$D_6/\langle r\rangle\cong\mathbb{Z}/2\mathbb{Z}$$
H: Adjunctions in Category Theory My question is how can we see that if $F$ and $F'$ are both left adjoints of $G$, there is natural isomorphism between $F$ and $F'$? So how can we show that adjoints are unique upto isomorphism? AI: If $F \vdash G$, $F' \vdash G$, then we have units/counits $\epsilon:I \rightarrow FG, \eta:GF \rightarrow I$, and $\epsilon':I \rightarrow F'G, \eta':GF' \rightarrow I$. Then $\epsilon F':F' \rightarrow FGF'$, while $F\eta' :FGF' \rightarrow F$. Composing these gives a natural transformation $F \rightarrow F'$. Now you, of course, has to find the inverse map and show it actually is an inverse. The nice thing about this approach is it's already natural! For me, the key is to understand how to translate your information into the "algebra of functors" and work with it there.
H: Sum of two subspaces Let $$U_1 := \{(\gamma_1, \gamma_2, \gamma_3) \in \Re^3 : \gamma_1 + \gamma_2 + \gamma_3 = 0\}$$ $$ U_2 := \{ (\lambda, \lambda, \lambda) : \lambda \in \Re \}$$ be subspaces. Show that $U_1 + U_2 = \Re^3$. The sum of the two subspaces $U_1 + U_2$ is by definition $\{u_1 + u_2 : u_1\in U_1, u_2 \in U_2\}$. If $v\in \Re^3 $ is some arbitrary element, then this means that $v = u_1 + u_2$ for some $u_1$ and $u_2$. But I don't see how you can create every element of $\Re^3$ on that way. By adding $(\lambda, \lambda, \lambda)$ I can create all multiples of a given $(\gamma_1, \gamma_2, \gamma_3)$, but since $\gamma_1 + \gamma_2 + \gamma_3 = 0$, the gammas are always in a fixed proportion, so that two gammas always equal the remaining one or all are equal to 0, and I don't see how I can create elements from $\Re^3$ in which all three gammas are different. Is the statement $U_1 + U_2 = \Re^3$ true? Can someone give me a hint please or tell me whether I am doing something wrong? AI: You can create every element $(x,y,z)\in\mathbb R^3$ it the following way. $$(x,y,z) = (x-b,y-b,2b-x-y)+(b,b,b)$$ where $b=\frac{x+y+z}{3}$
H: What does the author (Ahlfors) mean here (modular $\lambda$ function) In Ahlfors' complex analysis text, page 281 it says By reflection the region $\Omega'$ that is symmetric to $\Omega$ with respect to the imaginary axis is mapped onto the lower half plane, and thus both regions together correspond to the whole plane, except for the points $0$ and $1$. Here $\Omega$ is the region bounded by the lines $\Re \tau=0,\Re \tau=1$ and the circular arc of $\|\tau-1/2\|=1/2$ in the upper half plane. By "correspondence" he means under the modular $\lambda$ function. Now, both regions together is IMHO the region $\Omega \cup \Omega'$ along with the positive imaginary axis. I don't think that real values are taken on this region except those in the open interval $(0,1)$. Thus, in my opinion both regions together should correspond to $$\mathbb C \setminus \left[ (-\infty,0] \cup [1, +\infty) \right] $$ rather than $$\mathbb C \setminus \{0,1 \} $$ as the author claims. Am I right? If not, please tell me why. Thanks. AI: You are right, but Ahlfors is also right. It's caused by a not perfectly clear formulation. You are (naturally) thinking of the "region" $\Omega \cup \Omega' \cup i\cdot (0,\infty)$, understanding "region" as "connected open set". In this part of the theory, however, Ahlfors considers the "fundamental region" of $\lambda$, which is the set $\overline{\Omega} \cup \Omega'$, as witnessed by Theorem 8. Every point $\tau$ in the upper half plane is equivalent under the congruence subgroup $\mod 2$ to exactly one point in $\overline{\Omega}\cup \Omega'$. That fundamental region (to add further confusion, the closure $\overline{\Omega}$ should be understood as the closure in the upper half plane, so $0,1 \notin\overline{\Omega}$) is mapped bijectively to $\mathbb{C}\setminus \{0,1\}$ by $\lambda$. In that view, "both regions together correspond to the whole plane, except for the points $0$ and $1$."
H: Is half of the area of a Fibonacci triangle a congruent number? A Fibonacci triangle is a triangle with integer area and sides whose lengths are Fibonacci numbers. An example of a Fibonacci triangle is the triangle whose sides have lengths (5,5,8). Is half of the area of a Fibonacci triangle a congruent number ? (A congruent number is a positive integer which is the area of a right-angled triangle with rational sides.) For the triangle (5,5,8) the area is $n=12$, and $\frac{n}{2}=6$ is a congruent number. Indeed, the right-angled triangle with sides $(3,4,5)$ is a half of the Fibonacci triangle $(5,5,8)$. AI: (Corrected - thanks to Daniel Fischer) If a triangle is made from 3 Fibonacci numbers $a \le b \le c$, then since we must have $c < a + b$ (in any triangle), it must be that the Fibonacci numbers are actually $F_n, F_n, F_k$ where $1\le k \le n+1$ and the triangle is isosceles. Divide the isosceles triangle along its line of symmetry. Since the area is assumed to be an integer, and $A = \frac12 b \times h$, it follows that the height must also be a rational number, so that half the area is just the area of right angled triangle with rational sides, and must then automatically be a congruent number.
H: A question on combinatorics Bob and Laura have bought an apartment, and are going to carpet the floor in the kitchen. The kitchen has size $2 \times n$, where $n$ is a positive integer. In how many ways can Bob and Laura complete it using carpet pieces of two types: of size $1 \times 2$ and of size $2 \times 2$? Solve by recursion. Can anybody please help with this question, and possibly explain how to solve it? AI: Let $a_n$ be the number of ways to carpet a $2\times n$ kitchen. Think of the kitchen as a strip of length $n$ running from left to right. The righthand end of the strip can be carpeted in any one of three different ways: it can be a $2\times 2$ piece, it can be a pair of $1\times 2$ pieces oriented parallel to the strip, or it can be a single $1\times 2$ piece oriented crosswise to the strip. The first two possibilities are extensions of a carpeting of a $2\times(n-2)$ kitchen; the third is an extension of a carpeting of a $2\times(n-1)$ kitchen. That is, each of the $a_{n-2}$ carpetings of a $2\times(n-2)$ kitchen can be extended to $2$ different carpetings of a $2\times n$ kitchen, and each of the $a_{n-1}$ carpetings of a $2\times(n-1)$ kitchen can be extended to one carpeting of a $2\times n$ kitchen. Moreover, every carpeting an a $2\times n$ kitchen arises uniquely in one of these three ways. You should now be able to write down a recurrence expressing $a_n$ in terms of $a_{n-1}$ and $a_{n-2}$. It will be a homogeneous recurrence with constant coefficients, and you should have some techniques for solving such recurrences. I’ll stop here for now; if you get the recurrence but can’t solve it, leave a comment indicating where you’re stuck.
H: Asymptotic behavior of a solution to the quadratic equation I have a quadratic equation of real $x$, $$ x^2 - 4(1+2y)x + 8(y+1) = 0 $$ for $ x>0, y>0$ and the solution is $$ x(y) = 4y + 2 - \sqrt{4(1+2y)^2 - 8(y+1)} $$ $$ = 4y + 2 - 2\sqrt{4y^2 + 2y -1} $$ I found the solution approaches to 1 for large $y$, if I plot $x$ vs $y$. How can I show the solution is going to 1, for large $y$? Want to show $$ \lim_{y \rightarrow \infty} x(y) = 1 $$ Matlab script : y = linspace(0.00001, 500, 1000); xy = 4y +2 - 2(4y.^2 + 2y - 1).^(1/2); plot(y, xy) AI: $\begin{align} \sqrt{4y^2 + 2y -1} &=\sqrt{4y^2 + 4y+1-2y -2}\\ &=\sqrt{(2y+1)^2-2y -2}\\ &=\sqrt{(2y+1)^2-(2y+1) -1}\\ &=(2y+1)\sqrt{1-\frac1{2y+1} -\frac{1}{(2y+1)^2}}\\ \end{align} $ Let $z = 2y+1$. Then since $\sqrt{1+w} =1+\frac{w}{2}+O(w^2) $ as $w \to 0$, $\begin{align} 4y + 2 - 2\sqrt{4y^2 + 2y -1} &= 2z-2z\sqrt{1-\frac1{z} -\frac{1}{z^2}}\\ &= 2z-2z(1-\frac12(\frac1{z} -\frac{1}{z^2})+O(\frac1{z^2})\\ &=2z-(2z-1+O(\frac1{z})\\ &=1+O(\frac1{z})\\ \end{align} $ Therefore $\lim_{y \to \infty} 4y + 2 - 2\sqrt{4y^2 + 2y -1} = 1 $.
H: How to show the space is totally bounded? the space the Real line with bounded metric (i.e. d/(1+d), d: euclidean) we know that totally boundedness means that there exists a finite epsilon-net. we first approached to question by directly try to find finitely many points s.t. their metric balls cover the space. then, tried , by contradiction, but failure. AI: HINT: Let $d_b$ be the bounded metric, so that $$d_b(x,y)=\frac{\|x-y\|}{1+\|x-y\|}$$ for all $x,y\in\Bbb R$. Suppose that $m,n\in\Bbb Z$ and $m\ne n$; then $d_b(m,n)\ge\ldots\;$?
H: Recovering ring information from localizations I'm curious if such a statement is true: Let $R$ be a commutative ring with unity. Then if $x \in R$ and $x = 0$ in $R_p$ for all primes $p$, where $R_p$ is the localization of $R$ at prime $p$, then $x = 0$ in $R$. AI: Yes this is true. You can even restrict to maximal ideals. To say that $r\in\ker(R\rightarrow R_\mathfrak{m})$ for a maximal ideal $\mathfrak{m}$ means that there is some $s\in R\setminus\mathfrak{m}$ with $sr=0$. It follows that $s\in\mathrm{Ann}_R(r)$. If this is true for every maximal ideal $\mathfrak{m}$ of $R$, then the ideal $\mathrm{Ann}_R(r)$ cannot be contained in any maximal ideal, and therefore must be equal to $R$. So $1\in\mathrm{Ann}_R(r)$, i.e. $0=1\cdot r=r$.
H: Prove $x^n-1$ is divisible by $x-1$ by induction Prove that for all natural number $x$ and $n$, $x^n - 1$ is divisible by $x-1$. So here's my thoughts: it is true for $n=1$, then I want to prove that it is also true for $n-1$ then I use long division, I get: $x^n -1 = x(x^{n-1} -1 ) + (x-1)$ so the left side is divisible by $x-1$ by hypothesis, what about the right side? AI: So first you can't assume that the left hand side is divisible by $x-1$ but for the right hand side we have that $x-1$ divides $x-1$ and by the induction hypothesis we have that $x-1$ divides $x^{n-1}-1$ so what can you conclude about the left hand side.
H: Help me evaluate this integral $\displaystyle \int \frac{\ln x}{x^2} \mathrm dx$ I just can't seem to figure this one out. I tried integrating by parts but I'm stuck. AI: Integration by parts is not u-substitution. You split the integrand into two parts, $u$ and $dv$. Choose $u$ so that it will simplify when you take it's derivative. In this case, $u$ simplifies when it is ln(x) and the derivative is 1/x . $u = \ln(x)$ $du = (1/x)dx$ $v = -1/x$ $dv = (1/(x^2))dx$ $uv - \int{vdu} = (\ln(x))\cdot(-1/x)- \int{(-1/(x^2)) dx} = (-\ln x)/x - (1/x) = -( ln x+1)/x$
H: Clarification on Rules Differentiation and First Principles Derivatives My grade 11 class has just started differential calculus, the one area seemingly glazed over in our book. We have covered some simple rules of differentiation, like f(x) = n x^(n-1), and have applied these with subsequent rules like the constant rule, constant multiple rule, and sum or difference rule, child's play to you, I am sure. Anyway, we also covered the derivative, defined as: The limit as h approaches zero = (f(x+h)-f(x))/h... We were told by our teacher that tossing any equation into that formula will result in the spitting out of our derivative function... We were also told that using the rules of differentiation will break things down into a derivative function; therefore, should using these to methods of derivation not produce the exact same results? Apparently not: Take x^2 + 1 From the rules of differentiation we get 2 x right? But from the above formula we get 2 x+h Now, clearly just by the presence of the h, which cannot magically come into the formula via the rules of differentiation, the rules and the formula do/produce different things... Does one produce the function of the tangent line? Does one produce the function of the gradients of possible tangents? Which one of the above is the derivative? The specific tangent? Or the function to produce the slopes of said tangents? Which is yielded by which? Sorry, I understand how stupid and trivial this all seems but I want to set a proper calculus foundation for the future and cannot risk misunderstanding something, and if I do, then I need rectification! Thank you to any and all who may answer I will certainly appreciate the help. AI: The problem is with your calculation of the limit: $$\begin{align} \lim_{h\to 0}\frac{f(x+h)-f(x)}h&=\lim_{h\to 0}\frac{\big((x+h)^2+1\big)-\big(x^2+1\big)}h\\\\ &=\lim_{h\to 0}\frac{(x^2+2xh+h^2+1)-(x^2+1)}h\\\\ &=\lim_{h\to 0}\frac{2xh+h^2}h\\\\ &=\lim_{h\to 0}(2x+h)\\\\ &=2x\;, \end{align}$$ not $2x+h$. In fact the formula comes from applying the definition to the general case, so it cannot give a different result.
H: What is the easiest way to find the inverse Laplace of F(s)? $$ F(s)= \frac{1}{(s-1)^2(1-1/s^2)} $$ Do I have to multiply by $s^2/s^2$ and then use partial fractions or is there a way to use the convolution theorem? AI: This can be rewritten as (another form is also possible): $$F(s) = \dfrac{s^2}{(s-1)^3 (s+1)}$$ Using Partial fractions, this is: $$F(s) = \dfrac{s^2}{(s-1)^3 (s+1)}= -\dfrac{1}{8 (s+1)} + \dfrac{1}{8 (s-1)}+\dfrac{3}{4 (s-1)^2}+\dfrac{1}{2 (s-1)^3}$$ Now, you can find the inverse laplace transform using a table or using the definitions. We arrive at: $$F(t) = \dfrac{1}{8} e^{-t} (2 e^{2 t} t^2+6 e^{2 t} t+e^{2 t}-1)$$ A second form is possible (of course the ILT is identical): $$F(s) = = \dfrac{s^2}{(s-1)^3 (s+1)}= \dfrac{s^2}{(s-1)^2 (s^2-1)}$$
H: Compute the limit $\lim_{n \to \infty} \frac{n!}{n^n}$ I am trying to calculate the following limit without Stirling's relation. \begin{equation} \lim_{n \to \infty} \dfrac{n!}{n^n} \end{equation} I tried every trick I know but nothing works. Thank you very much. AI: By estimating all the factors in $n!$ except the first one, we get: $$0 \leq \lim_{n \rightarrow \infty} \frac{n!}{n^n} \leq \lim_{n \rightarrow \infty} \frac{n^{n-1}}{n^n} = \lim_{n \rightarrow \infty} \frac{1}{n} = 0$$
H: How many transitive relations on a set of $n$ elements? If a set has $n$ elements, how many transitive relations are there on it? For example if set $A$ has $2$ elements then how many transitive relations. I know the total number of relations is $16$ but how to find only the transitive relations? Is there a formula for finding this or is it a counting problem? Also how to find this for any number of elements $n$? AI: There is no simple formula for this number (but see http://oeis.org/A006905 for the values for small $n$). The case $n=2$ is small enough that you can list out all 16 different relations and count the ones that are transitive. (You should get 13 of them.)
H: Show that the set $\{\frac{1}{n^2}: n \in N \} \cup \{0\}$ is closed I want to show that $Y=\{\frac{1}{n^2}: n \in N \} \cup \{0\}$ is closed. I think I can do so by arguing that $Y$ contains its upper and lower bounds, $1$ and $0$, and hence is closed. But I think I might need to argue that $Y^c$ is open, which implies that $Y$ is closed. Do you have any suggestions? AI: It’s not enough to show that $Y$ contains its upper and lower bounds: $[0,1)\cup(2,3]$ contains its upper and lower bounds but is not closed. Show instead that its complement is open; you can actually write the complement explicitly as a union of pairwise disjoint open intervals (counting open rays as open intervals).
H: How to prove that $(\sec x-\cos x)(\csc x -\sin x)=\tan x/(1+\tan^2x)$? I've tried solving from both sides but can't seem to get them equal. Any tips on which side to start on and what to do after? AI: Outline: $$\sec x - \cos x = \frac 1 {\cos{x}} - \cos x = \frac{1 - \cos^2 x}{\cos x} = \frac {\sin^2 x}{\cos{x}}$$ Likewise, $$\csc x - \sin x = \frac{\cos^2 x}{\sin x}$$ Now divide and simplify, noting that $$1 + \tan^2 x = \sec^2 x = \frac 1 {\cos^2{x}}$$ Combine this with the fact that $\tan x = \sin x / \cos x$.
H: How to find $f(2013)$ if $f(5)=45$ and $f(m)+f(n)= f(m+n)$ for all $m,n\in\mathbb N$? $f: \mathbb N\to\mathbb N$, $f(m)+f(n)=f(m+n)$ for all $m,n\in\mathbb N$, and $f(5)=45$. Find $f(2013)$. I messed up my original posting, its fixed now. I changed $m+ n$ to $f(m+n)$. AI: The question now says that $f$ is linear (which is very different!). Hint: $kf(1)=f(1)+\ldots +f(1)=f(1+\cdots +1)=f(k)$ and so if $f(5)=45$ then $5f(1)=f(5)=45$. Can you find $f(1)$ from this? (This was the contradiction that appears using the previous, unedited question.) No such $f$ exists. If $\mathbb{N}$ includes $0$ then $f(m)=m$ for all $m\in\mathbb{N}$ as shown in the comments above. If $\mathbb{N}$ does not contain $0$ then $2f(1)=f(1)+f(1)=1+1=2$ and so $f(1)=1$. Now, $f(m)+1=f(m)+f(1)=m+1$ and so $f(m)=m$ for all $m\in\mathbb{N}$. This contradicts $f(5)=45$.
H: Laplace transform of two functions I know how to do a basic laplace transform, but how does one deal with transforming complex combination of functions? For example, how would we handle: $$\mathcal{L}\left( \ \sqrt{\frac{t}{\pi}}cos(5t) \right) = ... $$ From a table of laplace transforms it is known that: $$\mathcal{L}\left( \ \frac{cos(5t)}{\sqrt{\pi t}} \right) = \frac{e^{-5/s}}{\sqrt{s}}$$ This table value must be of some use to solve this problem, but how? EDIT: Can we use $\mathcal{L}\left( f(t) g(t) \right) = \mathcal{L}\left( f(t)\right) \mathcal{L}\left( g(t)\right) $? AI: Hint: Use the property $$ L(t f(t)) = -F'(s). $$ Added Note that, $$\frac{\sqrt{t}}{\sqrt{\pi}}{\cos(5t)} = t \frac{\cos(5t)}{\sqrt{\pi t}}.$$ Now, take Laplace transform of both sides of the above equality $$ \mathcal{L}\left\{ \frac{\sqrt{t}}{\sqrt{\pi}}{\cos(5t)} \right\}= \mathcal{L}\left\{ t \frac{\cos(5t)}{\sqrt{\pi t}} \right\}=-\frac{d}{ds}\frac{e^{-5/s}}{\sqrt{s}}=\dot\,. $$
H: Construction of two graphs I would like to know if it is possible to construct two graphs $G,H$ such that $\|G\|=\|H\|, e(G)=e(H)$ (means that the two graphs have the same number of vertices and edges) and $\chi(G)>\chi(H)$ where $\chi()$ represents the chromatic number, such that there are more ways to colour $G$ than $H$ when the number of available colours is large. It would be great if you have some ideas. AI: The first graph contains a copy of $K_4$, so it requires $4$ colors.
H: Why arithmetic series never sums to a fraction Sometimes with series we find a solution in a form of a fraction which does not a priori obviously take only integer values. On the other hand from the sum it is pretty obvious that the sequence of partial sums can take only integer values. (arithmetic and geometric series are examples) Someone might come along and say: Alright this is the formula for the sum. Now we need to prove that it is always an integer. Another person might say: Actually, no. We have already proved that this formula amounts to adding the series which obviously takes integer values. Who is right? I guess my question is: after you prove $1+2+3+4+...+n= \frac{n(n+1)}{2}$ do you need to prove that the formula on the right hand side is always an integer, or does it just follow? AI: Perhaps a better example for the kind of question you are asking is the formula $C(n,k)=\frac{n!}{k!(n-k)!}$. It is not a-priori trivial that the RHS is an integer, but it is proven to be so if one proves that this formula counts the number of combinations of $k$ elements from $n$ elements. It is then a very rigorous proof that the RHS is indeed an integer.
H: Evaluate $\tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)$ without using a calculator Evaluate without using a calculator: $\displaystyle{\tan\left(2\sin^{-1}\left(\sqrt{5} \over 5\right)\right).}$ So I built my triangle hyp=$5$, adj=$2\sqrt{5}$, opp=$\sqrt{5}$. $$ \tan\left(2\theta\right) = 2\sin\left(\theta\right)\cos\left(\theta\right)\,, \qquad\qquad \tan\left(2\theta\right) = 2\,{\sqrt{5} \over 5}\,{2\sqrt{5} \over 5} = 4. $$ I used my calculator and it says the answer is $4/3$. I'm not sure what I'm doing wrong AI: Using the identity $\tan{2A}=\dfrac{2\tan A}{1-\tan^2 A}$, $\tan \theta=\dfrac{\sqrt{5}}{\sqrt {20}} =\dfrac{1}{2}. \text{Hence} \tan\left(2\sin^{-1}\frac{\sqrt{5}}{5}\right)=\dfrac{2\times \dfrac{1}{2}}{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{4}{3}$
H: Eigenvector and Its Span Let $V$ be a vector space over the field $F$ and let $T$ be a linear transformation from $V$ to $V$. Let $v\in V$ such that $v\neq 0$, let $W=span\{v\}$. Prove that if $T(W) \subset W$, then $v$ is an eigenvector for $T$. So, my idea for this problem was since $v\in V, v=1.v\in span\{v\}=W$. So $v\in W$. Then, since $W=span\{v\}$, can this be inferred that $T(v)\in T(W)$? If this is true, then I know how to do the rest, but I just want to know that if I can say that? Thank you for any input! AI: Yes, $T(v)\in T(W)$. You can convince yourself of this by looking at the definition of $T(W)$ which is $T(W):=\{T(w)\colon w\in W\}$.
H: Path counting discrete I need help understanding a simple concept. Lets say you're given a problem where you start at (0,0) on a 2D grid and want to count the number of paths to (8,4). Would I be following the combinations problem where (m,n) = (m + n) choose m? And would this be the same as (m + n) choose n? To build off that problem, lets say I'm not able to move through (3,2) to (4,2). Would this mean that I would solve the same problem (12 choose 8) - (5 choose 3) because I don't want to move right at the point (3,2)? Thanks guys! AI: If you’re allowed only to move one unit to the right or one unit up, then you must take a total of $m$ steps to the right and $n$ up to get from $\langle 0,0\rangle$ to $\langle m,n\rangle$. You can take those steps in any order and still reach your destination, so the $m$ steps to the right can occupy any $m$ positions in the string of $m+n$ steps. Thus, there are indeed $\binom{m+n}m=\binom{m+n}n$ possible paths using only these two types of step. In particular, there are $\binom{12}8$ such paths from the origin to $\langle 8,4\rangle$. If you’re not permitted to take the step from $\langle 3,2\rangle$ to $\langle 4,2\rangle$, you must subtract from $\binom{12}8$ the number of paths that use this step. Each of those paths consists of a path from the origin to $\langle 3,2\rangle$, the forbidden step, and a path from $\langle 4,2\rangle$ to $\langle 8,4\rangle$. There are $\binom53$ paths from the origin to $\langle 3,2\rangle$. Getting from $\langle 4,2\rangle$ to $\langle 8,4\rangle$ requires $8-4=4$ steps to the right and $4-2=2$ steps up, so there are $\binom{4+2}4=\binom62$ such paths. Each of them can be combined with any of the $\binom53$ paths from the origin to $\langle 3,2\rangle$ to make a forbidden path, so there are altogether $\binom53\binom64$ forbidden paths and $$\binom{12}8-\binom53\binom64=495-10\cdot15=345$$ acceptable paths.
H: Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$ Find all solutions of the equality $y^2=x^3+23$ for integers $x,y$ I guess, that x cannot be even. because we can apply mod 4 test say $x=2k$ then $y^2=8k^3+23\equiv3\mod(4)$ but, this is not possible, since the square of an integer is congruent to $0$ or $1$ $\mod (4)$ so $x$ is of the form $4k+3$ or $4k+1$ $x=4k+3$ is also not possible for the same reason but for the last case the test fails. any hints ? AI: There are no integer solutions. You noted already that $x$ cannot be even, because the equation would then fail mod $4$; so we need only consider $x$ odd. Then $x^3+23$ is even, so must be a multiple of $4$, whence $x \equiv 1 \bmod 4$. We cannot obtain a contradiction from further congruence considerations. But since $23 = 3^3 - 2^2$ we try adding $4$ on both sides, getting $$ y^2 + 4 = x^3 + 27 = (x+3) (x^2-3x+9), $$ and observe that $x \equiv 1 \bmod 4$ implies $x^2 - 3x + 9 \equiv 1 - 3 + 9 \equiv 3 \bmod 4$. Since also $x^2-3x+9 > 0$ we deduce that $x^2-3x+9$ has a prime factor $p \equiv 3 \bmod 4$. But this is impossible by quadratic reciprocity: $y$ would be a square root of $-4 \bmod p$. QED (In fact mwrank reports that there aren't even any rational solutions, but that's not an elementary proof.) [Added later: See Keith Conrad's Examples of Mordell's Equation for further examples of elementary but nontrivial proofs that certain equations of this form $y^2 = x^3 + k$ have no integer solutions, and also some examples where a nonempty list of solutions of $y^2 = x^3 + k$ can be proved complete.]
H: Removing the square root in $\sum_{n=1}^{\infty}\frac{\sqrt{n+3}}{4n^2+n+4}$ I know I can use the comparison test to examine this series, but say I wanted to take the limit of this series, how do I handle square root in the series? $$\sum_{n=1}^{\infty}\dfrac{\sqrt{n+3}}{4n^2+n+4}$$ For example, if there wasn't a square root I would just divide all the terms in the numerator and denominator by the highest power of $n$. But I just don't see how I do that algebraically with a square root. AI: For example $$\frac{\sqrt{n+3}}{4n^2+n+4}\le\frac{\sqrt2\sqrt n}{4n^2+n+4}\le\frac{\sqrt2}4\frac{\sqrt n}{n^2}=\frac1{2\sqrt2}\frac1{n^{3/2}}$$ and now apply the $\;p-$ series criterion: the series converges.
H: Is $O(n \log n)$ always smaller than $O (m)$ for $n-1 < m < n^2$? I am writing an algorithm that needs to finish in $O(m)$. The problem is for a graph $G( V, E )$, where $m = \|E\|$ and $n = \|V\|$. $m$ can be in the range of $n-1$ to $n^2 - 1$. If I do some processing that takes $O\left( (n-3) \log(n-3)\right)$ time, will that still satisfy being smaller than $O(m)$ or not? Thanks AI: $m > n - 1$ implies that $(n-3) \log(n-3) < (m-2) \log (m-2)$. That is $O(m \log m)$, but not $O(m)$.
H: Kronecker's Theorem - what can be deduced if $f$ is reducible? I am happy with the statement of Kroncker's Theorem as follows. Let $\:$ $ f \in K[X]$. Then there exists a simple field extension $L = K(\alpha)$ of $K$ with $f(\alpha) = 0$. If $f$ is irreducible over $K$, then the extension $L$ is unique in sense that the natural map $$K[X]/fK[X] \rightarrow L, \: \: \: \: \: \: g + fK[X] \rightarrow g(\alpha)$$ constitutes an isomorphism. However, I am trying to prove something at present and need to understand what can be said if $f$ is actually reducible. So I am wondering to what extent we can reason the other way, i.e. if we have some polynomial $f$ of degree $n$ that is not irreducible, then is it impossible that it generates a field of size $p^n$ via the quotient group $F_p[X]/fF_p[X]$ and hence can we conclude that $f$ must be irreducible? AI: If $f$ is reducible, then it does not generate a field at all. For example, if $f = X^2$ then there is no polynomial $g$ such that $Xg = 1$ modulo $f$. This generalises to other $f$ and does not depend on the field $K$. However, the irreducible factors of $f$ do generate a field.
H: Prove that free modules are projective Prove that free modules are projective. Help me. AI: Definition: $M$ is projective iff for every surjective morphism $f:A\rightarrow B$, and every morphism $g:M\rightarrow B$ there is morphism $h:M\rightarrow A$ such that $fh=g$. Now if $M$ is free with base $\{x_i\}$ then we can define $g$ setting $x_i\mapsto a_i\in f^{-1}(g(x_i))$.
H: Show $\int_{\mathbb{R}^n}\Delta_x \Phi(x-y)f(y)dy = \int_{\mathbb{R}^n}\Delta_y \Phi(x-y)f(y)dy.$ I read in an article about Laplace's equation that $$-\int_{\mathbb{R}^n}\Delta_x \Phi(x-y)f(y)dy = -\int_{\mathbb{R}^n}\Delta_y \Phi(x-y)f(y)dy.$$ Could someone explain to me why this is? I understand that one can move the differentiation from one factor of the convolution to the other... AI: $ \nabla_x \Phi(x-y) = - \nabla_y \Phi(x-y) $ by the chain rule. Apply this identity twice.
H: What is this formula called? summation Does people know what this formula called? I want to google its properties and read about it more - if it has official name AI: It is called a "weighted average" or "weighted mean". The numbers $\mu_i$ are called the "weights".
H: equivalence relation-showing that an operation is well-defined Define $f: \mathbb{Z}_n \to \mathbb{Z}_n$ as $f([a]) = [a^2]$. Show that $f$ is a well-defined function. I am confused as to how I could show this. AI: You need to show that if $[a]=[b]$ in $\Bbb Z_n$ (i.e., $a\equiv b\pmod n$, i.e. $n\,\|\,b-a$), then $[a^2]=[b^2]$. For this, show first that $[a]=[b]$ implies $[a^2]=[ab]$.
H: How do you prove a piece of the Short Five Lemma? Let $\alpha, \beta, \gamma$ be a homomorphism of short exact sequences, in that order. Then if $\alpha, \gamma$ are injective, then so is $\beta$. Let the sequeces be: $$ \begin{matrix} 0 & \to & A & \xrightarrow{\psi} & B & \xrightarrow{\phi} & C & \to & 0 \\ \ & \ & \downarrow^{\alpha} & \ & \downarrow^{\beta} \ & \ & \downarrow^{\gamma} \\ 0 & \to & A' & \xrightarrow{\psi'} & B' & \xrightarrow{\phi'} & C' & \to & 0 \end{matrix} $$ Then since $\psi, \psi', \alpha$ are injective and $\beta \psi = \psi' \alpha$, we have that $\beta$ is injective on $\psi(A)$. AI: I was reluctant to use the book, but ended up resorting to it. Here's my understanding of the proof. Suppose $\beta(b) = 0$. Then $\gamma \phi(b) = \phi'\beta(b) = \phi'(0) = 0$, so $\phi(b) = 0$ by injectivity of $\gamma$. Then $\ker \beta \subset \ker \phi = \psi(A)$. So our $b = \psi(a)$ for some $a \in A$. By commuting diagrams again $\beta \psi (a) = 0 =\psi' \alpha (a) $, and by injectivity $a = 0$, so $b = \psi(0) = 0$. We're done. There was just more usefulness for the diagram that I didn't see!
H: Prove that the set of Real numbers R can be partitioned into a denumerable collection of uncountable sets Prove that the set of Real numbers R can be partitioned into a denumerable collection of uncountable sets My thoughts on the problem were to show the break down of the set of real numbers into subsets such as rational, irrational, and integers. Im not sure if im over complicating this. The wording is confusing to me because i have never done something like this in class. AI: How about the union of the intervals $[n, n+1); n $ in $\mathbb Z^+ $ ? and $(n-1,n]$ for $n$ in the negative integers less than or equal to $-1$, and $(-1,0)$. Basically, the partition is : { $ \cup (n-1,n ]; n \leq -1 \cup (-1,0) \cup [0,1) \cup [1,2) \cup....\cup [n, n+1) \cup.....$}
H: Code book for Hamming Code For a natural nuber $d$ there are $N=2^d-1$ non-zero vectors in $\mathbb F_2^d$. Now I take these as the columns of a $d\times N$ matrix $S$. The kernel of $S$ is then the code book for a Hamming code $c:\mathbb F_2^{N-d}\rightarrow \mathbb F_2^N$ (Is this a clear result or is there anything to prove?) $S$ is the Syndrome, we can use it to check whether a vector $x\in\mathbb F^N$ is a code word by cheking $Sx=0$ or not. My question is why is each of these Hamming codes a 1-error correcting-code? AI: Suppose you have a valid message $x \in F_2^N$, i.e. $Sx = 0$. Suppose you modify it in one place to obtain $y$. Then $Sy = Sz$ where $z = y + x$. So we can restrict the question to messages with the form of $z$, i.e. messages which consists entirely of zeros apart from a single one. Let the position of the error be $1 \leq n \leq N$. Then $Sz$ is a binary representation of $n$, because all the columns of $S$ are different. This allows us to find the error and correct it. There are only $2^d$ possible values of $Sx$. One of these (zero) is interpreted as an indication that there are no errors, while the rest are interpreted as single errors. Therefore, if we receive a message with more than one error in it, we will certainly decode it incorrectly. This proves that the code is a perfect 1-error-correcting code.
H: Jordan Normal Form I'm asked to find the Jordan Normal form of $A\in M_5(\mathbb{C}^{5x5})$ with the characteristic polynomial: $p(A)=(\lambda-1)^3(\lambda+1)^2$ and minimum polynomial $m(A)=(\lambda-1)^2(\lambda+1)$ I got so far: $$m_A(x)=(x-1)^2(x+1)\;\;\;:\;\;\;\;\begin{pmatrix}1&1&0&0&0\\ 0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&-1&0\\ 0&0&0&0&-1\\ \end{pmatrix}$$ The above matrix i got it by substitution of $m(A)$ in a $A\in M_5(\mathbb{C}^{5x5})$ matrix. And i think it's correct(if it isn't please let me know) but i don't know why does it works? Can you please give me a hint of why does this works? AI: Knowing the Jordan normal form theorem, you can just start out from the Jordan normal form. Assume that it belongs to a basis $e_1,..,e_5$. The characteristic polynomial tells you the diagonal elements, and in this case the minimal polynomial could determine the sizes of the blocks. Because $(\lambda+1)$ in $m(A)$ has exponent $1$, there cannot be a $2\times 2$ block around diagonal $-1$. Similarly, there cannot be a $3\times 3$ block around diagonal $1$, but if only $1\times 1$ blocks it contained, $(\lambda-1)$ in $m(A)$ would also have exponent $1$. So, there must be a $2\times 2$ and a $1\times 1$ block of $1$. We can divide the space into $U:={\rm span}(e_1,e_2,e_3)$ and $V:={\rm span}(e_4,e_5)$, these are $A$-invariant, and we have $$m(A\|_U)=(\lambda-1)^2\,,\quad\quad m(A\|_V)=(\lambda+1)\,.$$ This second one tells use that $A\|_V+I_V=0$, that is, $A\|_V=-I_V$ (where $I_V$ is the identity of subspace $V$). Since the nilpotent $N=\pmatrix{0&1&0\\0&0&1\\0&0&0}$ has $N^2\ne 0$ but $N^3=0$, we have that $m(\pmatrix{a&1&0\\0&a&1\\0&0&a})=(\lambda-a)^3$. This justifies the reasoning text above.
H: Existence of uncountable set of uncountable disjoint subsets of uncountable set "Can you to any uncountably infinite set $M$ find an uncountably infinite family $F$ consisting of pairwise disjoint uncountably infinite subsets of $M$?" Intuitively, I feel like it should be possible for the real numbers at least: you simply split the real numbers into two intervals, and since there are uncountably many points where you can do that, there are uncountably many ways to split the reals into two disjoint subsets. But this question isn't asking specifically about the real numbers, so how can I prove this more generally? AI: Your approach does not work even for $\Bbb R$: you’ve merely shown that there are uncountably many different ways to split $\Bbb R$ into two disjoint uncountable sets. Your task is to split $M$ into uncountably many pairwise disjoint uncountable sets. HINT: $\|M\times M\|=\|M\|$ (assuming the axiom of choice).
H: proving something is a well-defined function 1) Define ~ S4 as follows: for f, g element of S4 f~g if and only if f(4) = g(4) this is easily seen to be an equivalence relation on S4 (you don't have to show this) let X = S4/ ~ be the set of all equivalence classes under ~. Define * X as [f] * [g] = [f o g]. Is * a well-defined operation on X? If so, prove this fact; if not explain why not. 2) Show that G is a group and N is a normal subgroup of G, that G/N is abelian if and only if aba^-1b^-1 exists in N for all a,b element of G New to abstract algebra and trying to teach myself. Any help with these two problems would be greatly appreciated! AI: First of all, '$\in$' is read as 'element of'. 1) No, it is not well-defined on $S4/\sim$. Find an $f$ for $g_1:=$ constant $4$ and $g_2:=$ identity ($x\mapsto x$) functions such that $\ g_1\circ f\not\sim g_2\circ f$. 2) Rather straightforward. The equivalence relation determined by $N$ is $x\sim y$ iff $xy^{-1}\in N$. If you haven't checked yet, verify that $\sim$ is indeed an equivalence relation, and --provided $N$ is a normal subgroup-- that it preserves group operation, i.e. the group operation in the quotient set $G/\sim\,$ is well defined. And the exercise itself is then obvious: $$aba^{-1}b^{-1}=(ab)(ba)^{-1}\in N \iff ab\sim ba \iff [ab]=[ba] \iff [a][b]=[b][a]\,.$$
H: Solving a integro-differential equation I'm trying to solve the following problem, but my solution doesn't seem to be correct. Could I be rewriting the integral incorrectly? $$ \frac{dy}{dt} + 25 \int_0^t{y(t-w)e^{-10w}dw} = L[4]; y(0)=0$$ $$ sL[y(t)] - y(0) + 25(L[y(t)]L[e^{-10t}])=L[4] $$ Let $L[y(t)] = Y(t)$: $$ sY(t) + 25(Y(t)\frac{1}{s+10})=\frac{4}{s} $$ $$ sY(t) + 25(Y(t)\frac{1}{s+10})=\frac{4}{s} $$ $$ sY(t) + 25Y(t)\frac{25}{s+10}=\frac{4}{s} $$ $$ sY(t) + \frac{2525Y(t)}{s+10}=\frac{4}{s} $$ $$ sY(t) + \frac{625Y(t)}{s+10}=\frac{4}{s} $$ $$ Y(t) (s + \frac{625}{s+10}) = \frac{4}{s} $$ $$ Y(t) =\frac{4(s+10)}{s(s^2+10s+625)} $$ I'm not sure where I am going wrong on this one. It is worth noting that when rewriting the integral I am using the following definition: $$(fg)(t)=\int_0^tf(w)g(t-w)dw =f(t)g(t)$$ AI: We have: $$\mathcal{L} (y' + (25 y(t)e^{-10t}) = 4)$$ This yields: $$sy(s) - 0 + 25 y(s) \dfrac{1}{s+10} = \dfrac{4}{s}$$ Factoring, we have: $$y(s) \left(s + \dfrac{25}{s+10}\right) = \dfrac{4}{s}$$ This gives us: $$y(s) = \dfrac{4(s+10)}{s(s+5)^2} = \dfrac{8}{5 s} -\dfrac{4}{(s+5)^2} - \dfrac{8}{5 (s+5)}$$ Next, find $\mathcal{L}^{-1} (y(s))$
H: Examples of canonical projections that are not epimorphisms and canonical injections that are not Although in $\mathsf{Set}$, canonical projections from a product are surjective and canonical injections to a coproduct are in fact injections, there seems to be nothing forcing this to be the case elsewhere, and indeed Wikipedia indicates without proof that they needn't be epic/monic. Unfortunately, I'm a rank beginner and have not yet knowingly read about a category less rich in monomorphisms and epimorphisms than $\mathsf{Set}$ (though Aluffi has already introduced a couple examples that are richer in them), so I don't know where to try to find a counterexample. Any hints? To clarify: I am seeking canonical injections to coproducts that are not monomorphisms and/or canonical projections from products that are not epimorphisms. AI: Take a category that looks like this: $X \leftarrow P \rightarrow Y \rightrightarrows Z$. It has four objects, $X,Y,Z,P$, the only nonidentity morphisms are the ones I indicated (there are four), and the two composites $P \to Y \to Z$ which are the same. $P$ is clearly the product of $X$ and $Y$. But the projection $P \to Y$ is not an epimorphism. The opposite of this category gives a counterexample for injection.
H: Solving ODE, reduce to exact differential equation Solve the equation: $xdy=(x^5+x^3y^2+y)dx$ My attempt at a solution: I've tried to check if this was an exact differential equation or if I could reduce it to one, so $xdy=(x^5+x^3y^2+y)dx \iff xdy-(x^5+x^3y^2+y)dx=0$. If I call $M(x,y)=-(x^5+x^3y^2+y)$ and $N(x,y)=x$, the equation is exact if and only if $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x} \iff -(2x^3y+1)=1$, which is clearly false. I've tried to reduce this to an exact differential equation: I suppose there exists a function $u$ that depends only on $x$ or on $y$ such that $(uM)dx+(uN)dy=0$ Case 1: if $u$ is $u(x)$, then $u$ must satisfy $\dfrac{u_x}{u}=\dfrac{M_y-N_x}{N} \iff \dfrac{u_x}{u}=\dfrac{-1-2yx^3-1}{x}$ But the member of the right doesn't depende only on $x$, so I can't find $u(x)$ that satisfies the required conditions. Case 2: I suppose there exists $u$ that satisfies $(uM)dx+(uN)dy=0$ but that only dependes on $y$. So $\dfrac{u_y}{u}=\dfrac{N_x-M_y}{M}$. Here I had the same problem, the member of the right doesn't depend only on $y$. What other method could I apply to solve the ODE? Should I propose a function $u$ that depends on both $x$ and $y$? I don't know how to find $u$ if this is the case. Here's the solution with Amzoti's answer: Suppose $y=vx \implies y'=v'x+v$, replacing this into the ODE, we get $xdy=(x^5+x^3y^2+y)dx \iff x\dfrac{dy}{dx}=x^5+x^3y^2+y \iff x(v'x+v)=x^5+v^2x^5+vx \iff v'x+v=x^4+v^2x^4+v \iff v'x=x^4(1+v^2) \iff v'=x^3(1+v^2)$. But $v'=x^3(1+v^2)$ is a separable equation, dividing by $1+v^2$ and multiplying by $dx$, we get $\dfrac{dv}{1+v^2}=x^3dx$, integrating we get that $arctan(v)=\dfrac{x^4}{4}+c, c \in \mathbb R$. Then, the implicit solution of the original ODE is $arctan(\dfrac{y}{x})==\dfrac{x^4}{4}+c , c \in \mathbb R$ AI: Hint: Let: $$y = v x \rightarrow y' = v + x v'$$ Substitute this into the ODE and solve. It will reduce it to a separable equation and you can use integration of both side. It may also be possible to get this in Riccati form.
H: Direct sum of projective module Let $\left\{ P_{i}\right\} _{i\in I}$ be a family of $R$-module. I know that if each $P_{i}$ is projective then $\oplus_{i\in I}P_{i}$ is projective. Is the converse true, i.e if $\oplus_{i\in I}P_{i}$ is a projective module, is $P_{i}$ projective module for all $i\in I$? AI: Yes its true. $P$ is projective iff there is free module $F$ such that $F=P\oplus Q$. Now if $\bigoplus P_i$ is projective then there is free module $F$ such that $F=\bigoplus P_i\oplus Q=P_i\oplus(\bigoplus_{j\neq i}P_i\oplus Q)$. This means that $P_i$ is projective.
H: Primitive Roots Proofs I am stuck on how to prove these two questions: (1) Let r be a primitive root of the prime $p$ with $p$ congruent to $1$ modulo $4$. Show that $-r$ is also a primitive root. (2) Let n be a positive integers possessing a primitive root. Using this primitive root, prove that the product of all positive integers less than $n$ and relatively prime to $n$ is congruent to $-1$ modulo $n$. AI: 1) Let $p=4k+1$. Since $r$ is a primitive root of $p$, it follows that $r$ is a quadratic non-residue of $p$. Thus by Euler's Criterion, we have $r^{2k}\equiv -1\pmod{p}$. It follows that $(-r)^{2k}\equiv -1\pmod{p}$. Multiply by $-r$. We get that $(-r)^{2k+1}\equiv r\pmod{p}$. Since $r$ is a power 0f $-r$, and every $a$ relatively prime to $p$ is congruent to a power of $r$, it follows that every $a$ relatively prime to $p$ is congruent to a power of $-r$. This implies that $-r$ is a primitive root of $p$. 2) The result is easy to prove for $n=1$ and $n=2$, so we may assume that $n\ge 3$. Let $g$ be a primitive root of $n$. Then the $\varphi(n)$ numbers in the interval from $1$ to $n$ that are relatively prime to $n$ are congruent (modulo $n$) in some order to $g^1,g^2,g^3,\dots, g^{\varphi(n)}$. Thus their product is congruent to $g^N$, where by the usual formula for the sum of the first $\varphi(n)$ positive integers we have $2N=\varphi(n)(\varphi(n)+1)$. Since $n\ge 3$, the number $\varphi(n)$ is even, so $\varphi(n)+1$ is odd. We have $g^N=(g^{\varphi(n)/2})^{\varphi(n)+1}$. Since $g$ is a primitive root of $n$, we have $g^{\varphi(n)/2}\equiv -1\pmod{n}$. Raising this to the odd power $\varphi(n)+1$, we see that $g^N\equiv -1\pmod{N}$, which is what needed to be shown.
H: Is it possible to flip tails indefinitely? If someone makes the argument that it is impossible to flip tails $n$ times on a two-sided coin, then we can argue there is a $1$ in $2^n$ chance. There is not a definable point at which it becomes impossible. Is the following statement true or false? It is possible to flip tails indefinitely on a two-sided coin. AI: Possible, yes, but extremely unlikely, in the sense that the probability is $0$. That does not mean that it is impossible, especially since "probability" in such circumstances is a human model of phenomena. That is, for example, there is not actually any "force of nature" that "prevents" flipping heads for an arbitrarily long time... a.k.a. "forever". But one should bet against it, as in the surely-googleable "gambler's ruin" phenomenon/scenario. "Probability zero" and "impossible" are very different notions, though for day-to-day purposes they are similar in function. At extremes, the presumed equivalence is much less useful for reasoning about either the mundane world or mathematical things. (One should certainly keep in mind that formulation of mathematical ideas does not directly lend enforcement power over physical phenomena...)
H: Double complement law proof I'm having difficulty proving that $S-(S-A) = A$ iff $A \subseteq S$, where "$-$" is the set difference. For the "only if" part, I found that I can prove that $S \subset A \Longrightarrow S-(S-A) \neq A$ by letting $x$ be an element in $A$ which is not in $S$. Then I show that $S-(S-A)$ does not contains $x$ from which it follows that $S-(S-A) \neq A$. For the "if" part, even though it is conceptually very intuitive, I can't find a way to make a formal proof. AI: $A\subset S$ is not the negation of $A\subseteq S$: you must also consider the possibility that there is some $x\in A\setminus S$. It’s not hard to see that that also implies that $S\setminus(S\setminus A)\ne A$, but you do have to consider the case if you take this approach to proving the theorem. You could prove this direction more easily by simply observing that $S\setminus(S\setminus A)\subseteq S$, so if $S\setminus(S\setminus A)=A$, then $A\subseteq S$. For the other direction, you could try to prove directly that if $A\subseteq S$, then $S\setminus(S\setminus A)=A$. Suppose that $x\in A$; then $x\in S$, and $x\notin S\setminus A$, so $x\in S\setminus(S\setminus A)$, and therefore $A\subseteq S\setminus(S\setminus A)$. Now suppose that $x\in S\setminus(S\setminus A)$. Then $x\in S$ and $x\notin S\setminus A$, so ... ?
H: Power Set Difference Proof P(A-B)=P(A)-P(B) I need help proving or disproving this! If anyone could show me the sequence of steps I can use to help me show the left is equal to the right I would be grateful. $$\mathcal P(A-B)=\mathcal P(A)-\mathcal P(B)$$ AI: HINT: $\varnothing\in\wp(A\setminus B)$, $\varnothing\in\wp(A)$, and $\varnothing\in\wp(B)$.
H: What is the probability of picking Exactly 1 red marble and than not 1 red marble? without rep. A urn has 3 red marbles, 2 blue marbles, 1white, 1 black 1 brown. What is the probability of getting exactly 1 red marble than not 1 red marble? What is the probability of getting at least 1 red marble? I could not find a way to solve this question without listing the sample spaces. It's $2\bigg(\dfrac{3}{8} \cdot \dfrac{2}{7}\bigg)$ for the 1st one and for the 2nd one I asked what is the probability of not getting a red marble and took the compliment and get, $1-\dfrac{5}{8} \cdot \dfrac{4}{7}$ my question is why do I mulitply by 2 on the 1st question? I know there are 2 ways to get exactly 1 red marble but there are also 2 ways to not get a red marble right? How can I do this problem using combinations because this is a little shady to me. AI: You need to clearly specify how you draw. Without this, there is no answer. If you draw two marbles without replacement and are asking the chance that the first is red and the second is not, you just have the multiplication law: $\frac 38$ for the first red, then $\frac 57$ that the second is not (given that the first was red).
H: Separating Partial Differential Eq I have a PDE: $$ \frac{\partial^2\phi(r,\theta)}{\partial r^2} + \frac{1}{r}\frac{\partial\phi(r,\theta)}{\partial r} + \frac{1}{r^2}\frac{\partial^2\phi(r,\theta)}{\partial\theta^2} + C^2\phi(r,\theta)=0 $$ I need to separate the PDE (just functions of r,theta) and show the relationship between the separation constants and $C^2$. I need to use solution of $\phi(r,\theta)$ = $f(r)g(\theta)$. When I do that, then divide by solution, I do not see how I can separate $g$ from $1/r^2$ without having the $C^2$ change to $C^2r^2$. Any ideas? AI: Well, if you suppose $\phi(r,\theta) = R(r) \Theta(\theta)$, then $$ R'' \Theta + \frac{1}{r} R'\Theta + \frac{1}{r^2} R \Theta'' + C^2 R \Theta = 0 $$ Multiplying by $r^2$, you have $$ r^2 R'' \Theta + r R'\Theta + R \Theta'' + r^2 C^2 R \Theta = 0 $$ Dividing by $R \Theta$ $$ \frac{1}{R}\left(r^2 R'' + r R'\right) + \frac{\Theta''}{\Theta} + r^2 C^2 = 0 $$ Then $$ \frac{1}{R}\left(r^2 R'' + r R'\right) + r^2 C^2 = - \frac{\Theta''}{\Theta} $$ And given that the left hand side depends only on $r$ and the right hand only on $\theta$, you have $$ \frac{1}{R}\left(r^2 R'' + r R'\right) + r^2 C^2 = - \frac{\Theta''}{\Theta} = \lambda $$ where $\lambda$ is a constant. Then \begin{align} r^2 R'' + r R' + (C^2 r^2 - \lambda) R &= 0 \\ \Theta'' + \lambda \Theta &= 0 \end{align} The first one is a scaled version of Bessel equation, while the second one is the harmonic oscillator. EDIT Here is the transcript of the chat I with the OP, where a full answer is given.
H: What are the reasons for using a semi-circle in upper half plane of $\mathbb{C}$ for contour integration? Why is it that when one in considering contour integration of a real function, such as $$ \int_{-\infty}^{\infty} \frac{dx}{1+x^2}$$ the contour in the complex plane used is the following: Furthermore, what are the general strategies for choosing a contour when integrating a real function using the method of residues? For example, what contour would be used for the function $$\int_{0}^{\infty} \frac{dx}{1+x^2}.$$ Or for the function $$\int_{0}^{\pi} \frac{d \theta}{(a+b \cos\theta)^2}.$$ (Notice how the limits of integration changed). I am also curious how the position of the residues in $\mathbb{C}$ affect a contour. My complex analysis book tends to pull a contour out of thin air with out any appeal to intuition. Any help would really be great! Thank you. AI: I think it's hard to make a general statement, but in many cases, the contour that you've shown above is good because the portion of the semicircle lying along the real axis gives us the integral we want (when we take $R$ to be very large), and the circular portion of the contour can be easily integrated on and shown to tend to zero for large $R$. It's not necessarily that you couldn't choose something else, but we understand (in many cases like these) how to integrate along circular contours well. There is no one size fits all method, though. Different contours will lend themselves well to different situations. For example, if you a residue which lies on the real axis, you might have to use a contour which uses a different fraction of a circle - say, in some cases a third circle. In other cases, you might have to avoid a branch cut, and can use an annular region. Experience and clever choices will dictate what contours are best. In one of your specific cases, for example, the integral $$\int_{0}^{\infty} \frac{dx}{x^{2}+1}$$ can be integrated by using the contour given, and then observing that $\frac{1}{x^{2}+1}$ is an even function.
H: Find point between two intervals which equals the average value. Find the point in the interval [5,9] at which the function $f(x)=17e^{3x}$ equals its average value on that interval. So I've got my function as follows $f(x)=\frac{1}{4}(\frac{17}{3}(e^{27}-e^{15})$ So I've simplify it to $\frac{17e^{12}}{12}$ Now I believe I have to solve for $c$ so I've replace $x$ in my original function with $c$ so it looks like $17e^{3c}=\frac{17e^{12}}{12}$ So how would I go on from here, or am I doing the whole thing wrong. Please Help!! AI: The average value is indeed $\frac{1}{12}(17)(e^{27}-e^{15})$. It does not simplify in a very useful way, though you may want to write it as $$\frac{1}{12}(e^{15})(e^{12}-1).$$ Now you want to solve the equation $$17e^{3c}=\frac{1}{12}(17)(e^{15})(e^{12}-1).$$ We might as well cancel the $17$'s. After that, take the logarithm of both sides. The logarithm can be written as $15+\ln(e^{12}-1)-\ln(12)$. Divide by $3$ to get $c$.

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