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H: Limit point versus limit and implications for convergence I understand that similar questions to this have been asked here (1) and here (2) but I believe my question is different because with regards to (1) I am not asking about the set $X$ that contains the points of my sequence and with regard to (2) because I have the distinct definitions of limit and limit point in mind. That said: My Question: My textbook, Fundamental Ideas of Analysis by Michael Reed, defines a limit point as: Definition: Let $(a_n)$ be a sequence of real numbers. A number $d$ is called a limit point if, given $\epsilon > 0$ and an integer $N$, there exists $\color{red}{\mathrm{ an }} \ n \ge N$ so that $\|a_n - d\| \le \epsilon$. The definition of a limit is as follows: Definition: We say that a sequence $(a_n)$ converges to a limit $a$ if, for every $\epsilon \ge 0$, there is an integer $N(\epsilon)$ so that $\|a_n - a\| \le \epsilon \color{red}{\text{ for all }} n \ge N$. From these definitions, and given this question: If a sequence converges it has exactly one limit point. Is the converse true? If not, provide a counterexample Can I conclude that the converse is indeed true simply by appealing to the definitions, or do I need a tighter proof. And if I do, I'm a little stuck on how to refine this proof. AI: The sequence $2,1/2,3,1/3,4,1/4 \cdots, n,1/n \cdots$ has the only limit point $0$, but not converges.
H: How do I integrate this expression: $\int{2x\,dx\over x^3+x^{2/3}}$? How do I integrate this expression: $\displaystyle\int{2x\,dx\over x^3+x^{2/3}}$? AI: Let $u=x^{1/3}$, then $u^3=x$ so $dx=3u^2 du$ and now your integrand is a rational function.
H: Two problems with prime numbers Problem 1. Prove that there exists $n\in\mathbb{N}$ such that in interval $(n^2, \ (n+1)^2)$ there are at least $1000$ prime numbers. Problem 2. Let $s_n=p_1+p_2+...+p_n$ where $p_i$ is the $i$-th prime number. Prove that for every $n$, there exists $k\in\mathbb{N}$ such that $s_n<k^2<s_{n+1}$. I've found these two a while ago and they interested me. But don't have any ideas. AI: Problem 2: For any positive real $x$, there is a square between $x$ and $x+2\sqrt{x}+2$. Therefore it will suffice to show that $p_{n+1}\geq 2\sqrt{s_n}+2$. We have $s_{n}\leq np_n$ and $p_{n+1}\geq p_n+2$, so we just need to show $p_n\geq 2\sqrt{np_n}$, i.e., $p_n\geq 4n$. That this holds for all sufficiently large $n$ follows either from a Chebyshev-type estimate $\pi(x)\asymp\frac{x}{\log(x)}\,$ (we could also use PNT, but we don't need the full strength of this theorem), or by noting that fewer than $\frac{1}{4}$ of the residue classes mod $210=2\cdot3\cdot5\cdot7$ are coprime to $210$. We can check that statement by hand for small $n$. There have already been a couple of answers, but here is my take on problem 1: Suppose the statement is false. It follows that $\pi(x)\leq 1000\sqrt{x}$ for all $x$. This contradicts Chebyshev's estimate $\pi(x)\asymp \frac{x}{\log(x)}$
H: Finding integer solutions to $m$ and $n$ How can we find the positive integer solutions to the variables $m$ and $n$, if we know $r$, that satisfy the equation: $$r = \frac{\sqrt{3(m-n)^2 n^2}}{2},$$ where $m$ and $n$ are coprime, and $0 < n < m$. AI: Let's suppose $r\ge 0$ (because of the square root) then squaring we get : $$\tag{1} 4r^2=3(m-n)^2n^2$$ If $r$ is supposed integer then we need $3\|r$ i.e. $r=3k$ with $k$ a nonnegative integer (because $m$ and $n$ are integer) and your equation becomes : $$4\cdot 3k^2=(m-n)^2n^2$$ but this can't have a positive solution since the number of $3$ at the left is odd while the number of $3$ at the right is even. This implies that $k=0,\ n=0,\ m=0$. If $r$ is not supposed integer then your equation becomes simply : $$r'=\frac {2r}{\sqrt{3}}=(m-n)n\quad \text{(since $\ 0<n<m$)}$$ Since we want $m$ and $n$ integer $r'$ must be integer and may be : $r'=1.n$ (if $m-n=1$ corresponding to the trivial solution $n=r',\ m=r'+1$) $r'=p.n$ with $p$ and $n$ coprime ($p=m-n$) : i.e. computing $r'$ you got an integer that can't be power of prime ; decompose it in powers of primes : $$r'=\prod_{i=1}^N p_i^{k_i}$$ and consider all the partitions in two classes possible of these $N$ primes, one will define $n$ and the other $p$ (after that deduce $m=n+p$). Not sure it will really help...
H: Pi-subgroup in group Let $\mathbb{P}$ is a set primes numbers, $\pi \subseteq P$ and $\pi ^{\prime }=P-\pi $ Let $O_{\pi }\left( G\right) =\left\langle N~;~N\trianglelefteq G\text{ and }% N\text{ is }\pi \text{-subgroup}\right\rangle $ Is it true that $O_{\pi }\left( G\right) =1\Longrightarrow G$ is $\pi ^{\prime }-$group? if it is true, could not it have other $H$ $\pi $-subgroup (not normal) in $G$? AI: No. Let $\pi=\{2\}$ and $G$ be non-abelian of order 6. Then $O_\pi(G) = 1$, but $\|G\|$ is divisible by 2.
H: Affine transformations of measures and the resulting densities Define the transformation of a measure $v(X)$ for $X \subseteq {\mathbb R}^d$ by function $A: {\mathbb R}^d \rightarrow {\mathbb R}^d$ as $$ A(v)(X) = v(A^{-1}(X)) $$ Then the density function $v(x)$ for $x \in {\mathbb R}^d$ is not transformed in the same way (except in certain special cases). Now I would guess that if A is an affine function, that we can say $$ A(v)(x) = \frac{1}{\mbox{det}\|A\|} v(A^{-1}(x)) $$ Unfortunately I lack the knowledge of measure theory (if that is what is required, maybe it's just calculus?) to work out a proof, and Google and Wikipedia have not been very helpful. My main question is whether this is right, and if not, what the correct equation is. Secondly, what would the proof look like, and would there be some book on some subject that actually discusses it in this form? PS. I'm no expert in measure theory. If it helps, the measures can be simplified to whatever category of well behaved measures are available. For my purposes they only need to be probability measures on ${\mathbb R}^d$. Perhaps it is required to approximate a uniform measure for sufficiently small neighbourhoods? AI: By definition, the image measure $A(v)$ satisfies, for bounded measurable $g$: $$\int g(x) A(v)(dx)=\int g(Ax) v(dx)=\int g(Ax) v(x)\,dx.\tag1$$ Here I assume that the measure $v$ admits a density (called $v(x)$) with respect to Lebesgue measure. Now perform the change of variables $w=A(x)$. The inverse map is $x=A^{-1}(w)$ and $dx=\|A^{-1}\| dw$. Substituting into (1), we get $$\int g(x) A(v)(dx)=\int g(w) v(A^{-1}w) \|A^{-1}\|\,dw.$$ Thus the density of the measure $A(v)$ with respect to Lebesgue measure is $v(A^{-1}w) \|A^{-1}\|$, as you expected. We assumed that $A$ is invertible to make this work.
H: Existence of Analytic Functions Based of f(1/n) I have the following question, and I don't even know where to begin: Do functions $f$ or $g$ exist which are analytic at the point $z=0$ and satisfy the conditions: $f(\frac{1}{n})=f(\frac{-1}{n})=\frac{1}{n^2}$, and $g(\frac{1}{n})=g(\frac{-1}{n})=\frac{1}{n^3}$? AI: take $f(z)=z^2$. For $g$, define $h(z):=g(z)-z^3$. Then $h(n^{-1})=0$, and by uniqueness theorem, $g(z)=z^3$. But it doesn't match with the second condition, hence $g$ cannot exist.
H: What is $\|X\|$ and $1n$? I'm doing a course in calculus but I'm quite young so haven't learnt all the mathematical terms one should perhaps no before learning calculus, and these 2 terms keep coming up and I have no idea in the world of what they are: "$\|X\|$" and "1n". I am really sorry that this probably seems ridiculously elementary but I really want to know what they are. What are they? AI: $\|x\|$ is the absolute value of $x$. $\ln(x)$ (that's ell, not one) is the natural logarithm of $x$.
H: A question about integral operator I have a question: Prove or disprove that: for every $f\in L^{1}\left(\mathbb{R}\right)$, $$\sup\left\{ { \int_{\mathbb{R}}\frac{\sqrt{n}}{\sqrt{\left\|x-y\right\|}\left(1+n^{2}\left(x-y\right)^{2}\right)}f\left(y\right)dy:n\in\mathbb{N}}\right\} <\infty,$$ for Lebesgue almost every $x\in\mathbb{R}$. I failed in my attempt to disprove this statement (I think so!!!). Can everybody help me? AI: Let $$f_n(x):=\int_{\Bbb R}\frac{\sqrt n}{\sqrt{\|x-y\|}(1+n^2\|x-y\|^2)}f(y)dy,$$ assuming WLOG that $f\geqslant 0$. We use the substitution $t=n(x-y)$, hence $dt=-ndy$ to get $$f_n(x)=\int_{\Bbb R}\frac 1{\sqrt{\|t\|}}\frac 1{1+t^2}f\left(x-\frac tn\right)dt. $$ By Fubini's theorem for non-negative functions, and since $f$ is integrable, we get that $f_n$ is integrable, and in particular almost everywhere finite. Approximate $f\in L^1$ by $g$, continuous with compact support, such that $\lVert f-g\rVert_{L^1}\leqslant 1$. Then, as the integral $\int_{\Bbb R}\frac 1{\sqrt{\|t\|}}\frac 1{1+t^2}dt<\infty$, to show the result when $f$ is continuous and bounded. With the latest formula, it's easier to see it.
H: Abelian subgroup of a group of order $2002$ Another unsolved question from my studying for quals - Show that if $G$ is a group of order $2002=2\cdot 7 \cdot 11 \cdot 13$, then $G$ has an abelian subgroup of index $2$. I know it has to do with the direct product of the normal subgroups of $G$, but I'm not sure how to relate that to what I need. AI: If $\,n_p=\,$ number of Sylow p-subgroups of $\,G\,$ , then $\,n_{13}=1=n_{11}\,$, thus letting $\,Q_p\,$ be a Sylow p-sbgp., we have that $$Q_{11}\,,\,Q_{13}\triangleleft G\Longrightarrow Q_{11}Q_{13}\triangleleft G\Longrightarrow P:=Q_{11}Q_{13}Q_7\triangleleft G$$ the last one being a subgroup of order $\,7\cdot 11\cdot 13\,=1001$ in $\,G\,$. Abelianity follows from the fact that the only group of order $\,1001\,$ is the cyclic one, as $\,P\cong C_{11}\times C_{13}\times C_7\,\,,\,\,C_m:=\,$ the cyclic group of order $\,m\,$ Added: Please note the comment by Jack Schmidt below for an alternative approach: by the Schur-Zassenhaus theorem, if we have $\,N\triangleleft G\,$ s.t. $\,\left(\|N\|,\left\|G/N\right\|\right)=1\,$ then $\,G\cong\,N\rtimes G/N\,$ . In our case, take $\,N:=Q_{11}\,$ and apply the above, then take $\,G/N\,\,,\,\left\|G/N\right\|=2\cdot 7\cdot 13=182\,$ , which has a unique Sylow 7-subgroup, which we know is $\,\,\,\overline Q_7=Q_7N/N\,$ , and again apply the S-Z theorem to obtain $$G/N\cong \overline Q_7\rtimes \left(G/N\right)/\overline Q_7$$ with $\,\,\left(G/N\right)/\overline Q_7\cong G/Q_7Q_{11}\,\,$ , and so on. Note that we can directly apply Sylow theorems to deduce that $\,\,G/Q_7Q_{11}\,$ has a unique Sylow 13-sbgp. which we cannot do, at least directly or easily, with the original group using only Sylow theorems, which is what Jack writes there.
H: Local Path Connectedness and Cones I am trying to prove the following: Let $X$ be a topological space. $CX$ is locally path connected if and only if $X$ is locally path connected, where $CX=X\times I/(X\times\{0\})$ I cannot seem to make much headway in either direction of the problem. Assuming $CX$ is locally path connected means there exists a basis $\mathcal{B}$ such that every element of $\mathcal{B}$ is path connected. It's easy to see that $\mathcal{B}\cap X=\{X \cap B \: \| \; B \in \mathcal{B}\}$ is a basis for $X$. I cannot quite see why the elements of $\mathcal{B}\cap X$ should be path connected. Let $\hat{B} \in\mathcal{B}\cap X$ and let $p,q \in \hat{B}$. $p$ and $q$ correspond with $(p,1)$ and $(q,1)$, respectively, in $B$ since $X\approx X\times\{1\}$. In $B$ there is a path $\alpha(t):[0,1] \rightarrow B$ such that $\alpha(0)=(p,1)$ and $\alpha(1)=(q,1)$. Perhaps the solution is obvious, but I cannot see how to turn a path in $B$ into a path in $\hat{B}$ or even if that should be possible. If $\alpha(t)$ was not entirely in $X \times \{1\}$, is projecting it onto $X\times \{1\}$ the solution? I have also hit a wall in assuming $X$ is locally path connected. As such, there exists some path connected basis $\mathcal{C}$ of $X$. My first thought given $x,y \in CX$ was to find an open cone of the form $U\times I/(X\times \{0\})$, where $U$ is an open subset of $X$, such that $x,y \in U\times I/(X\times \{0\})$. I'm not even sure this is a basis of $CX$ and I cannot see why it should be. Overall, I have no idea how to continue. Are any of my ideas leading in the right direction? If not, could anyone give me a recommendation in the right direction? AI: Here is a quick sketch of the argument. I can elaborate further if you wish. Assume first that $X$ is locally path connected. Perhaps you know, and if not I encourage you to prove, that a finite product of locally (path) connected spaces is locally (path) connected and the image of a locally (path) connected space with a quotient map is locally (path) connected. Putting these two facts together shows that $CX$ must be locally path connected. By the way, you are right in questioning whether your proposed basis really is one. In fact it is not (clearly you cannot get open sets that "don't touch the base of the cone" from your family). For the converse, assume $CX$ is locally path connected. Your idea of projecting paths to the base of the cone is good, but has a slight problem. If the path contains the apex of the cone you obviously can't project it in a continuous way, so we have to fix this. I feel it is somewhat easier to work with another characterization of local path connectedness, i.e. that every point have a basis of path connected neighbourhoods. So take a point $x\in X$, which corresponds to $(x,1)\in CX$. Let $U$ be a neighbourhood of $x$ in $X$. Then (slightly abusing notation) $U\times (1/2,1]$ is a neighbourhood of $x$ in $CX$ and it doesn't contain the apex of the cone. By assumption, there exists a path connected neighbourhood $V$ of $x$ below $U\times(1/2,1]$. Projecting $V$ onto $X$ then gives us a path connected neighbourhood of $x$ in $X$ below $U$ (remember, we are basically in a product space now and projections are open maps).
H: Proving $\|A^A\|=\|2^A\|$ for infinite $A$. Possible Duplicate: Cardinal power $\kappa^\kappa$. When is it equal to $2^\kappa$? How can one prove that $\|A^A\|=\|2^A\|$ for infinite $A$? (summary of proof or providing link with proof will also suffice.) Thanks! AI: We have $\|2^A\|\leq \|A^A\|$ and by Cantor-Bernstein theorem we are reduced to show that $\|A^A\|\leq \|2^A\|$. To see that, we use the fact that $\|A\times A\|=\|A\|$, applying Zorn lemma to $S:=\{(B,f), B\subset A, f\colon B\times B\to B\}$ with the partial order $(B_1,f_1)\leq (B_2,f_2)$ if $B_1\subset B_2$ and $f_{2\mid B_1}=f_1$. We also have $$\|2^A\|=\|2^{A\times A}\|=\|(2^A)^A\|$$ which gives what we wanted.
H: What is a conformal mapping with $\infty$ in its image? Suppose we have a bijection $f$ between two open sets in $\mathbb{C}\cup\{\infty\}$, for example $U=\{\|z\|<1\}$ and $\Omega$, with $\infty\in\Omega$. Let $f(0)=\infty$. What do we mean by saying that "$f$ is a conformal mapping" in this case? What conditions are imposed on the behaviour of $f$ near $0$? (Maybe continuity and analyticity in $U\setminus\{0\}$?) AI: $f$ is conformal if it is holomorphic and has a nonzero derivative. In your case, it means that $z\mapsto \frac{1}{f(z)}$ is holomorphic and has nonzero derivative on $U$ (because the usual atlas for Riemannian sphere is $\{(\mathbf C,z\mapsto z), (\mathbf (\mathbf C\cup\{\infty\})\setminus\{0\}, z\mapsto 1/z)\} $).
H: Original article on the Grothendieck group Is there someone who knows the title of the original publication of Grothendieck on the construction of the Grothendieck group? Thanks in advance. AI: What seems to be clear is that the origin of the Grothendieck group is Grothendieck's work on Riemann-Roch's theorem around 1956. According to Weibel's The Development of Algebraic K-theory before 1980 (which is certainly worth reading!), the original source is Classes de faisceaux et théorème de Riemann-Roch, some mimeographed notes from 1957 who were later published (in 1971) in Théorie des intersections et théorème de Riemann-Roch (SGA6). In his Lecture Notes on Algebraic K-theory, Rognes links to Borel and Serre, Le théorème de Riemann-Roch, Bulletin de la Société Mathématique de France 86 (1958) “for the published exposition.”
H: If $a^2$ divides $b^2$, then $a$ divides $b$ Let $a$ and $b$ be positive integers. Prove that: If $a^2$ divides $b^2$, then $a$ divides $b$. Context: the lecturer wrote this up in my notes without proving it, but I can't seem to figure out why it's true. Would appreciate a solution. AI: By the fundamental theorem of arithemtic, you can write $a$ and $b$ as a product of primes, say $$ a=p_1^{\alpha_1}\cdots p_r^{\alpha_r},\qquad b=p_1^{\beta_1}\cdots p_r^{\beta_r} $$ where $\alpha_i,\beta_i\geq 0$. Allow the exponents to possibly be $0$ if such a prime $p_i$ occurs in the factorization of one integer but not the other. So $a^2=p_1^{2\alpha_1}\cdots p_r^{2\alpha_r}$ and $b^2=p_1^{2\beta_1}\cdots p_r^{2\beta_r}$. Since $a^2\mid b^2$, by unique factorization, necessarily $2\alpha_i\leq 2\beta_i$ for each $i$. That implies $\alpha_i\leq\beta_i$ for all $i$, and so $a\mid b$.
H: how to prove optimality of this greedy algo I need some suggestions on how to prove the below greedy algorithm is optimal. Problem: There are $n$ fires on a road. Each fire $i$ is given as an interval where it starts and ends $[s(i), f(i)]$. An extinguisher can cover a fire of length $m$. Find the minimum number of extinguishers to put out all fires. Algo: c = 0 # the count of needed extinguishers L = 0 # the last point where fire is put out for i=1 to n: if (s(i) < L) then s(i) = L end if x = (f(i) - s(i))/m # number of extinguishers needed for fire i c += x # update the total count L = f(i) + x*m end for For the proof of optimality, I think I have to prove that the size of an optimal solution is at most equal to the size of the greedy solution. But how can I exactly prove that? The optimal solution can be a result of any not-necessarily-greedy algorithm. Thank you. AI: Let $i_0$ be such that $s(i_0)$ is minimal (among ${s(i)}_i$). Some fireman must cover $s(i_0)$ so we know that the leftmost fireman covers a segment $[a,b]$ with $a \leq s(i_0)$. On the other side, if you have a fireman covering a segment $[a,a+m]$ with $a<s(i_0)$ then you can move this fireman right to $[s(i_0),s(i_0)+m]$ and you still have a solution with the same number of firemen. In particular, you can take an optimal solution and shift all firemen covering areas to the left of $s(i_0)$ to the right. Therefore, there is an optimal solution where the leftmost fireman covers $[s(i_0),s(i_0)+m]$. Place a fireman there, remove all the fires or parts of fire he covers and you're left with the same problems, only with less fires, or at least smaller ones. Obviously, after deciding where one fireman is, we still want to solve the remaining problem optimally, so we continue in the same manner. That's the solution you get from this code.
H: How to obtain more than two abcissas for one ordinate? I'm reading a book on trigonometry and the author shown that some of the examples contain one abcissa for one ordinate and he showed that a parabola has two abscissas for every positive ordinates, I got curious: Is it possible to have more than two abscissas for one and only one ordinate? Is it possible to have more than two abscissas for every element of a specific set of numbers? AI: The answer to both questions is easily yes if you don’t require the function to be continuous. Let $$f(x)=\begin{cases} 2,&\text{if }x\in\{0,1,2\}\\ \tan^{-1}x,&\text{otherwise}\; \end{cases}$$ The function $\tan^{-1}x$ is one-to-one and takes values between $-\frac{\pi}2$ and $\frac{\pi}2$, so every ordinate except $1$ corresponds to at most one abscissa, while $1$ corresponds to three abscissae. The same basic idea can be used to provide a positive answer to the second question. Let $Y$ be the set of ordinates that are to have more than two abscissae. For each $y\in Y$ pick a set $X_y$ of three real numbers, making sure that the sets $X_y$ are pairwise disjoint and that $\left\|\Bbb R\setminus\bigcup_{y\in Y}X_y\right\|=\|\Bbb R\setminus Y\|$. (This is always possible.) Let $$g:\Bbb R\setminus\bigcup_{y\in Y}X_y\to\Bbb R\setminus Y$$ be any bijection, and let $$f(x)=\begin{cases} y,&\text{if }x\in X_y\\ g(x),&\text{otherwise}\;; \end{cases}$$ this function $f$ has the desired properties. The answer to the first question is still yes even if you require the function to be continuous; for instance, take $$f(x)=\begin{cases} x,&\text{if }x\le 0\\ 0,&\text{if }0\le x\le 1\\ x-1,&\text{if }x\ge 1\;. \end{cases}$$ Apart from the trivial example of a constant function, however, you cannot get such an example if $f$ is required to be a polynomial. If $F$ is any finite set of real numbers, you can easily modify the last example to get a continuous function that has a whole interval of abscissae for each ordinate in $F$ and is one-to-one everywhere else. It's also possible to modify it to get examples for countably infinite sets $F$, but that’s just a bit tricky. If you want $F$ to be a closed interval, you can do it with a polynomial of degree $5$. For instance, to get $F=[a,b]$ you can use a fifth degree polynomial with two local minima at height $a$ and two local maxima at height $b$. More generally, with a polynomial of degree $4n+1$ you can get $F$ to be the union of $n$ pairwise disjoint closed intervals.
H: Complex analytic functions and their zeros. I am having trouble with the following statement found in a textbook: "Let $U$ be a connected open set. Let $f$ be a complex analytic function on $U$ and not constant. Either $f$ is locally constant and equal to $0$ in a neighbourhood of a zero $z_0$, or $z_0$ is an isolated zero." I just cannot see how this is the case. For instance, isn't the identity function analytic, not isolated, and not locally constant at $z_0 = 0$? Edit: While the accepted answer to this question is a good proof of why the above statement is true, the actual trouble I ran into was not properly understanding the terminology. I mistook the meaning "isolated" in this case. If this may serve to help anyone in future, I say this: ensure you properly understand the precise meaning of all terminology before trying to consider a proof. AI: The zeroes of a non-constant analytic function on ${\mathbb C}$ are isolated. Let $f$ be an analytic function defined in some domain $D \subset {\mathbb C}$ and let $f(z_0)=0$ for some $z_0 \in D$ . Because f is analytic, there is a Taylor series expansion for $f$ around $z_0$ which converges on an open disk $\|z-z_0\|<R$ . Write it as $f(z) = \Sigma_{n=k}^{\infty} a_n (z-z_0)^n$ , with $a_k \ne 0$ and $k > 0$ ($a_k$ is the first non-zero term). One can factor the series so that $f(z) = (z-z_0)^k \Sigma_{n=0}^{\infty} a_{n+k} (z-z_0)^n$ and define$ g(z) = \Sigma_{n=0}^{\infty} a_{n+k} (z-z_0)^n$ so that $f(z) = (z-z_0)^k g(z)$ . Observe that g(z) is analytic on $\|z-z_0\|<R$ . To show that $z_0$ is an isolated zero of $f$ , we must find $\epsilon > 0$ so that $f$ is non-zero on $0<\|z-z_0\|<\epsilon$ . It is enough to find $\epsilon>0 $so that g is non-zero on $\|z-z_0\|<\epsilon$ by the relation $f(z) = (z-z_0)^k g(z)$ . Because $g(z)$ is analytic, it is continuous at $z_0$ . Notice that $g(z_0)=a_k \ne$ 0 , so there exists an $\epsilon > 0$ so that for all $z$ with $\|z-z_0\| < \epsilon$ it follows that $\|g(z) - a_k\| < \frac{\|a_k\|}{2}$ . This implies that $g(z)$ is non-zero in this set.
H: A question on Lie sub-group Well, definition of Lie subgroup what I know is, a Lie subgroup of a Lie group $G$ is an abstract subgroup $H$ which is an immersed submanifold via the inclusion map so that the group operations on $H$ are $C^{\infty}$. Could any one make me understand the following with an example? "Because a Lie subgroup is an immersed submanifold, it need not have the relative topology. In particular, the inclusion map $i:H\rightarrow G$ need not be continuous." AI: I can't help you understand that statement, because I don't understand it: an immersion is a differentiable map with injective derivative, and is in particular continuous. A standard example of a Lie subgroup such that the inverse of the inclusion is not continuous is the image of a line with irrational slope in the two-dimensional compact torus (quotient of the plane by $\mathbb{Z}^2$). As Paul Garrett mentions in a comment above, this kind of example is almost certainly what Tu had in mind when he wrote the statement you quote. One reason for making the definition this way instead of with a more restrictive definition of Lie subgroup is to be flexible enough to allow Lie subgroups of a Lie group to correspond to Lie subalgebras of its Lie algebra.
H: Given a tetrahedron, how to find the outward surface normals for each side? Say I have a triangle in $3$D space. I can get the surface normal by calculating the vector cross product of two of the edges. But, lets say I make this a tetrahedron. How can I work out the outward surface normal for each side? AI: For a given face $f,$ let its surface normal be $\vec{n}.$ The outward direction normal will be either $\vec{n}$ or $-\vec{n}.$ To determine which, let the $4$th vertex of the tertrahedron (the apex opposite to the $f$) be $v,$ and let one of the vertices of the face $f$ be $p.$ Thanks to Rahul Narain's hint: now, consider the $2$ vectors $\vec{n}$ and $\vec{pv} = v - p.$ If they're both on the same side (with respect to $f$), then $\vec{n} \cdot \vec{pv} $ is positive$^\dagger$ and $\vec{n}$ is facing inwards. If $\vec{n}$ is facing outward, then $\vec{n} \cdot \vec{pv} $ is negative. Based on the sign of $\vec{n} \cdot \vec{pv} $ you can know whether $\vec{n}$ or $-\vec{n}$ is the outward vector you're after. $^\dagger$ recall the dot product $a\cdot b = \\| a \\| \\| b \\| \cos \theta_{ab}.$
H: Which Interpolation should I use to create a curve? I'm pretty weak in the field of mathematics, but a strong programmer. I am looking for a mathematical solution that, given two points on a line will give me a curve between them, including those two points within the curve itself. For instance, if I have a set of points { (0, 3) (1,10) } I'd like a mathematical way to generate points between the two (I believe this is called interpolate) to create a curve that will contain { (0,3) (1,10) } Will Linear Interpolation give me this? Thank you AI: You could do that, but if a straight line is acceptable, the best-of-breed algorithm for calculating which pixels to plot is Bresenham's algorithm. It is easy to program, produces good-looking output, and it is extremely efficient. If you are interested in curved curves, you have a lot of choices. People often use cubic splines, because they are graceful, fit together well, and the algorithm is easy to write and runs quickly. There is a variation of Bresenham's algorithm for circular arcs instead of straight lines. If this isn't enough information, you should consider posting another question that elaborates on what you are looking for.
H: From $\sum_p \frac{\log p}{p^s} = \frac{1}{s-1} + O(1)$ conclude that $\sum_p \frac{1}{p^s} = \log \frac{1}{s-1} + O(1)$ I'm reading a book on analytic number theory. It asks me to prove: $$ \sum_p \frac{\log p}{p^s} = \frac{1}{s-1} + O(1) \tag{A}$$ and conclude, via integration, that: $$ \sum_p \frac{1}{p^s} = \log \frac{1}{s-1} + O(1) \tag{B}$$ Now, I know how to prove $(A)$ via Abel Summation. However, when it comes to $(B),$ I have the problem that although: $$\frac{d}{dx} \log(x-1) = \frac{1}{x-1}$$ and $$\frac{d}{ds} p^{-s} = (-\log p)p^{-s}$$ I have the problem that when I integrate over $O(1)$, I get $\infty$, not $O(1)$. What am I doing wrong? How do I get from $(A)$ to $(B)$? AI: What if you don't integrate over all of $(s,\infty)$? Say, fix $S$ large enough, and... $$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\int_s^S\frac{1}{\sigma-1}+O(1)d\sigma=\log\frac{1}{s-1}+O(1) $$ whereas $$\int_s^S\sum_{p}\frac{\log p}{p^\sigma}d\sigma=\sum_p\int_s^S\frac{\log p}{p^\sigma}d\sigma=\sum_{p}\left(\frac{1}{p^s}-\frac{1}{p^S}\right)=\sum_p\frac{1}{p^s}~+O(1) $$ for $s<S$ as $s\to1^+$. (Some uniform convergence stuff needs to be checked so that the interchange is justified.)
H: Beneficial to touch theoretically deeper texts earlier in core areas e.g. analysis, algebra? Desired future direction: Dynamical System(Chaos), PDE More beneficial to read theoretically deep, modern and masterpiece texts earlier, (e.g. levels like UTX/GTM/GSM/LNM/CSAM) ? Especially in core areas (e.g. Analysis, Algebra, Topology etc.) i.e. Read briefly less for intro-level, then quickly drop to the deeper/modern texts, although basic skill's not trained enough, but to be trained again/better in 'deeper-stage'. (Sometimes it seems learning new material in deeper/complex 'environment/context' gives better/faster maturity.) e.g: Calculus/Analysis: Apostol -> Hardy, Courant, Stein, Rudin, Amann Linear Algebra: Axler -> Dym, Grueb, Lax, Halmos Algebra: van der Waerden -> Lang, Hungerford, Jacobson, Bourbaki Topology: Kelley -> Milnor AI: "for topic of calculus, I spent long time on many books of intro level" definitely sounds like a mistake. I would not advise anyone to read many introductory calculus textbooks. (Unless you serve on an undergraduate committee or plan to write a textbook of your own.) It's hard to find anything wrong with reading deep and modern masterpieces early. Some amount of struggle and confusion is healthy. There is a simple way to find whether such advanced reading is beneficial: try to explain what you read to someone else (e.g., to an imaginary blog reader :).
H: Finding direction of steepest descent on a hyperplane subject to orthogonal constraint Given a linear objective function $f(\vec{x})=\sum_ia_ix_i$, the direction in which $f$ varies the greatest is known to be $\vec{\nabla}{f}$. Now given a non-zero vector $\vec{v}$, I am interested in finding the direction $\vec{d}$ in which $f$ varies the greatest, subject to the constraint that $\vec{v}^{T}\vec{d} = 0$. How do I go about finding $\vec{d}$ subject to such an orthogonal constraint? AI: Project $\nabla \mathbf{f}$ onto the orthogonal space of $\mathbf{v}$, that's the direction $\mathbf{d}$, which can be computed by $\mathbf{d}=\nabla\mathbf{f}-\mathbf{v}\mathbf{v}^T\nabla\mathbf{f}$ ($\mathbf{v}$ should be normalized first).
H: Bounding an expression involving digamma function Let $\psi$ be the digamma function. I have a conjecture that $$\frac ax > \log(x) - \psi(x)$$ holds for all $x > 0$ if (and only if) $a \ge 1$. I do not know how to prove it. Please help. AI: Let $$f(x) = \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x}.$$ Then for $x > 0$, we have $$f'(x) = \frac{1}{x^2} - \frac{1}{4\sinh^2(x/2)} = \frac{1}{x^2}\left(1 - \left(\frac{x/2}{\sinh (x/2)} \right)^2 \right) > 0. $$ This shows that $f(x)$ is increasing. Also, it is easy to find that $f(0+) = 0$ and $\lim_{x\to\infty} f(x) = \frac{1}{2}$. Thus we find that $0 < f(x) < \frac{1}{2}$ for $0 < x < \infty$ and hence the Laplace transform $\mathcal{L}f(s) = \int_{0}^{\infty} f(x) \, e^{-sx} \; dx$ of $f(x)$ satisfies $$ 0 < \mathcal{L}f(s) < \int_{0}^{\infty} \frac{1}{2} \, e^{-sx} \; dx = \frac{1}{2s}. $$ But we also have $$ \begin{align} \mathcal{L}f(s) &= \int_{0}^{\infty} \left( \frac{1}{2}\coth\left(\frac{x}{2}\right) - \frac{1}{x} \right) e^{-sx} \; dx\\ &= \left[ \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \right]_{0}^{\infty} \\ &\qquad + s \int_{0}^{\infty} \left( \log \sinh \left(\frac{x}{2} \right) - \log x + \log 2 \right) e^{-sx} \; dx \\ &= s \int_{0}^{\infty} \left( \frac{x}{2} - \log (sx) + \log s + \log \left(1 - e^{-x} \right) \right) e^{-sx} \; dx \\ &= \frac{1}{2s} + \gamma + \log s - s \int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{e^{-(n+s)x}}{n} \; dx \\ &= \frac{1}{2s} + \gamma + \log s - \sum_{n=1}^{\infty} \frac{s}{n(n+s)} \\ &= \frac{1}{2s} + \log s - \psi_0(1+s), \end{align}$$ where we have used the fact that $\int_{0}^{\infty} e^{-x} \log x \; dx = -\gamma$ and $$ \psi_0(1+x) = -\gamma + \sum_{n=1}^{\infty} \frac{x}{n(n+x)}. $$ Finally, since $\psi_0(1+s) = \frac{1}{s} + \psi_0 (s)$, we have $$ \frac{1}{2s} < \log s - \psi_0(s) < \frac{1}{s}. $$ Therefore the inequality holds if $a \geq 1$ and only if $a > \frac{1}{2}$. Now we prove that $a = 1$ is the sharp bound. We are to calculate the limit $$\alpha = \lim_{x\to 0} x(\log x - \psi_0(x)).$$ But by our previous calculations, $$\begin{align}\alpha &= \lim_{s\to 0} s(\log s - \psi_0(s)) \\ &= \lim_{s\to 0} s\left( \frac{1}{2s} + \mathcal{L}f(s) \right) \\ &= \frac{1}{2} + \lim_{s\to 0} s \int_{0}^{\infty} f(x) \, e^{-sx} \; dx \\ &= \frac{1}{2} + \lim_{s\to 0} \int_{0}^{\infty} f(u/s) \, e^{-u} \; du \qquad (u = sx) \\ &= \frac{1}{2} + \int_{0}^{\infty} \lim_{s\to 0} f(u/s) \, e^{-u} \; du \\ &= \frac{1}{2} + \int_{0}^{\infty} \frac{1}{2} \, e^{-u} \; du \\ &= 1. \end{align}$$ Thus if $\frac{a}{x} > \log x - \psi_0 (x)$ is true for all $x > 0$, then we must have $a \geq \alpha = 1$, as desired.
H: Clarification on the Hausdorff property If $X$ is a Hausdorff space then for points $a,b \in X$ there are disjoint open sets $U$ and $V$ such that $a \in U$ and $b \in V$. So, take a set of points $\{a_1, \ldots , a_n\}$ and another point $x$. Then for each $a_i$ there are disjoint open sets $U_i$ and $V$ containing $a_i$ and $x$, respectively. My question is: is $V$ disjoint from all the $U_i$, or does this only hold pairwise? Are there conditions we can impose to guarantee an open set separating $x$ from all points in some given finite set? AI: Let $\{a_1, ..., a_n\}$ be a finite collection of points. Let $x$ be a point different from all the $a_i$'s. For each $i$, by being Hausdorff, there exists $U_i$ and $V_i$ such that $a_i \in U_i$, $x \in V_i$ and $U_i \cap V_i = \emptyset$. Then $V = \bigcap_{1 \leq i \leq n} V_i$ is an open set (being a finite intersection of open set) containing $x$ which is disjoint from all the $A_i$'s.
H: How to evaluate $\lim\limits_{n\to+\infty} \prod\limits_{k=1}^n (1+k/n^2)$? I've got a limit which puzzle me several days. The question is $$ \lim_{n\to+\infty} \prod_{k=1}^n\left(1+\frac{k}{n^2}\right).$$ Can you help me? Thank you in advance AI: Intuitively, we have $$\log\left( 1 + \frac{k}{n^2} \right) = \frac{k}{n^2} + O\left(\frac{1}{n^2}\right) \quad \Longrightarrow \quad \log \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) = \frac{1}{2} + O\left(\frac{1}{n}\right)$$ and therefore the log-limit is $\frac{1}{2}$. Here is a more elementary approach: Let $P_n$ denote the sequence inside the limit. Then just note that $$ P_n^2 = \left[ \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right) \right]^2 = \prod_{k=1}^{n} \left( 1 + \frac{k}{n^2} \right)\left( 1 + \frac{n-k}{n^2} \right) = \prod_{k=1}^{n} \left( 1 + \frac{1}{n}+\frac{k(n-k)}{n^4} \right). $$ Now fix $m$ and let $n \geq m$. Since $k (n-k) \leq \frac{1}{4}n^2$, we have $$ \frac{k(n-k)}{n^4} \leq \frac{1}{4n^2} \leq \frac{1}{4mn}.$$ Thus we have $$ \left( 1 + \frac{1}{n} \right)^n \leq P_n^2 \leq \left( 1 + \frac{1+(1/4m)}{n} \right)^n. $$ Thus taking $n \to \infty$, $$e \leq \liminf_{n\to\infty} P_n^2 \leq \limsup_{n\to\infty} P_n^2 \leq e^{1+1/(4m)}.$$ Since $m$ is now arbitrary, we have $P_n^2 \to e$, or equivalently, $P_n \to \sqrt{e}$.
H: Can one solve a recurrence that contains a function? I'd like to solve a recurrence, so I've been reading about solving recurrences, and all the ones I've seen solved involve only previous terms of the recurrence, and constants. My recurrence is $$t(n) = \frac{t(n-1)(1-\ln(n-1))}{\ln n}$$ and if I could get help solving it, I'd be much obliged; but I'd especially like to know if there are any general principles for solving recurrences of this nature (with functions in them). AI: When you have $t(n)=t(n-1)f(n)$ you can just write $t(n)=t(1)\prod_{i=2}^n f(i)$. In this case I started $i$ at $2$ because trying $n=1$ leads to division by zero. So $t(n)=t(1)\prod_{i=2}^n \frac {1-\ln (i-1)}{\ln i}$
H: How do I prove that any sequence converging to a non-zero value is bounded away from 0? Please prove that any sequence converging to a non-zero value is bounded away from 0. AI: This is not true. Suppose $a_1 = 0$ and $a_n = 2$ for all $n > 1$. $\lim_{n \rightarrow \infty} a_n = 2$ but $\{a_n : n \in \mathbb{N}\}$ is not bounded away from $0$. However if you assume that the sequence $(a_n)_{n \in \mathbb{N}}$ is nonzero, then this is true. Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers. Suppose that $\lim_{n \rightarrow \infty} a_n = L$ where $L \neq 0$. By the definition of the limit, there exists a natural number $N$ such that for all $n > N$, $\|a_n - L\| < \frac{\|L\|}{2}$. Let $\alpha = min \{\|a_0\|, ..., \|a_n\|, \|\frac{L}{2}\|\}$. Then you have that for all $n$, $\|a_n\| > \alpha$.
H: Attaching a topological space to another I'm self-studying Mendelson's Introduction to Topology. There is an example in the identification topology section that I cannot understand: Let $X$ and $Y$ be topological spaces and let $A$ be a non-empty closed subset of $X$. Assume that $X$ and $Y$ are disjoint and that a continuous function $f : A \to Y$ is given. Form the set $(X - A) \cup Y$ and define a function $\varphi: X \cup Y \to (X - A) \cup Y$ by $\varphi(x) = f(x)$ for $x \in A$, $\varphi(x) = x$ for $x \in X - A$, and $\varphi(y) = y$ for $y \in Y$. Give $X \cup Y$ the topology in which a set is open (or closed) if and only if its intersections with $X$ and $Y$ are both open (or closed). $\varphi$ is onto. Let $X \cup_f Y$ be the set $(X - A) \cup Y$ with the identification topology defined by $\varphi$. Let $I^2$ be the unit square in $\mathbb{R}^2$ and let $A$ be the union of its two vertical edges. Let $Y = [0, 1]$ be the unit interval. Define $f : I^2 \to Y$ by $f(x, y) = y$. Then $I^2 \cup_f Y$ is a cylinder formed by identifying the two vertical edges of $I^2$. I don't understand how $I^2 \cup_f Y$ can be a cylinder. The set is equal to $(I^2 - A) \cup Y$. Which is a union of a subset of $\mathbb{R}^2$ and $[0, 1]$. How can this be a cylinder? The book has an exercise that constructs a torus in a similar manner. I'm hoping to be able to solve it once I understand this example. I looked up some examples online. While I understand the definitions and theorems of identification topologies, I have no clue how geometric objects are constructed. AI: Yeah so this is one of those things where imho intuition (given excellently by us2012) is way more important than the details. For the details, fix the cylinder as $[0,1] \times S^1$, with coordinates $(h, \theta), \theta \in [0, 2\pi)$ (how high on the cylinder you are, and where you are on the circle at that height). For ease of notation, describe points in $I^2 \cup_f A$ as just $(x,y) \in I^2$, modulo $(0,y) \sim (1, y)$. Then we have (inverse, continuous) maps $I^2 \cup_f A \rightarrow [0,1] \times S^1$ given by $(x,y) \mapsto (x, 2\pi y)$, and $[0,1] \times S^1 \rightarrow I^2 \cup_f A$ given by $(h, \theta) \mapsto (h, \frac{\theta}{2\pi})$. I'll leave checking these are inverse and continuous as an exercise.
H: Deducing formula for a linear transformation The question I'm answering is as follows: Let $ T: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a linear transformation such that $ T(1,1) = (2,1) $ and $ T(0,2) = (2,8) $. Find a formula for $ T(a,b) $ where $ (a,b) \in \mathbb{R}^2 $. Earlier we proved that $\{(1,1), (0,2)\}$ spans $\mathbb{R}^2$. I used this when trying to find a formula for $T$. My working is: $T(a(1,1) + b(0,2)) = aT(1,1) + bT(0,2) $ Because $T$ is linear. Thus: $ T(a(1,1) + b(0,2)) = a(2,1) + b(2,8) = T(a,b)$ Is this correct? It seems a bit too easy and so I'm wondering if I missed anything. AI: You’ve gone astray at the very end: $a(1,1)+b(0,2)=(a,a)+(0,2b)=(a,a+2b)$, so what you’ve calculated is $T(a,a+2b)$, not $T(a,b)$. You know that $T(0,2)=(2,8)$, and $(0,1)=\frac12(0,2)$, so $T(0,1)=\frac12T(0,2)=\frac12(2,8)=(1,4)$. You also know that $(1,0)=(1,1)-(0,1)$, so $T(1,0)=T(1,1)-T(0,1)=(2,1)-(1,4)$. Finally, you know that $(a,b)=a(1,0)+b(0,1)$, so $T(a,b)=aT(1,0)+bT(0,1)$. Can you finish it from here?
H: Computing the differential of the map $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ Let $\phi: M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ be the map $$(B_1,B_2)\mapsto [B_1,B_2]$$ which takes two $2\times 2$ matrices to its Lie bracket. Then why does $d\phi_{(B_1,B_2)}:M_2(\mathbb{C})\times M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ send $$(D_1,D_2)\mapsto [B_1,D_2]+[D_1,B_2]?$$ $$ $$ Taking $B_1=(g_{ij})$ and $B_2=(h_{ij})$, I do not think taking the partials of the Lie bracket $[B_1,B_2]=$ $$ \left[ \begin{array}{cc} g_{12} h_{21} - g_{21} h_{12} & -g_{12} h_{11} + g_{11} h_{12} - g_{22} h_{12} + g_{12} h_{22} \\ g_{21} h_{11} - g_{11} h_{21} + g_{22} h_{21} - g_{21} h_{22} & g_{21} h_{12} - g_{12} h_{21} \\ \end{array}\right] $$ is a clever way to figure out the map. AI: Taking differentials is all about looking at what happens to your map upon a very small perturbation. So compute the bracket $$[B_1 + \epsilon D_1, B_2 + \epsilon D_2]$$ and look at the coefficient of $\epsilon$.
H: The differential of $\psi: GL_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ which sends $g\mapsto gAg^{-1}$ Suppose $\psi: GL_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ defined by sending $$ g\mapsto gAg^{-1}. $$ Then why is it that $d\psi:T_eGL_2(\mathbb{C})=M_2(\mathbb{C})\rightarrow M_2(\mathbb{C})$ is defined to be $$ C\mapsto [C,A]? $$ It is sort of related to this link, but I am not sure if the same strategy as in that link will work here. AI: The same strategy works (with $2$ replaced by $n$). The saving grace here is that $\text{GL}_n(\mathbb{C})$ naturally embeds into $\mathcal{M}_n(\mathbb{C})$ and its tangent space at any point can be canonically identified with $\mathcal{M}_n(\mathbb{C})$. So write $g = 1 + \epsilon C$ and compute the $\epsilon$ term of $gAg^{-1}$...
H: Summation of an Arithmetic, Parametric Sequence I'm trying work on my ability to break complex patterns down, and in this case I'm trying to model the denominators of Lacsap's Fractions: I managed to get the sequence that represents the denominators: But now I want to convert it to a series in order to get rid of the tedious term D_n-1. I guess I'm just not quite sure how to apply the sum identity to a parametric sequence: The red is the answer that I've found works, but I don't know how to get to it. When I simply substitute k, D_1, and D_k with snippets from the sequence, I get it wrong. AI: I think you wanted to write $$D_{n,r}=D_{n-1,r}+(n-r)$$ though you could have written $$D_{n,r}=D_{n-1,r-1}+r$$ with the combined effect that (assuming that you start from $D_{0,0}=0$) $$D_{n,r}=\sum_{i=1}^{n-r} i + \sum_{j=1}^{r} j = \frac{(n-r)(n-r+1)}{2} + \frac{r(r+1)}{2}$$ and hence your result. Another way to look at it is to note that $D_{n,0}=D_{n,n}=\frac{n(n+1)}{2}$ and that $D_{n,r}=D_{n-r,0}+D_{r,r}$. For the numerator you have $N_{n,r}=\frac{n(n+1)}{2}$.
H: Number of solutions of $x^2_1+\dots+x^2_n=0,$ $x_i\in \Bbb{F}_q.$ Let $\mathbb F$ be field with $q$ element and $\operatorname{char}(\mathbb F)\neq2$. I want to know about the number of solutions of this equation: $$x^2_1+x^2_2+\dots+x^2_n=0 \text{ where } x_i\in \mathbb F.$$ Any suggestion? AI: Partial answer that works, when $q\equiv1\pmod4$. This is a minor simplification to the use of recurrences as described in the first sentence of Gerry Myerson's answer (that I somehow missed as my gaze was drawn to the character sums :/). For the cases $q\equiv-1\pmod4$ I recommend doing it his way. When $q\equiv1\pmod4$ we know that the field contains a primitive fourth root of unity, call it $i$. Let $A(n)$ be the number of solutions for the equation in $n$ variables. We know (see also comment's by Gerry Myerson) that $A(1)=1$ and $A(2)=2q-1$. We can write $$x_1^2+x_2^2=(x_1+ix_2)(x_1-ix_2)=z_1z_2,$$ where $z_1=x_1+ix_2$ and $z_2=x_1-ix_2$. The mapping $(x_1,x_2)\mapsto(z_1,z_2)$ is clearly a bijection, so $A(n)$ is equal to the number of solutions of $$ z_1z_2=-(x_3^2+x_4^2+\cdots+x_n^2). $$ Here the r.h.s. is equal to zero for exactly $A(n-2)$ vectors $(x_3,x_4,\ldots,x_n)$, so that gives us $(2q-1)A(n-2)$ solutions. OTOH, if the r.h.s. is non-zero, we can let $z_2$ be any non-zero element ($(q-1)$ choices), and can then uniquely solve for $z_1$. This gives us $(q-1)(q^{n-2}-A(n-2))$ solutions. Altogether we get a recurrence relation $$ A(n)=(2q-1)A(n-2)+(q-1)\left(q^{n-2}-A(n-2)\right)=(q-1)q^{n-2}+qA(n-2). $$ It is easy to prove by induction on $n$ that this has the solution $$ A(2k+1)=q^{2k} $$ for odd values of $n=2k+1$, and $$ A(2k)=q^{2k-1}+q^k-q^{k-1} $$ for even values of $n=2k$.
H: Function is continuous or not If $f$ is a function defined on $[0, 1]$ for which $f'(\frac{1}{2})$ exists but$f'(\frac{1}{2})$ is not an element in $[0,1]$ then f is discontinuous at $x=\frac{1}{2}$. I need to find the statment is true or false and my conclusions are the statement is true since ,if an inverse function exits and the domain of the inverse fuction should be the range of the function,in this case it fails. Is my analogy correct please help? AI: Nothing in the problem says anything about $f$ having an inverse. Consider the function $f(x)=3x$. What is $f'\left(\frac12\right)$? Is $f$ continuous on $[0,1]$? Added: In fact, a function $f$ that is defined on an interval $[a,b]$ and has a derivative at a point $c\in(a,b)$ is always continuous at $c$. This is a standard result that is mentioned in just about all elementary calculus texts and proved in some.
H: There is a stright formula for calculate n Power Sets P(A) of a A set with m members By using Cantor's Theorem: if We have a Set A, with n members the Cardinality of the Power Set is $P(A)=2^n$. There is Formula to calculate n Power Sets $P(...P(P(A))...)$ of a set A with m elements without doing the simple iteration $(((2^2)^2)..)^2^m$ n times? Many thanks AI: The correct formula is $$\large{2^{\left(2^{\left(2^{\dots(2^n)}\right)}\right)}}\;.$$ That is, you start with $2^n$ for the cardinality of $\wp(A)$ when $\|A\|=n$, then get $2^{2^n}=2^{(2^n)}$ for the cardinality of $\wp(wp(A))$, and so on. I know of no simpler form for this iterated power.
H: The double cover of Klein bottle I try to find out all the double covers of Klein bottle. Since the Euler characteristic is multiplicative with respect to covering space, there are only two candidates, that is, torus and Klein bottle itself. It is not hard to construct a mapping from torus to Klein bottle. My question is, is it possible to realize the Klein bottle as a double cover of itself? AI: For any $c>0$ the relations $$(x,y)\sim (x, \,y+ k c)\quad(k\in{\mathbb Z}), \qquad (x,y)\sim \bigl(x+\ell,\,(-1)^\ell y\bigr) \quad(\ell\in{\mathbb Z})$$ define a Klein bottle $K_c$ of "length" $1$ and "width" $c$ as a quotient of the $(x,y)$-plane , and with a rectangle $[0,1]\times[0,c]$ as fundamental domain. The identity map $\,\iota: \,{\mathbb R}^2\to{\mathbb R}^2$ realizes $K_2$ as a double cover of $K_1$: Each point $(x,y)_{\sim1}\in K_1$ has two preimages in $K_2$, namely $(x,y)_{\sim2}$ and $(x,y+1)_{\sim2}$.
H: Explicitness in numeral system Prove that for every $a\in\mathbb{N}$ there is one and only one way to express it in the system with base $\mathbb{N}\ni s>1$. Seems classical, but I don't have any specific argument. AI: Suppose that some positive integer $N$ has two representations base $s$. Then we can write $$N=\sum_{k=0}^ma_ks^k=\sum_{k=0}^nb_ks^k\;,\tag{1}$$ where $0\le a_1,\dots,a_m,b_1,\dots,b_n\le s-1$ are integers, and without loss of generality we may assume that $m\le n$. Suppose first that $m<n$. Then $$\begin{align} \sum_{k=0}^ma_ks^k&\le\sum_{k=0}^m(s-1)s^k\\ &=\sum_{k=0}^ms^{k+1}-\sum_{k=0}^ms^k\\ &=\sum_{k=1}^{m+1}s^k-\sum_{k=0}^ms^k\\ &=s^{m+1}-1\\ &<s^n\\ &\le\sum_{k=0}^nb_ks^k\;, \end{align}$$ contradiction the assumption $(1)$. Thus, we assume that $m=n$. Now make the further assumption that $N$ is the smallest positive integer with two different base $s$ representations. Without loss of generality assume that $a_n\le b_n$, and let $M=N-a_ns^n$. Then $$M=\sum_{k=0}^{n-1}a_ks^k=\sum_{k=0}^{n-1}b_ks^k+(b_n-a_n)s^n\;.\tag{2}$$ We just proved that this is impossible if $b_n-a_n\ne 0$, so $(2)$ must simplify to $$M=\sum_{k=0}^{n-1}a_ks^k=\sum_{k=0}^{n-1}b_ks^k\;.$$ But then $M$ is a number with two base $s$ representations that is smaller than $N$, contradicting our choice of $N$. It follows that no such $N$ exists and hence that every positive integer has a unique representation base $s$.
H: Difference between graph homomorphism and graph isomorphism I am still not getting how graph isomorphism and homomorphism differ. Can anyone show me two graphs that are homomorphic, but not isomorphic? Also, Wikipedia says that when the function and the inverse function are both homomorphic, two graphs are isomorphic. How is it? AI: A homomorphism can be from a bigger to a smaller graph. Here’s a concrete example: a / \ / \ 0 b c / \ \| \| / \ d-----e 1-----2 G H The map $h:V(G)\to V(H)$ given by $h(a)=h(d)=0$, $h(b)=h(c)=1$, and $h(e)=2$ is a graph homomorphism: it sends edges $ab,db$, and $ac$ to $01$, $ce$ to $12$, and $de$ to $02$. Clearly, however, $G$ and $H$ are not isomorphic, since they don’t even have the same number of vertices. Notice that in general if $h:V(G)\to V(H)$ is a graph isomorphism, then $h(u)h(v)$ is an edge of $H$ if and only if $uv$ is an edge of $G$. If $h$ is merely a graph homomorphism, however, $h(u)h(v)$ can be an edge of $H$ even if $uv$ isn’t an edge of $G$. Thus, if $K$ is the graph shown below, the map that takes $a,b,c,d$, and $e$ to $0,1,2,3$, and $4$, respectively, is a homomorphism but not an isomorphism. 0 / \ / \ 1-----2 \| \| 3-----4 K
H: How to calculate the rotation matrix between 2 3D triangles? I need to calculate the rotation matrix and the translation vector between 2 given triangles in Euclidean space. This is really just a rotation and a translation, the lengths of sides of the triangle are equal. Coordinates of the points are real. $P_1, P_2, P_3$ are the 3 points of the 1st triangle, $Q_1, Q_2, Q_3$ are the 3 points of the 2nd. My idea was: I translate the center of the triangles to the point of origin with $t_1 := \frac{P_1+P_2+P_3}{3}$, $P'_i = P_i-t_1$ and $t_2 := \frac{Q_1+Q_2+Q_3}{3}$, $Q'_i = Q_i-t_1$ and then I only need to solve the set of 9 linear equations with 9 unknown (the 9 entries of the rotation matrix $R$): $$ Q'_i = R.P'_i $$ Afterwards I could calculate the translation with $$ t = t_2 - R.t_1 $$ The problem is: after I translate the triangles to the point of origin the points are linearly dependent ($P_3 = -P_1 - P_2$) So in fact I only have 6 equation with 9 unknown, which allows multiple solutions for $R$. But I need "one" precise solution, because the triangles are just parts of two identical models, which I need the rotation for. I think the approach will work, if I use 4 points instead of 3. But can I somehow calculate the rotation matrix with "only" 3 points? AI: In principle you need two rotations and a matrix-multiplication. Call the recentered matrices Q ad S and P as T. Then find the rotation s, which rotates S to "triangular" form of its coordinates, so that the coordinates of the first point becomes $[x_{s,1},0,0]$, of the second becomes $[x_{s,2},y_{s,2},0]$. This can be done with your ansatz of using unknowns, if you assume three rotations (for each pair of planes 1 rotation, and gives a 3x3-matrix whose 3'rd column is zero and describes thus a triangle in a plane) Then do the same with another rotation t which rotes T in the same manner. Then $T = S \cdot s \cdot t^{-1} = S \cdot r$ where r is the matrix for the complete rotation. A bit of pseudocode (code in my proprietary MatMate-tool): S = Q - Meanzl(Q) // do the translation to the origin T = P - Meanzl(P) // get the rotation-matrices which rotate some matrix to triangular shape ts = gettrans(S,"tri") tt = gettrans(T,"tri") // make one rotation-metrix. the quote-symbol means transposition tr = ts * tt' // difference should be zero CHK = T - S*tr Example: We assume S and T being centered. Let $ \small \text{ S =} \begin{bmatrix} 14.469944&22.964690&-7.581631\\ -15.275348&5.923432&23.720255\\ 0.805404&-28.888122&-16.138624 \end{bmatrix} $ and $ \small \text{ T =} \begin{bmatrix} 22.808501&2.515200&16.361035\\ 8.393637&-5.071089&-27.109127\\ -31.202138&2.555889&10.748092 \end{bmatrix} $ Now you can solve for a rotation in y/z-plane, which makes the entry in $S_{1,3}=0$. The rotation-parameters are some cos/sin-values. Apply this and you get $\small \text{ S}^{(1)} = \begin{bmatrix} 14.469944&24.183840&0.000000\\ -15.275348&-1.811476&24.381471\\ 0.805404&-22.372364&-24.381471 \end{bmatrix}$ Now you can solve for a rotation in x/y-plane, which makes the entry in $S_{1,2}=0$. The rotation-parameters are some other cos/sin-values. Apply this and you get $\small \text{S}^{(2)} = \begin{bmatrix} 28.182218&0.000000&0.000000\\ -9.397482&12.178056&24.381471\\ -18.784736&-12.178056&-24.381471 \end{bmatrix}$ After that a third set of rotation-parameters cos/sin-values make S triangular. It looks then like this $\small \text{ S}^{(3)} = \begin{bmatrix} 28.182218&0.000000&0.000000\\ -9.397482&27.253645&0.000000\\ -18.784736&-27.253645&0.000000 \end{bmatrix}$ Because the center of your original triangle was moved to the origin, the last column (the z-coordinates) are zero/not needed, since 3 points can always be placed in a plane. From the three rotation with their cos/sin-values you can construct a 3x3-rotation-matrix, say s. The same can be done using the matrix T leading to a rotation-matrix t. If S and T describe in fact the same triangle, only rotated, the results are equal: $T^{(3)}=S^{(3)}$. Then you can use the nice fact, that the inverse of a rotation-matrix is just its transpose, such that with $\small \text{ tr =} \begin{bmatrix} 0.205215&0.860645&0.466022\\ 0.946329&-0.295966&0.129867\\ 0.249696&0.414359&-0.875190 \end{bmatrix}$ we get $ T = S \cdot tr $ In principle this can also be solved using the concept of pseudoinverses: we demand $ tr = S^{-1} \cdot T $ . But because S has reduced rank the inverse means to divide by zero. Using SVD-decomposition (see wikipedia) the pseudoinverse can be computed if the inverse of the diagonal matrix of the SVD-factors is used (where zeros are simply left zero). This should lead to the same solution.
H: Is it possible to have $D=\Bbb P$ Let $f:\Bbb R\to \Bbb R$ and $D=\{x\in \Bbb R: f $ is discontinuous at $x\}$. My problem is : Is it possible to have $D=\Bbb P$ where $\Bbb P$ is the set of irrationals in $\Bbb R$. I know the answer is negative, but, how to prove it?? My attempt: First, I proved that $\Bbb P$ is not a countable union of closed sets in $\Bbb R$.Then, I read somewhere that $D$ is an $F_{\sigma}$ set (but don't know how to prove it). If one could prove the second part, the problem is solved, but How to do it?? Thanks in advance!! AI: Denote $$G_{k}:=\bigcup\{U\subset\mathbb{R}:U\,\,\mathrm{is}\,\,\mathrm{open}\,\,\mathrm{and}\,\,\|f(x)-f(y)\|<\frac{1}{k}\,\,\mathrm{for}\,\,\mathrm{all}\,\,x,y\in U\}$$ for all $k\in\mathbb{N}$. Show that $D^{c}=\bigcap_{k=1}^{\infty}G_{k}$, making $D^{c}$ (i.e. the continuity points of $f$) a $G_{\delta}$-set and thus $D$ a $F_{\sigma}$. If you need some help concerning the steps I may expand this answer or give hints in the comment section. And by the way, irrationals is not $F_{\sigma}$ because rationals is not $G_{\delta}$. If rationals were $G_{\delta}$, then as a countable completely metrizable topological space ($G_{\delta}$ subsets of a complete metric space are completely metrizable) it has an isolation point by Baire category theorem. But since rationals have no isolation points, this is a contradiction. Hence rationals is not $G_{\delta}$ and thus irrationals is not $F_{\sigma}$.
H: $f'+\lambda f$ acts as $f'$ without integral Problem Suppose that $f(x)$ is a differentiable function and $f'(x)$ is the derivative of $f(x)$. Without aids of integral, can we prove that $f'(x)+\lambda f(x)$ has intermediate property? Intermediate property A (real) function $f(x)$ having intermediate property means that if $a,b\in f(\Bbb R)$ and $a<c<b$, then $c\in f(\Bbb R)$, where $f(\Bbb R)=\{\;f(x):x\in\Bbb R\;\}$ With aids of integral Let $g(x)=f'(x)+\lambda f(x)$. For $f(x)$ is continuous, we have $f(x)$ is Riemann-integrable. Let $G(x)=f(x)+\int_0^xf(t)dt$, we have $G'(x)=g(x)$; therefore, we can apply Darboux's theorem to $G(x)$, and we've done. AI: Let $a\lt b$, $g(t)=f'(t)+\lambda f(t)$, and $z$ in the interval between $g(a)$ and $g(b)$. One wants to show that there exists $x$ in $[a,b]$ such that $g(x)=z$. If $\lambda=0$, this is the usual case. Assume that $\lambda\ne0$. One can assume without loss of generality that $g(a)\lt z\lt g(b)$. Consider the function $h$ defined by $h(t)=\mathrm e^{\lambda t}(f(t)-z/\lambda)$. Then $h'(t)= \mathrm e^{\lambda t}(g(t)-z)$, hence $h'(a)\lt0\lt h'(b)$. The usual intermediate value theorem for derivatives applied to $h'$ ensures that $h'(x)=0$ for some $x$ in $(a,b)$. Thus, $g(x)=z$.
H: How to call simple expressions in a paper which do not need any proof? I am using an expression in a paper which doesnt need a proof at all such as $$\sum_{i=k}^K\binom{K}{i}p^i(1-p)^{K-i}$$ is a decreasing function of $k$ and $$\sum_{i=0}^{k-1}\binom{K}{i}p^{K-i}(1-p)^i$$ is an increasing function of $k$. However I connot write it in the text. Because I refer to this property in a non trivial proposition. Question: How should I put this thing in a format? should I call it as a statement or definition or corollary? Thank you very much. AI: I suggest you use the term Remark.
H: Decomposition of Modules Every finitely generated module $M$ over a principal ideal domain $R$ is isomorphic to a direct sum of cyclic modules. If $M=M_n(\mathbb Z)$, the matrices of order $n$ over the integers. What is the decomposition in sum of cyclic modules of $M_n(\mathbb Z)$? AI: As was suggested above, the additive group on the $n\times n$ matrices is just a sum of $n^2$ copies of $\mathbb{Z}$, as isn't hard to see: it's torsion-free, in particular. I could imagine you were thinking about how to make the multiplicative structure work, but the theorem only applies to modules, not algebras. The algebra $M_n$ is not a sum of cyclic $\mathbb{Z}$-algebras, since all of the latter are commutative.
H: One problem with Euler's function Let $n=100$. Is $\phi(n)$ the smallest number $\lambda$ such that if $\gcd(a,n)=1$ then $a^{\lambda}\equiv_{n}1$ ? Generalize this observation. I suppose the answer is no. I've read something about $\mathrm{ord}_n(a)$ but group theory rather goes beyond my discrete mathematics course. I don't know what exactly author of this task had on mind, but maybe what is the simplest way to find such smallest $\lambda$? AI: Let λ be the smallest positive number such that $a^λ \equiv 1 \pmod n$ where (a,n)=1. By Euler's totient theorem, $a^{\phi(n)}\equiv 1 \pmod n$ Let $\phi(n)=rλ+s$ where 0 ≤s<λ So,$1≡a^{\phi(n)}=a^{rλ+s}=(a^λ)^r\cdot a^s≡a^s$ So, there comes another s<λ, $a^s≡1 \pmod n$, which is impossible as λ is the smallest=>s=0. => $λ\| \phi(n)$ Unfortunately, we need to iterate the divisors of $\phi(n)$ to find the smallest positive λ which is called $\mathrm{ord}_na$. We know, n has a primitive root if it is of the form $2, 4, p^a , or 2p^a,$ where p is an odd prime and integer a≥0. So, we don't need to traverse up to $\phi(n)$ if n is not of any the above form. For example, $\phi(100)=40$. As 100=4*25, it can not have a primitive root, no $a$ co-prime to 100 can have order 40,$ ord_{100}a$ must be one of $1,2,4,5,8,10,20$. According to Carmichael Function, we can off-course, make reduction of the traversal. The reduction will increase with the number of unique prime divisors. Using Carmichael Function, $ord_{100}a$ must divide lcm$(\phi(4)\cdot \phi(25))=lcm(2,20)=20$.
H: Holomorphism of $f$ and $f^n$ Given $f^3$ and $f^7$ holomorphic functions, I want to show that $f$ is holomorphic. Can I say that if $f$ is not holomorphic then $f^3$ is not holomorphic which mean $f$ is must be holomorphic? AI: As David pointed out, $f(z)=\frac{f^7(z)}{f^6(z)}$ is holomorphic at all points where $f^6(z)\neq0$. Using Riemann's theorem on removable singularities it's enough to show that $\lim_{z\to0}f(z)$ exists. Now, assume that $f^6(z_0)=0$. Since $f^6$ is holomorphic at $z_0$, you can write $f^6(z)=(z-z_0)^kg(z)$ where $g$ is holomorphic and $g(z_0)\neq0$. Also, $f^6(z_0)=0$ implies that $f^7(z_0)=0$, so you can write $f^7(z)=(z-z_0)^mh(z)$. Now consider $f^{42}$: it is holomorphic, as a product of holomorphic and hence $(z-z_0)^{7k}g(z)=(z-z_0)^{6m}h(z)$. Since the order of the zero at $z_0$ is constant, we have $7k=6m$, so $m>k$. Hence you can write $f(z)=(z-z_0)^{m-k}\frac{h(z)}{g(z)}$, and $g$ doesn't vanish at $z_0$, so the limit of $f(z)$ at $z_0$ exists (and is equal to $0$).
H: Uniquely Decodable Code Given a code $c:S\rightarrow T^$ let $n_i$ denote the number of symbols in S that are encoded by strings of length $i$ in $T^$ If we want to construct a uniquely decodable code for 12 symbols using binary words of length less than or equal to 4. Make a list of all the sets of parameters $n_1,n_2,n_3,n_4$ for which a suitable code exists. I'm sure I need to use the Kraft-Mcmillian number here and I think this is pretty simple but I'm a bit lost as where to go. Thanks for any help. AI: According to the Kraft-McMillan theorem, a necessary and sufficient condition for $c$ to be uniquely decodable is that $$\sum_{k=1}^4 n_k\left(\frac12\right)^k\le1\;.\tag{1}$$ Thus, you need to find all of the $4$-tuples $\langle n_1,n_2,n_3,n_4\rangle$ of non-negative integers such that $(1)$ holds and $n_1+n_2+n_3+n_4=12$. One way to do it systematically is to observe that $\langle 0,0,0,12\rangle$ is a solution and then try to transfer $4$-bit strings to shorter strings. For instance, $\langle 0,0,1,11\rangle$ is still okay, because $\frac18+\frac{11}{16}\le1$. What about $\langle 0,0,2,10\rangle$? The lefthand side of $(1)$ is then $\frac28+\frac{10}{16}=\frac78\le1$, so that’s okay, too. So are $\langle 0,0,3,9\rangle$ and $\langle 0,0,4,8$: the sums are $\frac38+\frac9{16}=\frac{15}{16}\le1$ and $\frac48+\frac8{16}=1$. But $\langle 0,0,5,7\rangle$ isn’t acceptable: $\frac58+\frac7{16}=\frac{17}{16}>1$. There are ways to speed things up, but a good way to discover some of them is to continue the brute-force procedure until you start seeing some patterns. For instance, you might notice just in what I did above that every time I transferred one symbol from a $4$-bit to a $3$-bit code, the sum on the lefthand side of $(1)$ increased by $1/16$. Can you see why? Better yet, can you see how to make use of this in transferring symbols between codes of other lengths?
H: Confusion about chain of subspaces This is part of Problem 11.1.6 of Dummit and Foote. The problem reads Let $V$ be a vector space of finite dimension. If $\phi$ is any linear transformation from $V$ to $V$ prove that there is an integer $m$ such that the intersection of the image of $\phi^m$ and the kernel of $\phi^m$ is $\{0\}$. On my homework, I wrote If $x\in Im(\phi^n)$, then for some $a\in V$, $x=\phi^n(a)=\phi^{n-1}(\phi(a))$, which implies $x\in Im(\phi^{n-1})$. Thus we have a sequence of vector subspaces ordered by inclusion $$ \cdots\subseteq Im(\phi^n)\subseteq\cdots\subseteq Im(\phi^2)\subseteq Im(\phi) $$ My professor said my argument is wrong, and I agreed after he explained why. The problem is that this was a few months ago, and I've forgotten his reasoning and can't locate my hardcopy of the homework. If this argument isn't good enough, I would've just said $\phi(V)\subseteq V$, therefore $\phi^m(V)\subseteq\phi^{m-1}(V)$ and have the equivalent conclusion of the existence of a chain (which stabilizes due to the finite dimensionality of the vector space) and continued the rest of the argument (which I'm omitting). Is this approach okay? Maybe it's something really obvious and 4:30AM is not the right time to think about this. Thanks a lot. AI: Your argument that the subspace $\mathrm{Im}(\phi^n)=\phi^n(V)$ weakly decreases with $n$ is correct, so it stabilises to some subspace $W$. However one need not have $W=\{0\}$; indeed this would mean that $\phi$ is nilpotent which is certainly not always the case. Nonetheless you can complete the argument: $W$ is manifestly $\phi$-stable, and the restriction of $\phi$ to $W$ has zero kernel (otherwise the chain wouldn't have stabilised at $W$), so it is invertible $W\to W$. But then $(\phi\|_W)^n:W\to W$ is invertible, and $W\cap\ker\phi^n=\{0\}$
H: Help calculating combination of combinations I have a problem which I thought was really easy to solve but now I am here =) I need to construct a final combination of a content based on combinations of various sub-contents. A sub content is a group of items which can be varied in combinations. The example below lists each groups possible combinations not the items in the group that make up the combinations. The number of sub-contents can vary to any integer. The number of combinations in a sub-content can also vary to any integer. Example. Sub-content:x -> 4 combinations Xa,Xb,Xc,Xd Sub-content:y -> 10 combinations Ya,Yb,Yc,Yd,Ye,Yf,Yg,Yh,Yi,Yj Sub-content:z -> 3 combinations Za,Zb,Zc where These three different sub-contents will construct the whole content and should result in 4 * 10 * 3 = 120 combinations, right? Combination 1 should be Xa,Ya,Za and combination 5 should be Xa,Yb,Za etc 1 2 3 4 5 .. 14 .. 120 X a b c d a b Y a a a a b d Z a a a a a a Seems like an easy enough task to calculate but I cant manage to get it right. I need a way to calculate which sub-contents a specific combination (1-120) is based upon Thankful for any assistance and excuse me for not knowing the correct terms. AI: Suppose you have a number $n$, $1\le n\le120$. Divide $n-1$ by 4. You have Xa, Xb, Xc, or Xd according as the remainder, $r$, is 0, 1, 2, or 3. Then take $m=(n-1-r)/4$ and divide it by 10. You have Ya, Yb, ..., Yj according as the remainder $s$ is 0, 1, ..., 9. Then take $(m-s)/10$ and divide it by 3. You have Za, Zb, or Zc according as the remainder is 0, 1, or 2.
H: Stabilizer of a point and orbit of a point I really need help with this topic I have an exam tomorrow and am trying to get this stuff in my head. But the book is not explaining me these two topics properly. It gives me the definition of a stabilizer at a point where $\mathrm {Stab}_G (i) = \{\phi \in G \mid \phi(i) = i\}$, and where $\mathrm{Orb}_G (i) = \{\phi(i) \mid \phi \in G\}$. I do not know how to calculate the stabilizer nor the orbit for this. I am also given an example Let $G = \{ (1), (132)(465)(78), (132)(465), (123)(456), (123)(456)(78), (78)\}$ and then $\mathrm{Orb}_G (1) = \{1, 3, 2\}$, $\mathrm{Orb}_G (2) = \{2, 1, 3\}$, $\mathrm{Orb}_G (4) = \{4, 6, 5\}$, and $\mathrm{Orb}_G (7) = \{7, 8\}$. also $\mathrm{Stab}_G (1) = \{(1), (78)\},\\ \mathrm{Stab}_G (2) = \{(1), (78)\},\\ \mathrm{Stab}_G (3) = \{(1), (78)\},\text {and}\\ \mathrm{Stab}_G (7) = \{(1), (132)(465), (123)(456)\}.$ If someone could PLEASE go step by step in how this example was solved it would be really helpful. Thank you AI: You are not going to survive at all if you can't compute something after having its definition. Really spend time undestanding definitions, and do it a lot sooner than the night before the exam. Doing the whole thing will not help you in the long run, but doing some samples is fair enough! I'm assuming the notation is doing composition this way:$ (fg)(x)=f(g(x))$. Then $\mathrm{Orb}(1)=\{\phi(1)\mid \phi\in G\}=\underline{\{1,3,3,2,2,1\}}=\{1,2,3\}$. Each one of the numbers between the underlined braces is, in the order you listed them, the result of applying each element of $G$ to 1. For example $(132)(465)1=(132)1=3$. If you cannot apply the permutations to a single number, then you indeed have a lot more studying to do. For $\mathrm{Stab}_G(1)$, you just need to pick out all the elements of $G$ that don't move 1. Obviously $(1)$ and $(78)$ do not move 1. The first is just the identity permutation, and the latter does not move 1 at all, since 1 does not appear. Checking the others, you see that they move 1 either to 2 or to 3. All of the others are like this: completely routine computation to see if you can read and understand the notation and definitions.
H: Co-ordinate geometry problem I've been on this for about an hour, researched everywhere, but I cannot find a viable solution. Question: The straight line $mx = 5y + 4$ has the same gradient as the line $7x + 6y + 5 = 0$, Find the value of $m$. I've tried: $mx - 5y + 4 = 7x + 6y + 5$ but can't get it right :(. Any viable solution for this? I tried to change it to the $y = mx + b$ format, I got this: $5y = mx - 4$ and $6y = 7x - 5$, this is the part I'm stuck on. AI: Your equations should (your equation $6y=7x-5$ is slightly off) be $$ \tag{1}5y=mx-4 $$ and $$\tag{2} 6y=-7x-5.$$ You want $y$ isolated on the left hand side in each of these equations. So, divide both sides of equation $(1)$ by $5$: $$\tag{3} y={m\over5}x-{4\over5}; $$ and divide both sides of equation $(2)$ by $6$: $$\tag{4} y={-7\over6}x-{5\over6}. $$ The slope (gradient) of the line given by $(3)$ is $m/5$ and the slope of the line given by $(4)$ is $-7/6$. From this information, you should be able to solve for $m$.
H: Is there any known formula for $A^n \mathbf{x}^n + A^{n-1} \mathbf{x}^{n-1} + \ldots +A \mathbf{x} + I$? Good afternoon. My question is, if there is any known formula for the expression $$A^n \mathbf{x}^n + A^{n-1} \mathbf{x}^{n-1} + \ldots+ A \mathbf{x} + I,$$ where $A$ is a matrix and $\mathbf{x}$ is a column vector. By $\mathbf{x}^n$, I denote the element-wise $n$-th power of the column vector $\mathbf{x}$. What I am looking for is something like a closed-form formula. It seems to be likely that something like that exists, since the expression resembles the geometric sum. Thanks a lot in advance. AI: Edit: The following answer assumes that all components of the vector ${\bf x}$ are $=x$, $x\in{\mathbb R}$. Let $[1\ 1\ \ldots\ 1]'=:{\bf e}$. For $x\in{\mathbb R}$ your "powers" ${\bf x}^k$ can be written as $${\bf x}^k=\bigl[{\rm diag}(x,x,\ldots ,x)\bigr]^k {\bf e}\ ,$$ and as $A$ commutes with $[{\rm diag}(x,x,\ldots, x)\bigr]$ we have $$A^k{\bf x}^k=(xA)^k {\bf e}\qquad(k\geq 0)\ .$$ By the formula for the sum of a geometric series it follows that $$\sum_{k=0}^n A^k{\bf x}^k=\bigl(I-(xA)^{n+1}\bigr)(I- xA)^{-1}\ {\bf e}\ ,$$ which makes sense for all $x$ of sufficiently small absolute value.
H: Determining a function through three equations I have the following assignment question, and I'm having trouble even getting started: Consider the set of functions $\mathcal{F}=\{f,g\}$, with $f:\mathbb{R}^2\to\mathbb{R}$, and $g:\mathbb{R}\to\mathbb{R}$, defined by $f(x,y)=xy$ and $g(x)=x+1$. Let $A=cl_\mathcal{F}(B)$, with $B=\{0\}$. I.e., $A$ is the closure of $B$ under the operations in $\mathcal{F}$. Then clearly $A=\mathbb{N}$. How would you go about showing that the following three equations, uniquely determine a function $h$? $h(f(x,y))=f(h(x),h(y))$, $h(g(x))=h(x)+2$, and $h(0)=0$. What is $A$ good for? Is it necessary to show what we're being asked to show? And also, we're on the topic of primitive recursive functions and such, so I don't understand how this would fit into that topic anyway. AI: Using the second two equations, we have that $h(0)=0$ and $h(x+1)=h(x)+2$. Thus we can use induction to find the value of $h(x)$ for any value $x\in\Bbb N$. In fact, it is easy to see that if $x\in \Bbb N$ we have $h(x)=2x$. Now the first condition $h(xy)=h(x)h(y)$ provides a contradiction, however, since we can take $y=1$ and find that $h(x)=2h(x)$, i.e. $2x=4x$ for $x\in\Bbb N$. Thus no function $h$ exists satisfying all three conditions, which is why André Nicolas asked if there was perhaps a typo.
H: Calculating the divisor, known to be small, of two Stirling approximations of the logarithmic Gamma function without overflows Earlier, I asked a question on MathOverflow regarding how one might analytically approximate a function of the form: $f(n) = \prod_{i=1}^{n-1} (1-ai)$ for $a \ge 0$, $(ai) < 1$, and $n > 10^5$ or $10^6$. Robert Israel answered the question in a very nice way, observing that: $f(n) = \dfrac{a^{n-1} \Gamma(1/a)}{\Gamma(1-n+1/a)}$ ...and that one could use Stirling's series for an asymptotic approximation of $\ln(\Gamma(1-n+ 1/a))$. My question here is: how might one actually compute values for this function (using, for example, Mathematica) for very large $n$ ($n > 10^9$ or so) and very small $a$ ($a < 10^{-30}$), without overflows/underflows, provided that the known output for $f(n)$ falls in the range of, say, $10^{-12} \le f(n) \le 1$? Are there any simple strategies for achieving this? AI: Since this is a common problem many software implemented a lngamma function. In the case of Mathematica it is named LogGamma[]. If this fails you may use an expansion of $\ln(\Gamma(x))$. You may find these at Wikipedia. Of course once the LogGamma obtained you shouldn't compute directly the exponential ! Instead evaluate the $\log$ of the complete expression and take the exponential of the (simplified) result.
H: Show $f$ is primitive recursive, where $f(n) = 1$ if the decimal expansion of $\pi$ contains $n$ consecutive $5$'s Let $f:\mathbb{N}\to\mathbb{N}$ be given by $f(n)=1$ if the decimal expansion of $\pi$ contains $n$ consecutive $5$'s, and $f(n)=0$ otherwise. How would you go about showing such a function is primitive recursive? I suppose you should be able to construct it out of other p.r. functions, via composition or the basic recursion scheme that the family of p.r. functions is closed under (I don't know if such an algorithm has a name). However, I do not know how to deal with the decimal expansion of anything (much less an irrational) as something that would come from p.r. functions. Particularly because if it's irrational, you'd have to express it as an infinite series (i.e. an infinite number of sums). AI: It's a trick question -- it is easy to show that your function must have a primitive recursive definition, but very hard to say with certainty what that primitive recursive definition is. Either the decimal expansion of $\pi$ contains arbitrary long sequences of consecutive 5's, and in that case $f(n)$ is the function that always returns $1$. That is obviously primitive recursive. -- Or the decimal expansion of $\pi$ contains $N$ consecutive 5's somewhere, but never contains $N+1$ consecutive 5's, for some finite $N$. In that case $$ f(n)=\begin{cases}1 & \text{if }n\le N \\ 0\ &\text{otherwise} \end{cases}$$ which is also obviously primitive recursive. The lesson to take home is that "is primitive recursive" is a property only on the contents of the abstract set $\{(x,y)\mid f(x)=y\}$, and does not depend on how you describe or define that contents. If this feels too fuzzy, consider this simpler example: Define the number $K$ to be $42$ if the Riemann Hypothesis is true and $57$ otherwise. It is extremely difficult to prove stuff about $K$ explicitly, but we can say with absolute certainty that it's an integer, is less than $100$ and is not a perfect square. To prove these things we don't need to know whether RH is true; it suffices to show that no matter whether it is or not, the property we're after will be true. It's the same with "is a primitive recursive function".
H: Convergence in $L^1$ problem. Problem: Let $f \in L^1(\mathbb{R},~\mu)$, where $\mu$ is the Lebesgue measure. For any $h \in \mathbb{R}$, define $f_h : \mathbb{R} \rightarrow \mathbb{R}$ by $f_h(x) = f(x - h)$. Prove that: $$\lim_{h \rightarrow 0} \\|f - f_h\\|_{L^1} = 0.$$ My attempt: So, I know that given $\epsilon > 0$, we can find a continuous function $g : \mathbb{R} \rightarrow \mathbb{R}$ with compact support such that $$\int_{\mathbb{R}} \|f - g\|d\mu < \epsilon.$$ We can then use the inequality $\|f - f_h\| \leq \|f - g\| + \|g - g_h\| + \|g_h - f_h\|$ to reduce the problem to the continuous case, so to speak, since the integral of the first and last terms will be $< \epsilon$. But now I'm stuck trying to show that $$\lim_{h \rightarrow 0} \\|g - g_h\\|_{L^1} = 0.$$ I tried taking a sequence $(h_n)_{n \in \mathbb{N}}$ converging to $0$ and considering $g_n := g_{h_n}$, but I don't have monotonicity and the convergence doesn't seem to be dominated either, so I don't know what to do. Any help appreciated. Thanks. AI: By your construction, $g$ is continuous and compactly supported. Let $K$ be the support of $g$, and let $K_h=K\cup (h+K)$. Then we have $$ \int\|g(x)-g(x-h)\|\,\mathrm{d}x\leq \|K_h\|\\|g-g_h\\|_{L^\infty(K_h)}. $$ For all $h>0$ sufficiently small we have $K_h\subset K_1$, and you can invoke uniform continuity of $g$.
H: What is actually "relatively consistent"? Gödel's incompleteness theorem states that: "if a system is consistent, it is not complete." And it's well known that there are unprovable statements in ZF, e.g. GCH, AC, etc. However, why does this mean that ZF is consistent? What does "relatively consistent" actually mean? AI: We have to hedge quite a bit in what you said. First, there are many complete theories: the theory of all groups of order 7 is complete, as an example. However, Gödel showed that any "nice enough" (read: "recursively enumerable") consistent first-order theory capable of encoding basic arithmetic is incomplete. (There is a complete theory of number theory, usually denoted $\mathrm{Th}(\mathbb{N})$, which consists of all sentences that are true in the standard model of number theory; unfortunately, this is not a "nice enough" theory — it's impossible to program a computer to print out all and only the axioms of this theory!) Also, these statements that you gave, $\mathsf{(G)CH}$, $\mathsf{AC}$, are not provably unprovable in $\mathsf{ZF}$. However, it is provable that they are unprovable in $\mathsf{ZF}$ only if $\mathsf{ZF}$ is itself inconsistent. This is what a relative consistency result means: it is a result that says the consistency of one theory implies the consistency of the other. Kurt Gödel proved that if $\mathsf{ZF}$ is consistent, then so if $\mathsf{ZFC}+\mathbf{V}=\mathbf{L}$ (and also that $\mathsf{ZFC}+\mathbf{V}=\mathbf{L}$ implies $\mathsf{GCH}$). Paul Cohen proved that if $\mathsf{ZF}$ is consistent, then both $\mathsf{ZF}+\neg\mathsf{AC}$ and $\mathsf{ZFC}+\neg\mathsf{CH}$ are consistent. More elementarily, with $\mathsf{ZF}$ one can construct the natural numbers, and so the consistency of $\mathsf{ZF}$ implies the consistency of $\mathsf{PA}$. These are fundamentally relative consistency results, and cannot be improved to straight consistency results. This stems from Gödel's Second Incompleteness Theorem. Because of this the acceptance of the consistency of $\mathsf{ZF(C)}$ is an article of faith (though one I'm happy to believe in). I recommend that you look at the first page of George Boolos's Gödel's second incompleteness theorem explained in words of one syllable (Mind, vol.103, pp.1-3). It is a quite entertaining look at the meaning of Gödel's Second Incompleteness Theorem.
H: Base of shortest connection of two skew straight lines During tutoring (12th grade, regular Math class), I had to explain how to find the two points $s$ and $s'$ that are the base of the perpendicular connection between two skew straight lines $g$ and $h$. Where $g$ and $h$ are given respectively as: $$ g \colon \mathbb R \to \mathbb R^3; t \mapsto \vec x = \vec x_0 + \vec v_g t $$ What I did is to calculate a normal vector to both $\left(\vec n \propto \vec v_g \times \vec v_h \right)$. Then move $h$ along $\vec n$ so that $h'$ intersects $g$. Then I would just calculate the intersection $s$ of $g$ and $h'$, and move the intersection back down to $h$, yielding $s'$. Is there any shorter way? AI: The difference $$\vec x_g + \vec v_g s - \vec x_h - \vec v_h s'$$ must be orthogonal to both $\vec v_g$ and $\vec v_h$. This gives a pair of linear equations for $s$ and $s'$. Moreover, the determinant of the system is $\\|\vec v_g\\|^2 \\|\vec v_h\\|^2 - (\vec v_g\cdot \vec v_h)^2$, which is $0$ iff the two lines are parallel (or coincident).
H: Notation question, Stieltjes integral I'm reading a signal processing textbook and they frequently write out formulas where if I have a Stieltjes integral: $$\int_a^b f(t) \, dX(t)$$ They'll write an explicit formula for $dX(t)$, such as $$dX(t) = g(t)$$ Is the best way to interpret this notation to read my integral as the limit of the following sum? $$\sum f(t^*) (g(t_i) - g(t_{i-1}) )$$ The notation that I would expect to see would be to write $X(t) = g(t)$ and then interpret the integral as the limit of the above sum. AI: "$dX(t)=g(x)$" makes no sense. One can write $dX(t)=g(t)\,dt$, with $t$ rather than $x$ as the variable, and with "$dt$" after it, provided $X$ is differentiable and $X'=g$. But the Riemann-Stieltjes integral makes sense even in many cases where $X$ is not differentiable, and is the limit of the sum you wrote provided you put $X(t_i)-X(t_{i-1})$ where you had $g(t_i)-g(t_{i-1})$. Later edit: In view of the later revision of the question, I will add this: "$dX(t) = g(t)$" doesn't make sense without "$dt$", i.e. one has $dX(t)=g(t)\,dt$ provided $X'=g$. But the Riemann-Stieltjes integral can make sense even in many cases where $X$ is not differentiable. You need the limit of the sum with $X(t_i)-X(t_{i-1})$, not with $g(t_i)-g(t_{i-1})$.
H: About absolute convergence and completeness in rings EDIT: Let R be a commutative ring with unit ring and $I$ a maximal ideal in R. The completion of R with respect to $I$ is the inverse limit of the factor rings $R / I^k$ under the usual quotient maps. A ring is said to be complete with respect to a maximal ideal if the map to its completion with respect to that ideal is an isomorphism. See this page for additional info on convergence and ring completions. Let $R$ be a ring such that it is complete w.r.t. some ideal $I$ and let $(x_n)_n$ be a sequence in $I$. I was told that $$\sum_{n=1}^\infty x_n\quad\mbox{converges}\iff x_n\mbox{ converges to }0.$$ The direction $\implies$ is trivial, the other direction is harder. How can I prove this part? AI: I've just figured it out and for those interested, here is my proof: Note that in any complete ring, a series converges if and only if the sequence of its partial sums $(s_n=\sum_{k=1}^nx_k)_n$ converges. Since $(x_k)$ converges to $0$, there is a $M\in\mathbf N$ such that $x_n\in I^k$ for each $n\ge M$. So by choosing $N = M$,$$s_j-s_n=\sum_{i=1}^jx_{j+i}\in I^k$$ for each $j,n\ge N$.
H: What is the set operation corresponds to the last Venn diagram below? Let the universal set be $U$ which is represented as the rectangular frame. Set $A$, $B$ and $C$ are the 3 circles on the left top, right top and the bottom, respectively. The first diagram represents the $U$. The second diagram represent the complement of $A \cap B\cap C$. Etc... But what is the set operation corresponds to the last Venn diagram below? AI: If the first one is the 'universe' $U$, the last is $U\setminus U = \emptyset$. The blank in the pictures means that we are subtracting something from the 'universe' $U$. As an example, the second figure, the blank is $A\cap B\cap C$, and therefore the orange remaining is $U\setminus (A\cap B\cap C)$.
H: Does $\sum\ln(n) / n^3$ converge? By Direct Comparison Test: $$\ln(n)/n^3 < n/n^3 = 1/n^2 $$ Converges as that is a convergent a p-series By n'th term Test: $$(\ln(n))' = 1/n$$ $$(n^3)' = 3n^2$$ $$\lim_{n\to\infty} (1/n) / 3n^2 = \lim_{n\to\infty} (3n^2)/n =\lim_{n\to\infty} 3n = \infty$$ so by $n$'th term it diverges. What am I doing wrong? AI: You made an algebraic mistake. You have the quotient $$ 1/n\over 3n^2.$$ If you wish to "move the numerator $1/n$ downstairs", you need to take its reciprocal, ${1\over 1/n}=n$, first and do it: $$ {\color{maroon}{1/n}\over 3n^2}={1\over \color{maroon}n\cdot 3n^2}. $$ You did this correctly. However, when you "moved the denominator $3n^2$ upstairs", you failed to use its reciprocal $1/3n^2$. Done correctly, you would have obtained $$ {1/n\over\color{maroon}{ 3n^2}}={(1/n)\cdot(\color{maroon}{1/3n^2})\over1 }. $$ Note both methods give $1\over 3n^3$ as a result; there is no need to do both. Perhaps the simplest thing to do in order to simplify your expression is to write $$ {1/n\over 3n^2}={1\over n}\cdot{1\over 3n^2}={1\over n\cdot3n^2}={1\over 3n^3}. $$
H: Can I shift this integral along the complex plane? Does $$\int_{-\infty}^\infty \text{e}^{\ a\ (x+b)^2}\ \text dx=\int_{-\infty}^\infty \text{e}^{\ a\ x^2}\ \text dx\ \ \ \ \ ?$$ hold, even if the imaginary part of $b$ is nonzero? What I really want to understand is what the phrase "By analogy with the previous integrals" means in that link. There, the expression $\frac{J}{a}$ is complex but they seem to imply the integral can be solved like above anyway. The reusult tells us that the integral is really independend of $J$, which is assumed to be real here. I wonder if we can also generalize this integral to include complex $J$. In case that the shift above is possible, this should work out. But even if the idea is here to perform that substitution, how to get rid of the complex $a$ to obtain the result. If everything is purely real or imaginary, then this solves the rest of the problem. AI: Let us write $b= r+it$. The real part of $b$ does not matter as you have already proven yourself. So wlog $r=0$. For shifting along the imaginary axis, we have to employ the residue theorem. We have $$ \begin{align} \int_{-\infty}^\infty f(x+i t) \,dx&- \int_{-\infty}^\infty f(x)\, dx\\ &=\int_{-\infty-it}^{\infty-it} f(x) \,dx- \int_{-\infty}^\infty f(x)\, dx \\ &= 2\pi i \sum \text{Res}(f)+ \int_{\infty-it}^{\infty} f(x) \,dx - \int_{-\infty-it}^{-\infty} f(x) \,dx \end{align},$$ where $\sum \text{Res}(f)$ is the sum over the residues of $f$ in the area $z\in \mathbb{C}$ with $-t<\text{Im}\, z<0$. So the two integrals are the same if there are no residues and if the two integral at $\pm \infty$ vanish (both of which is the case for your example as long as $\text{Re}\,a <0$).
H: Puzzle : There are two lengths of rope ... You have two lengths of rope. If you set fire to the end of either of them, the rope will burn in exactly one hour. They are not the same length or width as each other. They also are not of uniform width (they might, for example, be wider in the middle than at the end), thus burning half of the rope is not necessarily going to take $30$ minutes. By burning the ropes, how do you measure exactly $45$ minutes worth of time? Note : I don't know whether a solution even exists at all. AI: HINT You can burn a rope at both ends.
H: Two-Sample Confidence Interval for Normal Distributions Let's say I have two independent random samples $X_1, X_2, \dots, X_n$ and $Y_1, Y_2, \dots, Y_n$ from normal distributions with real, unknown means $\mu_x$ and $\mu_y$ and known standard deviations $\sigma_x$ and $\sigma_y$. How would I go about deriving a $100(1 - \alpha)$% confidence interval for $\mu_x - \mu_y$? This is straight forward (in my mind) assuming the standard deviations are equal, but what if they are unequal? AI: Alright, you say known variances. So it's an exercise on a point of theory, not a realistic problem. And you actually assume the two sample sizes are equal. Start by recalling something from the one-sample problem: $$ \bar{X} = \frac{X_1+\cdots+X_n}{n} \sim N\left(\mu_X,\frac{\sigma^2_X}{n} \right) $$ $$ \bar{Y} = \frac{Y_1+\cdots+Y_n}{n} \sim N\left(\mu_Y,\frac{\sigma^2_Y}{n} \right) $$ You don't explicitly state that the two samples are independent. If they are, they we have $$ \bar X - \bar Y \sim N\left(\mu_X-\mu_Y,\frac{\sigma^2_X+\sigma^2_Y}{n}\right) $$ (If we had unequal sample sizes $n$ and $m$, then the variance would be $\dfrac{\sigma^2_X}{n}+\dfrac{\sigma^2_Y}{m}$.) So $$ \frac{((\bar X-\mu_X) - (\bar Y-\mu_Y))\sqrt{n}}{\sqrt{\sigma^2_X+\sigma^2_Y}} \sim N(0,1). $$ So the probability that $$ -A < \frac{(\bar X-\mu_X) - (\bar Y-\mu_Y)}{\sqrt{ \frac{\sigma^2_X+\sigma^2_Y}{n} }} <A \tag{1} $$ is the desired confidence when the number $A$ is suitably chosen. Now do a bit of algebra to rearrange the inequalities $(1)$: $$ \bar X - \bar Y - A\sqrt{\frac{\sigma^2_X+\sigma^2_Y}{n}} < \mu_X-\mu_Y < \bar X - \bar Y + A\sqrt{\frac{\sigma^2_X+\sigma^2_Y}{n}} $$ That's the confidence interval.
H: Need help to show $R/I$ is not necessarily flat over $R$ Let $R$ be a ring with unit and $I$ an ideal in $R$. I want to show that $R/I$ is need not be flat over $R$, but I do not know how to come up with a counter-example. Any hint is appreciated. AI: Consider $R = \mathbf Z$ and take the ideal $I = 2\mathbf Z$. Now take a look at the exact sequence $0\to 2\mathbf{Z}\to\mathbf{Z} \to \mathbf{Z}/2\mathbf{Z} \to 0$. Can you see what happens when you tensor this exact sequence over $\mathbf{Z}$? You should be able to prove that the resulting sequence cannot be exact.
H: (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed. Let $X$ be a metric space. Let $\{p_n\}$ be a sequence in $X$. Let $E$ be a set of all subsequential limits of $\{p_n\}$. How do i prove that $E$ is closed in ZF? Is there a well-ordering of convergent subsequences? I can't think of one.. AI: Let us denote $\{ p_n : n \in \mathbb{N} \}$ by $B$. Clearly $x \in \overline{ B }$ iff $x = p_n$ for some $n$, or $( \forall m ) ( \exists n ) ( 0 < d (x,p_n ) \leq \frac{1}{m} )$. In the latter case we can construct, via induction and without any choice, a subsequence converging to $x$. (In fact, the latter condition is easily seen to be equivalent to $x$ being a subsequential limit of $(p_n)_{n \in \mathbb{N}}$.) Suppose $x \in \overline{ B }$ is not a subsequential limit. This means that $x = p_n$ for some $n$. But also that there is an $m \in \mathbb{N}$ such that $d ( x , p_n ) > \frac{1}{m}$ for all $n$. Therefore $x$ is an isolated point of $B$, and thence it is also an isolated point of $\overline{ B }$. It thus suffices to show that $\overline{ B } \setminus \{ x : x\text{ is an isolated point of }B \}$ is closed, and this follows from the following: Claim: Suppose $F \subseteq X$ is closed and $A \subseteq F$ is a set of isolated points of $F$. Then $F \setminus A$ is closed. Proof: It suffices to show that $X \setminus ( F \setminus A)$ is open. If $x \in X \setminus ( F \setminus A )$ there are two cases: Either $x \notin F$, in which case $x \in X \setminus F$, and this is a neighbourhood of $x$ disjoint from $F \setminus A$. Otherwise $x \in A$, but as $x$ is isolated there is a neighbourhood $U$ of $x$ such that $F \cap U = \{ x \}$, and therefore $( F \setminus A ) \cap U = \emptyset$.) $\dashv$ Edit: There is a minor issue in what I have done above. I seem to have assumed that the sequence $( p_n )_{n \in \mathbb{N}}$ is one-to-one (silly me). This means that some of what I said above is not quite true in general. The relevant facts we still have are: If $x \in \overline{B}$ then either $x \in B$ or there is a subsequence of $( p_n )_{n \in \mathbb{N}}$ converging to $a$. If there is a subsequence of $( p_n )_{n \in \mathbb{N}}$ converging to $x$, then either $( \forall m ) ( \exists n ) ( 0 < d (x,p_n ) \leq \frac{1}{m} )$, or $( \forall N ) ( \exists n \geq N ) ( p_n = x )$. The set we wish to show is closed is therefore $\overline{B} \setminus A$ where $$A = \{ x \in B : x\text{ is an isolated point of }A\text{ and }( \exists N ) ( \forall n \geq N ) ( p_n \neq x ) \}.$$ As $A$ is a set of isolated points, by the above $\overline{B} \setminus A$ is closed.
H: Is the inverse to a monoidal equivalence also monoidal? Let ${\cal C,D}$ be two categories, and let $$ F:{\cal C} \to {\cal D}, ~~~~~~~~~~~~~~~~~ G:{\cal D} \to {\cal C}, $$ be an equivalence of categories. Let us now further assume that ${\cal C}$ can be endowed with a monoidal structure $\otimes$, and that ${\cal D}$ can also be endowed with a monoidal structure $\bullet$. Finally, let us assume, for each $X,Y \in C$, that we have isomorphisms $$ J_{X,Y}: F(X \otimes Y) \to F(X) \bullet F(Y), $$ that give $F$ the structure of a monoidal functor. Now it seems to me that this automatically gives $G$ the structure of a monoidal functor: Any two objects in ${\cal D}$ will be isomorphic to $F(X)$, and $F(Y)$, for some $X,Y \in {\cal C}$, and $$ G(F(X) \bullet F(Y)) \simeq G(F(X \otimes Y)) \simeq X \otimes Y \simeq G(F(X)) \otimes G(F(Y)), $$ giving $G$ the structure of a monoidal functor. Thus, I would conclude that given a equivalence of monoidal categories as simple categories, such that one of the functors in the equivalence is a monoidal functor, then we always get an equivalence of monoidal categories. Am I correct here, or have I missed some categorical subtlety? AI: I think you want to be a little more careful, as a priori there might be multiple ways of writing a given object as F(X). This isn't a really problem though because if you have some object A, you know basically how to write it as F(X): just let X = G(A)! So we just have $$G(A \bullet B) \cong G(F(G(A))\bullet F(G(B))) \cong GF(G(A) \otimes G(B)) \cong G(A) \otimes G(B).$$ Again, as pointed out in the comments you need to check that this map really does satisfy the hexagon relations, but that's straightforward. (Bonus exercise: Use a similar argument to see that the left adjoint of a strong monoidal functor is lax monoidal, while the right adjoint of a strong monoidal functor is oplax monoidal.)
H: Multiplicative group of integers modulo n definition issues It is easy to verify that the set $(\mathbb{Z}/n\mathbb{Z})^\times$ is closed under multiplication in the sense that $a, b ∈ (\mathbb{Z}/n\mathbb{Z})^\times$ implies $ab ∈ (\mathbb{Z}/n\mathbb{Z})^\times$, and is closed under inverses in the sense that $a ∈ (\mathbb{Z}/n\mathbb{Z})^\times$ implies $a^{-1} ∈ (\mathbb{Z}/n\mathbb{Z})^\times$. The question is the following: Firstly, are $a$ and $b$ referring to each equivalence class of integers modulo $n$? Secondly, by $a^{-1}$, what is this referring to? If $a$ is the equivalence class, I cannot see (or I am not sure) how I can make inverse set. AI: To the first question, just quoting your intro, "$a,b\in(\mathbb{Z}/n\mathbb{Z})^{\times}$". So if $\mathbb{Z}/n\mathbb{Z}$ is a set of equivalence classes, then yes, $a$ and $b$ are equivalence classes. To the second question, $a^{-1}$ is an equivalence class such that $aa^{-1}$ is the equivalence class of $1$. In general, there is no quick formula to find an element of $a^{-1}$ given $a$. Instead, since a representative $A$ of $a$ and $n$ are relatively prime, the Euclidean algorithm provides solutions to the equation $$AB + nt = 1$$ Then mod $n$, $AB\equiv 1$. So the Euclidean algorithm will lead you to a representative of $a^{-1}$. Now, to back-peddle a little bit, actually there is a rather simple formula for a representative of $a^{-1}$, given a representative $A$ of $a$. Take $B=A^{\varphi(n)-1}$, where $\varphi$ is Euler's totient function. Then $AB=A^{\varphi(n)}\equiv1\mod{n}$. The problem with this "simple" formula is that it's not computationally efficient. For example, if $n=97$, the Euclidean algorithm quickly tells us that $2^{-1}=49$. But this formula would give us $2^{96}$, which is tiresome to compute even if we reduce at every multiplication.
H: Collatz-ish Olympiad Problem The following is an Olympiad Competition question, so I expect it to have a pretty solution: For a positive integer $d$, define the sequence: \begin{align} a_0 &= 1\\ a_n &= \begin{cases} \frac{a_{n-1}}{2}&\quad\text{if }a_{n-1}\text{ is even}, \\ a_{n-1}+d &\quad\text{if }a_{n-1}\text{ is odd.} \end{cases} \end{align} Find all values of $d$ such that $a_n=1$ for some $n>0$. It is obvious that $d$ must be odd, or else the sequence is monotone increasing. Also, I have numerically observed that all odd values of $d$ seem to work. Can anyone provide a hint as to how to even begin to prove this? Thank you! AI: Here is an outline of an argument. Our guess was that the sequence goes back to $1$ for any $d = 2m-1$. First, it seems preferable to work with the sequence $$b_0 = 1, \quad b_n = \begin{cases} \dfrac{b_{n-1}}2 & (b_{n-1}\text{ even}) \\ \frac{b_{n-1}+d}2 & (b_{n-1} \text{ odd})\end{cases}$$ instead of $a_n$ (the statement to be proven remains the same). Hint: Show that it is enough to show $b_n\equiv 1 \pmod d$ for some $n$. What does the sequence $b_n$ look like mod $d$? I have included a more detailed outline below (but I fear it is giving away way too much!) Claim 1: We have $0<b_n<d$ for all $n$. - This allows us to work mod $d$ from now on, i.e. we have $b_n = 1$ if and only if $b_n\equiv 1 \pmod d$. - Claim 2: $\frac12 \equiv m \pmod d$, where $d=2m-1$ as above. - Now notice that the sequence is really simple when considered $\pmod d$! Use this to show - Claim 3: $b_n \equiv m^n \pmod d$ - Claim 3 implies that $b_k \equiv 1 \pmod d$ for some $k$ (why?), so by Claim 1 it follows that $b_k=1$ for this $k$. $\square$
H: What is the difference between $d$ and $\partial$? After seeing the following equation in a lecture about tensor analysis, I became confused. $$ \frac{d\phi}{ds}=\frac{\partial \phi}{\partial x^m}\frac{dx^m}{ds} $$ What exactly is the difference between $d$ and $\partial$? AI: As mentioned $d$ means total and $\partial$ partial derivative and are not the same. Total derivative also counts $x$ dependencies in other variables. For instance: $$ f(x,v) = x^2 + v(x) \\ \frac{\partial f}{\partial x} = 2x \\ \frac{\partial f}{\partial v} = 1 \\ \frac{d f}{d x} = 2x + \frac{\partial v(x)}{\partial x} $$ Your formula most probably uses Einstein notation and is only a shorter way to write $$ \frac{d\phi}{ds}= \sum_m \frac{\partial \phi}{\partial x^m}\frac{dx^m}{ds} $$
H: Instant insanity question My question is regarding the necessary conditions that a graph must fulfill to satisfy instant insanity problem. Now take for example the left, right, front and back face colors of the four cubes be(the colors are red, black, green and white): Front Back Right Left R W R G G W W B G B G W B R B R Now as you can see on front and back, green and white show up twice respectively which violates the problem. But if we make their graph, their graphs fulfill all the conditions: 1.the graph can be dissociated into two edge disjoint subgraphs 2.each subgraph has vertices of degree 2 3.each subgraph has all the edges representing the four cubes once Note: I stack the cubes one above the other. This is different from the link i gave in which cubes are kept on the side of each other AI: I understand you have to play around with the cubes in order to get the answer. You just know what the fronts and backs; and lefts and rights of each cube are going to be. "To solve the game, the first graph represents the front and back faces, and the second graph represents the top and bottom faces. Align the cubes according to which edges you have in your two graphs. There is some choice on how to do this; for example, on the game above, cube 1 has the front and back faces blue and green, and the top and bottom faces red and yellow. However, we do not yet know whether blue is on the front or back. You may have to play around a little to arrange your cubes appropriately." I think all of the cubes are fine the way they are except for the second cube. If you play around with the second cube, then you'll see that if you if you flip it (front becomes back, and back becomes front, but right stays right and left stays left) then you have a valid answer that fits your graph. EDIT: I think I noticed something. If you think of each of the two graphs as a path, in which you can only walk in one direction, then fill in the answers that way. So you went from R to W, W to G, G to B, and B-R for the blue graph. See how that is one continuous path, and you can fill out your chart in that order. Same thing with the orange graph, even if lines 1 and 2 are not connected. Think of the path as going from red to green to white to black back to red. So you would fill out R to G, then W to B, then G to W, and then B to R in your chart. I don't know if this is actually true, but just a theory.
H: Are monomorphisms of rings injective? Let $R$ and $S$ be rings and $f:R\to S$ a monomorphism. Is $f$ injective? AI: Yes, $f$ is injective: Let's assume the map of the underlying sets is NOT injective. Then there are distinct $x,y\in R$ such that $f(x)=f(y)$. Next, consider the monomorphisms $g_1,g_2\colon \mathbf{Z}[X]\to R$ by $g_1(X)=x$ and $g_2(X)=y$. Note that these monomorphisms are different and $f\circ g_1= f\circ g_2$. This implies $f$ is NOT a monomorphism, which leads to a contradiction.
H: Sobolev spaces for functions mapping to $\mathbb R^n$ Hi I would like to know if there is a definition of Sobolev space as a set of mappings $f:U\subset \mathbb R^n \rightarrow \mathbb R^m$ (instead to $\mathbb R$). In particular what should be $H^{1,2}(U)_{\mathbb R^m}$. Does it make sense to define it as the set of maps $f=(f_1,...,f_m)$ such that $f_i \in H^{1,2}(U)$? What would be a natural norm? Would it be a Hilbert space as in the case where $f$ goes to $\mathbb R$? thank you AI: Yes, it's defined coordinate-wise. It's just the direct sum of $m$ copies of $H^{1,2}$. The direct sum of Hilbert spaces is automatically a Hilbert space: just add the inner products together. Usually denoted $H^1(U;\mathbb R^m)$ or $W^{1,2}(U;\mathbb R^m)$. The notation $H^{1,2}$ appears redundant to me: $H$ already indicates a Hilbert space, so $p=2$ is understood.
H: Proving a continuity inequality I am new to the mechanics of the $\epsilon$-$\delta$ definition of continuity of a function at a point, and so I am having trouble selecting my $\delta$ for the following question: Suppose a function $f$ is continuous at a point $c$ and $f(c) > 0$. Prove that there is a $\delta > 0$ such that for all $x \in \mathrm{domain}(f)$, $$ \|x - c\| \le \delta \ \Rightarrow \ f(x) \ge \frac{f(c)}{2} $$ I tried starting with the definition of continuity i.e. If $f$ is continuous at $c$ then letting $\epsilon > 0$ be given there exists a $\delta > 0$ such that $\|x -c\| \le \delta$ implies $\|f(x) - f(c)\| \le \epsilon$ However, I haven't found a way to work in the condition that $f(c) > 0$. Should I try a contrapositive proof? Edit: I see! Thank you. So this is saying that there is always a $\delta$ such that $f(x)$ is no more than $f(c)/2$ away from $f(c)$ and as the answerer pointed out, indeed any distance $z$ away from the limit. AI: Let $\epsilon = \frac{f(c)}{2}$. Since $f(c) > 0$, you have that $\epsilon > 0$. By continuity, there exists a $\delta$ such that for all $x$ such that $\|x - c\| < \delta$, you have that $\|f(x) - f(c)\| < \epsilon$. This implies that $f(c) - \epsilon < f(x) < f(c) + \epsilon$ we only care about the left side. Remembering that $\epsilon = \frac{f(c)}{2}$, you get $$f(c) - \frac{f(c)}{2} = \frac{f(c)}{2} < f(x).$$ for $\|x - c\| < \delta$, as desired.
H: How to find the maxium number of edge-disjoint paths using flow network Given a graph $G=(V,E)$ and $2$ vertices $s,t \in V$, how can I find the maximum number of edge-disjoint paths from $s$ to $t$ using a flow network? $2$ paths are edge disjoint if they don't have any common edge, though they may share some common vertices. Thank you. AI: Hint: if each edge has a capacity of one unit, different units of stuff flowing from $s$ to $t$ must go on edge-disjoint paths.
H: In the definition of a group, is stating the set together with the function on it redundant? In the definition of a group (or other similar structures) $$(G,)$$ as being the pair of a set $G$ together with binary operation $$:G\times G\rightarrow G$$ isn't the first component of the pair basically just redundant information? Is this done for pedagogical reasons? Or has is any logical advantages, like as to extract the set from the group by sucking out the first component. I don't know how to formally steal away the codomain from a function, althoght I think it also just involves some ordered pairs. AI: To be absolutely precise a group really is an order pair $(G, \cdot)$. (If you work in model theory, you may even want $(G,\cdot, e)$ to indicate the distinguished constant.) The point is that you want to precisely define what is a group. But you notice that saying a group is just a set $G$, or the group is just a function $\cdot : G \times G \rightarrow G$ is not correct. Groups are not functions. In fact saying that a group is a set $G$ together with a function $\cdot : G \times G \rightarrow G$ is not even correct. (This is something called a structure in the language of groups.) Group even have to satisfy certain axioms. So a group really is a pair $(G, \cdot)$ which satisfies all the group axioms. The above is a discussion about what a group really is. However, there is a distinction between the definition of a group and how much information you need so that most people would understand you in context. If you said "$\cdot : G \times G \rightarrow G$ is a group", I am sure most people would understand this to mean that $\cdot$ is the multiplication on a set $G$ which satisfy all the group properties.
H: Parameters giving maximal-length Collatz-like sequence In a recent question the following recursive sequence was considered: $$ a_{n+1} = \cases{\frac{a_n}{2} & $a_n$ is even \\ a_n +d & $a_n$ is odd}, \quad a_1 = d + 1 $$ where $d$ is an odd positive integer. Such a sequence will eventually reach $1$, and let $m(d)$ be the value of the index, where the sequence element equals one, $a_m = 1$. Here is a visualization of $m(d)$: It is apparent from the plot that $m(d) \leqslant 3 \left\lfloor \frac{d-1}{2} \right\rfloor$. Selecting those $d$ which saturate the bound: OEIS tells us that this is sequence A001122 of "primes with primitive root $2$". This begs for an explanation. References, intuitive insights, solutions are welcome. AI: Instead of the sequence $a_n$, let us look at the following related sequence: $$b_0 = 1, \qquad b_{k+1} = \begin{cases} \dfrac{b_{k}}2 & (b_{k}\text{ even}) \\[6pt]\dfrac{b_{k}+d}2 & (b_{k} \text{ odd})\end{cases}$$ Since $d$ is odd, we can write it in the form $d = 2n-1$ for some natural number $n\ge 1$. By a simple induction argument, we see that $0<b_k<d$ for all $n$. So that $$b_k = 1\quad \iff\quad b_k\equiv 1 \pmod d$$ Since $2n = d + 1 \equiv 1 \pmod d$, it follows that $\frac 12 \equiv n \pmod d$ and hence that the sequence $b_k$ satisfies $$ b_k \equiv n^k \pmod d$$ But then the smallest integer $k\ge 1$ such that $b_k=1$ is given by $k=\mathrm{ord}_{\mathbb Z_d^\times}(n)$, the order of $n$ in $\mathbb Z_d^\times$. Remember that $n \equiv 2^{-1}$, so $\mathrm{ord}_{\mathbb Z_d^\times}(n) = \mathrm{ord}_{\mathbb Z_d^\times}(2)$. How does this relate to the initial sequence $a_n$? The sequence $b_k$ meets each element of $\mathbb Z_d^\times$ at most once and therefore – for the sequence $a_k \pmod d$ – we must have that each odd element in $\mathbb Z_d^\times$ (when represented by $\mathbb Z_d^\times \subset \{1, \dots, p-1\}$) is visited exactly twice or never, each even element at most once. Hence we obtain the following formula for the length $m(d)$ of the sequence $a_k$, until reaching $1$: Formula for $m(d)$: \begin{align} m(d) &= 2\cdot (\# \text{ of odd elements in }\langle 2 \rangle\subset \mathbb Z_d^\times ) + (\# \text{ of even elements in }\langle 2 \rangle \subset\mathbb Z_d^\times ) \end{align} In particular, this shows that if $2$ is a primitive root of $\mathbb Z_d^\times$ for $d$ prime, then $$m(d) = 2 \frac{d-1}2 + \frac{d-1}2 = 3\left\lfloor \frac{d-1}{2}\right\rfloor$$ and if $2$ is not a primitive root, or $d$ is not a prime, then $m(d) < 3\left\lfloor \frac{d-1}{2}\right\rfloor$. $\quad \square$
H: Is a vector of coprime integers column of a regular matrix? Given a vector $a_1=(k_1,\ldots,k_n)^T$ of coprime integers. Are there $a_2,\ldots,a_n \in \mathbb{Z}^n$ such that that the matrix $A := (a_1,\ldots,a_n) \in \mathbb{Z}^{n \times n}$ is regular, i.e. $A \in GL_n(\mathbb{Z})$ ? In case $n=2$ this is true because there are integers $l_1,l_2$ s.t. $k_1l_1 + k_2l_2 = 1$. Then $a_2=(-l_2 , l_1)^T$ will do. AI: Solution from "Problems From The Book": Yes, by induction. Let $g= \gcd(k_1,\dots,k_{n-1})$, and do it for $k_1/g, \dots, k_{n-1}/g$, call this matrix $A'$. Add to $A'$ the following row: $(k_n /g , 0, \dots ,0)$. (almost all zeroes) $g$ must be coprime to $k_n$, so we can find integers $x,y$ such that $xg+yk_n = 1$. Now add it the following column: $(yk_1,yk_2, \cdots, y k_{n-1}, xg)^{T}$. (almost proportional to first column) It can be verified that the matrix has determinant $\pm 1$. To make the entries integral, multiply the first column by $g$ and divide the last column by $g$.
H: Showing a certain form is exact I'm trying to solve the following: Let $f: S^{2n - 1} \rightarrow S^n$ be a smooth map, and let $\omega$ be an n-form on $S^n$ such that $\int_{S^n} \omega = 1$. Show that $f^\omega$ is exact, and if $f^\omega = d\alpha$, then $\int_{S^{2n-1}} \alpha \wedge d\alpha$ is independent of the choice of $\omega$ and $\alpha$. I don't really know how to go about answering this. If I knew $f^\omega$ was closed, then it would be exact since $S^{2n - 1}$ has no nth cohomology. For the second part, if n was odd, then we would have $d(\alpha \wedge \alpha) = -2 \alpha \wedge d\alpha$, so $\alpha \wedge d\alpha$ is exact and hence 0 independent of $\omega$ and $\alpha$. AI: Note that $\omega$ is an $n$-form on $S^n$, an $n$-dimensional manifold. Therefore $d\omega = 0$ as it is a $(n + 1)$-form on $S^n$. So $\omega$ is closed. Since pullbacks commute with exterior differentiation, we have that $$d(f^\ast \omega) = f^\ast d\omega = 0,$$ so $f^\ast \omega$ is closed as well. Now apply what you said about $S^{2n-1}$ having trivial cohomology in degree $n$. Now to show that $$\int_{S^{2n-1}} \alpha \wedge d\alpha$$ is independent of the choice of $\omega$ and $\alpha$, let $\omega' \in \Omega^n(S^n)$ be another $n$-form on $S^n$ such that $$\int_{S^n} \omega' = 1$$ and let $\alpha' \in \Omega^{n-1}(S^{2n-1})$ be such that $d\alpha' = f^\ast \omega'$. We will show that $$\int_{S^{2n-1}} \alpha \wedge d\alpha = \int_{S^{2n-1}} \alpha' \wedge d\alpha',$$ which implies the integral is independent of the choices. Since $H^n(S^n; \mathbb{R}) \cong \mathbb{R}$, by deRham's theorem there is some $\tau \in \Omega^{n-1}(S^n)$ such that $$\omega' = \omega + d\tau.$$ Now $$d(\alpha' - \alpha - f^\ast \tau) = f^\ast(\omega' - \omega - d\tau) = 0$$ and $H^n(S^{2n-1}; \mathbb{R}) \cong 0$, so again by deRham's theorem there exists some $\eta \in \Omega^{n-2}(S^{2n-1})$ such that $$\alpha' = \alpha + f^\ast \tau + d\eta.$$ Hence we have that \begin{align} \alpha' \wedge d\alpha' & = (\alpha + f^\ast \tau + d\eta) \wedge (d\alpha + f^\ast d\tau) \\ & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + f^\ast(\tau \wedge (\omega + d\tau)) + d(\eta \wedge (d\alpha + f^\ast d\tau)) \\ & = \alpha \wedge d\alpha + \alpha \wedge d(f^\ast \tau) + d(\eta \wedge (d\alpha + f^\ast d\tau)), \end{align} where in going to the third line we used the fact that $f^\ast(\tau \wedge (\omega + d\tau)) = 0$ since $\tau \wedge (\omega + d\tau) = 0$ as it is a $(n + 1)$ form on $S^n$. Now, since \begin{align} \alpha \wedge d(f^\ast \tau) & = -d(\alpha \wedge f^\ast \tau) + d\alpha \wedge f^\ast \tau \\ & = -d(\alpha \wedge f^\ast \tau) + f^\ast (\omega \wedge \tau) \\ & = -d(\alpha \wedge f^\ast \tau), \end{align} where once again $f^\ast (\omega \wedge \tau) = 0$ since $\omega \wedge \tau$ is an $(n+1)$-form on $S^n$, we get that $$\alpha' \wedge d\alpha' = \alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau)).$$ So by Stokes' theorem, \begin{align} \int_{S^{2n-1}} \alpha' \wedge d\alpha' & = \int_{S^{2n-1}} (\alpha \wedge d\alpha + d(-\alpha \wedge f^\ast \tau + \eta \wedge (d\alpha + f^\ast d\tau))) \\ & = \int_{S^{2n-1}} \alpha \wedge d\alpha, \end{align*} showing that the integral is independent of the choices. Remark: The number $$H(f) = \int_{S^{2n-1}} \alpha \wedge d\alpha$$ is called the Hopf invariant of the map $f$. Here we showed that it only depends on $f$. You can also show that it only depends on the homotopy class of $f$, which isn't too difficult.
H: What kind of problem is picking sets of sets of things that don't overlap? I was working out some combinations for a game, when I hit upon a problem that seems unsolvable. I'm wondering if some existing mathematics (probably in graph theory or combinatorics) covers this particular class of problem. To express it without any game-specific terminology: I have a set of 6 numbers {1,2,3,4,5,6} from which I will create 3 number combinations. These combinations are always ordered and never contain duplicates, such as {123,134,145,156...}. Simple combinations tells me there are 20 such configurations. Now from those configurations, I wish to select sets of 4 each, such that no two numbers recur in any of the 4 triplets. An example would be {123,145,246,235}. My goal was to sort all 20 combinations into 5 sets of 4, but I don't think it's possible. I always end up with 2 pairs that repeat, either in 2 separate sets or both in one set, like {123,145,236,245} (23 and 45 occur twice). Note that being split still counts, so {234,245} is still a failure because 24 repeats. I figure this problem must be analogous to a node walking problem or something like that, but I don't know the math well enough to prove that no solution exists or to find similar problems by searching. AI: This sounds a lot like block design. This has actually found some very practical uses!
H: Examples of bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$ Could any one give an example of a bijective map from $\mathbb{R}^3\rightarrow \mathbb{R}$? Thank you. AI: First, note that it is enough to find a bijection $f:\Bbb R^2\to \Bbb R$, since then $g(x,y,z) = f(f(x,y),z)$ is automatically a bijection from $\Bbb R^3$ to $\Bbb R$. Next, note that since there is a bijection from $[0,1]\to\Bbb R$ (see appendix), it is enough to find a bijection from the unit square $[0,1]^2$ to the unit interval $[0,1]$. By constructions in the appendix, it does not really matter whether we consider $[0,1]$, $(0,1]$, or $(0,1)$, since there are easy bijections between all of these. Mapping the unit square to the unit interval There are a number of ways to proceed in finding a bijection from the unit square to the unit interval. Usully one starts with the "interleaving" technique I mentioned in the comments, mapping $$\langle 0.a_1a_2a_3\ldots,\\\qquad 0.b_1b_2b_3\ldots\rangle$$ to $$0.a_1b_1a_2b_2a_3b_3\ldots.$$ To see why this doesn't quite work, consider the numbers $\frac 12=0.50000\ldots$ and $\frac9{22} = 0.40909\ldots$. Both of these are the image of the pair $\langle\frac12, 0\rangle$, so we have not described a bijection. This problem can be fixed. (In answering this question, I tried many web searches to try to remember the fix, and I was amazed at how many sources I found that ignored the problem, either entirely, or by handwaving. I never did find it; I had to remember it. Sadly, I cannot remember where I saw it first.) First, we will deal with $(0,1]$ rather than with $[0,1]$; bijections between these two sets are well-known, or see the appendix. For real numbers with two decimal expansions, such as $\frac12$, we will agree to use only the expansion that ends with nines rather than the one that ends with zeroes. So for example we represent $\frac12$ as $0.4999\ldots$. Now instead of interleaving single digits, we will break each input number into chunks, where each chunk consists of some number of zeroes (possibly none) followed by a single non-zero digit. For example, $\frac1{200} = 0.00499\ldots$ is broken up as $$004\ 9\ 9\ 9\ldots,$$ and $0.01003430901111\ldots$ is broken up as $$01\ 003\ 4\ 3\ 09\ 01\ 1\ 1\ldots.$$ This chunking process is well-defined since we are ignoring representations that contain infinite sequences of zeroes. Now instead of interleaving digits, we interleave chunks. To interleave $0.004999\ldots$ and $0.01003430901111\ldots$, we get $$0.004\ 01\ 9\ 003\ 9\ 4\ 9\ldots.$$ This is obviously reversible. It can never produce a result that ends with an infinite sequence of zeroes, and similarly the reverse mapping can never produce a number with an infinite sequence of trailing zeroes, so we win. A problem example similar to the one from a few paragraphs ago is resolved as follows: $\frac12 = 0.4999\ldots$ is the unique image of $\langle 0.4999\ldots, 0.999\ldots\rangle$ and $\frac9{22} = 0.40909\ldots$ is the unique image of $\langle 0.40909\ldots, 0.0909\ldots\rangle$. This is enough to answer the question posted, but I will give some alternative approaches. Continued fractions According to the paper "Was Cantor Surprised?" by Fernando Q. Gouveâ, Cantor originally tried interleaving the digits himself, but Dedekind pointed out the problem of nonunique decimal representations. Cantor then switched to an argument like the one Robert Israel gave in his answer, based on continued fraction representations of irrational numbers. He first constructed a bijection from $(0,1)$ to its irrational subset (see this question for the mapping Cantor used and other mappings that work), and then from pairs of irrational numbers to a single irrational number by interleaving the terms of the infinite continued fractions. Since Cantor dealt with numbers in $(0,1)$, he could guarantee that every irrational number had an infinite continued fraction representation of the form $$x = x_0 + \dfrac{1}{x_1 + \dfrac{1}{x_2 + \ldots}}$$ where $x_0$ was zero, avoiding the special-case handling for $x_0$ in Robert Israel's solution. Cantor-Schröder-Bernstein mappings The Cantor-Schröder-Bernstein theorem takes an injection $f:A\to B$ and an injection $g:B\to A$, and constructs a bijection between $A$ and $B$. So if we can find an injection $f:[0,1)^2\to[0,1)$ and an injection $g:[0,1)\to[0,1)^2$, we can invoke the CSB theorem and we will be done. $g$ is quite trivial; $x\mapsto \langle x, 0\rangle$ is one of many obvious injections. For $f$ we can use the interleaving-digits trick again, and we don't have to be so careful because we need only an injection, not a bijection. We can choose the representation of the input numbers arbitrarily; say we will take the $0.5000\ldots$ representation rather than the $0.4999\ldots$ representation. Then we interleave the digits of the two input numbers. There is no way for the result to end with an infinite sequence of nines, so we are guaranteed an injection. Then we apply CSB to $f$ and $g$ and we are done. Appendix There is a bijection from $(-\infty, \infty)$ to $(0, \infty)$. The map $x\mapsto e^x$ is an example. There is a bijection from $(0, \infty)$ to $(0, 1)$. The map $x\mapsto \frac2\pi\tan^{-1} x$ is an example, as is $x\mapsto{x\over x+1}$. There is a bijection from $[0,1]$ to $(0,1]$. Have $0\mapsto \frac12, \frac12\mapsto\frac23,\frac23\mapsto\frac34,$ and so on. That takes care of $\left\{0, \frac12, \frac23, \frac34,\ldots\right\}$. For any other $x$, just map $x\mapsto x$. Similarly, there is a bijection from $(0,1]$ to $(0,1)$.
H: Limit of series $1 - 1/9 - 1/15- 1/21-\cdots$ In looking at a paper online I came across the following proposition: $$1 - \frac{1}{9} - \frac{1}{15}- \frac{1}{21}-\cdots = 0$$ After wasting a lot of time, I rewrote it, $$1 -\left(\frac{1}{9} + \frac{1}{15}+\cdots\right)= 1 - \left(\frac{1}{3}\sum_{k = 2}^\infty\frac{1}{p_k}\right) \rightarrow -\infty. $$ So it seems to me that the r.h.s. diverges. Is this correct? Edit: in case someone doesn't see Sasha's question and my response here are more terms in the sequence. The denominators are $3 p_k, p_k$ being the $k$th prime. $$1- 1/9 - 1/15 - 1/21 - 1/33 - 1/39 - 1/51-\cdots$$ AI: Assuming that the 9, 15, and 21,... stand for $3p_k$, then yes, I would say that this series diverges to $-\infty$. http://en.wikipedia.org/wiki/Divergence_of_the_sum_of_the_reciprocals_of_the_primes
H: Characters of affine algebraic groups and the determinant Let $G$ be an affine algebraic group (i.e. a $k$-variety which is also a group and the group multiplication and inversion are morphisms of varieties). A character of $G$ is a morphism of algebraic groups $\chi:G\to k^\times$. You may assume for the following that $k^\times$ acts algebraically on $G$, but I do not know if that is required at all. It is well-known that I can always embed $G$ as a closed, algebraic subgroup of some $\mathrm{Gl}_n$. Now given any (nonzero) character $\chi$, can I embed $G$ in such a way that $\chi=\det_n\|_G$, the restriction to $G$ of the determinant on $\mathrm{Gl}_n$? If no, can you characterize the characters that do satisfy this property? Because certainly, some characters of $G$ arise in this manner. AI: Let $G$ be a closed subgroup of $\newcommand{\GL}{\operatorname{GL}}\GL_n$ and let $\chi : G \to k^\times$ be a character. Then consider the subgroup $$G_\chi = \left\{ ~ \begin{bmatrix} g & . & . \\ . & \det(g)^{-1} & . \\ . & . & \chi(g) \end{bmatrix} ~:~ g \in G ~\right\} \leq \GL_{n+2}$$ By definition of character, this is a closed subgroup of $\GL_{n+2}$ and clearly the determinant of the “$g$” matrix is now $\chi(g)$. $G_\chi \cong G$ by definition of homomorphism. Hence every (linear) character is the determinant of a faithful representation.
H: Solving a recurrence of polynomials I am wondering how to solve a recurrence of this type $$p_1(x) = x$$ $$p_2(x) = 1-x^2$$ and $$p_{n+2}(x) = -xp_{n+1}(x)+p_{n}(x).$$ I am wondering, how could one solve such a recurrence. One way would be to pretend $x$ is fixed and solve it using the well known method for linear recurrences. My problem with this is that it gets rather messy and besides when solving for the initial terms one gets a fraction that is not well defined for all $x$. I would therefore like to ask if there is an easier way to solve it or perhaps whats the proper way to apply the theory of linear recurrences. Edit. The recurrence is indeed of second order. AI: related problem: (I), (II), (III). Here is the solution of the corrected recurrence relation computed by maple $$ \left( -3\,x-\sqrt {{x}^{2}+4} \right)\left( -2\, \left( -x-\sqrt{{x}^{2}+4} \right)^{-1} \right) ^{n}{\frac {1}{\sqrt {{x}^{2}+4}}}\left( -x-\sqrt {{x}^{2}+4} \right)^{-1} $$ $$+ \left( 3\,x-\sqrt {{x}^{2}+4} \right)\left( -2\,\left( -x+\sqrt {{x}^{2}+4} \right)^{1} \right) ^{n}{\frac {1}{\sqrt {{x}^{2}+4}}} \left( -x+\sqrt {{x}^{2}+4}\right)^{-1}$$ If you are interested in how you can solve it. Assume your solution $p_n = r^n\,, $ for some $r$ to be determined and substitute in your recurrence relation which gives $$ r^{n+2} + x r^{n+1} - r^n = 0 \Rightarrow r^n(r^2+xr-1)=0 \Rightarrow (r^2+xr-1)=0$$ $$ r_1 = -\frac{1}{2}\,x+\frac{1}{2}\sqrt {{x}^{2}+4}\,, \,\,\, \, r_2 = -\frac{1}{2}\,x - \frac{1}{2}\,\sqrt {{x}^{2}+4} $$ Now, you construct the general solution as $$ p_n(x) = c_1 {r_{1}}^n + c_2 {r_{2}}^n = c_1 {\left( -\frac{1}{2}\,x+\frac{1}{2}\sqrt {{x}^{2}+4} \right)}^n + {\left( -\frac{1}{2}\,x - \frac{1}{2}\,\sqrt {{x}^{2}+4} \right)}^n $$ To determine the constants $c_1$ and $c_2$, you just need to exploit the initial conditions $p_0(x)$ and $p_1(x)$ and solve for $c_1$ and $c_2$.
H: A Countable Set of Real Numbers Let $S$ be a subset of $\mathbb{R}$. Let $C$ be the set of points $x$ in $\mathbb{R}$ with the property that $S \cap (x - r, x + r)$ is uncountable for every $r > 0$. Show that $S - C$ is finite or countable. Thanks for any help. AI: Instead of considering arbitrary neighborhood $(x - r, x + r)$ for $x \in \mathbb{R}$ and $r > 0$, you can consider just those open intervals where $x \in \mathbb{Q}$ and $r \in \mathbb{Q}$. These form a countable basis for the topology on $\mathbb{R}$. Let $(U_n)_{n \in \mathbb{N}}$ denote a countable enumeration of these open intervals. Then you have that $C$ is the set of all $x \in \mathbb{R}$ such that $S \cap U_n$ is uncountable for all $n$ such that $x \in U_n$. Hence $S - C$ is the union of over all $n \in \mathbb{N}$ of $S \cap U_n$ such that $S \cap U_n$ is countable, i.e. $S - C = \bigcup_{n \text{ st } \|S \cap U_n\| = \aleph_0} S \cap U_n$ A countable union of countable sets is countable. You are done. By the way, the points in $C$ are usually called condensation points. If $S$ happens to be closed, the result above is a step in proving the Cantor Bendixson Theorem. This theorem states that every closed set is the union of a perfect set and a countable set. Perfects sets have cardinality $2^{\aleph_0}$. Hence the Cantor Bendixson Theorem states that no closed subset of $\mathbb{R}$ is a counterexample for the continuum hypothesis.
H: If $a + b \equiv 0 \pmod p$, and $a + b \equiv 0 \pmod q$, why does $a + b \equiv 0 \pmod {pq}$? Put a bit more coherently, given $p$ and $q$ as distinct prime numbers, and thus $(p,q)=1$, if $$p^{(q-1)} + q^{(p-1)} \equiv 1 \pmod p$$ and $$p^{(q-1)} + q^{(p-1)} \equiv 1 \pmod q,$$ why does that lead to $$p^{(q-1)} + q^{(p-1)} \equiv 1 \pmod {pq}?$$ The textbook I'm working with jumps to that conclusion as if it were obvious, but it's not. Not to me, at least. AI: Your question should be rephrased as: Let $p$ and $q$ be distinct prime numbers. If $p \| x$ and $q \| x$, is it true that $pq \| x$? The fundamental theorem of arithmetic should answer this immediately.
H: How to show that this is an Algebraic System Define $\oplus$ on $\mathbb{R} \times \mathbb{R}$ by setting $$(a,b) \oplus (c,d)=(ac-bd, ad+bc).$$ How to show that $(\mathbb{R} \times \mathbb{R}, \oplus)$ is an algebraic system. I don't understand the difference between algebraic structure andalgebraic system I read this article but I didnt understand it properly. Can anybody help me me understand by giving hints? AI: $\oplus$ on $\mathbb{R} \times \mathbb{R}$ is a just multiplication of the complex numbers on $\mathbb{C}$. It is not a group since $(0,0) \in \mathbb{R} \times \mathbb{R}$ does not have a inverse under $\oplus$. However $\mathbb{R} \times \mathbb{R} - \{0\}$ is a group under $\oplus$. $(\mathbb{R} \times \mathbb{R}, \oplus)$ however is a commutative monoid. It is associative, has an identity element, but not every element has an inverse.
H: (ZF)subsequence convergent to a limit point of a sequence Arthur's answer; (ZF) Prove 'the set of all subsequential limits of a sequence in a metric space is closed. Let $\{p_n\}$ be a sequence in a metric space $X$. Let $B=\{p_n\|n\in\mathbb{N}\}$ and $p$ be a limit point of $B$. It first seemed obvious that there exists a subsequence convergent to $p$, but i realized that I can construct a 'infinite subset' of $\{p_n\}$ and form a new sequence convergent to $p$, but can't construct a 'subsequence'. Help me how to construct a subsequence. Additional Question; What is the precise definition of convergent? I think the definition should mention ordering of $\mathbb{N}$, but every definition i saw doesn't mention this. For example, let $p_n = 1/n$. Say $G$ is the usual ordering of $\mathbb{N}$ (i.e. Well-ordered by $\in$) If i follow this ordering, $\{p_n\}$ is convergent to 0. But if i follow $G^{-1}$, it's convergent to 1. Or is it defined by the point where is the most, unofficially speaking, 'dense'? AI: Let $B = \{p_n : n \in\Bbb N\}$. Let $p$ be a limit point of $p$. Define by recursion (possible in ZF): Suppose you have $p_{n_1}, p_{n_2}, ..., p_{n_k}$ such that $n_1 < n_2 < ... < n_{k}$ and $d(p, n_j) < \frac{1}{j}$ for all $1 \leq j \leq k$, then find the least $l > n_k$ such that $d(p, p_l) < \frac{1}{k + 1}$. Let $l = n_{k + 1}$. This does not require any choice. You just start computing $d(p, p_l)$ for $l > n_k$. By assuming $p$ is a limit point of $B$, you sure to find an $l$. Then $(p_{n_k})_{k \in\Bbb N}$ is a subsequence converging to $p$. The definition of convergence does use the ordering of $\mathbb{N}$. The definition of $(p_n)_{n \in \mathbb{N}}$ converging to $p$ is : For all $\epsilon > 0$, there exists a $N \in \mathbb{N}$ such that for all $n > N$, $d(p_n, p) < \epsilon$. The part "$n > N$" refers to the usual well-ordering of ordering of $\mathbb{N}$.
H: Determine the nature of a critical point (Lagrange multipliers) Let $F \colon \mathbb R^2 \to \mathbb R$ be the function $$ F(x,y):=xye^x + ye^y - e^x+1 $$ and denote with $C$ the set of zeroes of $F$, i.e. $C:=\{(x,y) \in \mathbb R^2 : F(x,y)=0\}$. Let also $f \colon \mathbb R^2 \to \mathbb R$ be a function which is $C^2$ in a neighbourhood of $(0,0)$ and such that $$ \nabla f(0,0) = (-2,2), \qquad \qquad H_f(0,0) = \begin{pmatrix} 2 & 1 \\ 1 & 0 \end{pmatrix}. $$ Question: is $(0,0)$ a minimum for $f$ on $C$? Well, I show you what I have done so far. First of all, some routine calculations yield $$ \nabla F(0,0) = (-1,1) \qquad \qquad H_F(0,0) = \begin{pmatrix} -1 & 1 \\ 1 & 2 \end{pmatrix} $$ Indeed, the fact that $$ \nabla f(0,0) = 2\nabla F(0,0) $$ does agree with the theory of Lagrange multipliers: the gradients are parellel, so I think that $(0,0)$ is an extremum for $f$ on $C$. The problem is how to classify it without any information on $f$: we have only its hessian matrix, which is - I suppose - the key to solve this. Both $H_f(0,0)$ and $H_F(0,0)$ are indefinite. How can we establish the nature of the critical point $(0,0)$? Thanks in advance for your help. AI: Suppose you have a parametrization $x=X(t), y=Y(t)$ of the curve $C$ near $(0,0)$, with $X(0)=0$ and $Y(0)=0$. Thus if $V$ and $A$ are the corresponding velocity and acceleration vectors, $$\dfrac{d}{dt} f(X(t),Y(t)) = X' f_x(X,Y) + Y' f_y(X,Y) = (\nabla f)\cdot V$$ $$\dfrac{d^2}{dt^2} f(X(t),Y(t)) = X'' f_x(X,Y) + Y'' f_y(X,Y) + (X')^2 f_{xx}(X,Y) + 2 X' Y' f_{xy}(X,Y) + (Y')^2 f_{yy}(X,Y) = (\nabla f)\cdot A + V^T H(f) V $$ Similarly for $f$ replaced by $F$, but in this case $\dfrac{d}{dt} F(X(t),Y(t)) = \dfrac{d^2}{dt^2} F(X(t),Y(t)) = 0$. Now since $\nabla f(0,0) = 2 \nabla F(0,0)$, $(\nabla F(0,0)) \cdot V(0) = 0$ implies $\dfrac{d}{dt} f(X(t),Y(t))\|_{t=0} = (\nabla f(0,0)) \cdot V(0) = 0$. And $$\dfrac{d^2}{dt^2} f(X(t),Y(t))\|_{t=0} = \dfrac{d^2}{dt^2} f(X(t),Y(t))\|_{t=0} - 2 \dfrac{d^2}{dt^2} F(X(t),Y(t))\|_{t=0} = V(0)^T (H(f) - 2 H(F)) V(0)$$ Now $V(0)$ is orthogonal to $\nabla F(0,0) = (-1,1)$, so it is a multiple of $(1,1)$, and $$ (1,1) \left( \pmatrix{2 & 1\cr 1 & 0\cr} - 2 \pmatrix{-1 & 1\cr 1 & 2\cr}\right) \pmatrix{1\cr 1\cr} = -2$$ so the second derivative is negative, and you have a local maximum.
H: Topology of Branch Cuts and Elliptic Integrals In reading these notes (elliptic curves starting from elliptic integrals) I came across a couple claims about the topology of some complex surfaces. On page 4, they discuss the integral $$\phi(x) = \int_0 ^x \frac{dt}{\sqrt{1 - t^2}}$$ In order to define it on all of $\mathbb C$, you have to use a branch cut; they glue two copies of $\mathbb C$ together along $[-1,1]$, in the same crossing-over manner as you do when dealing with $\sqrt z$ (at least, I think). They then claim that this surface $C$ is homeomorphic to a cylinder. However, I'm having trouble seeing a way to explicitly bend $C$ into a cylinder. I think I might be missing some intuition on what a complex cylinder looks like. I think I understand why $C$ would be homotopically equivalent (not sure if that's the best term) to a cylinder, because there's one set of loops from going around the branch cut, and if you avoid integrating across the branch cut the other ones are all null-homotopic. But why glue two copies of $\mathbb C$ together at all if you're going to avoid integration across the branch cut? I also don't quite get they say that $C$ can also be defined as $\{ (x,y) \in \mathbb C ^2 : x^2 + y^2 = 1 \}$; is it just that we can integrate $dx/y$ on $C$, because that differential on $C$ looks like the differential in $\phi$? They make similar claims a page later about $$ \psi (x) = \int_0^x \frac{dt}{\sqrt{t(t-1)(t-\lambda)}}$$ but I think my issues are basically the same. AI: Here is an explanation of what is going on; I haven't comparted it directly with the notes that you link to, but they will surely say something essentially equivalent: The differential $dt/\sqrt{1-t^2}$ is not unambiguously defined on the $t$-plane, because the denominator $\sqrt{1-t^2}$ is not. If we make a cut along $[-1,1]$ then we obtain a region homeomorphic to a cylinder (as you can see by looking at it --- but if this is your point of confusion, feel free to say so in the comments), on which we can define two branches of $dt/\sqrt{1-t^2}$. If we want to "glue" these two branches together into a single differential on a single Riemann surface, then we have to consider the Riemann surface $C:= \{(x,y) \, \| \, y^2 = 1 - x^2\}$; on this surface, we identify $x$ with $t$, so that $y$ is then one the of the two choices of $\sqrt{1-t^2}$. The differential can then be written as $dx/y$. If we define the map $\pi: C \to \mathbb C$ via $(x,y) \mapsto x$ then the restriction of $\pi$ to the preimage of $\mathbb C \setminus [-1,1]$ is $2$-to-$1$, and indeed this preimage is the disjoint union of two copies of $\mathbb C\setminus [-1,1]$; this is just a consequence of the fact that over the cut plane, we can choose $\sqrt{1-t^2}$ in two distinct well-defined ways. The entire curve $C$ is however connected: it consists of the two copies of $\mathbb C \setminus [-1,1]$ that lie over $\mathbb C\setminus [-1,1]$, glued together along a circle (the preimage under $\pi$ of $[-1,1]$). Gluing two cylinders along a common boundary circle just gives another cylinder, and so $C$ is indeed homeomorphic to a cylinder. Another way to think of it is as being homeomorphic to a sphere with two points removed; this is topologically the same as a cylinder.
H: Showing truncation is a ring homomorphism Let $R$ be a commutative ring with $1$ and let $n$ be any positive integer. Denote the ring of formal power series by $R[[x]]$ and define a map as follows: $f: R[[x]] \rightarrow R[x]/(x^{n})$ by sending $g$ to $g+(x^{n})$. It is clear this map preserves sums, but why preserves products? i.e a ring homomorphism. AI: Let $I$ be the ideal of $R[[x]]$ generated by $x^n$. One can easily see that $I$ is the set of all series in $R[[x]]$ whose first non-zero coefficient is that of $x^m$ with $m\geq n$. There is a canonical map $q:R[[x]]\to R[[x]]/I$, mapping each series to its coset module $I$, and it is a ring homomorphism. Now, the inclusion gives us a ring morphism $\phi:R[x]\to R[[x]]$. If $J\subseteq R[x]$ is the ideal of $R[x]$ generated by $x^n$, then clearly $\phi(J)\subseteq I$, so $\phi$ induces a well-defined ring morphism $\bar\phi:R[x]/J\to R[[x]]/I$. Let us check that $\bar\phi$ is an isomorphism. Its kernel is $\phi^{-1}(I)/J$, but this is zero because $\phi^{-1}(I)$ is actually equal to $J$. On the other hand, $\bar\phi$ is surjective: if $g\in R[[x]]$ is any series, there exist $a_0$, $\dots$, $a_{n-1}\in R$ and $h\in R[[x]]$ such that $g=a_0+a_1x+\cdots+a_{n-1}x^{n-1}+x^nh$, so that $g\cong a_0+a_1x+\cdots+a_{n-1}x^{n-1}\mod I$ and therefore $$\bar\phi(a_0+a_1x+\cdots+a_{n-1}x^{n-1}+J)=g.$$ This does all you want, because what you are calling truncation is the composition $\bar\phi^{-1}\circ q$.
H: Trying to prove that $\lim_{n\rightarrow\infty}(\frac{\Gamma '(n+1)}{n!} -\log(n))=0$ In my attempt to prove that $\Gamma'(1)=-\gamma$, I've reduced the problem to proving that $\lim_{n\rightarrow\infty}(\frac{\Gamma '(n+1)}{n!} -\log(n))=0$. Where $\gamma$ is the Euler-Mascheroni constant, and $\log$ denotes the natural logarithm. I've been messing with it for a while without achieving much of anything. The first derivative of the Gamma function does have a recursive formula which can be found through iterated integration by parts, but that was what I used to get where I am, and applying it again just takes me back to where I started. My book lists a ton of equivalent definitions for the Gamma function, but only gives the integral definition for its derivatives and I've just had considerable trouble doing much with that integral. I should note that at first I was trying to prove that $\frac{\Gamma'(n+1)}{n!}\sim\log(n)$, but now I'm pretty sure that showing that their difference in the limit is zero would be sufficient, since what I'm ultimately interested in is showing that $\lim_{n\rightarrow\infty}(-\sum_{k=0}^n\frac{1}{k} +\frac{1}{n!}\Gamma'(n+1)) =-\gamma$ Hopefully someone can help me with this. Thanks. AI: We have $\log(\Gamma(n+2))-\log(\Gamma(n+1))=\log(n+1)$, so by the mean value theorem, $\frac{\Gamma'}{\Gamma}(s)=\log(n+1)$ for some $s\in[n+1,n+2]$. Now, $\frac{\Gamma'}{\Gamma}$ is increasing, so repeating the argument on the interval $[n,n+1]$, we get $\log(n)\leq\frac{\Gamma'}{\Gamma}(n+1)\leq \log(n+1)$. The result now follows, as $\lim_{n\to\infty}\log(n+1)-\log(n)=0$.
H: Questions about $su(2)$. Edit: In physics, it seems that people usually study $su(2)$ but not only $sl(2)$? Why people study $su(2)$ but not only $sl(2)$? AI: The form of the question is somewhat iffy, but may be intended to ask a question that has a sense that I understand: since most representations are on complex vector spaces, a representation of a Lie algebra is indistinguishable from a representation of its complexification. The complexifications of su(2) and sl(2,R) are both sl(2,C)... often unhelpfully written merely as sl(2), not acknowledging the real-versus-complex game. Perhaps another iteration of the question would make clearer what other/remaining issues might be... Edit: The Lie groups $SU(2)$ and $SL_2(\mathbb R)$ are easier to distinguish: the former is compact and the latter is not, which has as consequence that the irreducible unitary repns of the former are all finite-dimensional (indexed by highest weight), while the irreducible unitaries of $SL_2(\mathbb R)$ are all infinite-dimensional (apart from the trivial repn). Appreciating this point about the Lie groups is probably necessary in order to appreciate distinctions about the Lie algebras which otherwise might seem capricious.
H: Does there exist $m,a\in \mathbb{N}$ such that $m(m+1)=a^7$? Let $m$ be a natural number. Can $m(m+1)$ be written as a seventh power of a natural number? If it is true, is it possible to generalize? I can use only Euclidean Division. I am not suppose to use the Fundamental theorem of arithmetic. I think the answer is no, then I have assumed that $m(m+1)=a^{7}$ and I would like to get a contradiction. I know that $m(m+1)=2(1+2+\cdots+m)$, but I got stucked here. I have tried to use the Euclidean Division and write $m=aq+r$, for $0\leq r <a$, and this doesn't help me as well. I would appreciate your help. AI: Step 1 At first notice that $m$ and $m+1$ are co-prime. Step 2 use Fundamental theorem of arithmetic to prove $$m=k^7$$ and $$m+1=l^7$$ Then substract $$1=l^7-k^7>1$$ for nontrivial $l,k$
H: Show the existence of a polynomial $f$ such that $T^{-1} = f(T)$ for an invertible linear operator $T.$ Came across this question and have had difficulty approaching it so any help is greatly appreciated. Let $T$ be an invertible linear operator on a finite dimensional vector space $V$ over a field $F$. Prove that there exists a polynomial $f$ over $F$ such that $T^{-1} = f(T)$. AI: Let $p(x)$ be the characteristic polynomial of $T$. Since $T$ is invertible, $0$ is not an eigenvalue of $T$ so $p(x)$ has a non-zero constant term. Write $p(x) = x^n + a_{n-1} x^{n-1} + \ldots + a_0$. But any linear transformation satisfies its characteristic polynomial, i.e. $p(T) = 0$. Thus $$ T^n + a_{n-1} T^{n-1} + \cdots + a_1 T + a_0 = 0 $$ which gives $$ T(\frac{-1}{a_0}(T^{n-1} + a_{n-1}T^{n-2} + \cdots + a_1)) = Id $$ so $T^{-1} = \frac{-1}{a_0}(T^{n-1} + a_{n-1}T^{n-2} + \cdots + a_1)$.
H: 3D graphing in matlab I'm having trouble graphing using Matlab, and was hoping someone could help. I just want to hand the program a function like $z=x^2-3xy+2$, or whatever, and have Matlab generate a 3D graph. From the tutorials I've seen online, it seems like this is difficult. Can I not just hand the program that equation and the domain that I want, and have it generate a graph? What code will generate that graph? Thanks! AI: As an example, lets plot the function in the interval $x=[-2;2], y=[-2;2]$. As mentioned in the comments, this can be achieved in Matlab by the code [X,Y] = meshgrid(-2:.2:2, -2:.2:2); Z = X.^2 - 3X.Y + 2; surf(X,Y,Z) It can also be done in Maple by the following code plot3d(x^2 - 3xy, x=-2..2, y=-2..2); which is probably more intuitive. At last, one can use Wolfram Alpha.
H: What is the importance of integral domains? What is the importance of integral domains? In abstract algebra Thanks for your help AI: Definition: An Integral domain is a commutative ring( with unity) $(R,+,\cdot)$ for which the following property holds If $a,b \in R$ such that $a,b \neq 0$, then $ab\neq 0$. To see some motivation behind the definition of an Integral domain, notice that when we were little kids, while solving polynomial equations in $\mathbb Z$, after factoring a polynomial, we make the jump from $$x(x-2)(x-3)=0$$ $$\text{to}$$ $$x=0 \text{ or } x=2 \text{ or } x=3$$ This is a convenient property to have, since it almost always makes it easy to find ALL solutions of a polynomial equation.And this property holds for all elements if and only if the ring is an integral domain. This is quite easy to verify. Another motivation is the following question: Given a ring $R$, when does there exist a field $F$ and an injective ring homomorphism $\phi:R \to F$? Informally, we are asking when a given ring $R$ can be described as a subring of a field. The answer turns out that this is possible if and only if $R$ is an integral domain. One direction of this is easy. Suppose that $R$ can be embedded into a field as above. For any two elements $a,b\neq 0 \in R$, if $ab=0$, then $\phi(ab)=0$ too. Since $\phi$ is a homomorphism, this means that $\phi(a)\phi(b)=0$, Since $\phi$ is injective, neither $\phi(a)$ or $\phi(b)$(Both elements of $F$) are zero. Since $F$ is a field, this means they must be invertible. Multiplying on the left by $(\phi(a))^{-1}$, we see that $\phi(b)=0$, a contradiction. So, $ab \neq 0$. So, $R$ is an integral domain. For the other direction, see Field of Fractions. This is a construction which not only constructs a field containing our Integral Domain, but also constructs the smallest such one. For example, the Field of Fractions of $\mathbb Z$ is $\mathbb Q$, the field of Rational numbers. This might clarify which the construction is called the field of fractions, since $\mathbb Q$ is what we usual think of as fractions. Thus, the notion of an abstract Integral domain is closely modeled after the Integers.
H: Can you construct a field with 6 elements? Possible Duplicate: Is there anything like GF(6)? Could someone tell me if you can build a field with 6 elements. AI: $\def\x{\otimes}$There is not. Suppose $\langle F, +, \x\rangle$ is a field where $F$ has six elements. Then $\langle F, +\rangle$ is an abelian group; it must be $Z_6$, which is the only abelian group with six elements. So take $F=\{0,1,2,3,4,5\}$ and $+$ to be addition modulo 6. By Lagrange's theorem, every element of $\langle F, +\rangle$ has an order that divides 6, so any element $f$ of this group has the property that $f+f+f+f+f+f = 0$. Now we consider multiplication. We don't know yet what $1\x1$ is—it might not be $1$—so let's call it $i$, and consider $2\x 3$: $$\begin{eqnarray} 2\x 3 & = & (1+1)\x(1+1+1) \\ & = & 1\x 1 +1\x 1 +1\x 1 +1\x 1 +1\x 1 +1\x 1 \\ & = & i+i+i+i+i+i\\ & = & 0 \end{eqnarray}$$ But this cannot happen in a field: $ab=0$ implies $a=0$ or $b=0$, and that fails here. So there is no way to define $\x$ to make $\langle F, +, \x\rangle$ into a field.

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