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H: What is the name of the logical puzzle, where one always lies and another always tells the truth? So i was solving exercises in propositional logic lately and stumbled upon a puzzle, that goes like this: Each inhabitant of a remote village always tells the truth or always lies. A villager will only give a "Yes" or a "No" response to a question a tourist asks. Suppose you are a tourist visiting this area and come to a fork in the road. One branch leads to the ruins you want to visit; the other branch leads deep into the jungle. A villager is standing at the fork in the road. What one question can you ask the villager to determine which branch to take? I intuitively guessed the answer is "If I asked you this path leads to the ruins, would you say yes?". So my questions are: What is the name and/or source of this logical riddle? How can i corroborate my answer with mathematical rigor? AI: The specific problem is shown in My Best Mathematical and Logic Puzzles, by Martin Gardner. It was in his Scientific American column long ago. The solution is on page 40 with a reference in 1957. The problem itself is the fourth one and is on the page 2. The full definition of the problem is, as published on This Side of the Pond: The Fork in the Road Here’s a recent twist on an old type of logic puzzle. A logician vacationing in the South Seas finds himself on an island inhabited by the two proverbial tribes of liars and truth-tellers. Members of one tribe always tell the truth; members of the other always lie. He comes to a fork in a road and has to ask a native bystander which branch he should take to reach a village. He has no way of telling whether the native is a truth-teller or liar. The logician thinks a moment and then asks one question only. From the reply, he knows which road to take. What question does he ask? Added: the point is that you get only one bit of information. You need that bit to tell you which road leads to the village, so you can't learn whether the native tells the truth or lies. The key to many of these puzzles is finding a way to get the information you need without getting anything more. In this case (and in many truth-teller/liar puzzles) the secret is arranging two negations so you get the answer you need.
H: How can one express $\sqrt{2+\sqrt{2}}$ without using the square root of a square root? I was trying to review some analysis, and came across problem 3 from page 78 of Walter Rudin's Principles of Mathematical Analysis. As part of the problem, I wanted to try to write $\sqrt{2+\sqrt{2}}$ without using the square root of a square root. In other words, I wanted to express the number in the form $q_1+q_2{q_3}^s$, where $q_1, q_2, q_3, s \in \mathbb{Q}$ (or perhaps something similar). I'm aware that this is probably a duplicate of another question, but I wasn't able to find it (I wasn't sure what to search for)... Many thanks in advance! Edit (my work so far): I tried expressing it as the solution to the quartic $x^4-4x^2+2=0$, but this seemed futile... Edit 2 (the original problem): The original problem from the text states: If $s_1=\sqrt{2}$, and $$S_{n+1}=\sqrt{2+\sqrt{s_n}} (n=1,2,3,\ldots),$$ prove that $\{s_n\}$ converges, and that $s_n<2$ for $n=1,2,3,\ldots$. The problem is from the 3rd chapter of the book, which talks about sequences and series. Rudin provides numerous theorems on this topic, such as the comparison, ratio, and root tests for convergence. AI: Short answer: you can't. Call your element $\alpha$. You want $\alpha$ to live in a field extension of the form $\mathbb{Q}(\beta^{1/4})$ for some $\beta \in \mathbb{Q}$. It's clear that $L = \mathbb{Q}(\alpha)$ is a degree 4 field extension of $\mathbb{Q}$ (you've written down a minimal polynomial). One can check easily that $L$ is Galois over $\mathbb{Q}$ (i.e. you can check that $\sqrt{2 - \sqrt{2}} \in L$). But $\mathbb{Q}(\beta^{1/4})$ (for $\beta$ squarefree) is not.
H: Map bounded if composition is bounded Let $X,Y,Z$ Banach spaces and $A:X\rightarrow Y$ and $B:Y\rightarrow Z$ linear maps with $B$ bounded and injective and $BA$ bounded. Prove that $A$ is bounded as well. If I knew that $B(Y)$ is closed I'd have a bounded linear map $B^{-1}:B(Y)\rightarrow Y$ by the bounded inverse theorem. Therefore $A=B^{-1}BA$ is bounded. How to prove the claim if $B(Y)$ is not closed. AI: This is extended version of Nate Eldredge's hint: Take $\{x_n\}$ such that $\lim_{n\to\infty}x_n= x$,$\lim_{n\to\infty}A(x_n)= y$. Show that $\lim_{n\to\infty}B(A(x_n))= B(A(x))$. Recall that $B$ is injective. Apply closed graph theorem.
H: Pythagorean Triplets with "Bounds" I am interested in the algebraic/geometric way of finding the pythagorean triplets such that $$a^2 + b^2 = c^2$$ $$a + b + c = 1000$$ I do the obvious $$a + b = 1000 - (a^2 + b^2)^{1/2}$$ $$a^2 + b^2 = 1000^2 -2(1000)a - 2(1000)b +2ab + a^2 +b^2$$ $$2a + 2b - \frac{2ab}{1000} = 1000$$ $$a + b -\frac{ab}{1000} = 500$$ I have no idea what to do at this point. I can't separate the variables, and any geometric solution is beyond my reach. The only other thing I can think of is writing $$a + b = 500 + \frac{ab}{1000} = 1000 - c$$ But this is going backwards AI: The following is copy and pasted directly from Yahoo Answers All Pythagorean triples are generated by ${m^2+n^2, m^2-n^2, 2mn}$, where $m$ and $n$ are positive integers, and $m\gt n$. You need $a+b+c=1000$, yielding $m(m+n)=500$. So, $m$ and $(m+n)$ are factors of $500$. $m+n\gt m$, so $m(m+n)\gt m^2$, so $m\lt \sqrt{(500)}$, and $m+n\lt2m$ (since $m\gt n$), so $m(m+n)\lt2m^2$, so $m\gt \sqrt{(250)}$. The prime factorisation of 500 is $2 \cdot 2 \cdot 5 \cdot 5 \cdot 5$, so the only factor of $500$ between $15.8$ and $22.4$ is $20$. Thus, $m=20$, $n=5$, giving the answer required: $m^2+n^2$ $= 425$; $m^2-n^2 = 375$; $2mn = 200$.
H: Legendre symbol, second supplementary law $$\left(\frac{2}{p}\right) = (-1)^{(p^2-1)/8}$$ how did they get the exponent. May be from Gauss lemma, but how. Suppose we have a = 2 and p = 11. Then n = 3 (6,8,10), but not $$15 = (11^2-1)/8$$ n is a way to compute Legendre symbols from Gauss lemma: $$\left(\frac{a}{p}\right) = (-1)^n$$ AI: I don't know how your source arrived at the exponent, but I'll tell you one of my favorite elementary ways of getting there. added ex post facto: this is probably the elementary way that Andre alluded to Let $s = \frac{p-1}{2}$, and consider the $s$ equations $$\begin{align} 1&= (-1)(-1) \\ 2&=2(-1)^2 \\ 3&= (-3)(-1)^3 \\ 4&= 4 (-1)^4 \\ & \quad\quad \ldots\\ s&= (\pm s)(-1)^s \end{align}$$ Where the sign is always chosen to have the correct resulting sign. Now multiply the $s$ equations together. Clearly on the left we have $s!$. On the right, we have a $2,4,6,\dots$ and some negative odd numbers. But note that $2(s) \equiv -1 \mod p$, $2(s-1) \equiv - 3 \mod p$, and so on, so that the negative numbers are the rest of the even numbers mod $p$, but disguised. So the right side contains $s! (2^s)$ (where we intuit this to mean that one two goes to each of the terms of the factorial, to represent the even numbers $\mod p$). We only have consideration of $(-1)^{1 + 2 + \ldots + s} = (-1)^{s(s+1)/2}$ left. Putting this all together, we get that $2^s s! \equiv s! (-1)^{s(s+1)/2} \mod p$, or upon cancelling factorials that $2^s \equiv (-1)^{s(s+1)/2}$. And $s(s+1)/2 = (p^2 - 1)/8$, so we really have $2^{(p-1)/2} \equiv (-1)^{(p^2 - 1)/8}$.
H: Prove $\cos^2x\sin^4x = \frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))$ I need help with a problem some may consider odd but here it is. I have the following trig identity I been working on and I managed to get it to. $$\cos^2x\sin^4x=\frac{3}{16}-\frac{\cos(2x)}{4}+\frac{3\cos(2x)}{16}+\frac{1}{8}(1+\cos(4x))+\frac{1}{32}(\cos(6x)+\cos(2x))$$ However I am not sure how to make this simplify to $\frac{1}{32}(2-\cos(2x)-2\cos(4x)+\cos(6x))=\cos^2x\sin^4x$ and thus I am stuck. AI: $\cos^2x\sin^4x=\cos^2x(1-\cos^2x)^2=\cos^6x-2\cos^4x+\cos^2x$ Now $\cos(2x)=2\cos^2x-1$, $\cos(4x)=\cos(2\cdot2x)=2\cos^2x-1=2(2\cos^2x-1)^2 - 1 = 8\cos^4x-8\cos^2x+1$ and $\cos(6x)=\cos(2\cdot3x)=4\cos^3(2x)-3\cos(2x)= 32\cos^6x + 18\cos^2x - 48\cos^4x - 1 $ Let $A\cdot\cos(6x)+B\cdot\cos(4x)+C\cdot\cos(2x)+D=\cos^6x-2\cos^4x+\cos^2x$ $Or, A(32\cos^6x - 48\cos^4x + 18\cos^2x- 1)+B(8\cos^4x-8\cos^2x+1)+C(2\cos^2x-1)+D=\cos^6x-2\cos^4x+\cos^2x$ Comparing the coefficients of different powers of $\cos x$, 6th power=>$A=\frac{1}{32}$ 4th power=>$-48A+8B=-2$ =>$B=-\frac{1}{16}$ 2nd power=>$18A-8B+2C=1$=>$C=-\frac{1}{32}$ constants (power $0$)=$-A+B-C+D=0$=>$D=A+C-B=\frac{1}{16}$ So, $\cos^2x\sin^4x$ $=\frac{1}{32}\cdot\cos(6x) - \frac{1}{16}\cdot\cos(4x) -\frac{1}{32}\cdot\cos(2x) + \frac{1}{16} $ $=\frac{1}{32}(\cos(6x) - 2\cos(4x) -\cos(2x) + 2) $. Alternatively, we know $e^{iy}=\cos y+i\sin y$ => $e^{-iy}=\cos y - i\sin y$ So, $\cos y=\frac{e^{iy}+e^{-iy}}{2}$ and $\sin y=\frac{e^{iy} - e^{-iy}}{2i}$ $\cos^2x\sin^4x$ $=(\frac{e^{ix}+e^{-ix}}{2})^2\cdot (\frac{e^{ix} - e^{-ix}}{2i})^4$ $=\frac{1}{64} (e^{ix}+e^{-ix})^2\cdot(e^{ix} - e^{-ix})^4$ $=\frac{1}{64} (e^{ix}+e^{-ix})^2\cdot(e^{ix} - e^{-ix})^2 \cdot(e^{ix} - e^{-ix})^2$ $=\frac{1}{64} ((e^{ix}+e^{-ix})\cdot(e^{ix} - e^{-ix}))^2 \cdot(e^{ix} - e^{-ix})^2$ $=\frac{1}{64} (e^{2ix} - e^{-2ix})^2 \cdot(e^{ix} - e^{-ix})^2$ $=\frac{1}{64} (e^{4ix} + e^{-4ix} -2 ) \cdot(e^{2ix} + e^{-2ix} - 2)$ $=\frac{1}{64} (e^{6ix} + e^{-6ix} -2(e^{4ix} + e^{-4ix}) -(e^{2ix} + e^{-2ix}) +4)$ $=\frac{1}{64} (2\cos6x -2(2\cos4x) -2\cos2x + 4)$ $=\frac{1}{32} (\cos6x -2\cos4x -\cos2x + 2)$ This is probably how the problem came to being. But if we know the RHS, the task becomes far easier. $\cos(6x) - 2\cos(4x) -\cos(2x) + 2$ $=\cos(6x) -\cos(2x) - 2(1-\cos(4x))$ $=-2\sin4x\sin2x -2\cdot2\sin^22x$ applying $ \cos 2C - \cos 2D=-2\cdot\sin (C+D) \sin(C-D)$ and $\cos2A=1-2\cdot \sin^2A$ formula $= 2\cdot\sin2x(-\sin4x+2\sin2x)$ $= 2\cdot\sin2x(-2\cdot\sin2x\cos2x+2\sin2x)$ (applying $\sin2A=2\cdot\sin A\cos A$ formula) $=4(\sin2x)^2(1-\cos2x)$ $=4(2\cdot\sin x \cos x )^2(2\sin^2x)$ (applying $\sin2A$ and $\cos 2A$ formula) $=32\cos^2x\sin^4x$
H: Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$. Simplify the difference quotient $\frac{f(x+h)-f(x)}{h}$ where a) $f(x)=2x+3,$ b) $f(x)=\frac{1}{x+1},$ c) $f(x)=x^2.$ I believe that if anyone can help me out with the first one, the other two might come clearer to me. But I started out this problem by plugging in $f(x)$ and got: $$\dfrac{(2x+3)+f(h)-(2x+3)}{h}$$ I have no idea what to do after this. Because the $2x+3$'s can cancel out and leave me with just $\frac{f(h)}{h}$ but that doesn't make sense to me. Please help. EDIT: The first one is solved and now for the second one, this is what I got: $$\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}$$ Now, to subtract fractions the denominator has to be the same and they are the exact same except for the first fraction has an $h$ while the other one does not. How would I go about this? For the third problem, this is all of my work: $$\begin{align}\dfrac{(x+h)^2-x^2}{h}&= \dfrac{x^2+2hx+h^2-x^2}{h}\\ &= \dfrac{2hx+h^2}{h}\\ &= \dfrac{h(2x+h)}{h}\\ &= 2x+h\end{align}$$ Is this all correct? AI: For the second one, you do the same thing: $$\dfrac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}=\dfrac{1}{h}\left(\frac{1}{(x+h)+1}-\frac{1}{x+1}\right)=\dfrac{1}{h}\dfrac{(x+1)-(x+h+1)}{(x+1)(x+h+1)}=\dfrac{1}{h}\dfrac{-h}{(x+1)(x+h+1)}=\ldots$$ Added: Now, you use the distributivity: $$(x+1)(x+h+1)=(x+1)x+(x+1)h+(x+1)1=(x^2+x)+(xh+h)+(x+1)=\ldots$$
H: Map from $\operatorname{Ext}^1(M,M)$ to $H^1(G, \operatorname{End}(M))$ The setting is as follows: $(R,m)$ is a local ring (assume noetherian, complete, if you need) and $\rho\colon G\to \operatorname{Aut}(M)$ is a group representation on the free, finite-rank $R/m^n$-module $M$, for some $n\geq 1$. Let $\operatorname{End}(M)$ be the adjoint representation on which $G$ acts by conjugation. To a class $c\in H^1(G,\operatorname{End}(M))$ one can associate (via an infinitesimal deformation) an exact sequence $0\to M\to E\to M\to 0$ of $(R/m^n)[G]$-modules. The class of this extension in $\operatorname{Ext}^1(M,M)$ is well defined (does not depend on the cocycle representing $c$). Question: What is a map in the inverse direction? Note, that if $n=1$, every sequence $0\to M\to E\to M\to 0$ splits as $k:=R/m$-vector spaces. And any such splitting $s$ yields a 1-cocycle $g\mapsto g \cdot s \cdot g^{-1}-s$. But, e.g. for $n>1$ the sequence of $\mathbb{Z}_2/(4)$-modules $0 \to \mathbb{Z}_2/(2)\to \mathbb{Z}_2/(4)\to \mathbb{Z}_2/(2)\to 0$ does not split. Or am I missing something? AI: Given an extension $$ 0 \to M \to N \to M \to 0 $$ apply the functor $\text{Hom}_{R/\mathfrak{m}^n}(M, -)$ to get $$ 0 \to \text{End}(M) \to \text{Hom}(M, N) \to \text{End}(M) \to 0 $$ (using freeness to get exactness on the right). Now take $G$-invariants to get a map $$ \text{End}_G(M) \to H^1(G, \text{End}(M)). $$ The image of the identity endomorphism under this map is your desired cocycle.
H: An exercise about convergence of series Possible Duplicate: Convergence/divergence of $\sum\frac{a_n}{1+na_n}$ when $\sum a_n$ diverges. Let $a_n$ be a non-negative sequence such that the series $\sum a_n$ does not converge. Could the series $\sum a_n/(1+na_n)$ be convergent? AI: Hint: Try $a_n=1$ if $n$ is a power of $2$ and $0$ otherwise.
H: Local invariants of the discrete Galois module associated to a $p$-ordinary newform Let $f=\sum_{n=1}^\infty a_nq^n$ be a $p$-ordinary newform of weight $k\geq 2$, level $N$, and character $\chi$, and let $\rho_f:G_\mathbf{Q}\rightarrow\mathrm{GL}_2(K_f)$ be the associated $p$-adic Galois representation, where $K_f$ is the finite extension of $\mathbf{Q}_p$ obtained by adjoining the Fourier coefficients of $f$. Let $\mathscr{O}_f$ be the ring of integers of $K_f$, and $A_f$ a cofree $\mathscr{O}_f$-module of corank $2$, i.e., $(K_f/\mathscr{O}_f)^2$, on which $G_\mathbf{Q}$ acts by $\rho_f$ (so we've chosen an integral model of $\rho_f$). My question involves the local invariants of $A_f$. Specifically, let $F$ be a number field, and let $v$ be a finite prime of $F$ not dividing $p$ or the conductor of $\rho_f\vert_{G_F}$. Fix a decomposition group $G_v$ of $v$ in $G_F\leq G_\mathbf{Q}$. Is it true that $H^0(G_v,A_f)$ is finite? I'm really interested in whether or not $\ker(H^1(G_v,A_f)\rightarrow H^1(I_v,A_f))$ vanishes ($I_v\leq G_v$ the inertia group), but with my hypotheses on $v$, the vanishing of this kernel is equivalent to the finiteness of $H^0(G_v,A_f)$ (because the kernel in question is divisible of the same $\mathscr{O}$-corank as $H^0(G_v,A_f)$). This vanishing seems to be implicit in a couple papers I've been looking at, and I'm not sure why it's true. AI: If the invariants were infinite, they would be divisible, and so they would correspond to an invariant line in $V_f$ (the representation on $K_f^2$ attached to $\rho$). Let $\ell$ be the rational prime lying undre $v$. The char. poly. of $\mathrm{Frob}_{\ell}$ acting on this rep'n is exactly the $\ell$th Hecke polynomial, so by Ramanujan--Petterson, the eigenvalues of $\mathrm{Frob}_{\ell}$ are Weil numbers of weight $(k-1)/2$. In particular, they are not roots of unity (provided the weight $k > 1$). The eigenvalues of $\mathrm{Frob}_v$ are powers of the eigenvalues of $\mathrm{Frob}_{\ell}$ (since $\mathrm{Frob}_v$ is a power of $\mathrm{Frob}_{\ell}$), and so they cannot be $1$. Consequently, $H^0(G_v,V_f) = 0$. QED (If $k = 1$ this argument breaks down, and of course the statement is false.)
H: How do I integrate this distribution? I have a multinomial multivariate normal distribution of the form: $$\exp\left[-\frac{1}{2\sigma^2}(({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right]$$ I wish to integrate with respect to $\boldsymbol \beta$. I have found a form of the Gaussian integral from wikipedia to be as following: $$\int\limits_{-\infty}^\infty\exp\left[-\frac{1}{2}\sum\limits_{i,j=1}^{n}{\bf A}_{ij}x_ix_j\right] d^nx=\sqrt{\frac{(2\pi)^n}{\det A}} $$ I do not know how to work out this integral or use this 'rule', but have come out with: $$\int\limits_{-\infty}^\infty\exp\left[-\frac{1}{2\sigma^2}(({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right] d^n\beta = \sqrt{\frac{(2\pi)^n}{\det \Sigma^{-1}}} $$ This probably is not right? How do I do the integral? How is the working out done? AI: You wrote $$\exp\left[-\frac{1}{2\sigma^2}({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right]$$ If you let the new value of $\Sigma$ be $\sigma^2\Sigma$, then you have $$\exp\left[-\frac{1}{2}({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu)\right].$$ There's no reason to separate out that scalar, and it's not conventionally done. The finite-dimensional case of the spectral theorem says every real symmetric matrix can be diagonalized by an orthogonal matrix, and you have $$ \Sigma = G^T \begin{bmatrix} \lambda_1 \\ & \lambda_2 \\ & & \lambda_3 \\ & & & \ddots \end{bmatrix} G. $$ Since $\Sigma$ is a variance (a "variance-covariance matrix" if you like), all of the $\lambda$s are non-negative, and since $\Sigma$ is nonsingular, all of them are positive. So let $\Sigma^{1/2}$ denote the matrix $$ \Sigma^{1/2} = G^T \begin{bmatrix} \sqrt{\lambda_1} \\ & \sqrt{\lambda_2} \\ & & \sqrt{\lambda_3} \\ & & & \ddots \end{bmatrix} G. $$ and then $\Sigma^{1/2}$ is a positive-definite symmetric matrix, and $(\Sigma^{1/2})^2=\Sigma$, and we let $\Sigma^{-1/2}$ denote the inverse. And since $\Sigma^{1/2}$ is symmetric, we have $(\Sigma^{1/2})^T\Sigma^{1/2}=\Sigma$. Then we have $$ ({\boldsymbol \beta}-\mu)^T\Sigma^{-1}({\boldsymbol\beta}-\mu) = \Big( \Sigma^{-1/2}({\boldsymbol\beta}-\mu) \Big)^T \Big( \Sigma^{-1/2}({\boldsymbol\beta}-\mu) \Big) = \gamma^T\gamma, $$ where $\gamma=\Sigma^{-1/2}({\boldsymbol\beta}-\mu)$. Then $$ \begin{align} \int_{\mathbb{R}^n} \cdots\cdots d\beta = \int_{\mathbb{R}^n} \cdots\cdots \|\det\Sigma^{1/2}\| \, d\gamma & = \|\det\Sigma^{1/2}\|\int_{\mathbb{R}^n} \cdots \cdots \\[10pt] & = \|\det\Sigma^{1/2}\|\int_{\mathbb{R}^n} \exp\left[ \frac{-1}{2} \gamma^T\gamma \right]\,d\gamma. \end{align} $$ This integral becomes $$ \int_{\mathbb{R}^n} \exp\left(\frac{-1}{2} \gamma_1^2 \right)\cdots\exp\left(\frac{-1}{2} \gamma_n^2 \right) \, d\gamma_1\cdots d\gamma_n. $$ Then it becomes the $n$th power of $$ \int_\mathbb{R} \exp\left(\frac{-1}{2}\gamma^2\right)\,d\gamma. $$ (And it's not hard to show that $\det(\Sigma^{1/2}) = \left(\det\Sigma\right)^{1/2}$.)
H: Fermat's Last Theorem: rational solutions iff integer solutions Problem Statement: In Fermat's Last Theorem $$x^n + y^n = z^n$$ $x,y,z$ are considered integers. But upon closer inspection it is seen that it is also true for any rational numbers $x,y,z$. And that FLT is not applicable only when $x,y,z$ are irrational. Query : Why is it that then it is always and only mentioned that Fermat's theorem is true when $x,y,z$ are integers and not rational numbers ? Is my perception correct? Can this be proven or disproved ? AI: As MTurgeon points, the two problems are equivalent. More exactly, for some $n$, the equation $x^n+y^n=z^n$ has non trivial integer solutions if and only if $x^n+y^n=z^n$ has non trivial rational solutions. Anyhow, many of the techniques used to attempt a proof, both in general and in the particular cases, work for the integer version. For example, the case $n=3$ relays on the fact that $\mathbb{Z}(\omega)$ is an UFD, the case $n=4$ is based on the fact that one gets a contradiction by building a smaller positive solution. Since the two problems are equivalent, and in the study the integer version is easier to approach, it is typically posted as an equation over the integers.
H: Kernel of Group Action this is my first post here. I have a question regarding a proof in Algebra by Hungerford: Let $G$ be a group and $H$ a subgroup of $G$. Let $S$ be the set of all cosets of $H$, where $G$ acts on $S$. Chapter II Collorary 4.9: "$\dots$ the kernel of $G \rightarrow A(S)$ is a normal subgroup of $G\dots$" Question: Why is the kernel normal? I looked through the chapter but there were no references, so I figure this is probably "obvious" and hence not included. I came up with an attempt at the proof: Let $K$ be the kernel of the action. $ex = x \implies e\in K \implies K$ contains identity. Let $a,b\in K$. $(ab)x=a(bx) = ax=x \implies ab\in K \implies K$ is closed. $a\in K \implies ax=x\implies x=a^{-1}x\implies a^{-1}\in K\implies$ all elements in $K$ has an inverse. Hence $K$ is a group. Let $g\in G$. $(gKg^{-1})x=(gK)(g^{-1}x)=g(K(g^{-1}x))=g(g^{-1}x)=x\implies gKg^{-1}\subset K$ Substitute $g$ for $g^{-1}$ and we get $K\subset gKg^{-1}$. Hence $gKg^{-1}=K$ and $K$ is normal. Is this approach correct? Also, suppose we were to consider the isotropy group of an element $x\in G$ instead. i.e. $G_x=\lbrace g\in G\mid gx=x\rbrace$ $G_x$ will be a subgroup of $G$ using a similar argument, but now it is not guaranteed to be a normal subgroup because in $(gG_xg^{-1})x=g(G_x(g^{-1}x))$, $G_x(g^{-1}x)=g^{-1}x$ is not necessary true. Is this argument right? Thanks for your time! AI: Your proof is fine. Alternatively, notice that an action is really given by a group homomorphism $G\rightarrow \operatorname{Perm}(S)$ into group of permutations of $S$. Then the kernel of the action is nothing but the kernel of this map, and a kernel of a homomorphism is always normal.
H: Dense subset of $C[0,1]$ In $C[0,1]$ the set $\{f(x): f(0)\neq 0\}$ is dense? I know only that polynomials are dense in $C[0,1]$, could any one give me hint how to show this set is dense?thank you. AI: Yes. Take $f\in \mathcal{C}[0,1]$ so that $f(0) = 0$. Now define $$f_n(x) = f(x) + {1\over n}, \qquad n\in\mathbb{N}.$$ We have $f_n\to f$ uniformly, whilst $f_n(0) \not= 0$ for all $n\in \mathbb{N}.$
H: Finding positive integer solutions to $n = ax^2 +by^2 - cxy$ How can I find the positive integer solutions to $x$ and $y$, given that $n$, $a$, $b$ and $c$ are all positive integers, in an equation of the form: $$n = ax^2 + by^2 - cxy.$$ Specifically, I want to find the positive integer solutions to the following equation, given $n$: $$n=3 x^2+20 y^2-16 x y.$$ AI: $$ n = (3x - 10y)(x-2y) {}{} $$ Alright, changing variables with $$ u = x - 4 y, \; v = x - 3 y $$ so that $$ x = 4 v - 3 u, \; y = v - u, $$ we have $$ 3 x^2 - 16 x y + 20 y^2 = -u^2 + 4 v^2. $$ So, factoring $n,$ we find all possible ways to write $$ n = -u^2 + 4 v^2 = (2 v + u )(2 v - u) $$ as an even square minus an odd square, definitely including both $u,-u$ for each success, also both $v,-v.$ Then, for each success (finitely many) we switch back to the original variables with $ x = 4 v - 3 u, \; y = v - u, $ and choose the solutions that include your conditions on positivity, whatever you might have meant by that. Note that there are no solutions if $n \equiv 1,2 \pmod 4.$ When $n<0,$ there are no solutions when $ \|n\| \equiv 2,3 \pmod 4. $ EDIT, Wednesday morning, before Gerry wakes up in Australia, another completely cosmetic change: switch to $-n$ and find all solutions to $$ -n = s \, (s + 4 t) $$ which is just to find ALL ways of writing $$ -n = FG $$ such that $F$ and $G$ differ by a multiple of $4.$ Then use $$ x = -3s - 2 t, \; y = -s-t, $$ from $$ s = -x+2y, \; t = x-3y. $$ So There. Well, if $n=0$ I guess there really are infinitely many solutions. But, as soon as $n \neq 0,$ we get finitely many solutions because $$ \|s\| \leq \|n\|, \; \; \|s+4t\| \leq \|n\|. $$
H: I don' t understand why the ratio of two meromorphic functions is meromorphic Let $D\subseteq\mathbb C $ be a connected set and consider $f,g\in\mathcal M(D)$.The ratio $r=\frac{f}{g}$ has the pole set $P(r)\subseteq P(f)\cup Z(g)$ (where $Z()$ is the set of all zeros). Why is $P(r) $ discrete in $D$ ? I think that disjoint union of discrete sets isn't discrete, consider for example $$\{0\}\cup\big\{\frac{1}{n} : n\in\mathbb N\big\}$$ AI: The pole set or zero set of a nonzero meromorphic function is not only discrete, it is also closed (meromorphic functions are continuous !), which eliminates your example. The closed and discrete subsets of $\mathbb C$ can also be described by the property that their intersection with any compact subset of $\mathbb C$ is finite. Then it is obvious that the reunion of two such subsets is again closed and discrete. Suppose $S$ is closed and discrete, and let $K$ be a compact subset of $\mathbb C$. Since $S$ is discrete, for each element $x \in S \cap K$, there is an open $U_x$ of $K$ containing $x$ such that $U_x \cap S = \{ x \}$. Since $S$ is closed, $K \setminus S$ is an open of $K$. Then, we have that $K \setminus S$ together with all the $U_x$ form an open cover of $K$. Since $K$ is compact, we can extract a finite subcover. But for each $x \in S \cap K$, $U_x$ was the only open containing $x$ so this finite cover still has to contain every $U_x$, thus there were finitely many $x \in S \cap K$ in the first place. Suppose that for every compact $K$, $S \cap K$ is finite. Then $S$ is closed : if $x \notin S$, then there are finitely many elements of $S$ in the disk of radius $1$ centered at $x$, so their distances to $x$ has a minimum $d>0$, and there is no elements of $S$ in the open disk of radius $d$ centered at $x$. $S$ is discrete : If $x \in S$, repeat the argument where $S$ is replaced with $S \setminus \{ x \}$
H: For $L_1,L_2,L_3$ , $L_1 \cap L_2 = L_3$, if $L_1,L_2,L_3$ such that $L_1 \cap L_2 = L_3$, if $L_1$ and $L_3$ are CFLs, so $L_2$ is CFGLas well I'm trying to answer this question: Is it true that for three languages $L_1,L_2,L_3$ such that $L_1 \cap L_2 = L_3$, if $L_1$ and $L_3$ are context free languages, so $L_2$ is context free languages as well. I know that context free languages are not closed under intersection, but it doesn't mean that there isn't an example for such language. I didn't come with any good idea for languages that fit the question, Any idea? AI: Since there exist context free languages $L_1$ which are contained in non-context free languages $L_2\supseteq L_1$, the answer is no.
H: Lie algebra of a Lie subgroup Let $G$ be a Lie group and $H$ a Lie subgroup of $G$, i.e. a subgroup in the group theoretic sense and an immersive submanifold. Let $\mathfrak{g}$ and $\mathfrak{h}$ be the associated Lie algebras. Now, the Lie algebra $\mathfrak{h}$ is given by: $$ \mathfrak{h} = \{ X\in \mathfrak{g} : \exp_G(tX)\in H, \text{ for } \vert t \vert < \varepsilon \text{ for one } \varepsilon > 0 \} $$ Thus, this definition varies from the usual one, $$ \mathfrak{h} = \{ X\in \mathfrak{g} : \exp_G(tX)\in H \quad \forall t\in \mathbb{R}\} $$ by restricting the values for $t$ to a small interval. The only thing that we know, is that $H$ is a Lie subgroup of $G$, but how does this property allow for the restriction of the $t$ values? AI: If $X\in\mathfrak g$ is such that $\exp(tX)\in H$ for small values of $t$, then in fact $\exp(tX)$ is in $H$ for all values of $t$. Indeed, suppose you know that $\exp(tX)\in H$ for $t\in(-t_0,t_0)$ and let $t\in\mathbb R$. There exists $n\in\mathbb N$ such that $t/n\in(-t_0,t_0)$, and then $$\exp(tX)=\exp(n\cdot\tfrac tnX)=\exp(\tfrac tnX)^n,$$ and the latter is in $H$ because $\exp(\tfrac tnX)$ is and $H$ is a subgroup.
H: Log likehood functions - Expected value Let $X_1,X_2,\ldots,X_n$ be a random sample from a Bernoulli($θ$) distribution with probility function $$P(X=x)= (θ^x)(1-θ)^{1-x},\qquad x=0,1;\ 0 < θ < 1.$$ $dl/dθ = [n \overline{x}/θ] \cdot (n-n\overline{x})/(1-θ)$ <-- Is it this that's wrong? :/ Got help with this too (Perhaps you can tell stats isn't my best subject) Show that $E[(dl(θ)/dθ)] = 0$ Apologies, exam in a couple of days in a mad scattered panic! When I first did this I integrated by accident, then I diffentiated and got a very strange answer and wasn't sure how to bring it to zero. AI: Let $X_1, X_2, \cdots, X_n$ be a random sample from $\mathrm{Bernoulli}(\theta)$ distribution. The log-likehood function for the sample mean $\bar{X}$ is given by $$l = l(\theta \| \bar{x}) = \log \mathcal{L}(\theta \| \bar{x}) = \log \mathbb{P}_{\theta}(\bar{X} = \bar{x}) = \log \left[ \binom{n}{n\bar{x}}\theta^{n\bar{x}}(1-\theta)^{n-n\bar{x}} \right], $$ thus we have $$ \frac{dl}{d\theta} = \frac{n\bar{x}}{\theta} - \frac{n-n\bar{x}}{1-\theta}.$$ It is easy to see, from binomial distribution, that $\mathbb{E}(\bar{X}) = \theta$. Thus $$\begin{align} \mathbb{E}\left[\frac{dl}{d\theta}(\bar{X})\right] &= \mathbb{E}\left[\frac{n\bar{X}}{\theta} - \frac{n-n\bar{X}}{1-\theta}\right] = \frac{n\mathbb{E}(\bar{X})}{\theta} - \frac{n-n\mathbb{E}(\bar{X})}{1-\theta} \\ &= \frac{n\theta}{\theta} - \frac{n-n\theta}{1-\theta} = n - n = 0. \end{align}$$
H: Number of field homomorphisms from an extension field of $\mathbb Q$ to $\mathbb C$ Take $\mathbb{Q}$ $\subset$ $K$ $\subset$ $\mathbb{C}$ with $[K:\mathbb{Q}]$ finite. How would you show that the number of field homomorphisms from $K$ to $\mathbb{C}$ is equal to $[K: \mathbb{Q}]$? I guess it is clear that any element of $\mathbb{Q}$ would map to itself, so you would need to look at the generators of $K$ (i.e. write elements of $K$ as linear combinations of elements of $\mathbb{C}$ with coefficients in $\mathbb{Q}$). Can you give me some suggestions on where to start? Thank you. AI: One way to prove this is by first using the primitive element theorem to show that $K = \mathbb{Q}(\alpha)$ for some algebraic number $\alpha$, whose minimum polynomial over $\mathbb{Q}$ has degree $n = [K:\mathbb{Q}]$. Then you can in fact describe the $n$ monomorphisms $\phi_i : K \hookrightarrow \mathbb{C}$ for $i = 1, \dots, n$ by the effect they have on $\alpha$. In particular if the minimal polynomial of $\alpha$ over $\mathbb{Q}$, say $p_\alpha(x)$ factorices over $\mathbb{C}$ as $$ p_\alpha(x) = (x - \alpha_1)\cdots (x - \alpha_n) $$ with $\alpha_1 = \alpha$, then the monomorphisms $\phi_i$ are given by $\phi_i(\alpha) = \alpha_i$. For example if you have the extension $K = \mathbb{Q}(\sqrt{5})$ then the minimal polynomial of $\sqrt{5}$ over $\mathbb{Q}$ is $p(x) = x^2 - 5 = (x- \sqrt{5})(x + \sqrt{5})$. Then there are only two monomorphisms given by $\phi_1(\sqrt{5}) = \sqrt{5}$ and $\phi_2(\sqrt{5}) = -\sqrt{5}$. Then since the elements of $K = \mathbb{Q}(\sqrt{5})$ are of the form $a + b\sqrt{5}$ they are given by $$ \phi_1(a + b\sqrt{5}) = a + b\sqrt{5} \quad \quad \phi_2(a + b\sqrt{5}) = a - b\sqrt{5} $$ Now, if you already now the description of the monomorphisms, the proof that indeed these are monomorphisms and that every monomorphism $K \hookrightarrow \mathbb{C}$ is of this form is not difficult and you should try to complete it on your own.
H: Changing from quadratic formula to standard form. The graph of a quadratic function has $x$-intercepts $-1$ and $3$ and a range consisting of all numbers less than or equal to $4$. Determine an expression for the function. This is my problem. I know what the graph looks like, but I only know quadratic formula and every time I input that into WolframAlpha, it combines my terms to create a different graph. I looked up the answer to this problem to check, and the answer is $-x^2+2x+3$. The only form I know is $y=af(bx+c)+d$. So how would I get that answer from this question without using $y=af(bx+c)+d$? AI: You want a quadratic $$f(x)=a(x+1)(x-3)\,\,\,s.t.\,\,\,f(x)\leq 4\,\,\,\forall x\in \Bbb R\Longrightarrow$$ $$ax^2-2ax-3a-4\leq 0\,\,\,\forall x\in \Bbb R\Longrightarrow \Delta=4a^2+4a(3a+4)\leq 0\Longrightarrow $$ $$4a(a+1)\leq 0\Longrightarrow -1\leq a\leq 0$$ So choose any $\,a\,$ as above and $\,f(x)=a(x+1)(x-3)\,$ fulfills your conditions, say $$a=-1\Longrightarrow f(x)=-x^2+2x+3$$ Another way: Since the vertex of a parabola $\,y=ax^2+bx+c\,\,,\,a\neq 0\,$ , is the point $$\left(-\frac{b}{2a}\,,\,-\frac{\Delta}{4a}\right)\,\,,\,\Delta=b^2-4ac$$ and the wanted parabola obviously opens downwards (as we want a maximum point) and the vertex is its maximal point, we get with $\,f(x)=a(x+1)(x-3)=ax^2-2ax-3a\,$: $$-\frac{\Delta}{4a}\leq 4\Longrightarrow -4a^2+12a\geq 16a\,\,(a<0\,\,\text{since the parabola opens downwards!})\Longrightarrow$$ $$4a(a+1)\leq0$$ and we get again the same solution as above. Added: Since you want the parabola to be less than or equal to $\,4\,$ you have to take the left extreme value, i.e. $\,a=-1\,$...can you see it?
H: Showing elements with certain properties are in a normal subgroup I'm preparing for an exam, and this little guy just had me stumped: Let $G$ be a group, $N\subset G$ a normal subgroup of $G$, $x,y,z \in G$, and $x^3 \in N$, $y^5 \in N$, $zxz^{-1}y^{-1} \in N$. Show that $x,y\in N$. What I want to do is to show for that for $g \in G$ that $gxg^{-1} \in N$. Another approach I tried is to just multiply the 3 given elements in $N$ in certain ways and use that $N$ is closed under multiplication to show that $x \in N$. No luck so far. This can't be that difficult right? AI: Adding on Jack's hint (well, in fact solution!): we have in the quotient $$G/N:\,\,\,(xN)^3\in N\,\,,\,(yN)^5\in N\,\,,\,(zxz^{-1}N)=(ynN)=yN\Longrightarrow$$ the elements $\,xN\,,\,yN\,\in G/N\,$ are conjugated (by $\,zN\,$) and thus they have the same order...but $\,(3,5)=1\,$ , so it must be that $$xN=yN=N=\,\text{the unit element in the quotient}\,\Longleftrightarrow x\,,\,y\in N$$
H: Zero sections of any smooth vector bundle is smooth? Could any one give me hint how to show that the zero section of any smooth vector bundle is smooth? Zero section is a map $\xi:M\rightarrow E$ defined by $$\xi(p)=0\qquad\forall p\in M.$$ AI: Smoothness can be checked locally on $M$ and locally $E$ is trivial. Can you use these two facts to conclude what you want?
H: Applications of Operator Algebras to modern physics I think that recently I've started to lean in my interest more towards operator algebras and away from differential geometry, the latter having many applications to physics. But while taking physics courses, it was also brought to my attention that operator theory is a very integral part of quantum mechanics. Are there applications of operator algebras in particular to quantum or relativity, or other fields of modern physics? What about applications of operator theory? In a related vein, I've been trying to find some research problems at my level or only slightly higher so that I can get a flavor, either for these applications, or for the abstract subject itself, as it's studied today. I was wondering if anybody knows either of any such problems, or a source where I could find such problems. To be honest, being a first year graduate student going on second, I'm not even quite sure where I'd look to find research problems that I could definitely guarantee are open, let alone ones I could reach right now. In fact, I'm told that sometimes professors even mistake solved questions as being open. If you have good examples of solved problems that are recent and representative of what I might face in the future, those would be helpful too. Thanks. AI: The only applications to general relativity that I know of (my field!) is via Connes' noncommutative geometry, which is...complicated. Connes' work is freely (and legally!) available online. See also the math overflow thread Applications of Noncommutative Geometry, which may be interesting. Operator algebras pop-up in Algebraic Quantum Field Theory too, which may be fun to look at. Actually, trying to discuss quantum fields on curved spacetime requires operator algebras. The Physics.SX thread "Why are von Neumann Algebras important in quantum physics?" is also relevant.
H: Find $a$ for which figure's area is maximum Suppose that we have following interval $(-5,2)$,we should find such $a$, which takes all possible values from this interval,creates following inequality systems $$5+a-\|2y\|\ge 0$$ $$\|x\|\leq \frac{\|a-2\|}{2}$$ we are working in $OXY$ cordinantes system,we have to find maximum value of area of figures,which can be defined by all solutions of inequality systems and find possible values of $a$,for which this area is maximum,or shortly we have system of inequality,we have different solution of this system for different value of $a$ from interval $(-5,2)$ and we have different figure created by these different set of solutions,we have to find maximum area between this figures and also this value of $a$ for which this area is possible,i have one idea and dont know if i am correct or wrong,let see first one we have $$5+a-\|2y\|\geq 0\Longrightarrow -\|2y\|\geq -5-a$$ or after dividing by $-2$ $$\|y\|\leq \frac{5+a}{2}$$ so it means $$-\frac{5+a}{2}<y<\frac{5+a}{2}$$ for $a\in (-5,2)$, we could write it as $\,-5<y<0\,\,,\,-\frac{6}{2}<y<\frac{6}{2}$ so $-5<y<0$ for second $[x]\le[a-2]/2$ $[x]\le7/2$ $[x]\le0$ but last one means that $x=0$ so what i am doing wrong?i think that maximm value could be achieved when $a$ is $-5$ and are is $35/2=17.5$ but i am not sure AI: Let's take an example, it may clarify things a bit for you. Suppose $a=0$. Then $\|y\|\le5/2$, and $\|x\|\le1/2$, so the figure we are talking about is the rectangle bounded by the horizontal lines $y=5/2$ and $y=-5/2$, and the vertical lines $x=-1/2$ and $x=1/2$. This rectangle has sides 5 and 1, and area 5. Now try it for some other value of $a$, like $a=1$ or $a=-1$, and see what you get. Then try to get a formula that works for all values of $a$. EDIT: So, let's finish this one off. We're told $-5\lt a\lt2$, so $a-2$ is always negative, so $\|x\|\le(2-a)/2$, so $${a-2\over2}\le x\le{2-a\over2}$$ Also, $5+a$ is always positive, so $\|y\|\le(5+a)/2$, so $$-{5+a\over2}\le y\le{5+a\over2}$$ So the area in question is a rectangle with sides $2-a$ and $5+a$, hence, area $$(2-a)(5+a)=10-3a-a^2$$ Maximizing the quadratic is an exercise left to the reader.
H: Finding the number of factors of product of numbers If $a,b,c,d\in\mathbb{N}$ be distinct. Each of which has exactly five factors, can we determine the number of factors of the product of $a,b,c,d$? Edit This is the solution given the in the back of the book I am reading. Does not make sense to me. If a, b, c and d have five factors each , they are all the fourth powers of prime numbers. Hence their product will have a total of $(4+1)(4+1)(4+1)(4+1) = 625$ factors. Ans : $625$ AI: Yes! If you're familiar with how to find the # of factors of a number, then it becomes a matter of prime factorizing each of $a,b,c,d$. Since $5$ is prime itself, it follows that each of the four integers must be of the form $p_i^4$ for some prime $p$ and $i=1,2,3,4$. Now we are given that $a,b,c,d$ are all distinct; namely, each of the $p_i$'s must be distinct. Thus, $abcd= p_1^4 \cdot p_2 ^4 \cdot p_3 ^ 4 \cdot p_4 ^4$, which by the formula, has $(4+1)(4+1)(4+1)(4+1)=625$ factors.
H: Constructing sets from connected sets I feel that this is probably really obvious, but I don't know how to get started. Is it true that every set in a metric space is the union of connected, pairwise-separated sets? And does this generalize to topologies easily? AI: It does indeed generalize to arbitrary topologies, although it's a slightly weird definition, because it's defined a little backwards, in terms of what it isn't rather than in terms of what it is. A separation of a topological space $\langle X, {\mathcal T}\rangle$ is two nonempty sets, $U$ and $V$, whose union is $X$, such that each is disjoint from the closure of the other. And we say that the space $X$ is disconnected if there is a separation of it, and connected if not. So for example, $\Bbb R$ with the usual topology is the prototypical example of a connected space, but ${\Bbb R}\setminus \{0\}$ is disconnected: a separation of it is $U=(-\infty,0), V=(0, \infty)$. Similarly, $\Bbb Q$ with the usual topology is disconnected, since a separation is $U=(-\infty,\sqrt 2)\cap{\Bbb Q}, V=(\sqrt 2, \infty)\cap{\Bbb Q}$. It's easy to show that a space $X$ is disconnected exactly when there is a nonempty subset of $X$ that is both open and closed. For example, in $\Bbb R$ with the usual topology, the only sets both open and closed are $\Bbb R$ and $\emptyset$, so once again we have the $\Bbb R$ is connected when it has the usual topology. On the other hand, when $\Bbb R$ is given the half-open interval topology (open sets are unions of half-open intervals $[a, b)$), the set $[0, 1)$ is both open and closed, and so $U=[0,1), V=(-\infty, 1)\cup[1,\infty)$ forms a separation and the space is disconnected. Subsets of a topological space $X$ can be identified as connected or disconnected, if they are connected spaces themselves in the subspace topology inherited from $X$. Every subset $S$ of a space can be considered to be a disjoint union of connected components, which are just the maximal connected subsets $S$, and it's exactly the connected components you are asking about. A connected subset $S$ has exactly one component, and a disconnected subset has more.
H: Play a slot machine for uncountable number of times There is a slot machine. You insert one coin, it destroy the coin and return countable number of coins. Then you can pick any one of the returned coin and put in the slot machine again. Formally, each coin is an ordinal. let $f(a)$ be the coin you insert into the machine, $g(a)$ be the set of coins returned after inserting $f(a)$. Certain conditions must hold for the function $f$ and $g$ for all ordinal $a, b \neq 0$. $f(a)=f(b)$ iff $a=b$ $f(a)\in (\bigcup_{i<a} g(i)) \backslash \{ f(i) \| i<a \}$ $g(a)\cap g(b) = \emptyset$ if $a\neq b$. You start with 1 coin $0$. so $f(0) = 0$. Prove no matter what is the strategy for inserting the coins, there is an ordinal $\alpha < \omega_1$ where you lose all your coins. In other words, for any $f,g$ satisfying the above requirements, $$ \{f(i) \| i<\alpha\} = \bigcup_{i<\alpha} g(i)$$ I learned about this an year ago. Now I have a hard time coming up with the proof. I can only find a simpler version of this problem as Ross–Littlewood paradox. AI: Let us suppose towards contradiction that you play according to a strategy that allows play to continue for $\omega_1$ many stages. At each countable stage $\alpha$, you have countably many coins $A_\alpha$ in your possession. One of them will be played right at that stage, and others perhaps will be played later. Let $B_\alpha\subset A_\alpha$ be the collection of those coins that will eventually be played at some countable stage, and let $\beta_\alpha$ be the supremum of the stages at which those coins will be played. Thus, every coin that you have at stage $\alpha$, if it will ever be played at a countable stage, will be played by stage $\beta_\alpha$. Now, let $\gamma_0=1$, and $\gamma_{n+1}=\beta_{\gamma_n}$, and $\gamma=\sup_n\gamma_n$. This ordinal $\gamma$ is a countable limit ordinal since it is the supremum of a strictly increasing countable sequence of countable ordinals. Suppose that you have a coin $c$ to play at stage $\gamma$. You earned it at some earlier stage, before some $\gamma_n$. But in that case, if you were ever to play $c$, then you would have already played it by stage $\gamma_{n+1}$, strictly before $\gamma$. Thus, at stage $\gamma$ you must have no coin to play. Contradiction. The argument amounts essentially to the fact about countable ordinals, that every function $f:\omega_1\to\omega_1$ has a closure point, an ordinal $\gamma$ such that $\alpha\lt\gamma\to f(\alpha)\lt \gamma$. Here, $f(\alpha)$ is the supremum of the stages where the coins that existed at stage $\alpha$ are played, if they are played at all. If $\gamma$ is closed under this function, then at stage $\gamma$, you can have no coins to play, since every such coin would have been born at an earlier stage $\alpha\lt\gamma$, and so if the strategy called for it to be played at $\gamma$ it would have already been played by stage $f(\alpha)$, which is strictly before $\gamma$. One can show the existence of closure points in this general sense by simply iterating the function as I did above: $\gamma_0$ is arbitrary, $\gamma_{n+1}=\sup f[\gamma_n]+1$ and then $\gamma=\sup_n \gamma_n$ is the desired closure point, since any $\alpha<\gamma$ has $\alpha<\gamma_n$ for some $n$ and hence $f(\alpha)<\gamma_{n+1}<\gamma$.
H: Prove: symmetric positive definite matrix I'm studying for my exam of linear algebra.. I want to prove the following corollary: If $A$ is a symmetric positive definite matrix then each entry $a_{ii}> 0$, ie all the elements of the diagonal of the matrix are positive. My teacher gave a suggestion to consider the unit vector "$e_i$", but I see that is using it. $a_{ii} >0$ for each $i = 1, 2, \ldots, n$. For any $i$, define $x = (x_j)$ by $x_i =1$ and by $x_j =0$, if $j\neq i$, since $x \neq 0$, then: $0< x^TAx = a_{ii}$ But my teacher says my proof is ambiguous. How I can use the unit vector $e_1$ for the demonstration? AI: Let $e_1 = \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}$, and so on, where $e_i$ is a vector of all zeros, except for a $1$ in the $i^{\mathrm{th}}$ place. Since $A$ is positive definite, then $x^T A x > 0$ for any non-zero vector $x \in \Bbb R^n$. Then, $e_1^T A e_1 > 0$, and likewise for $e_2, e_3$ and so on. If the $i^{\mathrm{th}}$ diagonal entry of $A$ was not positive, $a_{ii} < 0$, then $e_i^T A e_i = 0\cdot a_{11}\cdot 0 + 1\cdot a_{12}\cdot 0 + \cdots + 1\cdot a_{ii}\cdot 1 + \cdots + 0\cdot a_{nn} \cdot 0$, since $e_i$ has zeros everywhere but in the $i^{\rm th}$ spot. Thus, what would happen if $a_{ii}$ was negative?
H: Summation equation for $2^{x-1}$ Since everyone freaked out, I made the variables are the same. $$ \sum_{x=1}^{n} 2^{x-1} $$ I've been trying to find this for a while. I tried the usually geometric equation (Here) but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need: 1, 3, 7, 15, 31, 63 If my math is correct. AI: Use the equation for the sum of a geometric series: $$\sum_{i=1}^n a\cdot r^{i-1}=\frac{a(r^n-1)}{r-1}$$ where $a$ is the initial value of the sequence $u_n=a\cdot r^{n-1}$ and $r\ne1$. In your specific case the equation becomes: $$\frac{1\cdot(2^n-1)}{2-1}=2^n-1$$ So the sum of the first $n$ terms is $2^n-1$
H: Inequality of weights on a graph If $\sum_{j=1}^nx_j^2=1$ with $x_j\!\in\!\mathbb{R}$, why does it follow that $\sum_{j=1}^nx_j^4\geq\frac{1}{n}$. I'm trying to understand the following excerpt from Brandes & Erlebach's Network Analysis, p.407: AI: By generalized mean inequality (see e.g. Wikipedia, PlanetMath or AoPS) $$ \sqrt[4]{\frac{\sum_{i=1}^n x_i^4}n} \ge \sqrt[2]{\frac{\sum_{i=1}^n x_i^2}n}\\ \frac{\sum_{i=1}^n x_i^4}n \ge \left(\frac{\sum_{i=1}^n x_i^2}n\right)^2 $$ In your case $\sum x_i^2=1$ so you have $$\frac{\sum_{i=1}^n x_i^4}n \ge \frac1{n^2}\\ \sum_{i=1}^n x_i^4 \ge \frac1{n}.$$
H: Simplifying $\|a+b\|^2 + \|a-b\|^2$ I want to simplify $\|a+b\|^2 + \|a-b\|^2$ where $a, b \in \mathbb{C}$. I've used Wolfram Alpha to get $$ \|a+b\|^2 + \|a-b\|^2 = 2\left(\|a\|^2 + \|b\|^2\right) $$ I'm trying to understand the steps involved in arriving at this result: $$\begin{eqnarray} \|a+b\|^2 + \|a-b\|^2 &=& \|(a+b)^2\| + \|(a-b)^2\| \\ &=& \| a^2 + 2ab + b^2 \| + \| a^2 - 2ab + b^2 \| \end{eqnarray} $$ But I'm at a loss as to how to continue from here; I find it hard to work symbolically with absolute values. AI: $\|z\|^2=zz'$ where $z'$ stands for the complex conjugate of $z$. $$\|a+b\|^2+\|a-b\|^2=(a+b)(a'+b')+(a-b)(a'-b')=aa'+ab'+ba'+bb'+aa'-ab'-ba'+bb'=2aa'+2bb'$$ and you're pretty much there.
H: Does this sequence have this interesting property relating to the prime factorization of the index? Define a sequence as $a_0 = 0$ and $a_n$ equals the number of divisors of $n$ (including 1 and $n$) that are greater than $a_{n-1}$. This is sequence A152188 in OEIS, by the way. (For example, the first few terms are: 0, 1, 1, 1, 2, 1, 3, 1, 3, 1, 3, 1, 5, 1, 3, 2, 3, 1, 5, 1, 5, 2, 2, 1, 7). Here are some basic results that I know to be true: $a_n$ for prime $n$ equals $1$, but a value of $1$ does not imply primality (e.g., $a_9 = 1$ but $9$ is the square of a prime). $a_n$ can never exceed $\lfloor n/2\rfloor$ for $n > 1$. $a_n$ for even $n$ greater than $2$ can never equal $1$. I have tried to prove that $a_n = 1$ implies that $n$ is either prime or has a prime factorization consisting entirely of $p^j$ for some prime $p$ and an integer $j$ (there is only one prime in the factorization). A computer-based test I ran found no counterexamples, and interestingly, $a_n = 1$ implies the primality of $n$ every time with the exception of a few numbers early on (up until I quit testing after looking at about 10 million terms). Can anyone prove this property? AI: If $n$ is composite, it has at least two divisors $\ge \sqrt{n}$. Now $a_{n-1} < \tau(n-1) < \sqrt{n-1}$ (where $\tau$ is the number-of-divisors function) for all sufficiently large $n$: in fact for any $\epsilon > 0$, $\tau(n) \le n^\epsilon$ for sufficiently large $n$. So if $n$ is sufficiently large, $a_n = 1$ implies $n$ is prime. It appears from http://oeis.org/A035033 that $1260$ is the greatest $n$ such that $\tau(n) \ge \sqrt{n}$, but I don't have a reference to a rigourous proof of that.
H: What is the sum of $\sum\limits_{i=1}^{n}ip^i$? What is the sum of $\sum\limits_{i=1}^{n}ip^i$ and does it matter, for finite n, if $\|p\|>1$ or $\|p\|<1$ ? Edition : Why can I integrate take sum and then take the derivative ? I think that kind of trick is not always allowed. ps. I've tried this approach but I made mistake when taking derivative, so I've asked, mayby I should use some program (or on-line app) for symbolic computation. AI: $$\sum_{k=1}^n kp^k=\frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\tag{}$$ Proof by induction: $n=1$: $ p=\frac{p(p^{1+1}-(1+1)p^1+1)}{(p-1)^2}=\frac{p(p^{2}-2p+1)}{(p-1)^2}=p$ $$ \begin{eqnarray} (n+1)p^{n+1}+\sum_{k=1}^n kp^k&=&(n+1)p^{n+1} + \frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\\ &=&\frac{(n+1)p^{n+1}(p^2-2p+1)}{(p-1)^2} + \frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2}\\ &=&\frac{np^{n+3}+p^{n+3}-np^{n+2}-2p^{n+2}+p}{(p-1)^2}\\ &=&\frac{p((n+1)p^{n+2}-(n+1+1)p^{n+1}+1)}{(p-1)^2}\\ &=&\sum_{k=1}^{n+1} kp^k \end{eqnarray} $$ If $p=1$, we expect $\sum_{k=1}^n k\cdot 1^k= \frac12 n(n+1)$: Since the RHS of $()$ gives $\frac00$, when we insert $p=1$, we apply L'Hospital's rule two times: $$ \lim_{p\to 1} \frac{p(np^{n+1}-(n+1)p^n+1)}{(p-1)^2} =\lim_{p\to 1} \frac{n(n+2)p^{n+1}-(n+1)^2p^{n}+1}{2(p-1)}\\ =\frac12 \lim_{p\to 1} (n(n+1)(n+2)p^{n}-n(n+1)^2p^{n-1})\\ =\frac12 n(n+1) \underbrace{\lim_{p\to 1} p^{n-1}((n+2)p-(n+1))}_{=1}\\ $$ If $n\to \infty$, the series converges due to ratio test ($\lim_{n\to \infty}\left\|\frac{(k+1)}{k}p\right\|<1$), when $\|p\|<1$. You'll get $$ \sum\limits_{k=1}^\infty kp^k = \frac p{(p-1)^2}+\underbrace{\lim_{n\to \infty} \frac{np^{n+2}-(n+1)p^{n+1}}{(p-1)^2}}_{=0} = \frac p {(p-1)^2} $$
H: Sorting flags of different countries Question: In how many ways can you sort 8 of 12 flags (4 flags for each country) so there will be at least one flag from each country? Final answer: 4620 I'm not sure whether it's possible, but I'd like to solve this question by reducing the impossible ways from the total options. I think that there are ${3 \choose 1}\dfrac{8!}{4! 4! 0! }$ impossible ways of sorting. How can the total options be calculated? I assume that the flags of the same country are indistinguishable. Thanks! AI: I don't see how to get the total options without adding up contributions from individual partitions of $8$. I'd do it like this: $$ 3!\frac{8!}{4!3!1!}+3\frac{8!}{4!2!2!}+3\frac{8!}{3!3!2!}=4620\;. $$
H: Chain rule for multivariable functions confusion Suppose $f=f(x,y(x))$. Then applying the chain rule we get $\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}$. From this it seems that it always holds that $\frac{\partial f}{\partial y}\frac{\partial y}{\partial x}=0$. Where's the mistake? AI: As usual when there's confusion about partial derivatives, everything is readily cleared up if we remedy the deficiency in our notation for them by marking which variables are being held fixed: $$ \def\part#1#2#3{\left.\frac{\partial #1}{\partial #2}\right\|_{#3}} \part fxz=\part fxy\part xxz+\part fyx\part yxz=\part fxz=\part fxy+\part fyx\part yxz\;, $$ so there's no such implication, since $$ \part fxz\ne\part fxy\;, $$ unless of course you choose $z=y$, in which case indeed $$ \part yxz=\part yxy=0\;. $$
H: Trace of the multiplication operator Let $V$ be vector space, $\dim V=N$. Define the multiplication operator $L_{\mathbf{b}}$ as $L_{\mathbf{b}}:\omega\to \mathbf{b}\wedge\omega$, where $\omega\in\wedge V$ ($\wedge V$ is the entire exterior algebra) and $\mathbf{b}\in V$. I want to compute the trace of $L_{\mathbf{b}}$. If $N=1,2$ choosing a basis in $V$ we can obtain a basis of $\wedge V$ and with some calculations we obtain $tr(L_{\mathbf{b}})=0$. However this method will be very complicated for larger value of $N$ because $\dim(\wedge V)=2^N$. So I search for a coordinate-free (basis-free) calculations applying the following definition of trace. If $A=\sum_{k=1}^N \mathbf{v}_k\otimes\mathbf{f}_k^{}\in V\otimes V^{}$ then $tr(A)=\sum_{k=1}^N \mathbf{f}_k^{*}(\mathbf{v}_k)$. AI: You don't need to choose any particular basis; all you need is the fact that $\wedge V$ is graded and $L_b$ raises the degree by $1$, so if you choose any basis that respects the gradation, then $L_b$ sends any basis vector to a different subspace, so the diagonal elements in such a basis all vanish.
H: Periodic solution of differential equation let be the ODE $ -y''(x)+f(x)y(x)=0 $ if the function $ f(x+T)=f(x) $ is PERIODIC does it mean that the ODE has only periodic solutions ? if all the solutions are periodic , then can all be determined by Fourier series ?? AI: No, it doesn't. But in in special cases it has at least one periodic solution. The equation is known as Hill's equation, and the theory of its solutions, whether periodic or not, is known as Floquet theory. The referenced wikipedia page deals with first order systems; you'll have to rewrite your second order equation to a first order system to use the theory directly. If you want to learn more, the little book Hill's equation by Magnus and Winkler (in the Dover series) is an excellent resource.
H: Evaluate $f(x_0)+f(y_0)$ Let $$f(x)=3(x-2)^{\frac{2}{3}}-(x-2),~0\leq x\leq 20$$ Let $x_0$ and $y_0$ be the points of the global minima and maxima, respectively, of $f(.)$ in the interval $[0,20]$. Evaluate $f(x_0)+f(y_0)$ Note that $$f'(x)=2(x-2)^{-\frac{1}{3}}-1=0$$ $$=>x=10$$ and $$f''(10)=-\frac{2}{3}(10-2)^{-\frac{4}{3}}=-10.67<0$$ So,at $x=10,~f(x)$ is max. But Note that $$f(10)=4$$ but $$f(0)=6.76$$ I think I make some mistake but I can't find it out .Please solve the problem AI: Hint: After finding all the local maximum and local minimum, you also need to find $f(0)$ and $f(20)$. Then compare all the values of the local maximum and local minimum with $f(0)$ and $f(20)$, then the largest one is the maximum value, and the smallest one is the minimum value.
H: Two number partition problems Let $p_k(n)$ be a number of ways to express $n$ as a sum of $k$ positive integers. For example $p_2(3)=1$. Problem 1. Prove that following recurrences are correct: $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ $p_k(n)=p_k(n-k)+p_{k-1}(n-k) + ... + p_1(n-k)$ Problem 2. Find $p_2(n)$ and $p_3(n)$. Unfortunately I was always weak in showing recurrences and I have no idea for 1. I tried to use Ferrers diagrams to see sth but I haven't seen anything. For 2 it is easy to observe that $p_2(n)=\lfloor n/2 \rfloor$ and using first recurrence, that I failed to prove, I got $\displaystyle p_3(n)=\sum_{k=0}^{\lceil n/2 \rceil-1} \left\lfloor\frac{n-3k-1}{2}\right\rfloor$ but I don't know if it's possible to simplify this sum. Maybe some combinatorial interpretation, with Ferrers diagrams for example, will lead to more explicit form for $p_3(n)$? AI: For Problem 1: a) Either you have a part that's $1$, then the remaining parts can be any partition of $n-1$ into $k-1$ parts, or you don't, then you can subtract $1$ from all $k$ parts and are left with a partition of $n-k$ into $k$ parts. b) Subtract $1$ from every part; then you have $n-k$ left, in anything from $1$ to $k$ parts.
H: How to address mistakes in published papers? I have recently discovered some mistakes in a published maths article. I have contacted the author pointing out politely my concerns, but I got no specific answer, just a "polite" one, that the aspects I am addressing are clarified in some of his other papers (without mentioning which are these papers). Although my remarks where specific and could have been answered with a simple counter-argument (if there was any), the author avoided a direct answer. What is also concerning is that these errors are "copied" by others who base their research on this published paper. My question is how do you signal these errors to the community? What is the right process if the authors avoids to take credit? AI: The most polite way of pointing out a mistake in a previously published paper is first to pass it to the editor of that journal along with your well written claims. In this case the editor can take action and make it public such that other people is also aware of this mistake and dont use those results anymore. It might be the case that the editor doesnt take any action or it doesnt bother him much. In such a case the report should go to the chief editor of that journal indicating all the story. It is quite likely that one of them will solve the problem. However, according to my experience, it is also possible to address the mistakes in the same journal paper most probabily in the "correspondence" part. Human beings can make mistakes and it is disappointing for a mathematician to have mistakes in his paper. Very great mathematicians can even make mistakes therefore I think any claims should carry the best of politeness.
H: What is the sum of $\sum\limits_{i=1}^{n}i^k p^i$? (I've asked similar question, but this is much more complicated, I think) What is the sum of $\sum\limits_{i=1}^{n}i^k p^i$? Interpretation (why is it important ) : $f(k,n,p)=\sum\limits_{i=1}^{n}i^k p^i$ if n goes to plus infinity, then $f(1,n,1/2)$ is average length or series of heads while tipping symmetric coin, and $f(2,n,1/2)-f(1,n,1/2)^2$ is the variance of that length, so for another k we get k-th moment AI: Well, starting again with $f(p)=\sum_{i=1}^n p^i=\frac {p(p^n-1)}{p-1}$ Consider the theta operator $\Theta=p\frac d{dp}$ then your answer is $\Theta^k(f(p))$ (compute the derivative, multiply by $p$, repeat $k$ times).
H: A property of outer measure for bounded sets of real numbers. I have a bounded set $E$ of real numbers. I'm in the process of showing that there is a set $G$ that is a countable intersection of open sets $G_i$ such that $E\subseteq G$ and $E,G$ have the same outer measure. What I have so far: I know that outer measure is invariant under unions of disjoint sets, so I am characterizing $E$ as a union of open intervals, closed intervals, half-open intervals, and unions of countably many singletons as well as unions of collections of uncountably many singletons. Open intervals of $E$ can be replaced by $G_i$ for some $i$. Closed intervals of $E$, say $[a,b]$ can be written as $\bigcap _{i=1}^\infty (a-\frac{1}{i},b+\frac{1}{i})$, half-open intervals $[a, b)$ can be written as $\bigcap _{i=1}^\infty(a-\frac{1}{i},b)$ and a similar statement can be made for half-open intervals of the form (a,b]. I'm stuck on the singletons since I have to make sure I use only countably many sets. Any suggestions? I know that the outer measure of a collection of uncountably many points is equal the measure of the smallest open interval containing the points, but I'm having the hardest time given a point determining whether I want it to belong to a countable or uncountable set. AI: Since $A$ is bounded, its outer measure is finite, since $$\inf\left\{\sum_{j=1}^{+\infty}\lambda(I_j),A\subset\bigcup_{j=1}^{+\infty}I_j\right\}$$ is finite. Therefore, for each integer $n$, we can find an open set $G_n$ containing $A$ such that $\lambda(G_n)-\lambda^(A)\leqslant \frac 1{2^n}.$ Let $G:=\bigcap_{n\geqslant 1}G_n$. We can assume WLOG that $\{G_n\}$ is decreasing. Then $$\lambda(G)=\lim_{n\to +\infty}\lambda(G_n)\leqslant \lambda^(A).$$ Since $\lambda^(A)\leqslant \lambda(G_n)$ for all $n$, we have that $\lambda^(A)=\lambda(G)$.
H: Regular value: intuition about surjectivity condition Let $f:M\rightarrow N$ be a smooth function between two smooth manifolds. A $\textit{regular point}$ is a point $p\in M$ for which the differential $df_p$ is surjective. What does the surjectivity condition for the differential mean intuitively? What is then so "regular" about this point? AI: What is regular about a regular point $p$ is that the fibre $f^{-1}(f(p))$ is itself a manifold in a suitable neighbourhood of $p$, and that manifold has dimension $dim (M)-dim (N)$. About the simplest example is obtained by taking $M=\mathbb R^2, N=\mathbb R$ and $f(p)=f(x,y)=y^2-x^3$. We have $df_{(a,b)}(u,v)=-3a^2u+2bv$ for $p=(a,b)$ and this linear form is surjective (=non-zero) unless $p=(a,b)=0$. As a consequence the fibers of $f$, aka the contour curves $y^2-x^3=c$ are smooth submanifolds (of dimension $1$) of $M=\mathbb R^2$ for $c\neq0$. However the contour curve through $p=(0,0) $ is the subset $C\subset \mathbb R^2$ given by $y^2-x^3=0$, which is not a manifold in any neighbourhood of $(0,0)$. Differential geometers generally recoil in horror before beasts like $C$, whereas algebraic geometers study them under the name of singular varieties.
H: How can I get matrices for practicing Jordan normal form? I would like to practice the algorithm for the transformation from a matrix to its jordan normal form (with change of basis). To do so, I wrote this script that generates random $n \times n$ matrices, with $n \in \{2,3,4,5\}$ import random, numpy n = random.randint(2,5) matrix = [] for i in xrange(n): line = [] for j in xrange(n): line.append(random.randint(-10, 10)) matrix.append(line) A = numpy.matrix(matrix) print("Here is your %i x %i Matrix" % (n, n)) print(A) This way of generating random matrices isn't very good for the following reasons: The numbers can get really ugly (example) Sometimes it is not possible to calculate the decomposition of the matrix (example) Do you know either pages with many examples of "good" matrices up to $5 \times 5$ or do you know how to change my script? AI: You could generate a Jordan normal form $J$ randomly, then an invertible matrix $X$ randomly and have the algorithm compute $Y=XJX^{-1}$ for you. $Y$ would be your exercise.
H: Show that for $n, m \in \omega$, the ordinal and cardinal exponentiations $n^m$ This is an exercise from Kunen's book. Show that for $n, m \in \omega$, the ordinal and cardinal exponentiations $n^m$ are equal. What I've tried: I want to prove by using induction on $m$. For $m=0$, the ordinal exponentiation $n^m=n^0=1$ by the definition; and the cardinal exponentiation $n^m=n^0=\|n^0\|=1$. Now we assume for $m=k$ the case is right. Then for the $m=k+1$, ordinal exponentiation $n^{k+1}=n^k\times n$ and the cardinal exponentiation $n^{k+1}=\|n^{k+1}\|=\|n^k\times n\|$. Then I don't how to go on. Could anybody help me? Thanks ahead:) AI: First show that addition and multiplication also behave the same with ordinal and cardinal exponentiation when restricting them to natural numbers. Start with addition, that should be quite easy. Then use this to show that multiplication also has this property. From there you can use this to continue from where you got stuck.
H: Notations for groups of order $p^3$ Are there any relatively common notations for the two non-isomorphic nonabelian groups of order $p^3$ where $p$ is a prime number? I remember reading some notations like $p_+^{1+2}$ and $p_-^{1+2}$. Where can I find the definition of these notations? AI: These groups are the so-called extra-special groups of order $p^{3}.$ For each prime $p$ there are exactly two isomorphism types of such groups. The case $p=2$ has been discussed in another answer. When $p$ is odd, the two isomorphism types are distinguished by the fact that one has exponent $p,$ the other has exponent $p^{2}.$ I believe an explanation of the notation you refer to may be found in the "Atlas of Finite Groups".
H: Transforming an inhomogeneous Markov chain to a homogeneous one I fail to understand Cinlar's transformation of an inhomogeneous Markov chain to a homogeneous one. It appears to me that $\hat{P}$ is not fully specified. Generally speaking, given a $\sigma$-algebra $\mathcal A$, a measure can be specified either explicitly over the entire $\sigma$-algebra, or implicitly by specifying it over a generating ring and appealing to Caratheodory's extension theorem. However, Cinlar specifies $\hat{P}$ over a proper subset of $\hat{\mathcal{E}}$ that is not a ring. AI: We give the condition that $\widehat P$ is a Markow kernel, and we have that $$\widehat P((n,x),\{n+1\}\times E)=P_{n+1}(x,E)=1,$$ hence the measure $\widehat P((n,x),\cdot)$ is concentrated on $\{n+1\}\times E\}$? Therefore, we have $\widehat P((n,x),I\times A)=0$ for any $A\subset E$ and $I\subset \Bbb N$ which doesn't contain the integer $n+1$.
H: Calculate distance, knowing actual and perceived size What's equation would I use to calculate distance to an object, if I know it's actual and perceived size? Say there is a line, and I know it's actual length is 65 (units), I perceive it as 62 (units). I found various angular size equations on wikipedia and calculators like http://www.1728.org/angsize.htm however I don't know angle. AI: Suppose objects $X$ and $Y$ appear to be the same size. $X$ is actually of size $s_X$ and at distance $d_X$, and $Y$ is actually of size $s_Y$ and at distance $d_Y$. Then: $${s_X\over s_Y} = {d_X \over d_Y}$$ or equivalently: $${d_X\over s_X} = {d_Y\over s_Y}$$ Any three of these will determine the fourth. For example, a U.S. quarter dollar coin is about 2.5 cm in diameter. Held at a distance of 275 cm, it appears to just cover the full moon. We can conclude that the distance to the moon, divided by the diameter of the moon, is equal to about 275 cm / 2.5 cm = 110. If we somehow know that the distance to the moon is around 385,000 km, then this tells us that the diameter of the moon is around 3500 km, or vice versa, but we cannot get either one without the other. Similarly, we know from the existence of total solar eclipses that $${d_☉\over s_☉} = {d_☾\over s_☾} $$ where ☉ is the sun and ☾ is the moon. So we know that ${d_☉\over s_☉}\approx 110$ also. Again, if we know that the distance to the sun is about $150·10^6$ km, we can conclude that the sun's diameter is about $1.36·10^6$ km, and if we know the diameter of the sun instead we can estimate its distance. But without one we can't get the other.
H: Why is the probability that a continuous random variable takes a specific value zero? My understanding is that a random variable is actually a function $X: \Omega \to T$, where $\Omega$ is the sample space of some random experiment and $T$ is the set from which the possible values of the random variable are taken. Regarding the set of values that the random variable can actually take, it is the image of the function $X$. If the image is finite, then $X$ must be a discrete random variable. However, if it is an infinite set, then $X$ may or may not be a continuous random variable. Whether it is depends on whether the image is countable or not. If it is countable, then $X$ is a discrete random variable; whereas if it is not, then $X$ is continuous. Assuming that my understanding is correct, why does the fact that the image is uncountable imply that $Pr(X = x) = 0$. I would have thought that the fact that the image is infinite, regardless of whether it is countable or not, would already imply that $Pr(X = x) = 0$ since if it is infinite, then the domain $\Omega$ must also be infinite, and therefore $$Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}} = \frac{\text{# outcomes of the experiment where X = x}}{\|\Omega\|} = \frac{\text{# outcomes of the experiment where X = x}}{\infty} = 0$$ What is wrong with my argument? Why does the probability that a continuous random variable takes on a specific value actually equal zero? AI: The problem begins with your use of the formula $$ Pr(X = x) = \frac{\text{# favorable outcomes}}{\text{# possible outcomes}}\;. $$ This is the principle of indifference. It is often a good way to obtain probabilities in concrete situations, but it is not an axiom of probability, and probability distributions can take many other forms. A probability distribution that satisfies the principle of indifference is a uniform distribution; any outcome is equally likely. You are right that there is no uniform distribution over a countably infinite set. There are, however, non-uniform distributions over countably infinite sets, for instance the distribution $p(n)=6/(n\pi)^2$ over $\mathbb N$. For uncountable sets, on the other hand, there cannot be any distribution, uniform or not, that assigns non-zero probability to uncountably many elements. This can be shown as follows: Consider all elements whose probability lies in $(1/(n+1),1/n]$ for $n\in\mathbb N$. The union of all these intervals is $(0,1]$. If there were finitely many such elements for each $n\in\mathbb N$, then we could enumerate all the elements by first enumerating the ones for $n=1$, then for $n=2$ and so on. Thus, since we can't enumerate the uncountably many elements, there must be an infinite (in fact uncountably infinite) number of elements in at least one of these classes. But then by countable additivity their probabilities would sum up to more than $1$, which is impossible. Thus there cannot be such a probability distribution.
H: Need help with Unbiased estimator Let $X_1,X_2,X_3,\ldots,X_n$ be a random sample from a $\mathrm{Bernoulli}(\theta)$ distribution with probabilty function $P(X=x) = (\theta^x)(1 - \theta)^{(1 - x)}$, $x=0,1$; $0<\theta<1$. Is $\hat\theta(1 - \hat\theta)$ an unbiased estimator of $\theta(1 - \theta)$? Prove or disprove. I tried $x=\theta(1-\theta)$, $\bar x=\hat\theta(1-\hat\theta)$, $E[\bar x)=x$ $E[\bar x(1-\bar x)]=E[\bar x]-E[\bar x-1)$ but I'm not sure what to do now or how to prove it. I have an exam tomorrow so any help is really appreciated! Hopefully this is the last stats question I'll have to ask! AI: $\newcommand{\var}{\operatorname{var}}$ $\newcommand{\E}{\mathbb{E}}$ Your notation is confusing: you use $x$ to refer to two different things, and you seem to use the lower-case $\bar x$ to refer to the sample mean after using capital letters to refer to random variables initially. Remember that the variance of a random variable is equal to the expected value of its square minus the square of its expected value. That enables us to find the expected value of its square if we know it variance and its expected value. I surmise that by $\hat\theta$ you mean $(X_1+\cdots+X_n)/n$. That makes $\hat\theta$ an unbiased estimator of $\theta$. So $\E(\hat\theta) = \theta$ and $$ \var(\hat\theta) = \var\left( \frac{X_1+\cdots+X_n}{n} \right) = \frac{1}{n^2}\var(X_1+\cdots+X_n) = \frac{1}{n^2}(\var(X_1)+\cdots+\var(X_n)) $$ $$ =\frac{1}{n^2}\cdot n\var(X_1) = \frac 1 n \var(X_1) = \frac 1 n \theta(1-\theta). $$ Now we want $\mathbb{E}(\hat\theta(1-\hat\theta))$: $$ \mathbb{E}(\hat\theta(1-\hat\theta)) = \mathbb{E}(\hat\theta) - \mathbb{E}(\hat\theta^2) = \theta - \Big( \var(\hat\theta) + \left(\E(\hat\theta)\right)^2 \Big) = \theta - \left( \frac{\theta(1-\theta)}{n} + \theta^2 \right) $$ $$ = \frac{n\theta - \theta(1-\theta) - n\theta^2}{n} = \frac{n-1}{n}\theta(1-\theta). $$ From this you can draw a conclusion about whether $\hat\theta(1-\hat\theta)$ is an unbiased estimator of $\theta(1-\theta)$. (By the way, $\hat\theta(1-\hat\theta)$ is the maximum-likelihood estimator of $\theta(1-\theta)$.)
H: Strongly complete profinite group Let $G$ be a profinite group (or equivalently a compact and totally disconnected topological group ) with the property that all of its normal subgroups of finite index are open sets. Does this imply that all of its subgroups of finite index are open sets ? (if all subgroups of finite index from $G$ are open sets, than $G$ is called strongly complete ; this motivates the title of this post) AI: Yes. Lemma: Let $H$ be a subgroup of finite index in a group $G$. Then $H$ contains a normal subgroup of finite index, namely $\bigcap_{g \in G} gHg^{-1}$. Proof. $G$ acts on the left cosets $G/H$ by translation. Since $\|G/H\|$ is finite, the kernel of this action has finite index (dividing $\|G/H\|!$), and it is precisely the above intersection. $\Box$ So every subgroup of finite index is a union of cosets of a normal subgroup of finite index. Hence if the latter are open, then so are the former.
H: If every infinite subset has a limit point in a metric space $X$, then $X$ is separable (in ZF) I can prove this in ZFC, but don't know how to prove this in ZF. Following is the argument of this in ZFC. Fix $0<r\in \mathbb{R}$ and $x_0\in X$. Let $A_i = \{x\in X\mid d(x,x_j)\ge r,\, j<i\}$. Suppose $A_i≠\emptyset$ for every $i\in \omega$. Then by AC, we can choose $x_i$ for each $A_i$ to derive contradiction. I want to prove this in ZF. Help. Additional Question: Does infinite set have a countable subset in ZF? I guess it's true, but I only know the proof in ZFC. (If it is true, I think it would be really useful in many proofs in ZF.) AI: Unfortunately, the claim is not provable in ZF. It is relatively consistent with ZF that there is an infinite Dedekind-finite set of reals, an infinite set of reals $X$ with no countably infinite subset. This set of reals forms a metric space with the usual metric, and it cannot be separable, because it has no countably infinite subset (and no finite subset of an infinite metric space can be dense). Meanwhile, such an $X$ has your limit point property. To see this, suppose $Y\subset X$ is infinite, but has no limit point in $X$. In particular, every point $y\in Y$ is isolated in $Y$, and therefore we may pick a rational interval neighborhood $(q_x,r_x)$ of $y$ containing no other points from $Y$ except for $y$ itself. We do not need AC to pick this interval, since we may enumerate the rational intervals and pick the first one with this property. Further, each $y\in Y$ gives rise to a distinct such interval. Thus, we have an injective map from $Y$ to the set of rational intervals, and from this it follows that $Y$ is countable, contradicting our assumption that $X$ has no countably infinite subset. Note that $X$ also serves as a metric space that is limit-point compact but not compact. We've already shown that it is limit-point compact. But $X$ is not compact because we may simply pick any real $z\notin X$ that is a limit point of $X$, and then use the rational approximations to $z$ to produce an open cover of $X$ with no finite subcover, consisting of intervals straddling these approximations. Thus, ZF (if consistent) does not prove that limit-point-compact metric spaces must be compact.
H: what is the easiest way to represent $ \sqrt{1 + x} $ in series How to expand $ \sqrt{1 + x}$. $$ \sum_{n = 0}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!} = 1 + \sum_{n = 1}^\infty {{\left ( 1 \over 2\right )!}x^n \over n! \left({1 \over 2 }- n\right )!}$$ How can I simplify $ \left({1 \over 2 }- n\right )! $? AI: Per OP's request: One way to deal with the binomial coefficients in your series would be to use the duplication formula for the factorial: $$\binom{2n}{n}=\frac{4^n}{\sqrt \pi}\frac{\left(n-\frac12\right)!}{n!}$$ and the reflection formula $$\left(-n-\frac12\right)!\left(n-\frac12\right)!=(-1)^n\pi$$ The other way is to use the following definition, valid for any complex $x$: $$\binom{x}{n}=\frac1{n!}\prod_{k=0}^{n-1}(x-k)$$
H: Why does the bilinear form vanish between two direct factors? I extract a problem from the book I am reading: Let $R$ be a field, $A$ be a semisimple split $R$-algebra (associative with $1$). Let $A = \oplus_{n=1}^t A_n$ be a decomposition of $A$ into simple algebras. $(,): A \times A \rightarrow R$ is a non-degenerate symmetric $R$-bilinear form. Then $(A_i, A_j) =0$ for $i \neq j$. Is this statement true? If it is, why? If not, how can I find a conunterexample? In the first place, what is the definition for a split algebra? I think of split extension of fields, but there seems to be no realtion. Thanks very much. AI: An $R$-algebra $A$ being split means that ${\rm End}_A M = R$ for any simple $A$-module $M$, to the best of my knowledge. In any case, statement is false though. If any of the $A_i$ are not distinct, then we can twist the bilinear form a little, while preserving the desired properties, and end up with a counterexample. An explanation follows, but it would be worth considering it yourself first with say, the case $A=S\oplus S$ for some simple $R$-algebra $S$. Take for example $A=Re_1\oplus Re_2$ to be the semisimple $R$-algebra, where $e_1,e_2$ are orthogonal idempotents. This is split because the simple $A$-modules are just $e_1A$ and $e_2A$ and so it is apparent that the endomorphisms of these are just multiples of the identity. Now define a bilinear form on $A$ by $(x,y)=x_1y_2+x_2y_1$ where $x=x_1e_1+x_2e_2$ and $y=y_1e_1+y_2e_2$ with $x_1,x_2,y_1,y_2\in R$. Then this bilinear form is symmetric and nondegenerate since $(x,e_2)=x_1$ and $(x,e_1)=x_2$. By construction, the simple subalgebras are isotropic (i.e. $(e_i,e_i)=0$) but the inner product between them is nonzero (i.e. $(e_1,e_2)=1$). By the way, a more concrete way to see all of this is that this inner product is given by $$\begin{bmatrix}x_1 & x_2\end{bmatrix}\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}\begin{bmatrix}y_1 \\ y_2\end{bmatrix},$$ and nondegeneracy and symmetry follow from the fact the matrixis invertible and symmetric.
H: When a is less than c in $ \int_a^b \frac {dx} {x^4 - c^4} $ $$ \int_a^x \frac {dx} {x^4 - c^4} = \frac {1} {4c^3} \ln \left(\frac {x-c} {x+c} \right)_a^x - \frac {1} {2c^3} \tan^{-1} \Bigl(\frac {x} {c} \Bigr)_a^x $$ $$ =\frac {1} {4c^3} \Bigl[ \ln \Bigl(\frac {x-c} {x+c} \Bigr) - \ln \Bigl(\frac {a-c} {a+c} \Bigr) \Bigr]- \frac {1} {2c^3} \Bigl[\tan^{-1} \Bigl(\frac {x} {c} \Bigr)-\tan^{-1} \Bigl(\frac {a} {c} \Bigr)\Bigr]$$ When $x$ is less than $c$, or when $a$ is less than $c$, the number in the natural log becomes negative. Then, should the general answer have absolute sign instead of parenthesis as below? Is this a more proper or general answer? $$ =\frac {1} {4c^3} \left[ \ln \left\|\frac {x-c} {x+c} \right\| - \ln \left\| \frac {a-c} {a+c} \right\| \right] $$ AI: Yes. Notice that $$ \int \frac{dx}{x+c} = \ln \left\| x+c \right\| + \text{constant} $$ (where "constant" actually means a piecewise constant function that is constant on $(-\infty,-c)$ and also on $(-c,\infty)$). Similar remarks apply with $x-c$ in place of $x-c$. The partial-fraction decomposition of $1/(x^4-c^4)$ includes both $\text{constant}/(x+c)$ and $\text{constant}/(x-c)$, so those two integrals are where these logarithms come from.
H: Maximal Separable Subextension is Finite? Consider the following statement: "Let $L/K$ be an algebraic field extension. Then the maximal separable sub-extension is finite." Here is what seems to be a proof: "Let $M/K$, $K \subset M \subset L$ be a finite separable subextension of maximal degree. Let $x \in L$ be separable over $K$. So it will be separable over $M$. Consider the field tower $K \subset M \subset M(x)$. Each step of this tower is separable and finite and so $M(x) / K$ is finite and separable. But then $x \in M$ by the maximality assumption. This shows that $M/K$ is in fact the maximal separable sub-extension." Could you confirm the validity of the above proof? Thanks. AI: The quoted result is false: take for instance $K = \mathbb{Q}$ and $L = \overline{\mathbb{Q}}$. The first step of the purported proof is faulty: evidently there need not be a finite separable subextension of maximal degree!
H: Evaluating $\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$ I believe this integral $$\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$$ can not be computed exactly. However is there a method or transformation to express this integral in terms of the cosine integral or similar? I am referring to the integrals here. $a$ is real number; with the change of variable this integral becomes $$ \int_0^a\cos(u\sin t) \ \mathrm dt $$ with $$ x=a\sin t, $$ So, the new integral is $$ \int_0^{\pi /2}\cos(ua\sin t) \ \mathrm dt $$ AI: From $$\int_0^a \frac{\cos(ux)}{\sqrt{a^2-x^2}}\mathrm dx$$ you were able to transform it into $$\int_0^{\pi/2}\cos(au\sin\,t)\mathrm dt$$ which is expressible in terms of the Anger function $\mathscr{J}_\nu(z)$, which is equivalent to the more familiar Bessel function of the first kind $J_\nu(z)$ for integer orders: $$\int_0^{\pi/2}\cos(au\sin\,t)\mathrm dt=\frac12\int_0^\pi\cos(au\sin\,t)\mathrm dt=\frac{\pi}{2}J_0(au)$$
H: Find the number of real roots of the polynomial Find the number of real roots of the polynomial $$f(x)=x^5+x^3-2x+1$$ If I use Descarte's Rule then I get $$f(x)=x^5+x^3-2x+1$$ there can't be more than two positive real roots. Again $$f(-x)=-x^5-x^3+2x+1$$ there can't be more than one negative real root. AI: It's obvious there is one negative real root, because the graph goes through $(0,1)$, and the end behavior of the graph is down on the left hand side. Of the remaining four roots, at least two and at most four are complex (because complex roots come in pairs.) The derivative shows that the only positive real local minimum occurs at $x=\sqrt{0.4}$, but checking this in the equation shows that it is still above the $x$ axis. So the graph does not cross the x-axis on the positive real numbers, and the remaining four roots are complex.
H: Winning strategy for a matchstick game There are $N$ matchsticks at the table. Two players play the game. Rules: (i) A player in his or her turn can pick $a$ or $b$ match sticks. (ii) The player who picks the last matchstick loses the game. What should be the conditions on $N$ so that a winning strategy can be derived for the first player? What should be the strategy of first player so that he or she always wins this game provided $N$ is such that a wiinning strategy can be derived? I have solved this problem by hit and trial for small numbers one or two, but is there a general solution? Edit : Suppose rules of the game are changed and now a player in his or her turn can pick any number of matchstick upto p < N , then how many sticks should first player pick to ensure a win AI: With the current edit, the game goes like this: There are $N$ matchsticks on a table, two players (Left and Right) and a pre-ordained number $p$. Play alternates between Left and Right, with Left starting, and on a player's turn that player may remove up to $p$ matchsticks from the table. The person who takes the last matchstick loses. So you ask how Left can guarantee a win. The answer is that he can't always guarantee a win. For example, suppose $p = 1$ and there are 3 matchsticks. He picks up 1, Right picks up 1, and left picks up the last one, losing. But Left can always win unless $N = kp + (k + 1)$ for $k \geq 0$. Why is this? If $N = 1$, Left loses. Ok. If $N = 2$, Left picks one up. If $N = 3$, Left picks 2 up. This works until $N = p + 1$, when Left picks up as many as he can take, $p$. At $p + 2$, no matter how many Left picks up, there are a number of sticks that allow Right to win (Right just has to use the strategy that Left would have used). But at $p + 3$ sticks, all that Left has to do is pick up $1$, landing Right in the losing position $p+2$. This works until $2p+2$, when Left can pick up exactly $p$. At $2p + 3$, Left loses as no matter what move Left does, Right can use the strategy we described above for Left. For example, suppose $N = 14$ and $p = 5$. Then Left wants to move the game to a 'losing position' $kp + (k + 1)$. Here, that means moving to $2(5) + (2 + 1) = 13$, by picking exactly one up. No matter how many Right picks up, Left's next move will be to $1(5) + (1 + 1) = 7$ (i.e. after Left moves, there will only be $7$ sticks on the table). As this is $6$ away from $13$, Right can't move there. Left's next move will be to exactly $1$ stick on the table. So $N = kp + (k+1)$ are the 'losing positions' of the game.
H: Prove that $\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5}≥0 $. Given $x, y, z $ are 3 positive reals such that $xyz≥1$. Prove that $$\frac{x^5-x^2}{x^5+y^2+z^2}+\frac{y^5-y^2}{x^2+y^5+z^2}+\frac{z^5-z^2}{x^2+y^2+z^5}≥0.$$ This question is so complicated. I failed many times to get the proof. Can anyone help me please? Thank you. AI: This is problem №3 from IMO 2005. Here you can find its solution.
H: How many circles are needed to cover a rectangle? TRUE OR FALSE Suppose that a rectangle in $R^{2}$ can be covered by (allowing overlaps) $25$ discs of radius $1$, then it can also be covered by $101$ discs of radius $0.5$. Of course, though it is a true or false question, I would like the logic on it and possible a general proof. The answer is true but I don't see any specific logic. P.S. The question is from the entrance exam 2012 to graduate program at TIFR, Mumbai. AI: This is really funny - but this is not actually a math question. Here's the trick: If 25 circles can cover the full rectangle, then 25 circles of half-radius can cover the half-scale rectangle (i.e. cutting both side lengths in half). Tile this 4 times, and you see that yes, 100 half-radius circles can cover the whole rectangle. I just realized that there are 101 circles. I suppose you wear the 101$^{st}$ as a hat.
H: Boundary Question in $\mathbb{R}^{2}$ (Manifolds) Given a subset $A$ of $\mathbb{R}^{n}$, a point $x \in \mathbb{R}^{n}$ is said to be in the boundary of A if and only if for every open rectangle $B\subseteq\mathbb{R}^{n}$ with $x\in B$, $B$ contains both a point of $A$ and a point of $\mathbb{R}^{n}\setminus A$. My question is from Spivak's Calculus on Manifolds: Construct a set $A \subseteq [0,1]\times [0,1]$ such that $A$ contains at most one point on each horizontal and vertical line but has boundary equal to $[0,1]\times[0,1]$. AI: Since $\mathbb Q^2$ and the set of primes are both countably infinite we can write $$\mathbb Q^2 = \{ (x_p,y_p) : p \text{ prime} \}$$ where $p \mapsto (x_p,y_p)$ is a bijection. Now let $$A := \{(x_p + \sqrt{p}/2^p, y_p + \sqrt{p}/2^{p}) : p \text { prime} \} \cap [0,1]^2.$$ To show that $A$ contains at most one point on every vertical or horizontal line it suffices to show that the maps $p \mapsto x_p + \sqrt{p}/2^p$ and $p \mapsto y_p + \sqrt{p}/2^p$ are injective. Suppose $x_p + \sqrt{p}/2^p = x_q + \sqrt{q}/2^q$ for primes $p$ and $q$. Then $\sqrt{p}$ and $\sqrt{q}$ are linearly dependent over $\mathbb Q$ which is only possible if $p = q$. Since $A$ contains at most one point on every vertical or horizontal line we already know that every open set in $[0,1]^2$ contains some points outside of $A$. Therefore, it remains to show that $A$ is dense in $[0,1]^2$ (or, equivalently, in $(0,1)^2$). If $(x,y)$ is any point in $(0,1)^2$ then, since $\mathbb Q^2 \cap (0,1)^2$ is dense in $(0,1)^2$, there is a subsequence $(p_k)$ of the primes s.t. $(x_{p_k},y_{p_k})$ is a sequence in $(0,1)^2$ which approaches $(x,y)$. But then also $(x_{p_k} + \sqrt{p_k}/2^{p_k},y_{p_k} + \sqrt{p_k}/2^{p_k}) \to (x,y)$ as $k \to \infty$. For large $k$ this is a sequence in $A$, thus $A$ is dense in $(0,1)^2$.
H: Two hard number partition problems For every positive integer $n$, let $p(n)$ denote the number of ways to express $n$ as a sum of positive integers. For instance, $p(4)=5$. Also define $p(0)=1.$ Problem 1. Prove that $p(n)-p(n-1)$ is the number of ways to express $n$ as a sum of integers each of which is strictly greater than $1$. Problem 2. Prove that the total number of components in all partitions of $n$ is equal to $\displaystyle\sum_{k=0}^{n-1}p(k)\cdot \tau(n-k)$, where $\tau(m)=\sum_{d\|m} 1$ is the number of positive divisors of $m$. I came up with a solution for first problem, but I'm not sure if everything is OK with it, so I would be very grateful if you tell me what I should correct here: Let $p_k(n)$ denote number of ways to express $n$ as a sum of exactly $k$ positive integers. So $p_k(n)=\left\|\left\{ \langle a_1,...,a_k \rangle : \sum_{i=1}^k a_i=n \wedge 1\le a_i\le a_{i+1} \text{ for } 1\le i\le k-1 \right\}\right\|$. It's quite easy to observe that $p_k(n)=p_{k-1}(n-1)+p_k(n-k)$ by dividing above set into two disjoint sets: -set of tuples $\langle 1,...,a_k \rangle$ -set of tuples $\langle a_1,...,a_k \rangle$, where $a_1>1$ So let's count $p(n)-p(n-1)$ knowing that $p(n)=\sum_{k=1}^n p_k(n)$. $$p(n)-p(n-1)=p_n(n)+\sum_{k=1}^{n-1}p_k(n)-p_k(n-1)=\\=1+\sum_{k=1}^{n-1}p_{k-1}(n-1)+p_k(n-k)-p_k(n-1)=\\=1+ p_0(n-1)+p_1(n-1)-p_1(n-1)+p_1(n-1)+p_2(n-2)-p_2(n-1)+\dots+p_{n-3}(n-1)+p_{n-2}(2)-p_{n-2}(n-1)+p_{n-2}(n-1)+p_{n-1}(1)-p_{n-1}(n-1)=\\=1+0-p_{n-1}(n-1)+\sum_{k=1}^{n-1}p_k(n-k)=\sum_{k=1}^{n-1}p_k(n-k)$$ On the other hand, let $r(n)$ be number of partitions that we are looking for: $r(n)=\left\| \left\{ \langle a_1,...,a_k \rangle : 1\le k\le n-1, \ \sum_{i=1}^k a_i=n \wedge 2\le a_i\le a_{i+1} \text{ for } 1\le i\le k-1 \right\} \right\|$ and for a given $k$: $r_k(n)=\left\| \left\{ \langle a_1,...,a_k \rangle : \sum_{i=1}^k a_i=n \wedge 2\le a_i\le a_{i+1} \text{ for } 1\le i\le k-1 \right\} \right\|$ is a number of partitions that we are looking for but consisting only of $k$ integers. Since there is a simple bijection between tuples $\langle a_1,...,a_k \rangle \leftrightarrow \langle a_1-1,...,a_k-1 \rangle $, where $2\le a_i\le a_{i+1}$ for $1\le i\le k-1$ and still $1\le a_i-1$, then $\sum_{i=1}^k(a_i-1)=n-k$ and tuple $\langle a_1-1,...,a_k-1 \rangle$ is some partition of $n-k$. So $r_k(n)=p_k(n-k)$. Hence cases $r_k(n)$ are pairwise disjoint for $1\le k\le n-1$, it follows that: $\displaystyle r(n)=\sum_{k=1}^{n-1}r_k(n)=\sum_{k=1}^{n-1}p_k(n-k)=p(n)-p(n-1)$ as desired. Is that correct? I don't know how to approach second problem. Too hard it seems to be. AI: Generating function proofs The trick is to note the generating function for $p(n)$ satisfies this nice formula: $$P(x)=\sum_{n=0}^\infty p(n)x^n = \prod_{m=1}^\infty \frac{1}{1-x^m}$$ Proving that is a counting argument, noting that $\frac{1}{1-x^m}=\sum_{j=0}^\infty x^{jm}$ By the same reasoning, if $q(n)$ is the number of partitions of $n$ with all values greater than $1$, then: $$\sum_{n=0}^\infty q(n)x^n = \prod_{m=2}^\infty \frac{1}{1-x^m} = (1-x)P(x)$$ Then you get, by equating coefficients, that $q(n)=p(n)-p(n-1)$. The second problem is a bit harder. Let $p(k,n)$ be the number of partitions of $n$ into exactly $k$ positive numbers. The number you are looking for is: $$\sum_{k=1}^n kp(k,n)$$ We can write a new generating function for $p(k,n)$ as: $$F(x,y) = \sum_{n=0}^\infty \sum_{k=0}^\infty p(k,n)x^ny^k$$ By the same counting argument as before, we get the result: $$F(x,y) = \prod_{m=1}^\infty \frac{1}{1-x^my}$$ Now, differentiate both sides with respect to $y$, we get: $$\sum_{n=0}^\infty\sum_{k=0}^\infty kp(k,n)x^ny^{k-1} = \frac{dF}{dy} = F(x,y) \sum_{m=1}^\infty \frac{x^m}{1-x^my}$$ Now we are going to evaluate both sides at $y=1$. The left side at $(x,1)$ evaluates to: $$\sum_{n=0}^\infty x^n \sum_{k=1}^n kp(k,n)$$ So the coefficient of $x^n$ is the sum that we wanted above. On the other hand, on the right hand side, $\frac{x^m}{1-x^m}$ contributes a coefficient of $1$ for $x^n$ if $m\|n$ and zero otherwise, so $$\sum_{m=1}^\infty \frac{x^m}{1-x^m} = \sum_{n=1}^\infty \tau(n)x^n$$ Notice also that $F(x,1)=P(x)=\sum p(n)x^n$. So $$F(x,1)\sum_m \frac{x^m}{1-x^m} = P(x)\sum_m \tau(m) x^m$$ But the coefficient of $x^n$ in that multiplication is exactly what we want: $$\sum_{k=0}^n p(k)\tau(n-k)$$ We can eliminate $k=n$ since $\tau(0)=0$, so we get the result. There is almost certainly a direct counting argument in here, I'm just not sure what it is...
H: Algorithm for planarity test in graphs I am implementing a graph library and I want to include some basic graph algorithms in it. I have read about planar graphs and I decided to include in my library a function that checks if a graph is planar. I found on the web many efficient algorithms, but they all have the same drawback: they are very hard to implement. So here is my question: does there exist an algorithm for planarity checking which is easy to understand and to implement? AI: Several criteria for planarity are listed here: http://en.wikipedia.org/wiki/Planar_graph. Kuratowski's Theorem gives one possible algorithm, although it is quite slow. http://en.wikipedia.org/wiki/Planar_graph#Kuratowski.27s_and_Wagner.27s_theorems Hopcraft and Tarjan devised a linear-time algorithm. http://en.wikipedia.org/wiki/Planarity_testing#Path_addition_method You may find good answers on Stack Overflow. https://stackoverflow.com/questions/9173490/python-networkx https://stackoverflow.com/questions/1854711/how-to-check-if-a-graph-is-a-planar-graph-or-not
H: No closed form for the partial sum of ${n\choose k}$ for $k \le K$? In Concrete Mathematics, the authors state that there is no closed form for $$\sum_{k\le K}{n\choose k}.$$ This is stated shortly after the statement of (5.17) in section 5.1 (2nd edition of the book). How do they know this is true? AI: The very next paragraph of the book says: "Near the end of this chapter, we'll study a method by which it's possible to determine whether or not there is a closed form for the partial sums of a given series involving binomial coefficients, in a fairly general setting. This method is capable of discovering identities (5.16) and (5.18), and it also will tell us that (5.17) is a dead end." As Byron pointed out, the specific answer is on P228. The method is called Gosper's algorithm. The next section tells you about Zeilberger's algorithm, which can do more. The book "A = B" is available free online and is all about such mathematics, including more powerful stuff than what is shown in "Concrete Mathematics".
H: Holomorphic function in an annulus I'm trying to do a question but I have a doubt on holomorphic functions, here is the problem. Let $A = \{z ∈ \mathbb{C} : \frac{1}{R}< \|z\| < R\}$. Suppose that $f : A → \mathbb{C}$ is holomorphic and that $\|f(z)\| = 1 $ if $\|z\| = 1$. Show that $f(z) = {\left(\overline{f(\bar{z}^{−1})}\right)}^{-1}$ when $ \|z\| = 1 $ and deduce that this holds for all $z ∈ A$. Now, my problem is with the final deduction... I think I need something about the zeroes of f as the right-hand side of the equality could not be defined at some point. Am I right or is there a way to prove that a function with these properties can never be zero in A? Thank you all. AI: $f(z) \cdot \overline{f(1/\,\overline{z})}$ is well-defined and holomorphic on $A$ and equal to $1$ for $\|z\|=1$. Then it is constant on $A$, since level sets of non-constant holomorphic functions are discrete.
H: Kernel of adjoint of Lie algebra Let $G$ be a Lie group and $\mathfrak{g}$ its Lie algebra. The adjoint representation of the Lie algebra $\mathfrak{g}$ is defined as: $$ \text{ad: } \mathfrak{g} \rightarrow \text{End}(\mathfrak{g}), X \mapsto [X,\cdot] $$ Now, it holds true that $$ \text{ker ad} = \mathfrak{z}(\mathfrak{g}) = \{X \in \mathfrak{g} : [X,Y] = 0 \quad\forall\; Y \in \mathfrak{g}\}.$$ On the other hand, the definition of the kernel of this homomorphism is (at least in my mind) $$ \text{ker ad} = \{ X \in \mathfrak{g} : [X,\cdot] = \text{id}, \text{ i.e. } [X,Y] = Y \quad \forall \;Y \in \mathfrak{g} \}, $$ since the group identity in the endomorphism group is the identity-map. Evidently, the two sets are not the same, but where is my mistake? AI: As KotelKanim mentionned in the comments, the adjoint homomorphism is a Lie algebra homomorphism, and as such, the kernel is the preimage of the zero vector. The second "definition" of the kernel you gave is what you would expect if Lie algebras were groups - which is not the case. So that much is settled. Let me just point out two facts: The fact that the adjoint homomorphism really is a Lie algebra homomorphism (and not merely a linear map) is because of the Jacobi identity. In my opinion, this is the strongest motivation for this axiom of a Lie algebra. (Here, I assume the Lie algebra is finite-dimensional.) What you have shown is that, any Lie algebra with trivial center admits a faithful representation. However, much more is true. Indeed, Ado's theorem states that any Lie algebra over a field of characteristic zero admits a faithful representation. Later, the restriction on the characteristic was removed, and so we get the following (and perhaps surprising) result: Every Lie algebra admits a faithful representation. (For more information, see e.g. this Wikipedia article).
H: Solving linear system of equations when one variable cancels I have the following linear system of equations with two unknown variables $x$ and $y$. There are two equations and two unknowns. However, when the second equation is solved for $y$ and substituted into the first equation, the $x$ cancels. Is there a way of re-writing this system or re-writing the problem so that I can solve for $x$ and $y$ using linear algebra or another type of numerical method? $2.6513 = \frac{3}{2}y + \frac{x}{2}$ $1.7675 = y + \frac{x}{3}$ In the two equations above, $x=3$ and $y=0.7675$, but I want to solve for $x$ and $y$, given the system above. If I subtract the second equation from the first, then: $2.6513 - 1.7675 = \frac{3}{2}y - y + \frac{x}{2} - \frac{x}{3}$ Can the equation in this alternate form be useful in solving for $x$ and $y$? Is there another procedure that I can use? In this alternate form, would it be possible to limit $x$ and $y$ in some way so that a solution for $x$ and $y$ can be found by numerical optimization? AI: $$\begin{equation} \left\{ \begin{array}{c} 2.6513=\frac{3}{2}y+\frac{x}{2} \\ 1.7675=y+\frac{x}{3} \end{array} \right. \end{equation}$$ If we multiply the first equation by $2$ and the second by $3$ we get $$\begin{equation} \left\{ \begin{array}{c} 5.3026=3y+x \\ 5.3025=3y+x \end{array} \right. \end{equation}$$ This system has no solution because $$\begin{equation} 5.3026\neq 5.3025 \end{equation}$$ However if the number $2.6513$ resulted from rounding $2.65125$, then the same computation yields $$\begin{equation} \left\{ \begin{array}{c} 5.3025=3y+x \\ 5.3025=3y+x \end{array} \right. \end{equation}$$ which is satisfied by all $x,y$. A system of the form $$\begin{equation} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} b_{1} \\ b_{2} \end{pmatrix} \end{equation}$$ has the solution (Cramer's rule) $$\begin{equation} \begin{pmatrix} x \\ y \end{pmatrix} =\frac{1}{\det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} } \begin{pmatrix} a_{22}b_{1}-a_{12}b_{2} \\ a_{11}b_{2}-a_{21}b_{1} \end{pmatrix} = \begin{pmatrix} \frac{a_{22}b_{1}-a_{12}b_{2}}{a_{11}a_{22}-a_{21}a_{12}} \\ \frac{a_{11}b_{2}-a_{21}b_{1}}{a_{11}a_{22}-a_{21}a_{12}} \end{pmatrix} \end{equation}$$ if $\det \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22}% \end{pmatrix}% \neq 0$. In the present case, we have $$\begin{equation} \begin{pmatrix} \frac{1}{2} & \frac{3}{2} \\ \frac{1}{3} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 2.6513 \\ 1.7675 \end{pmatrix} \end{equation}$$ and $$\det \begin{pmatrix} \frac{1}{2} & \frac{3}{2} \\ \frac{1}{3} & 1 \end{pmatrix} =0$$
H: Is my solution to this permutation question correct? Question: Suppose there are m girls and n boys in a class. What is the number of ways of arranging them in a line so that all the girls are together? (Biggs, Discrete Mathematics 2nd ed, Exercise 10.7.5) My solution Say $m=4$ and $n=3$ The number of ways they can be arranged are: G - a girl, B - a boy position 1: [G G G G][B B B] position 2: [B][G G G G][B B] position 3: [B B][G G G G][B] position 4: [B B B][G G G G] So in this case the group of girls can be placed in $4$ different positions. On each position the group of girls can internally be arranged in $4!$ different ways, and the boys can be arranged in $3!$ different ways. Hence the total number of arrangements are $4!3!4$ Converting the answer back to variables, I end up with: $m!n!(n + 1)$ The book provides no solution, so I would really like to know if I came up with the right answer.. AI: The basic idea and the answer are correct, but for a complete solution, you should show it in abstract context, so start with arbitrary (unspecified) $m,n$ and try to show that you get $(n+1)!m!$ as a result.
H: Expressing $\sin^4x-\sin^6x$ in another way I am slightly confused on how one would subtract $\sin^4x-\sin^6x$. I know that $\sin^2x=(1/2)(1-\cos2x)$, so $\sin^4x$ would logically be $[(1/2)(1-\cos2x)]^2=(1/4)(1-2\cos(2x)+\cos^2(2x)$ However the value of $\sin^6x$ eludes me. Would it be $(1-\cos2x)^3$? I did that and got $1-\cos2x-2\cos2x-2\cos^2(2x)+\cos^2(2x)-\cos^3(2x)$ I am not sure if that is correct. AI: I have noticed the other answers either don't answer your question (and get many upvotes none-the-less) or are much more complicated than necessary (even more complicated than you understand based on the comment by Raymond). So, here is the answer to your actual question. You dropped the 1/2. $$\sin^6 x = (\sin^2 x)^3 = \left(\frac{1 - \cos{2x}}{2}\right)^3 = \frac{1}{8} (1 - \cos{2x})^3$$ so you end up missing $1/8$ in the end. Also, you messed up the sign on one term. In general $$(A + B)^3 = A^3 + 3A^2B + 3AB^2 + B^3$$ so here we have \begin{align} (1 - \cos(2x))^3 &= \bigg(1 + (-\cos(2x))\bigg)^3 \\ &= 1 + 3(-\cos(2x)) + 3(-\cos(2x))^2 + (-\cos(2x))^3 \\ &= 1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x). \end{align} Therefore, your final answer is \begin{align} \sin^4x - \sin^6x &= \frac{1}{4}\bigg(1 - 2\cos(2x) + \cos^2(2x)\bigg) \\ &- \frac{1}{8}\bigg(1 - 3\cos(2x) + 3\cos^2(2x) - \cos^3(2x)\bigg) \\ &= \frac{1}{8}\bigg(1 - \cos(2x) - \cos^2(2x) + \cos^3(2x)\bigg) \end{align}
H: Exponentiation and Set of Functions Notation. Given arbitrary sets $A$ and $B$, the notation $A^{B}$ is mostly clear from context to mean $A^{B} = \{f : f : B \rightarrow A\}$. However, when these sets are ordinal or cardinals, especially $\omega$, the notation is not consistent even among subfields of logic. For example $2^\omega$ in one sense can denote ordinal exponentiation. Hence $2^\omega = \lim_{n < \omega} 2^n = \omega$. However, you can also consider $2^{\aleph_0}$. By using the cardinal $\aleph_0$, some people may consider it clear that $2^{\aleph_0}$ denotes the cardinal of the set $\{f : \aleph_0 = \omega_0 = \omega \rightarrow 2\}$. In the above paragraph $2^{\aleph_0}$ is a cardinal, (in ZFC) it is a special ordinal. However, it descriptive set theory, you may want to consider not the cardinal but Cantor Space (or Baire Space), i.e. the set of functions from $\omega \rightarrow 2$. When you want the set of functions as oppose to the ordinal, is there a notation for that. In recursion theory, I have found that $2^\omega$ or $\omega^\omega$ most frequently refers to Cantor or Baire space, and not the ordinal or cardinal. In Mostovachis book, he uses $\text{}^\omega2$ to denote Cantor Space. Does anyone know of any establish custom to distiguish between ordinal exponentiation, cardinality of the set of functions between ordinals, and the actual set of functions between ordinals. I was thinking perhaps the left right exponent like $\text{}^\omega2$ and $2^\omega$ could be used as distinction, but from reference to recursion theory and Moschivakis's book, it seems that this is not the case. Thanks for any help you can provide. AI: My experience tells me that there is no "good" way of discerning the uses except your own experience in guessing the correct context. Especially in the ordinal-cardinal exponentiation. It takes some time to get used to, but that's just one of the problems working with a limited set of characters (while trying to be as succinct as possible). Sometimes one has to read a sentence or two, holding in one's head a few possible interpretations until the correct one has been chosen. Sometimes one is wrong and later comes the "Ohhhhhh, so that was ordinal exponentiation" but with some experience this is almost completely eliminated. [Good] writers will usually try to ensure the context is set when setting the notation "consider Cantor space, denoted by $2^\omega$, ..." or something of this kind. That been said, when $\aleph$ numbers are in the exponent this is always the cardinal version of exponentiation. There is also a common notation of $^AB$ as the set of all functions from $A$ into $B$, in this aspect $^\omega\omega$ can be seen as the Baire space (as its underlying set is exactly this). Note also that $\left\|^AB\right\|=\|B\|^{\|A\|}$. Mildly related: Do $\omega^\omega=2^{\aleph_0}=\aleph_1$?
H: How likely is it for a randomly picked number to be larger than all previously chosen numbers? Suppose we pick a uniformly distributed number on the range [a,b]. Then we continue to pick more numbers on the same range. Let n(t) be the number of times we have found a number bigger than any previously found, after sampling t total numbers. The initial number picked is not counted. Solve: $$\lim_{t\to\infty}\sqrt[n(t)]{t}$$ For reference, the problem I'm trying to solve is the geometric mean of the amount of time it takes to find the next bigger number, relative to the amount of time it took for the previous bigger number. So say it finds a new best value at the 4th try, 11th try, and 29th try, I get: $$\sqrt[3]{4\dfrac{11}{4}\dfrac{29}{11}}$$ which simplifies into the top equation. Experiments seem to indicate the amount of time it takes to find the next number multiples by about 3 for each number found, but I'm curious if there might be tools to solve this analytically. Interestingly, the value seems to be the same even if I take the randomly chosen number and plug it into a function (i.e. I'm checking to see if f(x) is greater than any previously seen f(x) for random values of x). Related questions: Is there any way to guess at the probability of finding a new biggest number? Can the geometric mean be shown to be a constant across all (or some) functions? Intuitively I feel like it should be. I don't have too extensive a background in math, so I'm hoping this isn't an unsolved problem. AI: First let's consider a fixed $t$. After we have picked $t$ numbers, we will almost surely have picked $t$ different real numbers, so we can assume that this is always the case without disturbing any probabilities or expectations. Now, instead of picking $t$ numbers one by one, we could start by picking an unordered set of $t$ numbers, and then afterwards decide a random order to present them in. Since the $t$ picked numbers are being presented in a random order (and it ought to be clear that no order is more probable than any other), it is easy to see that the $t$th number is a new maximum with probability $\frac1{t}$. On the other hand given whatever the $t$th number is, the first $t-1$ numbers are also presented in a uniformly random order, so the $(t-1)$th number was a new maximum with probability $\frac 1{t-1}$, independently of whether the $t$th one is. What this argument shows is that the $k$th number is a new maximum with probability $\frac 1k$, and that the new-maximumness at different positions in the sequence are all independent events. So the expectation of $n(t)$ is simply $\sum_{k=1}^t \frac 1k$. For large $t$ this sum approaches $\gamma+\log t$, where $\gamma$ is a constant known as the Euler-Mascheroni constant. Finally un-fixing $t$, your limit ought to be (insert handwavy appeal to the law of large numbers here): $$\lim_{t\to \infty} \sqrt[\gamma+\log t]{t} = \lim_{t\to\infty} t^{\frac1{\gamma+\log t}} = \lim_{t\to\infty} e^{\frac{\log t}{\gamma+\log t}} = \lim_{t\to\infty} e^1 = e$$
H: Generalized homomorphism theorem Let $f: G \rightarrow G'$ be a group homomorphism with kernel $H$. Then we know as an elementary fact of abstract algebra that there is an injective group homomorphism $f^:G/H \rightarrow G'$ such that $f^(x+H) = f(x)$. More generally, let $\psi: G \rightarrow G/H$ be a surjective homomorphism with kernel $H$. Then again we have an injective homomorphism $f_{\psi}^* : G/H \rightarrow G'$ defined by $f^_{\psi}(x+H)=f(y)$ where $\psi(y) = x+H$. Could you verify the validity of this reasoning? Thanks. Edited: To give you an example about the motivation of this question, we know that given a homomorphism $f:G \rightarrow G'$ with kernel $H$ there exists always a homomorphism $f^: G/H \rightarrow G'$ such that $f = f^* \circ \phi$ where $\phi:G \rightarrow G/H$ is the canonical mapping. Now, what if instead of the canonical mapping $\phi$ we have another surjective homomorphism $\psi:G \rightarrow G/H$ with kernel $H$? Does then exist a homomorphism $f^_{\psi} : G/ H \rightarrow G'$ such that $f = f^_{\psi} \circ \psi$? AI: By the first isomorphism theorem there is a unique $\psi^:G/H\to G/H$ satisfying $\psi^\circ\phi=\psi,$ which must be an isomorphism since it is injective. Let $f^_\psi:=f^\circ(\psi^)^{-1}.$ Then $$f^_\psi\circ\psi = \left(f^\circ(\psi^)^{-1}\right)\circ\psi = f^\circ\left((\psi^)^{-1}\circ\psi\right) = f^*\circ\phi=f.$$
H: Proof about Field conjugation isomorphisms I'm having an awful time making sense of a proof and I was hoping someone could help. Theorem: Let $\alpha$ and $\beta$ be algebraic over a field $F$ with $deg(\alpha, F) = n$, as elements of a field extension $E$ of $F$. Define the map $\psi_{\alpha,\beta}:F(\alpha)\to F(\beta)$ by $\psi_{\alpha,\beta}(c_{0} + c_{1}\alpha + ... + c_{n-1}\alpha^{n-1}) = c_{0} + c_{1}\beta + ... + c_{n-1}\beta^{n-1}$. Then $\psi_{\alpha,\beta}$ is an isomorphism if and only if $\alpha$ and $\beta$ are conjugates over $F$. Proof: $(\Rightarrow)$ Assume the map is an isomorphism. Let $irr(\alpha, F) = a_{0} + a_{1}x + ... + x^{n}$. Then $a_{0} + a_{1}\alpha + ... +\alpha^{n} = 0$, so $0 = \psi_{\alpha,\beta}(0) = \psi_{\alpha,\beta}(a_{0} + a_{1}\alpha + ... +\alpha^{n}) = a_{0} + a_{1}\beta + ... +\beta^{n}$, which implies that $\beta$ is a root of $irr(\alpha, F)$. The author then uses the same argument to show that $\alpha$ is a root of $irr(\beta, F)$, which I haven't read carefully yet or worked through. But I am already in objection. How do we obtain $\psi_{\alpha,\beta}(a_{0} + a_{1}\alpha + ... +\alpha^{n}) = a_{0} + a_{1}\beta + ... +\beta^{n}$? The definition only gives us a formula for up to degree $n-1$ polynomials. My partial justification: Since $a_{0} + a_{1}\alpha + ... +\alpha^{n} = 0$, we know $-a_{0} - a_{1}\alpha - ... -a_{n-1}\alpha^{n-1} = \alpha^{n}$, so $\psi_{\alpha,\beta}(\alpha^{n}) = \psi_{\alpha,\beta}(-a_{0} - a_{1}\alpha - ... -a_{n-1}\alpha^{n-1}) = -a_{0} - a_{1}\beta - ... -a_{n-1}\beta^{n-1}$ But I can conclude that this last expression equals $\beta^{n}$ only if I assume the result which I am trying to prove: that $\beta$ is a root of $irr(\alpha, F)$. Thus, I am stuck. AI: If the mapping $\psi=\psi_{\alpha,\beta}$ is an isomorphism, then it respects multiplication, so $$ \psi(\alpha^n)=\psi(\alpha)^n=\beta^n. $$ In other words, while proving "$\Rightarrow$" you are allowed to assume that $\psi$ is an isomorphism. By definition it satisfies $\psi(\alpha)=\beta$.
H: Calculating $\lim_{n \to +\infty}\int_0^1 (n + 1)x^{n}(1 - x^3)^{1/5}\,dx$ This question is from a bank of past master's exams. I have been asked to evaluate $$\lim_{n \to +\infty}\int_0^1 (n + 1)x^{n}(1 - x^3)^{1/5}\,dx.$$ I did this problem in a hurried manner, but here's what I think. Since $x^n$ is decreasing in $n$ for fixed $x$ in the closed unit interval, it seems like the integrand, which we may denote by $f_n$, converges pointwise to zero. If I can show that the integrand in fact converges uniformly to zero, by showing $$M_n = \sup_{x\in[0,1]}\|f_n(x)\|\rightarrow 0,$$ then the question is simply a matter of commuting the limit with the integral. Now, $f_n(x)$ is continuous and differentiable on $[0 , 1]$, so it achieves its supremum, which can be found by differentiating and finding the critical points. I found this critical point to be $x=(\frac{5n}{5n+3})^{1/3}$. The denominator exceeds the numerator in this expression for all $n$, so the critical point is between 0 and 1. It also seems clear to me that this is a local maximum. At this point, $f_n$ achieves the value $$M_n = (n+1)(\frac{5n}{5n+3})^{n/3}(1 - \frac{5n}{5n+3})^{1/5}$$ which goes to infinity. So, my intuition failed me at some point. What is the proper solution to this limit? More importantly, is there a better approach to this type of problem? AI: First, integrate by parts: $$\small I_n:=\int_0^1(n+1)x^n(1-x^3)^{1/5}dx=[x^{n+1}(1-x^3)^{1/5}]_0^1-\int_0^1x^{n+1}\frac 15(1-x^3)^{-4/5}(-3x^2)dx,$$ hence $$I_n=\frac 35\int_0^1x^{n+3}(1-x^3)^{-4/5}dx.$$ What you did before shows that now we have uniform convergence to $0$ of the integrand.
H: Confused about which Hölder spaces are Banach If $\Omega$ is an open set in $\mathbb{R}^n$, is the Hölder space $C^{k, \alpha}(\Omega)$ Banach? Or is it only that $C^{k, \alpha}(\overline{\Omega})$ is Banach, like with ordinary continuous functions? If not, why is that??? Norms are $$\|f\|_{C^{0,\alpha}} = \sup_{x,y \in \Omega ,\ x \neq y} \frac{\| f(x) - f(y) \|}{\|x-y\|^\alpha},$$ $$\|f\|_{C^{k, \alpha}} = \\|f\\|_{C^k}+\max_{\| \beta \| = k} \| D^\beta f \|_{C^{0,\alpha}}$$ where $$\|f\|_{C^k} = \max_{\| \beta \| \leq k} \, \sup_{x\in\Omega} \|D^\beta f (x)\|$$ Thanks for any help AI: Note that all $g\in C^{k,\alpha}(\Omega)$ and $\beta\in\mathbb{N}_0^k$ we have $$ \Vert \partial^\beta g\Vert_{C_b(\Omega)}\leq\Vert g\Vert_{C^{k,\alpha}(\Omega)}\tag{1} $$ $$ \|\partial^\beta g\|_{C^{0,\alpha}}\leq \Vert g\Vert_{C^{k,\alpha}(\Omega)}\tag{2} $$ Let $\{f_n:n\in\mathbb{N}\}$ be a Cauchy sequence in $C^{k,\alpha}(\Omega)$, then for all $\varepsilon>0$, we have $N\in\mathbb{N}$ such that $m,n>N$ implies $$ \Vert f_n-f_m\Vert_{C^{k,\alpha}(\Omega)}<\varepsilon\tag{3} $$ Fix multi-index $\beta\in\mathbb{N}_0^k$ such that $\|\beta\|\leq k$. From $(1)$ and $(3)$ it follows that for all $\varepsilon>0$, we have $N\in\mathbb{N}$ such that $m,n>N$ implies $\Vert \partial^\beta f_n-\partial^\beta f_m\Vert_{C_b(\Omega)}<\varepsilon$. This means that $\{\partial^\beta f_n: n\in\mathbb{N}\}$ is a Cauchy sequence in $C_b(\Omega)$. Since $C_b(\Omega)$ is complete, then $\{\partial^\beta f_n: n\in\mathbb{N}\}$ converges to some function in $C_b(\Omega)$. By $\varphi$ we denote limit of $\{ f_n:n\in\mathbb{N}\}$ in $C_b(\Omega)$. Now we use the following standard result Let $\{f_n:n\in\mathbb{N}\}\subset C(\Omega)$ be a family of differentiable functions, such that 1) the sequence $\{\partial_{x_i}f_n:n\in\mathbb{N}\}$ converges in $C_b(\Omega)$ to $g\in C_b(\Omega)$. 2) for some point $x\in\Omega$ the sequence $\{f_n(x_0):n\in\mathbb{N}\}$ is convergent Then $\{f_n:n\in\mathbb{N}\}$ converges in $C_b(\Omega)$ to some differentiable function $f\in C_b(\Omega)$, and moreover $\partial_{x_i} f=g$. Using this result and induction by multi-indices one can show that for all $\beta\in\mathbb{N}_0^k$ with $\|\beta\|\leq k$ the sequence $\{\partial^\beta f_n:n\in\mathbb{N}\}$ uniformly converges to $\partial^\beta \varphi$. This means that $$ \lim\limits_{n\to\infty}\Vert f_n-\varphi\Vert_{C^k(\Omega)}=0\tag{4} $$ From $(2)$, $(3)$ and deinition of $\|\cdot\|_{C^{0,\alpha}(\Omega)}$ it follows that for all $\beta\in\mathbb{N}_0^k$ with $\|\beta\|= k$ and all $x,y\in\Omega$ such that $x\neq y$ we have $$ \frac{\|\partial^\beta f_n(x) - \partial^\beta f_m(y)\|}{\|x-y\|^\alpha}<\varepsilon $$ Let's take $m\to\infty$ in this inequality, then we get $$ \frac{\|\partial^\beta f_n(x) - \partial^\beta \varphi(y)\|}{\|x-y\|^\alpha}<\varepsilon $$ Since $\beta\in\mathbb{N}_0^k$ with $\|\beta\|= k$ and $x,y\in\Omega$ such that $x\neq y$ are arbitrary we can say that $$ \max\limits_{\|\beta\|=k}\|\partial^\beta f_n-\partial^\beta \varphi\|_{C^{0,\alpha}(\Omega)}<\varepsilon $$ Thus for all $\varepsilon>0$ we have $N\in\mathbb{N}$ such that $n>N$ implies $$ \max\limits_{\|\beta\|=k}\|\partial^\beta f_n-\partial^\beta \varphi\|_{C^{0,\alpha}(\Omega)}<\varepsilon $$ This means that $$ \lim\limits_{n\to\infty}\max\limits_{\|\beta\|=k}\|\partial^\beta f_n-\partial^\beta \varphi\|_{C^{0,\alpha}(\Omega)}=0\tag{5} $$ From $(4)$ and $(5)$ it follows that $\{f_n:n\in\mathbb{N}\}$ converges to $\varphi$ in $C^{k,\alpha}(\Omega)$. Since we showed that arbitrary Cauchy sequence in $C^{k,\alpha}(\Omega)$ is convergent, then $C^{k,\alpha}(\Omega)$ is complete. Proof of the completeness of $C^{k,\alpha}(\overline{\Omega})$ is already discussed in comments.
H: Uniform continuity of a function transforming Cauchy sequences into Cauchy sequences Problem: $f:(0,\infty )\rightarrow \mathbb{R}$ defined as: $f(x)=x^{2}$ Can anyone show me how to prove that $f$ transforms Cauchy sequences of elements of $(0,\infty )$ into Cauchy sequences, but $f$ is not uniformly continuous? The purpose of this problem is to show how essential the boundedness of the set on which $f$ is defined is. $f$ is definitely not uniformly continuous because for the two sequences: $\left \{ x_{n} \right \},\left \{ y_{n} \right \}$ defined by: $x_{n}=n+\frac{1}{n}$ and $y_{n}=n$. We have: $\left \| x_{n}-y_{n} \right \| \to 0$ as $n \to \infty $, but $\left \| f(x_{n})-f(y_{n}) \right \| \to 2$ as $(n \to \infty )$. Can anyone, please, show me how to prove that $f$ transforms Cauchy sequences of elements of $(0,\infty )$ into Cauchy sequences? Thanks AI: Let $\{x_n\}$ a Cauchy sequence. It's bounded, since for $n,m$ large enough, say, $n,m\geq n_0$ we have $\|x_n-x_m\|\leq 1$ hence $\|x_n\|\leq \max\{1,\|x_1\|,\dots,\|x_n\|\}=:M$. We have for $k,j$ integers that $$\|x_k^2-x_j^2\|=\|x_k-x_j\|\|x_k+x_j\|\leq \|x_k-x_j\|(\|x_k\|+\|x_j\|)\leq 2M\|x_k-x_j\|,$$ and since $M$ doesn't depend on $j$ or $k$, the sequence $\{x_k^2\}$ is Cauchy. The statement: If $f\colon \Bbb R\to \Bbb R$ is uniformly continuous and $\{x_k\}$ is Cauchy, then $\{f(x_k)\}$ is Cauchy. is true, and this counter-example shows that the converse doesn't hold.
H: Question on smoothness of the projective closure of a smooth variety I was taught in the previous thread "Is the projective closure of a smooth variety still smooth?", that the projective closure $\bar X\subseteq \mathbb{P}^n$ of a smooth closed subscheme $X\subseteq \mathbb{A}^n$ (over a basefield) needs not to be smooth again. (-1-) I wonder, if one can test the smoothness of $\bar X$ on the closed ''complement'' $\mathbb{P}^{n-1}$ of $\mathbb{A}^n$ in $\mathbb{P^n}$: Is $\bar X$ smooth if $X$ and $\bar X\cap \mathbb{P}^{n-1}$ are smooth? If there is a counterexample, it would be nice to have an irreducible and reduced one, a good old variety. In the thread ''Example of a non-smooth intersection'', I was taught that the converse is not true: If $\bar X$ is smooth, $\bar X\cap \mathbb{P}^{n-1}$ does not have to be, even if $\bar X\cap \mathbb{P}^{n-1}$ is reduced. (-2-) Let $\bar X$ be smooth and suppose that $\bar X\cap \mathbb{P}^{n-1}$ is reduced. Is it true that in this case all the irreducible components of $\bar X\cap \mathbb{P}^{n-1}$ are smooth? AI: Yes, if $X$ is smooth and $\bar X\cap \mathbb P^{n-1}$ is smooth in the scheme theoretic sense, then $\bar X$ is indeed smooth. In other words if $s\in \bar X$ is singular, so is $s\in \bar X\cap H $ for any hyperplane $s\in H\subset \mathbb P^{n}$ . Beware however that you have to take scheme-theoretic intersections. For example if $\bar X$ is the curve $z^2y=x^3$ in $\mathbb P^{2} $, then that curve is smooth in the affine plane $z\neq1$ and its intersection with the line at infinity $H$ given by $z=0$ is the single point $s=[0:1:0]$. So, naïvely one might think that since a single point is a smooth variety one may conclude by the above that $\bar X$ is smooth. In reality $\bar X$ has a singularity at $s$. The mistake was to not see that the intersection of $\bar X$ with $H$ is not the reduced point $s$ but $s$ with a nilpotent structure. Edit: detailed calculation Indeed the intersection of $\bar X$ with the line at infinity $H$ is best computed in the affine plane $\mathbb A^2_{x,z}=Spec (k[x,z])$ (=the points of $\mathbb P^{2}$ where we may choose $y=1$), whose coordinates are $(x,z)=[x:1:z]$. The point $s$ then has coordinates $x=0,z=0$, $\bar X$ has equation $z^2=x^3$, $H$ has equation $z=0$ and the ideal in $k[x,z]$ of the intersection $\bar X \cap H$ is $(z^2-x^3,z)=(z,x^3)$. So the intersection $\bar X \cap H$ is the affine subscheme $Spec (k[x,z]/(z,x^3)) \subset \mathbb A^2_{x,z}$, which is clearly isomorphic to $A=Spec(k[x]/(x^3)$, hence non reduced and thus singular. Geometrically $\bar X$ is a cusp cut by every projective line in three points. The line at infinity however cuts it in what was classically known as a "triple point" before the introduction of schemes. Second edit Let me add a few words to the second sentence of my answer in order to address Daniel's comment. The problem is local at $s$, so we may assume that $s\in X$ is a singularity and we must show that $X\cap H $ is singular at $s$ too. For simplicity assume that $X$ is a hypersurface. It thus has equation $f(x_1,...,x_n)=0$, with $f(x_1,...,x_n)=q_2(x_1,...,x_n)+q_3(x_1,...,x_n)+...$ where $q_i(x_1,...,x_n)$ is homogeneous of degree $i$. The crucial point is that there is no linear term $q_1$: this is equivalent to $X$ being singular at $s$. If the hyperplane $H$ has equation $x_n=l(x_1,...,x_{n-1})$ ( $l$ linear) the intersection $X\cap H$ is given by $q_2(x_1,...,l(x_1,...,x_{n-1})+q_3(x_1,...,l(x_1,...,x_{n-1}))+...=0$ in the affine coordinates $x_1,...,x_{n-1}$ and is thus also singular since it begins with a quadratic term.
H: Limit of a function as accumulation point In the context of this answer to another question about representing I thought of the following possible description of the limit of a function: $\lim_{x\to a}f(x)=y$ iff $(a,y)$ is an accumulation point of $f$ (interpreted as a set of pairs) and there's no $y′≠y$ so that $(a,y′)$ is also an accumulation point of $f$. Now my question is: Is that claim correct? AI: This would work in a compact space, in which values $y'$ that don't tend to the limit have to have some other accumulation point. However, if the space is not compact, there need no be a second accumulation point. For instance, the function $$ f(x)=\begin{cases}1/x&1/x\in\mathbb N\\0&\text{otherwise}\end{cases} $$ has no limit for $x\to0$, but according to your description it would, since $(0,0)$ is the only accumulation point of the set of pairs.
H: What kind of "mathematical object" are limits? When learning mathematics I tend to try to reduce all the concepts I come across to some matter of interaction between sets and functions (or if necessary the more general Relation) on them. Possibly with some extra axioms thrown in here and there if needed, but the fundamental idea is that of adding additional structure on sets and relations between them. I've recently tried applying this view to calculus and have been running into some confusions. Most importantly I'm not sure how to interpret Limits. I've considered viewing them as a function that takes 3 arguments, a function, the function's domain and some value (the "approaches value") then outputs a single value. However this "limit function" view requires defining the limit function over something other then the Reals or Complexes due to the notion of certain inputs and outputs being "infinity". This makes me uncomfortable and question whether my current approach to mathematics is really as elegant as I'd thought. Is this a reasonable approach to answering the question of what limits actually "are" in a general mathematical sense? How do mathematicians tend to categorize limits with the rest of mathematics? AI: Do you by any chance have a computer science background? Your ideal of reducing everything (even operations like limits) to function and sets has a flavor of wanting mathematics to work more or less like a programming language -- this is a flavor that I (being a computer scientist) quite approve of, but you should be aware that the ideal is not quite aligned with how real mathematicians write mathematics. First, even though everything can be reduced to sets and functions -- indeed, everything can be reduced to sets alone, with functions just being sets of a particular shape -- doing so is not necessarily a good way to think about everything all of the time. Reducing everything to set theory is the "assembly language" of mathematics, and while it will certainly make you a better mathematician to know how this reduction works, it is not the level of abstraction you'll want to do most of your daily work at. In contrast to the "untyped" assembly-level set theory, the day-to-day symbol language of mathematics is a highly typed language. The "types" are mostly left implicit in writing (which can be frustrating for students whose temperament lean more towards the explicit typing of most typed computer languages), but they are supremely important in practice -- almost every notation in mathematics has dozens or hundreds of different meanings, between which the reader must choose based on what the types of its various sub-expressions are. (Think "rampant use of overloading" from a programming-language perspective). Mostly, we're all trained to do this disambiguation unconsciously. In most cases, of course, the various meanings of a symbol are generalizations of each other to various degrees. This makes it a particular bad idea to train oneself to think of the symbol of denoting this or that particular function with such-and-such particular arguments and result. A fuzzier understanding of the intention behind the symbol will often make it easier to guess which definition it's being used with in a new setting, which makes learning new material easier (even though actual proofwork of course needs to be based on exact, explicit definitions). In particular, even restricting our attention to real analysis, the various kinds of limits (for $x\to a$, $x\to \infty$, one-sided limits and so forth) are all notated with the same $\lim$ symbols, but they are technically different things. Viewing $\lim_{x\to 5}f(x)$ and $\lim_{x\to\infty} f(x)$ as instances of the same joint "limit" function is technically possible, but also clumsy and (more importantly) not even particularly enlightening. It is better to think of the various limits as a loose grouping of intuitively similar but technically separate concepts. This is not to say that there's not interesting mathematics to be made from studying ways in which the intuitive similarity between the different kind of limits can be formalized, producing some general notion of limit that has the ordinary limits as special cases. (One solution here is to say that the "$x\to \cdots$" subscript names a variable to bind while also denoting a net to take the limit over). All I'm saying is that such a general super-limit concept is not something one ought to think of when doing ordinary real analysis. Finally (not related to your question about limits), note that the usual mathematical language makes extensive use of abstract types. The reals themselves are a good example: it is possible to give an explicit construction of the real numbers in terms of sets and functions (and every student of mathematics deserves to know how), but in actual mathematical reasoning numbers such as $\pi$ or $2.6$ are not sets or functions, but a separate sort of things that can only be used in the ways explicitly allowed for real numbers. "Under the hood" one might consider $\pi$ to "really be" a certain set of functions between various other sets, but that is an implementation detail that is relevant only at the untyped set-theory level. (Of course, the various similarities between math and programming languages I go on about here are not coincidences. They arose from programming-language design as deliberate attempts to create formal machine-readable notations that would "look and feel" as much like ordinary mathematical symbolism as they could be made to. Mathematics had all of these things first; computer science was just first to need to name them).
H: Fourier transform of a measure I'm a bit confused - How is the Fourier transform of a measure on a compact abelian group defined? specifically the Fourier transform of a measure on $\mathbb{T}$ the unit circle in the complex plain. AI: If $\mu$ is a measure on the compact abelian group $G$ and $\gamma$ is in the dual group, $$\hat{\mu}(\gamma) = \int_G (-g, \gamma)\ d\mu(g)$$ In the case ${\mathbb T}$, the dual group is $\mathbb Z$, acting on $\mathbb T$ by $(n, \omega) = \omega^n$.
H: What is the number of all possible values of $[Z^{6}]$? Its given that $$[Z]=3$$ $$[Z^{2}]=11$$ $$[Z^{3}]=41$$ Then, what is the number of all possible values of $[Z^{6}]$ where $[\;\cdot\;]$ is floor function. AI: Hint: $$ [Z^k]=b\quad\Longleftrightarrow\quad b\leq Z^k<b+1 \quad\Longleftrightarrow\quad b^{1/k}\leq Z<(b+1)^{1/k} $$
H: Finite modules over finite local rings Let $R$ be a finite commutative local ring with identity. If $M$ is a finite $R$-module it is necessarily projective? AI: Let $p$ be a prime number. Let $R = \mathbb{Z}/p^2\mathbb{Z}$. Let $M = R/pR$. Since the number of elements of $M$ is $p$, $M$ cannot be free. Hence $M$ cannot be projective.
H: Product of right cosets equals right coset implies normality of subgroup I cannot see how to find a way to prove that if $H$ is a subgroup of $G$ such that the product of two right cosets of $H$ is also a right coset of $H,$ then $H$ is normal in $G.$ (This is from Herstein by the way.) Thank you. AI: Hint: if $HaHa^{-1}$ and $Ha^{-1}Ha$ are right cosets they must be $H$ because they contain the identity. (I have updated my hint to involve both $HaHa^{-1}$ and $Ha^{-1}Ha$ because $aHa^{-1}\subseteq H$ is not by itself equivalent to $aHa^{-1}=H$ when $H$ is infinite; see counterexamples here, here, here.)
H: Greatest common denominator of measurements In a couple months, I'll do the Millikan experiment. Then, I'll end up with a number of charge measurements and their errors $$((q_i, \Delta q_i))_{i \in \mathbb N}.$$ The idea is that all those $q_i$ can be represented as a multiple of a fixed elementary charge $e$ like $$q_i = n_i e.$$ Now I do not know $e$ in advance. Is there some method to get $(e, \Delta e)$ out of the sequence of measurements? I found this question on StackOverflow, but it does not have any answers that solve the problem in a stable way. AI: This is how you would do it if there were no experimental error: Let $q_0$ be your smallest measured value for the charge and enumerate the other charges $q_1,\dotsc,q_n$. By assumption, $q_k=en_k$ for some positive real $e$ and $n_k$ an integer. Thus, $\frac{q_k}{q_0}$ is rational for all $k$. Write all of these rationals is reduced form and let $\ell$ be the least common multiple of all the denominators. Take $e=q_0/\ell$. Then, by construction, $q_k/e$ is an integer for all $k$ and $e$ is the smallest positive real number such that this is true. The problem with this, however, is that, when you calculate $\frac{q_k}{q_0}$, you are only getting an estimate, and so $\ell$ is going to be completely off. In practice, I think you just have to hope that $\frac{q_k}{q_0}$ is (approximately) an integer for all $k$, and if that's not the case, go back to the lab and take more data until this is the case. Of course, as pointed out, if $e$ works, so will $e/2$, and $e/3$, etc. The best you will ever be able to do here is calculate the probability that you measured your value of $e$ (call it $e_0$) was indeed the correct value of $e$ under the assumption the true value of $e$ is $e_0/2$, $e_0/3$, etc.
H: How to calculate the degree of this Gauss map? In reviewing the familiar Poincare-Hopf theorem I come across the following question: Suppose $x$ an isolated 0 of $V$. Pick up a disk around $x$ in its neighborhood. Calculate the degree of the map $$u:\partial D \rightarrow S^{m-1},u(z)=\frac{V(z)}{\|V(z)\|}$$ where $V(z)$ is the the map $M\rightarrow TM$ which represents a vector field. I am confused because while in case 1 and 2 it is clear the degree $S^{1}\rightarrow S^{1}$ must be 1 since they are orientation preserving maps combined with rotation, I do not know how to calculate case 3. The author claimed the index is -1. But how? A hint would be mostly welcome. AI: Draw a circle around the $0$. Then, at each point of this circle there will be a vector. As you travel around the circle counterclockwise, the direction of the vector at that point will change. The number of times this vector rotates about the origin, counting orientation, is the degree of the Gauss map. Try keeping track of the vector in case two as you travel around the $0$. You should notice that the vector rotates clockwise exactly once, and hence the degree is $-1$.
H: Dedekind Domains and Affine Varieties Let $k$ be an algebraically closed field, and let $B$ be a finitely generated $k$ algebra that is also a Domain. Then $B$ is the affine coordinate ring of some affine variety $Y$; this part is straight out of Hartshorne and is not terribly difficult to understand. However, I am having trouble making sense of the following extension of this statement. If $B$ is Dedekind, then $Y$ is both non-singular and has dimension 1. This shows up in Hartshorne (page 41, Lemma 6.5, last paragraph). AI: Let $P$ be a non-zero prime ideal of $B$. Since $P$ is a maximal ideal, ht($P$) = 1. Hence dim $B$ = 1. It remains to prove that $B_P$ is regular. Since $B_P$ is Noetherian, integrally closed and dim $B_P = 1$, $B_P$ is a discrete valuation ring(e.g. Atiyah-MacDonald). Hence $B_P$ is regular.
H: For what values of $q$ would $3n - q^2 \equiv 0\pmod{4q}$? For what values of $q$ would $3n - q^2 \equiv 0\pmod{4q}$, or for what values of $q$ would $3n - q^2$ divide by $4q$ and leave no remainder, where $n$ is a positive integer and $q$ is a positive divisor of $n$. AI: If $n = q k$, $3n - q^2 = (3k - q) q$. So you need $3k - q$ to be divisible by $4$, i.e. $q \equiv -k \mod 4$. Now since $q k = n$, that says $n \equiv -q^2 \mod 4$. The squares mod $4$ are $0$ and $1$, so we have the following possibilities: if $n \equiv 1$ or $2 \mod 4$, it is impossible if $n \equiv 0 \mod 4$, $q \equiv n/q \equiv 0$ or $2 \mod 4$ (thus either $n \equiv 0 \mod 16$ or $n \equiv 4 \mod 8$). if $n \equiv 3 \mod 4$, it is true for any $q$ dividing $n$.
H: Definition of quadratic equation? What is a quadratic equation and what is its simplified and cannonic form? AI: A quadratic equation (in one variable) is a polynomial equation $P(x)=0$, where $P(x)$ is a polynomial of degree 2. The canonical form of a quadratic equation is $ax^2+bx+c=0$. The simplified form I'm not sure about, but I'm guessing $x^2+bx+c=0$ (obtained from canonical by dividing both sides by the coefficient of $x^2$).
H: Why is there no foliations of the 2-sphere, or a genus two surface? I'm trying to see why there is no (one-dimensional) foliation of $S^2$ or an orientable surface of genus two. Originally I was thinking that such a foliation could give me a non-vanishing vector field, which would be a contradiction, but now I have learned that line fields don't necessary lift to vector fields. Is it still something that depends on Euler characteristic $0$ so that the torus is the only orientable surface with a foliation? AI: A line field $L$ on a manifold $M$ need not come from a vector field. The trouble is that you might not be able to orient the lines in the various tangent spaces in a coherent way. However, if $L$ is not orientable, then there exists a $2$-fold cover $\tilde{M}$ of $M$ and a lift $\tilde{L}$ of $L$ to to $\tilde{M}$ which is orientable. This implies that $\tilde{M}$ has a nonvanishing vector field, and thus has Euler characteristic $0$. Since Euler characteristic is multiplicative in covers, this implies that $M$ has Euler characteristic $0$. In response to your question on Chris Gerig's answer : A compact manifold of any dimension supports a $1$-dimensional foliation if and only if its Euler characteristic is $0$. In fact, it has a nonvanishing vector field. Indeed, choose a vector field with isolated zeros. Poincare-Hopf tells you that the signs of the zeros add to $0$. It is then a fun exercise to see that you can "move the zeros together and make them collide and cancel" to get a nonvanishing vector field. Something even more amazing is true. It is true that an $n$-manifold that supports an $(n-1)$-foliation has Euler characteristic $0$ (choose a metric, and after passing to a double cover if necessary, you can find a unit vector field which is orthogonal to the foliation). Shockingly, the converse is also true! This is a very deep theorem of Thurston from the 1970's (before he got interested in $3$-manifolds).
H: Testing if a geometric series converges by taking limit to infinity If the limit as n approaches infinity of a geometric series is not zero, then that means the series diverges. This makes intuitive sense to me, because it is an infinite series and we keep adding nonzero terms, it will go to infinity. However, if the limit as n approaches infinity does equal zero, series that is not enough information to tell whether the series converges or diverges. I would think that if the limit as n approaches infinity is zero, and the series is a continuous function, then the series would converge to some real number. Why is this not the case? The only counter-example I can think of is a series like: a^n = { (-1)^n }, but that is not a continuous function. AI: $\lim_{n \rightarrow \infty} \frac{1}{n} = 0$ but $\sum_{n = 1}^\infty \frac{1}{n}$ does not converge. The associated function $\frac{1}{x}$ is continuous from $[1, \infty)$. Also a geometric series is the series associated with the sequence $a_n = pr^n$ for some $r$ and $p$. Geometric series converges if $\|r\| < 1$.
H: How do I proceed with these quadratic equations? The question is $$ax^2 + bx + c=0 $$ and $$cx^2+bx+a=0$$ have a common root, if $b≠ a+c$, then what is $$a^3+b^3+c^3$$ AI: The value of $a^3+b^3+c^3$ is not determined. Just choose $a=c$. But leaving out the condition $a\ne c$ is probably an oversight, so assume from now on that $a\ne c$. If $q$ is a common root, then $aq^2+bq+c=cq^2+bq+a=0$. Subtracting, we find that $(a-c)q^2-(a-c)=0$. Since $a\ne c$, we get $q^2=1$. We cannot have $q=-1$, for that implies that $b=a+c$. We are left with $q=1$, which gives $a+b+c=0$. Conversely, if $a+b+c=0$, then $1$ is a common root of the two equations. That still leaves many possibilities for the value of $a^3+b^3+c^3$. For example, let $a=1$, $b=-3$, $c=2$. Then $a^3+b^3+c^3=-18$. Let $a=1$, $b=-4$, $c=2$. Then $a^3+b^3+c^3=-36$. However, we can say something interesting about $a^3+b^3+c^3$ in the case $a\ne c$. Use the general identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-bc-ca-ab),\tag{$1$}$$ which can be verified by multiplying out. Since in our case $a+b+c=0$, we conclude that $$a^3+b^3+c^3=3abc.$$ Perhaps this is the intended answer.
H: Writing inequalities in interval notation? When writing the solution to an inequality in interval notation, say $x < 5$ as $(-\infty, 5)$, how can the $x$ be "involved"? Is it correct to just write $(-\infty, 5)$, or should it be $x=(-\infty, 5)$, or perhaps $x\in(-\infty, 5)$? AI: One defines the interval $(-\infty, a)$ precisely as $\{x:x<a\}$. So saying $$x\in (-\infty,a)$$ is saying the same as saying $$x<a$$ Note that writing $x=(-\infty, a)$ would only by correct if you're saying "All elements of $x$ are solution" (viz, $x$ would be denoting an set, not a real number). It is usual to write $S=(-\infty,a)$, where $S$ stand for "solution set", that is, all elements of $S$ are solutions.
H: Hellinger-Toeplitz theorem use principle of uniform boundedness Suppose $T$ is an everywhere defined linear map from a Hilbert space $\mathcal{H}$ to itself. Suppose $T$ is also symmetric so that $\langle Tx,y\rangle=\langle x,Ty\rangle$ for all $x,y\in\mathcal{H}$. Prove that $T$ is a bounded directly from the uniform boundedness principle and not the closed graph theorem. This is problem III.13 in the Reed-Simon volume 1. Hints are welcome. AI: Hint: Consider the family of linear functionals $f_x$ defined by $f_x(y) = \langle Tx,y \rangle$, as $x$ ranges over the unit ball of $\mathcal{H}$.
H: Sequences from the positive integer numbers $(a,b,c,d,e)$ How to find how many sequences from the positive integer numbers $(a,b,c,d,e)$,such that : $$abcde \le a+b+c+d+e \le 10$$ AI: I would try to solve this logically. If 10 is the maximum number of a + .... + e, then that means the numbers must be limited to 6. Now let's find out which sets of numbers excluding 1 gives products less than 10: 2,2 2,3 2,4 \|\| 2,2,2 2,5 3,3 Any other sets of numbers multiplied will give values greater than 10 for abcde. So you can try (1,1,1,2,2), (1,1,1,2,3), (1,1,1,2,4), (1,1,1,2,5), (1,1,1,3,3), and lastly (1,1,2,2,2). We can't forget single numbers by themselves either. (1,1,1,1,1), (1,1,1,1,2), (1,1,1,1,3), (1,1,1,1,4), (1,1,1,1,5), and (1,1,1,1,6). Now make sure the sums are all greater than the product. 1,1,1,2,2:4:7-good 1,1,1,2,3:6:8-good 1,1,1,2,4:8:9-good 1,1,1,2,5:10:10-good 1,1,1,3,3:9:9-good 1,1,2,2,2:8:8-good and obviously, all of the (1,1,1,1,x) sets also work. So now we have to make combinations of each type of set, since (1,1,1,1,2) is different than (1,1,1,2,1). For sets in the form of (a,a,a,a,a), there are 5!/5! = 1 combinations. For sets in the form of (a,a,a,a,b), there are 5!/4! = 5 combinations. For sets in the form of (a,a,a,b,b), there are 5!/(3!2!) = 10 combinations. For sets in the form of (a,a,a,b,c), there are 5!/3! = 20 combinations. There are 1 set of the first type, 5 sets of the second type, 3 sets of the third type, and 3 sets of the last type. This leaves us with 11 + 55 + 310 + 3*20 = 1 + 25 + 30 + 60 = 116.
H: $x$-axis is meager set on $\mathbb{R}^2$ Subset $A$ of metric space $X$ is meager on $X$, iff $\text{IntCl}A=\emptyset$. But, why $x$-axis is meager set on $\mathbb{R}^2$? My attempt (please don't kill me): $\text{IntCl}\mathbb{R}=\text{Int}\mathbb{R}=\mathbb{R}\neq \emptyset$ Thank you! AI: The interior of $\mathbb{R}$ as a subset of $\mathbb{R}^2$ is very different from its interior as a standalone topological space. Given any point $P$ on the $x$-axis, every open neighbourhood of $P$ contains a point not on the $x$-axis. So $P$ is not in the interior of the $x$-axis. To put it another way, every point of the $x$-axis is on the boundary of the $x$-axis.
H: Oscillation of $f$ (at $x_0$) equals zero iff $f$ is continuous. I'm reading a proof of that claim mentioned in the title and I have some difficulties understanding it. Statement: $$O_f(x_0)=0 \iff f \ \ \text{is continuous on} \ \ x_0 $$ (Where oscillation, $O_f(x):= \displaystyle\lim_{n\to\infty}\text{diam}f(D(x,\frac{1}{n}) $ and where $D(x,\frac{1}{n})$ is an open ball with center $x$ and radius $\frac{1}{n}$) Proof: Suppose that $O_f(x_o)=0$. Let $V$ be some neighborhood of $f(x_0)$. Exist $\epsilon>0$ which for him $D(f(x_0,\epsilon))\subseteq V$. From $O_f(x_o)=0$, we know that there is $k\in\mathbb{N}$, that from that $k$ and on, $\text{diam}f(D(x,\frac{1}{n}))<\epsilon$ we choose $n>k$. For all $x\in D(x_0,\frac{1}{n})$ we have: $$d(f(x),f(x_0))\leq \text{diam}f(D(x,\frac{1}{n})<\epsilon$$ and therefor(which I can't understand): $$f(D(x_0,\frac{1}{n}))\subseteq D(f(x_0),\epsilon)$$ That last claim completes the proof(of first direction) So, again -- how from $d(f(x),f(x_0))\leq \text{diam}f(D(x,\frac{1}{n})<\epsilon$ implies that $f(D(x_0,\frac{1}{n}))\subseteq D(f(x_0),\epsilon)$? Thank you! AI: The diameter of a set in a metric space is the supremum of the distances between all its points, so in your definition of oscillation we have this: $$\mathcal O_f(x_0)=\lim_{n\to\infty}\sup_{x,y\in D\left(x_0,1/n\right)}d\left(f(x),f(y)\right)$$ and since the following inequality is true $$d\left(f(x),f(x_0)\right)\leq \textrm{diam} f\left(D\left(x_0,\frac{1}{n}\right)\right)$$ it also holds for the supremum, from which the inclusion you ask about follows. Added to answer the OP's doubt below: Because we took any $\,x\in D(x_0,1/n)\,$ and we got $$d(f(x),f(x_0))\leq \text{diam} f(D(x_0,1/n))<\epsilon\Longrightarrow$$ and this means $\,f\,$ maps any element of $D(x_0,1/n)\,$ , namely $\,x\,$ , to the element $\,f(x)\,$ which is within a distance of $\,\epsilon\,$ from $\,f(x_0)\,$ , i.e. to the open ball $\,D(f(x_0),\epsilon)\,$...! In other words, all the elements in $\,D(x_0,1/n)\,$ are mapped by $\,f\,$ to elements within a distance of $\,\epsilon\,$ from $\,f(x_0)\,$ , i.e. are mapped into the ball $\,D(f(x_0),\epsilon)\,$
H: Two corollaries in Lang's Algebraic Number Theory. I'm having difficulty understanding the relationship between two corollaries in Lang's Algebraic Number Theory, on page 16 for those with the book. They can also be found in his Algebra. The first is: Let $A$ be a ring integrally closed in its quotient field $K.$ Let $L$ be a finite Galois extension of $K$ and let $B$ be the integral closure of $A$ in $L.$ Let $\mathfrak{p}$ be a maximal ideal of $A.$ Let $\varphi:A\rightarrow A/\mathfrak{p}$ be the canonical map, and let $\psi_1,\psi_2$ be two homomorphisms of $B$ extending $\varphi$ in an algebraic closure of $A/\mathfrak{p}.$ Then, there exists and automorphism $\sigma$ of $L$ over $K$ such that $\psi_1=\psi_2\circ\sigma.$ The second, under the same assumptions: Let $\mathfrak{B}$ be the only prime of $B$ lying above $\mathfrak{p}.$ Let $f(X)$ be a monic polynomial in $A[X].$ Assume that $f$ is irreducible in $K[X]$ and that it has a root $\alpha$ in $B.$ Then the reduced polynomial $\overline{f}$ is a power of an irreducible polynomial in $\overline{A}[X].$ Lang goes on to state that it follows from the first corollary that any two roots of $\overline{f}$ are conjugate under an isomorphism of $\overline{B}$ over $\overline{A},$ the reductions mod the the respective primes. I'm not sure how this follows immediately. Also, it seems to me that this could be deduced from the fact that the Galois group of $L$ over $K$ acts transitively on the roots of $f,$ and under these assumptions all such mappings descend to mappings on $\overline{B}$ over $\overline{A}.$ Could someone help me see if my reasoning is off, and also explain the relationship between these two corollaries? If it's helpful, these are corollaries to the proposition that the decomposition group surjects onto the Galois group of $\overline{B}$ over $\overline{A}.$ AI: Proposition Let $A$ be an integral domain. Let $K$ be the field of fractions of $A$. Let $L$ be an algebraic(not necessarily finite) extension of $K$. Let $B$ be the integral closure of $A$ in $L$. Let $\mathfrak{p}$ be a maximal ideal of $A$. Let $\bar A = A/\mathfrak{p}$. Assume that $\mathfrak{P}$ is the only prime ideal of $B$ lying over $\mathfrak{p}$. Let $f(X) \in A[X]$ be a monic polynomial. Let $\bar f(X) \in \bar A[X]$ be the reduction of $f(X)$ mod $\mathfrak{p}$. Assume that $f(X)$ is irreducible in $K[X]$ and has a root $\alpha$ in $B$. Then $\bar f(X)$ is a power of an irreducible polynomial in $\bar A[X]$. Proof: Let $\varphi$ be the canonical homomorophism $A \rightarrow \bar A$. Let $\Omega$ be the algebraic closure of $\bar A$. Let $\omega_1$ be a root of $\bar f(X)$ in $\Omega$. Assume $g(X) \in A[X]$ and $g(\alpha) = 0$. Since $f(X)$ is irreducible in $K[X]$, there exists $h(X) \in K[X]$ such that $g(X) = f(X)h(X)$. Since $f(X)$ is monic, $h(X) \in A[X]$. Hence $\bar g(X) = \bar f(X) \bar h(X)$. Hence $\bar g(\omega_1) = 0$. Therefore there exists a homomorphism $\psi_1:A[\alpha] \rightarrow \Omega$ extending $\varphi$ such that $\psi_1(\alpha) = \omega_1$. Let $P_1$ be the kernel of $\psi_1$. By the lying-over theorem, there exists a prime ideal $Q_1$ of $B$ lying over $P_1$. By the assumption, $Q_1 = \mathfrak{P}$. Since $B$ is integral over $A$, $\mathfrak{P}$ is a maximal ideal and $B/\mathfrak{P}$ is algebraic over $\bar A$. Hence $B/\mathfrak{P}$ is isomorphic to a subfield of $\Omega$. Hence $\psi_1$ can be extended to a homomorphis $\Psi_1:B \rightarrow \Omega$. Let $\omega_2$ be another root of $\bar f(X)$ in $\Omega$. By the similar argument as above, there exists a homomorphis $\Psi_2:B \rightarrow \Omega$ extending $\varphi$ such that $\Psi_2(\alpha) = \omega_2$. Since Ker($\Psi_1$) = Ker($\Psi_2$), there exists an isomorphism $\sigma:\Psi_1(B) \rightarrow \Psi_2(B)$ over $\bar A$ such that $\Psi_2 = \sigma\Psi_1$. Hence $\omega_1$ and $\omega_2$ are conjugate over $\bar A$. Hence $\bar f(X)$ is a power of an irreducible polynomial in $\bar A[X]$. QED
H: How to resize an image? I am not sure about the title of this question, so if someone knows an appropriate one, please rename it. It's a programming related question (but doesn't involve any programming). I posted it on stack overflow but didn't get any responses so I am trying here. I need to map a piece of rectangle (x, y, width, height) of an (original)image onto a canvas by resizing the original image. Here's a picture that explain it better. Here's another one, bit different, so you get a better idea: I can scale and move the image however I want. How do i figure out the scale at which to resize the original image. I had an idea, which was to figure the bigger dimension of the given rectangle, and use that to figure out the scale, so: Given: imageWidth, imageHeight, rectWidth, rectHeight, canvasWidth, canvasHeight (let's ignore the offsets for now) But that doesn't work in some cases. So I wondering what's the best way to do this. , AI: If I understand your question correctly then you need to do this: scaleVert:=(rectHeight/rectWidth > canvasHeight/canvasWidth); if scaleVert then scale:=canvasHeight/rectHeight else slace:=canvasWidth/rectWidth; The value of scale is ratio by which you have to scale your image. The Boolean variable scaleVert is used to test whether you have to fit your rectangle vertically (i.e., the height of rectangle after scaling will be the same as the height of the canvas) or horizontally. Notice that if rectHeight/rectWidth = canvasHeight/canvasWidth then it does not matter whether you scale vertically or horizontally - you obtain the same result. (The two rectangles are similar.) If the rectangle is "thinner" (i.e., the ratio rectHeight/rectWidth is higher) you have to scale vertically and otherwise you have to scale horizontally.

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