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H: How to rationalize multiple terms with fractional exponents I'm trying to derive the derivative of $f(x) = x^{2/3}$ using the limit definition: $$f'(x)=\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ $$=\lim_{h \to 0} \frac{(x+h)^{2/3} - x^{2/3}}{h}$$ I suspect I have to rationalize the numerator in order to cancel an $h$ from the numerator and denominator, but I'm not sure how to rationalize the numerator. I've tried multiplying by the conjugate and even tried to render the numerator a difference of cubes and then using $A^3 -B^3 = (A - B)(A^2 +AB + B^2)$ to rationalize, but to no avail. My two questions are: How do I rationalize the numerator? Is there a general formula for rationalizing multiple terms with rational exponents? Is there something I can read or study to learn more about this? AI: The numerator is a difference of cube-roots, not a difference of cubes. So rationalizing the numerator: $$ \frac{(x+h)^{2/3}-x^{2/3}}{h}= \frac{(x+h)^2-x^2}{h[(x+h)^{4/3}+x^{4/3}(x+h)^{2/3}+x^{4/3}]} $$ gives the factor of $h$ you want in the numerator. Hence \begin{align} \require{cancel} \lim_{h\to 0} \frac{(x+h)^{2/3}-x^{2/3}}{h} &=\lim_{h\to 0} \frac{\cancel{h}(2x+h)}{\cancel{h}[(x+h)^{4/3}+x^{4/3}(x+h)^{2/3}+x^{4/3}]}\\ &=\frac{2x}{3x^{4/3}}=\frac23 x^{-1/3}. \end{align} In general, $A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\dots+B^{n-1})$, so $$ \alpha^{m/n}-\beta^{m/n}=\frac{\alpha^m-\beta^m}{\alpha^{(n-1)m/n}+\alpha^{(n-2)m/n}\beta^{m/n}+\dots+\beta^{(n-1)m/n}} $$ which allows you to deal with the derivative of $x^{m/n}$.
H: Fatou's Lemma applied to simple functions Show that the sequence of measurable functions $f_i: \mathbb{R} \rightarrow \mathbb{R}$ defined via \begin{align}f_i(x)= \begin{array}{cc} \{ & \begin{array}{cc} -1 & i \leq x \leq i+1 \\ 0 & o.w. \\ \end{array} \end{array} \end{align} (o.w. stands for otherwise of course) satisfies lim inf $\int_{\mathbb{R}} f_i < \int_{\mathbb{R}}$ lim inf $f_i$ So I am approaching this using Fatou's Lemma. In order to apply Fatou's Lemma however I define \begin{align} g_i(x):=-f_i(x) \end{align} to ensure nonnegtivity. To invoke Fatou's Lemma do I use the limit of the $g_i(x)$? Then swap back in for the $f_i$ and the inequality switches but where do I get STRICT inequality? by applying Fatou I can only get a less than or equal to. Am I going about this the right way? Thanks in advance!! AI: Limit of $f_i(x)$ is $0$ at every point: for any $x$ the inequality $i\leq x\leq i+1$ fails after some stage. Hence RHS of the given inequality is $0$. Also $\int f_i(x)=-1$ by direct calculation. Hence LHS $=-1$. Since $-1 <0$ we are done.
H: If $\|a_n\| \to \|a\|$ and $\|\frac{a_n}{\|a_n\|}-\frac{a}{\|a\|}\|\to0$ can we conclude $a_n\to a$? If $\|a_n\| \to \|a\|$ and $\|\frac{a_n}{\|a_n\|}-\frac{a}{\|a\|}\|\to0$ can we conclude $a_n\to a$? I am not sure. I tried various algebraic manipulations but could not figure it out. This problem came up when trying to prove something, and I can prove it for normalized sequences, now i want to generalize it for all sequences. AI: (I'll assume that $a,a_n\neq 0$, as otherwise the second quantity makes no sense.) We see that $$\frac{x}{\|x\|}=\begin{cases}1&\text{if }x>0 \\ -1&\text{if }x<0.\end{cases}$$ So, $$\left\|\frac{a_n}{\|a_n\|}-\frac{a}{\|a\|}\right\|$$ is $0$ if $a$ and $a_n$ lie on the same side of $0$, and $2$ otherwise. Your statement that this goes to $0$ means that $a_n$ lies on the same side of $0$ as $a$ for large $n$; can you see why this helps?
H: Solve $ny(x)^2=\sqrt{1+y'(x)^2}$ and determine the range of $x$ where $y(x)$ is real-valued I have the following differential equation: $$ny(x)^2=\sqrt{1+y'(x)^2}$$ I know that $n$ is a real number, and that the intial condition is $y(a)=b$, where $a$ and $b$ are also real numbers. The questions I have are: What is the function $y(x)$? In what range of $x$ is the function $y(x)$ real-valued, and how does that range depend on $n$, $a$, and $b$? AI: HINT To start with, notice that \begin{align} ny^{2} = \sqrt{1+(y')^{2}} & \Longleftrightarrow y' = \pm\sqrt{n^{2}y^{4} - 1} \Longleftrightarrow \int\frac{\mathrm{d}y}{\sqrt{n^{2}y^{4} - 1}} = \pm\int 1\mathrm{d}x \end{align} Then WA gives the following result. Hopefully this helps.
H: Is there an ideal scoring term for determining two numbers? Let's say I have a known vector of two numbers: c(A,B) Is there a scoring term, or a combination of scoring terms, that can measure the unique closeness of a random vector c(a,b) to the known vector? In other words, is there a scoring term that can be used to determine that a is very close to A AND that b is very close to B? Heres an example of a scoring term, $S$, I thought of: $$S = \lvert a - A \rvert + \lvert b - B \rvert$$ It's not ideal because if our vector is c(1,5), then the vectors c(2,4) and c(0,4) would give the same scoring metric value of $2$. I would like the scoring metric to give me more information. Such as, for this example, $a$ is approaching $A$ from above for c(2,4)and $a$ is approaching $A$ from below for c(0,4). I would also be interested to see how a scoring term, or combination of scoring terms, could be generalized to a vector of length $n$. AI: A natural choice would be the Euclidean distance to $(A,B)$: $$c(a,b) = \sqrt{(a-A)^2+(b-B)^2}.$$ This generalizes to $n$ dimensions.
H: From 7 inputs and 1 output, approximate a possible function? I'm trying to approximate a car insurance quote algorithm/function: It takes 7 input variables that I can change (Vehicle Cost, Post Code, Gender, Persons Age, Licence type, Licence age, and excess) and outputs a single numerical solution (Cost/week). I've already created a web scraper to collect data for me for one provider. I'm looking for some direction on the possibilities of this; where to start; topics are relevant. Its a really interesting scenario that I can learn alot from Eg n(0): Vehicle cost = $10,000 Post Code = 1025 Gender = M Age [Person] = 30 Licence Type = Full Licence age = 10 Output = $45 / week Function approximation = ??? Thank you AI: You need a bunch of data, not just one point. Several of the inputs are discrete, not continuous-sex, license type, post code. The others are continuous. The simplest assumption is that the cost is a linear function of each of the continuous variables and the functions are either added or multiplied. You can change each of the variables to find the linear function it creates and see if it is linear. Then you can change two of them and try to see how they are combined. The sex and license type don't have many options. The simplest is a fixed difference depending on the response. The post code may well impact the constants in the effects of the continuous variables. It is research. Collect data, make a theory that explains it, and collect some more data to test the theory.
H: Math operator for safe division Is there a math operator for a safe division, returning a pre-set value, usually 0, when divide by zero is encountered? If not, in a computer science or mathematics paper would one just say before an equation that all divisions are "safe"? AI: This is equivalent to defining an operation $\widetilde{\div}:\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ with $a\widetilde{\div}b=a/b$ when $b\neq0$ and $a\widetilde{\div}b=0$ when $b=0$. There is no commonly used operation that I know of that does this, but you could define one such as this very easily. If you are going to use this in a paper I would make sure to clarify that you are using a different definition of division but it shouldn’t cause any other issues so long as it is clear what your definition is.
H: If $a+b+c=\pi$ and $\cot t=\cot a+\cot b+\cot c$, show $\sin^3t=\sin(a-t)\sin(b-t)\sin(c-t)$ I have a problem from a mathematics book: If $\alpha + \beta +\gamma = \pi \tag{1}$ and $$\cot \theta = \cot\alpha + \cot \beta + \cot \gamma, 0 < \theta < \frac{\pi}{2}\tag{2}$$ show that $$\sin^{3}\theta = \sin(\alpha - \theta)\sin(\beta - \theta)\sin(\gamma - \theta)\tag{3}$$ The only way I can think of doing this is by using brute force (multiplying out all the terms on the right of equation (3) ). There must be some solution to this that is not as complicated. I have tried using the identity that if $\alpha + \beta +\gamma = \pi$ then $\tan\alpha+\tan\beta+\tan\gamma=\tan\alpha\tan\beta\tan\gamma$, but it does not fit as (2) uses cotangents instead of tangents. Somehow, the $\alpha$, $\beta$ and $\gamma$ all disappear in (3). My question is: how can I solve this, and is there any way other than brute force to solve this? AI: This is the Brocard angle of a triangle, conventionally denoted by $\omega$ rather than $\theta$. Anyway, by expanding the compound angle formula for sine, you get \begin{align} \frac{\sin(\alpha-\theta)}{\sin\theta}&=\frac{\sin\alpha\cos\theta-\cos\alpha\sin\theta}{\sin\theta}\\ &=\sin\alpha(\cot\theta-\cot\alpha)\\ &=\sin\alpha(\cot\beta+\cot\gamma)\\ &=\sin\alpha\cdot\frac{\cos\beta\sin\gamma+\sin\beta\cos\gamma}{\sin\beta\sin\gamma}\\ &=\frac{\sin\alpha}{\sin\beta\sin\gamma}\cdot(\cos\beta\sin\gamma+\sin\beta\cos\gamma)\\ &=\frac{\sin\alpha}{\sin\beta\sin\gamma}\cdot\sin(\beta+\gamma)\\ &=\frac{\sin^2\alpha}{\sin\beta\sin\gamma} \end{align} and similar. So taking their product gives equation (3).
H: Characteristic functions and convergence of complex sequence I'm trying to solve the following question, but I have no idea why the hint was given as it was: My attempt: I'm not really able to make use of the hint so far, so I'm a bit lost: The assumptions give that $e^{i t x_n} \rightarrow c(t) + i s(t) \equiv z(t)$ for some real functions $s(t), c(t)$ with $\cos(x_n t) \rightarrow c(t)$, $\sin(x_n t) \rightarrow s(t)$. $z(t)$ cannot be zero for any $t$ since $\|\|e^{itx_n}\| - \|z(t)\|\| \leq \|e^{itx_n} - z(t)\|$ and $\|e^{itx_n}\| = 1$ so that $\|z(t)\| = 1$ $$E(e^{i t x_n U}) = f(x_n) \equiv \begin{cases} \frac{e^{itx_n} - 1}{itx_n} & x_n \ne 0 \\ 1 & x_n = 0 \end{cases}$$ Dominated convergence gives us that $$E(c(tU) + i s(tU)) = \lim_{n \rightarrow \infty} f(x_n)$$ Unfortunately I am not sure where to go from here. AI: The hint is suggesting that the sequence of random variables $(x_nU)$ is converging in distribution to some limit $L$. If you manage to prove this, it remains to deduce that the sequence $(x_n)$ itself is converging to something. This would be trivial if $(x_n)$ were bounded, in which case you conclude that $\frac12x_n=E(x_nU)\to E(L)$. But boundedness of $(x_n)$ is not given. Instead, try proving that the sequence $(x_n)$ is Cauchy, i.e. that $$\lim (x_{n_k}-x_{m_k}) = 0$$ for any increasing sequences $(n_k)$ and $(m_k)$. By hypothesis $$\lim_k\exp (itx_{n_k}U)=\lim_k\exp(itx_{m_k}U)$$ almost surely. Therefore the ratio $$\exp(itx_{n_k}U) / \exp(itx_{m_k}U) = \exp it(x_{n_k}-x_{m_k})U$$ tends to $1$ almost surely. Apply dominated convergence to conclude $$E[\exp it(x_{n_k}-x_{m_k})U]\to 1$$ for every $t$, which implies $(x_{n_k}-x_{m_k})U$ converges in distribution to the constant $0$. From here you can argue the sequence $(x_n)$ is Cauchy.
H: Understanding the roots of the irreducible factors of the 15-th cyclotomic polynomial modulo $7$ We consider the 15-th cyclotomic polynomial over $\mathbb{Z}$ first: $$\Phi_{15} = x^8 - x^7 + x^5 - x^4 + x^3 - x + 1.$$ If we reduce it modulo $7$, we obtain two irreducible factors of $\Phi_{15}$ over $\mathbb{F}_7[x]$: $$\Phi_{15} = (x^4+2x^3+4x^2+x+2)(x^4+4x^3+2x^2+x+4).$$ Let us name the first factor $f$ and the second factor $g$. Let us also choose $\alpha \in \mathbb{F}_{7^4}$ with minimal polynomial $\min_{\mathbb{F}_7}(\alpha) = f$. Then my teacher immediately concluded $$f = (x-\alpha^1)(x-\alpha^{7})(x-\alpha^{4})(x-\alpha^{13})$$ and $$g = (x-\alpha^2)(x-\alpha^{14})(x-\alpha^8)(x-\alpha^{11}).$$ I noticed that the powers ($\{ 1,7,4,13\}$ and $\{2,14,8,11\}$) of the different factors are different cosets in $(\mathbb{Z}/15 \mathbb{Z})^\times$ where the equivalence relation is defined by $a \sim b \: :\Leftrightarrow \: a = b \cdot 7^k$ for some $k$. My question: Does this go back to a general result regarding cyclotomic polynomials? Or is this merely a coincidence? AI: If $p$ is a prime and $p\nmid n$ then the polynomial $X^n-1$ has distinct roots $1,\alpha,\alpha^2,\ldots,\alpha^{n-1}$ in some extension $k=\Bbb F_{p^t}$ of $\Bbb F_p$. The roots of the cyclotomic polynomial $\Phi_n$ are the $\alpha^j$ where $\gcd(j,n)=1$. The Galois group of $k$ over $\Bbb F_p$ is generated by the Frobenius map $F:x\mapsto x^p$. Its orbits on the roots of $\Phi_n$ have the form $$\{\alpha^j,\alpha^{pj},\alpha^{p^2j},\ldots,\alpha^{p^{t-1}j}\}$$ which corresponds to the irreducible factor $$(X-\alpha^j)(X-\alpha^{pj})(X-\alpha^{p^2j})\cdots(X-\alpha^{p^{t-1}j})$$ of $\Phi_n$ which is irreducible over $\Bbb F_p$. This shows that $t$ is the least positive integer with $p^t\equiv1\pmod n$. The exponents in this factor are the integers $j$, $pj$, $p^2j,\ldots,p^{t-1}j$ considered modulo $p$; this is an equivalence class in $\Bbb Z/n\Bbb Z$ under $a\sim b\iff a\equiv p^sb\pmod n$ for some $s$.
H: Are two distance regular graphs with the same intersection array also cospectral for their Laplacian matrices? So we know that two DRGs with the same intersection array must be co-spectral on their adjacency matrices, i.e. their adjacency matrices have the same set of eigenvalues. But is this necessarily true also for Laplacian matrices? Over here, the Laplacian matrix is defined as $L(G)=D(G)-A(G)$, where $D(G)$ and $A(G)$ represent the degree matrix and adjacency matrix, respectively. Other than the Laplacian, what about the other matrices, such as the signless Laplacian matrix $\|{L}\|(G)=D(G)+A(G)$? AI: Well, distance-regular graphs with the same intersection array are in particular $k$-regular graphs with the same degree $k$. So for each graph, the Laplacian matrix $L(G)$ is just $kI - A(G)$. If the adjacency matrix has eigenvalues $\lambda_1, \dots, \lambda_n$, then the Laplacian matrix has eigenvalues $k - \lambda_1, \dots, k - \lambda_n$. Therefore, the two graphs must also be co-spectral with respect to their Laplacian matrices, and (by the same argument) with respect to their signless Laplacian matrices.
H: Need help with limit proving. Is it safe to say that $\frac{\sqrt{n+1}}{\sqrt{n}}\rightarrow1$ if $$\lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n})=0$$? Because I want to prove that $\sqrt{n+1}\sim\sqrt{n}\ $when $n\to\infty$, but I don't know how to approach from $\frac{\sqrt{n+1}}{\sqrt{n}}\rightarrow1$. So I was thinking whether by showing $\lim_{n\to\infty}(\sqrt{n+1}-\sqrt{n})=0$ will work. Thx guys. Edited: I forgot to mention I have to use the $\mathcal{E}-N$ argument to prove this limit. AI: For, $\frac{\sqrt{n+1}}{\sqrt{n}}=\sqrt{1+\frac{1}{n}} $ $\lim_{n\to\infty}\frac{\sqrt{n+1}}{\sqrt{n}}=\lim_{n\to\infty} (\sqrt{1+\frac{1}{n}}) = 1 $ Also,$\sqrt{n+1}-\sqrt{n} = \frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}} $ So,$\lim_{n\to\infty} (\sqrt{n+1}-\sqrt{n}) = \lim_{n\to\infty} (\frac{1}{\sqrt{n+1}+\sqrt{n}}) = 0 $ Then, dividing bothsides by $\sqrt{n} $ , So,$\lim_{n\to\infty} (\frac{\sqrt{n+1}}{\sqrt{n}}-1)=0 \implies \lim_{n\to\infty} (\sqrt{1+\frac{1}{n}}) = 1 $ Edit: clearly, $\frac{\sqrt{n+1}}{\sqrt{n}}=\sqrt{1+\frac{1}{n}} $ Now by Archimedean property, for any $\epsilon>0 $ , there exists $m\in \mathbb{N} $, $\frac{1}{n}< \epsilon $ for all $n \ge m $. Hence , $\sqrt{1+\frac{1}{n}}-1 < \sqrt{1+\epsilon}-1 < (1+\epsilon)-1= \epsilon $ , for all $ n \ge m $ So, $\sqrt{1+\frac{1}{n}}-1 < \epsilon $, for all $ n \ge m $.
H: Proof of Equality of Null Spaces Let ${\bf A}, {\bf B} \in \mathbb{C}^{n \times n}$ be a pair of matrices such that: $$ {\rm Null}\left({\bf A}\right) \subseteq {\rm Null}\left({\bf B}\right) $$ Furthermore, it is known that the null spaces of both $\bf A$ and $\bf B$ have the same dimension. Is it true then that ${\rm Null}\left({\bf A}\right) = {\rm Null}\left({\bf B}\right)$? If so, how can it be proven? AI: The more general question is "if $W$ is a subspace of a finite-dimensional vector space $V$ and $\dim W = \dim V$, does this imply $W=V$?" This is true: take a basis for $W$. It is a linearly independent set in $V$ with the same cardinality as the dimension of $V$, so it is also a basis for $V$.
H: Question on the given proof of $(\det A)(\det B) \leq [(\operatorname{tr} AB)/n]^n$ I was reading through this paper and on page 6, there is a lemma proving $(\det A)(\det B) \leq [(\operatorname{tr} AB)/n]^n$ for two positive semideifinite matrices $A$ and $B$. I get every single line until the part that concluded $\operatorname{tr} AB = \sum \lambda_l \mu_l $. Especially, what's bugging me is the very last equation. Before the equality sign we have $$ \sum S_{li} (S^T)_{il} (D_1)_{ii} (D_2)_{ll} $$ and after the equality sign we have $$ \sum \delta^{il} (D_1)_{ii} (D_2)_{ll} $$ Since S is a symmetric orthonormal matrix, $\ S_{li} = (S^T)_{il}$, so I believe the first term should just be $ s^2_{li} (D_1)_{ii} (D_2)_{ll} $. But it doesn't make sense that $ s^2_{li} = \delta^{il} $, since the orthonormal matrix just confirms that the 2-norm of any 'row' or 'column' is 1, not that the single $ s^2_{li} $ is $1$ for $l=i$ and $0$ otherwise. There must be something I am missing. Can anyone help me filling the gap between those two line, please? P.S. The paper is about elliptic equation, but the lemma is for general positive semidefinite matrices, so I put tag just for these type of questions. AI: $S=R^TQ$ is orthogonal, not symmetric as they claimed, but that is inconsequential to the proof. Anyway, it is not true that $\operatorname{tr}(SD_1S^TD_2)=\operatorname{tr}(D_1D_2)$. For example, $D_1=\operatorname{diag}(1,2)=D_2$, and $S=\begin{pmatrix}0&1\\1&0\end{pmatrix}$. Then $SD_1S^T=\operatorname{diag}(2,1)$ and so we end up with $4$ on LHS and $5$ on RHS. A much easier proof of the result they want is simply to use $\sqrt{A}B\sqrt{A}$ is symmetric positive semidefinite, has determinant $\det A\det B$ and trace $\operatorname{tr}AB$. The result follows from AM-GM on its eigenvalues.
H: $P(X_1 > 0 \mid X_1 + X_2 > 0)$ for IID $X_1, X_2 \sim \mathcal{N}(0,1)$ Given IID $X_1, X_2 \sim \mathcal{N}(0,1)$, we want to determine $P(X_1 > 0 \mid X_1 + X_2 > 0)$. This is what I think is the approach for this problem: \begin{align} P(X_1>0\mid X_1 + X_2 > 0) = P(X_1 > 0 \mid X_1 > -X_2) \\ P(X_1 \leq x \mid X_1 > (-X_2=x_2)) = \int_{-x_2}^x \frac{1}{\sqrt{2\pi}}\exp(-\frac{1}{2}x_1^2) \, dx_1 \end{align} I integrate the above to get $P(X_1 \leq x \mid X_1 > (-X_2=x_2))$, and then I integrate $P(X_1 \leq x_1\mid X_1 > (-X_2=x_2))$ over all $x_2$: $$ P(X_1 \leq x_1\mid X_1 > X_2) = \int_{-\infty}^\infty P(X_1 \leq x_1\mid X_1 > (-X_2=x_2)) \, dx_2 $$ Is this a correct approach? I feel that there is something simpler than this. AI: The formula for conditional probability is $$P(A \| B) = \frac{P(A \cap B)}{P(B)}$$ This then means that the answer is $$\frac{P(X_1 > 0 \cap X_1 + X_2 > 0)}{P(X_1 + X_2 > 0)}$$ Clearly, the denominator is $1/2$. Then for the numerator, the integral is $$\iint_{R} p(x_1) p(x_2) \,dx_1 \,dx_2$$ where $R$ is the region satisfying $x_1>0, x_1 + x_2 > 0$ and $p(x)$ is the equation for the probability distribution. Plugging everything in, the integral comes out to $\frac{3}{8}$, which means the final probability comes out to $$\frac{\frac{3}{8}}{\frac{1}{2}} = \frac{3}{4}$$
H: Why bother with the space $\mathcal{L}^1$ for integration when we can abstractly deal with the completion of a semi-normed space I'm studying the Bochner-Lebesue integral, and while I understand the general construction, I have a few questions about the way it is being presented. Typically, the story goes like this: We start with a measure space $(X,\mathcal{A}, \mu)$, and a Banach space $E$ (over $\Bbb{R}$ or $\Bbb{C}$). Then, we can define the space $S$ of simple functions $X\to E$, and for such simple functions, we can define an integral $I(\cdot):= \int_X (\cdot) \, d\mu : S \to E$ in the usual way. Then, we can define a semi-norm $\lVert\cdot \rVert_1$ on $S$ by setting $\lVert \phi \rVert_1 := \int_X\|\phi\|\, d\mu$ (this is to be thought of as integration when the Banach space is $E=\Bbb{R}$, which is of course well-defined). Thus, we have a semi-normed space $(S, \lVert \cdot \rVert_1)$. At this point, we note that $S$ need not be complete, which of course very undesirable for analysis. So, all the presentations I've seen start by defining $\mathcal{L}^1$ as the space of functions $X\to E$ which are the almost-everywhere pointwise limit of Cauchy sequences in $S$. Then, one proves that under these hypotheses, we can extend the integral to a map (pardon the reuse of notation) $I(\cdot)\equiv \int_X(\cdot)\, d\mu:\mathcal{L}^1 \to E$, and also extend the seminorm $\lVert \cdot \rVert_1$ to $\mathcal{L}^1$, such that integration is still a continuous map (with operator norm $\leq 1$), and that finally, $(\mathcal{L}^1, \lVert \cdot \rVert_1)$ is a complete semi-normed space containing the simple functions $S$ as a dense subspace. Therefore, by taking the quotient space of $\mathcal{L}^1/\{\phi\in \mathcal{L}^1: \, \lVert \phi\rVert_1 = 0\}$, and calling this $L^1$, this becomes a Banach space (because by taking this quotient, the semi-norm induces a norm, which is easily verified to be complete). Finally, it is a simple matter of linear algebra to see that we can "transfer" the integration map in the sense that we get a map $\tilde{I}:L^1 \to E$, such that $I = \tilde{I}\circ \pi$ ($\pi$ being the quotient map $\mathcal{L}^1 \to L^1$). The result is that we have an integration operator $\tilde{I}$, defined on a Banach space $L^1$, which naturally reduces to what we'd like it to be on simple functions. Now, my question is that why do we bother to introduce the space $\mathcal{L}^1$ along the way. My thinking is that just as every metric space has a completion, which is uniquely determined up to isometry, we can do a similar thing for semi-normed spaces, in the form of this theorem: Theorem Let $(S, \lVert \cdot \rVert)$ be a semi-normed space (over real or complex field), and let $S_0$ be the subspace of elements with $0$ semi-norm. Then, there exists a completion of $S$, i.e a pair $(V,\gamma)$, where $V$ is a Banach space (over the same field), and $\gamma:S\to V$ is a map such that $\gamma$ is linear $\ker(\gamma) = S_0$ $\text{image}(\gamma)$ is a dense subspace of $V$ $\gamma$ preserves seminorms and norms; i.e for all $s\in S$, $\lVert\gamma(s) \rVert_V = \lVert s \rVert_S$. Also, this completion is determined up to isomorphism (i.e if we had another such pair, then we can make a nice commutative diagram and then obtain an isomorphism of Banach spaces simply by extending the relevant maps from the dense subspaces to the whole space). So, when we have the space of simple functions $S$, we could apply this theorem to get the Banach space $V$ (which is up to isomorphism the same as $L^1$ constructed above), and using similar linear algebra trickery, we can induce an integral $\tilde{I}$ on a dense subspace of $V$, and then extend by continuity to the whole space. My questions/concerns: I realize that by the uniqueness aspect of the completion, both these methods give us the same final outcome: a Banach space, and some type of notion of integral, and of course, the first approach is much more concrete and easier to appreciate on first glance. However, I recently read up about completions of metric (semi-)normed spaces, which is why I thought of the second approach. So I guess my question boils down to: is there anything we gain significantly (besides a bit of concreteness) by realizing $L^1$ as a certain quotient space of functions, rather than just thinking of $L^1$ as an abstract completion of the space of simple functions? Is it perhaps because thinking of Banach spaces as being (almost) a space of functions, rather than some abstract construction (like equivalence classes of Cauchy sequences) makes it significantly easier to analyze the space in some sense (hence the term "functional" analyis)? If this is the case, I'd appreciate if you could elaborate on why specifically thinking in terms of function spaces makes the analysis easier/clearer/preferable (I'm not too sure what word I should use here). AI: I think the reason people are interested in Bochner integrable functions is because people are interested in Banach space valued functions, rather than the properties of the Banach space of Bochner integrable functions. For example, you might want to do harmonic analysis or probability theory on Banach valued functions. Otherwise, why even define real valued $L^p$ the way we do? Why not just define it as the completion of compactly supported continuous functions under the $L^p$ norm? It is because we might more interested in the elements of the $L^p$ spaces than the spaces themselves. If you look at the literature, there are people who study the abstract properties of these spaces. But I think there are far more people who are interested in Banach valued functions.
H: Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question: Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$. My try: $$\begin{align}\\ &x^2+y^2-4x=0\\ &\implies (x-2)^2+y^2=4\\ &\implies \|z-2\|=2\\ \end{align}\\ $$ Now, $w=\frac{2z+3}{z-4}$ $$\begin{align}\\ &\frac w2=\frac{2z+3}{2z-8}\\ &\implies\require{cancel}\frac{w}{w-2}=\frac{2z+3}{\cancel{2z}+3-\cancel{2z}+8}\\ &\implies\frac{w}{w-2}=\frac{2z+3}{11}\\ &\implies\frac{2z}{11}=\frac{w}{w-2}-\frac{3}{11}\\ &\implies\frac{2z}{\cancel{11}}=\frac{8w+6}{\cancel{11}(w-2)}\\ &\implies z=\frac{4w+3}{w-2}\\ &\implies z-2=\frac{2w+7}{w-2}\\ \end{align}\\ $$ $$\therefore\left\|\frac{2w+7}{w-2}\right\|=2\\ \implies 2w+7=2w-4 $$ Now, what to do? Where is my fault? Or how to do it? Is there any other possible ways? AI: You can't replace modulus like that in the last line ( because complex numbers with equal magnitude may not be actually equal. For e.g., $2i$ and $-2i$ have magnitude $2$ but aren't equal.) Use $w = u+iv$ So, $\|2(u+iv)+7\| = 2\|(u+iv)-2\|$ Evaluating and squaring both sides, $\Rightarrow (2u+7)^2 + 4v^2 = 4(u-2)^2 + 4v^2 $ $\Rightarrow 4u^2 + 28u +49 = 4u^2 - 16u + 16 $ $\Rightarrow 44u + 33 = 0 $ $\Rightarrow 4u +3 =0 $
H: Does $ \lim_{n \to \infty}\sum_{k = 1}^n \zeta\Big(k - \frac{1}{n}\Big)$ equal the Euler-Mascheroni constant? Let $\zeta(s)$ be the Riemann zeta function and $\gamma$ be the Euler-Mascheroni constant. I observed the following result empirically. Looking for a proof or disproof. $$ \lim_{n \to \infty}\sum_{k = 1}^n \zeta\Big(k - \frac{1}{n}\Big) = \gamma $$ Also, I searched for different summation formulas for the Euler-Mascheroni constant using the Riemann zeta function but could not find it anywhere. Is there any reference to this sum in literature? Update: Applying the method of @Simply Beautiful Art, we can show that $$ \sum_{k = 1}^n \zeta\Big(k + \frac{1}{m}\Big) = \gamma + n + m + \mathcal O(n^{-1} + m^{-1}) $$ AI: We have the simple asymptotic expansion as $s\to1$ given by: $$\zeta(s)=\frac1{s-1}+\gamma+\mathcal O(s-1)\tag{$s\to1$}$$ For the first term of your sum, you have $$\zeta\left(1-\frac1n\right)=-n+\gamma+\mathcal O(n^{-1})$$ and for the rest of the terms, \begin{align}\sum_{1<k\le n}\zeta\left(k-\frac1n\right)&=\sum_{1<k\le n}\left(1+\sum_{m>1}\frac1{m^{k-\frac1n}}\right)\tag1\\&=n-1+\sum_{1<k\le n}\sum_{m>1}\frac1{m^{k-\frac1n}}\tag2\\&=n-1+\mathcal O(2^{-n})+\sum_{k>1}\sum_{m>1}\frac1{m^{k-\frac1n}}\tag3\\&=n-1+\mathcal O(2^{-n})+\sum_{m>1}\sum_{k>1}\frac1{m^{k-\frac1n}}\tag4\\&=n-1+\mathcal O(2^{-n})+\sum_{m>1}\sqrt[n]m\frac{m^{-2}}{1-m^{-1}}\tag5\\&=n-1+\mathcal O(2^{-n})+\sum_{m>1}\sqrt[n]m\left(\frac1{m-1}-\frac1m\right)\tag6\\&=n+\mathcal O(2^{-n})+\sum_{m>1}(\sqrt[n]m-1)\left(\frac1{m-1}-\frac1m\right)\tag7\\&=n+\mathcal O(n^{-1})\tag8\end{align} where $(1):$ Definition of $\zeta$. $(2):$ Summing over $1$. $(3):$ Extending $k$ from $(1,n]$ to $(1,\infty)$, with $\mathcal O(2^{-n})$ error. $(4):$ Rearranging the series. $(5):$ Geometric series. $(6):$ Partial fractions. $(7):$ Using telescoping series and $1=\sum_{m>1}\left(\frac1{m-1}-\frac1m\right)$. $(8):$ Asymptotic expansion using $\sqrt[n]m=\exp(n^{-1}\ln(m))=1+\varepsilon n^{-1}\ln(m)$, where $\|\varepsilon\|\le\sqrt[n]m$, which gives the series bounded by $n^{-1}$ times another series with dominating term $\mathcal O(m^{\frac1n-2}\ln(m))$ and thus converges. Adding these results together, we find that $$\sum_{k=1}^n\zeta\left(k-\frac1n\right)=\gamma+\mathcal O(n^{-1})$$
H: Parameterization of a curve within a path integral? I have a question about the following problem: Find an appropriate parametrization for the given piecewise-smooth curve in $\mathbb{R}^{2}$, with the implied orientation. The curve $C$, which goes along the circle of radius 3, from the point $(3, 0)$ to the point $(−3, 0)$, and then in a straight line along the x-axis back to $(3, 0)$. My question is in regards to the segment connecting the points $(-3, 0)$ and $(3, 0)$. Using the formula $c(t) = (1 - t)<x_{0}, y_{0}> + \ t<x_{1},y_{1}>$, I arrived at the following solution: $c(t) \ = \ (1 - t)<-3 , 0> + \ t<3, 0>$ $=\ <3t - 3, 0> + <3t, 0>$ $=\ <6t - 3, 0>, t \in [0, 1]$ However, in the text I was working from, I was given this solution: $<6t - 9, 0>, t\in [1,2]$ How would I go about obtaining this solution for this particular segment? AI: You can obtain the parametrization $\langle 6t - 9, 0 \rangle, t \in [1,2]$ from the parametrization $\langle 6t - 3, 0 \rangle, t \in [0,1]$ by the change of variable $t \leadsto t-1$. In the second parametrization, $t$ runs from $0$ to $1$, and so the new $t$ (that in $t-1$) must run from $1$ to $2$ (so that $t-1$ runs from $0$ to $1$). You might do this because you might want the parameter in the parametrization for the line segment to run from $1$ to $2$, so that the parameter in the parametrization for the circular segment can run from $0$ to $1$, giving a combined parametrization for the entire curve $C$ with the parameter $t$ varying from $0$ to $2$.
H: Absolutely continuous functions that fix zero and satisfies $f'(x)=2f(x)$ A past question from a qualifying exam at my university reads: Let $f$ be a continuous real-valued function on the real line that is differentiable almost everywhere with respect to the Lebesgue measure and satisfies $f(0)=0$ and $$ f'(x)=2f(x)$$ almost everywhere. Prove that there exists infinitely many such functions, but that only one of them is absolutely continuous. I have tried modifying the function $e^{2x}$, but I cannot satisfy all the conditions given. Once one shows that there are infinitely many such functions, then if we pick 2 such functions $f_1$ and $f_2$ and fix $a>0$, we can apply the fundamental theorem of Calculus for Lebesgue Integrals on $[0,a]$ and see that if both are absolutely continuous, then $$ f_1(x)=\int_0^x 2f(t)dt=f_2(x) $$ So this would imply that the are the same function on $[0,\infty)$. I'm not sure how to proceed with the whole real line. AI: It well known that there is a non-constant continuous function $g$ with $g'=0$ a.e. [ You can search for 'Cantor Function']. We may suppose $g(0)=0$ (by considering $g(x)-g(0)$). The function $g^{n}(x)e^{2x}$ satisfies the stated property for any positive integer $n$ so there are infinitely many solutions. Now let $f$ and $g$ be two solutions which are absolutely continuous. I will let you verify that $h(x)=e^{-2x} (f(x)-g(x))$ is absolutely continuous on any finite interval and its derivative is $0$. Now $h(b)-h(a)=\int_a^{b} h'(t) dt=0$ whenever $a <b$. Since $h(0)=0$ we get $h(x)=0$ for al $x$ and hence $f(x)=g(x)$ for all $x$. PS: Use the inequality $ \|(FG)(b_i)-(FG)(a_i)\| \leq \|F(b_i)-F(a_i)\|\|G(b_i)\|+\|F(a_i)\|\|G(b_i)-G(a_i)\|$ to show that the product of two bounded absolutely continuous functions is absolutely continuous.
H: Prove that there are infinite number of mapping from $\mathbb{R}$ onto $\mathbb{Q} ?$ Prove that there are infinite number of mapping from $\mathbb{R}$ onto $\mathbb{Q} ?$ My attempt : If i take $f(x) = [x]^{x} $ where $[.]$ denote the greatest integer function .then it will be satisfied the onto mapping Im confusing that how to prove this AI: Let $f_n(x)=x$ for $x$ rational, $f_n(\sqrt 2)=n$ and $f_n(x)=0$ for any irrational number other than $\sqrt 2$. These are distinct maps from $\mathbb R$ onto $\mathbb Q$.
H: how (a!)/(b!) = (b + 1)×(b + 2)×⋯×(a − 1)×a I was solving a problem in which i need to figure out the prime factorization of $\frac{a!}{b!}$ and i did that by computing (a!) and then (b!) by looping ((1 to a) & (1 to b)) and then derived n by dividing them ($n = \frac{a!}{b!}$) and then prime factors of n, but it gives me to TLE(Time limit exceeded) so i refered editorial and in editorial they describe an alternate method , they says factorization of number $\frac{a!}{b!}$ is this same as factorization of numbers $(b + 1)\times(b + 2)\times \cdots\times (a - 1)\times a$. I am unable to figure out how $\frac{a!}{b!}$ == $(b + 1)\times(b + 2)\times \cdots\times (a - 1)\times a$ ? AI: Assume $a > b$. Then: $$\frac{a!}{b!} = \frac{a(a-1)(a-2) \cdots (b+1) \cdot (b)(b-1)(b-2) \cdots (3)(2)(1)}{b(b-1)(b-2) \cdots (3)(2)(1)}$$ $$= \frac{a(a-1)(a-2) \cdots (b+1) \cdot \require{cancel} \cancel{(b)(b-1)(b-2) \cdots (3)(2)(1)}}{\cancel{(b(b-1)(b-2) \cdots (3)(2)(1)}}$$ Do you see how this works?
H: Proving $\|x\|+\|y\|+\|z\| ≤ \|x+y-z\|+\|y+z-x\|+\|z+x-y\|$ for all real $x$ We need to prove that for all real $x,y,z$ $$\|x\|+\|y\|+\|z\| ≤ \|x+y-z\|+\|y+z-x\|+\|z+x-y\|$$ Source ISI entrance examination sample questions I don't know how to solve in mod form so I thought about squaring and removing the mod, but the mod on RHS still persists. I believe there can be a different approach to it, I would like to have some hints on how to do approach it. AI: Hint: $$2x=(x+y-z)+(x+z-y) \implies x=\frac{x+y-z}{2}+\frac{x+z-y}{2}$$ Now repeat the process for $y$ and $z$ and use the triangle inequality trice
H: On a variant of Farkas Lemma The version on Mordecai Avriel’s book Nonlinear Programming is: Let A be a given $m\times n$ real matrix and $b$ a given n vector. The inequality $b^Ty≥ 0$ holds for all vectors $y$ satisfying $Ay ≥ 0$ if and only if there exists an m vector $\rho ≥ 0$ such that $A^T\rho=b$. It seems that a more wildly accepted version is this: Let A be an $m \times n$ matrix and $b$ an $m$-dimensional vector. Then, exactly one of the following two statements is true: There exists an $x \in \Bbb R^n$ such that $Ax = b$ and $x \ge 0$. There exists a $y \in \Bbb R^m$ such that $A^Ty \ge 0$ and $b^Ty < 0$. I can’t understand how these two versions imply each other. Besides, the explanation beneath seems questionable: An illustration of the Farkas lemma is given in Figure 3.2 for a 3 ×2 matrix A. The vectors A1, A2, A3 are the row vectors of the matrix A. Consider the set Y consisting of all vectors y that make an acute angle with every row vector of A. The Farkas lemma then states that b makes an acute angle with every y ∈ Y if and only if b can be expressed as a nonnegative linear combination of the row vectors of A. In Figure 3.2, b1 is a vector that satisfies these conditions, whereas b2 is a vector that does not. In my opinion, the shadowed region is exactly “consisting all vectors y that make an acute angle with every row vector of A”, however, it is not all vectors that “can be expressed as a nonnegative linear combination of the row vectors of A”. For example, $A_1=A_1+0A_2+0A_3$ can be expressed as a nonnegative linear combination of the row vectors of A, but $A_1$ makes an obtuse angle with $A_3$. My questions are: Is this variant of Farkas Lemma, and its illustration correct? And if so, How does it relate to the other version of Farkas Lemma? AI: The two versions of Farkas’ lemma you give are in fact equivalent—this is just because the statements “exactly one of A or B holds” and “A holds if and only if B does not” are logically equivalent. As for the illustration, the condition in question is whether $b$ makes an acute angle with all members of $Y$, not whether $b$ is in $Y$. Indeed, by definition, $A_1$ makes an acute angle with all vectors which make an acute angle with each $A_i$ ($1\le i \le 3$).
H: Convergence of the sequence $ a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\int_{1}^{n} \frac{1}{x} d x$ Let $$ a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}-\int_{1}^{n} \frac{1}{x} d x$$ for all $n \in \mathbb{N} .$ Show that $\left(a_{n}\right)$ converges. Actually $ a_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}$ is the Reimann upper sum of $f(x)=\frac{1}{x}$ on the interval $[1,n]$ relative to a partition such that each subinterval of unit length, But I couldn't prove the convergence. Can we use the fact $\ln (n+1) \leq 1+\frac{1}{2}+\ldots+\frac{1}{n} \leq 1+\ln n$ AI: Rearrange the integral as a telescopic sum and use the fact that $\ln (1+x) = x+\mathcal O(x^2)$: $$\begin{split} \left (\sum_{k=1}^n \frac 1 k\right) -\int_1^n\frac{dx}x &= 1 + \left (\sum_{k=2}^n \frac 1 k\right)-\ln n\\ &= 1 + \sum_{k=2}^n \left(\frac 1 k -\ln k +\ln (k-1) \right)\\ &= 1 + \sum_{k=2}^n \left(\frac 1 k +\ln \left(1 - \frac 1 k \right) \right)\\ &= 1 + \sum_{k=2}^n \mathcal O \left(\frac 1 {k^2} \right)\\ \end{split}$$ The latter sum converges.
H: Well-ordering on natural numbers Let $\omega=\{0,1,2,3,\ldots\}$. We say that $\omega$ is a well-ordered set. But I can't understand why. By the definition of well-ordering, there should be no infinite descending chain, but if I start from infinity, how can I reach 0 in finite descents? Or is this not allowed? Is this nonsensical to choose infinity? Then what about $\omega+1$? Is this set well-ordered? How can I reach to $0$ from $\omega$? AI: Every descending chain in $\omega$ has to start somewhere - specifically, it must start with a finite value $n$. One cannot "start from infinity" within $\omega$ since every element of $\omega$ is finite. $\omega + 1$ is well-ordered in the obvious way. Suppose we have a descending chain beginning with $\omega$. Then the next element in the chain must be some $n < \omega$: that is, some finite $n$. After that point, it's clear there can only be finitely many steps until reaching zero.
H: Weak compactness of nonnegative part of unit ball of $L^1$ Let $(\Omega,\mathfrak{A},\mu)$ be a measure space and let $B$ be the unit ball of $L^1(\Omega,\mathfrak{A},\mu;\mathbb{R})$. Suppose that $$ B_+=\big\{u\in B\;\|\;u\geq 0\;\:\text{$\mu$-a.e.}\big\} $$ is weakly compact. My question: Does the weak compactness of $B_+$ imply the weak compactness of $B$? I believe that it is the case, but I have no clue where to start. Any help/hint is highly appreciated. AI: If $(u_i)$ is a net in $B$ then $(u_i^{+},u_i^{-})$ is a net in $B_{+} \times B_{+}$ which is weakly compact. Hence there is a convergent subset and this gives a convergent subnet of $(u_i)$.
H: Groups up to isomorphism Does there exist an answer to the next question: How many groups of order $n$ ($\|G\|=n$, $n\in \mathbb{N}$) are exist up to isomorphism? (The groups are not necessarily Abelian) I am curious about this question. If the answer ($\forall n\in\mathbb{N}$) does not exist,does the answer exist for $n=$? $6$? $10$? (The question about $n=1,2,3,4$ (and primes or p*q) is easy because the groups are abelian). AI: Let $$g(n)=\text{number of isomorphism classes of groups of order }n$$ Then we can rephrase your question as: Do we know the value of $g(n)$ for any $n\in\mathbb{N}$? There are algorithms to calculate $g(n)$. The simpliest is as follows: for a given set $G$ of size $n$ we can simply go through all possible functions $G\times G\to G$ and verify how many of them produce groups (by checking axioms) and then filter out non-isomorphic (we can again go through all possible homomorphisms). The complexity of this is huge though, since there are $n^{n^2}$ potential candidates for a group operation. There are other algorithms, however all of them have complexity beyond being practically useful. Unfortunately we don't know a closed formula for $g(n)$. We do know a bit, for special $n$. If $p$ is prime then $$g(pq)=1\text{ if }q\text{ is another prime such that }p<q\text{ and }p\text{ does not divide }q-1$$ $$g(p)=1$$ $$g(p^2)=2$$ $$g(p^3)=5$$ $g(p^m)$ gets eventually very complicated, and soon (for $m\geq 5$ I think?) turns out to be too hard to solve precisely at the moment. Some approximations exist though, e.g. Higman-Sims asymptotic formula: $$g(p^m)\sim p^{\frac{2}{27}m^3+O(m^{8/3})}$$ This gets even harder for naturals which are not prime powers. If the answer ($\forall n\in\mathbb{N}$) does not exist,does the answer exist for $n=5$? $6$? $7$? $g(5)=g(7)=1$ since these are primes. $g(6)=2$ (namely $\mathbb{Z}_6$ and $S_3$) but it is not trivial (although not hard as well).
H: Volume of convex body as an integral of its radial function Let $C$ be a compact convex set in $\mathbb{R}^d$. Let the origin $O$ by in the internal of $C$. The gauge function $\gamma_C(.) : \mathbb{R}^d \to [0, \infty]$ of $C$ is defined as $$ \gamma_C(x) = \inf\{t : x \in t \cdot C\}. $$ The radial function is defined as a reciprocal of gauge function or, equivalently, $$ 1 / \gamma_C(x) = \sup\{x : t \cdot x \in C\} $$ I'm considering the following integral: $$ I(C) = \int_{S^{d - 1}} \|1 / \gamma_C(u)\| du. $$ I have two following notions: Intuitively, it looks like a volume of $C$: we integrate over all directions and summarize all distances from the origin $O$ to the boundary of $C$. However, if we consider the body $a \cdot C$ for some constant $a > 0$, we will have $1 / \gamma_{a \cdot C}(u) = a \cdot 1 / \gamma_C(u)$, and: $$ I(a \cdot C) = \int_{S^{d - 1}} \|a \cdot 1 / \gamma_C(u)\| du = a \int_{S^{d - 1}} \|1 / \gamma_C(u)\| du = a \cdot I(C). $$ From this equality, we have that $I(C)$ is not a volume of $C$ because $Vol(a \cdot C) = a^3 \cdot Vol(C)$. So, which of (1) and (2) is wrong? And, if (1) is wrong, the next question: can we construct a volume of $C$ from the integral of its radial function? AI: 1 is wrong. For example consider the unit disk in $\mathbb{R}^2$. Then the integral is $\int_0^{2\pi} 1= 2\pi$. The volume, however, is of course $\pi$. The problem is, like you point out in 2, that the volume is not actually linear with respect to scaling. The integral treats the area of a disc like a rectangle on the circle, but it’s actually much more like a triangle. (With both sides depending on the scaling factor.) The problem is that the infinitesimal volume element in spherical coordinates is not $d\rho du$ where $du = vol_{S^{d-1}}$ is the volume form of $S^{d-1}$ and $\rho$ is the radius. You need to scale with the Jacobi determinant when you change coordinates.
H: Lots of doubts abot the surface area of a cylinder. I recently started to study parametric surfaces, and I come across this exercise that I try to solve but I have a lot of doubts reganding the correctness of my resolution, and also I don't find similar examples on the internet. I need to find the surface area of the cylinder $$x^{2} + y^{2} = 4x$$ bounded by z=0 and z+ x =4. The cylinder is centered at (2,0) with radius 2. I made the parametrization $$<2+rcos(t) , rsen(t), 2-rcos(t)>$$ with r between 0 and 2, and t between 0 and 2π (First doubt : Is the parametrization right?) Then, if everything is ok, I would proceed to do the formula of a surface area (I would not write the whole formula because I am very bad at MathJax). But you know, the double integral of the norm of the vector ("u") being "u" the cross product of the partial derivatives of the parametrization. The vector u in this case is (r,0,r) and the norm is $$\sqrt{2}r $$ Then, if everything is right, the area of the surface is the double integral $$\int_0^{2π}\int_0^2\sqrt{2}*r^2 dθdr $$ Is this resolution right? If not, can you help me? Thanks. PS : I know that the cylinder bounded by the plane is half of the full cylinder. This is the main reason that I think this resolution is wrong. AI: Let us solve for $$(x-2)^2+y^2=4,\\z\in[0,2-(x-2)].$$ in polar coordinates. We have $$A=\int_0^{2\pi}\int_0^{2-2\cos\theta}\,dz\,2\,d\theta=2\int_0^{2\pi}(2-2\cos\theta)\,d\theta=8\pi$$ (the cosine term cancels out).
H: How to prove that any matrices have their own generalized inverse. Let $A$ be a matrix with a form $(m.n)$, and $X$ be a matrix with a form of $(n,m)$. If $AXA = A$, $X$ is called a generalized inverse of $A$. How can we prove that any matrices have their own generalized inverse? \begin{eqnarray} \\ \end{eqnarray} \begin{eqnarray} \\ \end{eqnarray} ----What I have found----------------------------------------- When $A$ is a row echelon form matrix $F=F(m,n;r)=\begin{pmatrix} E_{ r } & O \\ O & O \end{pmatrix}$, $Y=\begin{pmatrix} E_{ r } & Y_{ 2 } \\ Y_{ 3 } & Y_{ 4 } \end{pmatrix}$ is always $F$'s generalized inverse. AI: One approach is to use the fact that $A$ has a rank factorization $A = CF$ (such a factorization can be attained, for instance, using row-reduction). Because $C$ has full column-rank, it has a left-inverse $C^g$. Because $F$ has full row-rank, it has a right inverse $F^g$. If we defined $X = A^g = F^g C^g$, then we find that $$ AXA = (CF)(F^gC^g)(CF) = C(FF^g)(C^gC)F = CF = A. $$ Proof of the existence of the one-sided inverses: Suppose that $F$ is $m \times n$ with rank $m$ ($m \leq n$). Because $F$ has rank $m$, $F$ has a set of $m$ linearly independent columns (say that the columns $i_1,\dots,i_m$ are linearly independent; we could see that such columns must exist because the reduced row echelon form exists). Let $M$ denote the matrix whose columns are $e_{i_1},\dots,e_{i_m}$, where $e_i$ is the $i$th column of the identity matrix of size $n$. The matrix $FM$ is square and invertible, since its columns are linearly independent. The matrix $F^g = M(FM)^{-1}$ is a right-inverse of $F$, which is to say that $FF^g = I$. For a matrix $C$ of full column-rank, we see that $C^T$ has full row-rank and therefore has right-inverse $[C^T]^g$. It follows that $$ [[[C^T]^g]^T C]^T = C^T [C^T]^g = I, $$ so that $[[C^T]^g]^T C = I$. That is, $[[C^T]^g]^T$ is a left-inverse to $C$. Alternatively: to construct $F^g$, we could have used the fact that $FF^T$ must be invertible, from which it follows that $F^T(FF^T)^{-1}$ is a right-inverse to $F$. This coincides with the Moore-penrose pseudoinverse.
H: Is annihilator of principal ideal comparable? Is annihilator of principal ideal comparable or intersection is zero? It seems to me there is no reason to believe this is true. But I couldn't found a counter example yet. Is it true under some conditions? AI: Let $R=\mathbb{Z}[x]/(12,2x)$. Then $(2)$ is a principal ideal with annihilator $(6,x)$. We have: \begin{eqnarray} 2&\not\in& (6,x),\\ x&\not\in& (2),\\ (2)\cap(6,x)=(6)&\neq& (0). \end{eqnarray} Edit, a simpler example: Let $R=\mathbb{Z}/24\mathbb{Z}$. Then $(4)$ is a principal ideal with annihilator $(6)$. \begin{eqnarray} 4&\not\in& (6),\\ 6&\not\in& (4),\\ (4)\cap(6)=(12)&\neq& (0). \end{eqnarray}
H: What are the subdifferentials $\partial f(0)$ and $\partial f(1)$? Let $ f: \mathbb{R} \to \mathbb{R} $ given by \begin{equation} f(x) = \left\{ \begin{array}{rl} x \log x -x & \text{if } x \geq 0\\ \infty & \text{if else}\\ \end{array} \right. \end{equation} What are the subdifferentials $\partial f(0)$ and $\partial f(1)$? Definition: The set of all subgradient of $f$ at $x$ is called the subdifferential of $f$ at $x$: $$ \partial f(x)=\{\alpha \in \mathbb{R}: f(y)\geq f(x) +\langle\alpha, y-x\rangle,\ \forall \ y\in \mathbb{R} \}$$ AI: The function is differentiable for $x>0$ with $ f'(x) = \log(x) $. Since the function is differentiable at $x=1$ we have $\partial f(1) = \{f'(1)\}$ and then $ \partial f(1) = \{ 0 \} $. I'll assume that you are setting $f(0)=0$. Suppose $\alpha \in \partial f(0)$, then $$ f(y) \geq f(0) + \langle \alpha, y-0 \rangle = \alpha y $$ for all $y\in \mathbb{R}$. That holds for $y \leq 0$, assume now $y>0$, then we have $$ y\log(y) - y \geq \alpha y \Rightarrow \log(y) \geq 1+\alpha $$ but since $\lim_{y\to 0}\log(y) = -\infty $ for any $\alpha$ we can find $y$ such that $\log(y)<1+\alpha$. Therefore: $$ \partial f(0) = \emptyset $$
H: Is the dimension of a Noetherian local ring equal to its associated graded ring? For a noetherian local ring $A$ with maximal ideal $\mathfrak{m}$, let $I$ be a primary ideal in $A$, the associated graded ring is $$ \bigoplus_{n=0}^{\infty} I^n/I^{n+1}$$ AI: Yes. First we deal with the case $I=\mathfrak m$. If we define $f(n)=l(A/\mathfrak{m}^n)$ where $l$ denotes the length of a module, then for large $n$, $f(n)$ equals a polynomial of degree $d=\dim A$ in $n$. Replacing $A$ by its associated graded ring with $I=\mathfrak m$ does not change $f(n)$, and so does not change $d$. Now consider general $I$. Then $\mathfrak{m}\supseteq I\supseteq \mathfrak{m}^r$ for some positive integer $r$. Therefore $g(n)=l(A/I^n)$ satisfies $f(n)\le g(n)\le f(rn)$. Again, $g(n)$ is a polynomial for large $n$, and these inequalities imply that it has the same degree $d$ as $f(n)$. If we consider the graded ring $R=\bigoplus(I^n/I^{n+1})$, this is also a Noetherian local ring with maximal ideal $\mathfrak{m}' =\mathfrak{m}/I\oplus (I/I^2)\oplus (I^2/I^3)\oplus\cdots$ and having $I'=0\oplus (I/I^2)\oplus (I^2/I^3)\oplus\cdots$ as a primary ideal. Then $l(R/I'^n) =g(n)$ also and $R$ also has dimension $d$.
H: $(\forall n \in \mathbb{Z}):n^{3} \equiv n$ (mod $6$) [This is not a duplicate, since I am seeking for an alternative proof for this problem] This is a problem from Proofs and Fundamentals, by Ethan D. Bloch. Show that, for all $n \in \mathbb{Z}$, $n^{3} \equiv n$ (mod $6$). I wrote my proof as follows and I would really appreciate if you could check it: Proof: Let $n \in \mathbb{Z}$. Note that $n^{3} - n = (n-1) \cdot n \cdot (n+1)$, so the former is the product of three consecutive integers. By the Division Algorithm, we know that $n = 3q + r$, with $q,r \in \mathbb{Z}$ and $r \in \{0,1,2\}$. Since we have three consecutive integers, we know that each is either equal to $3k$, or $3k + 1$, or $3k + 2$ for some $k \in \mathbb{Z}$. Let’s suppose that $n = 3k$ for some $k \in \mathbb{Z}$ (the argument is similar for the rest two cases, so for the sake of briefness I will omit here). It follows that $n-1 = 3k - 1 = 3k - 3 + 2 = 3(k-1) +2$ and $n+1 = 3k +1$. We have to possibilites on the parity of $k$. If $k$ is even, then $k=2j$ for some $j \in \mathbb{Z}$. Hence $n = 3 \cdot 2 \cdot j = 6k$. Hence $n^{3}-n=6k(n-1)(n+1)$ and $k(n-1)(n+1) \in \mathbb{Z}$. Therefore $n^{3}-n \equiv 0$ (mod $6$) and then $n^{3} \equiv n$ (mod $6$). If $k$ is odd, then $k = 2i + 1$ for some $i \in \mathbb{Z}$. Hence $n + 1 = 3(2i+1) +1 = 6i + 4 = 2(3i+2)$. It follows that $n^{3} - n = (n-1) (3k) (2(3i +2)) = 6k(n-1)(3i+2)$ and $k(n-1)(3i+2) \in \mathbb{Z}$. Thus $n^{3}-n \equiv 0$ (mod $6$). Therefore $n^{3} \equiv n$ (mod $6$). This completes our proof. $\square$ I would also like to know that if there is other way to prove this result? (In the book, Bloch gives a hint that says to note that $n \equiv 0$ (mod $6$), or ..., or $n \equiv 5$ (mod $6$) and then in each case to compute $n^3$ (mod $6$), but I don’t know how this will work as a proof... any idea?). Thank you for your attention! AI: Your proof is correct but you can simplify it. Once you correctly note that $n^3-n$ is always the product of three consecutive integers, just observe that any three consecutive integers must include a multiple of $3$ and a multiple of $2$, so their product must be a multiple of $6$.
H: Non-real numbers in system of equations Given $a^2+b^2=1$, $c^2+d^2=1$, $ac+bd=0$ To prove $a^2+c^2=1$,$b^2+d^2=1$,$ab+cd=0$ Now this can be easily done by trigonometric substitution if it was given that $a,b,c,d$ are real numbers. I have a solution using matrix which I think is valid even if the given numbers are non real. https://artofproblemsolving.com/community/c6h475289p2661966 I am curious to know if there other ways to it Algebraical or with any other method. AI: This says that $$\pmatrix{a&b\\c&d}\pmatrix{a&c\\b&d}=\pmatrix{1&0\\0&1}$$ implies $$\pmatrix{a&c\\b&d}\pmatrix{a&b\\c&d}=\pmatrix{1&0\\0&1}.$$ This follows from the fact that (over any commutative ring) any square matrix with a left inverse has a right inverse (necessarily the same as the left inverse). But if you want an "equational" proof avoiding concepts from linear algebra observe that $$a^2+c^2=1+a^2c^2-(1-a^2)(1-c^2)=a+a^2c^2-b^2d^2=1+(ac+bd)(ac-bd)=1$$ and similarly for $b^2+d^2=1$. Also $$ab+cd=ab(c^2+d^2)+(a^2+b^2)cd=(ac+bd)(ad+bc)=0.$$
H: Shift invariance and Krylov subspaces Let $A \in \mathbb{R}^{n \times n}$ a matrix, and $r_0 \in \mathbb{R}^{n}$. Also, let $(\sigma)_i$ a sequence of complex scalars. Consider the Krylov subspace $K_n(A,r_0)=\text{span} \{r_0,A r_0, \ldots, A^{n-1}r_0 \}$. I want to show that $$K_n(A + \sigma_j I,r_0) = K_n(A+\sigma_i I,r_0)$$ i.e. the so-called "shift invariance of Krylov subspaces" I consider $x \in K_n({A+\sigma_i I})$, therefore $$x= \sum_k^{n-1} \alpha_k (A+\sigma_i I)^k r_0$$ I search for scalars $\beta_k$ such that $$\sum_k^{n-1} \alpha_k (A+\sigma_i I)^k r_0 = \sum_k^{n-1} \beta_k (A+\sigma_j I)^k r_0$$ and therefore I obtain $$\sum_k^{n-1} \Bigl( \alpha_k (A+\sigma_i I)^k - \beta_k (A+\sigma_j I)^k \Bigr) r_0$$ But now I don't know how to find $\beta_k$. How can I move? Also other ways are really appreciated AI: It suffices to show that for any $\sigma$, we have $K(A + \sigma I, r_0) \subset K(A,r_0)$. To that end, we note by the binomial theorem that for any $0 \leq k\leq n-1$, we have $$ (A + \sigma I)^k = \sum_{j=0}^k \binom kj \sigma^{k-j}A^jr_0 \in \operatorname{span}\operatorname{span}\{A^k r_0 : 0 \leq k \leq n-1\} = K(A,r_0). $$ Since this holds for all $k$, conclude that $\operatorname{span}\{(A + \sigma I)^ k : 0 \leq k \leq n-1\} \subset K(A,r_0)$, which is what we wanted. With that, note that $$ K(A + \sigma_j I,r_0) \supset K([A + \sigma_j I] + (\sigma_i - \sigma_j)I,r_0) = K(A + \sigma_i I,r_0). $$ By symmetry (i.e. by switching $i$ and $j$), we see that the reverse inclusion holds as well.
H: Importance of the 'prime' condition Prime avoidance theorem: Let $A$ be a ring (commutative with unity) and $p_1,...,p_n\subset A$ prime ideals. Let $a\subset A$ be an ideal such that $a\subset (p_1\cup p_2\cup\cdot\cdot\cdot\cup p_n)$, then $a\subset p_k$ for some $1\leq k\leq n$. Now, I have no problem in proving this theorem. I want to illustrate with an example, the importance of the prime condition in the theorem. That is, I want to show that there exist ideals $a_1,...,a_n\subset A$ such that $a\subset (a_1\cup a_2\cup\cdot\cdot\cdot\cup a_n)$, but $a\not \subset a_k$ for all $1\leq k\leq n$. Some of the things that I noticed is that, we cannot find the example in a principal ideal domain nor can we get the example if we take $n=2$. I wasn't able to make much progress beyond this. Thank you in advance! AI: Let $A=\mathbb {Z}[X,Y], a=(2,X,Y), a_{1}=(2,X^2,Y), a_{2}=(2,X,Y^2), a_{3}=(2,X+Y,X^2,Y^2,XY)$ Then $a$ is contained in the union of $a_1,a_2,a_3$ (you can check it by computing in $A/(2,X^2,Y^2,XY)$ , which is a ring with $4$ elements) but not contained in any $a_i$. Source of the counter example: https://fr.wikipedia.org/wiki/Lemme_d%27%C3%A9vitement_des_id%C3%A9aux_premiers
H: Area of a triangulation of a non-planar polygon Does the area of a triangulated simple non-planar polygon in 3D space depend on the triangulation, or is it the same for any triangulation of the points? I would suppose it is not the same, but when I try to come up with simple examples it looks like it's the same for symmetry reasons. What is a counter example, or is the area constant for any possible triangulation? AI: Take two equilateral triangles with a common edge and unit area. If you change the angle they form, the total surface remains $2$. At the same time, the area formed by joining the other diagonal varies between $0$ and $2$.
H: Dual space of continuous functions on an open set of $\mathbb{R}^m$ Let $V \subset \mathbb{R}^m$ be an open subset and define $$ C(V) := \{f:V \rightarrow \mathbb{C}\| f\text{ is continuous}\}. $$ We can make $C(V)$ into a topological vector space as follow. Let $Q_1 \subseteq Q_2 \subseteq \cdots \subset V$ be compact sets such that $\cup_{n=1}^{\infty} Q_n = V$ and consider the semi-norm $\\|\cdot\\|_n$ defined by $$\\|f\\|_n := \\|f\\|_{L^{\infty}(C(Q_n))}.$$ These semi-norms separate elements in $C(V)$ and the topology induced by these semi-norms makes $C(V)$ into a topological vector space. My question: What is the topological dual of C(V)? Remark: It is easy to see that to every continuous linear functional on $C(Q_n)$, there corresponds a continuous linear functional on $C(V)$ because if $\phi \in C(Q_n)^{}$ and $f\in C(V)$, then $\tilde{\phi}(f) := \phi(f\|_{Q_n})$ is an element of $C(V)^$. We know the dual space $C(Q_n)^$ from Riesz representation theorem and $C(Q_n)^$ consists of complex Borel measures on $Q_n$. Is there any other linear functional $\psi \in C(V)^$ such that $\psi \notin C(Q_n)^$ for any $n$? Thanks! AI: Every continuous linear functional $\phi$ on $C(V)$ is continuous with respect to one of the given seminorms $\\|\cdot\\|_{Q_n}$ (supremum over $Q_n$). Using Tietze's extension theorem you then define a continuous linear functional $\tilde\phi$ on $C(Q_n)$ by $\tilde\phi(g)=\phi(f)$ where $f\in C(V)$ is any extension (this is well defined because $\|\phi(f)\|\le c\\|f\\|_{Q_n}$)). Then represent $\tilde\phi$ by a signed measure. The conclusion is that the dual of $C(V)$ is given by all signed measures with compact support.
H: Trace norm equality in proof Consider the following theorem in Murphy's '$C^$-algebras and operator theory': Why is the marked equality true? This seems to boil down to showing that $$tr(uw^) = tr(w^u)$$ Myrphy already showed that $tr(uv) = tr(vu)$ when one of the operators $u,v$ is trace class or they both are Hilbert Schmidt operators, but neither of these conditions seem to apply here. I'm also a little bit confused why Murphy uses the trace for operators that are not necessarily trace-class (the trace was defined for such operators). What am I missing? AI: As you have noted, that equality is no problem when $u$ is trace class. When $u$ is not trace class, then the statement $\Vert u \Vert_1 = \Vert u^ \Vert_1$ should be read as "neither is $u^$". I.e. $u$ is trace class iff $u^$ is. This is now clear, for example, by the characterisation of trace class operators as the product of Hilbert–Schmidt operators (Theorem 2.4.13), because the adjoint of a Hilbert–Schmidt operator is Hilbert–Schmidt (as can be seen from the explicit formula in Example 2.4.1.).
H: Why are we using combination in this problem instead of permutation? What is the number of non-negative integers of at most $4$ digits whose digits are increasing? The answer to this problem is $10\choose 4$. But, I want to know why are we using combination instead of permutation when the order matters in this question? AI: Note that $\binom {10}{4}=210$ is only correct if we count leading $0$s as distinct digits, and so allow $123 \equiv 0123$, but disallow $12 \equiv 0012$. If any number of leading $0$s are allowed (in which case the phrase "non-negative integers" also allows $0$ itself) then the answer is $\binom 9 4 + \binom 9 3 + \binom 9 2 + \binom 9 1 + \binom 9 0 = 256$.
H: Serge Lang - Introduction to linear algebra, Linear Mappings I have these problems in Serge Lang's Introduction to linear algebra's Linear Mappings section. (a) What is the dimension of the subspace of $R_n$ consisting of those vectors $A = (a_1, ... ,a_n)$ such that $a_1 + ... + a_n = 0$? I did the following. A is a set of linearly dependent vectors. $a_1 + ... + a_{n-1} = -a_n$. So I think that the subspace can be generated by $(a_1, ... ,a_{n-1})$ Consequently the dimension of the subspace is $n - 1$. Am I right? (b) What is the dimension of the subspace of the space of $n$ x $n$ matrices $(a_{ij})$ such that $a_{11} +···+a_{nn} = \sum_{i=1}^n a_{ii} = 0?$ The second one should somehow be similar to the first one. But I don't understand how to solve it. And also I don't see how can I make use of linear mappings in these problems. The answers in the book's answer section are (a) $n - 1$ (b) $n^2 - 1$. Any help is appreciated. Thanks in advance. AI: For part $(a)$ Think of the linear map $T:\mathbb{R}^n\to \mathbb{R}$ given by $T(a_1,a_2,...,a_n)=a_1+a_2+..+a_n$ Notice that $T$ is non-zero linear transformation and the co-domain space is of dimension $1$ What is $\operatorname{Ker}(T)$ ? What happens if we apply the Rank-Nullity Theorem ? Similarly for $(b)$ ,take the mapping $U:M_n(\mathbb{R})\to \mathbb{R}$ defined by $U(A) =\operatorname{tr}(A)$ .
H: Find the determinant of a matrix $A$, such that $A^4 + 2A = 0$ Find the determinant of an revertible, $6 \times 6$ matrix $A$, such that $A^4 + 2A = 0$ This question seems odd to me, because when I tried to solve it: $A^4 + 2A = O$ $(A^3+2I)A = O\det$ $\|(A^3+2I)\|\|A\| = \|O\|$ But I know that $\|A\|$ is revertible, thus $\|A\| != 0$, which must mean that $\|(A^3+2I)\| = 0$. But if I try: $\|(A^3+2I)\|\|A\| = \|O\|$ $0\|A\| = 0$ which means it true for every $6 * 6$ matrices, but the question specified $1$. What am I getting wrong here, and how should I solve this? edit: Here is another thing I have tried: $A^4+2A=0$ $(A^3+2I)A = 0$ $A^3+2I=0$ $A^3=-2I$ $A^3=-2AA^-1$ $A^2=-2A^-1\det$ $\|A^2\|/2=-2(1/\|A\|)$ $\|A\| \|A\| / 1 = -2/\|A\|$ $\|A\|^3 = -2 $ $\|A\|=-1.2599$ AI: $A^4+2A = 0 \implies A^4 = -2A \implies \det(A^4) = \det(-2A^4) \implies \det(A)^4 = (-2)^6\det(A)$. Let $\det(A) = x$, then the last equation reduces to $x^4 = (-2)^6x$. So $$x(x^3+2^6) = 0$$ Since $A$ is invertible, the only possibility is $x = -2^2$.
H: Variance of sum of independent random variables - case of undefined density When the density $f_{X, Y}$ is not defined for independent random variables $X, Y$, is it possible to say anything beyond \begin{align} Var(X + Y) &= Var(X) + Var(Y) + 2 \int_{\Omega} (X - E(X))(Y-E(Y)) dP, \end{align} in terms of simplifying the last integral? AI: If $X,Y$ are independent random variables defined on the same probability space then the rule: $$\mathsf{Var}(X+Y)=\mathsf{Var}X+\mathsf{Var}Y$$ is valid if $\mathbb EX^2$ and $\mathbb EY^2$ are both finite. For that we do not need a density. Observe that because of independence:$$\begin{aligned}\int\left(X-\mu_{X}\right)\left(Y-\mu_{Y}\right)dP & =\int\left(X-\mu_{X}\right)dP\times\int\left(Y-\mu_{Y}\right)dP\\ & =\left(\mu_{X}-\mu_{X}\right)\left(\mu_{Y}-\mu_{Y}\right)\\ & =0 \end{aligned} $$
H: Linear function preserving the Gram determinant In Euclidean space $X$ the Gram's determinant of a system of vectors $x_1,...,x_k\in X$ is called the determinant of $k\times k$ matrix $ [\langle x_i,x_j \rangle]$: $ G(x_1,..,x_k)=\det[\langle x_i,x_j \rangle]. $ In $n$ dimensional Euclidean space $X$, let $f: X\rightarrow X$ be a linear mapping and let $k\in \{2,...,n-1\}$ be a fixed number. I wish to prove not using exterior algebra that if $ G(f(x_1),...,f(x_k))=G(x_1,...,x_k) $ for each $x_1,...,x_k\in X$, then $f$ is orthogonal mapping. Thanks AI: You can solve this problem using polar decomposition. Write $f = o \circ p$ where $o \colon X \rightarrow X$ is orthogonal and $p \colon X \rightarrow X$ is self-adjoint and positive semi-definite. To show that $f$ is orthogonal, it is enough to show that $p = \operatorname{id}$. Choose an orthonormal basis of eigenvectors $v_1, \dots, v_n$ of $p$ with $p(v_i) = \lambda_i v_i$ where $\lambda_i \geq 0$ and choose two distinct indices $1 \leq i < j \leq n$. Since $k \leq n-1$, we have $k - 1 \leq n - 2$ and so we can choose distinct indices $l_1, \dots, l_{k-1}$ such that $l_1,\dots,l_{k-1},i,j$ are all distinct. Then $$ G(v_i,v_{l_1},\dots,v_{l_{k-1}}) = 1 $$ while $$ G(f(v_i),f(v_{l_1}),\dots,f(v_{l_{k-1}})) = G(p(v_i), p(v_{l_1}), \dots, p(v_{l_{k-1}})) \\ = G(\lambda_i v_i, \lambda_{l_1} v_{l_1}, \dots, \lambda_{l_{k-1}} v_{l_{k-1}}) = \lambda_i^2 \cdot \prod_{r = 1}^{k-1} \lambda_{l_r}^2 $$ so $$ \lambda_i^2 \prod_{r=1}^{k-1} \lambda_{l_r}^2 = 1. $$ In particular, $\lambda_{l_r} > 0$ for all $r$ and since the eigenvalues are non-negative, we get $$ \lambda_i \prod_{r=1}^{k-1} \lambda_{l_r} = 1. $$ The same holds with $i$ replaced by $j$ and so we get $$ \lambda_i \prod_{r=1}^{k-1} \lambda_{l_r} = \lambda_j \prod_{r=1}^{k-1} \lambda_{l_r} \implies (\lambda_i - \lambda_j) \prod_{r=1}^{k-1} \lambda_{l_r} = 0 \implies \lambda_i = \lambda_j. $$ Thus, we deduce that $\lambda_1 = \dots = \lambda_n$ and $\lambda_i^k = 1$ which implies that $\lambda_i = 1$ for all $1 \leq i \leq n$ so $p = \operatorname{id}$.
H: Is this Factorization? I'm doubtful about the some parts of the solution to this question: Suppose that the real numbers $a, b, c > 1$ satisfy the condition $$ {1\over a^2-1}+{1\over b^2-1}+{1\over c^2-1}=1 $$ Prove that $$ {1\over a+1}+{1\over b+1}+{1\over c+1}\leq1 $$ The solution says that noticing $a\geq b\geq c$ leads to ${ a-2\over a+1 }\ge{ b-2\over b+1}\ge{ c-2\over c+1}$ and $ { a+2\over a-1 }\le{ b+2\over b-1}\le{ c+2\over c-1} $, But I don't understand how. If $a\geq b\geq c$, it is quite visible that $a-2\geq b-2\geq c-2$ and $a+1\geq b+1\geq c+1$, but this does not essentially lead to ${ a-2\over a+1 }\ge{ b-2\over b+1}\ge{ c-2\over c+1}$. Also after this, the solution uses Chebyshev's Inequality: $$ 3\left( \sum_{cyc}{a^2-4\over a^2-1}\right) \leq \left(\sum_{cyc} {a-2\over a+1} \right)\left(\sum_{cyc} {a+2\over a-1} \right) $$ And then states that the LHS is $0$ by hypothesis. But as Chebyshev's Inequality uses ascending ordered real numbers, this should be false and I also don't know how LHS became $0$. I think the solution in the book is either flawed or has a typographical error. Any explanation will be accepted thankfully! AI: Because for $a\geq b$ we have $$\frac{a-2}{a+1}-\frac{b-2}{b+1}=\frac{3(a-b)}{(a+1)(b+1)}\geq0.$$ The rest is similar. Also, the sum $\sum\limits_{cyc}{a^2-4\over a^2-1}=\sum\limits_{cyc}\left(\frac{a-2}{a+1}\cdot\frac{a+2}{a-1}\right)$ is the smallest and it gives a proof of Chebyshov by Rearrangement. Another way: $$1-\sum_{cyc}\frac{1}{a+1}=\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a+1}\right)=$$ $$=\sum_{cyc}\left(\frac{1}{3}-\frac{1}{a+1}-\frac{1}{4}\left(\frac{1}{3}-\frac{1}{a^2-1}\right)\right)=\sum_{cyc}\frac{(a-2)^2}{4(a^2-1)}\geq0.$$
H: Why is the blow up of a submanifold of $\mathbb{P}^n$ again projective I saw somewhere that the blow up (at any point) of a submanifold of $\mathbb{P}^n$ is still projective. I have the feeling that this is a consequence of the Kodaira embedding theorem, any thoughts? AI: Yes, this follows from Kodaira embedding. If $L$ is a positive holomorphic line bundle on $X$ and $\hat{X} = \operatorname{Bl}_{\{x\}} X$ is a one-point blow-up with projection $\sigma: \hat{X} \to X$ and exceptional divisor $E = \sigma^{-1}(x)$ then $\mathcal{O}_{\hat{X}} ( -E) \otimes \sigma^L^k$ is positive for $k$ large enough, so $\hat{X}$ is again projective by Kodaira embedding. To see that last bit notice that the curvature of the pullback connection on $\sigma^L$ is positive on all vector except those in $TE$. But in a small coordinate neighborhood $U$ centered on $x$, $\hat{U} = \sigma^{-1}(U)$ is given by the incidence variety $\{(y, l) \in U \times \mathbb{P}^{n-1} \mid y \in l\}$ which has a projection $p$ to $\mathbb{P}^{n-1}$. One can see that $p^* \mathcal{O}_{\mathbb{P}^{n-1}}(1)$ has a section whose zero set is exactly $E$, so $p^* \mathcal{O}_{\mathbb{P}^{n-1}}(-1) \cong \mathcal{O}_{\hat{U}}(E)$ and $p^* \mathcal{O}_{\mathbb{P}^{n-1}}(1) \cong \mathcal{O}_{\hat{U}}(-E)$ . Pulling back the canonical positive metric on $\mathcal{O}_{\mathbb{P}^{n-1}}(1)$ we get a metric on $\mathcal{O}_{\hat{U}}(-E)$ that is positive on vectors in $TE$. Then you use a bump function of unity that is $1$ on a neighborhood of $\{0\} \times \mathbb{P}^{n-1}$ to extend this to a metric on $\mathcal{O}_{\hat{X}}(-E)$ and pick a large $k$ such that the overall metric on $\mathcal{O}_{\hat{X}} ( -E) \otimes \sigma^*L^k$ is positive
H: How to simplify $\frac x{\|x\|} Is there a simplification for this relation? $$ \frac{x}{\left\| x\right\| } $$ where $x=a+i b$, $a$ and $b$ are reals. AI: By polar form $x=\|x\|e^{i\theta}$ we obtain $$\frac{x}{\left\| x\right\| }=e^{i\theta}$$ with $\theta = \operatorname {atan2} \left(b, a \right)$ (see atan2).
H: If every two-dimensional (vector) subspace of a normed space is an inner product space, then so is that normed space Let $\big( X, \lVert \cdot \rVert \big)$ be a (real or complex) normed space. Suppose that, for every two-dimensional (vector) subspace $Y$ of $X$, the norm on $Y$ (i.e. restriction of the norm of $X$ to $Y$) satisfies the parallelogram identity and thus is induced by a certain inner product on $Y$. Can we conclude from this that the norm of $X$ is also induced from an inner product on $X$? That is, if every two-dimensional (vector) subspace of a normed space is an inner product space, then can we prove that that normed space is itself an inner product space? AI: Since $\|a+b\|^2+\|a-b\|^2=2(\|a\|^2+\|b\|^2)$ is trivial when $a\parallel b$, and otherwise $a,\,b$ span a $2$-dimensional subspace of $X$, the identity is true in general, and the inner product on $X$ is then defined by a suitable polarization identity.
H: Other absolute value definitions in $\mathbb R$ I know these definitions for the absolute value (or module): given a real number $x$, then $$\bbox[yellow] {\|x\|=\begin{cases}x & \text{if } x\geq 0\\ -x& \text{if } x< 0\end{cases}}$$ or $$\bbox[yellow] {\|x\|=\max\{x,-x\}}$$ Are there other definitions in $\mathbb R$ (for example using $\text{sgn}\, x$)? PS: The question is referred to high school students. AI: First of all, what you are asking is not really notation. The word "notation" refers to how we write a particular concept. The concept of "absolute value" has really only one notation: the vertical bars. That is, $\|x\|$ is the standard notation for the concept "absolute value of $x$". What you are asking is the definitions of $\|x\|$, and in particular, you are listing two equivalent definitions of $\|x\|$. I can think of two more equivalend definitions for $\|x\|$: $\|x\| = \mathrm{sign}(x) \cdot x$ $\|x\| = \sqrt{x^2}$
H: Epsilon delta for infinite limits The limit: $$\lim_{x \to \infty} f(x) = -\infty$$ iff $$\big\{ \forall M>0, \exists N >0 \ s.t\ \forall x > N \implies f(x) <-M \big\}$$ This means for all $x \in (N,\infty)$ , $f(x)$ lies in $(-\infty,M)$, however this doesn't account for when $f(x)$ is not defined so it is not $<-M$ AI: As often, this definitions forgets the clause $\forall x\in\text{Dom}(f)$.
H: prove that this function isn't Lipschitz continuous I'm trying to prove that this function isn't Lipschitz continuous: $f:[-1,1]\to \Bbb R:x\mapsto x^{2}\sin1/x^{2}$ when $x$ not equal to zero and $0$ if $x=0$. Now I just can't find the right $x$ and $y$ to prove it. Take $M$ from $\Bbb R^+$. Chose then $x$ and $y$ (I can't find them) and it follows that: $$\|f(x)-f(y)\|....>M\|x-y\|$$ but without $x$ and $y$ I can't start the proof at all, can someone just help me with the $x$ and $y$, I would like to finish to prove on my own. AI: It is easy to see that this function is differentiable at every point. If it is Lipschitz it would follow that its derivative is bounded. But $f'(x)=2x\sin (\frac 1 {x^{2}})-(2/x) \cos (\frac 1 {x^{2}})$. This is unbounded along the sequence $(\frac 1 {\sqrt {2n \pi}})$. If $x$ and $y$ are sufficiently close to $\frac 1 {\sqrt {2n\pi}}$ then $\|f(x)-f(y)\| $ would exceed $M\|x-y\|$.
H: Finding Expected Value of Conditional Poisson Distribution Consider a random variable X ~ Poisson (1). Namely, $P(x=k) = \frac{e^{-1}}{k!}$ , k=0,1,2,... I'm trying to solve for $\mathbb{E}\{X\|X\geq 1\}$. My approach: Given that $X\geq1$ then we know that at least one person has arrived (using arrival / no arrival terminology). Using the fact that Poisson is memoryless then $\mathbb{E}\{X\}=\mathbb{E}\{Y+1\}$ where $Y$ has the same distribution as above Poisson distribution. The solution afterwards is really simple. Is that approach correct? AI: You can use the equality: $$\mathbb EX=\mathbb E[X\mid X\geq1]P(X\geq1)+\mathbb E[X\mid X=0]P(X=0)$$leading to:$$1=\mathbb E[X\mid X\geq1](1-P(X=0))=\mathbb E[X\mid X\geq1](1-e^{-1})$$ hence to:$$\mathbb E[X\mid X\geq1]=\frac{1}{1-e^{-1}}=\frac{e}{e-1}$$
H: In an $n\times(n+1)$ nonnegative matrix, there is a positive pivot at which the row sum is greater than the column sum Let $A$ be a matrix of shape $n\times (n+1)$ with non-negative real entries, which has at least one positive entry in each column. Show that there is an $a_{ij} > 0$ such that the sum over the $i$-th row is higher than sum over the $j$-th column. I found this exercise in a book about combinatorics and algebra. Thus, I tried methods known from linear algebra and combinatorics to solve this problem. My main approaches are: I use induction on $n$ and assume the opposite in the induction step, i.e. that for all $a_{ij} > 0$ the sum over the $i$-th row is less than or equal to the sum over the $j$-th column. Now, I observed that if there is a $a_{ij} > 0$ such that the sum over the $i$-th row is equal to the sum over the $j$-th row, then the matrix gained by deleting the $i$-th row and $j$-th column does still fulfill the prerequests. Finally, by doing this step as often as possible, we obtain a matrix such that if $a_{ij} > 0$, then the sum over the $i$-th row is less than the sum over $j$-th column or in the other case, can conclude a contraction since the matrix has the shape $0 \times 1$. Then, I tried to conclude a contradiction also for the first case, which I can't do. AI: By permuting the rows and columns of $A$, we may assume that the matrix has descending row sums $r_1\ge r_2\ge\cdots\ge r_n$ and descending column sums $c_1\ge c_2\ge\cdots\ge c_{n+1}$. Now consider the following condition: $$ a_{ij}=0 \text{ whenever } r_i>c_j.\tag{1} $$ Assume for the moment that it is true. There are two possibilities: $r_i\le c_{i+1}$ for every $i$. Then $\sum_{i=1}^nr_i\le\sum_{i=1}^nc_{i+1}=\sum_{j=2}^{n+1}c_j$. It follows that the first column of $A$ is zero, which is a contradiction to the assumption that each column of $A$ contains a positive entry. $r_k>c_{k+1}$ for some $k$. Among all such indices $k$, we pick the largest one. Then $r_1\ge\cdots\ge r_k>c_{k+1}\ge\cdots\ge c_n$ and condition $(1)$ implies that $$ a_{ij}=0 \text{ for all } i\le k \text{ and } j\ge k+1.\tag{2} $$ Hence $k<n$, or else the last column of $A$ will be zero, which is a contradiction. Since $k$ is the largest possible index such that $r_k>c_{k+1}$, we have $r_i\le c_{i+1}$ for all $i>k$. In turn, $$ \sum_{j>k+1}c_j =\sum_{i>k}c_{i+1} \ge\sum_{i>k}r_i \color{red}{\ge}\sum_{i>k}\sum_{j>k+1}a_{ij} =\sum_{j>k+1}\sum_{i>k}a_{ij} \stackrel{\text{by } (2)}{=}\sum_{j>k+1}\sum_{i=1}^na_{ij} =\sum_{j>k+1}c_j. $$ Thus equality holds in the second inequality above. That is, $$ a_{ij}=0 \text{ for all } i>k \text{ and } j\le k+1.\tag{3} $$ But then $(2)$ and $(3)$ imply that the $(k+1)$-th column of $A$ is zero, which is a contradiction. Hence condition $(1)$ does not hold. Since $r_1\ge\frac{1}{n}\sum_{i,j}a_{ij}>\frac{1}{n+1}\sum_{i,j}a_{ij}\ge c_n$, the set $\{(i,j): r_i>c_j\}$ is non-empty. The falseness of $(1)$ thus implies that $a_{ij}>0$ and $r_i>c_j$ for some $(i,j)$.
H: Let $f(n) = an^2 + bn + c$ be a quadratic function. Show that there's an $n ∈ N$ such that $f(n)$ is not a prime number. Let $f(n) = an^2 + bn + c$ be a quadratic function, where $a, b, c$ are natural numbers and $c ≥ 2$. Show that there is an $n ∈ N$ such that $f(n)$ is not a prime number. I figure this might have something to do with factorising the $f(n)$. So, let $c=n$, then: $f(n) = c(ac+b+1)$ The solutions say immediately after this step that $c(ac+b+1)$ is divisible by $c ≥ 2$, but I do not understand why or how this is so. Any clarifications or further explanation on how to prove this would be much appreciated! This is from a grade 11 Maths Specialist textbook under the topic of "for all" and "there exists" proofs. AI: $f(n)$ is a function from the natural numbers to the natural numbers. Given $n$, we compute $f(n) = an^2+bn+c$. The question is to prove : there exists a natural number $n$ such that $f(n)$ is not prime. So, we must prove that one of $f(1),f(2),f(3),...$ is not prime. Therefore, all we need is ONE value of $n$, or one natural number so that when I evaluate the function at that natural number, I get a composite number. The proof then shows that $f(c)$ is not prime. Now, this mean that for ONE value of $n$, we have that $f(n)$ is not prime : that is, when we set $n=c$. How is $f(c)$ shown to be not prime? We find a factor of $f(c)$ which is not $1$ or $f(c)$ : this shows that $f(c)$ is not prime. By factorization, substituting $n=c$ gives $f(c) = ac^2+bc+c = c(ac+b+1)$, and $ac+b+1$ is a natural number, so $c$ divides $f(c)$, but $c \geq 2$ by assumption, and $ac+b+1 \geq ac \geq c \geq 2$, so $f(c)$ is not prime because there exists a factor of it which is not $1$ or itself(either $c$ or $ac+b+1$ will work). Thus, we see that there exists an $n$ such that $f(n)$ is not prime.
H: Evaluate right hand limit $\lim_{x\to 2^+} (x^2 + e^{\frac{1}{2-x}})^{-1}$ $$\lim_{x\to 2^+} \frac{1}{x^2 + e^{\frac{1}{2-x}}} = \frac{1}{\lim_{x\to 2^+} x^2 + \lim_{h\to 0} e^{\frac{1}{h}}} =\frac{1}{4+\lim_{h\to 0} e^{\frac 1h}}$$ How do I solve further ? AI: Only a small mistake, it should be $$L=\frac{1}{4+\lim_{h\to 0} 2^{-1/h}}=1/4.$$ Remember, $h$ is always positive and $$\lim_{x\to 2^+}f(x) \implies \lim_{h \to 0} f(2+h).$$ Similarly, $$\lim_{x \to 2^-}f(x)=\lim_{h \to 0} f(2-h)$$
H: Can an orthogonal matrix that represents a linear transformation from $\mathbb{R}^n \to \mathbb{R}^n$ have no eigenvalues? I know that if the matrix is normal and represents a transformation in a unitary space then it can be unitarily diagonalized, so it must have eigenvalues. Plus, its characteristic polynomial is never a constant, so it has roots according to the Fundamental Theorem of Algebra. But what if we talk about orthogonal matrices that represent transformations in a Euclidean vector space? Can they have no eigenvalues? AI: Yes, see for instance $n=2$ and the matrices in the form $\begin{pmatrix}\cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha\end{pmatrix}$ for $\alpha\notin\pi\Bbb Z$.
H: Find the last two digits of $7^{100}-3^{100}$ Find the last two digits of $7^{100}-3^{100}$ From Euler's theorem one gets that $\phi(100) = 40 \Rightarrow 7^{40} \equiv 1 \pmod{100}, 3^{40} \equiv 1 \pmod{100}.$ I couldn't really work this out without using a calculator to compute the powers. How can I continue from here? I could write the expressions as $7^{100} \equiv 7^{40}\cdot7^{60} \equiv 7^{60} \pmod{100}$, but I would still need to deal with the $7^{60}...$ AI: Alternatively, notice:$$(10-3)^{100}-3^{100} = \sum_{k=0}^{99}\binom{100}{k}10^{100-k}(-3)^k \equiv0 \pmod {100}$$
H: Inverse of multiple projections of the same point I have a vector $\vec{x}$ in 3D space that is unknown. I do know $\vec{p_1}$, $\vec{p_2}$, $\vec{p_3}$ which are orthogonal projection vectors of $\vec{x}$ onto lines $P_1$, $P_2$, $P_3$, all going through the origin and not parallel to each other. From $\vec{p_1}$,$\vec{p_2}$ and $\vec{p_3}$ I can calculate the unit vectors in the direction of the lines and with those I can calculate the linear transformation matrices $A_1$, $A_2$ and $A_3$ of the projections with $\vec{p_1} = (\vec{x} \vec{e_{p1}}) \vec{e_{p1}}$ So I have: $\vec{p_1} = A_1 \vec{x}$ $\vec{p_2} = A_2 \vec{x}$ $\vec{p_3} = A_3 \vec{x}$ Since no single projection matrix A is invertible, how would I calculate $\vec{x}$? $\vec{x}$ is definitely uniquely determined. For 2D I can show that $\vec{p_1} + \vec{p_2} = (A_1+A_2) \vec{x}$ is solvable for $\vec{x}$, if $\vec{p_1}$ and $\vec{p_2}$ are not parallel. But the proof in 3D is much harder, so maybe there is an easier way. AI: An easy way is to assemble the matrices as $$\begin{bmatrix}A_1\\A_2\\A_3\end{bmatrix}x=\begin{bmatrix}p_1\\p_2\\p_3\end{bmatrix}$$ then use least squares method to find $x$. This gives the solution $$x=(A_1^TA_1+A_2^TA_2+A_3^TA_3)^{-1}(A_1^Tp_1+A_2^Tp_2+A_3^Tp_3)$$ If you're not familiar with that method, then suppose $$x=\alpha_1p_1+\alpha_2p_2+\alpha_3p_3$$ $$\therefore p_1=A_1 x=\alpha_1p_1+\alpha_2A_1p_2+\alpha_3 A_1p_3$$ where $A_1p_j$ can be calculated. Repeat for the other projections $p_i=A_ix=\sum_j\alpha_jA_ip_j$ to get three equations in $\alpha_i$.
H: Choose correct answers on eigen value: A = [0 0 L 0 1 1 0 L 0 0 0 1 L 0 0 M M N N M 0 0 L 1 0] a) eigenvalue are purely real b) 0 is the only eigenvalue c) eigenvalues are n-th roots of unity Exp(2πi/n) for i = 0,1,...,n-1 d) none of these What can be the easiest way to get answers in these type of questions? AI: I think that the best option is really to pick nice values for $N$, $M$, and $L$ and try to compute. This matrix doesn't appear to have enough structure to answer this type of question quickly (although I might be missing something). Sketch: 1.) Compute eigenvalues for $L=1$, $M=0$, and $N=1$. This gets rid of two possibilities (note that the rows sum to $2$). 2.) Compute the eigenvalues for $L=0$, $M=0$, and $N=1$. This gets rid of one more case (note that this choice eliminates most of the terms in the characteristic equation and Descartes' rule of signs can be applied to the polynomial to count the real roots).
H: Derive the change of coordinates without using differentials Please forgive me for using the following lousy notations. I'm just a beginner in DG. Given two charts $(\psi,\{x_i\}_{i=1}^n),(\phi,\{y_i\}_{i=1}^n)$ around a point $p$ in an n-dimensional manifold $M$, I'd like to derive the change of coordinates $$\frac{\partial(f\circ\psi^{-1})}{\partial x_i}(\psi(p))=\sum_{j=1}^n\frac{\partial [r_j\circ(\phi\circ\psi^{-1})]}{\partial x_i}(\psi(p))\frac{\partial(f\circ\phi^{-1})}{\partial y_j}(\phi(p))$$ for $f\in C^\infty(M)$. Here $r_j$ denotes the $j$-th coordinate function that takes out the $j$-th component of a Euclidean vector. After consulting several textbooks, I found some solutions in terms of differentials or push-forwards. But what if I want to obtain the result by applying the chain rule directly? I couldn't come up with the details. Can anyone help me? Thank you. AI: That equation you wrote IS the multivariable chain rule, applied to the composition formula $$f \circ \psi^{-1} = (f \circ \phi^{-1}) \circ (\phi \circ \psi^{-1}) $$
H: Is this formula equal to 1? Is this formula equal to $1$? $$\frac{\left\| \eta \right\| }{\sqrt{\eta } \sqrt{\eta ^}}$$ where $\eta$ is complex. And if so, how can I prove that? AI: By definition for $\eta=a+bi\in \mathbb C$ $$\|\eta\|=\sqrt {a^2+b^2}$$ $$\eta\cdot \eta^ =(a+bi)(a-bi)=a^2+abi-abi+b^2=a^2+b^2=\|\eta\|^2\in \mathbb R$$ then what is true is that $$\sqrt{\eta\cdot \eta^}=\|\eta\|\in \mathbb R$$ but, as already pointed out, $\sqrt{\eta}\sqrt{\eta^}\neq \sqrt{\eta\cdot \eta^}$ since in general $\sqrt{\eta}$,$\sqrt{\eta^}\in \mathbb C$.
H: Inequality in proof concerning trace class norm Consider the following fragment from Murphy's '$C^$- algebras and operator theory': Why is the marked inequality true? I.e. why is $\Vert w'' \Vert \leq \Vert v \Vert?$ AI: $$\\|w''\\|=\\|w^{\prime}vw\\|\le\\|w^{\prime}\\|\\|v\\|\\|w\\|=\\|v\\|$$ P.S. $\\|w\\|=1$ for partial isometries since $\\|wx\\|\le\\|x\\|$ for all $x$, with equality on some non-trivial subspace. A partial isometry on a Hilbert space is defined by $(w-1)Y=0$ on some closed subspace $Y$ and $wY^\perp=0$. Even if one works purely in $C^$-algebra terms, so a partial isometry is defined by $(w^w)^2=w^w$, then $\\|w\\|^2=\\|w^w\\|=\\|(w^w)^2\\|=\\|w\\|^4$.
H: Is the product of a maximal subgroup and a cyclic subgroup a group? $G$ is a finite group. I wanted to show that for a maximal subgroup $M$ of $G$, if $g\in G\setminus M$, then $M\langle g\rangle=G$. It it true? My argument is that if $M\subset M\langle g\rangle\neq G$, then it is a contradiction to the maximality of $M$ in $G$. But if neither $M$ nor $\langle g\rangle$ is not normal in $G$, then $M\langle g\rangle$ is not necessarily a subgroup of $G$. Hence we cannot get the contradiction from that. Could you tell me how to fix my proof? Or could you give me a counterexample if it is not true? Any help would be appreciated. AI: If by $M\langle g\rangle$ you mean the subgroup generated by $M$ and $g$, then yes, you are correct. But if you mean the set product then it clearly cannot work, because the cardinality of it is at most $\|M\|$ times $o(g)$, the order of $g$. If $o(g)=2$ and $\|G:M\|>2$, like almost every maximal subgroup and many group elements, then the set product cannot be all of $G$.
H: Is this answer on eigenvector diagonalisation wrong? This is the question and answer from MIT OCW 18.06 on eigenvectors and diagonalisation: Two things I don't understand: Shouldn't the eigenvector for $\lambda = -0.3$ be $\begin{bmatrix} -1 \\ 1 \end{bmatrix}$ because the second column is the free variable? Assuming the answer is right in 1, wouldn't $S^{-1} = \frac{1}{3} \begin{bmatrix} -1 & -1 \\ -4 & 9 \end{bmatrix}$ because $\begin{bmatrix} a & b \\ c & d \end{bmatrix}^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$? AI: It doesn't matter. $(1,-1)$ is as good an eigenvector as $(-1,1)$ or $(3,-3)$ are. $ad-bc=-13$, therefore a minus sign got absorbed in $+\frac1{13}$.
H: Quadratic closure of $Z_2$ Find or disprove the existence of a minimal quadratically closed field extension of $Z_2$ I.e. $$\forall b,c \in S, \exists x\in S, x^2+ bx+c = 0$$ Note: Because $S$ is a field we can reduce every polynomial to a monic one by multiplying by $a^{-1}$. Attempt 1 In $Z_2$, $x^2+x+1$ is irreducible, by introducing a new element, $b$, such that $b^2 + b + 1$(Equivalent to $Z_2[b]/(b^2+b+1)$). I thought the field was quadratically closed. but, $x^2+x+b$ was irreducible. I tried to add more elements but it became hard to test all the quadratics. Attempt 2 I tried proving $$\forall a, a +a=0 \Rightarrow \exists b,c \in S, \forall x\in S, x^2+ bx+c \neq 0$$ By constructing $b$ and $c$, I didn’t manage to. AI: Let $F$ be an algebraic closure of $\Bbb Z_2$. Now consider the following sequence of sets of subfields of $F$: $\mathcal X_1 = \{\Bbb Z_2\}$, and for any $n>1$ we set $\mathcal X_n$ to contain all elements of $\mathcal X_{n-1}$, as well as all quadratic extensions of all of the fields in $\mathcal X_{n-1}$ (still as subfields of $F$). Now take the union $\mathcal X$ of all these families. It contains $\Bbb Z_2$. For any field $E\in \mathcal X$, any quadratic extension of $E$ is also in $\mathcal X$. Each element of $\mathcal X$ can be reached from $\Bbb Z_2$ doing some finite number of quadratic extensions. And finally, any two fields of $\mathcal X$ are both subfields of some other common superfield $\mathcal X$. The union of all of the fields in $\mathcal X$ (as subfields / subsets of $F$) gives you the smallest quadratically closed subfield of $F$.
H: If $xyz=32$, find the minimal value of If $xyz=32;x,y,z>0$, find the minimal value of $f(x,y,z)=x^2+4xy+4y^2+2z^2$ I tried to do by $A.M.\geq M.G.$: $\frac{x^2+4y^2+2z^2}{2}\geq\sqrt{8x^2y^2z^2}\to x^2+4y^2+2z^2\geq32$ But how can I maximaze 4xy? AI: Your application of AM-GM is wrong. The statement for AM-GM states that for positive integers $a_1, a_2, \dots, a_n$: $$\frac{a_1 + a_2 + \dots + a_n}{n} \geq \sqrt[n]{a_1a_2\dots a_n}$$ with equality at $a_1 = a_2 = \dots = a_n$. Observe that $$\frac{x^2+2xy+2xy+4y^2+z^2+z^2}{6} \geq \sqrt[6]{16x^4y^4z^4} = 16 \implies x^2 + 4xy + 4y^2 + 2z^2 \geq 96.$$ Equality occurs when $x^2 = 2xy = 4y^2 = z^2 \implies (x,y,z)=(4,2,4)$.
H: Bijection between topology and topological space I've been working with a class of topological spaces $(X,\tau)$ with the following property: There exists a bijection $f:X\to \tau\setminus\varnothing$ such that for all $x\in X$, $x\in f(x)$ (And, $\|X\|>1$ to avoid discrete topological spaces). Is there a name for such a topological space? Are the reals under the standard topology an example of this topology? AI: Any space where $\|X\| = \|\tau\| = \|U\|$ for every non-empty open $U$ and (ETA) $\|\{U \in \tau: x \in U \}\| = \|X\|$ for every $x \in X$ will have such a bijection, by a simple transfinite construction (assuming AC). That includes the reals. Index $X = \{x_\alpha: \alpha < \kappa\}$ and $\tau \setminus \emptyset = \{U_\alpha : \alpha < \kappa\}$ by $\kappa = \|X\|$. Construct $f$ recursively by letting $f(x_\lambda)$ be $U_\alpha$ for the least $\alpha$ such that $x_\lambda \in U_\alpha$ but $U_\alpha \neq f(x_\beta)$ for any $\beta < \lambda$. This is always possible as $\|\lambda\| < \kappa$ so there must be open sets containing $x_\lambda$ that haven't been used yet. $f$ is obviously injective and has each $x \in f(x)$ so it only remains to show it is surjective. If some $U_\mu$ is not in the image of $f$, then at every stage of the construction when $x_\lambda \in U_\mu$, $U_\mu$ was not chosen and so $f(x_\lambda) = U_\alpha$ for some $\alpha < \mu$ but that is not possible by cardinality (there are $\kappa$ points in $U_\mu$)
H: Negate this statement: There exist $x, y ∈ \Bbb{R}$ such that $x < y$ and $x^2 > y^2$ Negate this statement: There exist $x, y ∈ \Bbb{R}$ such that $x < y$ and $x^2 > y^2$ From my understanding, "there exists" becomes "for all" "and" becomes "or" by De Morgan's laws equality signs "reverse" So, our negated statement is: For all $x, y ∈ \Bbb{R}$, $x ≥ y$ or $x^2 ≤ y^2$ However, the textbook's answer says that the negated statement is: I do not understand why the negated "implies" $x^2 ≥ y^2$ at the very end. Wouldn't it be $x^2 ≤ y^2$ (to "reverse" the equality sign)? Any clarification would be much appreciated! This is from a grade 11 Maths Specialist textbook on proofs involving the use of negation. AI: Your wording is a little clumsy, but you are right. The negation is $$\forall x,y\in \mathbb R: x\ge y\lor x^2\le y^2.$$ Given that $$a\implies b\equiv \lnot a\lor b$$ we can also write $$\forall x,y\in \mathbb R: x<y\implies x^2\le y^2.$$ I also disagree with the book.
H: $\|p(y)\| > \|y\| + \|\delta y\|$ Let $p(y)$ be a complex polynomial of degree at least two and $\delta$ some complex number. I want to show $$\|p(y)\| > \|y\| +\|\delta y\|$$ for $\|y\| > \kappa$ sufficiently large. I have tried applying the triangle or reverse triangle inequalities but I can't seem to get the absolute values to match up. AI: As in the comment, since $p(y)$ is a polynomial of degree greater than or equal to two, $\left\| \frac{p(y)}{y} \right \|\to \infty$ as $\|y\| \to \infty$. We can choose sufficiently large $\|y\|$ so $\left\| \frac{p(y)}{y} \right \| > 1+\|\delta\|$ and the result follows.
H: proving stochastic process is independent Guys can anyone help me with this question? On a probability space let be filtration $F = (F_n)_{n \in N_0}$ and a real valued adaptive stochastic process $(X_n)_{n \in N_0}$ for all the Borelsets $ A \in B(R) $ we have $P[X_{n+1} \in A \| F_n ] = P [X_{n+1} \in A ]$ P-almost surely I must prove that the family $(X_n)_{n \in N_0}$ with respectt to $P$ is independent. I have no idea how to solve it. I will be thankful for any help AI: To prove family $\{X_n\}_{n \in \mathbb N}$ is independent, it is sufficient and necessary (via definition) to prove that family $\{X_n\}_{ n \in \{1,...,k\}}$ is independent (for any $k \in \mathbb N$). Hence take any $k$ and borel sets $A_1,...,A_k$. We need to show $$ \mathbb P(\bigcap_{j=1}^k \{X_j \in A_j\}) = \prod_{j=1}^k \mathbb P(X_j \in A_j)$$ How can we use our assumption? We're start with left side and condition on $F_{k-1}$, getting: $$ \mathbb P(\bigcap_{j=1}^k \{X_j \in A_j\}) = \mathbb E \mathbb E[ \prod_{j=1}^k 1_{ \{X_j \in A_j\}}\| F_{k-1}] = \mathbb E[\prod_{j=1}^{k-1}1_{\{X_j \in A_j\}} \mathbb E[ 1_{\{X_k \in A_k\}} \| F_{k-1}]] $$ Now use our assumption getting $\mathbb P(X_k \in A_k)$ in the middle which can be throw out of expectation (since it's a constant), getting: $$\mathbb P(\bigcap_{j=1}^k \{X_j \in A_j\})= \mathbb P(X_k \in A_k) \mathbb E[ \prod_{j=1}^{k-1} 1_{\{X_j \in A_j\}}] = \mathbb P(X_k \in A_k) \mathbb P(\bigcap_{j=1}^{k-1} \{X_j \in A_j\})$$ Can you proceed inductivelly now? You need to condition the rest by $F_{k-2}$ then what's left by $F_{k-3}$ and so on, everytime using your assumption, at last getting what we need to prove, that is $$ \mathbb P( \bigcap_{j=1}^k \{X_j \in A_j\}) = \prod_{j=1}^k \mathbb P(X_j \in A_j)$$
H: Let $X \sim P_1$, $Y \sim P_2$, can we find $f$ such that $f(X,Y) \sim P_1 P_2$? Let $X \sim P_1$, $Y \sim P_2$ be two random variables with respective probability laws $P_1,P_2$. We define $P(A)=P_1(A)P_2(A)$ for all $A$'s in the sigma algebra. $P$ is a probability law. Can we find a function $f$ such that $f(X,Y)\sim P$ ? AI: Let $A,B$ be disjoint. Then $$P(A \cup B) = P_1(A \cup B) P_2(A \cup B) = (P_1(A) + P_1(B))(P_2(A) +P_2(B) = P(A) + P(B) + P_1(A)P_2(B) + P_2(A)P_1(B)$$ So, apart from some special cases, this is not a probability measure in general.
H: cardinality of centralizer of an element Suppose $G$ is finite group of order $p^dn$ where $d$ and $n$ are positive integers and $p$ is a prime that does not divide $n$. Show that $G$ contains an element of order $p$ such that the cardinality of its conjugacy class divides $n$ Cardinality of conjugacy class of an element divides the order of group. So it is sufficient to prove that for some element $g$ of order $p$, $p^d\mid C(g)$ where $C(g)$ is the centralizer. Please give a hint. Please do not give solution. Thanks! Edit: I realized that the above attempt does not make use of (i thought it does make use of) the fact that $p$ does not divide $n$. AI: Hints: sylow p-subgroup always has an element of order $p$. Center of p-group is always non trivial, there is an element of order p which commutes all the elements of sylow p-subgroup. Can you continue?
H: What am I doing wrong? Differentiation. I was given the following function: $$ f(t) = \frac{t}{t^2 + 1} $$ And I attempted to derive it and came up with this answer: $$ \frac{-2t^2}{(t^2 + 1)^2} $$ So I was incorrect obviously with this answer, and I am unsure now if you can obtain the correct answer using the chain rule. Perhaps I am just applying it incorrectly? I am able to get the correct result using the quotient rule. I was under the impression that I could rearrange the function so that it is in the form $$f(g(x))$$ as follows: $$ f(t) = t (t^2 + 1)^{-1} $$ AI: $t (t^2 + 1)^{-1}$ is not of the form $f(g(x))$. $t (t^2 + 1)^{-1}$ is multiplication, it is $t \cdot (t^2 + 1)^{-1}$, we multiply $t$ by $(t^2 + 1)^{-1}$ $f(g(x))$ is function composition, it is $(f\circ g)(x)$, it is the function $f$ evaluated at $g(x)$.
H: Question on semialgebra This exercise comes from A First Look At Rigorous Probability (Exercise 2.7.19): Let $\Omega$ be a finite non-empty set, and let $\mathcal{J}$ consist of all singletons in $\Omega$, together with $\emptyset$ and $\Omega$. Show that $\mathcal{J}$ is a semialgebra. Definition is semialgebra is here. I don't think this is true. Consider the following Suppose $\Omega = \{ \{1,2,3,4\}, 1, 2\}$. Then $\mathcal{J} = \{ \Omega, \emptyset, \{1\}, \{2\}\}$. $\Omega \setminus \{1\} = \{ \{1,2,3,4\}, 2\}$, which cannot be written as a disjoint union of the elements in $\mathcal{J}$. What's wrong with what I did? AI: Note that a singleton with respect to $\Omega$ is every set that contains a single element $w \in \Omega$. So in your example, $\{1,2,3,4\}$ is actually a singleton, and hence, it belongs to $\mathcal J$. You can prove that it is indeed a semi-álgebra. First, let $A,B \in \Omega$ , the it is clear that $A\cap B$ is either a singleton or empty, hence, $A\cap B \in \mathcal J$. Secondly, let $A, B \in \mathcal J$, then it’s clear that $A-B$ is either empty, a singleton or the union of singletons. We then conclude that $\mathcal J$ is a semi-algebra.
H: Show that $f\in \mathcal{L}^p$ and $\|\|f\|\|_p \leqslant M.$ Let $(X, \mathcal{B}, \mu)$ be a measured space, $1<p<\infty$, and $q$ the conjugate exponent of $p$ : $\left(\dfrac{1}{p}+\dfrac{1}{q}=1\right)$. Show that $f \in \mathcal{L}^p \implies$ $\displaystyle{\|\|f\|\|_p=\sup\left\{ \left\|\int_X f(x)g(x)d\mu(x) \right\| ; \|\|g\|\|_q \leqslant1\right\}}$. We assume here that the measure $\mu$ is $\sigma$-finite. Let $f : X\to \mathbb{C}$ a mesurable function. we assume that there exists $M\geqslant0$ such that $\displaystyle{\left\|\int_X f(x)g(x)d\mu(x) \right\| \leqslant M}$ for all $g \in \mathcal{L}^p$ with $\|\|g\|\|_q\leqslant 1$. Show that $f\in \mathcal{L}^p$ and $\|\|f\|\|_p \leqslant M.$ My attempt: on the one hand \begin{align} \left\|\int_X f(x)g(x)d\mu(x) \right\| & \leqslant \int_X\|f(x)g(x)\|d\mu(x) \\ & \leqslant \left(\int_X\|f(x)\|^pd\mu(x) \right)^{\frac{1}{p}} \left(\int_X\|g(x)\|^q d\mu(x) \right)^{\frac{1}{q}}\\ & =\|\|f\|\|_p\|\|g\|\|_q \\& \leqslant \|\|f\|\|_p \end{align} on the other hand Let $g_0(x)= \dfrac{\|f(x)\|^p}{\|f(x)\|}$ if $f(x)\not=0$, $g_0(x)=0$ if $f(x)=0$. so $g_0(x)f(x)=\|f(x)\|^p$, for all $x\in X$, then $\|g_0\|^q= \|f\|^{(p-1)q}=\|f\|^p$, so $g_0 \in \mathcal{L}^q$, since $f\in \mathcal{L}^p$. let's pose now $g=\dfrac{g_0}{\|\|g_0\|\|_q}$, we notice that $\|\|g\|\|^q=1$, and $$\int_Xf(x)g(x)d\mu(x) =\int_X f(x)\dfrac{g_0(x)}{\|\|g_0\|\|_q^q}d\mu(x)=\int_X\dfrac{\|f(x)\|^p}{\|\|g_0\|\|_q^q}= \dfrac{\|\|f\|\|_p^p}{\|\|f\|\|_p^{\frac{p}{q}}} = \|\|f\|\|_p.$$ 2. I got stuck here ! Any help is highly appreciated. AI: For part 2., one can do without $\sigma$-finiteness and in place assume that $\{\|f\|\neq0\}$ is $\sigma$-finite. Assume $1<p<\infty$. For any $E\in\mathscr{F}$ with $E\subset\{\|f\|\neq0\}$ and $\mu(E)<\infty$, we will show that $\\|f\mathbb{1}_E\\|_p\leq M_f$. This would imply that $f\in\mathcal{L}_p$ and that $\\|f\\|_p\leq M_f$, because of the assumption that $\{\|f\|\neq0\}$ is $\sigma$-finite. The reverse inequality follows from H"older's inequality. Let $f_n$ be a sequence of simple functions such that $\|f_n\|\leq\|f\|$ and $f_n\rightarrow f$. Then $h_n=f_n\mathbb{1}_E$ belongs to $\mathcal{L}_s$ for all $s>0$, $\|h_n\|\leq \|f\|\mathbb{1}_E$ and $h_n\rightarrow f\mathbb{1}_E$. If $$ \phi_n=\frac{\overline{f}}{\|f\|}\frac{\|h_n\|^{p-1}}{\\|h_n\\|^{p-1}_p} \mathbb{1}_E, $$ then $\\|\phi_n\\|_q=1$ and $$ \\|f\mathbb{1}_E\\|_p\leq\liminf_n\\|h_n\\|_p=\liminf_n\int \|\phi_n h_n\| \,d\mu\leq \liminf_n\int \phi_n f \,d\mu\leq M_f. $$
H: Proving the continuity of the inverse of a continuous, stricly monotonic function Please tell me if the following is correct. We have a continuous, strictly monotonic, increasing function on some closed and bounded interval $I$, and $x_0\in I$. Let $g(y)$ be its inverse, and $f(x_0)=y_0$. I want to show that $\|g(y)-g(y_0)\|<\epsilon\implies\|y-y_0\|<\delta$. $$\begin{align} \|g(y)-g(y_0)\|<\epsilon&\Leftrightarrow x_0-\epsilon<g(y)<x_0+\epsilon\\ &\Leftrightarrow f(x_0-\epsilon)<y<f(x_0+\epsilon)\\ &\Leftrightarrow f(x_0-\epsilon)-f(x_0)<y-f(x_0)<f(x_0+\epsilon)-f(x_0)\\ &\Leftrightarrow -(y_0-f(x_0-\epsilon))<y-y_0<f(x_0+\epsilon)-y_0\\ \end{align}$$ If I consider only the $y$s that are extremely close to $y_0$, then I think that I can set $\delta = \min(y_0-f(x_0-\epsilon), f(x_0+\epsilon)-y_0)$. For the case where the function is decreasing, I just flip the inequality symbols from line one to two, and take $\delta=\min(f(x_0-\epsilon)-y_0,y_0-f(x_0+\epsilon))$. AI: I don't prove that $f$ is bijective, as it's very easy. Just injectivity requires the monotonicity and surjectivity requires IVT. We know since $f$ is continuous on $[a,b]$ then for any $\varepsilon >0$ one can find $\delta >0$ such that $\|x-c\|<\delta$ implies $\|f(x)-f(c)\|<\varepsilon$. Now there is a unique $y,y_0$ such that $\|x-c\|=\|g(y)-g(y_0)\|$ and this solves the problem. (Just change the job of epsilon and delta in your approach)
H: Justification for the calculation of expectation on the first indicator random variable of a series of mutually dependent experiments. I am not asking for a proof of linearity of expectation, which is available in multiple posts. Instead I would like to develop a more robust understanding as to why the calculation of the expectation of the number of $BB$ when pairing up $N$ balls ($W$ and $B,$ such that $W + B = N$) can be performed by considering simply the probability of getting the first pair $BB,$ which is obviously $\frac{B}{B+W}\frac{B-1}{B+W-1}$ times the number of pairs, as in this answer by @lulu, which I am sure it is correct: This sort of thing is best handled by indicator variables, exploiting the fact that expectation is linear, regardless of any possible dependence between the variables. $$E=E\left[ \sum X_i\right]=\sum E\left[X_i\right]=\frac {B+W}2\times \frac B{B+W}\times\frac {B-1}{B+W-1}=\boxed {\frac {B(B-1)}{2(B+W-1)}}$$ My intuitive check on it is that each pair of balls is equally likely to be the first one to be chosen, but this is in conflict with the fact that the probability of getting a second $BB$ pair after a first $BB$ will necessarily go down. AI: Suppose that you have a deck of cards. The probability of the second card being a king is the same as the probability of the first card to be a king, they are completely symmetric. Similarly, the probability of any card being a king is the same and equal. However, once the first card is dealt and opened, the conditional probability given its face value, changes the probability of the second card to be king. The same here. The indicator $X_i$ looks only on the $i$th experiment. If you don't know anything that happened before, this is exactly the same as $X_1$. You are correct that $X_2\vert X_1$ is different from $X_2$, but as you calculate the unconditional $X_2$, it is the same. The conditional would go up or down based on the result of the first pair, but when computing all options you'll get back to the unconditional distribution. A stronger intuition can be made using the following idea: for each pairing here is another pairing where all balls are paired the same but pair #1 and pair #i are switched. A full calculation: $\Pr(X_1=1)=\tfrac{{B \choose 2}}{{n \choose 2}}=\tfrac{B(B-1)}{n(n-1)}$ since you need to choose two black balls to form the first pair. We all agree on that. Now, for $X_2$: $$\Pr(X_2=1)=\tfrac{{B \choose 2}}{{n \choose 2}}\cdot\tfrac{{B-2 \choose 2}}{{n-2 \choose 2}}+\tfrac{{W \choose 2}}{{n \choose 2}}\cdot\tfrac{{B \choose 2}}{{n-2 \choose 2}}+\tfrac{{B \choose 1}{W \choose 1}}{{n \choose 2}}\cdot\tfrac{{B-1 \choose 2}}{{n-2 \choose 2}}=\tfrac{B(B-1)(B-2)(B-3)}{n(n-1)(n-2)(n-3)}+\tfrac{W(W-1)B(B-1)}{n(n-1)(n-2)(n-3)}+2\tfrac{B(B-1)W(B-2)}{n(n-1)(n-2)(n-3)}=\tfrac{B(B-1)}{n(n-1)}\left[\tfrac{(B-2)(B-3)}{(n-2)(n-3)}+\tfrac{W(W-1)}{(n-2)(n-3)}+2\tfrac{W(B-2)}{(n-2)(n-3)} \right] $$ Now note that $(B-2)(B-3)+W(W-1)+2W(B-2)=(B-2)(W+B-3)+W(W-1+B-2)=(B-2)(n-3)+W(n-3)=(n-3)(W+B-2)=(n-2)(n-3)$ so the $[\ldots]=1$ and $\Pr(X_2=1)=\Pr(X_1=1)$
H: Convergence in measure implies alternating sequences converges to zero in measure This problem came up in studying for a qual. Suppose $\{f_{n}\}$ is a sequence of measurable functions that converges in measure to $f$. Prove that the sequence $\{g_{n}\}$, where $g_{n} = (−1)^{n}f_{n}$, converges in measure to a function $g$ if and only if $f= 0$ almost everywhere. I have the converse direction. For the forward direction, I'd like to make use of the the fact that the $\{f_n\}$ admits a subsequence, say $\{f_{n_k}\} \to f$ point-wisely. Similarly, $\{g_{n_j}\} = \{(-1)^{n_j}f_{n_j} \} \to g$ point-wisely. Perhaps I can refine the subsequences and somehow use the fact that $(-1)^n$ diverges to arrive to the conclusion? AI: The sequence $(g_{2n})$ converges in measure to $f$ and the sequence $(g_{2n+1})$ converges in measure to $-f$ hence $g=f$ and $g=-f$ almost everywhere, by uniqueness of the limit in measure. Hence, $f=0$ almost everywhere.
H: Prove that $ \operatorname {dom} G \setminus \operatorname {dom} H \subseteq \operatorname {dom} ( G \setminus H ) $. im trying to learn some basics from set theory, and I got stuck in this proof. Prove that $$ \operatorname {dom} G \setminus \operatorname {dom} H \subseteq \operatorname {dom} ( G \setminus H ) \text . $$ AI: $\newcommand{\dom}{\operatorname{dom}}$Suppose that $x\in\dom G\setminus\dom H$. Then $x\in\dom G$, so there is some $y$ such that $\langle x,y\rangle\in G$. And $x\notin\dom H$, so there is no $z$ such that $\langle x,z\rangle\in H$. In particular, $\langle x,y\rangle\notin H$, so $\langle x,y\rangle\in G\setminus H$. One more small step, which I’ll leave to you, and you’re done.
H: If $(M,g)$ is a Riemannian manifold and $S$ is a regular level set of $f:M\to \Bbb R$ then $\text{grad}f\|_S$ is nowhere vanishing I have a question reading a proof of the following theorem. Theorem. Let $M$ be an oriented smooth manifold, and suppose $S\subset M$ us a regular level set of a smooth function $f:M\to \Bbb R$. Then $S$ is orientable. Proof. Choose a Riemannian metric $g$ on $M$, and let $N=\text{grad}f\|_S$. The hypotheses imply that $N$ is nowhere tangent vector field along $S$, so the result follows. I know that $N$ is normal to $S$ at each point of $S$, but how do we know that $N$ does not vanish at each point of $S$? If $(U,x^1,\dots,x^n)$ is a chart of $M$ near a point in $S$, then $g$ can be written as $g=g_{ij}dx^i \otimes dx^j$ where $(g_{ij})$ is a positive definite real symmetric matrix. Then $\text{grad}f\|_S$ equals $g^{ij} \dfrac{\partial f}{\partial x^i} \dfrac{\partial }{\partial x^j}$ where $(g^{ij})=(g_{ij})^{-1}$. Since $S$ is a regular level set, some $\dfrac{\partial f}{\partial x^i}$ does not vanish at each point of $S$, but how do we know that $g^{ij} \dfrac{\partial f}{\partial x^i}$ does not vanish for some $j$, at each point of $S$? AI: Arguing by contradiction, if $$\tag{1} g^{ij} \frac{\partial f}{\partial x^i} = 0$$ for all $j$, then for each $k$, multiply the above by $g_{jk}$ and sum over $j$, $$\sum_{j=1}^ng_{jk} g^{ij} \frac{\partial f}{\partial x^i} = 0\Rightarrow \delta_{ik} \frac{\partial f}{\partial x^i} = 0.$$ Thus $\frac{\partial f}{\partial x^k} = 0$ for all $k$, which is a contradiction. (This is just linear algebra: (1) holds for all $j$ if and only if $$ g^{-1} \nabla_0 f = 0, $$ where $\nabla_0 f = \left( \frac{\partial f}{\partial x^1} , \cdots, \frac{\partial f}{\partial x^n}\right)^t$. To show that $\nabla_0 f$ is nonzero we, of course, multiply $g$ to the left).
H: Surjective ring morphism $f:R\to R$ satisfies Ker$(f^{n+1})\subset $ Ker$(f^n)$ then $f$ is injective. As in the title, the set-up of the problem is as follows: $f: R\to R$ is a surjective ring homomorphism and $R$ is a commutative ring. Suppose that for some $m\in \mathbb{N}$, Ker$(f^{m+1})\subset$ Ker$(f^m)$. Prove that $f$ is injective. Here are my thoughts so far: We know by the first isomorphism theorem (since $f$ is surjective) that there is an isomorphism $\phi_n: R \to R/\text{Ker}(f^n)$ for any $n\in \mathbb{N}$. Now consider the map $$ R \longrightarrow^{f^{m+1}} R \longrightarrow^{\pi_m}R/\text{Ker}(f^n)\longrightarrow^{\phi_m^{-1}} R. $$ Now the above composition is an isomorphism by our hypotheses and I would like to conclude that the map agrees with $f$ but I don't see why this should be true (in fact I know that it shouldn't be in general but I feel that I am on the right path). I would appreciate a hint or some guidance on how to make my solution more complete. This is not a HW question (it is a problem on this practice qualifying exam). AI: edited to hide a full solution in case you just wanted a hint: I think it might be more straightforward to work from first principles. While I think your solution could work, I find it easier just to use the definition of the kernel. The key is that your assumption gives you an inclusion $\text{ker}(f^{m+1})\subset\text{ker}(f^m)$, but there is always the reverse inclusion $\text{ker}(f^m)\subset\text{ker}(f^{m+1})$ since $f$ is a ring homomorphism and must map 0 to 0. When you have these two inclusions together, you get that $f^m(R)\xrightarrow{f}f^{m+1}(R)$ is not just surjective but also injective. Once you convince yourself of this, try to conclude that if this restriction of $f$ is injective then so is $f$ itself. By assumption, $\text{ker}(f^{m+1}) = \text{ker}(f^m)$, so the restriction $f\|_{f^m(R)}$ is injective since $f\circ f^m(x) = 0$ implies that $f^m(x) = 0$ for every $x\in R$. However, $f$ is surjective, so all of its iterates must be surjective as well. Then the restriction $f\|_{f^m(R)}$ is actually just $f$ itself. Therefore, $f$ is injective.
H: Why is $\text{tr}(\sqrt{\sqrt A B \sqrt A})=\text{tr}(\sqrt{A B }) $ for positive semidefinite matrices $A,B$ Let $A,B$ be two real positive semidefinite matrices of same size. Denote the trace operator by tr. We define the square root of a positive semidefinite matrix $A$ by the unique matrix $\sqrt A$ such that $A= \sqrt A ^\top \sqrt A$. Why is $\text{tr}(\sqrt{\sqrt A B \sqrt A})= \text{tr}(\sqrt{A B}) $ true ? Is $\text{tr}(\sqrt{\sqrt A B \sqrt A + \text{Id}})=\text{tr}(\sqrt{A B + \text{Id} }) $ also true, where Id is the identity matrix of same size? AI: $\sqrt{A} B \sqrt{A}$ and $AB$ are similar matrices, since $AB = \sqrt{A} (\sqrt{A} B \sqrt{A}) \sqrt{A}^{-1}$. So one square root of $AB$ (presumably the one you want) is $\sqrt{A} \sqrt{\sqrt{A} B \sqrt{A}} \sqrt{A}^{-1}$. By similarity it has the same eigenvalues, and thus the same trace, as $\sqrt{\sqrt{A} B \sqrt{A}}$. In the same way, one square root of $AB + I$ is $ \sqrt{A} \sqrt{\sqrt{A} B \sqrt{A} + I} \sqrt{A}^{-1}$, and this has the same trace as $\sqrt{\sqrt{A} B \sqrt{A}+I}$.
H: Calculate :$\int_{0}^{2\pi }e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$ Calculate: $$\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$$ My try: $\displaystyle\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt\\ \displaystyle \int_{\|z\|=R}e^{\mathfrak{R\textrm{z}}}\cos(\mathfrak{I\textrm{z}}+3(-i\log z)dz\\ \displaystyle \int_{\|z\|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}}^{(\mathfrak{I\textrm{z}}+3(-i\log z))i}dz\\ \displaystyle \int_{\|z\|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}}z^{3}dz\\ \displaystyle \int_{\|z\|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}R}z^{3}dz$ and there is nothing here that is not holomorphic, therefore according to Cauchy theorem it must be exactly $0$. AI: The idea is right here but the proof steps could use a little polishing. More specifically, note that the integral is connected to a nicer looking form: $$\int_{0}^{2\pi}e^{R\cos t}\cos(R\sin t+3t)dt=\Re\int_{0}^{2\pi}e^{R\cos t+iR\sin t+3it}dt=\Re\int_{0}^{2\pi}e^{Re^{i t}}e^{i3t}dt$$ This can be transformed to a contour integral on the unit circle via the substitution $z=e^{it}$, under which we conclude that $$\int_{0}^{2\pi}e^{Re^{it}}e^{i3t}dt=\frac{1}{3i}\oint_{\|z\|=1}z^3e^{Rz}dz=0$$ and indeed, the integral presented above is identically zero.
H: Is there a preference between writing complex numbers as $z=a+bi$ or $z=a+ib$? This is probably just a minor notational issue, but I am unsure whether I should write $z=a+bi$ or $z=a+ib$ when denoting complex numbers. Though the former notation seems more common, Euler's identity tends to be written as $$ e^{i\pi}+1=0 $$ where the exponent is written as $i\pi$, not $\pi i$. I don't doubt that they mean the same thing, but I was wondering if one of the notations is more readable than the other. Perhaps someone can explain this apparent discrepancy between how we write $z$, and how we write $e^{i\pi}$. AI: In my experience, $z=a+bi$ is used when b is a constant (for example $4+3i$) and $z=a+ib$ when b is a variable (for example $x+iy$). When b has both variable and constant components, the i appears between them (for example $7+3ix$)
H: Quick way to determine characteristic subgroups Reminder: A characteristic subgroup of a group $G$ is a subgroup which is stable under all elements of $\mathrm{Aut}(G)$. This is a stronger property than being normal. A while ago there was a question here, which unfortunately got deleted, about a group having two different but isomorphic characteristic subgroups. (It was not this question I found via search, although it basically asks the same thing.) Of the examples proposed in the comments to that deleted question, the following got stuck in my head: $$G := Q_8 \times \mathbb Z/8$$ where $Q_8$ is the quaternion group with eight elements. (I'll write elements of $G$ like $(-i, 3)$, i.e. multiplicatively on the left and additively on the right, sorry if that is non-standard.) The commenting user there claimed $G$ has two (and I think they meant: at least two) distinct characteristic subgroups of order $2$. Thinking about this, I saw three elements of order $2$ in $G$, namely $$(1,4), (-1,0), (-1,4)$$ That the first one is fixed by all automorphisms follows easily from the fact that it is of the form $g^4$ for all elements $g\in G$ with $\mathrm{ord}(g)=8$. But I needed more work to convince myself that the second and third are also fixed by all automorphisms. Sure it's true for both as soon as it is for one of them; and if an automorphism did not fix them, it would map them to each other; but still to actually show this cannot happen, the best I came up with was writing out what such an automorphism mapping $(-1,0) \mapsto (-1,4)$ could possibly do on all $(\pm i, j, k,0)$ and eventually getting a contradiction from that. That was not satisfying and to be completely honest, I'm not even sure my proof is correct. So I wondered if there is a shorter / smarter way to see this, which also applies to more general situations. How would you determine all characteristic subgroups of the above $G$? As a proposal, is there a way to find the characteristic subgroups of $G_1 \times G_2$ if one knows both the characteristic subgroups of $G_1, G_2$ and the homomorphisms in either direction $\mathrm{Hom}(G_i, G_j)$ (that these hom-sets are important I infer from When is a centerless group characteristic in direct product with $\mathbb{Z}^n$? and common mathematical sense; and in our example, they are very easy to see). One thing I noticed in this example is that the centre of $G$ (which of course is characteristic) is $\pm1 \times \mathbb Z/8$, which contains all elements of order $2$; but this centre does have an automorphism mapping $(-1,0) \mapsto (-1,4)$, which is why I gave up this route and started trying with elements involving $\pm i,j,k$. Would there, in this example or a general setting, have been any way to use information about the centre to conclude? AI: I'll start by answering your original question: why are the three involutions in $Q_8\times C_8$ 'different'? The answer to this is easy: one involution is the power of an element of order $8$, and another involution is the one in the derived subgroup. The third is neither of these. This now means that it's a little easier to determine all characteristic subgroups of $G$. We know that any automorphism must fix all involutions, and thus we merely need to consider overgroups of them. There are obvious characteristic subgroups: each involution, and their direct product. The set of all elements of order dividing $4$, the cyclic subgroup of order $8$, and so on. Since the automorphisms of the $Q_8$ factor extend, you can use them to eliminate some subgroups. But in general I believe it is quite difficult to determine the full automorphism group. In this case, the outer automorphism group has order $192=2^4\cdot 3$, although I checked that in GAP. We can see a group of order $6$ acting on $Q_8$, and a group of order $4$ on the $C_8$ factor. So there's an extra $2$ that we have missed. That one isn't so easy to see. It must preserve the $C_8$ factor, but cannot preserve the $Q_8$ factor as if it preserved both, it could be chosen (via multiplying by the other automorphisms) to centralize both, a contradiction. Thus it maps the $Q_8$ to a diagonal subgroup. For your question about determining the automorphisms of $G_1\times G_2$ from having all of the homomorphisms among and between the factors, I don't think that's clear. It doesn't even seem to work in the case of a direct product of isomorphic simple groups. If $G_1=G_2=C_p$ then you make up a whole $\mathrm{GL}_2(p)$ of outer automorhisms. But if $G_1=G_2=A_5$, for example, you only obtain a $D_8$ of outer maps.
H: How are you able to factorise a determinant like this? I'm aware of many different properties of determinants, like two rows being equal implies determinant is zero, or if a row is multiplied by a constant, then you factor out that constant. I'm not aware of being able to do this though. What property am I missing that allows you to simplify a determinant like this? AI: Rule: if a matrix $B$ is obtained from $A$ , by adding or subtracting one column to another column (part of elementary column operation), Then, these two matrices $A$ and $B$ have equal dets.
H: "Finite-Dimensional-Type-Spectral-Theorem" for Orthogonal Projections Let $H$ be a Hilbert space, not assumed separable, and $p$ and $q$ (bounded - not sure if that is important) orthogonal projections. Question 1: Is it the case that $p$ has an orthonormal eigenbasis for $H$/is diagonalisable? Question 2: Is it the case that $p$ and $q$ commute if and only if they share an eigenbasis/are simultaneously diagonalisable? If the answer to question 1 is no, what are some reasonable assumptions on $H$ that guarantee such an eigenbasis exists (it seems to me that separability is enough. I don't think assumptions (such as compactness) on $p$ and $q$ are much good to me.). If the answer to question 1 is no, do these assumptions give a positive answer to question 2? Thanks for your help. I am a little concerned when I work in infinite dimensions. For example I would just say, OK, $$H=\operatorname{ran }p \oplus \ker{p},$$ each are closed and so each are Hilbert spaces and so each have onb... and just union those, bingo-bango, jobs a good 'un... but I am concerned there is an error there. AI: Yes, there's nothing tricky about this. As you say, $H$ is the orthogonal direct sum of the image of $p$ and the kernel of $p$. Both of these are closed (the image is closed since it is the kernel of $1-p$), and so can pick orthonormal bases for each of them and their union is an orthonormal basis for $H$ which diagonalizes $p$. If $p$ and $q$ commute, then $pq$, $p(1-q)$, $(1-p)q$, and $(1-p)(1-q)$ are all orthogonal projections which are pairwise orthogonal and their sum is $1$. So again, $H$ is the orthogonal direct sum of their ranges, and picking an orthonormal basis for each of their ranges, you get an orthonormal basis for $H$ which simultaneously diagonalizes $p$ and $q$.
H: Correspondence between binary numbers and $ab = \Pi_{i=1}^m p_i^{k_i}$ s.t. $a, b$ have no common factors I'll note that $p_i$ is prime and $k_i \in \mathbb Z^+.$ I read in a textbook that the number of integers $ab = \Pi_{i=1}^m p_i^{k_i}$ s.t. $a, b$ have no common factors is $2^m$ which is the number of binary numbers of length $m$. I am trying to understand how they got $2^m$. Initally I thought I understood how they came about the answer by considering $a = 2^i3^j5^k7^l, \ b = 11^a13^b17^c$ in which case $a$ is of the form $x\cdot x\cdot x \cdot x \cdot1\cdot1\cdot 1$ where $x$ is a factor of $a$. This form of $a$ is similar to a binary number. But then the latter is a sequence and the former is a product meaning order in $a$ doesn't matter. So I don't think I undeerstand what they did. Can someone, please, elaborate on how such products are counted? Thanks. AI: Notice first that in order to not have common factors, then either $p_i^{k_i}\|a$ or $p_i\not \| a,$ you can not have any other option otherwise you are separating that prime into the two numbers $a,b$ and so they will have a factor in common. Now, label the primes as either $0$ or $1$ if your label is $1$ then you make it belong to $a$ and otherwise is in $b.$ Notice that this labelling is a bijection in between the different tuples $(a,b)$ having the desired property and the number of labellings, but the labellings are just binary words of size $m$ so there are $$\underbrace{2\times 2\times \cdots \times 2}_{m\text{ times}}=2^m.$$ Inspired in your example: $ab = 2^i3^j5^k$ you have $a=1,b=2^i3^j5^k$ so $000$ $a=2^i,b=3^j5^k$ so $100$ $a=3^j,b=2^i5^k$ so $010$ $a=5^j,b=2^i3^j$ so $001$ $a=2^i5^j,b=3^j$ so $101$etc... Also, notice that there is a symmetry in the problem, meaning that if you do not care who is $a$ and who $b$ then you are counting some things twice and the result has to be $$2^{m-1}.$$
H: Let $G$ be a group and let $a, b$ be elements of $G$, show that $\|ab\|=l.c.m (\|a\|,\|b\|)$ Let $G$ be a group and let $a, b$ be elements of $G$ such that: i) $\langle a\rangle\cap\langle b\rangle={\{1}\}$ ii) $ab=ba$ II) $\|a\|=m, \|b\|=n$ Show that $\|ab\|=l.c.m (\|a\|,\|b\|)$ Idea: Note that $a^m=1,\; b^n=1,$ let's do $k=l.c.m (\|a\|,\|b\|)$ (least common multiple), $\|ab\|=w,$ hence $(ab)^k=a^kb^k=1,$ then $k$ divides $w$, How can I prove that $ w $ divides $ k $? Thanks so much for reading. AI: You know that $1=(ab)^w=a^wb^w$ so $\lvert a\rvert,\lvert b\rvert\mid w$, hence $lcm(\lvert a\rvert,\lvert b\rvert)\mid w$. Also, $(ab)^{lcm(\lvert a\rvert,\lvert b\rvert)}=a^{lcm(\lvert a\rvert,\lvert b\rvert)}b^{lcm(\lvert a\rvert,\lvert b\rvert)}=(a^{\lvert a\rvert})^{\frac{lcm(\lvert a\rvert,\lvert b\rvert)}{\lvert a\rvert}}(b^{\lvert b\rvert})^{\frac{lcm(\lvert a\rvert,\lvert b\rvert)}{\lvert b\rvert}}=1$, so $w\mid lcm(\lvert a\rvert,\lvert b\rvert)$. Thus $w=lcm(\lvert a\rvert,\lvert b\rvert)$.
H: Proving relation between exponential generating functions whose coefficients count certain types of graphs Let's there be two exponential generating functions: $A(x)=\sum^\infty_{n=1}\frac{a_n}{n!}x^n$ $B(x)=\sum^\infty_{n=1}\frac{b_n}{n!}x^n$ Sequence ${\{a_n\}}^\infty_{n=1}$ defines number of all possible simple graphs on n labelled vertices. Sequence ${\{b_n\}}^\infty_{n=1}$ defines number of all possible simple connected graphs on n labelled vertices. I am trying to prove the following relationship between these two generating functions: $A(x)=e^{B(x)}-1$ The expression for $a_n$ is easy to derive: $a_n=2^{\frac{n(n-1)}{2}}$, but I don't know how to demonstrate the equation above. I found the following hint for this problem: It can be shown that $\frac{(B(x))^k}{k!}$ is the exponential generating series for the labelled graph with exactly k components. AI: A further HINT: The hint tells you that $$e^{B(x)}=\sum_{k\ge 0}\frac{\big(B(x)\big)^k}{k!}\,,$$ so $$e^{B(x)}-1=\sum_{k\ge 1}\frac{\big(B(x)\big)^k}{k!}\,.$$ A simple graph on $n$ vertices has at least one component, so . . . You may find Problem $413$ of this web page or Section $4$ of this PDF helpful.
H: Weird Sum Identity I came across a weird sum identity that I would like to prove: $$ \sum_{i=1}^{k-2}\frac{(-1)^i}{(i-1)!(k-2-i)!(n-k+2+ij)}=\frac{-\Gamma\left(1+\frac{n-k+2}{j}\right)}{j\Gamma\left(k-1+\frac{n-k+2}{j}\right)}. $$ I would like to know how one can prove this. It seems like direct computations, induction, combinatorial interpretation and other simple proof techniques won't work. However, software like Wolfram Alpha can evaluate it right away so there must be some easy way to go at it. Because of the look of the summands, I even tried using the Wilf-Zeilberger algorithm, but it fails in this case. So the last thing I tried is showing that, if $S_m$ represents the partial sums up to m, then $S_m-S_{m-1}$ is equal to the summand. Then the sum we consider telescopes and we get $S_{k-2}$, which is just what we want. The problem with this approach is that $S_m$ is really nasty if $m\neq k-2$ (it includes hypergeometric series that vanish at $m=k-2$, which can be seen by plugging $S_m$ in Wolfram Alpha). Any help proving this result will be appreciated AI: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\bbox[#ffd,10px]{\sum_{i = 1}^{k - 2}{\pars{-1}^{i} \over \pars{i - 1}!\pars{k - 2 - i}!\pars{n - k + 2 + ij}} = -\,{\Gamma\pars{1 + \bracks{n - k + 2}/j} \over j\,\Gamma\pars{k - 1 + \bracks{n - k + 2}/j}}}:\ {\Large ?}}$. With $\ds{a \equiv \pars{n - k + 2}/j}$: \begin{align} &\bbox[#ffd,10px]{\left.\sum_{i = 1}^{k - 2}{\pars{-1}^{i} \over \pars{i - 1}!\pars{k - 2 - i}!\pars{n - k + 2 + ij}} \,\right\vert_{\ k\ \in\ \mathbb{N}_{\large\ \geq\ 3}}} \\[5mm] = &\ {1 \over j\pars{k - 2}!}\sum_{i = 0}^{k - 2}{\pars{k - 2}! \over i!\pars{k - 2 - i}!}\,\pars{-1}^{i}\,{i \over i + a} \\[5mm] = &\ {1 \over j\pars{k - 2}!}\sum_{i = 0}^{k - 2}{k - 2 \choose i} \pars{-1}^{i}\,\pars{1 - {a \over i + a}} \\[5mm] = &\ {1 \over j\pars{k - 2}!}\bracks{\delta_{k,2} - a\sum_{i = 0}^{k - 2}{k - 2 \choose i}\pars{-1}^{i} \int_{0}^{1}t^{i + a - 1}\,\dd t} \\[5mm] = &\ {\delta_{k,2} \over j} - {a \over j\pars{k - 2}!}\int_{0}^{1}t^{a - 1} \sum_{i = 0}^{k - 2}{k - 2 \choose i}\pars{-t}^{i}\,\dd t \\[5mm] = &\ {\delta_{k,2} \over j} - {a \over j\pars{k - 2}!}\int_{0}^{1}t^{a - 1} \pars{1 - t}^{k - 2}\,\dd t \\[5mm] = &\ {\delta_{k,2} \over j} - {a \over j\pars{k - 2}!}\,{\Gamma\pars{a}\Gamma\pars{k - 1} \over \Gamma\pars{a + k - 1}} \\[5mm] = &\,\,\, \bbox[10px,#ffd,border:1px groove navy]{ -\,{1 \over j}\,{\Gamma\pars{1 + a} \over \Gamma\pars{k - 1 + a}}}\,, \qquad a \equiv {n - k + 2 \over j} \\ & \end{align} because $\ds{k = 3,4,5,\ldots}$
H: Proving a result related to $100^{th}$ roots of unity. If we take the $100^{th}$ roots of unity ie. all complex roots of the equation $z^{100}-1=0$ and denote them as $\alpha_{1},\alpha_{2},...,\alpha_{100}$ then we are required to prove that $$\alpha_{1}^r+\alpha_{2}^r+...+\alpha_{100}^r=0$$ for $r\neq100k$ where $k$ is an integer. I tried using the Euler form by denoting $$\alpha_{t}=e^{\frac{i2t\pi}{100}}$$ and trying to evaluate it as a GP but it was pretty lengthy and I couldn't simplify it out to zero so now I am confused. Any help would be appreciated. AI: So, the roots of $$z^n-1=0$$ are $$a_k=e^{2i\pi k/n},0\le k<n$$ $$\sum_{t=0}^{n-1}a_k^r=a_0\dfrac{a_1^{nr}-1}{a_1^r-1}=0$$ if $n\nmid r$
H: What were the steps taken to get from point A to point B in this forced vibrations problem? I'm working through the derivation of the forced response (vibration) of a cantilevered beam. I have a basic understanding of the derivation until this point. screenshot of derivation I can see how the summation(s) are equivalent to an integral - but I don't see why $q_n$ and/or $\omega$ wouldn't be included within that integral, since q is dependent on $n$. I can rationalize how to get the same result if $\omega$ and $q$ are regarded as constants (not dependent on $n$) but that doesn't seem to be the case (?). I have a feeling the "orthogonality conditions" they mention have something to do with it, but I haven't been able to discern how from my research. In general, I have limited knowledge of linear algebra, and I haven't had to use vector calculus concepts of like orthogonality since taking the course a couple years ago. Any help is appreciated :) Full document (screenshot is from pgs. 28 + 29): http://www1.aucegypt.edu/faculty/mharafa/MENG%20475/Continuous%20Systems%20Fall%202010.pdf AI: Supposing the orthogonality conditions are $$ \int_0^l W_m(x)W_n(x)dx=\delta_{mn}b. $$ Now, multiplying (84) by $W_m(x)$ $$ \sum_{n=1}^\infty \omega_n^2W_m(x)W_n(x)q_n(t)+\sum_{n=1}^\infty W_m(x)W_n(x)\frac{dq_n^2(t)}{dt}=\frac{f(x,t)}{\rho A}W_m(x) $$ then integrate over $x\in[0,l]$ $$ \sum_{n=1}^\infty \omega_n^2q_n(t)\int_0^lW_m(x)W_n(x)dx+\sum_{n=1}^\infty \frac{dq_n^2(t)}{dt}\int_0^lW_m(x)W_n(x)dx=\frac{1}{\rho A}\int_0^lf(x,t)W_m(x)dx $$ then using orthogonality $$ \sum_{n=1}^\infty \omega_n^2q_n(t)\delta_{mn}b+\sum_{n=1}^\infty \frac{dq_n^2(t)}{dt}\delta_{mn}b=\frac{1}{\rho A}\int_0^lf(x,t)W_m(x)dx $$ or, considering the Kronecker delta $$ \omega_m^2q_m(t)b+\frac{dq_m^2(t)}{dt}b=\frac{1}{\rho A}\int_0^lf(x,t)W_m(x)dx $$ dividing by $b$ $$ \omega_m^2q_m(t)+\frac{dq_m^2(t)}{dt}=\frac{1}{\rho A b}\int_0^lf(x,t)W_m(x)dx $$ and using the definition of $Q_n(t)$ $$ \omega_m^2q_m(t)+\frac{dq_m^2(t)}{dt}=\frac{1}{\rho A b}Q_m(t) $$ Given the arbitrarity of the index $m$ (this relation holds for each $m$), you can rename the index $n$ and you have (85).
H: Is there an infinite amount of primes in base n made from an equal amount of even and odd digits. Is there an infinite amount of primes in base n made from an equal amount of even and odd digits? A list of primes that have this property is this sequence $$23,29,41,43,47,61,67,83,89,1009,1021,1049,1061,\small\dots$$ I think this is true because if you pick a big random number the number of digits in that number with an n or an m is about equal. but I'm sure most numbers don't have an exact number of 2's as 1's just really close to equal. So I'm guessing that the number of primes like this is either less and less frequent or more and more frequent. So my second question is what is the percentage of primes that have this property. is it almost 0% as the number of primes goes to infinity or does it go to 50%? AI: As twnly already mentioned in their answer, research by Mauduit and Rivat on a related problem proved (a generalized version of the result) that asymptotically half the primes have an even/odd sum of digits. I want to point out that this seemingly simple statement was an open problem for over 40 years, was proved less than a decade ago, and the proof was published in arguably the most prestigious mathematics journal in the world. Questions about digits of primes are very hard! The probability that a randomly chosen $2n$-digit integer has an equal number of odd and even digits is asymptotically $1/\sqrt{\pi n}$ (from asymptotics for central binomial coefficients). The probability that a randomly chosen $2n$-digit integer is prime is asymptotically $1/(2n\ln10)$ (from the prime number theorem). The natural conjecture would be that these two events are asymptotically independent, so that the probability that a randomly chosen $2n$-digit integer both is prime and has the same number of odd/even digits should be asymptotically $1/(n^{3/2}\sqrt\pi\ln10)$. In particular, the probability that a randomly chosen $2n$-digit prime has the same number of odd/even digits should also be asymptotically $1/\sqrt{\pi n}$, which in particular tends to $0$ as $n\to\infty$. But also, in particular, there should be infinitely many primes with this property—this heuristic predicts that the number of such primes is rather larger than the number of twin primes, for example.
H: Markov Chain: Conditional distribution at time $t$, given $t-1$ and $t+1$ For a Markov process given by $$x_t = \mu +\kappa(x_{t-1} - \mu) + \sigma \cdot \varepsilon_t $$ where $\varepsilon_t \sim N(0,1)$ and $\mu$, $\kappa$, $\sigma^2$ are the parameters, how would I find the conditional distribution $p\left(x_t \vert x_{t-1}, x_{t+1},\kappa ,\mu,\sigma^2\right)$? The paper I'm reading says it is given by: $$N\left(\mu + \kappa[(x_{t-1}-\mu)+(x_{t+1}-\mu)] / [1+\kappa^2], \quad\sigma^2/(1-\kappa^2)\right)$$ however I can't work out how this is derived. Many thanks in advance! AI: First notice that: $$ \begin{align} x_{t+1} &= \mu+\kappa(x_t-\mu)+\sigma \epsilon_{t+1}\\ & = \mu+\kappa( \mu+\kappa(x_{t-1}-\mu)+\sigma \epsilon_t -\mu)+\sigma \epsilon_{t+1}\\ & = \mu+\kappa( \kappa(x_{t-1}-\mu)+\sigma \epsilon_t)+\sigma \epsilon_{t+1}\\ & = \mu+\kappa( \kappa(x_{t-1}-\mu))+ \kappa \sigma \epsilon_t +\sigma \epsilon_{t+1} \end{align} $$ Using conditional expectation: $$ \begin{align} \mathbb{E}(x_t \| x_{t-1}, x_{t+1}) & = \mathbb{E}( \mu+\kappa(x_{t-1}-\mu)+\sigma \epsilon_{t} \| x_{t-1}, x_{t+1})\\ & = \mu+\kappa(x_{t-1}-\mu)+\mathbb{E}(\sigma\epsilon_{t} \| x_{t-1}, x_{t+1})\\ \end{align} $$ I imagine that the noises $\epsilon_t$ are all independent, right? Then: $$ \begin{align} \mathbb{E}(x_t \| x_{t-1}, x_{t+1}) & = \mu+\kappa(x_{t-1}-\mu)+\mathbb{E}(\sigma\epsilon_{t} \| \kappa \sigma \epsilon_t +\sigma \epsilon_{t+1} = x_{t+1} - (\mu+\kappa( \kappa(x_{t-1}-\mu))) )\\ \end{align} $$ Now all you need to do is to understand $\mathbb{E}( X \| \kappa X+Y )$ where $X,Y \sim N(0,\sigma^2)$ with $X,Y$ independent. But the sum of independent gaussians is a gaussian and you know that: $$ f_{X\|X+Y}(x\|y) = \frac{f_{X,Y}(x,y)}{f_X(x)} $$ Can you continue from here?
H: Simplification of an algebraic determinant After studying some analytic geometry, I came across this step in a solution, however, I am not how they managed to simplify the determinant in this way. When I tried to evaluate this, I got: $\frac{bc-ad}{2}+\frac{ad-bc}{2b-2d}$, but didn’t see how this got to the desired form. Many thanks. AI: Expand the first on the first line and the second on the second line. You obtain $$\frac12b\cdot\frac{ad-bc}{d-b}-\frac12\left(-d+d\cdot\frac{ad-bc}{d-b}\right)=\frac d2+(b-d)\frac{ad-bc}{2(d-b)}=\frac{d-ad+bc}{2}$$
H: Show a Cubic Diophantine Equation has no solutions in coprime integers Show that $x^3+2y^3 = 7(z^3+2w^3)$ has no solutions in coprime integers $x,y,z,w$. I've been stuck on this problem for a bit and haven't made any progress or found any strategies that might work. Any help would be greatly appreciated. AI: Oh: if there is any solution (not all zero), there is such a solution in coprime integers, just divide out by the greatest common divisor. This is the "infinite descent" in brief. We assume that we have a not-all-zero solution in coprime integers. it's just mod 7. Cubes are $0,1,-1 \pmod 7.$ Check, the only way to get this expression $0 \pmod 7$ is to have both $x,y \equiv 0 \pmod 7.$ Next, the left hand side is currently divisible by $343.$ The reult is that $z^3 + 2 w^3$ must also be divisible by $7,$ at which point we have all four $x,y,z,w \equiv 0 \pmod 7.$ All divisible by $7.$ This contradicts the assumption of coprime variables.
H: $\frac{2}{3}^{\text{th}}$ roots of $4 + 4\sqrt{3}i$ A textbook, in a section about roots of unity, poses the problem Find the values of $$ \left( 4\sqrt{3} + 4i \right)^{\frac{2}{3}} $$ The answer key says $$ w \in \left\{ 4e^{-100^\circ i}, \quad 4e^{20^\circ i}, \quad 4e^{140^\circ i} \right\} $$ and I know exactly how this was obtained, but the problem is if I let $w = 4e^{140^\circ i}$, for example, I get $$ w^\frac{3}{2} = 8e^{210^\circ i} \neq 4\sqrt{3} + 4i = 8e^{30^\circ i} $$ I think that the problem is $ \left( 4\sqrt{3} + 4i \right)^{\frac{2}{3}} $ is intended by the author to mean the solutions to the equation $$ w^{\frac{3}{2}} = 4\sqrt{3} + 4i $$ in which case, when we square both sides in order to solve $$ w^3 = (4\sqrt{3} + 4i)^2 $$ we pick up some false solutions. My question is, are the answers in the book incorrect, or do I misunderstand the meaning of $z^{\frac{a}{b}}$, or is the whole concept of $\frac{a}{b}^{\text{th}}$ roots of a complex number just badly defined? AI: From the given solutions, $w^{3/2}$ is one of $$8e^{-150°i},8e^{30°i},8e^{210°i},$$ where the arguments differ by $180°$. These numbers are $$\pm(4\sqrt{3}+4i).$$

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