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H: Doubt: "A group representation is exactly like a module over the group ring" It is traditional to say that a representation of a group $G$ over a field $F$ is "exactly like" a module over the group ring $F[G]$. I think it is inaccurate. I think a module over $F[G]$ encodes more than a representation of $G$. I will give an example of two modules over $F[G]$ giving rise to the same group representation of $G$. I will give the simplest example: Let $G$ be the trivial group of order $1$, and let $F=\mathbb{C}$. I can give the abelian group $M=\mathbb{C}$ a structure of a $\mathbb{C}[G]$-module in different ways. I can let each $z\cdot 1_G\in\mathbb{C}$ act on $m\in M$ either by $(z\cdot 1_G)\cdot m=zm$, or by $(z\cdot 1_G)\cdot m=\overline{z}m$. Is it true to say that there is a bijective correspondence between the following collections? The collection of pairs $(\rho,\theta)$, where $\rho$ is a group representation of $G$ over $F$, and $\theta$ is a field endomorphism of $F$, and The collection of modules over $F[G]$ AI: It is possible to define a $\mathbb{C}[G]$-module structure like this, but the structure is not compatible with the $\mathbb{C}$-vector space structure on $M$. When you define a representation, you start with a vector space, so the action of the base field is prescribed. So these two module structures are two different representation: the usual one; the one that starts with a different vector space over $\mathbb{C}$, that you might call $\overline M$, and the action you described. As for your more general question: you have two categories, the category of $\mathbb{C}[G]$-modules and the category of representations of $G$ (where objects are pairs $(V, \rho)$, $V$ a vector space, $\rho : G \to End(V)$). Then these two categories are equivalent.
H: What's the most basic yet interesting Algebraic Geometry result regarding this polynomial? Let $f(x,y,z) = x^a + y^b - z^c$, where $a,b,c \gt 0$. What is the most basic yet interesting result about this polynomial from Algebraic Geometry? AI: You might be interested in the paper Faltings plus epsilon, Wiles plus epsilon, and the Generalized Fermat Equation by H. Darmon, which contains a number of interesting results and conjectures about the integer solutions to the equation $f(x,y,z) =0$. For instance, Darmon and Granville conjecture that there are finitely many (primitive) triples of integers $(x,y,z)$ with $x^a+y^b = z^c$ and such that $$1/a+1/b+1/c<1.$$ Moreover, Darmon offers a bounty of $$300\left(\frac{1}{\frac1a+\frac 1b+\frac1c}-1\right)\:\text{dollars}$$ for any solution not in the list contained in the paper. (However, please don't search too hard, because Prof. Darmon is my doctoral advisor and I wouldn't want to cause his ruin. Of course, I am kidding: good luck finding any new solutions...)
H: Show that $X_{n}=\sum_{k=0}^{n} \frac{1}{k!}$ and $X_{n}=\sum_{k=0}^{n} \rho^{k}$ are Cauchy sequences. Show that $X_{n}=\sum_{k=0}^{n} \frac{1}{k!}$ and $X_{n}=\sum_{k=0}^{n} \rho^{k}$ are Cauchy sequences. I know that for each sequence I have to show the following: "$\forall \epsilon>0 \text{ }\exists \text{ } N_{\epsilon}=N(\epsilon) \in \mathbb{N}$ s.t. $\forall n \geq N_{\epsilon}$ , $ \left \| X_{n}-X_{m} \right \| < \epsilon $" 1) For the first I have done the following: $\left \| X_{n}-X_{m} \right \|=\left \| \sum_{k=0}^{n} \frac{1}{k!}-\sum_{k=0}^{m} \frac{1}{k!} \right \|= \left \| \sum_{k=n+1}^{m} \frac{1}{k!}\right \| $ After this I want to use the fact that $\frac{1}{k!}\leq \frac{1}{2k}$ for $k\geq 3$, suppose that $m\geq n$, and get: $= \left \| \sum_{k=n+1}^{m} \frac{1}{k!}\right \| \leq \left \| \sum_{k=n+1}^{m} \frac{1}{2k}\right \|=\left \| \frac{1}{2(n+1)}+\frac{1}{2(n+2)}+\cdots +\frac{1}{2(m)}\right \|$ then I have to bound this with a value that depends on n, on which I will impose that it is less or equal to $\epsilon$ and deduct the value of $N_{\epsilon}$ from there. my guess so far is $\frac{m-n}{n}<\frac{m}{n} <\epsilon$ but this wont work because it is too big and I it should be "sufficiently" small (smaller than $\epsilon$). 2) For the second sequence I have done the following: Since $0<\rho<1 \Rightarrow \sum_{k=0}^{\infty}\rho^{k}=\frac{1}{1-\rho}=l $, then we have that "$\forall \frac{\epsilon}{2}>0 \text{ }\exists \text{ } N'_{\frac{\epsilon}{2}}=N'(\frac{\epsilon}{2}) \in \mathbb{N}$ s.t. $\forall n \geq N'_{\frac{\epsilon}{2}}$ , $ \left \| \sum_{k=0}^{n} \rho^{k}-l \right \| < \frac{\epsilon}{2} $" Let $ \epsilon>0$, I choose $N_{\frac{\epsilon}{2}}=N'_{\frac{\epsilon}{2}}$ . We want to show that $N_{\frac{\epsilon}{2}}$ works. $\text{ if } n,m\geq N_{\frac{\epsilon}{2}} \Rightarrow$ $\left \| X_{n}-X_{m} \right \|=\left \| \sum_{k=0}^{n} \rho^{k}-\sum_{k=0}^{m} \rho^{k}\right \|< \left \| \sum_{k=0}^{n} \rho^{k}-l+l-\sum_{k=0}^{m} \rho^{k}\right \| \leq \left \| \sum_{k=0}^{n} \rho^{k}-l\right \|+\left \| \sum_{k=0}^{m} \rho^{k}-l \right \|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$. $\therefore N_{\frac{\epsilon}{2}} $ works. Could you give me any suggestions for the first sequence? Is the second demonstration correct? Thanks in advance for your help. AI: For the first we have that $\frac {1}{k!}<\frac {1}{k^2}$ for $k\geq 4$ and thus $\sum_{k=n+1}^{m} \frac {1}{k!}\leq \sum_{k=n+1}^{m} \frac {1}{k^2}$ and $\sum_{k=n+1}^{m} \frac {1}{k^2}$ converges and so the sequence $x_m=\sum_{k=n+1}^{m} \frac {1}{k^2}$ is Cauchy, and thus $y_m=\sum_{k=n+1}^{m} \frac {1}{k!}$ is upper bounded by a Cauchy sequence,so $x_m$ is Cauchy(do your thing with $ε>0$). For the second well done. But these sums converge and so there are Cauchy.You don't have to show that they are Cauchy with definition,unless you've benn asked too:)
H: Evaluating the Average value of f(x) Determine the average value of $f(x)$ over the interval from $x=a$ to $x=b$, where $f(x)=\frac{1}{x}$, $a=\frac{1}{10}$, and $b=10$ Can someone please explain the simplifying in this problem step-by-step? AI: Note that $$\ln\left(\frac{1}{x}\right) = - \ln{x}$$ In particular, we have that $$\ln{a} - \ln\left(\frac{1}{a}\right) = 2 \ln{a}$$ Hence, we multiply $10$ by $2$ to get $20$.
H: How much Differential Geometry is needed to appreciate Algebraic Geometry? I want to start self-studying Algebraic Geometry at some point in the near future. There are plenty of posts discussing prerequisites, but one thing I couldn't find a discussion of: How much Differential Geometry is needed to appreciate Algebraic Geometry? I know very little Differential Geometry. I read comments saying Differential Geometry is needed to motivate the equivalent of curvature in Algebraic Geometry. This is what's making me concerned. If Differential Geometry is needed, do you think Bishop's book "Tensor Analysis on Manifolds" sufficient? Is there a better (but still concise) book out there? AI: I don't think you need to learn a lot of differential geometry before beginning to learn algebraic geometry. It will definitely help with some things, but most of it you can learn along the way. When you encounter a new concept that you have a hard time understanding, or whose definition you have a hard time motivating, you can take a step aside and look at the analogous constructions on the differential geometry side. You don't need to master it all to begin with. Having a decent background in commutative algebra is probably much more important at the beginning. The spaces you'll encounter at first are quite simple from the topological point of view, and you don't need to be a differential geometer to think about them. However, to understand their subtleties, you'll need to have a good understanding of basic commutative algebra. Having a good background in complex analysis is also quite useful. Assuming you already know basic complex analysis, you might want to pick up a good book on compact Riemann surfaces (such as Rick Miranda's), because studying compact Riemann surfaces is as close as you'll come to doing algebraic geometry without even being aware of it. Nevertheless, a useful book on the "differential" side, in my opinion, is Bott & Tu's Differential Forms in Algebraic Topology. It's a fantastic introduction to the basic ideas of cohomology, and it requires relatively little background to read, which is definitely a plus.
H: Is $ \forall x(P(x) \lor Q(x)) \vdash \forall x P(x) \lor \exists xQ(x) $ provable? I know I should be able to determine whether the following holds, but I am not able to either find a model to show that this is false nor can I prove its correctness by using natural deduction. $ \forall x(P(x) \lor Q(x)) \vdash \forall x P(x) \lor \exists xQ(x) $ Could anybody help me here? AI: The entailment is true, as you can easily check using a tableaux. To construct a natural deduction proof, I would use the deduction theorem to transform the syntactic entailment into a material conditional, then proceed to construct the conditional statement by reductio ad absurdum. I don't know which natural deduction system you're using, but a sketch of a proof would look something like this. Hopefully you can translate my proof into whatever system you're using. $$ \begin{align} (1) & \quad \forall x (P(x) \vee Q(x)) && [\text{HYP}] \\ (2) & \quad \neg (\forall x (P(x)) \vee \exists x (Q(x))) && [\text{HYP}] \\ (3) & \quad \neg \forall x(P(x)) \wedge \neg \exists x(Q(x)) && [\text{DM} (2)] \\ (4) & \quad \neg \forall x(P(x)) && [\wedge\text{-elim}(3)] \\ (5) & \quad \neg \exists x(Q(x)) && [\wedge\text{-elim}(3)] \\ (6) & \quad \exists x (\neg P(x)) && [\neg\forall\text{-elim}(4)] \\ (7) & \quad \forall x (\neg Q(x)) && [\neg\exists\text{-elim}(5)] \\ (8) & \quad \neg P(a) && [\exists\text{-elim}(6)] \\ (9) & \quad \neg Q(a) && [\forall\text{-elim}(7)] \\ (10) & \quad \neg P(a) \wedge \neg Q(a) && [\wedge\text{-intro}(8,9)] \\ (11) & \quad P(a) \vee Q(a) && [\forall\text{-elim}(1)] \\ (12) & \quad \neg (\neg P(a) \wedge \neg Q(a)) && [\text{DM}(11)] \\ (13) & \quad \forall x(P(x) \vee \exists x(Q(x)) && [\text{RAA}(2,10,12)] \\ (14) & \quad \forall x (P(x) \vee Q(x)) \rightarrow (\forall x(P(x) \vee \exists x(Q(x)) && [\rightarrow\text{-intro}(1,13)] \end{align} $$ Since you don't have De Morgan's Laws, I will sketch a proof that $\neg (P \wedge Q)$ implies $\neg P \vee \neg Q$ using the primitive rules of natural deduction, leaving the proof that $\neg (P \vee Q)$ implies $\neg P \wedge \neg Q$ as an exercise. The proofs are similar, so hopefully the following will be useful. $$ \begin{align} (1) & \quad \neg (P \wedge Q) && [\text{HYP}] \\ (2) & \quad \neg (\neg P \vee \neg Q) && [\text{HYP}] \\ (3) & \quad \neg P && [\text{HYP}] \\ (4) & \quad \neg P \vee \neg Q && [\vee\text{-intro}(3)] \\ (5) & \quad P && [\text{RAA}(3,4)] \\ (6) & \quad \neg Q && [\text{HYP}] \\ (7) & \quad \neg P \vee \neg Q && [\vee\text{-intro}(6)] \\ (8) & \quad Q && [\text{RAA}(3,7)] \\ (9) & \quad P \wedge Q && [\wedge\text{-intro}(5,8)] \\ (10) & \quad \neg P \vee \neg Q && [\text{RAA}(1,9)] \\ (11) & \quad \neg (P \wedge Q) \rightarrow \neg P \vee \neg Q && [\rightarrow\text{-intro}(1,10)] \end{align} $$
H: Determine all monic irreducible polynomials of degree $4$ in $\mathbb{Z_2[x]}$ Determine all monic irreducible polynomials of degree $4$ in $\mathbb{Z_2[x]}$ Well these polynomials will be of the form - $a_0 + a_1x + a_2x^2 + a_3x^3 + x^4$ So we have four coefficients that can each have values of either $0$ or $1$. So we have $2^4 = 16$ monic polynomials of degree $4$ in $\mathbb{Z_2[x]}$. Now to determine the irreducible polynomicals is it necessary to write them all out and manually check if they are irreducible? Or is there some lemma I can apply here? AI: You will need to exclude the polynomials divisible by $X$, which are those without constant term $1$, the polynomials divisible by $X+1$, which are those whose coefficients sum to $0$ (mod $2$), the polynomial $(X^2+X+1)^2 = X^4 + X^2 + 1$.
H: Injective and Surjective Functions on Sets I'm fairly new to math proofs. I've been looking for some counterexamples to the following theorems, especially the second one. I haven't been able to think of a scenario. Are the following theorems true? If $f: S \to T$ is an injection and $A \subseteq S$, then $f^{-1}(f(A))=A$. If $f: S \to T$ is a surjection and $C \subseteq T$, then $f(f^{-1}(C))=C$. AI: Making use of the injectivity of the function, We first prove the inclusion $f^{-1}(f(A)) \subset A$. let $x \in f^{-1}(f(A))$ then we have $f(x)\in f(A)$ which implies $x \in A$. We now prove the inclusion $A \subset f^{-1}(f(A))$. let $x\in A$ then we have $f(x)\in f(A)$ which implies $x\in f^{-1}(f(A))$ A similar proof shows the second statement is also true.
H: $A^3 = I$. Find the possible Jordan Forms??? If $A^3 = I$, then I want to find the possible Jordan forms of the matrix. Since the minimal polynomial has degree at most three, each block is at most 3, and the eigenvalues are third roots of unity. Is that the answer, or are there some other components I am missing? AI: You're a little off there. Note that $A^3 = I$, which means that $A^3 - I = 0$, which means that the minimal polynomial $q_A(t)$ of $A$ divides $$ t^3 - 1= (t-1)(t-\omega)(t-\omega^2) $$ where $\omega=e^{2\pi i/3}$ is a root of unity. Note that the maximal degree of any factor of the minimal polynomial is $1$, which means that the maximum length Jordan block for any given eigenvalue is $1$.
H: Can sign of integration by parts equation be changed to negative? Integration by parts: $$\int u \:dv+\int v \: du=uv$$ Are there any circumstances under which the sign can be changed so that the following is true? $$\int u \:dv-\int v \: du=uv$$ AI: $\bullet$ For definite integrals, this is true if $\int_a^b v(t)u'(t) dt=0$ Example in $[-1,1]$ if we put $v(t)=t^2+1$ and $u(t)=\ln(t^2+1)$ then : $$[u(t)v(t)]_{-1}^1=0$$ and :$$\int_{-1}^1udv-\int_{-1}^1 v du = \int_{-1}^1 2t \ln(t^2+1)dt-\int_{-1}^12tdt = 0 $$since $t \mapsto t \ln(t^2+1)$ and $t\mapsto 2t$ are odd one $[-1,1]$ $\bullet$ For indefinite integrals it's true only if $v=0$ or $u$ is constant.
H: What does the phrase "invariant under f" mean? Let $f:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $f(x,y)=(\frac{x}{3}+11y^2,-2y)$. Let $A=\{(3y^2,y)\|y\in \mathbb{R}\}$. Show that $A$ is invariant under $f$. What is it asking? AI: See that $f(3y^2,y)=(\frac{3y^2}{3}+11y^2,-2y)=(12y^2,-2y)=(3(-2y)^2,(-2y))$ which is obiviously in the set A.
H: How to apply Fubini here Let $f$ and $g$ be integrable functions on the measure space $(X, \mathcal M, \mu)$ with the property that $$ \mu(\{f > t\} \triangle \{g > t\}) = 0 $$ for $\lambda$-a.e. $t \in \mathbb R$. Prove that $f = g\;$ $\mu$-almost everywhere. Here, $\lambda$ is Lebesgue measure and $\triangle$ denotes symmetric difference. My thoughts: I'm quite certain I should apply Fubini here to some $F: X \times \mathbb R \to \mathbb R$. But I'm not sure how to choose $F$ and, even if I did have an $F$ in hand, I don't know how to apply Fubini without some completeness or $\sigma$-finite condition on $(X, \mathcal M, \mu)$. Any hints would be much appreciated. AI: No need for Fubini. Let $D_t = f^{-1}(t,\infty) \triangle g^{-1}(t,\infty)$. Note that if $s<t$, then $D_t \subset D_s$. It follows that $\mu D_t = 0$ for all $t$ (not just ae. [$\lambda$]). Let $q_n$ be an enumeration of $\mathbb{Q}$. Define $\Delta_n = f^{-1}(q_n,\infty) \cap g^{-1}(-\infty, q_n]$. Let $\Delta = \cup_n \Delta_n$. Then we have $\Delta = \{ x \| f(x) > g(x) \}$. Also, note that $\Delta_n \subset D_{q_n}$. Suppose $\mu \Delta>0$, then $\mu \Delta_n >0$ for some $n$, which contradicts $\mu D_{q_n} = 0$. Hence $\mu \Delta = 0$. Exactly the same analysis applies with $f,g$ swapped, hence $f(x) = g(x)$ ae. [$\mu$].
H: Is there a single valued definition for $\sqrt[n]{z}$ on $\Bbb{C}$? Since the $n$th root of a complex number has $n$ possibilities, which one do you choose so that its restriction to $\Bbb{R}$ is the positive square root if $r$ is positive and $\sqrt{\|r\|}i$ if $r$ is negative. AI: Define $$\sqrt[n]{r e^{i\theta}} = \sqrt[n]{r} e^{i\frac{\theta}{n}} $$ where $\theta \in [0, 2\pi)$ Seems to do the trick.
H: Is it the case that every quotient ring ${F[x]}/{(p(x))}$ is a PID? Let $F$ be a field and $F[x]$ the univariate polynomial ring over $F$. Is it the case that every quotient ring ${F[x]}/{(p(x))}$, where $p(x) \in F[x]$, is a PID (principal ideal domain)? This is just a minor issue that's been bothering me for the last couple of hours. I'm aware that when the ideal $(p(x))$ is prime (i.e. $p(x)$ irreducible) then ${F[x]}/{(p(x))}$ is a PID. AI: If $p(x)$ isn't irreducible (and isn't the zero polynomial), then $F[x]/(p(x))$ isn't an integral domain, much less a PID. (And when $p(x)$ is irreducible, then $F[x]/(p(x))$ is a field.)
H: Is it theotically possible to create a machine that could randomly generate any elementary geometric problems/theorem? Refering back to this quesion: Is it true that all of the euclidean geometry problem in the IMO(international mathematical olympiad) could all be solve by the analytical geometry? , we know that any elemetary geometry problem could be solved by computer in finite amount of time. So, conversely, is it theotically possible to create a machine that could randomly generate any elementary geometric problems/theorem? AI: Suppose we formalize elementary geometry, say the plane version, using say Tarski's formalization (there are others). Then of course we can generate all well-formed sentences. As to the "randomly" part, we could use a pseudo-random number generator. If that is not good enough, I have no algorithmic suggestion. Note that the axioms of Tarski's geometry are a recursive set. Therefore, as noted in the answer by apt1002, we can not only algorithmically list all sentences, we can algorithmically list all proofs. (A proof is just a special kind of finite list of sentences.) Tarski's theory is complete. Therefore, in principle, to verify whether a sentence $\phi$ is a theorem, we just list all proofs. If $\phi$ appears at the end of a proof, then $\varphi$ is a theorem. If $\lnot\varphi$ appears at the end of a proof, then $\varphi$ is not a theorem. This procedure must terminare, so we have an algorithm. Of course it is a terrible algorithm. Tarski's algorithm for the theory of real-closed fields, and the various implementations since, are far more efficient.
H: $A=\{1,2,3,4,5\}$. How many functions $f : A \to A$ so that $(f\circ f)(1) = 3$ $A=\{1,2,3,4,5\}$ How many functions $f : A \to A$ so that $f$ is onto? $5!$ is this correct? How many functions $f : A \to A$ so that $(f\circ f)(1) = 3$? $f(1)=1, f(1)=2, f(1)=3, f(1)=4, f(1)=5$? I don't know what to do... How many onto functions $f : A \to A$ so that $(f\circ f)(1) = 3$? AI: Your first answer is correct. As for the second question, consider the following process of constructing a suitable function: $f(1)$ has to be something. If $f(1)$ is $1$, then $f(f(1))=1$, which is disallowed. So, set $f(1)=a$, where $a$ is some number in this set which isn't $1$. Our constraint is that $f(f(1))=3$. That is, $f(a)$ has to be $3$. So, set $f(a)=3$. We are now sure that our function, however we proceed to construct the rest of it, will satisfy the constraint $f(f(1)) = 3$. Proceed to extend the function to all other numbers in this set. Note that any function satisfying the constraint may be constructed in this fashion. Now, we ask the question: how many ways are there to proceed through this process? In our first step, setting $f(1) = a$, we have four choices of $a$. Having chosen an $a$, we have no choice but to set $f(a) = 3$. We have $3$ more distinct elements $t\in A$ whose value $f(t)$ must be chosen from a set of the $5$ possible options in $A$. Thus, there are $4\times 5 \times 5 \times 5 = 500$ possibilities. All right, apparently this process is confusing, so let's build an example function. First, we need to choose some $a \neq 1$. Let's say $a = 4$. Now, we have $f(1) = 4$, and $f(4) = 3$. With that settled, we have to figure out what $f(2),f(3),$ and $f(5)$ are. Let's say $f(2) = 1$, $f(3) = 3$, and $f(5) = 4$. We now have a suitable function. We could have picked any of $5$ values for each of $f(2),f(3)$ and $f(5)$.
H: Proving that $0^\infty$ is not indeterminate? Suppose that $f(x)>0$ for all $x$, and for some $a$ both $\lim_{x\to a} f(x) = 0$ and $\lim_{x\to a} g(x) = \infty$ are true. How would I go about proving: $$\lim_{x\to a} f(x)^{g(x)} = 0$$ I am stuck after attempting to evaluate $\lim_{x\to a} \ln(f(x)^{g(x)})$. After some work, I received $$\lim_{x\to a} \frac{g(x)}{-ln(f(x))}$$ but even though I know $\ln(f(x))$ exists for all $x$, it is hardly a guarantee of continuity or differentiability; $g(x)$ is in a similar situation. Thus, I am reluctant to apply L'Hôpital's rule. May I have a hint as to how I should proceed? AI: If $0 \lt \epsilon \lt 1$, show that if $f(x) \lt \epsilon$ and $g(x) > 1$, then $f(x)^{g(x)} \le \epsilon$. Then explain what that has to do with the problem.
H: Non-isomorphic simple graphs: order $n$, size $\displaystyle \frac{na}{2}$, degree sequence $(a,a,a,...,a) \in \mathbb{N}^n$ If a simple graph has order $n$, size $\displaystyle \frac{na}{2}$ and degree sequence $(a,a,a,...,a) \in \mathbb{N}^n$ then is it unique up to isomorphism? I thought of this question while examining the Petersen Graph, which is a simple graph (no multiple edges, no loops) satisfying all of the above with $n=10$ and $a=3$. I can easily find many $\textbf{non-simple}$ graphs satisfying all of the above which are not isomorphic to each other. I am looking for either a proof, or counterexample in the form of two non-isomorphic simple graphs. AI: No. Compare the circle on six vertices to the disjoint union of two triangles. Both have six vertices and degree sequence $(2,2,2,2,2,2)$. Check, that the size is determined by this two properties, so it must also be the same. In case you are asking about connected graphs, compare: The complete bipartite graph with three vertices per partition $K_{3,3}$ The disjoint union of two triangles with three edges added, such that each vertex of a triangle is connected to exactly one vertex of the other triangle. EDIT: Regarding the Petersen graph, you could create a non-isomorphic graph as follows: Take three disjoint triangles. Connect each two triangles by two non-incident edges. Now each triangle is left with exactly one vertex of degree two. Connect those vertices to a tenth vertex. This graph is not isomorphic to the Petersen graph as it contains triangles.
H: Block matrix characteristic polynomial Let $n \in \Bbb N^*$ and $A \in \cal M_n(\Bbb R)$ a square matrix. Let the block matrix $$B = \begin{pmatrix}A&A\\A&A \end{pmatrix} \in \cal M_{2n}(\Bbb R)$$ Calculate the characteristic polynomial $\chi_B$ using $\chi_A$. Proof that $A$ is diagonalisable $\iff$ $B$ is diagonalisable. AI: For the first part: we have $$ tI_{2n} - B = \pmatrix{tI_n - A & -A\\ -A & tI_n - A} $$ Noting that these matrices commute (or in particular, that the lower two matrices commute), we have $$ \begin{align} \det(tI_{2n} - B) &= \det((tI_n - A)^2-A^2)\\ &= \det[((tI_n - A)-A)((tI_n - A)+A)]\\ &= \det((tI_n - A)-A) \det((tI_n - A)+A)\\ &= \det((tI_n - 2A) \det(tI_n)\\ &= \det(2((t/2) - A)) \det(tI_n)\\ &= 2^n \det((t/2) - A) t^n\\ &= 2^n t^n \chi_A\left(\frac t2\right) \end{align} $$ Second part: I haven't put it all together nicely, but here are some potentially helpful thoughts Thought 1: If $B$ is diagonalizable, then $B$ has $2n$ linearly independent eigenvectors. Now, let $$ v = \pmatrix{v_1\\v_2} $$ be an eigenvector of $B$, with $v_1,v_2 \in \mathbb{C}^n$. What can we say about the vectors $v_1$ and $v_2$? Remember that $B$ has the eigenvector $0$ of multiplicity $n + \operatorname{null}(A)$. Thought 2: $$ \pmatrix{I & I\\ I& -I} \pmatrix{A & A\\ A & A}\pmatrix{I& I\\I & -I} = \pmatrix{4A & 0\\0 & 0} $$ Or, in particular, $$ \frac{1}{2}\pmatrix{I & I\\ I& -I} \pmatrix{A & A\\ A & A} \pmatrix{I& I\\I & -I} = \\ \left(\frac{1}{\sqrt{2}}\pmatrix{I & I\\ I& -I}\right)^{-1} \pmatrix{A & A\\ A & A} \frac{1}{\sqrt{2}}\pmatrix{I& I\\I & -I} = \pmatrix{2A & 0\\0 & 0} $$
H: Derivation of Symmetry Property of Metric Spaces I am given the following modified triangle inequality property of metric spaces, where for any $x_1$, $x_2$, $x_3 \in X$, we have $d(x_1, x_2) \le d(x_1, x_3)+d(x_2, x_3)$. I am tasked to show that the symmetry property follows from the use of this and the positive definite property. I am having difficulty moving beyond the definition of Euclidean distance in thinking about this problem. Can someone offer a suggestion to get started? AI: Let $x_3=x_1$ in the formula above. This gives $d(x_1,x_2) \le d(x_1,x_1) + d(x_2,x_1) = d(x_2,x_1)$. Exchanging the roles of $x_1,x_2$ gives $d(x_2,x_1) \le d(x_1,x_2)$. Hence $d(x_2,x_1) = d(x_1,x_2)$.
H: Symbol for "is closest to"? I am writing a paper on probabilities and we have to find a $k$ such that $P_n(k)$ is "closest to" $P_0$. $P_0$ is getting 4-of-a-kind in a five card hand in a standard 52 card deck. $P_n(k)$ is probability of getting $k$-of-a-kind in an $n$ card hand in some modified 88 card deck. I want to say that getting 5-of-a-kind (there are 11 suits) in a 5 card hand for our modified deck produces a probability "most similar" or "closest to" $P_0$. So would this be ok for a theorem?\ $P_5(k)$ is the probability closest to $P_0$ when $k=5$. That is, $\lvert P_0 - P_5(k)\rvert$ is minimized for $k=5$. AI: You could either say exactly what you said above, or, more formally: $\|P_i-P_n(k)\|$ is minimized for $i=0$.
H: showing a bijection is true for the assumption that it takes 4 points to have 1 intersection in a circle Problem: 15 points are taken on the circumference of a circle, and through any two of them a chord is drawn. Suppose that no three chords intersect at the same point inside the circle. How many points of intersections are between these chords? My attempt: I know that you need atleast 4 distinct points on the circumference to have an intersection. If i can show that a bijection is true for n points then I can easily say how many points of intersections occur for fifteen points. I know to prove a bijection I must also prove that it is injective and surjective however I get stuck when creating the function. surjective: if(fx) = f(y) then f=y so I have a function k(x) = x(1/4). where x >= 4. However my function fails when I input 5 since you cant have 1.25 intersections. any help? AI: Given any $4$ of our $15$ points, precisely one pair of lines determined by pairs of these points meet in the interior of the circle. Since intersection points are by assumption distinct, there are $\binom{15}{4}$ intersection points.
H: How to prove the $f(x) = \sqrt{x + \sqrt{x}}$ is injective. The function is $$ f(x)=\sqrt{x+\sqrt{x}} $$ I know that you need to set up the equation $$\sqrt{x_1+\sqrt{x_1}}=\sqrt{x_2+\sqrt{x_2}}$$ and you have to solve step by step until you get $x_1=x_2$. But I am having difficulty figuring out what to do. I have tried squaring both sides, factoring, then completing the square but nothing has worked. All answers appreciated. Please keep in mind that I need every step, thank you. AI: Can you use the fact that the (nonnegative branch of the) square root is strictly increasing? If so, one way to prove it would be to notice that if for some $0\leq x_1< x_2$ we have $\sqrt{x_1+\sqrt{x_1}}=\sqrt{x_2+\sqrt{x_2}}$ then $x_1+\sqrt{x_1}=x_2+\sqrt{x_2}$ and so $x_1-x_2=\sqrt{x_2}-\sqrt{x_1}$ but since we are assuming $x_1<x_2$, the left side of the equation is negative, and the right side is positive. Just in case, to prove that the square root is strictly increasing, let $x_1,x_2$ be nonnegative real numbers with $x_1<x_2$. If we suppose that $\sqrt{x_1}\geq \sqrt{x_2}$ then, since we are dealing with nonnegative numbers, squaring the inequality gives us $x_1\geq x_2$ which contradicts our hypothesis.
H: Find the Laurent series and residue of $\frac{z}{(\sin(z))^2}$ at $z_0 = 0$. Find the Laurent series for the given function about the indicated point. Also, give the residue of the function at the point. $$\frac{z}{(\sin(z))^2}\quad \text{at}\quad z_0 = 0 \quad\text{four terms of the Laurent series}$$ I am not sure how to approach this question. Can anyone help me with this? Thank you. AI: The Laurent series for $\sin z$ about $z_0 = 0$ is given by it's Taylor series about $z_0 = 0$ because sine is entire. Then, you have $z$ over the Taylor series of $\sin(z)$. Assuming you want to find the Laurent series to find the residue, I will first do everything approximately removing terms that won't contribute to the residue. i.e. $\frac{z}{\sin^2(z)} = \frac{z}{\left(z-\frac{z^3}{3!}+\ldots\right)\left(z-\frac{z^3}{3!}+\ldots\right)} = \frac{z}{(z^2 - \frac{z^4}{3}+\ldots)} = \frac{z}{z^2\left(1-\frac{z^2}{3}+\ldots\right)} = \frac{1}{z(1-w)}$ where $1-w = 1-\frac{z^2}{3}+\ldots$. Then $w$ is a small term because we are expanding about $z_0=0$. So we can expand $1-w$, $\frac{1}{1-w} = 1 + w + w^2 + \ldots$, hence we find $\frac{z}{\sin^2(z)} = \frac{1}{z}\left(1+w + w^2 + \ldots\right) = \frac{1}{z}(1+ (\frac{z^2}{3} + \ldots))$, where everything has been valid if we care only about the residue. Hence, $Res\left(\frac{z}{\sin^2(z)} \right) = 1$. If instead you actually do require a Laurent series, we can let $w = \frac{1}{z}$ and find the Laurent series of $\frac{(1/w)}{\sin^2(1/w)}$ so we have $\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)}. $ Note that $\sum_{n=0}^\infty (1/w)^{2n+1} = \sum_{m=-\infty}^0 w^{2m+1}$, we use this identity ($m=-n$) to find $\frac{(1/w)}{\left(\sum_{n=0}^\infty (-1)^n \frac{(1/w)^{2n+1}}{(2n+1)!}\right)^2} = \frac{1/w}{ \left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)\left(\sum_{n=-\infty}^0 (-1)^{-n} \frac{w^{2n+1}}{(2(-n)+1)!}\right)}$. Next we multiply the series $\frac{(1/w)}{ \sum_{n=-\infty}^0 \sum_{m=-\infty}^0 (-1)^{n+m}\frac{w^{2(n+m)+2}}{(2(-n)+1)!(2(-m)+1)!}}$ where I have used that $(-1)^n = (-1)^{-n}$ Now let $l=n+m$, yielding $\frac{(1/w)}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+2}}{(2(m-l)+1)!(2(-m)+1)!}} = \frac{1}{ \sum_{l=-\infty}^0 \sum_{m=-\infty}^l (-1)^{l}\frac{w^{2l+1}}{(2(m-l)+1)!(2(-m)+1)!}}$. And you find the residue when $l=0$, so you find the residue to be 1, again!
H: Clarification on solution to a problem: The problem is as follows: How many zeroes do we write when we write all the integers from 1 to 243 in base 3? The given solution starts as follows: The 1-digit numbers don't have any zeroes. The 2-digit numbers use 2 zeroes: 10 and 20. There are $3^2 = 9$ three-digit numbers starting with 1 and 9 starting with 2. For each leading digit, a zero appears in each digit in 9/3 = 3 of the numbers, so each has a total of 3+3 = 6 zeroes. Thus, the 3-digit numbers contain 26 = 12 zeroes. I do not understand how they arrived with the fact that "for each leading digit a zero appears in 9/3 = 3 of the numbers" and why they are multiplying the result by 2 to get 12 at the end. AI: Half of the three-digit numbers start with 1; the other half start with 2. Consider those that start with 1. The 'middle' digit (to the left of the leading 1) can be a 0 in three different cases: when the rightmost (least significant) digit is 0, 1 or 2. Similarly, the rightmost digit will be 0 for the three values (0, 1, 2) that the middle digit will take on. Thus, for the three-digit numbers starting with 1, we use a total of 6 zeroes. The same calculation applies for the three-digit numbers with a leading 2. This gives us a total of $26 = 12$ zeroes in the three-digit numbers.
H: About the symmetry of Riemann Tensor It is a problem in my homework. First I was asked to show $$ \nabla_a\nabla_bA_c-\nabla_b\nabla_aA_c=R_{a,b,c}^{\;\;\;\;\;d}A_d $$ where $A$ is a (0,1)-tensor and $R_{a,b,c}^{\;\;\;\;\;d}$ is the Riemann curvature tensor, which is defined by $$ \nabla_a\nabla_bV^c-\nabla_b\nabla_aV^c=R_{a,b,d}^{\;\;\;\;\;c}V^d $$ I proved it. Then the question says Hence, show that $R_{a,b,c}^{\;\;\;\;\;d} +R_{b,c,a}^{\;\;\;\;\;d}+R_{c,a,b}^{\;\;\;\;\;d}=0$ However, I don't know how it can be deduced from the above identity. I tried to look at $$ [\nabla_a,\nabla_b]A_c+[\nabla_b,\nabla_c]A_a+[\nabla_c,\nabla_a]A_b $$ But it seems not obvious that it is zero. Can anyone help? AI: It suffices to show that $$R_{a,b,c}^{d} A_{d} + R_{b,c,a}^{d}A_{d} + R_{c,a,b}^{d}A_{d} = 0$$ for any $A_d$. Expanding via the first formula, you get $$\left( \nabla_{a}\nabla_{b}A_{c} - \nabla_{b}\nabla_{a}A_{c} \right) + \left( \nabla_{b}\nabla_{c}A_{a} - \nabla_{c}\nabla_{b}A_{a} \right) + \left( \nabla_{c}\nabla_{a}A_{b} - \nabla_{a}\nabla_{c}A_{b} \right)$$ Now, if we take the first and last terms $$\nabla_{a}\nabla_{b}A_c - \nabla_{a}\nabla_{c}A_b = \nabla_{a}(\nabla_{b}A_{c} - \nabla_{c}A_b) = \nabla_{a}(\nabla_{b}c(A) - \nabla_{c}b(A))$$ $$ = \nabla_{a}(\nabla_{b}c - \nabla_{c}b)A = \nabla_{a}([b,c]A) = 0$$ Since $a,b,c$ are coordinate vector fields for which the Lie bracket vanishes or is constant. The other pairs of terms cancel similarly.
H: If $G_1$ and $G_2$ are simple groups, what can we say about normal subgroups of $G_1 \times G_2$? If $G_1$ and $G_2$ are simple groups, what can we say about normal subgroups of $G_1 \times G_2$? I remember when I was taking Algebra I this was brought up in the class but at the time the professor left it for us to think over it. Well, I remember I conjectured that $\{e\}, G_1 \times \{e\}, \{e\}\times G_2$ and $G_1 \times G_2$ are the only normal subgroups of $G_1 \times G_2$. I'm trying to see whether I am right or not. I haven't studied Sylow theorems yet and at the time the professor brought up this question I remember that we were studying direct products of groups. Is it possible to answer this question with elementary theorems in abstract algebra? Notice that $G_1$ and $G_2$ are not restricted to finite groups in my question. AI: Hints: If $N$ is a normal subgroup of $G_1\times G_2$, as the projections $p_i:G_1\times G_2\to G_i$ are surjective, they map $N$ onto a normal subgroup. For the case both $p_i(N)=G_i$, let $M:=\{g_1\mid (g_1,1)\in N \}$. Then $M$ is again a normal subgroup (of $G_1$). If $M=G_1$ we are ready soon. Finally, if $M=\{1\}$ (by symmetry it also means $\{g_2\mid (1,g_2)\in N\}=\{1\}$), conclude that $N$ is a graph of an isomorphism (e.g. $(g_1,g_2)\in N$ and $(g_1,g_2')\in N$ implies $g_2=g_2'$), in other words $N$ is just the diagonal of $G_1\times G_2$ applying the isomorphism $G_1\cong G_2$. On the other hand, unless $G$ is commutative, the diagonal of $G\times G$ for a simple group is not a normal subgroup.
H: If $Y: G\to H$ is a group homomorphism and $G$ is abelian, prove that $Y(G)$ is also abelian. What I got: Suppose $Y$ is a homomorphism and that $G$ is abelian. Then for all $a,b \in G$, $ab=ba$, and thus $Y(ab)=Y(ba)=Y(a)Y(b)=Y(b)Y(a)$. However, this seems too simple and I was confused on what $Y(G)$ meant (since it wasn't $Y(g)$, I'm not sure if $Y(G)$ is a group or a function.) I need my math.stackexchange buddies to tell me where I probably messed up. AI: Let $h_{1}, h_{2} \in Y(G)$. Then there exist elements $g_{1}, g_{2} \in G$ such that $Y(g_{1}) = h_{1}$ and $Y(g_{2}) = h_{2}$. So, $h_{1}\cdot h_{2} = Y(g_{1})\cdot Y(g_{2}) = Y(g_{1}\cdot g_{2}) = Y(g_{2}\cdot g_{1}) = Y(g_{2})\cdot Y(g_{1}) = h_{2}\cdot h_{1}$. Hence, $Y(G)$ is Abelian.
H: Work done by a force field $F$ via the line integrals Let $F:\mathbb R^2\to \mathbb R^2$ be the force field with $$F(x,y) = -\frac{(x,y)}{\sqrt{x^2 + y^2}}$$ the unit vector in the direction from $(x,y)$ to the origin. Calculate the work done against the force field in moving a particle from $(2a,0)$ to the origin along the top half of the circle $(x−a)^2+y^2=a^2$. Okay, I tried to use the line integral and I set $x=a+a\cos(t)$, $y= a\sin(t)$ and $t\in [0,\pi]$. Then the work should be $$\int_0^\pi F(r (t))(r)′dt$$ But I can't got the right answer!! AI: Your vector field is conservative: $\nabla \times F = 0$. Thus the integral is path independent. This should simply your calculation considerably—choose the easy straight line path from $(2a,0)$ to $(0,0)$ and integrate.
H: Indefinite integral...for calc 1? a calc 1 student I know was given this integral do via u-substitution, and even though I think it's an obvious typo, I know the answer is somewhat simple thanks to Wolfram, but am not sure how to arrive at it. $\int{x^2\sqrt{1+x}}$ $dx$. Any help would be appreciated! AI: let $u=1+x$ so that $du=dx$ and $x^2=(u-1)^2$ The integral now becomes $\int(u-1)^2(\sqrt{u}) du=\int(u^{5/2}-2u^{3/2}+u^{1/2})du=\frac{2}{7}u^{7/2}-\frac{4}{5}u^{5/2}+\frac{2}{3}u^{3/2}+c$ Remember to replace $u$ by $x+1$ in the last expression.
H: Consumer surplus for demand curve at the given sales level $x$ The square root is throwing me of doing the integration. Can someone please show the steps of integration. Answer is $\$26.19$ Ok then can someone show me how we got the integration here? AI: The height of the curve at $x = 50$ is $\sqrt{16 - 0.14\times50}=\sqrt{9}=3$. Thus, the consumer surplus will be given by $$ \int_0^{50}(p(x)-50)dx = \int_0^{50}\left[\sqrt{16 - 0.14x}-3\right]dx $$ As Tyler indicated, find this integral using the $u$-substitution of $u = 16 - 0.14 x$. So, we want the function whose derivative is $\sqrt{16 - 0.14x}$. If we naively just integrated the outside function, we would get $\frac{2}{3} (16 - 0.14x)^{3/2}$. Let's check what the derivative of this function would be. We find, using the chain rule, $$ \frac d{dx} \frac{2}{3}(16 - 0.14x)^{3/2} = \sqrt{16 - 0.14 x} \cdot(-0.14) $$ This is almost what we want, except that we have a $-0.14$ multiplying at the end. So, we're going to take what we got originally and divide by $-0.14$. We then have $$ \frac d{dx} \frac{1}{-0.14}\frac{2}{3}(16 - 0.14x)^{3/2} = \frac{1}{-0.14}\sqrt{16 - 0.14 x} \cdot(-0.14) = \sqrt{16 - 0.14 x} $$ And that's the derivative we wanted! So, we now know that $$ \int \sqrt{16 - 0.14 x} = -\frac{1}{0.14}\frac{2}{3}(16 - 0.14x)^{3/2} + C $$ Use this to find the integral you need.
H: Determine angle from radius of curvature This is another grade school problem that's giving me trouble (posting on someone else's behalf). I can see that a 36 inch semi-circumference yields a radius of 36/Pi or about 11.46 inches. However, I can't see how to use this information to calculate the angle. Given the width of the arch, I may be able to do this, but don't see an easy solution otherwise. Given that this is a grade school problem, I'm obviously missing something basic. Using the "eyeball theorem" (ha ha), it seems like that angle is 172 degrees (it's clearly not 85 or 100 obviously). AI: The formula is $r a = s$ where $r$ is the radius, $s$ is the arc length, and $a$ is the central angle in radians. So the angle is $36/12 = 3$ radians, which is about $172$ degrees (multiply by $\pi/180$).
H: Power series method to solve Airy’s differential equation Using power series method, solve Airy’s equation $$y′′+ xy = 0.$$ How do I start solving this? Thanks in advance! AI: Given: $$\tag 1 y''+ x y = 0$$ Solve this using Power Series. We assume: $$y = \sum_{m=0}^\infty a_mx^m$$ Thus we have: $$y'' = \sum_{m=2}^\infty m(m-1)a_mx^{m-2}$$ Substituting into the $(1)$, we get: $$\sum_{m=2}^\infty m(m-1)a_mx^{m-2} + \sum_{m=0}^\infty a_mx^{m+1} = 0$$ Aligning starting points for the series, we have: $$2a_2 + \sum_{m=1}^\infty ((m+1)(m+2)a_{m+2} + a_{m-1})x^m = 0$$ Now equate terms to zero and solve for the $a's$.
H: Does a function that has an exponential analog to $log(xy) = log(x) + log(y)$ exist? Similar to how $log(xy) = log(x) + log(y)$, does a nontrivial function exist that has the property $f(x^y) = f(x)f(y)$? How would one attempt to derive such a function? AI: Then $f(x^y) = f(y^x) \Rightarrow f(1^y) = f(y^1) \Rightarrow f(1) = f(y)$ for any $y$. f is a constant function, $0$ or $1$.
H: $1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ ... + \frac{1}{1+2+3+...+n} = ?$ How do I simplify the following series $$1 + \frac{1}{1+2}+ \frac{1}{1+2+3}+ \frac{1}{1+2+3+4} + \frac{1}{1+2+3+4+5} + \cdot\cdot\cdot + \frac{1}{1+2+3+\cdot\cdot\cdot+n}$$ From what I see, each term is the inverse of the sum of $n$ natural numbers. Assuming there are $N$ terms in the given series, $$a_N = \frac{1}{\sum\limits_{n = 1}^{N} n} = \frac{2}{n(n+1)}$$ $$\Rightarrow \sum\limits_{n=1}^{N} a_N = \sum\limits_{n=1}^{N} \frac{2}{n(n+1)}$$ ... and I'm stuck. I've never actually done this kind of problem before (am new to sequences & series). So, a clear and detailed explanation of how to go about it would be most appreciated. PS- I do not know how to do a telescoping series!! AI: $$ \sum_{n = 1}^{N}{2 \over n\left(n + 1\right)} = 2\sum_{n = 1}^{N}\left({1 \over n} - {1 \over n + 1}\right) = 2\left[1 - {1 \over 2} + {1 \over 2} - {1 \over 3} + \cdots + {1 \over N} - {1 \over N + 1}\right] $$ $$ \sum_{n = 1}^{N}{2 \over n\left(n + 1\right)} = 2\left(1 - {1 \over N + 1}\right)\quad\to\quad \color{#0000ff}{\Large 2}\quad\mbox{when}\quad N \to \infty $$
H: Basis for vector space given combination of vector components The following is the first step in a homework problem of mine: Find a basis for the vector space $S = \{(x,y,z,w) \in \mathbb{R}^4 \mid x - y - 2z + w = 0\}$. The actual problem involves computing the orthonormal basis using Gram-Schmidt, which I know how to do. As this is homework, I'm just looking for some information on methods that can be used to find the basis for this vector space, not the actual answer to this problem. Thanks! AI: You have a linear system with 1 equation in 4 unknowns. Hence you can think of this as having 3 free variables and one bound variable. For each of the free variables in turn, set it equal to 1 and the other free variables to zero. Solve for the bound variable(s). You will get a set of vectors, that is easy to prove is independent and of the right cardinality to be a basis.
H: What is $a_5$, given the recurrence $a_{n+1}=a_n+2a_{n-1}$ and we know that $a_0 = 4, a_2 =13$ I am having a very hard time figuring this out. So far I have been able to do the following: Writing the recurrence as a characteristic polynomial = $x^2-x-2=0$ so there are roots, $x=2, x=-1$. So there will be the general solution of $X_n = c_12^n + c_2 n(-1)^n$ I think the first constant will be 4..but I really am not sure. And from there on I am very confused with where to go. I have a big feeling this sort of question will be on my test in a few weeks so an explanation would be very helpful :) Thanks in advance! AI: Direct calculation is quite a bit easier. However, let us look into the characteristic equation method, since the technique will undoubtedly be needed later. The roots are $2$ and $-1$, so the general solution is $A(2^n)+B(-1)^n$. We need to find $A$ and $B$. Setting $n=0$, we find that $A(2^0)+B(-1)^0= 4$. So $A+B=4$. Setting $n=2$, we get $A(2^2)+B(-1)^2=13$, so $4A+B=13$. Solve the system of two equations in two unknowns. We get $A=3$ and $B=1$. Thus $$a_n=3(2^n)+(-1)^n.$$ Setting $n=5$, we get $96-1$, that is, $95$. Remark: More directly, $a_2=a_1+2a_0=13$, and therefore $a_1=5$. Now we can use the recurrence to find $a_3$ then $a_4$ then 4a_5$.
H: distribution of objects Find the number of ways to distribute $n$ distinct objects to $5$ distinct boxes, such that boxes $1$, $3$ and $5$ must hold an odd number of objects and boxes $2$ and $4$ must hold an even number of objects. AI: Is the value of $n$ odd or even?
H: calculus interval and concave up and down f(x)=3(x)^(1/2)e^-x 1.Find the interval on which f is increasing 2.Find the interval on which f is decreasing 3.Find the local maximum value of f 4.Find the inflection point 5.Find the interval on which f is concave up 6.Find the interval on which f is concave down Anyone can explain? I know the f'(x)=e^-x(3-6x)/2(x)^(1/2) AI: Given: $$f(x)=3(x)^{1/2}e^{-x}$$ Hints: $$f'(x) = \dfrac{3 e^{-x}}{2 \sqrt{x}} - 3 e^{-x} \sqrt{x}$$ $$f''(x) = \dfrac{3 e^{-x}}{4 x^{3/2}} - \dfrac{3 e^{-x}}{\sqrt{x}} + 3 e^{-x} \sqrt{x}$$ A plot shows: $(1.)$ If $f'(x) > 0$ on an interval I, then $f$ is increasing on that interval. $(2.)$ If $f'(x) < 0$ on an interval I, then $f$ is decreasing on that interval. $(3.)$ Find values of $x$ where $f'(x)=0$ or $f'(x)$ is undeﬁned. These will be candidates for possible maximum or minimum. Test these points further to determine which ones correspond to a maximum, a minimum or neither. $(4.)$ If $f$ has a point of inﬂection at $c$, then either $f''(c)=0$ or $f''(c)$ is undeﬁned. $(5.)$ The sign of $f''(x)$ tells us if $f$ is concave up or down. More speciﬁcally, if $f''(x) > 0$ on an interval I, then $f$ is concave up on that interval. $(6.)$ If $f''(x) < 0$ on an interval I, then $f$ is concave down on that interval.
H: How to derive duration of unemployment? The average monthly flow out of unemployment pool of $7.0$ million people each month is $3.1$ million. Put another way, the proportion of unemployed leaving unemployment equals $\frac{3.1}{7.0}$ or about $44 \%$ each month. Put yet another way, the average duration of unemployment - the average length of time people spend unemployed - is between $2$ and $3$ months. Now, I can't understand how the duration of unemployment is derived and claimed to be $2-3$ months. AI: I also have a hard time figuring out how exactly does one get to this claim of a 2-3 month interval. On way to obtain a result which would more or less match this 2-3 month contention would be to assume that the process of leaving unemployment is a Poisson process (see http://www.rle.mit.edu/rgallager/documents/6.262vdbw2.pdf). Then you would have that the time before one gets out of unemployement is a random variable $X$ following an exponential distribution $f_X(x) = \lambda \exp(-\lambda x)$, where $\lambda = 0.44$, the rate at which people get out of unemployment. Under the assumption of a Poisson process, given $X\sim \lambda \exp(-\lambda x)$, you have $\mathbb{E}(X) = \frac{1}{0.44} = 2.27$, which roughly matches the interval of 2-3 month. Now I realize that this is really farfetched and requires a lot of additional assumption which are not provided in the statement of your question. But it is the best I can think of to make sense of it...
H: Calculating the Jacobian Matrix I am working with the system: $$ u' = v $$ $$ v' = -w^{2}sin(3\pi+u)-cv $$ where $c$ and $w$ are positive constants. I'm computing the Jacobian matrix: $$J= \begin{pmatrix} F_u & F_v \\ G_u & G_v \\ \end{pmatrix} $$ $$ J= \begin{pmatrix} 0 & 1 \\ -w^2cos(3\pi+u) & -c \\ \end{pmatrix} $$ Is this correct? I think I am doing something wrong with calculating the derivative of $-w^{2}sin(3\pi+u)$. AI: How did you get that $G_v$? The rest looks good assuming $\omega$ is a constant.
H: If $G$ is solvable, is it true that for any $m,n\in\operatorname{cd}(G)$, there exists a prime $p$ such that $p\mid m,n$? Let $G$ be a finite group and let $\operatorname{cd}(G)$ be the set of degrees of irreducible characters of $G$. It is known that if for any $m,n \in \operatorname{cd}(G) \setminus \{1\}$, there exists a prime $p$ such that $p \mid m,n$, then $G$ is solvable. Does the converse hold? Thanks in advance. AI: No, take for example $S_4$, which has character degrees $1,2,$ and $3$.
H: To find Z-transform of given sequence How to find the $z$-transform of $\left[a^{n}\sin\left(bn\right)\right]/n!$ where "!" denotes factorial of a number and b is constant?? AI: Hints: i) $$\sin( bn )=\frac{1}{2i}(e^{ibn}-e^{-ibn}).$$ ii) Z-Transform of $a^ne^{ibn}$ is given by $$F(z) = \sum_{n=0}^{\infty} a^n (e^{ib})^nz^{-n}.$$ iii) The following is known as the geometric series which you need to find a closed form for $F(z)$ $$ \sum_{n=0}^{\infty} t^n = \frac{1}{1-t}. $$ I think you can finish it now.
H: Question about the Monte Carlo Algortihm I was reading the Monte Carlo algorithm for finding the area under a curve, say $y=f(x)$. The algorithm considers, $0\le f(x)\le M$ over the closed interval $a\le x\le b$. My question is,that why is it necessary for $f(x)\ge 0$ for the algorithm to work why can't it simply be $\|f(x)\|\le M$ ? AI: Monte Carlo integration works by randomly choosing a large number of points $(x,y)$ within a rectangle of height $M$ whose bottom side is the segment [a,b], and computing what fraction of those points lie in the region "under the curve", i.e. having $y < f(x)$. If you tried this with $\int_{-\pi}^{\pi} \sin x \ dx$ (which equals $0$), you wouldn't get any "hits" for $x < 0$, and you'd get an answer of $2$. Of course you can adjust the algorithm to take negative areas into account, but maybe the book was just trying to show the idea and didn't want to fool with the details.
H: Constructing a bijection between intervals So I am trying to solve questions below Let $A = \{(\alpha_1,\alpha_2,\alpha_3,\ldots): \alpha_i \in \{0,1\}, i \in N\}$, i.e., $A$ is the infinite cartesian product of the set $\{0,1\}$. Show that $A$ is uncountable. Prove that the intervals $[0,\infty)$ and $(-1,4)$ have the same cardinality by constructing a bijection between the two sets. For the second question I tried to construct a bijection between $[0,\infty)$ and $[-1,4)$, but couldn't go any farther. Thank you so much for any help! AI: For the first, this is a standard example for cantors diagonal counting, the second one would be much easier with Cantor Schröder Bernstein. If you want a bijection we construct at first a bijection from $[0,\infty)\to (0,\infty)$, use the identity on every non integer and for the integers add $1$. After this we can use the function $\frac{1}{x+1}$ to map $(0,\infty)$ on the set $(0,1)$. Finding a bijection between to open sets is now quite easy so you should be able to handle the rest.
H: Closed Compact Subset of Product Space Must Have Empty Interior Suppose that $\{X_{\alpha}\}_{\alpha\in A}$ is a nonempty family of topological spaces. Suppose also that there is an infinite index set $B\subseteq A$ such that $X_{\alpha}$ is not compact for any $\alpha\in B$. Furthermore, assume that $K\subseteq\prod_{\alpha\in A}X_{\alpha}$; $K$ is compact (in the product topology); $K$ is closed (in the product topology). $\mathbf{Claim:}\quad$ $K$ has no interior. How to show this? Should I assume that $K$ has nonempty interior and try to find an open cover of it that has no finite subcover (using the fact that infinitely many of the $\{X_{\alpha}\}_{\alpha\in A}$ are not compact) to derive a contradiction? Or a net in $K$ that does not have a cluster point? Or use the finite intersection property? I don't need a complete proof, just a hint to know where it is right place to begin. Thank you in advance for your help. AI: Hint: Suppose $\mathbf{x} = \langle x_\alpha \rangle_{\alpha \in A} \in \mathrm{Int} ( K )$ and take a basic open neighbourhood $U = \prod_{\alpha \in A} U_\alpha$ of $\mathbf{x}$ such that $U \subseteq K$. Then $U_\alpha = X_\alpha$ for some $\alpha$ where $X_\alpha$ is not compact. From this construct an open cover of $K$ with no finite subcover.
H: $a_1=1,a_{n+1}=\frac{n}{a_n}+\frac{a_n}{n}$. Prove that for $n\ge4$, $\lfloor{a_n^2}\rfloor=n$ Define a sequence $\left\lbrace a_{n}\right\rbrace$ by $\displaystyle{a_{1} = 1\,,\ a_{n + 1} = {n \over a_n} + {a_n \over n}.\quad}$ Prove that for $n \geq 4,\,\,\left\lfloor a_{n}^{2}\right\rfloor=n$ The substitution $b_{n} = a_{n}^{2}$ might be helpful, but I still haven't proved the assertion yet. AI: First,we use Mathematical induction have following inequality $$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$ we easy prove this function $$f(x)=\dfrac{x}{n}+\dfrac{n}{x} $$ is decreasing on $(0,n)$ becasue $$f'(x)=-\dfrac{n}{x^2}+\dfrac{1}{n}\le 0$$ since $a_{1}=1,a_{2}=2,a_{3}=2$,so $$\sqrt{3}\le a_{3}\le\dfrac{3}{\sqrt{2}}$$ Assmue that $$\sqrt{n}\le a_{n}\le\dfrac{n}{\sqrt{n-1}},n\ge 3$$ then $$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}}=\dfrac{n}{\sqrt{n-1}}>\sqrt{n+1}$$ $$a_{n+1}=f(a_{n})\le f(\sqrt{n})=\dfrac{n+1}{\sqrt{n}}$$ In fact,we can prove $$a_{n}\le \sqrt{n+1},n\ge 4$$ since $$a_{n+1}=f(a_{n})\ge f(\dfrac{n}{\sqrt{n-1}})=\dfrac{n}{\sqrt{n-1}},n\ge 3$$ so $$a_{n}\ge\dfrac{n-1}{\sqrt{n-2}},n\ge4$$ and note $$a_{n+1}=f(a_{n})\le f\left(\dfrac{n-1}{\sqrt{n-2}}\right)=\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}$$ it suffces prove that $$\dfrac{(n-1)^2+n^2(n-2)}{(n-1)n\sqrt{n-2}}\le\sqrt{n+2},n\ge 4$$ $$\Longleftrightarrow n^3-n^2-2n+1<(n^2-n)\sqrt{n^2-4}$$ $$\Longleftrightarrow 2n^3-6n^2+4n-1=2n^2(n-3)+4n-1>0$$
H: Are monics and epics in the category of finite dimensional vector spaces actually injective and surjective linear transformations? Consider the category of finite dimensional vector spaces with morphisms being linear transformations. Is it still true that monics and epics are actually injective and surjective linear maps, respectively? The converse is surely true since the category is concrete. I know this monics and epics are precisely the injective and surjective maps in the category of Sets and in the category of groups, but it is not necessarily true in the category of topological spaces, so I'm just curious if it is true or not in the category of f.d. vector spaces, and if so why? AI: If $f:V\to W$ is not injective, then there are two different maps $\ker(f) \to V$ yielding the same composition with $f$, namely the zero map and the inclusion map, so $f$ is not monic. If $f:V\to W$ is not surjective, then there are two different maps $W \to W$ yielding the same composition with $f$, namely the identity map and any projection onto $\operatorname{ran}(f)$, so $f$ is not epic.
H: If $G$ is a group, show that $x^2ax=a^{-1}$ has a solution if and only if $a$ is a cube in $G$ I was checking my old set of homework problems that I found this one: If $G$ is a group, show that $x^2ax=a^{-1}$ has a solution if and only if $a$ is a cube in $G$. One direction is easy. If $a$ is a cube in $G$ then there exists $y \in G$ such that $a=y^3$. Now with direct manipulations it is shown that $x=y^{-2}$ is a solution to $x^2ax=a^{-1}$. The other direction is harder though. I first attempted to isolate $x$ and then somehow show that it's necessary for $a$ to be a cube. But since I had no idea how to isolate $x$ I gave up. Then I decided to prove the existence of $y \in G: a=y^3$ by showing that a particular map is bijective so I can come up with a nice variable change to conclude that $a$ must be a cube, but this attempt failed as well. I'm tired now, so I think I'm giving up. I'm looking for a hint or a complete solution if a hint is not possible at this stage. AI: Starting with $x^{2}ax=a^{-1}$, try to manipulate the left-hand side to $xaxaxa= (xa)^3$ by judicious multiplications of the equation by $x, x^{-1}, a$, and $a^{-1}$. You will be pleasantly surprised by what ends up on the right.
H: Four lines are drawn on a plane with no two parallel.... I am stuck on the following problem: Four lines are drawn on a plane with no two parallel and no three concurrent.Lines are drawn joining the points of intersection of the previous four lines. Number of new lines obtained this way is : $\,\,3,\,\, 5, \,\, 2,\,\,12.$ My Attempt: I think the answer will be $2$ but the answer key says it will be $3$. Can someone explain where I went wrong? AI: Using the hints given by @AJStas, I can say that the number of new lines obtained this way will be $3.$
H: Prove the following $(a \cap(\neg b))\cup (a \cap c)=a\cap (\neg(b\cap(\neg c))) $ I want to prove the following: $$(a \cap(\neg b))\cup (a \cap c)=a\cap (\neg(b\cap(\neg c))) $$ What I tried to do so far is to minimize the LHS but I dont know if it enough: $$(a \cap(\neg b))\cup (a \cap c)=a\cap (\neg b \cup c)$$ what should I write? Distributive property and its right? now we I got is: $$a\cap (\neg b \cup c)=a\cap (\neg(b\cap (\neg c)))$$ now just to open the RHS? this is the right way? thanks. AI: We know that for all sets $A,B,C$ we have $(A \cap B) \cup (A \cap C) = A \cap (B \cup C)$. This is the distributive property. So then $(A \cap (\lnot B)) \cup (A \cap C) = A \cap ((\lnot B) \cup C)$. Then we need to show that $((\lnot B) \cup C) = (\lnot(B \cap (\lnot C))$. We know by DeMorgan's Laws that $(\lnot X) \cup (\lnot Y) = \lnot(X \cap Y)$. (This is DeMorgan's Law I.) We know by the Double Negation Law that $\lnot(\lnot X) = X$. Let's take $(\lnot(B \cap (\lnot C))$. By DeMorgan's Law I, $\lnot(B \cap (\lnot C) = (\lnot B) \cup (\lnot(\lnot C))$. By the double negation law, $(\lnot B) \cup (\lnot(\lnot C)) = (\lnot B) \cup C$. Therefore, $A \cap ((\lnot B) \cup C) = A \cap (\lnot(B \cap (\lnot C))$, and we are done.
H: Converges of integral, knowing the derivative has a limit $f: [0,\infty) \to \Bbb R$ we know that f>0 and f ' exists .$ $$\lim_{x \to \infty} (\ln f)'(x)=L<0$. Prove that $\int _0^\infty f$ converges Any hint would be helpful! AI: Hint: if $(\ln f)'(x) \le c < 0$ for all $x \ge x_0$, then $\ln f(x) \le \ln f(x_0) + \ldots $
H: Find the multiplicative inverse of $\,x^2+(x^3-x+2)$ in the quotient $\,F_3[x]/(x^3-x+2)$ Find the multiplicative inverse of $x^2+(x^3-x+2)$ in the quotient $F_3[x]/(x^3-x+2)$ . I've proved that $x^3-x+2$ is irreducible polynomial in $F_3[x]$, and that $x^2$ and $x^3-x+2$ are coprime integers, therefore gcd $(x^2, x^3-x+2)=1,$ and I can find $f,g$ in $F_3[x]$ such that: $1 = fx^2 + g(x^3-x+2)$ when I take modulo $(x^3-x+2)$, I get: $1=fx^2$ (every one of them with bar above to describe the modulo), it means that f is the multiplicative inverse and I need to find it. By Gauss I tried to divide: $x^3-x+2 / x^2 = xx^2 - x + 2$ and because we are in $F_3[x]$, I wrote $= x*x^2 + 2x+2$ , but I don't know how to continue from here, I got the "redundant" $2x+2$ and don't know how to get the multiplicative inverse... AI: You wrote you can find $f,g$ such that ... and that's what you need to do. This is just th e(extended) Euclidean algorithm as known from integers, but requiring polynomial divisions. Thius $$\begin{align} x^3-x+2 &= x\cdot x^2 -(x-2)\\ x^2 &=(x-2)\cdot x +2x\\ x-2 &= 2x\cdot \frac 12-2=2x\cdot 2+1\end{align}$$ (where only the last step is aware of us working in $\mathbb F_3$). From this with $p(x)=x^3-x+2, q(x)=x^2$ we obtain step by step: $$\begin{align}x-2 &= xq-p\\ 2x &= q-x(xq-p)\\ 1&=(xq-p)-2(q-x(xq-p))\end{align} $$ So $$(x+2)\cdot p(x)+(2x^2+x+1)\cdot q(x) = 1$$
H: How can I prove a coordinate ring is not isomorphic to a polynomial ring Let $Z$ be the plane curve $xy=1$. I would like to prove that $A(Z)$ is not isomorphic to a polynomial ring in one variable over $k$. I'm already prove that the coordinate ring is $A(Z)=k[x,y]/(xy-1)$, but I couldn't finish the question. I have a guess that $A(Z)\cong k[x,1/x]$, but I can't prove it formally yet and even I would I don't know what to say with this information. I really need help. Thanks a lot. AI: 1) The isomorphism $f:k[x,y]/(xy-1)\to k\left[x,\frac1x \right]$ is obtained from the morphism $F:k[x,y]\to k\left[x,\frac1x \right]:u(x,y) \mapsto u(x,\frac 1x)$ (which is clearly surjective) by passing to the quotient at the source by $ker(F)=(xy-1)$ (which last equality you have to check, of course). 2) In $k[x,1/x]$ the non-constant element $x\in k[x,1/x]\setminus k$ is invertible, whereas in $k[x]$ only the elements in $k^*$, the nonzero constants, are invertible. Hence the $k$-algebras $k[x,1/x]$and $k[x]$ are not isomorphic.
H: Prove a certain limit involving integrals and series Question: Let f>0 be an descending function. $f:[0,\infty)\to \Bbb R$. Prove that for all a>0 that: $$\lim_{a^+\to 0} a\sum_{n=1}^\infty f(na)= \int _0^\infty f(x)dx$$ What we know We figured using the series comparison theorem that $\sum_{n=1}^\infty f(na)$ converges and also that it's bounded from below by this integral. Maybe we somehow need to prove that the integral is the infimum of the sequence on the LHS? AI: Let $a=\Delta x$. The your sum becomes $$\lim_{\Delta x\to 0^+} \sum_{n=1}^\infty f(n\Delta x)\Delta x$$ Isn't this basically the definition of the Riemann integral for infinite domains?
H: Proving that if these quadratics are equal for some $\alpha$, then their coefficients are equal Let $$P_1(x) = ax^2 -bx - c \tag{1}$$ $$P_2(x) = bx^2 -cx -a \tag{2}$$ $$P_3(x) = cx^2 -ax - b \tag{3}$$ Suppose there exists a real $\alpha$ such that $$P_1(\alpha) = P_2(\alpha) = P_3(\alpha)$$ Prove $$a=b=c$$ Equating $P_1(\alpha)$ to $P_2(\alpha)$ $$\implies a\alpha^2 - b\alpha - c = b\alpha^2 - c\alpha - a$$ $$\implies (a-b)\alpha^2 + (c-b)\alpha + (a-c) = 0$$ Let $$Q_1(x) = (a-b)x^2 + (c-b)x + (a-c)$$ This implies, $\alpha$ is a root of $Q_1(x)$. Similarly, equating $P_2(\alpha)$ to $P_3(\alpha)$ and $P_3(\alpha)$ to $P_1(\alpha)$, and rearranging we obtain quadratics $Q_2(x)$ and $Q_3(x)$ with common root $\alpha$: $$Q_2(x) = (b-c)x^2 + (a-c)x + (b-a)$$ $$Q_3(x) = (c-a)x^2 + (b-a)x + (c-b)$$ $$Q_1(\alpha) = Q_2(\alpha) = Q_3(\alpha) = 0$$ We have to prove that this is not possible for non-constant quadratics $Q_1(x), Q_2(x), Q_3(x)$. EDIT: I also noticed that for distinct $a, b, c \in \{1, 2, 3\}$: $$Q_a(x) + Q_b(x) = -Q_c(x)$$ AI: Denote $$Q_1(x)=P_1(x)-P_2(x)=(a-b)x^2-(b-c)x-(c-a);$$ $$Q_2(x)=P_2(x)-P_3(x)=(b-c)x^2-(c-a)x-(a-b);$$ $$Q_3(x)=P_3(x)-P_1(x)=(c-a)x^2-(a-b)x-(b-c).$$ Then $\alpha$ is a real root of the equations $Q_i(x);$ so that $\Delta_{Q_i(x)}\geq 0 \ \ \forall i=1,2,3;$ where $\Delta_{f(x)}$ denoted the discriminant of a quadratic function $f$ in $x.$ Now, using this we have, $$(b-c)^2+4(a-b)(c-a)\geq 0;$$ $$(c-a)^2+4(a-b)(b-c)\geq 0;$$ $$(a-b)^2+4(b-c)(c-a)\geq 0.$$ Summing these up and using the identity $$\begin{aligned}(a-b)^2+(b-c)^2+(c-a)^2+2\left(\sum_\text{cyc}(a-b)(c-a)\right)\\=[(a-b)+(b-c)+(c-a)]^2=0;\end{aligned}$$ We obtain, $$\begin{aligned}0&\leq 2\left(\sum_\text{cyc}(a-b)(c-a)\right)=2\left(\sum_\text{cyc}(ca+ab-bc-a^2)\right)\\&=-\left[(a-b)^2+(b-c)^2+(c-a)^2\right];\end{aligned}$$ Possible if and only if $(a-b)^2+(b-c)^2+(c-a)^2=0\iff a=b=c.$ We are done
H: Sequence in $l^2$ that does not converge (but it should?) Let $l^2$ be the space of all real sequences $x = (x_1,x_2, x_3,\;...)$ for which $\sum_{n=1}^\infty \|x_n\|^2$ converges. It can be easily verified that map $$\langle x,y\rangle = \sum_{n=1}^\infty x_ny_n$$ defines an inner product on $l^2$ and that $x\to \|\|x\|\|=\sqrt{\langle x,x\rangle}$ defines norm on $l^2$, therefore one can speak about Cauchy's sequences. I am aware of the fact that if the sequence $\{x^j\}_{j\in \mathbb{N}}$, $x^j\in l^2$ , converges, it must be Cauchy's. Clearly, the seqence $$x^1 =(1,0,0,0\;...)\\ x^2 = (0,1,0,0,\;...)\\x^3 = (0,0,1,0, \; ...)\\ \vdots$$ is not Cauchy's, since $\|\|x^j-x^k\|\|=\sqrt{2}$ for all $j\neq k$. As consequence, it cannot be convergent. On the other hand, we can notice that $$\forall n \in \mathbb{N} \forall j\in \mathbb{N}. n>j \Rightarrow x^j_n = 0\text{,}$$ so - intuitively - kind of limit exists and equals to $x = (0,0,0,\;...)$. My question is: Is there any natural topology (or metric) for which the upper sequence really converges? AI: Yes, this converges to the zero sequence in the weak topology. Take a look here.
H: Find a solution of $x\frac{dy}{dx} = y^2 -y$ that passes through the points (1/2, 1/2) I do not understand how my instructor simplified the part marked with red circle. Did he make a mistake? Could anyone help me out here. AI: Nope, he is right $\frac{1}{y-1}-\frac{1}{y}=\frac{y-(y-1)}{y(y-1)}=\frac{1}{y(y-1)}$
H: Why is $\zeta'/\zeta(s) = -\sum_p \log p/(p^{s}-1)$? I'm going through the prime number theorem and there's a lemma which uses the fact that $\frac{\zeta'}{\zeta}(s) = -\sum_p \frac{\log p}{p^{s}-1}$. Can somebody explain this? Am I missing something obvious? Here's my work, $\log \zeta (s) = -\sum_p \log(1 - p^{-s})$, so $\frac{\zeta'}{\zeta}(s) = -\sum_p \frac{sp^{-s-1}}{p^{-s}-1}$. How did the logarithm on the top come in? AI: The logarithm in the numerator comes from the chain rule when differentiating with respect to $s$. You get $1-p^{-s}$ in the denominator from the derivate of the logarithm, and then you have to multiply by the derivative of the argument of the logarithm, i.e. the derivative of $1-p^{-s}$ with respect to $s$. Writing this as $1-e^{-s (\log p)}$ and using the chain rule once again gives the $\log p$ you mention.
H: I want to show$X : ((0,1],\mathcal B,\lambda)\to \mathbb R$ is random variable Let $F:\mathbb R \to [0,1]$ be a distribution function of a probability measure $P$ $(i.e.,F(x)=P((-\infty,x])) $. Then show that There is a random variable $X : ((0,1],\mathcal B,\lambda)\to \mathbb R$,(where $\mathcal B$ is the borel $\sigma $-algebra and $\lambda$ is Lebesgue measure) ,such that $P_{X}=P$ AI: This is a standard result that you should be able to find in any book on measure-theoretic probability. Look in chapter 1, section 2 here. It is Theorem 1.2.2.
H: Rudin Example 3.35B Why the $n$th root of $a_n$ is $1/2$? For Baby Rudin Example 3.35(b), I understand how the $\liminf$ and $\limsup$ of the ratio test were found, but I am not clear why $\ \lim \sqrt[n]{a_n } = \frac{1}{2} $. Please help. AI: The sequence in question is $$\frac{1}{2} + 1 + \frac{1}{8} + \frac{1}{4}+ \frac{1}{32}+ \frac{1}{16}+\frac{1}{128}+\frac{1}{64}+\cdots$$ In case the pattern is not clear, we double the first term, the divide the next by $8$, the double, then divide by $8$, and so on. The general formula for an odd term is $a_{2k-1}=\frac{1}{2^{2k-1}}$. The formula for an even term is $a_{2k}=\frac{1}{2^{2k-2}}$. In the first case, $\sqrt[n]{a_n}=\frac{1}{2}$. In the second, the limit is $\frac{1}{2}$. Since $n$th roots of both the even and odd terms converge to $\frac{1}{2}$, you have your desired result.
H: Convolution of distributions is not associative I need some help with this exercise: It proposes to show that convolution of distributions is not associative: If $T=T_1$ (distribution given by f=1), $S=\delta'$, and $R=T_H$ (we denote as $H$ the Heaviside function, in $\mathbb{R}$), then: $$T\ast(S\ast R)\neq(T\ast S)\ast R$$ I'm not very familiarized with convolutions of distributions, so any help will be very useful. Thanks in advance. AI: We have a couple of useful properties of convolution. First, $\delta\ast Z=Z$ for any distribution $Z$. Second, $(Z_1\ast Z_2)' = Z_1'\ast Z_2=Z_1\ast Z_2'$ for any distributions $Z_i$ for which the convolution is defined. Therefore, $S\ast R = \delta'\ast T_H = (\delta\ast T_H)'= T_H'= \delta$, then $T\ast \delta = T$. On the other hand, $T\ast S = T_1\ast \delta'= (T_1\ast \delta)' = T_1 ' = 0$, and obviously $0\ast R = 0$.
H: If $(a+b)(b+c)(c+a)=2$, then $(a^2+bc)(b^2+ca)(c^2+ab)\leq 1$ Prove that if $a,b$ and $c$ are non negative real numbers such that $(a+b)(b+c)(c+a)=2$, then we have $$(a^2+bc)(b^2+ca)(c^2+ab)\leq 1$$ AI: You can prove the inequality $$4(a^2+bc)(b^2+ca)(c^2+ab)\le (a+b)^2(b+c)^2(c+a)^2.$$Without loss of generality, assume that $a\ge b\ge c$. Since $$a^2+bc\le (a+c)^2$$ and $$4(b^2+ca)(c^2+ab)\le (b^2+ca+c^2+ab)^2,$$ it suffices to show that $$b^2+c^2+ab+ac\le (a+b)(b+c).$$ This inequality is equivalent to $c(a-b)\le 0$, which is clearly true. Solution 2: this inequality is equivalent to $$(a-b)^2(b-c)^2(c-a)^2+4bc\sum bc(b+c)+8a^2b^2c^2\ge 0.$$
H: properties of measures with density Let $\nu=\mu f$ be a $\sigma$-finite measure on $(\Omega,\mathcal A)$. Then $f$ is $\mu$-a.e. unique and $\mu$-a.e. real. Also: If $f(\omega)>0$ for all $\omega\in\Omega$, $\mu$ is also $\sigma$-finite. I was able to show the uniqueness, but I don't quite know how to show that $f$ is real almost everywhere. I think this has to do with the $\sigma$-finiteness of $\nu$, but I don't quite know how to work this. Also how do I show the $\sigma$-finiteness of $\mu$ if $f(\omega)>0$ for all $\omega\in\Omega$? AI: Write $\Omega=\bigcup_nE_n$, where $E_n\in\mathcal A$ and $\nu(E_n)$ is finite, and define $F_k:=\{f\geqslant k\}$. We have $$\int_{E_n}f\mathrm d\mu\geqslant k\mu(E_n\cap F_k).$$ Since the LHS is finite and $E_n$ has finite measure, we have for each $n$ that $\mu(E_n\cap F_k)\to 0$ as $k\to\infty$, hence $\mu(E_n\cap\{f= \infty\})=0$ for each $n$. For the last part, consider $S_n:=\{(n+1)^{-1}\leqslant f\lt n^{-1}\}$.
H: Is this matrix block matrix Hurwitz for some $c_1, c_2, c_3 > 0$? Given the next well partitioned real squared matrix $$ M = \begin{bmatrix}A & \frac{c_3}{c_1}BC^T \\ c_3c_2C & E- c_3^2\frac{c_2}{c_1}CC^T\end{bmatrix}, $$ where $A$ is Hurwitz (all the eigenvalues have negative real part), $CC^T$ positive definite, $E$ has its eigenvalues on the imaginary axis and is invertible (it represents an oscillator, in fact it is a block diagonal matrix with blocks similar to $\begin{bmatrix}0 & -\omega \\ \omega & 0 \end{bmatrix}$). Finally $c_1, c_2$ and $c_3$ are real positive constants. I know that for whatever selection of $c_1, c_2$ and $c_3$ lower right block matrix is always Hurwitz. If one of the non-diagonal blocks is equal to zero, then the eigenvalues of $M$ are the eigenvalues of the diagonal block matrices. My question here is, if I do $\frac{c_3}{c_1}$ sufficiently small (or $c_3c_2$) then one of the non-diagonal blocks will approach to zero, but on the other hand I am also varying the eigenvalues of the lower right block (which is always Hurwitz). Can I say that I can make $M$ Hurwitz? AI: We claim that $M$ is weakly stable (i.e. all its eigenvalues have non-positive real parts) when $E$ is invertible and $k=c_3^2c_2/c_1(>0)$ is sufficiently small, and it is stable (such that all its eigenvalues have negative real parts) for any sufficiently small positive $k$ if $C\left[I+(A-i\omega I)^{-1}B\right]C^Ty\ne0$ for every eigenpair $(i\omega, y)$ of $E$. Note that a matrix $M$ is (strictly) stable if and only if for any nonzero complex vector $x$, there exists a positive definite matrix $G$ (that may depend on $x$) such that $\operatorname{Re}(x^\ast GMx)<0$ (see Horn and Johnson, Topics in Matrix Analysis, p.107, theorem 2.4.11). Now, suppose $E$ is invertible. Let $\epsilon>0$. Perturb $E$ by $-\epsilon I$ in $M$ and call the resulting matrix $M_\epsilon$. Note that $(A-wI)^{-1}$ exists and is bounded for every complex number $w$ with nonnegative real part, because $A$ is stable. Also, the eigenvalues of $E-\epsilon I-zI$ are bounded away from zero whenever $\Re(z)\le0$. Therefore, when $k$ is sufficiently small, the matrix $$ E-\epsilon I-zI - kC\left[I+(A-wI)^{-1}B\right]C^T\tag{1} $$ is invertible for all complex numbers $z$ and $w$ with nonnegative real parts. So, for such $z$ and $w$, the system of equations \begin{align} y_1:= Ax_1 + \frac{c_3}{c_1}BC^Tx_2 &= wx_1,\tag{2}\\ y_2:=c_3c_2Cx_1 + (E-kCC^T)x_2 &= zx_2\tag{3} \end{align} has only the trivial solution $x^\ast=(x_1^\ast,x_2^\ast)=0$. In other words, when $x\ne0$ and $p$ or $q$ is small, $y_1\ne wx_1$ or $y_2\ne zx_2$. Therefore, there exist positive definite matrices $G_1,G_2$ such that $\operatorname{Re}(x_1^\ast G_1y_1+x_2^\ast G_2y_2)<0$, i.e. $\operatorname{Re}(x^\ast GM_\epsilon x)<0$ for $G=G_1\oplus G_2$. Hence $M_\epsilon$ is (strictly) stable and by continuity, $M$ is weakly stable. So, the remaining question is whether $M$ can possess a purely imaginary eigenvalue $\lambda=it$ with $t\in\mathbb R$. This occurs when $$ E-itI - kC\left[I+(A-itI)^{-1}B\right]C^T\tag{4} $$ is singular for some real $t$. Hence our assertion. Note that $(4)$ could really hold. For instance, when $A,B,C,E$ are $2\times2$, $C=I_2$ and $B=E-A$, if $y$ is an eigenvector of $E$ corresponding to the eigenvalue $i\omega$, then $\pmatrix{py\\ y}$ is an eigenvector of $M$ corresponding to the same eigenvalue.
H: Describe four different elements of a union of power sets of power sets $P(P(A))\bigcup P(P(P(A)))$ The empty is set one but other than that I'm not really sure what does a power set of a power set means. Any help would be appreciated. Edit: Is this how you read this for example: $P(P(\emptyset))= \{\emptyset\{\emptyset\}\}? $ AI: Let $X=P(P(A))\cup P(P(P(A)))$. Since $\emptyset\subseteq P(A)$, $\emptyset \in P(P(A))$ and thus $\emptyset\in X$. Since $\emptyset\subseteq A$, $\emptyset \in P(A)\implies\{\emptyset\}\subseteq P(A)\implies \{\emptyset\}\in P(P(A))\implies \{\emptyset\}\in X$. Again, $\emptyset\subseteq A\implies\emptyset\in P(A)\implies \{\emptyset\}\subseteq P(A)\implies\{\emptyset\}\in P(P(A))\implies \{\{\emptyset\}\}\subseteq P(P(A))\implies \{\{\emptyset\}\}\in P(P(P(A)))\implies \{\{\emptyset\}\}\in X.$ Finally, $\emptyset\subseteq A\implies\emptyset\in P(A)\implies \{\emptyset\}\subseteq P(A)\implies\{\emptyset\}\in P(P(A))$. But also, $\emptyset\subseteq P(A)$ and so $\emptyset\in P(P(A))$. Thus $\{\{\emptyset\}, \emptyset\}\subseteq P(P(A))\implies\{\{\emptyset\}, \emptyset\}\in P(P(P(A)))\implies \{\{\emptyset\}, \emptyset\}\in X$.
H: Caley graphs of gruops and symmetric generating sets There are several examples (of which Wikipedia show at least one) when the Caley graph (G,U) of a group G (where U generates the group) depend on the choice of generating set. Is requiring that the generating set U is symmetric enough to guarantee that the Caley graph only depend on the group (meaning that all Caley graphs with symmetric U are equal in some sense), or what other requirements on U would be enough to get uniqueness of the Caley graph? AI: Different generating sets $S,S'\subset G$ can have isomorphic Cayley graphs $\rm{Cay}(G,S)\cong \rm{Cay}(G,S')$ (this is the correct notion of equality of graphs). On the other hand, a group can have many different generating sets, and it should be obvious to you that if you take two generating sets $ S, S'\subset G $ with $\|S\|\neq \|S'\|$ then $\rm {Cay}(G, S)\not\cong\rm {Cay}(G, S') $. Your question is related to the so-called CI problem. See this short presentation for a survey and references.
H: Prove that if $\{K_n\}_{n\in\mathbb{N}}$ is a sequence of uncountable compact sets of $X$, then $\bigcap\limits_{n\in\mathbb{N}}K_n$ is uncountable The question I am interested in is the following: Let $\omega_1$ be the first ordinal with an uncountable number of predecessors. We consider $X=[0,\omega_1]$ supplied with order topology. Prove that if $\{K_n\}_{n\in\mathbb{N}}$ is a sequence of uncountable compact sets of $X$, then $\bigcap\limits_{n\in\mathbb{N}}K_n$ is uncountable. This suggests to me that an uncountable compact $K$ contains necessarily a subset of $X$ of the form $[\alpha,\omega_1]$: this result would indeed imply that the intersection of the sets $K_n$ contains such an uncountable set since $\omega_1$ cannot be the limit of a sequence of points in $[0,\omega_1[$. I can see that any $K_n$ contains $\omega_1$, otherwise $\cup_{\alpha<\omega_1}[0,\alpha]$ is an open cover of $K_n$ from which we cannot extract a finite subcover. But I can't figure out a way to prove that $K_n$ contains a set of the form $[\alpha,\omega_1]$. Is it true? AI: The set of limit ordinals in $[0,\omega_1]$ is an uncountable compact set and it does not contain any subsets of the form $[\alpha,\omega_1]$, so your argument doesn't work. For simplicity suppose that we have only two uncountable compact sets (the argument for the general case is a straightforward generalization), $K_0, K_1$. It suffices to show that for any $\xi<\omega_1$ there is a $\gamma \in K_0\cap K_1$ so that $\gamma>\xi$. Fix $\xi<\omega_1$, since $K_0$ is uncountable there is an $\alpha_0\in K_0$ with $\alpha_0>\xi$, by the same argument applied to $K_1$ there is a $\beta_0>\alpha_0\ \beta_0\in K_1$. Using the same argument we can construct recursively sequences $\{\alpha_n\}_{n\in\mathbb N}\subseteq K_0, $ and $\{\beta_n\}_{n\in\mathbb N}\subseteq K_1, $ such that $$\xi<\alpha_0<\beta_0<...<\alpha_n<\beta_n...$$ Observe that $\sup \alpha_n=\sup \beta_n$, take $\gamma=\sup \alpha_n$, it follows that $\gamma\in K_0\cap K_1$ and $\gamma>\xi$, which finishes the proof.
H: Complex Analysis and Probability Theory My question is a general one. I know that in complex analysis we find some very powerful theorems but given that my main area of study is Statistics and Probability, does complex analysis have applications in those areas? I know that it is helpful for example in the characteristic functions but do you think a deeper level of understanding can make a big difference? Will it make me a better Statistician? Thank you. AI: If you are interested in resolution of singularities of varieties and applications of analytic functions in statistical modelling you could have a look at Watanabe's Algebraic Geometry and Statistical Learning Theory. It is really a nice one.
H: Probability to find $1$ in a random sequence of bits Given the random sequence of $N$ bit formed by $0$ and $1$ obtained flipping a coin and associating for example $1$ to head and $0$ to tail, suppose, after having built the sequence, we randomically pick a number $k$ with a uniform probability and $1\le k\le N$, I have to calculate the probability that the number $k$ correspond to the position of a $1$ in the sequence. For example, if the sequence is $1010010100...$ and we pick $k=3$, in the sequence at position $3$ ther is a $1$. Same thing, if $k=6$ and so on. Thanks for any suggestion or answer. AI: HINT: Either the number $k$ corresponds to a position of a 1 in the sequence or the number $k$ corresponds to a position of a 0 in the sequence. Now how do you think the probability of the number $k$ corresponding to a position of a 1 in the sequence compares to the probability of the number $k$ corresponding to a position of a 0 in the sequence?
H: Non-Isomorph trees of a graph Please consider this graph How many non-Isomorph trees with 4 vertex has this graph? Is there any formula that show number of non-Isomorph trees with $n$ vertices? thanks AI: There are only two 4-node trees up to isomorphism: the straight line and the Y-shaped graph. The above graph contains both of them. For the number of trees with $n$ vertices, Wikipedia links here : http://oeis.org/A000055 .
H: Showing the polynomial $x^4 + x^3 + 4x + 1$ is irreducible in $\mathbb{Q}[x]$. Showing the polynomial $f(x) = x^4 + x^3 + 4x + 1$ is irreducible in $\mathbb{Q}[x]$. I have two question relating to this which I've bolded below. Attempt to answer () $f$ has degree $4 \ge 2$ so if $f$ has a root then $f$ is reducible. Well the rational root test says that if $\exists$ a rational root $\frac{p}{q}$ with $p, q$ coprime then $p\|a_0$ and $q\|a_n$. Hence the potential rational roots are $\pm 1$. Neither of these are roots so $f$ does not have a root. But () only says that if $f$ has a root then $f$ is reducible, it does not imply that if $f$ does not have a root then $f$ is irreducible. Is this correct? Now we can't apply Eisenstein's irreducibility criterion theorem as there is no prime $p$ such that $p\mid a_0, a_1,..., a_{n-1}$ $p \nmid a_n$ $p^2 \nmid a_0$ So where do we go from here? One other fact I am aware of is that if $f$ is primitive, which it is, $f$ irreducible in $Q[x] \iff f$ is irreducible in $Z[x]$. But I don't see how that can help me here. So how do I proceed from here? AI: it does not imply that if f does not have a root then f is irreducible. Is this correct? Yes, that is correct. For polynomials of degree $> 3$, the absence of roots in $\mathbb{Q}$ is only a necessary, but not a sufficient condition for irreducibility. Aside from trying a substitution $x \mapsto x - a$ to achieve a form where Eisenstein's criterion is applicable, you can consider the corresponding polynomial over $\mathbb{Z}/(p)$ for a prime $p$. If $f$ is reducible, so is its image $\overline{f} \in \left(\mathbb{Z}/(p)\right)[X]$, if it has the same degree as $f$ (since $f$ here is monic, that is the case). So if $\overline{f}$ is irreducible in any $\left(\mathbb{Z}/(p)\right)[X]$, then $f$ itself is irreducible. Note: if $\overline{f}\in \left(\mathbb{Z}/(p)\right)[X]$ is reducible, that does not imply that $f$ is reducible. If we look at the example over $\mathbb{Z}/(2)$, we have $\overline{f}(x) = x^4+x^3+1$. It is readily seen that that has no zero in $\mathbb{Z}/(2)$, so if it were reducible, both factors would have to be quadratic. Since the constant term is $1$, a factorisation would have the form $$x^4+x^3+1 = (x^2+ax+1)(x^2+bx+1).$$ But then the coefficients of $x^3$ and $x^1$ would be equal ($a+b$). So $\overline{f}$ is irreducible.
H: Set of m-vectors In my book (matrix computations by Gill) there's a term used quite a lot which I don't understand/find what is its meaning. The set of all $m$-vectors that are linear combinations s of the columns of the $m \times n$ matrix $A$ is called range space, column space or simply the range of $A$. Does "m-vectors" refer to vectors of dimension $m$? AI: It will be elements of $\mathbb{R}^m$, i.e. vectors of dimension $m$. See for example the definition on page 41 of Elementary Linear Algebra By Stewart Venit, Wayne Bishop, Jason Brown
H: is a number contained within this number series I hope I can ask this question with enough clarity. It is not exactly clear in my head., but here goes I have a number series, it starts at a random number and then increments by 8(always by 8) plus the last number. for example in the series: 36,80,132,192,260,336,420,512,612,720,836,960 We start at 36 (randomly chosen) We then add 8 to 36 to make 44 We then add 44 to 36 to make the next number in the series 80 Then we add 8 to 44 to make 52 and add that to 80 to make the next number in the series 132 So we have two sets of numbers one which is incrementing by 8 the other which is the addition of the incrementing numbers. 36, 44, 52, 60, 68, 76, 84, 92,100,108,116,124 36, 80,132,192,260,336,420,512,612,720,836,960 <== target series Now I want to test whether another number exists in that series for example we can see that 612 is valid for that series but 680 is not. Is the number 6992 a valid number for that series? What about 500345264? Is there a way I can test for validity without having to calculate every number in the series up until the one I want? Is there a way I can count how many iterations I'd need to go through to get to the number. For example 612 is the 8th iteration. This is part of a programming task (not school work but special interest) I will have tens of millions of these to calculate and I want the program to run as fast and as efficiently as possible. So any help will be appreciated. If it can't be done then so be it, but not being a mathematician I thought I better ask before I gave up. DC AI: As mentioned by Antonio, the recurrence relation that defines this sequence is: $$ a_n = a_{n-1} + a_0 + nb \,, $$ where in the example above $a_0 = 36$ and $b=8$. A little thought will show that, generally, $$ a_n = (n+1)a_0 + b\sum_{k=1}^n k = (n+1)a_0 + \frac{n(n+1)}{2}b \,. $$ This can be proved simply by induction but was deduced by considering the following: $$ a_0 = a_0 \\ a_1 = a_0 + a_0 + b = 2a_0 + b \\ a_2 = a_1 + a_0 + 2b = (2a_0 + b) + a_0 + 2b = 3a_0 + (1+2)b \\ a_3 = a_2 + a_0 + 3b = (3a_0 + (1+2)b) + a_0 + 3b = 4a_0 + (1+2+3)b \\ \ldots \\ a_n = (n+1)a_0 + b\sum_{k=1}^n k $$ Now that we have a formula to compute the $n$th number in the sequence without computing the previous numbers we can formulate a way to test any number to see if it is in the sequence by solving this for $n$, which results in the following quadratic equation: $$ \frac{b}{2}n^2 + \left(\frac{b}{2} + a_0\right)n + (a_0 - a_n) = 0\,, $$ where $a_n$ is the number you wish to test. If you solve for $n$ using the quadratic formula and one of the solutions is a non-negative integer then $a_n$ is in the sequence. Be careful implementing this to see if this is an integer because floating point arithmetic can be imprecise.
H: Prove that the cardinality of the reals and all binary funcions is not equal Let $S$ be the the set of all real functions that bring back only two values: 0 and 1 (Binary functions). If $f\in S$ then $f:\mathbb{R}\rightarrow \left\{0,1\right\}$. Prove that $\|\mathbb{R}\| \neq \|S\| $. I tried to start with a proof by contradiction that there's no one to to correspondence but I got stuck. I also assume that by the end of the proof we show that $\|S\|=\aleph_0 \ne C=\|\mathbb{R}\|$. Thanks in advance. AI: If you’ve not seen Cantor’s theorem before, this is a fairly hard problem. Suppose that $\varphi:\Bbb R\to S$ is an injection (one-to-one function). I claim that $\varphi$ cannot be a surjection (onto function). If true, this means that there is no bijection from $\Bbb R$ to $S$ and hence that $\|\Bbb R\|\ne\|S\|$. For each $r\in\Bbb R$ let $f_r=\varphi(r)$; $f_r$ is a function from $\Bbb R$ to $\{0,1\}$. To show that $\varphi$ is not surjective, we’ll find a function $g:\Bbb R\to\{0,1\}$ that is different from $f_r$ for every $r\in\Bbb R$. Or rather, I’ll tell you how to construct it and let you finish the details, including the verification that $g\ne f_r$ for every $r\in\Bbb R$. This will show that $g$ is not in the range of $\varphi$ and hence that $\varphi$ is not a surjection. The idea is simple but powerful: for each $r\in\Bbb R$ choose $g(r)\in\{0,1\}$ so that $g(r)\ne f_r(r)$. That doesn’t give you much choice, since $\{0,1\}$ has only two elements; in fact, it completely defines the function $g$.
H: Probability with bullets and walls There are two shooters with different guns and bullets. Each shooter shoots a bullet to a different target hanging on a wall. The hit of each bullet follows a normal distribution centered on its target. Each bullet removes a piece of the wall whose volume follows a normal distribution centered on ten times the volume of the bullet. What is the probabilty that a volume $V_1$ was extracted in a point $X_1$ on the wall by the shooters (i.e., $P(V_1,X_1)$)? What happens if we study two different points (i.e., $P(V_1, X_1 \ and \ V_2, X_2)$)? AI: Since your distributions are continuous densities, gaussian in this case, the probability of any point is 0. That is the answer to your questions as stated. To get probabilities, you need to specify a region on $P(V_1,X_1)$ whose area is greater than zero. For example, lets say you want to know $P(V_1\pm\Delta V,X_1\pm\Delta X)$ Where $\Delta X, \Delta V >0$. What you want is the probability that either the bullet from shooter 1 or shooter 2 or both hit a location within $X\pm\Delta X$ and remove $V\pm\Delta V$ volume. Let X be the modpoint of the interval where the wall will be hit by a bullet, and V be the midpoint of the volume interval for the piece removed. $B_i$ be the X-coordinate of bullet $i = 1,2$ $\mu_i$ be the X-coordinate of target $i$, $\sigma_{bi}$ be the standard deviation of the location of $B_i$,and $\sigma_{vi}$ the standard deviation of the volume removed. Then you need to get the probability for the folowing three situations: (1) $P(\|B_1-X\|\leq\Delta X\cap\|B_2-X\|>\Delta X\cap \|V_1-V\|\leq \Delta V)$ (2) $P(\|B_2-X\|\leq\Delta X\cap\|B_1-X\|>\Delta X\cap \|V_2-V\|\leq \Delta V)$ (3) $P(\|B_1-X\|\leq\Delta X\cap\|B_2-X\|\leq\Delta X\cap \|V_1+V_2-V\|\leq \Delta V)$ You can now calculate each using the normal CDF function.
H: Function from A to A Let $S = \{1,2,3,4,5\}$ how many functions from $S \to S$ and how many of these are $bijective$ Say we also had $A = \{1,2,3,4,5\}$ then the number of functions would be $B^A$ and the number of $bijective$ functions would be 5? But how you count the function from the same set? AI: The number of arbitrary functions is, as you said, $5^5 = 3125$. For bijective $\Leftrightarrow$ injective OR surjective in this case (why?) you have $5!$ since $f(i)$ may be any value in $\{1,2,3,4,5\}$ except for those "occupied" by $f(j), j < i$ so the total amount is $$5\cdot4\cdot3\cdot2\cdot1=5!=120$$ More generally: Given finite sets $A,B$ the number of functions $f: A\to B$ is $$\|B\|^{\|A\|}$$ and the number of bijective functions is $$\frac{\|B\|!}{(\|B\|-\|A\|)!} = \prod_{k=0}^{\|A\|-1} (\|B\|-k)$$
H: Inverse of Short-time Fourier transform The Gabor transform (i.e. Short-time Fourier transform with some Gaussian window) can be defined by (see http://en.wikipedia.org/wiki/Gabor_transform) : $G_x(t_0,\xi) = \int_{-\infty}^\infty e^{-\pi(t-t_0)^2}e^{-2i\pi \xi t}x(t)\,dt$ On this page it is mentioned that the Gabor transform is invertible, and that the original signal can be recovered by the following equation : $ x(t_0) = \int_{-\infty}^\infty G_x(t_0,\xi) e^{2i\pi t_0 \xi}\,d\xi$ Can somebody help me for a proof of that ? I thought we would need $G_x(u,\xi)$ for all $\xi$ and $u$ in a wide range in order to reconstruct $x(t_0)$, but here it seems that only $G_x(t_0,\xi)$ for all $\xi$ is needed. Is this true ? AI: Ok according to a book: the Windowed Fourier transform is obtained by the formula: $$Sf(u,\epsilon)=\int_\mathbb{R}f(t)g(t-u)e^{-i\epsilon t}dt$$ where $f$ and $g$ are the signal and window respectively: and the reconstruction formula is given by: $$f(t)=\frac{1}{2\pi}\int_\mathbb{R}\int_\mathbb{R}Sf(u,\epsilon)g(t-u)e^{i\epsilon t}d\epsilon du$$. So I reckon you will need both $u$ and $\epsilon$, to reconstruct $f$.
H: Divisors of $x^3 + x + 1$ in $Z_3[x]$ Divisors of $f(x) = x^3 + x + 1$ in $Z_3[x]$ Do I have to manually check whether each polynomial in $Z_3[x]$ with degree less than $3$ divides $f$ or is there a better way? That's $3^3 = 27$ polynomials to check (although a few are obviously trivial). AI: Hint: What is $f(1)$? That should get you a partial factorization, and make checking the rest quite a bit easier. Bear in mind that $\Bbb Z_3$ is a field, so every non-zero constant is a divisor.
H: Is this equivalent to continuity? I played around a little bit with the definition of continuity and I think I got the following relations that may be equivalent to continuity. Maybe there is somebody who can check this: $$ f(\overline{M}) \subset \overline{f(M)} \Leftrightarrow \overline{f^{-1}(A)} \subset f^{-1}(\overline{A}) \Leftrightarrow f^{-1}(A^°)\subset (f^{-1}(A))^°.$$ AI: It's not entirely clear how to interpret your $$f(\overline{M}) \subset \overline{f(M)} \Leftrightarrow \overline{f^{-1}(A)} \subset f^{-1}(\overline{A}) \Leftrightarrow f^{-1}(A^°)\subset (f^{-1}(A))^°,$$ I think it is meant to say: Let $X,Y$ be topological spaces, and $f\colon X \to Y$ a map. Then $f$ is continuous if and only if for all $M\subset X$ we have $f(\overline{M}) \subset \overline{f(M)}$. $f$ is continuous if and only if for all $A\subset Y$ we have $\overline{f^{-1}(A)} \subset f^{-1}(\overline{A})$. $f$ is continuous if and only if for all $A\subset Y$ we have $f^{-1}(A^°)\subset (f^{-1}(A))^°$. In that case, these are indeed correct characterisations of continuous maps. By definition, $f$ is continuous if and only if $f^{-1}(U)$ is open for all open $U\subset Y$. By taking complements, we have that $f$ is continuous if and only if $f^{-1}(F)$ is closed for every closed $F\subset Y$. So if $f$ is continuous, we have $M \subset f^{-1}(\overline{f(M)})$, the latter being closed, hence $\overline{M} \subset f^{-1}(\overline{f(M)})$, or equivalently $f(\overline{M}) \subset \overline{f(M)}$ for every $M\subset X$. Conversely, if $f(\overline{M}) \subset \overline{f(M)}$ for all $M\subset X$, let $F\subset Y$ closed, and consider $M=f^{-1}(F)$. Then $f(\overline{f^{-1}(F)}) \subset \overline{f(f^{-1}(F))} \subset \overline{F} = F$, whence $\overline{f^{-1}(F)} \subset f^{-1}(F)$, and that means $f^{-1}(F)$ is closed. That holds for all closed $F\subset Y$, hence $f$ is continuous. The arguments for the other cases are similar. If $f$ is continuous then $f^{-1}(A)$ is a subset of the closed set $f^{-1}(\overline{A})$, so $\overline{f^{-1}(A)} \subset f^{-1}(\overline{A})$ for all $A\subset Y$. Conversely, if $\overline{f^{-1}(A)} \subset f^{-1}(\overline{A})$ for all $A\subset Y$, then for closed $F\subset Y$ we have $\overline{f^{-1}(F)} \subset f^{-1}(\overline{F}) = f^{-1}(F)$, hence $f^{-1}(F)$ is closed. If $f$ is continuous, then $f^{-1}(A^°)$ is an open subset of $f^{-1}(A)$, hence $f^{-1}(A^°)\subset (f^{-1}(A))^°$ for all $A \subset Y$. Conversely, if $f^{-1}(A^°)\subset (f^{-1}(A))^°$ for all $A \subset Y$, then for open $U\subset Y$, we have $f^{-1}(U) = f^{-1}(U^°) \subset (f^{-1}(U))^°$, hence equality, and $f^{-1}(U)$ is open.
H: Explanation of $\mathbb{Z}/_5$ can someone explain me the following please?: $\mathbb{Z} /_5 = \{ [0]_5, [1]_5, [2]_5, [3]_5, [4]_5\}$ Is that correct? $[0]_5 = \{ \dotsc, -15, -10, -5, 0, 5, 10, 15, \dotsc\}$ $[1]_5 = \{ \dotsc, -14, -9, -4, 1, 6, 11, 16, \dotsc\}$ $[2]_5 = \{ \dotsc, -13, -8, -3, 2, 7, 12, 17, \dotsc\}$ and does $/_5$ mean that the set has 5 equivalence classes? AI: Here, $\mathbb Z/_5 = \mathbb Z_5$ is used to denote the group of integers under addition, modulo $5$. It consists of the equivalence classes (also known as residue classes $[r_i]$), which represent the integers $z_i$ which are all congruent to $r_i$ modulo 5, where $r_i$ denotes the remainder in the Euclidean Algorithm $$z_i = 5q + r_i,\;\;0\leq r_i \lt 5$$ You are correct that $\mathbb Z/_5 = \{[0]_5, [1]_5, [2]_5, [3]_5, [4]_5 \}$, which essentially means that the set of integers are partitioned into $5$ equivalence classes, every integer belonging to one, and only one, equivalence class. We can also represent your equivalence classes as follows: $[0]_5 = \{5k\mid k \in \mathbb Z\}$ $[1]_5 = \{5k + 1 \mid k \in \mathbb Z\}$ $[2]_5= \{5k + 2 \mid k\in \mathbb Z\}$ $[3]_5 = \{5k + 3 \mid k \in \mathbb Z\}$ $[4]_5 = \{5k + 4 \mid k\in \mathbb Z\}$ Indeed, for $\mathbb Z_n$, there will always be exactly $n$ such equivalence classes, one for each $r$ such that $0 \leq r \lt n$. So in your case, that means five equivalence classes.
H: Prove that if $f^2=g^2$ and $f(x)$ not zero then $f(x)=g(x)$ or $f(x)=-g(x)$ Suppose that $f$ and $g$ are continuous function on $R$ such that $f^2=g^2$ and $f(x)$ not zero. Show that it's either $f(x)=g(x)$ or $f(x)=-g(x)$ I tried to apply the definition of continuous function to $f$ and $g$, but I don't know how to show $f(x)=g(x)$ or $f(x)=-g(x)$ from $f^2=g^2$. I can't just take square root on both sides. AI: HINT: The only problem is going from plus to minus. OK a bit more info. I fear that you, and, moreover the question, are mixing up $f(x)$ the function with $f(x)$ the real number. To answer it properly we will rephrase the questions slightly differently. Suppose that $f$ and $g$ are continuous functions such that $f^2=g^2$, $f(x)\neq 0$ for any $x\in\mathbb{R}$. Then show that $g=f$ or $g=-f$. First we show that continuity is necessary for the result to hold. Consider $f$ defined by $$f(x):=\left\{\begin{array}{cc}1&\text{ if }x\geq 0\\ -1&\text{ if }x<0\end{array}\right.$$ Now consider $g$ defined by $g(x)=1$. Now $g^2=f^2$ because $g(x)^2=f(x)^2$ for all $x\in\mathbb{R}$ but the function $g$ is neither equal to the function $+f$ nor $-f$; although for every $x$, $g(x)=\pm f(x)$. Now this is where continuity comes in. Suppose that $f$ is positive and negative at say $x_+$ and $x_-$. By the Intermediate Value Theorem, $f$ has a root between $x_+$ and $x_-$. But by assumption $f(x)\neq 0$. Therefore $f$, and similarly $g$, are always positive or always negative. We know that $f^2=g^2$ so there is some point $x_0$ where $f(x_0)=+g(x_0)$ or $f(x_0)=-g(x_0)$. In the first case $f$ and $g$ have the same sign while in the second they differ... Can you finish it off?
H: Comparing models through partial isomorphisms Let $T$ be a theory of a language $\mathcal{L}$ with no function symbols. Let $\mathfrak{A}, \mathfrak{B} \models T$. For all finite sets $X \subseteq A$ and $Y \subseteq B$, there exists a function $F : X' \to Y'$ such that $$X \subseteq X' \subseteq A,$$ $$Y \subseteq Y' \subseteq B$$ and $F$ is an $\mathcal{L}$-isomorphism between the substructure of $\mathfrak{A}$ that is generated by $X'$ and the substructure of $\mathfrak{B}$ that is generated by $Y'$. Is there anything that can be said about models $\mathfrak{A}$ and $\mathfrak{B}$? Is there any set of sentences $\Sigma$ that is absolute between the two models? For example it is easy to see that they satisfy the same existential sentences. AI: As you have noticed, the condition implies that $\mathfrak{A}$ and $\mathfrak{B}$ satisfy the same existential sentences. It follows that they also satisfy the same universal sentences. And, of course, they will satisfy the same boolean combinations of such sentences. This appears to be as far as you can go, however. Note that the structures $( \mathbb{N} , < )$ and $(\mathbb{Z} , < )$ satisfy the given condition. However they differ on the sentence $( \forall x ) ( \exists y ) ( y < x )$, which is a sentence of the next logical complexity. Therefore the condition is not enough to imply that $\mathfrak{A}$ and $\mathfrak{B}$ satisfy the same sentences with even a single pair of alternating quantifiers.
H: Markup reverse to get to original amount? I am trying to figure out how to mark something up then reverse it. I am selling an item for $100$ and I need to cover a $15\%$ advertising expense. I want to net $100$ after advertising is paid for. How do I figure out what percent I should markup the $100$ item to pay for the $15\%$ advertising and still come out with $100$? Thanks for any help! Chris Edwards AI: Hint: You want to charge $x$ so that $85\%$ of $x$ is $100$. Can you write that as an equation?
H: Finding a limit results in division by 0. $$ \lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x^2}. $$ Tried multiplying by $\sqrt{1+x}+1$ but got $1/(x(\sqrt{1+x}+1))$ where substitution results in $1/0$ which is illegal (or maybe $1/0$ is Infinity?). Second approach is to divide by $x^2$ which leads to the $\sqrt{1/x^2 +1)}- 1/x^2$. How to solve it? AI: Your first approach was spot on, though your conclusion is off a bit: Multiplying the numerator and denominator by the conjugate, as you suggest, and canceling a common factor of $x$ from the numerator and denominator gives us, as you found, $$\lim_{x\to 0} \dfrac{1}{x(\sqrt{1+ x} + 1)}$$ which at first glance seems to result in a "form" of $\frac 10$. This is not "illegal" when we're evaluating limits. However, in this case that the limit does not exist, since as $x\to 0^-$, $f(x) \to -\infty$, whereas as $x\to 0^+$, $f(x)\to +\infty$. Conclusion: In this case, since the left-side and right-side limits to not agree, the limit does not exist. Note: It's not always the case that a limit of the form $\frac 10$ does not exist, in the sense that the limits from the left, and from the right, do not agree. For example, $\lim_{x \to 0} \dfrac 1{x^2} \to +\infty$, since as $x$ approaches $0$ from the right and from the left, the denominator is always positive (unlike your posted limit), and the denominator grows increasingly and incredibly small compared to $1$ in the numerator, and as a result, the limit approaches (positive) infinity (the value of the function explodes) as $x \to 0$. The non-existence of a limit, or a limit approaching infinity as $x$ approaches a zero from both directions, is not the same as (or because of) "division by zero". Evaluating a limit as $x \to 0$ is not the same as evaluating a function $f(x)$ at zero. Here, we are interested in what's happening as $x$ approaches zero from the right and from the left, and not what's happening exactly at zero.
H: If $G$ is a finite semi-group and $\forall x,y,z \in G: xy=yz \implies x=z$ then $G$ is an Abelian group If $G$ is a finite semi-group and $\forall x,y,z \in G: xy=yz \implies x=z$ then $G$ is an Abelian group. I have no idea where to start. I'm stuck! I can't prove even the existence of the identity element :\| AI: Pick an element $y$ of your semigroup and consider the successive powers $y, y^2, y^3, \dots$. By finiteness, there is eventually a repetition, i.e. numbers $1 \le m < n$ with $y^m = y^n$. Your cancellation rule lets us cancel off $y$s to eventually get $y = y^k$ for some $k>1$. For any $x$ in $G$, we have $yxy^k = yxy$ since $y^k = y$. Cancel off a $y$ from the right of the left-hand side and from the left of the right-hand side to get $yxy^{k-1} = xy$. Now cancel $y$s from the other side to get $xy^{k-1} = x$. Similarly, $y^{k-1}x = x$. This gives us that there is an identity element, namely $y^{k-1}$. Write $e = y^{k-1}$. Since this is a power of $y$, we automatically have that $y$ has an inverse, namely $y^{k-2}$ (if $k=2$, then $y$ is the identity, so is its own inverse). The identity is unique: if there are elements $e,f$ such that $ex = xe = x$ and $xf = fx = x$ for all $x$, then $e = ef = f$. So we can perform the above process to find an inverse for any element of $G$; this means we have a group. Finally, for any $x,y$ in $G$, we have $xyy^{-1} = y^{-1}yx$, so canceling $y^{-1}$, we get $xy = yx$, so $G$ is abelian.
H: Finding limit of $\frac{\sin\pi3^x}{x}$ as $x\to 0$ I have to find the limit of $\frac{\sin\pi3^x}{x}$ as $x\to 0$ using ONLY notable limits please help me. AI: $\sin x=\sin(\pi-x)\quad,\quad\lim_{x\to0}\frac{\sin x}x=1\quad,\quad\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=f'(a)\iff$ $$\lim_{x\to0}\frac{\sin(\pi\cdot3^x)}x=\lim_{x\to0}\frac{\sin(\pi-\pi\cdot3^x)}x=\lim_{x\to0}\frac{\sin(\pi-\pi\cdot3^x)}{\pi-\pi\cdot3^x}\cdot\frac{\pi-\pi\cdot3^x}{x}=$$ $$=\pi\cdot\lim_{x\to0}1\cdot\frac{1-3^x}x=-\pi\cdot\lim_{x\to0}\frac{3^x-3^0}{x-0}=-\pi\cdot(3^x)'_{x=0}=-\pi\cdot(3^x\cdot\ln3)_{x=0}=-\pi\cdot\ln3.$$
H: Trouble understanding Tensor product in context of Torsion Tensor I know that there are quite a few threads dealing with this question already. I have pored through them for quite some time and they have been informative. However there are still some clarifications that I seek. If this is too much of a duplicate, then I wouldnt mind if it is closed. I will sum up what I have understood.These are the definitions that I have seen: Tensor Product of two vector spaces $V,W$ over a field $\mathbb F$ is another vector space $T$ over $\mathbb F$ equipped with a bilinear mapping $\otimes:V \times W \to T$ ; $\otimes (v,w) \mapsto v\otimes w$ such that for any vector space $U$ over the same field and any bilinear map $f:V \times W \to U$, $ \exists !$ linear map $f^* :T \to U$ where $f^\circ \otimes =f $. An (r,s) tensor is a multilinear function $B_{i_1 \ldots i_r}^{j_1 \ldots j_s}$ on $(V^)^r \times (V)^s$ onto the field $\mathbb F$, denoted as $V \otimes \ldots \otimes V \otimes V^* \otimes \ldots \otimes V^$ . Now my question is the first is a definition of Tensor Product of Vector spaces, whereas the 2nd defines just a "tensor"??Also is the 2nd definition something that restricts itself to merely copies of a single vector space and its dual only?? Further a Torsion Tensor on a Riemannian Manifold M is defined as a vector valued two-form on $\chi(M) \times \chi (M) \to \chi(M) $ where $\chi (M)$ is the set of all $C^{\infty}$ vector fields on M. So in this case what kind of tensor is this??Its said to be (1,2) tensor , but I am not able to see that. The closest I would say is a (0,2) tensor. And does vector-valued mean this is the 1st definition in work?? Also am I right in saying the two definitions are similar only in the case of Finite Dimensional spaces whereas the 1st definition is the more general one?? Please help me differentiate between the two definitions and put the Torsion in proper context. AI: That's a lot of question marks, Vishesh. I hope I can help. In the two definitions above, the first defines the tensor product of vector spaces. The tensor product of the vector spaces $V$ and $W$ is written $V \otimes W$, where the underlying field is understood from the context. The second definition identifies tensors as multilinear maps whose domain is the vector space $$\underbrace{V^{} \times \cdots \times V^{}}_{r} \times \underbrace{V \times \cdots \times V}_{s}$$ and whose codomain is $\mathbb{F}$. The type of the tensor allows us to identify its domain, and in this case we say the tensor is of type $(r,s)$. The connection between these definitions can be seen by considering the simple case $V^{} \otimes V$. An element $T \in V^{} \otimes V$ is a $(1,1)$-tensor in the following way. Write $T = v^{} \otimes v$, and let $(w,w^{}) \in V \times V^{}$. Then $$T(w,w^{}) = v^{}(w)w^{}(v)$$ That is, $T \in V^{} \otimes V$ has a component in $V^{}$ which "eats" elements of $V$, and a component in $V$ which is "eaten" by elements of $V^{}$. Thus it makes sense to pass pairs of elements in $V \times V^{}$ into $T$ and get an element of $\mathbb{F}$, and the function you get is easily seen to be multilinear. This example shows how to identify mutilinear maps on $$\underbrace{V^{} \times \cdots \times V^{}}_{r} \times \underbrace{V \times \cdots \times V}_{s}$$ with elements of the tensor product space $$\underbrace{V \otimes \cdots \otimes V}_{r} \otimes \underbrace{V^{} \otimes \cdots \otimes V^{}}_{s}$$ explaining why we call those multilinear maps "tensors". The general construction of a tensor product can involve more vector spaces than $V$ or $V^{}$, but these are not the sort of tensors you will be interested in as you begin to study differential geometry. The torsion tensor is a $(1,2)$-tensor because it takes in two vector fields and spits out a vector field. Thus it must have the capacity to "eat" two vector fields, so it has two components in $\chi(M)^{}$. However, the end result is not an element of $\mathbb{F}$ but a vector field, so there must be an extra copy of $\chi(M)$ lying around, which is why the torsion tensor lives in $\chi(M) \otimes \chi(M)^{} \otimes \chi(M)^{*}$. In the infinite dimensional case, there are more multilinear maps than objects of the tensor product space, so you are correct that the correspondence breaks down there.
H: Is it trivial to say $\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k}$ Is it trivial to say $$\mathop {\lim }\limits_{n \to \infty } {(1 + {k \over n})^n} = e^{k},$$ considering the fact that we know $$\mathop {\lim }\limits_{n \to \infty } {(1 + {1 \over n})^n} = e?$$ AI: Since $\frac kn = \frac 1 {\frac nk}$ and $n = \frac nk \cdot k$ we have $$\lim_{n\to \infty} (1+\frac kn)^n = \lim_{n\to\infty} \left( (1 + \frac{1}{\frac nk})^{\frac nk} \right)^k = \left ( \lim_{u\to\infty} (1+\frac 1u)^u \right)^k = e^k$$ If $k<0$, you have $$(1+\frac 1u)^{-u} = \left((1+\frac 1u)^u\right)^{-1} \to e^{-1} \qquad (u\to\infty)$$
H: Banach algebra problem: $\left\\|e^{ta}\right\\|\leq Me^{-\omega t}$ Let $A$ be a unital algebra, and $a\in A$. Assume that $\sigma(a)\subset \{\lambda\in \mathbb{C}: Re\lambda < 0\}$. Show there exists $M,\omega >0$ such that $$\left\\|e^{ta}\right\\|\leq Me^{-\omega t} $$ for all $t>0$. Even if $\sigma(a)$ contains an eigenvalue $\lambda$ i find this difficult since then (for an eigenvector $x$) $$\left\\|e^{ta(x)}\right\\|= \left\\|e^{t\lambda x}\right\\| =\left\\|\sum_{k=0}^{\infty}t^k (\lambda x)^k/k!\right\\|... $$ But i cant seem to reduce this further to something of the above form, let alone if $\sigma(a)$ contains no eigenvalues. Is there some handy trick involved? The inequality does remind me of inequalities I have seen in theory for dynamical systems where $a$ is instead some functions $g$. Any hints of suggestions? AI: Hint: For a suitable contour $\Gamma$, $$ e^{ta} = \dfrac{1}{2\pi i} \oint_\Gamma e^{tz} (z - a)^{-1}\ dz$$
H: Can you factor out vectors? My prof introduced eigenvalues to us today: Let $A$ be an $n \times n$ matrix. If there a scalar $\lambda$ and an $n-1$ non-zero column vector $u$, then $$Au = \lambda u$$ then $\lambda$ is called an eigenvalue and $u$ is called an eigenvector. $$Au - \lambda u = 0$$ $$\implies (A - \lambda I)u = 0$$ $$\implies \det(A - \lambda I) = 0$$ How did he get from $Au - \lambda u = 0$ to $\implies (A - \lambda I)u = 0$? It looks like he factored out the vector, but I thought you could only factor out constants? If you can factor out vectors can you explain why? AI: The distributive laws apply to matrix (or matrix-vector) multiplication. $$\eqalign{(A+B) C &= AC + BC\cr A(C+D) &= AC + AD\cr}$$ whenever $A$,$B$,$C$,$D$ have the right dimensions for these to make sense.
H: subsets of a generating set of a free group Is it true that, given any free group $F$ and any set $S\subseteq F$ which generates $F$, there is a subset $T\subseteq S$ which is a free generating set for $F$? It would be great if you could give me some literature recommendations. AI: Notice that $\{2,3\}$ is a generator set of the free group $\mathbb{Z}$, but no proper subset generates the group.
H: Transitive relations and Subsets I have a question to prove: If relations R is transitive, than R^2 is transitive. In the answer the professor says that if R is transitive than: R^2 is a subset of R (I understand why, this is the definision) Therefore, R^2*R^2 is a subset of R^2. This is the part I don't understand. Is this like a multipication of two sides in an equation? Thank you. AI: You believe that $R^2=R\times R\subseteq R$. Therefore, $$R^2\times R^2=(R\times R)\times (R\times R)\subseteq R\times R=R^2$$
H: Partial derivative in higher dimension . Let us consider $g: \mathbb R^n \to \mathbb R^m$ , $f:\mathbb R^m \to \mathbb R^k$ be $C^1$ functions . Define $F= f\circ g$. I am having the problem in finding the partial derivative of $F$, $$\frac {\partial F}{\partial x_i} = \frac{\partial \{f_1(g(x)), .......f_k(g(x)) }{\partial x_i}$$ Let $g(x)=y \in \mathbb R^m$ which is not constant. $$\frac {\partial F}{\partial x_i} = \frac{\partial \{f_1(y), .......f_k(y)}{\partial y} . \frac{\partial y}{\partial x_i}$$ $$=\nabla f_{(g(x))} \cdot \sum_{i=1}^{m} \frac{\partial g_i(x)}{\partial x_i} e_i$$ This doesn't make sense , because the first one has $k$ components and the second term has $m$ components . Can someone point out where i am going wrong ? AI: $F:\mathbb R^n\rightarrow \mathbb R^k$ has $k$-components. Then $$\frac{\partial F_r}{\partial x_i}(x)=(\text{chain rule})= \sum_{j=1}^m\frac{\partial f_r}{\partial y_j}(g(x))\frac{\partial y_j}{\partial x_i}, $$ for all $r\in \{1,\dots,k\}$, where $x=(x_1,\dots,x_n)$ and $y=g(x)\in\mathbb R^m$.
H: Series representation of $m=1^k + 2^k + \ldots n^k$. Is there any simple form for following question? $m=1^k+2^k+\cdots+n^k$ AI: See Faulhaber's formula, for instance.
H: Solving roots of a sum of sinusoids Suppose I have a sinusoid with fundamental frequency $f_0$ and $N$ harmonics (all with distinct amplitudes $a_k$. Each harmonic also has it's own corresponding phase $\phi_k$ and offset $c_k$. $y(t) = \sum_{k=0}^{N} [a_k sin(2 \pi k f_0 t + \phi_k)+c_k]$ How could one find the zeros of such a function? I have a microprocessor computing necessary parameters (amplitude, phase, offset of each frequency component) as the signal is passed, and I'd like to figure out the zeros of the function ahead of time (before the signal actually hits 0). One approach was to approximate each sinusoid as a Taylor series of, say, 8 degrees for each sinusoid... but then solving the roots of that polynomial gets a bit complicated for a microprocessor. AI: You can combine all the $c_k$'s into a single $c$. Then I think (unless $N$ is small) you are in for a numeric solution. The function repeats every $\frac 1{f_0}$ seconds, so you just have to look over that span. Evaluate it on a mesh of points and use a root finder. The challenge will be places where it dips down toward the axis and turns around-does it cross (making two roots) or not? A function minimizer will help there. These are described in any numerical analysis text. The Taylor expansion is too local for your application-it will take too many terms to converge.
H: Fix some $\delta\in \mathbb R$ and let $f:[0,\infty)\rightarrow \mathbb R$ be given by the equation $$f(x)=\frac{\sin(x^2)}{x}+\frac{\delta x}{1+x}$$ Show that, $\lim_{n\rightarrow\infty}\int_{0}^{a}f(nx)\ dx=a\delta$ for each $\ a>0$. My attempt: $\lim_{n\rightarrow\infty}\ f(nx)=\delta$ and $\|f(nx)\|\le (\frac{1}{nx}+\delta)$ then the right-hand-side of the inequality is integrable but what if $a=\infty$ ? AI: You have $\int_0^a f(nx) dx = \frac{1}{n} \int_0^{na} f(t) dt$. It is straightforward to find that $ \int_0^{na} \frac{x}{1+x} dx = n a -\log(n a+1)$, and so $\lim_{n \to \infty} \frac{1}{n} \int_0^{na} \frac{x}{1+x} dx = a$. Note that $\frac{\sin x^2}{x} \le \begin{cases} 1, & x \in (0,1] \\ \frac{1}{x}, & 1 <x \end{cases}$. Hence for $n$ sufficiently large, we have $\int_0^{n a} \frac{\sin x^2}{x} dx \le \int_0^1 dx + \int_1^{n a} \frac{1}{x}dx = 1+\log(n \alpha)$, hence $\lim_{n \to \infty} \frac{1}{n} \int_0^{na} \frac{\sin x^2}{x} dx = 0$ Combining gives the desired result.
H: Example of Group which is not a direct product of its Sylow-Subgroups Can you please give me an example of a group which could not be written as the direct product of its Sylow-subgroups? AI: The minimal example is the symmetric group $S_3$.
H: Prove that the greatest integer function: $\mathbb{R} \rightarrow \mathbb{Z}$ is onto but not $1-1$ Statement: the greatest integer function int: $\mathbb{R} \rightarrow \mathbb{Z}$ is onto but not $1-1$ Proof: let $x \in \mathbb{R}$, then $int(x) \leq x$ and is an element of $\mathbb{Z}$. Since $\mathbb{Z}$ is an element of $\mathbb{R}$, the greatest integer function maps onto $\mathbb{Z}$. However, it is not one-to-one, because $int(0.2)=int(0.3)=0$ Is my proof valid? AI: Yes, your reasoning is correct, but you should say $\Bbb Z$ is a subset of $\Bbb R$. The element relation does not hold between $\Bbb Z$ and $\Bbb R$. (Some students confuse containment and membership early on.) It looks like you are using the fact that integers map to themselves to demonstrate the map is onto, and that's good. However, it might strengthen what you wrote if you stated that a little more explicitly. (Incidentally, the question should also be more like "determine if this function is 1-1 and/or onto," but I can tell that's really what you were interested in because of your solution.) (Obsolete due to intervening edits.)
H: Stuck trying to solve this differential equation The equation is: $$\dfrac{dy}{dx} = \dfrac{3y}{(3y^{2/3}-x)}$$ So I wrote this as: $$\dfrac{dx}{dy}= \dfrac{(3y^{2/3}-x)}{3y}$$ $$\therefore \dfrac{dx}{dy} + \dfrac{x}{3y} = y^{-1/3}$$ If I let $v=y^{\frac{2}{3}}$ then: $$\dfrac{dx}{dy} = \frac{2}{3}y^{-1/3}\dfrac{dx}{dv}$$ $$\therefore \dfrac{dx}{dv}+\dfrac{x}{2}y^{-2/3} = \dfrac{9}{2}$$ But when I try to solve this by multiplying through by an integrating factor I get stuck. So I was wondering if this method is a correct way to solve this equation? AI: Let $M(x, y) = 3 y$ and $N(x, y) = x-3 y^{2/3}$ so that the equation is $M+Ny'=0$ This is not an exact equation, because $M_y(x, y)= 3 \neq 1 = N_x(x, y)$. We have to find an integrating factor $\mu(y)$ such that $\mu(y) M(x, y)+\mu(y)N(x, y)y' = 0$ is exact. This means $\frac{\partial \mu M}{\partial y} = \frac{\partial \mu N}{\partial x} $ that is $$3 y \mu'+3 \mu = \mu$$ Isolating $\mu(y)$ $$\frac{\mu'}{\mu} = -\frac{2}{3 y}$$ and integrating both sides with respect to $y$ we find $$\mu(y) = y^{-2/3}.$$ Multiplying both sides of the original ODE $3 y+(x-3 y^{2/3}) y'= 0$ by $\mu$ we have $$3 y^{1/3}+\left(\frac{x}{y^{2/3}}-3\right)y' = 0.$$ Let $P(x, y) = 3 y^{1/3}$ and $Q(x, y) = \frac{x}{y^{2/3}}-3$ so that $P+Qy'=0$. This is an exact equation, because $P_y = y^{-2/3} = Q_x$. Defining $f(x, y)$ such that $f_x(x, y)= P(x, y)$ and $f_y(x, y) = Q(x, y)$, the solution will be given by $f(x, y) = K$, where $K$ is an arbitrary constant. Integrate $f_x(x, y)$ with respect to $x$ in order to find $f(x, y)$ $$ f(x, y) = \int 3 y^{1/3} \operatorname{d}x = 3 y^{1/3} x+g(y) $$ where $g(y)$ is an arbitrary function of $y$. Differentiating $f(x, y)$ with respect to $y$ in order to find $g(y)$ $$ \frac{\partial f(x, y)}{\partial y} = \frac{x}{y^{2/3}}+g'(y) $$ and substituting into $f_y(x, y)= Q(x, y)$ we find $$\frac{x}{y^{2/3}}+g'(y) = \frac{x}{y^{2/3}}-3$$ and finally $$g(y) = \int-3 \operatorname{d}y = -3 y.$$ Substituting $g(y)$ into $f(x, y)$ we have $f(x, y) = 3 y^{1/3} x-3 y$ and then the solution is $$ 3 y^{1/3} x-3 y = K$$

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