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H: Onto the notation and interpretation of queueing theory-related markov chains If in a server with probability $p$ one job arrives and independently with probability $q$ one job departs, could you please explain to me what is the quantity and let me know if I have understood? The queue is infinite. Note that during a time step, we might have both an arrival and transmission, or neither. $1$. $r=p(1-q)$: This denotes the probability that there is a job at the server right? i.e in Markov chain notation $\mathbb P(X_n=1\| X_{n-1}=0)$ $2.$ $s=q(1-p)$: This denotes the probability that there is no job at the server right? i.e in Markov chain notation $\mathbb P(X_n=0\| X_{n-1}=1)$ (I am not sure) $3.$ What is the quantity $r+s$ denotes? $4.$ What about $1-(r+s)$? Thanks for helping. AI: $r$ is the probability to have an arrival and no departure, i.e. $P(X_n=X_{n-1}+1)$. $s$ is the probability to have a departure and no arrival, i.e. $P(X_n=X_{n-1}-1)$. $r+s$ is the probability that the queue length changes, i.e. $P(X_n \neq X_{n-1})$. $1-(r+s)$ is the probability that the queue length doesn't change, i.e. $P(X_n=X_{n-1})$.
H: Product of Lebesgue integrals is bounded I'm attempting to show the following: If $f:[0,1] \to (0,\infty)$ is measurable, then $$ \int_0^1 f(x) \,dx \int_0^1 \frac{1}{f(x)} \,dx≥1$$ My first inclination is to use something like Fubini or Tonelli to turn the product into a single integral, but I'm not sure I can apply either theorem here. AI: By Cauchy-Schwarz's inequality, we have $$ 1=\left(\int_0^1 \frac{\sqrt{f(x)}}{\sqrt{f(x)}}dx\right)^2\leqslant\left(\int_0^1f(x)dx\right)\left(\int_0^1\frac{dx}{f(x)}\right) $$
H: Is the absolute value of a definite integral equal to the definite integral of the absolute value of the integrand? An interesting question occurred to me as I was reading some Physics: Is it true in general that $$\left\|\int_a^b f(x) \, dx\right\| = \int_a^b \|f(x)\| \, dx\, ?$$ If not, what properties must $f(x)$ satisfy for the above equality to be true? I'm not a mathematician, but my hunch is that the equality holds only for $f$ such that $f(x) > 0$ for every $x \in [a,b].$ This seems to work out with some simple examples I've tried, but I haven't been able to prove it rigorously so far. I suspect the Cauchy-Schwartz inequality might come in handy, but I do not know how to use it here, unless perhaps if I interpret the definite integral as a Riemann Sum. AI: Let $A\subset \operatorname{supp}f$ such that $\operatorname{arg} f$ is constant on $\operatorname{supp}f\backslash A$. If there exists some $A$ satisfying this property with measure zero, then $$\left\|\int_a^b f(x) \, dx\right\| = \int_a^b \|f(x)\| \, dx$$ Proof: $$\left\|\int_a^b f(x) \, dx\right\| = \left\| \int_{[a,b]\backslash\operatorname{supp}f} f(x) \, dx \, + \int_{\operatorname{supp}f\backslash A} f(x) \, dx \, + \int_A f(x) \, dx\right\| = \left\| \int_{\operatorname{supp}f\backslash A} f(x) \, dx\right\|$$ Because $\int_{[a,b]\backslash\operatorname{supp}f} f(x) \, dx = 0$ and, since $A$ has measure zero, $\int_A f(x) \, dx = 0$ Since $\operatorname{arg} f$ is constant on $\operatorname{supp}f\backslash A$, $\left\| \int_{\operatorname{supp}f\backslash A} f(x) \, dx\right\| = \int_{\operatorname{supp}f\backslash A} \left\|f(x)\right\| \, dx$ Therefore, because $\int_A \left\|f(x)\right\| \, dx = 0$, $$\left\|\int_a^b f(x) \, dx\right\| = \int_{\operatorname{supp}\backslash A} \left\|f(x)\right\| \, dx = \int_a^b \|f(x)\| \, dx$$
H: Why is $(A\times B)/(I\times J)=A/I\times B/J$? Why is it true that $(A\times B)/(I\times J)=A/I\times B/I$? Here, $A$ and $B$ are (commutative) rings (with $1$), and $I$ and $J$ are respective ideals. This looks similar to the chinese remainder theorem, but I'm not sure how this is (if it is at all) an application of it. AI: Consider the map $$ f\colon A\times B\to (A/I)\times(B/J),\qquad f(a,b)=(a+I,b+J) $$ and prove $f$ is a ring homomorphism; find its kernel; apply the homomorphism theorem.
H: Proving independence of the neighborhood axioms in topology I have not found any information on this topic on StackExchange or through a few minutes of searching with Google. This question is Exercise 8 in Section 2.1 on page 22 of Topology and Groupoids, by Brown. I am given the neighborhood axioms If $N$ is a neighborhood of $x$, then $x \in N$. If $N$ is a subset of $X$ containing a neighbourhood of $x$, then $N$ is a neighbourhood of $x$. The intersection of two neighbourhoods of $x$ is again a neighbourhood of $x$. Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$. and I am asked to prove that these four axioms are independent. It is my understanding that this can be accomplished by selecting selecting any subset of three axioms and supplying a purported neighborhood topology $\mathcal{N}$ that satisfies the all the axioms except the one that hasn't been selected. This requires $\binom{4}{3} = 4$ examples. I think I have one example. If I say that $N$ is a neighborhood of $x \in X$ if and only if $N = \{ x \}$, then I think this satisfies axioms 1, 3 and 4 but not 2. I haven't come up with any other examples yet, though. I looked briefly at Counterexamples in Topology by Steen and Seebach but it didn't look like they deal with this type of question. Does anyone know of any other examples that will work? Thanks. Edit: I think I have another example. Let $X$ = $[0, 1]$ and say that $N$ is a neighborhood of $x \in X$ if and only if $\frac{1}{2} \in N$, then I think that satisfies axioms 2, 3 and 4 but not 1. If I'm correct about both of these examples then I'm halfway to a solution. AI: For (1) you can let $X$ be any infinite set, and for each $x\in X$ let $$\mathscr{N}(x)=\{X\setminus F:F\text{ is a finite subset of }X\}\;;$$ I’ll leave it to you to verify that (2)-(4) are satisfied and (1) is not. For (3) let $X=\{0,1,2\}$. Let $\mathscr{N}(0)=\big\{\{0,1\},X\big\}$, $\mathscr{N}(1)=\big\{\{0,1\},\{1,2\},X\big\}$, and $\mathscr{N}(2)=\big\{\{1,2\},X\big\}$. For (4) you can again let $X=\{0,1,2\}$, but this time let $\mathscr{N}(0)=\big\{\{0,1\},X\big\}$, and $\mathscr{N}(1)=\mathscr{N}(2)=\big\{\{1,2\},X\big\}$.
H: Inequality involving size of random variable I’m following my professor’s notes, and I became confused by an inequality he used without justification, but I cannot see why it is true. Any help is appreciated. Let $\{T_t\}_t$ be a sequence of iid, positive, integer-valued random variables. Suppose $$P\{T \ge x \} = \Theta \left(\frac{1}{\sqrt x}\right). $$ What I do not understand is he then states: $$(1) \max_{1 \leq i \leq t} T_i = \Theta (t^2),$$ $$ (2) \sum_{i=1}^t T_i = \Theta (t^2), $$ both in probability. I would appreciate help on how he concluded both of these from the previous fact. Thank you! AI: Say $ a/\sqrt{x} \le P(T \ge x) \le b/\sqrt{x}$ for $x$ sufficiently large. For (1): $$ P\left(\max_{1\le i \le t} T_i < c t^2\right) = P(T < c t^2)^t $$ When $c > 0$, for $t$ sufficiently large this is between $(1 - a/(\sqrt{c} t))^t$ and $(1 - b/(\sqrt{c} t))^t$, which as $t \to \infty$ go to $\exp(-a/\sqrt{c})$ and $\exp(-b/\sqrt{c})$ respectively. Thus for any $\epsilon > 0$, taking $c_1$ and $c_2$ such that $\exp(-a/\sqrt{c_1}) < \epsilon$ and $\exp(-b/\sqrt{c_2}) > 1-\epsilon$, we find that $P\left(\max_{1\le i \le t} T_i < c_1 t^2\right) < \epsilon$ while $P\left(\max_{1\le i \le t} T_i < c_2 t^2\right) > 1-\epsilon$.
H: How to show Spec$\frac{\Bbb C [x, y]}{(xy)}$ is connected in the Zariski topology In exercise 3.6.E of Vakil's book, it is asked for the reader to find a ring for which its spec is connected and non-irreducible. Taking his hint, I thought of $A = \frac{\Bbb C [x, y]}{(xy)}$. It is not irreducible, because $V((x)) \cup V((y)) = V((xy)) = \operatorname{Spec}A$. To prove connectedness I tried to use the fact that if $ \operatorname{Spec}A= V(I) \cup V(J) = V(IJ)$ where $V(I) \cap V(J) = \emptyset$ would imply $I + J = A$. Can anyone give me some hint? AI: The comments give you a completely algebraic way to view connectedness: a scheme $X$ is disconnected if and only if there exists $e\in\Gamma(X,\mathcal{O}_X)$ such that $e\neq 0,1$ and $e^2 = e.$ Proving that no such $e$ exists will do the trick. Here's another method: Hint: View $\operatorname{Spec}\Bbb{C}[x,y]/(xy)$ as a closed subscheme of $\operatorname{Spec}\Bbb{C}[x,y] = \Bbb{A}^2_\Bbb{C}$ (the particular subscheme structure doesn't matter since we're concerned about connectedness, which is a purely topological property). Recall that if $U_1, U_2\subseteq X$ are connected and $U_1\cap U_2\neq\emptyset,$ then $U_1\cup U_2$ is connected.
H: Are nimbers the largest field of characteristic 2? Are nimbers the largest field* of characteristic 2? The nimbers are defined as follows: $$a+b=\operatorname{mex}(\{a+b'\}\cup\{a'+b\})$$ $$ab=\operatorname{mex}(\{ab'+a'b+a'b'\})$$ Where $a'<a,b'<b$, and $a$ and $b$ are arbitrary ordinals. If yes, is there a proof? If not, what field is larger? *Technically the nimbers don't form a field because a field's domain is a set, while the nimbers are a proper class AI: Remarkably, this is true, so long as you're willing to stretch the definition of a field to allow its elements to form a proper class. This is briefly mentioned here, and the definition of "largest" I am using is that Every small field (i.e., a field whose elements form a set) of characteristic two embeds into the field of nimbers. First, suppose $\Bbbk$ is an algebraically closed small field of characteristic $2$, then up to isomorphism it is given by the algebraic closure of $\Bbb F_2(X)$ for $X$ some set of independent variables (this is because every algebraically closed field is uniquely determined up to isomorphism by its characteristic and its transcendence degree over its prime field, which in the case of characteristic $2$ is given by $\Bbb F_2$). Now, since the "Field" $\mathbf{On}_2$ of nimbers forms a proper class, we can find a set $A$ of algebraically independent elements such that $\|A\|=\|X\|$ (otherwise, every subset of cardinality $\|X\|$ would be algebraically dependent, which would force the cardinality of $\mathbf{On}_2$ to be either countably infinite or of size $\|X\|$, which would force it to be a set). This allows us to embed $\Bbb F_2(X)\hookrightarrow\mathbf{On}_2$. Since $\mathbf{On}_2$ is algebraically closed, this map will factor through its algebraic closure, giving an embedding $\Bbbk\hookrightarrow\mathbf{On}_2$. If $\Bbbk$ is not algebraically closed, then it factors into its own algebraic closure $\bar{\Bbbk}$. By the above argument, this gives an embedding $\Bbbk\hookrightarrow\bar{\Bbbk}\hookrightarrow\mathbf{On}_2$. Therefore, in this sense of the word, it is somehow the "largest field" of characteristic two. However, the answer I linked also mentions that it's not unique with this property. The simplest counterexample is $\mathbf{On}_2(x)$ of rational functions with nimber coefficients. Since any field of characteristic $2$ embeds into $\mathbf{On}_2$, it will also embed into $\mathbf{On}_2(x)$. However, this field is not algebraically closed (you can't solve $t^2-x=0$ for $t$). If you assume global choice (as mentioned by another answer on the MO question I linked), the Field of nimbers is the unique algebraically closed large Field of characteristic $2$, so perhaps in this sense, it is the nicest "largest field" of characteristic $2$.
H: Proof that $(1 + t)^n = \sum\limits_{k = 0}^{n} \binom{n}{k} t^k$ I'm not understanding this proof from Combinatorics: Topics, Techniques, Algorithms by Peter J. Cameron. Here is his proof of this theorem: The theorem can be proven by induction on $n$. It is trivially true for $n = 0$. Assuming the result for $n$, we have $\displaystyle(1 + t)^{n + 1} = (1+t)^{n} \cdot (1 + t) = \left( \sum\limits_{k = 0}^{n} \binom{n}{k} t^k \right) \cdot (1 + t)$ the coefficient of $t^k$ on the right is $\displaystyle \binom{n}{k - 1} + \binom{n}{k}$ (the first term coming from $t^{k - 1} \cdot t$ and the second coming from $t^k \cdot 1$ and $\displaystyle \binom{n}{k - 1} + \binom{n}{k} = \binom{n + 1}{k}$ I'm very lost and don't see how this shows that the formula holds for $(1 + t)^{n + 1}$ AI: Well we have that \begin{align} (1+t)^{n+1} &= \left(\sum_{k=0}^{n}\begin{pmatrix}n \\ k \end{pmatrix}t^k\right)(1+t) \\ &= \sum_{k=0}^{n}\begin{pmatrix}n \\ k \end{pmatrix}t^k + \sum_{k=0}^{n}\begin{pmatrix}n \\ k \end{pmatrix}t^{k+1} \\ &= \begin{pmatrix}n \\ 0 \end{pmatrix} + \sum_{k=1}^{n}\begin{pmatrix}n \tag{1}\\ k \end{pmatrix}t^k + \sum_{k=1}^{n}\begin{pmatrix}n \\ k-1 \end{pmatrix}t^k + \begin{pmatrix}n \\ n \end{pmatrix}t^{n+1} \\ &= \begin{pmatrix}n \\ 0 \end{pmatrix} + \sum_{k=1}^{n}\left[\begin{pmatrix}n \\ k \end{pmatrix} + \begin{pmatrix}n \\ k-1 \end{pmatrix}\right]t^k + \begin{pmatrix}n \\ n \end{pmatrix}t^{n+1} \\ &= \begin{pmatrix}n \\ 0 \end{pmatrix} + \sum_{k=1}^{n}\begin{pmatrix}n + 1 \\ k \end{pmatrix} t^k + \begin{pmatrix}n \\ n \end{pmatrix}t^{n+1}, \tag{2} \end{align} where at $(1)$ we have used the fact that \begin{align} \sum_{k=0}^{n}\begin{pmatrix}n \\ k \end{pmatrix}t^{k+1} &= \sum_{k=1}^{n+1}\begin{pmatrix}n \\ k-1 \end{pmatrix}t^{k} \\ &= \sum_{k=1}^{n}\begin{pmatrix}n \\ k-1 \end{pmatrix}t^{k} + \begin{pmatrix}n \\ n \end{pmatrix}, \end{align} (which follows by relabelling the summands via $k \mapsto k-1$) and at $(2)$ we have used the identity \begin{align} \begin{pmatrix}n \\ k \end{pmatrix} + \begin{pmatrix}n \\ k-1 \end{pmatrix} = \begin{pmatrix}n +1 \\ k \end{pmatrix}. \end{align} Can you take it from here? Now, we can take this a step further by the following: by the above calculations \begin{align} (1+t)^{n+1} &= \begin{pmatrix}n \\ 0 \end{pmatrix} + \sum_{k=1}^{n}\begin{pmatrix}n + 1 \\ k \end{pmatrix} t^k + \begin{pmatrix}n \\ n \end{pmatrix}t^{n+1} \\ &= \begin{pmatrix}n \\ 0 \end{pmatrix} - \begin{pmatrix}n+1 \\ 0 \end{pmatrix} + \sum_{k=0}^{n+1}\begin{pmatrix}n + 1 \\ k \end{pmatrix} t^k + \begin{pmatrix}n \\ n \end{pmatrix}t^{n+1} - \begin{pmatrix}n + 1 \\ n+1 \end{pmatrix}t^{n+1} \\ &= \sum_{k=0}^{n+1}\begin{pmatrix}n + 1 \\ k \end{pmatrix} t^k,\end{align} where the last equality follows since $ \begin{pmatrix} N \\ 0 \end{pmatrix} = \begin{pmatrix} N \\ N \end{pmatrix} = 1$ for any $N \in \mathbb{N}$.
H: Contest math application for Wilson's theorem $$1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{23} = \frac{a}{23!}$$ Find the remainder when $a$ is divided by $13.$ I found this online and got stuck a bit. I approached the problem as such: From the expression we get $$a=\frac{23!}{1}+\frac{23!}{2}+\dots+\frac{23!}{23!}$$ so $$a \equiv \frac{23!}{13} \pmod{13}$$ from here we have that $$a\equiv(1\cdot 2\dots11\cdot12) \cdot(1\cdot2\dots 9\cdot10) \equiv(12!) \cdot(10!) \pmod{13}.$$ And now according to Wilson's theorem $$a \equiv(-1)(10!) \pmod{13}.$$ I cannot seem to get rid of the $10!$; what should I do here? AI: Since $12!=10!\times 11\times12\equiv-1\bmod13$, it follows that $10!\times-2\times-1\equiv-1\bmod13$, so $10!\times2\equiv-1\equiv12\bmod13$, so $10!\equiv6\bmod 13$.
H: Math Fraction Problem I am at the moment in the $5$th chapter of IGCSE mathematics and currently need help for this problem. Joseph needs $6\frac12$ cups of cooked rice for a recipe of nasi goreng. If $2$ cups of uncooked rice with $2 \frac12$ cups of water make $4 \frac13$ cups of cooked rice, How many cups of uncooked rice does Joseph need for his recipe? How much water should he add? AI: As a hint, I can see a couple of equations that would help you get started. Let $u$ represent the amount of uncooked rice, let $c$ represent the amount of cooked rice, and let $w$ represent the amount of water. All of these amounts are measured in cups. The given recipe states that $2$ cups of uncooked rice with $2\frac12$ cups of water yields $4\frac13$ cups of cooked rice. We must assume that uncooked rice and water should be scaled in tandem. For example, we could double the recipe, and our inputs would be $4$ cups of uncooked rice and $5$ cups of water. We can even make an equation out of that: $4u=5w$. Another equation relates the amount of uncooked rice to the amount of cooked rice. Note that $4\frac13=\frac{13}{3}$, so the equation would be $2u=\frac{13}{3}c$. What should you scale that equation by so you wind up with the needed amount of cooked rice? What does that tell you about the amount of water?
H: Making sense of a set notation I'm trying to make sense of this notation I came across in a post here: Let $\omega=\{(x,y)\in \mathbb{R}^2\,:\,0<x<y<2x<2\}$. Putting $0<x<y<2x<2$ into WolframAlpha gives me the solution interval as: $0<x<1$ and $x<y<2x$. What I'm trying to understand is how do you resolve $0<x<y<2x<2$ to $0<x<1$ and $x<y<2x$? I thought I knew about inequalities but honestly I've never seen one like this. I'm hoping someone can point me to a book or lecture or anything that I can read to educate myself Thanks. AI: A chain of inequalities is implicitly a logical "AND" of the simpler component inequalities. So "$0<x<y<2x<2$" means "$0<x$ and $x<y$ and $y<2x$ and $2x<2$". The first and last of these give you "$0<x$ and $x<1$", i.e., "$0<x<1$". The middle two give you "$x<y$ and $y<2x$", i.e., "$x<y<2x$". This is to be interpreted as $x$ ranges from $0$ to $1$; for a given value of $x$, $y$ ranges between $x$ and $2x$. This might be taken as limits on a double integral: $\int_0^1\int_x^{2x}f(x,y)\;dy\; dx$. The region described by the inequalities is the interior of a wedge between the lines $y=x$ and $y=2x$ in the first quadrant, cut off by the vertical line $x=1$.
H: Integrability of $\frac{1}{\|z\|^m}$ Suppose that we are in $\mathbb C^n$, and consider the unit ball $B_1(0)$ around the origin, I am interested in knowing the integrability of $\frac{1}{\|z\|^{2m}}$. Specifically, when is $$\int_{B_1(0)} \frac{1}{\|z\|^{2m}} \mathrm d\operatorname{vol}< \infty?$$ For $n = 1$, $\int_{B_1(0)} \frac{1}{\|z\|^{2m}} \mathrm d\operatorname{vol} = \int_{0}^{1}\int_{0}^{2 \pi} \frac{1}{r^{2m -1}} \mathrm dr \wedge \mathrm d\theta = 2 \pi r^{2-2m} \|^{2\pi}_{0}$. So it is integrable only when $m < 1$. What about when $n > 1$? AI: In this context, there's no difference between $\Bbb C^n$ and $\Bbb R^{2n}$. The surface area of a ball of radius $r$ in $\Bbb R^{2n}$ equals $C_{2n} r^{2n-1}$ for some constant $C_{2n}$. Then $$ \int_{B_1(0)} \frac1{\|z\|^{2m}}\,dvol = \int_0^1 \frac1{r^{2m}} C_{2n}r^{2n-1}\,dr = C_{2n} \int_0^1 \frac1r r^{2(n-m)}\,dr, $$ which converges precisely when $m<n$.
H: Prove that exist $b \gt 0$, so that $f$ may be defined at $x=0$ and be continuous. Given the function $$ f(x) = \begin{cases} (1 + 2^{\frac{3}{x}})^{bsin(x)} &\quad if \quad x\gt 0 \\ \\ \frac{arctan(9bx)}{x} &\quad if \quad x\lt 0 \\ \end{cases} $$ Prove that exist $b \gt 0$, so that $f$ may be defined at $x=0$ and be continuous. My procedure: (1) $$\lim_{x\to 0} \frac{arctan(9bx)}{x} = \lim_{x\to 0} \frac{arctan(9bx)-arctan(9b*0)}{x} = \frac d{dx}arctan(9bx)\|_{x=0}=\Bigl(\frac{1}{1+(9bx)^2}9b\Bigr)\|_{x=0}=9b=\lim_{x\to 0^{+}} \frac{arctan(9bx)}{x}=\lim_{x\to 0^{-}} \frac{arctan(9bx)}{x}$$ Then the limit $\lim_{x\to 0^{-}} \frac{arctan(9bx)}{x}$ exist. (2) $$\lim_{x\to 0^{+}} (1 + 2^{\frac{3}{x}})^{bsin(x)} = \infty^0 \;(indetermination)$$ The thing is I don´t really know how to calculate the second limit. Any hint in how to proceed with the limit?. Preferably without using L'Hopitals rule. AI: The problem is to compute the right-side limit. Assume henceforth $x>0$. $$\log(1+2^{3/x})^{b\sin x}=b\sin x\log(1+2^{3/x})=b\sin x\log(2^{3/x}(1+2^{-3/x}))=b\sin x\frac{3}{x}\log 2+o(x)$$ So the logarithm of the expression tends, as $x\downarrow 0$, to $3b\log 2$.
H: How many ways are there to give cookies and candies to these kids? The same kind of 5 cookies are given to 3 kids. At least one cookie is given to each child. Also, the same kind of 5 candies are given to the kids who received only one cookie. How many ways are there to give cookies and candies to these children? My Work: C(5,3) x C(5,3) = 10 x 10 = 100 ways C(5,3) (One for the cookies) C(5,3) (One for the candies) Did I do this correctly? I am not sure and want to double check if my reasoning is correct. AI: There are two different partitions for the cookies: $2, 2, 1$ and $3, 1, 1$. The partition $2, 2, 1$ can be distributed in three ways (each child can the one with 1 cookie). The candies must all be given to this child, so we have a total of three divisions. The partition $3, 1, 1$ can be distributed in three ways (each child can be the one with 3 cookies). For each of these three, there are four ways of distributing the candies to the other two children: each of the two children can be the child to receive more, and the split can be either $4,1$ or $3,2$. As such, there are a total of twelve divisions for this cookie partition. In total, there are 15 ways we can divide the cookies and candies
H: How to find the Laurent series expansion of $\frac{2}{z^2-4z+8}$ by long division? I'm trying to find the Laurent series expansion for $$ \frac{2}{z^2-4z+8} $$ using polynomial long division. However, I noticed that if I divide leading with the $8$ term, then I will only get positive power terms, namely $$ \frac{1}{4}+\frac{1}{8}z+\frac{1}{32}z^2 + ... $$ Whereas if I divide leading with the $z^2$ term, then I will get only negative power terms, as in $$ \frac{2}{z^2} + \frac{8}{z^3}+\frac{16}{z^4} + ... $$ What is going on? (I'm doing this to find the residues, i.e. the coefficient of the $\frac{1}{z}$ term.) AI: Well you wan't to find a function $$ f(z):= \sum_{n=-\infty}^\infty a_n z^n $$ defined around $0$ such that $f(z)(z^2-4z+8)=2$. If $$ f(z)=\frac{2}{z^2} + \frac{8}{z^3}+\frac{16}{z^4} + \cdots $$ then $f$ is not defined at $0$. However, if $$ f(z)= \frac{1}{4}+\frac{1}{8}z+\frac{1}{32}z^2 + \cdots $$ then $f$ is defined at $0$ and doing some algebra you can check that $f(z)(z^2-4z+8)=2$. Thus, your first approach is the correct one. Of course this means that the coefficient for $\frac{1}{z}$ is $a_{-1}=0$.
H: Break even points structure I am creating a Matchmaking ranking structure where players have a value to determine their skill level. If I win I get points, say $X$ points and if I lose then I win $Y$ points. I am playing around with the values in a spreadsheet to see long term where players will fit into. How can I calculate the breakeven point for $X$ and $Y$? For example, if I have a $50\%$ win ratio then $X = Y$ However, how do I calculate what will be the win percentage to retain your position if, for example, $X$ is $6$ and $Y$ say $-4$? AI: To retain position simply means that wins and losses should compensate, i.e. $6X-4Y=0$. You can solve this for $Y$ and get $Y=\frac{3}{2}X$. Win ratio is defined of the fraction of wins divided by the total number of games, i.e. $R=\frac{X}{X+Y}$. You can now insert the formula for $Y$ to get $R=\frac{X}{X+\frac{3}{2}X}=\frac{2}{5}=40\%$. That means, you should win 40% of all games top keep position.
H: Question about the elements of $\operatorname{Hom}_k(k,V)$ Let $V$ be a $\mathbb{K}$-vector space. I am trying to understand what the linear maps from $\operatorname{Hom}_\mathbb{K}(\mathbb{K},V)$ look like. Given some $v \in V$, the map $f_{v}:\mathbb{K} \rightarrow V$ given by $f_v(k)=kv$ is linear and so $f_v \in\operatorname{Hom}_{\mathbb{K}}(\mathbb{K},V)$. I think that all the maps from $\operatorname{Hom}_{\mathbb{K}}(\mathbb{K},V)$ look like this but I am not sure how to prove it. AI: Yes you are right. Take any $g \in \mathrm{Hom}(\Bbb{K}, V)$ and set $v:=g(1)$. Now check that $f_v=g$.
H: Finding the associated unit eigenvector Background Find the eigenvalues $λ_1<λ_2$ and two associated unit eigenvectors of the symmetric matrix $$A = \begin{bmatrix}-7&12\\12&11\end{bmatrix}$$ My work so far $$A = \begin{bmatrix}-7-λ&12\\12&11-λ\end{bmatrix}=λ^2-4λ-221=(λ+13)(λ-17)$$ Thus $$λ_1=-13$$ $$λ_2=17$$ To find the solution set for $λ_1$ $$\begin{bmatrix}6&12\\12&24\end{bmatrix}=\begin{bmatrix}1&2\\0&0\end{bmatrix}$$ $$x_1=-2x_2$$ $$x_2=x_2$$ $$=\begin{pmatrix}-2\\1\end{pmatrix}$$ and the solution set for $λ_2$ $$\begin{bmatrix}-24&12\\12&-6\end{bmatrix}=\begin{bmatrix}1&\frac{-1}{2}\\0&0\end{bmatrix}$$ $$x_1=\frac{1}{2}x_2$$ $$x_2=x_2$$ $$=\begin{pmatrix}\frac{1}{2}\\1\end{pmatrix}$$ However, I'm unsure how to get the associated unit eigenvectors. Would I plug these into the quadratic formula to find the solutions? For example, for $λ_2$ $$\sqrt{(\frac{1}{2})^2+1^2}=\sqrt{(\frac{1}{4})+1}=\pm\frac{2}{\sqrt{2}}=\begin{pmatrix}\frac{1}{\sqrt{2}}\\\frac{2}{\sqrt{2}}\end{pmatrix}$$ I know that I'm off here, but just took a guess. AI: As you correctly found for $\lambda_{1}=-13$ the eigenspace is $(−2x_{2},x_{2})$ with $x_{2}\in\mathbb{R}$. So if you want the unit eigenvector just solve: $(−2x_{2})^2+x_{2}^2=1^2$, which geometrically is the intersection of the eigenspace with the unit circle.
H: How do I integrate $\frac1{x^2+x+1}$? I have tried this: $$\frac1{x^2+x+1} = \frac1{\left( (x+\frac12)^2+\frac34\right)}$$ Now $u = x+\frac12$ $$\frac1{ u^2+\frac34 }$$ Now multiply by $ \frac34$ $$\frac1{ \frac43 u^2 + 1}$$ Now put the $\frac43$ outside the integral $$\frac34 \int \frac1{u^2+1}\,du=\frac34\arctan(u)=\frac34\arctan(x+1/2)$$ But the result is not the same result calculated by computers. What did I do wrong? Could someone please help me with this? I don't know where my wrong calculation is. The way should be correct to get to the result. Edit: So now $\int \frac{1}{x^2+x+1}=\int \frac{1}{(x+\frac{1}{2})^2}= \int \frac{4}{3} \frac{1}{\frac{4}{3}(x+\frac{1}{2})^2+1}$ u=x^2+1/2 $\int \frac{4}{3} \frac{1}{\frac{4}{3}(u)^2+1}$ $\int \frac{4}{3} \frac{1}{\frac{4u^2}{3}+1}$ $\int \frac{4}{3} \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$ Now it is: $\frac{4}{3} \int \frac{1}{\frac{2u^2}{\sqrt{3}}+1}$ $=\frac{4}{3} * arctan(2*(x^2+1/2)/(\sqrt{3}))$ Why is this still not the same as the computer calculated solution? AI: Note that $$\frac{1}{u^2+3/4}=\frac{1}{\frac34((2u/\sqrt 3)^2+1)}=\frac43 \frac{1}{v^2+1}$$ where $v=2u/\sqrt 3$. Can you finish now?
H: In triangle $\triangle ABC$, angle $\angle B$ is equal to $60^\circ$; bisectors $AD$ and $CE$ intersect at point $O$. Prove that $OD=OE$. In triangle $\triangle ABC$, angle $\angle B$ is equal to $60^{\circ}$; bisectors $AD$ and $CE$ intersect at point $O$. Prove that $OD=OE$. So I've already made a diagram(it is attached below), but I don't know how to prove it from there. Please help and explain your solution thoroughly because I have a test about this tomorrow and I want to understand this! Thank you! :D AI: Call the angle in $A$ $2\alpha$, then the angle in $C$ is $120-2\alpha$ and $OCA=60-\alpha$. So $DOE=COA=180-(\alpha +60-\alpha)=120$ so the $ODBE$ is cyclic. Now observe that $BO$ must be the bisector of $B$ and so $OED=OBD=30$ and also $ODE=OBE=30$. Since $OE$ and $OD$ are oblique sides of a isosceles triangle they must be equal.
H: Prove that the function integer part is continuous for any topology in Z. Prove that the function integer part $f:\mathbb{R}_l \to \mathbb{Z}, \quad f(x) =\lfloor x \rfloor $ is continuous for any topology in $\mathbb{Z}$. where $\mathbb{R}_l$ is the topology of the lower limit in $\mathbb{R}$. I know that if I prove that it is continuous in a finer topology in $\mathbb{Z},$ that is to say in the discrete topology, then it will be continuous in all topologies. But I can't prove it as the reverse image will be. I will appreciate your comments. AI: Verify that for any set $A \subseteq \mathbb Z$ we have $f^{-1}(A)=\cup_{n \in A} [n, n+1)$ which is open.
H: Proving that one can integrate a uniformly convergent series of functions term by term I aim to understand the following proof from Serge Lang's Introduction to Complex Analysis at a graduate level [ and I have the following definitions My question is: What does the last paragraph of the theorem in question actually mean? Does it mean that every $f_i$ is convergent, and if yes, how is this derived from the definition? AI: The last paragraph means that if for each $n$ we set $$ s_n:=\sum_{k=1}^n f_k, $$ then the sequence $(s_n)_{n=1}^\infty$ converges uniformly on $U$ to $s:=\sum_{k=1}^\infty f_k$. Then, by the first part of the Theorem $$ \lim_n\int_{\gamma} s_n = \int_\gamma s $$ which translates as $$ \lim_n\int_{\gamma} \sum_{k=1}^n f_k = \int_\gamma \sum_{k=1}^\infty f_k $$ and since $$ \lim_n\int_{\gamma} \sum_{k=1}^n f_k =\lim_n \sum_{k=1}^n \int_{\gamma}f_k = \sum_{k=1}^\infty \int_{\gamma} f_k $$ the desired result follows.
H: Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ I need help with the following question: Find all the integer solutions for: $3x^2+18x+95\equiv 0\pmod {143}$ My solution: First I know that $143=11\cdot 13$ then because $\gcd (11,13) = 1$ then $3x^2+18x+95\equiv 0\pmod {143}$ if, and only if $$3x^2+18x+95\equiv 3x^2+7x+7\equiv 0\pmod {11} \\ 3x^2+18x+95 \equiv 3x^2+5x+4\equiv 0\pmod {13}$$ I don't know how to solve those equations and I don't know how to combine it to the big solution for the real question (I know about the CRT, but I didn't realy understood how to use it, I'd love help with this). thanks in advance AI: Let's take one equation \begin{align} 3x^2+7x+7 & \equiv 0 \pmod{11}\\ 4(3x^2+7x+7) & \equiv 4(0) \pmod{11}\\ x^2+28x+28 & \equiv 0 \pmod{11}&& (\because 4(3) \equiv 1 \pmod{11})\\ x^2+6x+6 & \equiv 0 \pmod{11}&& (\because 28 \equiv 6 \pmod{11})\\ (x+3)^2-3 & \equiv 0 \pmod{11}\\ (x+3)^2-5^2 & \equiv 0 \pmod{11}&& (\because 5^2 \equiv 3 \pmod{11})\\ (x-2)(x+8) & \equiv 0 \pmod{11} \end{align} Since $11$ is prime so if $11 \| ab$, then $11$ divides at least one of them, so we get $$x\equiv 2 \pmod{11} \quad \text{ or } \quad x \equiv -8 \equiv 3\pmod{11}.$$ Likewise (you can work this out yourself) $$3x^2+5x+4 \equiv 0 \pmod{13} \implies x\equiv 2 \pmod{13} \, \text{ or } \, x \equiv \color{blue}{b}\pmod{13}. $$ So we have the following situation \begin{align} x&\equiv 2 \pmod{11} & x&\equiv 2 \pmod{11} & x&\equiv 3 \pmod{11} & x&\equiv 3 \pmod{11}\\ x&\equiv 2 \pmod{13} & x&\equiv \color{blue}{b} \pmod{13} & x&\equiv \color{blue}{b} \pmod{13} & x&\equiv 2 \pmod{13} \end{align} Now use CRT (hopefully you know how to apply it to simple systems like these) to solve these systems. For example the last system \begin{align} x & \equiv 3 \pmod{11}\\ x & \equiv 2 \pmod{13} \end{align} yields $$x \equiv 3(13)(6)+2(11)(6) \equiv \color{red}{80} \pmod{143}. $$ Likewise you will get a total of $\color{red}{4}$ incongruent solutions.
H: Proving $f$ is a real, uniformly continuous function on the bounded subset $E$ in $\mathbb{R}^1 \implies f$ is bounded on $E$. (Baby Rudin Chapter 4 Exercise 8) I am trying to prove: $f$ is a real, uniformly continuous function on the bounded subset $E$ in $\mathbb{R}^1 \implies f$ is bounded on $E$. My attempt: Suppose $E \subset \mathbb{R}$ is bounded and $f$ is a real, uniformly continuous function on $E$. For a contradiction, assume $f$ is not bounded on $E$. Then there exists a non-finite $f(p)$ for some $p \in E$. Let $r\in E$ be distinct from $p$. The boundedness of $E$ guarantees the existence of the real number $M$ such that $\|p-r\|< M$. Due to the usual properties of the absolute value function (and $p\ne r$), we conclude that $0<\|p-r\|<M$, that is, $M$ is positive. By the definition of uniform continuity, \begin{equation}\tag{8.1} \|p-r\|<M \implies \|f(p)-f(r)\| <\epsilon \end{equation} for some $\epsilon>0$. (We utilize the uniform continuity of $f$ in asserting that the $\delta$ in (8.1) would hold for our choice of $p$, given an arbitrary $r \in E$.) Note that the consequent (8.1) births a contradiction since the absolute difference of the non-finite $f(p)$ and $f(x)$ cannot be made smaller than $\epsilon$. My question: Is my proof correct in general? In particular, I am not sure if I should have taken $r \in \mathbb{R}$ instead of taking $r \in E$ above. The issue is that the definition of a bounded set mandates that $r$ be some real number (not necessarily in $E$) and the definition of uniform continuity requires that $r$ be in $E$. Is my proof totally incorrect? Can someone suggest how this proof can be improved / completed? Definition of a bounded set: Let $X$ be a metric space. $E$ is bounded if there is a real number $M$ and a point $q \in X$ such that $d(p, q)<M$ for all $p \in E$. Definition of uniform continuity: Let $f$ be a mapping of a metric space $X$ into a metric space $Y$. We say that $f$ is uniformly continuous on $X$ if for every $\epsilon>0$ there exists a $\delta>0$ such that $d_Y(f(p), f(q))$ for all $p$ and $q$ in $X$ for which $d_X(p, q)<\delta$. AI: If $f$ is not bounded there exists a sequence $(x_n)$ in $E$ such that $\|f(x_n)\|\to \infty$. Since $E$ is bounded so is the sequence $(x_n)$. Hence there is a convergent subsequence $(x_{n_i})$. Note that $\|x_{n_i}-x_{n_j}\| \to 0$ as $i,j \to \infty$. Now use the definition of uniform continuity to show that $\|f(x_{n_i})-f(x_{n_j})\| \to 0$ as $i,j \to \infty$. But then $(f(x_{n_i}))$ is Cauchy sequence. Any Cauchy sequence of real numbers is convergent, hence bounded. We have arrived at a contradiction to the fact that $\|f(x_{n_i})\|\to \infty$.
H: Is the braid group hyperbolic? The braid groups satisfy a number of properties that one would expect of a hyperbolic group, liking having a solvable word problem, and having exponential growth. Are the braid groups hyperbolic groups? If not, is there any obvious property of hyperbolic groups showing that they are not? AI: Outside of the one and two strand case they are not hyperbolic. One obstruction is that braid groups on $n>2$ strands, $B_n$, contain $\mathbb Z^2$ subgroups which can not happen in hyperbolic groups. First note that $B_3$ is a subgroup of $B_n$ for $n>2$. Order the strands and name them $b_1,b_2,b_3,...,b_n$. Let $\sigma$ be the element which braids $b_1$ around $b_2,b_3$ where $b_1$ goes around the back of those two and then cross in front and let $\rho$ be the element which braids $b_2,b_3$ (basically an element in $B_2$). These two elements "obviously" commute so we get $\langle \sigma, \rho \rangle \cong \mathbb{Z}^2$ as a subgroup of $B_n$ for $n>2$ (if you want a formal proof there is enough detail in A Primer on Mapping Class Groups by Farb and Margalit). In the case of $B_1,B_2$ we have the trivial group and $\mathbb{Z}$ respectively and these are hyperbolic but not in an "interesting" way. You mention that one of your motivations is that there do seems to be some similarities to hyperbolic groups and it turns out this is a really good observation which gets into some interesting math. I won't be giving much details but it seems good to mention some of these ideas. There is a different way to describe braid groups and it is as mapping class groups of punctured disk (basically the group of topological symmetries up to a natural identification). Very roughly you can think about punctures moving around each other in the disk as braids moving around each other. If you "plot" this with a time axis you can see the punctures moving "drawing" the braid. This becomes important because now you get tools from surface theory and their mapping class groups. The mapping class group acts naturally on something called the curve complex which, somewhat surprisingly, is $\delta$-hyperbolic, proved by Masur and Minsky, and infinite diameter (you can choose $\delta =19$). Lots of the hyperbolic behavior can actually be seen in this action on the curve complex. Further, Masur and Minsky developed a way of studying the geometry of mapping class groups through the curve complex of the surface and curve complexes of all its (essential) subsurfaces by "piecing together" the geometric information of the curve complexes in some consistent way.
H: What is this notation $\mathbb{Z}_3[x]_{x^2+1}$? What is this notation? $$\mathbb{Z}_3[x]_{x^2+1}$$ I know $\mathbb{Z}_3[x]$ denotes the polynomials with coefficients in $\mathbb{Z}_3$. What does the $x^2+1$ bit denote? AI: That is the notation used for the localization of the ring $\mathbb Z_3[x]$ at the multiplicative subset $S=\{1,(x^2+1),(x^2+1)^2,(x^2+1)^3,\dots\}$. That is, $\mathbb Z_3[x]_{x^2+1}=S^{-1}(\mathbb Z_3[x])$.
H: Estimating integral of $\frac{\log z}{z^2+a^2}$ over small upper-plane semi-circle I'm trying to calculate $$\int^\infty_0{\frac{\log z}{z^2+a^2}\mathrm{d}z}$$ I was able to calculate this using residue calculus and different half-plane and full keyhole contours, but have not been able to nail down an approximation of the inner semi-circle (or inner circle, depending on the contour) as the radius goes to zeros to show that the integral on the inner semi-circle goes to zero. The usual upper bound using the length of the curve and the maximum of the norm of the integrand didn't work in my mind. I also tried writing the integral over $[0, \pi]$ but did not get any insights from that either. AI: HINT: Note that we have for $a>\varepsilon$, $\|\varepsilon^2e^{i2\phi}+a^2\|\ge a^2 -\varepsilon^2$ and $$ \lim_{\varepsilon\to0} \frac{2\pi \varepsilon \log^2(\varepsilon)}{a^2-\varepsilon^2} =0 $$
H: Well defined function involving quotient spaces I wanna know how to prove that a function involving a cartesian product of quotient space is well-defined. Let's see this question: Bilinear form and quotient space In that question, I guess that $g$ is well-defined if $(u_1+U_0,v_1+V_0) = (u_2+U_0, v_2+V_0)$ implies $g(u_1+U_0,v_1+V_0) = g(u_1+U_0,v_1+V_0)$, that is, $f(u_1,v_1) = f(u_2,v_2)$, whenever we have $u_1-u_2 \in U_0$ and $v_1-v_2 \in V_0$. But I don't know how to get there, can you help me?? AI: Notice that if $u_1-u_2 \in U_0$ then $f(u_1-u_2,v)=0$ for any $v \in V$, that is, $f(u_1,v) = f(u_2,v)$ for any $v \in V$; and, if $v_1-v_2 \in V_0$ then $f(u,v_1-v_2) = 0$ for any $u \in U$, that is, $f(u,v_1) = f(u,v_2)$ for any $u \in U$. So $$f(u_1,v_1) = f(u_2,v_1) = f(u_2,v_2).$$
H: understand the proof of $\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}$ If $n \in \mathbb{N}^$, prove that \begin{align}\frac{2 n}{3} \sqrt{n}<\sum_{k=1}^{n} \sqrt{k}<\frac{4 n+3}{6} \sqrt{n}.\end{align} I am having trouble understanding the following proof of this problem: Let $a_{n}=(n+\lambda) \sqrt{n}, n \in \mathbb{N}^{}$, then \begin{align} \frac{\sqrt{n}}{a_{n}-a_{n-1}} &=\frac{\sqrt{n}}{(n+\lambda) \sqrt{n}-(n-1+\lambda) \cdot \sqrt{n-1}} \\ &=\frac{\sqrt{n}}{(\sqrt{n}-\sqrt{n-1})(2 n-1+\sqrt{n(n-1)})+\lambda(\sqrt{n}-\sqrt{n-1})} \\ &=\frac{n+\sqrt{n(n-1)}}{2 n+\sqrt{n(n-1)}-1+\lambda} \\ &=\frac{2}{3}+\frac{\sqrt{n(n-1)}-n+2(1-\lambda)}{6 n+3 \sqrt{n(n-1)}-3(1-\lambda)} \tag{1} \end{align} for $n \in \mathbb{N}^*$, we have $$\frac{1}{2}<n-\sqrt{n(n-1)} \leqslant 1 .$$ Substitute $\lambda=\frac{1}{2}, \frac{3}{4}$ into the formula (1), respectively, then we obtain that $$ \tag{2} \frac{2}{3}\left(n+\frac{1}{2}\right) \sqrt{n} \leqslant \sum_{k=1}^{n} \sqrt{k}<\frac{2}{3}\left(n+\frac{3}{4}\right) \sqrt{n}. $$ Can anyone explain to me how the inequality (2) was obtained. Any hlep would be appreciated. AI: If $\lambda = \frac 12,$ by using the inequality they provided : $$\dfrac{\sqrt{n}}{a_n-a_{n-1}}>\frac 23\iff \sqrt{n} >\frac 23\left(a_n-a_{n-1}\right ).$$ Sum this up for and then you will get one side by telescoping. You can do a similar thing for $\lambda=\frac 34$ as well.
H: How to solve $\log_2(x)+\log_{10}(x-7)=3$ using high-school math? A question given in a grade 12 "advanced functions" class, asks to solve $\log_2(x)+\log_{10}(x-7)=3$ with a hint to change bases. The given hint suggests the base of the second logarithm is 10, but when trying to massage the equation, how does one proceed after reaching $x(x-7)^{\log(2)}=8$ using only high school methods? It is possible to "cheat" by observing that $\log(1) = 0$ for any base, and that $\log_2(8) = 3$, so $x=8$ is a solution, but that only works because of the fortunate selection of constants. What is missing? AI: The only thing you need to prove is that $x=8$ is the only solution. Proof. $x\le 7$ is impossible because then the second $\log$ is not defined. If $7<x<8$ then the first $\log$ is $<3$ and the second $\log$ is $<0$, so the sum is $<3$. Finally if $x>8$ then the first $\log$ is $>3$ and the second $\log$ is $>0$, so the sum is $>3$. So the only possibility left is $x=8$. $\Box$
H: EGMO 2014/P3 : Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ We denote the number of positive divisors of a positive integer $m$ by $d(m)$ and the number of distinct prime divisors of $m$ by $\omega(m)$. Let $k$ be a positive integer. Prove that there exist infinitely many positive integers $n$ such that $\omega(n) = k$ and $d(n)$ does not divide $d(a^2+b^2)$ for any positive integers $a,b$ satisfying $a+b=n$. My progress: Really beautiful but hard problem! For $k=1$,we can take $n=2^{p-1}$, where p is an odd prime. Let's say for some $a+b=n$ and write $a=2^ke$ and $b=2^kf$ with $e, f$ odd and $0\le k<p-1$. If $d(n)\|d(a^2+b^2)$, then$$p\|d \left ( 2^{2k+1}\cdot \dfrac{e^2+f^2}{2} \right )=2^{2k+2}\cdot d\left ( \dfrac{e^2+f^2}{2} \right )$$. So $p\|d\left (\dfrac{e^2+f^2}{2}\right) $ .Now, for $p$ to divide $d\left (\dfrac{e^2+f^2}{2}\right) $, we should have $\left (\dfrac{e^2+f^2}{2}\right)=l^{p-1}\cdot x, l $ is a prime and $gcd(l,x)=1$. But note that both 2 and 3 does not divide $\left (\dfrac{e^2+f^2}{2}\right)$. But Max$(a^2+b^2)=4^{p-1}<5^{p-1}$ . So we are done for $k=1$ . I thought that this would be almost same for $k>1$ , but I am not able to prove. I have conjectured that for any $k$ we can take $n = 2^{p-1}j$ such that $j$ has only $k-1$ primes. But no progress!Please, if possible post hints rather than solution. Thanks in advance. AI: $\boxed{\text{Complete solution}}$ (The merit of the following solution is that it gives an explicit construction for $n$ with given $k$ satisfying the conditions.) Let $p_m$ denote the $m^{th}$ prime with $p_1=2,p_2=3,\ldots$ and so on. Take, for $k>1$, $$n=2^{p-1}p_2p_3\cdots p_k$$ for some suitable prime $p$ and work on it. Then $d(n)=2^{k-1}p$ and $\omega(n)=k$. The key observation is that $$d(n)\mid d(a^2+b^2)\implies p\mid d(a^2+b^2)\implies q^{p-1}\mid a^2+b^2$$ for some prime $q$. Now proceed considering different cases of $q$. Case 1 ($q>4$) Since $q^{p-1}\mid(a^2+b^2)$ then we have $$q^{p-1}\leq(a^2+b^2)\leq (a+b)^2=n^2=4^{p-1}p_2^2p_3^2\cdots p_k^2$$ Since $q>4$ we can choose sufficiently large prime $p$ such that $$q^{p-1}>4^{p-1}p_2^2p_3^2\cdots p_k^2$$ which is a contradiction! Hence for sufficiently large prime $p$, $n$ satisfies the condition. Case 2 ($q=3$) Since $-1$ is not a quadratic residue modulo $3$, $3^{p-1}\mid a^2+b^2$ implies $3^{p-1}\mid a^2,3^{p-1}\mid b^2$. This implies $3^{\frac{p-1}{2}}\mid a$ and $3^{\frac{p-1}{2}}\mid b$ which gives $$3^{\frac{p-1}{2}}\mid (a+b)=n$$ Take $p>3$ then we get $v_3(n)\geq 2$ but by construction $v_3(n)=1$. So for $p>3$, $n$, as constructed, satisfies the conditions. Case 3 ($q=2$) then we get $2^{p-1}\mid a^2+b^2$ and also by construction $2^{p-1}\mid n^2=(a+b)^2=a^2+b^2+2ab$. This implies $2^{p-2}\mid ab$. Then write $a=2^r\alpha$ and $b=2^s\beta$ where $r,s$ are both odd. Then $p-1=v_2(n)=v_2(a+b)=\mathrm{min}(r,s)$. Therefore $r\geq p-1$ and $s\geq p-1$. This implies $v_2(ab)=r+s\geq 2(p-1)$. Or $v_2(2ab)\geq 2p-1$. On the other hand $v_2(n^2)=2p-2$. So $v_2(a^2+b^2)=v_2(n^2-2ab)=\mathrm{min}(v_2(2ab),v_2(n^2))=2p-2$. Now try to prove why this will lead you to a contradiction! Remark: For establishing that there are infinitely many $n$ for a given $k$ we can consider numbers of the form $$2^{p-1}p_{m+2}p_{m+3}\cdots p_{m+k}$$ for $m\geq0$ and suitable primes $p$. The proof will be analoguous.
H: Function how to show whether it is pointwise or uniform convergent? My sequence is : $s_n=\frac{2}{3+n\|x\|}$ for real x and n $\in \mathbb{N}$ Now I have done this: $\lim_\limits{n \to 0} \frac{2}{3+n\|x\|}= \frac{2}{3}$ $\lim_\limits{ n \to \infty} \frac{2}{3+n\|x\|}= 0$ Now it is pointwise convergent to these 2. Now I want to show if it uniform convergent or not: $\left\|\frac{2}{3+n\|x\|}-0 \right\| \leqslant \left\|\frac{2}{nx}\right\| \leqslant \left\|\frac{2x}{nx}\right\|\leqslant \left\|\frac{2}{n}\right\| $ Now it is with this way because I have found a function for which this applies. Is this correct? Can I simply multiply the upper part by $2$ to make it independent from $x$ ? Edit: Now I have this For x= 1/n $\lim_\limits{ n \to \infty} sup \left\|\frac{2}{3+n\|1/n\|}-0 \right\| = 2/3 \neq 0 \ $ So it is not uniform convergent. Now my second function The sequence is : $s_n=\frac{2\|x\|}{3+n\|x\|}$ for real x and n $\in \mathbb{N}$ I have done this: $\lim_\limits{n \to 0} \frac{2x}{3+n\|x\|}= \frac{2\|x\|}{3}$ Is this now still pointwise convergent due to the function? $\lim_\limits{ n \to \infty} \frac{2\|x\|}{3+n\|x\|}= 0 ?$ $\lim_\limits{ n \to \infty} sup \left\|\frac{2\|x\|}{3+n\|x\|}-0 \right\| \leqslant \left\|\frac{2\|x\|}{nx}\right\| \leqslant \left\|\frac{2}{n}\right\| $ So this is now 100% uniform convergent, right? AI: Most easy is to use sufficient and necessary condition of uniform convergence $$\lim_\limits{n \to \infty}\sup_{x \in \mathbb{R}}\|f_n(x)-f(x)\|=0$$ In your example function $s_n(x)=\frac{2}{3+n\|x\|}$ is even on $\mathbb{R}$, and you can find, that sup reached in $x=0$. Hence?
H: Proving that the $p$-adic topology is a neighborhood topology This is (part of) Exercise 7 from Section 2.2 on page 22 of Topology and Groupoids, by Brown. I would appreciate any feedback regarding the quality of my proof. Exercise: Let $X = \mathbb{Z}$ and let $p$ be a fixed integer. A set $N \subseteq \mathbb{Z}$ is a p-adic neighborhood of $n \in \mathbb{Z}$ if $N$ contains the integers $n + mp^r$ for some $r$ and all $m = 0, \pm 1, \pm 2, \dots$ (so that in a given neighbourhood $r$ is fixed but $m$ varies). Prove that the $p$-adic neighborhoods form a neighborhood topology on $\mathbb{Z}$, the $p$-adic topology. Is this topology the same as the order topology? The discrete topology? The indiscrete topology? More information: The neighborhood axioms I am using are as follows: If $N$ is a neighborhood of $x$, then $x \in N$. If $N$ is a subset of $X$ containing a neighbourhood of $x$, then $N$ is a neighbourhood of $x$. The intersection of two neighbourhoods of $x$ is again a neighbourhood of $x$. Any neighbourhood $N$ of $x$ contains a neighbourhood $M$ of $x$ such that $N$ is a neighbourhood of each point of $M$. My attempt: First axiom: If $N$ is a neighborhood of $x$, then clearly $x \in N$ because we can let $m = 0$ in the expression $x + mp^r$. Second axiom: If $M$ is a neighborhood of $x$, and $M \subseteq N$, then clearly $N$ contains all the numbers $x + mp^r$ that $M$ contains, so $N$ must be a neighborhood of $x$ as well. Third axiom: If $M$ and $N$ are neighborhoods of $x$, then all $x + mp^{r_M} \in M$, and all $x + mp^{r_N} \in N$ where in each case $m \in \mathbb{Z}$ and $p, r_M$ and $r_N$ are all fixed integers. If $r_M = r_N$ then call the common exponent simply $r$, and both sets contain all $x + mp^r$ so their intersection must still be a neighborhood of $x$. If $r_M \neq r_N$, then assume without loss of generality that $r_M < r_N$. Then we may write $r_N = r_M + k$, where $k \in \mathbb{N}$, so that any number $x + mp^{r_N} \in N$ may be written as $x + mp^k p^{r_M}$ which is clearly an element of $M$. Therefore, all the $x + mp^{r_N} \in N$ are also in $M$, so the intersection $M \cap N$ is still a neighborhood of $x$. Fourth axiom: Assume that $N$ is a neighborhood of $x$. That means $N$ contains all numbers $x + mp^r$ for $m \in \mathbb{Z}$, fixed $p$ and fixed $r$. Now consider $M = \{ x + mp^r \colon m \in \mathbb{Z} \}$. In other words $M$ is only those numbers $x + mp^r$. Then $M$ is also a neighborhood of $x$ and $M \subseteq N$. We need to show that $N$ is a neighborhood of some $x + m_0 p^r \in M$. In other words, $N$ must contain all numbers $(x + m_0 p^r) + mp^r$. This is true because $(x + m_0 p^r) + mp^r = x + (m_0 + m)p^r$ which is clearly an element of $N$ because $m_0 + m$ is just another integer. Comparisons to the order, discrete and indiscrete topologies: The $p$-adic topology on $\mathbb{Z}$ is distinct from the order topology on $\mathbb{Z}$ (which is equivalent to the discrete topology on $\mathbb{Z}$) because the order topology allows for finite neighborhoods, while the $p$-adic topology does not. The $p$-adic topology on $\mathbb{Z}$ is also distinct from the indiscrete topology because the indiscrete topology allows only $\mathbb{Z}$ itself to be a neighborhood of any point, whereas the $p$-adic topology also allows smaller sets to be neighborhoods of a point. AI: This is fine. My only comment is that there’s no reason to split the argument for the third axiom into cases. Just let $r=\max\{r_M,r_N\}$, and observe that $$\{n+mp^r:m\in\Bbb Z\}\subseteq\{n+mp^s:m\in\Bbb Z\}$$ whenever $r\ge s$, since $n+mp^r=n+mp^{r-s}p^s$.
H: Question about the proof of $(\forall a)[a \in \mathbb{F} \rightarrow -(-a) = a]$ The proof of the proposition $(\forall a)[a \in \mathbb{F} \rightarrow -(-a) \in \mathbb{F}]$ is given on page 6 of the following link. https://www.math.ucdavis.edu/~emsilvia/math127/chapter1.pdf We have that $e$ denotes the additive identity and (A2) is the field axiom for commutativity of elements in a field. That is, (A2) is the axiom $$(\forall x)(\forall y)[x,y \in \mathbb{F} \rightarrow x + y = y + x]$$ And (A4) is the axiom $$(\forall a)[a \in \mathbb{F} \rightarrow (\exists (-a))(-a \in \mathbb{F} \rightarrow (-a) + a = a + (-a) = e)]$$ So, (A4) is the additive inverse property for elements in a field. Lemma: The additive inverse of an element of a field is unique. Proposition: $(\forall a)[a \in \mathbb{F} \rightarrow -(-a) \in \mathbb{F}]$ proof. Suppose that $a \in \mathbb{F}$. By the additive inverse property, $-a \in \mathbb{F}$ and $-(-a) \in \mathbb{F}$ is the additive inverse of $-a$; i.e $$-(-a) + (-a) = e$$ Since $-a$ is the additive inverse of $a$, $$(-a) + a = a + (-a) = e$$ From the uniqueness of additive inverses, we conclude that $-(-a) = a$. In the given proof, is axiom (A2) implicitly used to show that $-(-a) + (-a) = e$? I thought that we would have to show that $-(-a) + (-a) = (-a) + (-(-a)) = e$, by definition. AI: Yes, commutativity of addition is used implicitly to conclude that since $-(-a)+(-a)=e$, then also $-a+\big(-(-a)\big)=e$. The last sentence contains even more implicit argumentation. We know that $$a+(-a)=(-a)+a=e\;,$$ so we know that $-a$ is the additive inverse of $a$, but the same equalities show that $a$ is the additive inverse of $-a$. Uniqueness of that inverse then tells us that the only solution to the equation(s) $$x+(-a)=e=-a+x\tag{1}$$ is $x=a$. And since we know that $-(-a)$ is a solution to $(1)$, we conclude that $-(-a)=a$.
H: Finite ordinal Exponentiation I confused a little when i do arithmetic on ordinals especially multiplication is what i wrote right? : $(ω+1)$ = {$0,1,....ω$} $(ω+1)(ω+1)$ = sup({lexicographic Order($(ω+1)×(ω+1)$)}) = $ω²+1$ $(ω+1)(ω+1)(ω+1)$ = sup({lexicographic Order($(ω+1)×(ω+1)×(ω+1)$)})= $ω³+1$ $(ω+1)ⁿ$ = $ωⁿ+1$ and $(ω+k)ⁿ$ = $ωⁿ+k$ Im sorry for asking it because i couldn't find any library for python or website to check my calculation is right or not. Thanks. AI: Ordinal multiplication has the property that $\alpha\cdot(\beta+1)=(\alpha\cdot\beta)+\alpha$ for all ordinals $\alpha$ and $\beta$, so $$(\omega+1)\cdot(\omega+1)=\big((\omega+1)\cdot\omega\big)+(\omega+1)\;;$$ and $(\omega+1)\cdot\omega=\omega^2$, so $(\omega+1)\cdot(\omega+1)=\omega^2+\omega+1\ne\omega^2+1$. Then $$\begin{align} (\omega+1)^3&=(\omega+1)^2\cdot(\omega+1)\\ &=(\omega^2+\omega+1)\cdot(\omega+1)\\ &=(\omega^2+\omega+1)\cdot\omega+\omega^2+\omega+1)\\ &=\omega^3+\omega^2+\omega+1\;. \end{align}$$ Can you correct the rest of it from here?
H: Prove or disprove that $\sum_{n=1}^{\infty} \frac{x^{3/2}\cos(nx)}{n^{5/2}}$ is differentiable on $(0, \infty)$ Let $f(x) = \sum_{n=1}^{\infty} \frac{x^{3/2}\cos(nx)}{n^{5/2}}$ on $(0, \infty)$. (i) Is $f(x)$ differentiable on $(0, \infty)$? (ii) Does the series uniformly converge to $f$ on $(0, \infty)$? Any help or hints will be appreciated. $\\\\\\$ Edit: What I know about the differentiability of a sequence of functions is that If $(f_n(x))$ is defined on $I = [a,b]$ and (i) Each $f_n$ is differentiable on $I$. (ii) For some $x_0 \in I$, $(f_n(x_0))$ converges. (iii) $(f'_n)$ converges uniformly. Then $(f_n)$ converges uniformly, $f$ is differentiable on $I$ and $ f'(x) = \lim_{n \rightarrow \infty} f'_n(x)$ holds. The above case seems not to satisfy the conditions of the above theorem. At this point, I don't know how to proceed. AI: Hint for uniform convergence: If the series converged uniformly on $(0,\infty),$ then $$\sup_{0<x<\infty} \left \|\frac{x^{3/2}\cos (nx)}{n^{5/2}}\right \| \to 0$$ as $n\to \infty.$ Is this true?
H: Writing G as a product of groups $G$ is a connected, locally compact group satisfying the second axiom of countability, and $C$ is a discrete central subgroup of $G$ such that $G/C$ is compact. In the book I am reading (Varadarajan, Lie Groups, Lie Algebras, and their Representations, Lemma 4.11.1) it says that one can choose a compact set $D$ such that $G=C (\text{int} D)$. Why is this true? AI: Get yourself a compact neighbourhood $N=N^{-1}$ of the identity in $G$, and let $U=N^\circ$. Since $\bigcup_{n\geq 1}U^n$ is an open subgroup of $G$, and $G$ is connected, we have $\bigcup_{n\geq 1} U^n=G$. This means $CU^n=\{Cu_1\dots u_n\mid u_i\in U\forall i\}$, for $n\geq 1$, forms an open cover of $C\backslash G=G/C$ (since $C$ is central), which is compact. So there is a finite subcover. Now $(CU^n)_{n\geq 1}$ is nested, so there is some $n\in\mathbb{N}$ such that $CU^n=G/C$. Since $U\subseteq N$, we have $D=N^n$ is compact, $D^\circ\supseteq U^n$ and $C(D^\circ)=G$.
H: connected component and path connected component What are the connected components and path connected components of $\Bbb{Q}$ , $\Bbb{R}$ and $\Bbb{R}_\mathcal{l}$? Definition: A component $C$ of a topological space $X$ is a maximal connected subspace. I think every connected component of $\Bbb{R}_\mathcal{l}$ consists of just one point. Because, if we take any subset say $S$ with at least two distinct points of $\Bbb{R}$ then $S$ can't be connected. If $x,y\in \Bbb{R}$ be two distinct points, such that $x<y$ then $A=(-\infty,x)\cap S$ and $B=(y,\infty)\cap S$ are disjoint non-empty open sets in the subspace topology on $S$. Therefore, $S$ is disconnected. If the above proof is true for $\Bbb{R}_\mathcal{l}$. Similar proof for path component. Is this even true for $\Bbb{R}$? I mean $\Bbb{R}$ has components that consists of just one point. I believe $\Bbb{Q}$ has also connected components that consists of just one point. Any help or hint will be appreciable. Thanks! AI: You are right about $\Bbb R_\ell$ and $\Bbb Q$. For instance, $\{(a,b):a,b\in\Bbb R\setminus\Bbb Q\text{ and }a<b\}$ is a clopen base for $\Bbb Q$, so $\Bbb Q$ is totally disconnected (and even zero-dimensional). $\Bbb R$, however, is path-connected and hence also connected: for any $a,b\in\Bbb R$, the map $$f:[0,1]\to\Bbb R:x\mapsto a+(b-a)x$$ is a path from $a$ to $b$.
H: INMO : Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers Let $a$ and $b$ be two positive rational numbers such that $\sqrt[3] {a} + \sqrt[3]{b}$ is also a rational number. Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers. My first response was, isn't is obvious? but then I tried.. if $a=b$ then we prove it easily . So suppose $a\ne b$. Let $s=\sqrt[3]{a} + \sqrt[3]{b}$. then $s^3 = a + b + 3 s \sqrt[3]{a} \sqrt[3]{b}$ then the product $p=\sqrt[3]{a} \sqrt[3]{b}$ is a rational number. But what should I do next ? AI: From your work, we can write $$\frac{p}{\sqrt[3]{a}} + \sqrt[3]{a} = s,$$ so $$\sqrt[3]{a} = \frac{s\pm \sqrt{s^2-4p}}{2}.$$ Cube both sides and solve for $\sqrt{s^2-4p}$ (noting $s^2-p \neq 0$) to show it's rational, which then shows $\sqrt[3]{a}$ is rational.
H: Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ . Given that $x^2 + y^2 = 2x - 2y + 2$ , find the maximum value of $x^2 + y^2 + \sqrt{32}$ . What I Tried :- Since $x^2 + y^2 = 2x - 2y + 2$ , we have $2x - 2y + 2 + \sqrt{32}$ => $2(x - y + 1 + 2√2)$ . From this step I am not sure how to move forward . Also I tried to express $x^2 + y^2 + \sqrt{32} \leq S$ , so that in that way we can say that $x^2 + y^2 + \sqrt{32}$ is maximum at $S$ , but I couldn't do it . Can anyone help me ? Some hints or suggestions to this problem will be appreciated !! AI: Hint: $x^2+y^2=2x-2y+2$ is equivalent to $(x-1)^2+(y+1)^2=4$, which is the equation of a circle with the centre in $C=(1,-1)$ and radius $2$. To maximise $x^2+y^2+\sqrt{32}$ it is enough to first maximise $\sqrt{x^2+y^2}$, i.e. the distance from the origin $O=(0,0)$. Now, which point in that circle is the farthest from the origin? It will be the intersection of half-line $[OC)$ with the circle, i.e. the point $(x,y)=(1+\sqrt{2},-1-\sqrt{2})$, giving you the maximum value $x^2+y^2+\sqrt{32}=6+8\sqrt{2}$.
H: Showing that a function is Darboux Integrable using definition of integral I have been given a function $f:[0, 4]→ \mathbb{R}$ such that $f(x) = 0$ for all $x \ne 2$ and $f(2) = 2$, and told to show that $f$ is Darboux integrable in $[0,4]$. However, I don't understand how a function like this can be integrable, as it is mostly a constant function that equals $0$ and then a single disconnected point at $(2, 2)$. How is this function meant to be integrable? AI: Let $\{\mathcal P_n\}$ be a sequence of partitions of $[0,4]$ such that $\operatorname{gap}\mathcal P_n \rightarrow 0$ It is trivial to show that $\mathcal L(f,\mathcal P_n) = 0$ for any $n$. Then, you just need to show that $\mathcal U(f,\mathcal P_n) \rightarrow 0$, which you can do by using the fact that $\mathcal U(f,\mathcal P_n) \le 2 \cdot \operatorname{gap}\mathcal P_n$.
H: Standard symplectic form on a sphere is an area form in cylindrical coordinates. Consider a symplectic form $\omega_x(\xi,\nu)=\langle x,\xi\times \nu\rangle$ on $S^2$ where $x\in S^2$ and $\xi,\nu\in T_x S^2$ and a parametrization $\phi:U\to S^2$ where $U=(0,2\pi)\times(-1,1)$ and $\phi(\theta,x_3)=(\sqrt{1-x_3^2}\cos(\theta),\sqrt{1-x_3^2}\sin(\theta),x_3)$. I want to show that $\phi^\omega_x=\omega_{st}$ where $\omega_{st}=d\theta\wedge dx_3$. My logic is the following: It's enough to show that $\phi^\omega_x(v,w)=\omega_{st}(v,w)$ for all $v,w\in T_{p}U$ where $\phi(p)=x$. However, since $T_pU$ is spanned by two vectors $\partial_{\theta}$ and $\partial_{x_3}$, then it's enough to show that $$\phi^\omega_x(\partial_{\theta},\partial_{x_3})=\omega_{st}(\partial_{\theta},\partial_{x_3})=1$$ where $\phi^\omega_x(\partial_{\theta},\partial_{x_3})=\omega_x(d\phi_p(\partial_{\theta}),d\phi_p(\partial_{x_3})).$ By direct computation i.e. finding $d\phi_p$ and computing $\omega_x(d\phi_p(\partial_{\theta}),d\phi_p(\partial_{x_3}))$ explicitly by using a definition, we can check that $\phi^\omega_x(\partial_{\theta},\partial_{x_3})=1$. Therefore, $\phi^\omega_x=\omega_{st}$ as they agree on the basis vectors. I just want to check if this proof makes sense. Thank you! AI: Yes that's correct (assuming that you are checking $\phi^\omega_x(\partial_{\theta},\partial_{x_3})=\omega_{st}(\partial_{\theta},\partial_{x_3})$ for all $p$). Note that the above argument works for any top form (not only for surfaces), since $\wedge^n \mathbb (R^n)^$ is one dimensional, then two top forms $\omega_1, \omega_2 \in \wedge ^n\mathbb R^n $ are the same if and only if $$ \omega_1 (e_1, \cdots, e_n ) = \omega_2(e_1, \cdots, e_n)$$ for any fixed basis of $\mathbb R^n$.
H: Smaller binary relation on the set $\mathbb Q$ Given two binary relations $R$ and $S$ over sets $A$ and $B$,then $R$ is said to be contained in $S$ if $$\forall a,b: (a,b) \in R \implies (a,b) \in S$$ Moreover $R$ is considered to be smaller than $S$ if $R$ is contained in $S$,but $S$ is not contained in $R$,e.g.$$R ⊊ S$$ Wikipedia gives an example: On the rational numbers, the relation $>$ is smaller than $≥$, and equal to the composition $> ∘ >$. I don't understand the example,why $>$ is smaller than $≥$? And what is the composition relation given by $> ∘ >$? AI: I don't understand the example,why > is smaller than ≥? If $a>b$ then we also have $a\geq b$, so $>$ is contained in $\geq$. Additionally, $1$ is a rational number such that $1 \geq 1$ but $1\not>1$, so $>$ is smaller than $\geq$. And what is the composition relation given by >∘>? By transitivity, $a > b$ and $b > c$ implies $a > c$ so it follows that $> \circ >$ is contained in $>$. On the other hand, if $a>b$ where $a,b$ are rational, then $m = \frac{a+b}{2}$ is also rational and $a > m > b$, so $>$ is contained in $> \circ >$. Therefore, $> \circ >$ equals $>$.
H: A combinatorial lemma The following combinatorial lemma is from Benson's polynomial invariants of finite groups, lemma 1.5.1 used to prove a generalisation of Noether's degree bound. The polynomial in $n$ variables $x_1x_2\ldots x_n$ satisfies the identity$$ (-1)^nn!x_1x_2\ldots x_n = \sum_{I\subseteq\{1,\ldots,n\}} (-1)^{\|I\|}\left( \sum_{i\in I} x_i\right)^n, $$where $I$ runs over all subsets of $\{1,2,\ldots,n\}$. The author did not give a prove of this formula, and I have no idea how to prove this identity since I'm not familiar with combinatorics. Can someone points a reference to me or help me with a proof? AI: Call the sum on the right $S_n(x_1,\ldots,x_n)$. Then $S_n(x_1,\ldots,x_n)$ is a homogeneous polynomial of degree $n$. But if we set $x_k=0$ we get \begin{align} &S_n(x_1,\ldots,x_{k-1},0,x_{k+1},\ldots,x_n)\\ &=\sum_{J\subseteq\{1,\ldots,k-1,k+1,\ldots,n\}}(-1)^{\|J\|}\left(\sum_{i\in J}x_i\right)^n +\sum_{J\subseteq\{1,\ldots,k-1,k+1,\ldots,n\}}(-1)^{\|J\|+1}\left(\sum_{i\in J}x_i\right)^n=0 . \end{align} The first sum here accounts for the $I$ with $k\notin I$ and the second with $k\in I$. Therefore $x_k$ is a factor of $S_n(x_1,\ldots,x_n)$ for each $k$. This means that $S_n(x_1,\ldots,x_n)$ is a constant multiple of $x_1\cdots x_n$. But the only term in the sum containing the monomial $x_1\cdots x_n$ is the one with $I=\{1,\ldots,n\}$, and in this term the monomial has coefficient $(-1)^nn!$. Therefore $$S_n(x_1,\ldots,x_n)=(-1)^nn!x_1\cdots x_n.$$
H: If $a_n=100a_{n-1}+134$, find least value of n for which $a_n$ is divisible by $99$ Let $a_{1}=24$ and form the sequence $a_{n}, n \geq 2$ by $a_{n}=100 a_{n-1}+134 .$ The first few terms are $$ 24,2534,253534,25353534, \ldots $$ What is the least value of $n$ for which $a_{n}$ is divisible by $99 ?$ We have to find. $a_n \equiv 0\pmod{99}$ $$ a_n=100a_{n-1}+134 \\ \implies a_n-a_{n-1}\equiv 35 \pmod{99}$$ Now how do I proceed from here? I Concluded some results, verified for smaller values, which proved to be wrong. How does doing this get me the $n$? Or am I even correct to proceed like this? The (unofficial) solution isn't very good either: $a_{3}=253534 \\ a_{4}=25353534 \\ \therefore a_{n}=2 \underbrace{535353 \ldots 53}_{(n-1) \text { Times } 53}4$ Now, $a_n \rightarrow$ divisible by $99 \Rightarrow$ by $\ 9 \ \& \ 11$ both. Sum of digits $=6+8(n-1)$ To be divisible by 9 $\mathrm{n}=7,16,25,34,43,52,61,70,79,88, \ldots$ $a_{7}=2\underbrace{535353535353 }_{6 \text { Times } 53} 4$ But $a_{7} \rightarrow$ Not divisible by 11 . $a_{16}=2\underbrace{5353535353 \ldots \ldots 53 }_{15 \text { Times } 53}4$ Similarly, $a_{16} \rightarrow$ Not divisible by 11 . Now, $n=88$ $a_{88}=2 \underbrace{5353 \ldots \ldots 53}_{87 \text { Times } 53}4$ Divisibility by $11 \rightarrow\|(2+3+3 \ldots \ldots)-(5+5+\ldots+ 4)\|$ $$ \begin{array}{l} =\|263-439\| \\ =176 \end{array} $$ $\therefore$ Least $n=88$ Hints are more appreciated than the solution. AI: Your method can work quite well. For all $1 \lt i \le n$, note $a_i - a_{i-1}\equiv 35 \pmod{99}$ means each $a_i$ is congruent to $35$ more than the previous one of $a_{i-1}$. Thus, starting from $a_1$ and repeating this $n - 1$ times, you get $$a_n \equiv a_1 + (n - 1)35 \equiv 35n - 11 \equiv 0 \pmod{99} \tag{1}\label{eq1A}$$ Since $99 = 9(11)$, you can split \eqref{eq1A} into $$35n - 11 \equiv 0 \pmod{9} \implies 8n - 2 \equiv 0 \pmod{9} \implies 4n \equiv 1 \pmod{9} \tag{2}\label{eq2A}$$ $$35n - 11 \equiv 0 \pmod{11} \implies 2n \equiv 0 \pmod{11} \implies n \equiv 0 \pmod{11} \tag{3}\label{eq3A}$$ Note \eqref{eq3A} means $n = 11k, \; k \in \mathbb{Z}$. You can thus use this to determine $k$ from \eqref{eq2A}.
H: Is $\zeta(x)>\frac{1}{x-1}$ when $1 I had originally found that $\lfloor\zeta(\zeta(n))\rfloor$, where $\zeta(n)$ is the Riemann Zeta Function, seemed to be relatively close to $\left\lfloor\frac{1}{\zeta(n)-1}\right\rceil$ for $n \in \mathbb{Z} $ such that $n \geq 2$. This led me to realize that for real values of $x$ between $1$ and $2$, $\zeta(n) \approx \frac{1}{x-1}$. I posted a question about this here. Thanks to Nilotpal Kanti Sinha's response to that post, I was informed the similarities between the two expressions is because of the Laurent series expansion of the Riemann Zeta function: $$\zeta(s) = \frac{1}{s-1} + \sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n (s-1)^n$$ where $\gamma_n$ is the $n$-th Stieltjes constant. I had also noticed that on the same interval, though, it seems that $\zeta(n)>\frac{1}{x-1}$. I tried proving this by first realizing this would just be the same as proving that, if $1<x<2$, $$\sum_{n=0}^\infty \frac{(-1)^n}{n!} \gamma_n (x-1)^n > 0$$ I don't know how to continue, though, as I don't fully understand the Stieltjes constants. Is it possible to prove or disprove this? Is it possible to prove this inequality for all $x>1$? Additionally, can it be proven that the above infinite sum expression is strictly increasing on the interval $x \in (1,2)$? AI: The inequality is correct, even for all $x>0$ (except $x=1$ where $\zeta(x)$ is not defined). We start with the representation $$ \zeta(s) = \frac s{s-1} - s\int_1^\infty \{u\} u^{-s-1}\,du $$ valid for $\Re s>0$; here $\{u\}=u-\lfloor u\rfloor$ is the fractional part. (Note that this representation immediately implies that $\zeta(x) - 1/(x-1)$ is strictly increasing for $x>0$.) Using $0\le\{u\}<1$ in the integrand, and setting $s=x$ to be a positive real number, yields $$ 0 \le \int_1^\infty \{u\} u^{-x-1}\,du < \int_1^\infty u^{-x-1}\,du = \frac1x; $$ therefore $$ \zeta(x) = \frac x{x-1} - x\int_1^\infty \{u\} u^{-x-1}\,du > \frac x{x-1} - x \cdot\frac1x = \frac1{x-1} $$ and, for good measure, $$ \zeta(x) = \frac x{x-1} - x\int_1^\infty \{u\} u^{-x-1}\,du \le \frac x{x-1}, $$ both valid for all $x\in(0,\infty)\setminus\{1\}$.
H: If the sum of a set of numbers is bigger, will the average of the set of numbers be bigger too? If the sum of a set of real numbers $X$ is bigger than the sum of a set of numbers $Y$, will the average of $X$ be bigger than $Y$ and vice versa? What is a proof either way? AI: You are in fact asking if $$a>b\implies \frac ac>\frac bd$$ for some numbers $a,b,c,d$ (in your case $c,d$ are integers, but that does not change much). This clearly does not hold, unless $c=d$.
H: What is the chance of getting a number less than $0.01$ when using Math.random()? When using Math.random() in JavaScript in Google Firebase Functions I get $17$ decimal places like this: $$0.35361536181287034$$ I wanted to see what the chance is of getting any number where the first two decimals are $0$ like this: $$0.00361536181287034$$ Can I say with some accuracy that it is about $1$ in $100$ because there are $100$ possibilities for the first $2$ decimals $(00-99)$ ? AI: As of 2015, every major JavaScript engine1 implements Math.random() using the xorshift128+ algorithm, which passes the TestU01 suite. So the accuracy is pretty good. The true probability that Math.random() < 0.01 is likely somewhere within a few Float64 epsilons of $0.01$, let's say $\pm10^{-17}$. Assuming your application is going to call this function less than $10^{36}$ times during its lifetime, its random noise is going to swamp such a tiny bias anyway. So, short answer: yes!
H: Problem with understanding. This is an example in the book of "Abstract algebra by Dummit & Foote " . I didn't understand the indicate part of this proof. Please anyone help. How did they get class equation from that ? AI: If an element $g\in G$ has order $p$, then it generates a cyclic subgroup $\langle g\rangle$ of order $p$. The centraliser $C(g)$ of $g$ contains this subgroup, but isn't all of $G$, since $g\notin Z(G)$. By Lagrange's theorem, $C(g)=\langle g\rangle$ is the only option left. The class equation for $G$ reads $$ \|G\|=\|Z(G)\|+\sum [G:C(g)] $$ where the sum goes over a certain collection of non-central elements $g\in G$. It doesn't really matter how many of them there are, but let's say there are $k$ terms in the sum. We know $\|G\|=pq, \|Z(G)\|=1$, and for any relevant $g$ we have $[G:C(g)]=\frac{\|G\|}{\|\langle g\rangle\|}=q$. Inserting this into the above class equation gives the equality you're asking about.
H: Trace of point begin circle under certain condition Let $P$ be a point on a circle $(x-2)^2 + (y-2)^2 = 4$. Now let $Q$ be a point on the same line with $O$, the origin, $P$, and also on the first quadrant. If $Q$ satisfies $\overline{OP}\times\overline{OQ}=6$, the trace of $Q$ is a circle, $(x-3)^2 + (y-3)^2 = 9$. Is there a generalized theorem or an easy explanation? AI: You are inverting in the circle $x^2+y^2=6$ (see inversive geometry on Wikipedia). The inverse of a circleline is a circleline. Since your starting circle doesn't pass through the origin, it ends up as a circle.
H: Property of a positive Lebesgue measure set in $\mathbb{R}^2$ Let $A\subset \mathbb{R}^2$ be a closed set of positive Lebesgue measure. Can we find positive Lebesgue measure sets $A_1,A_2\subset \mathbb{R}$ such that $A_1\times A_2\subseteq A$? Note that the above is not true if $A$ is not assumed to be closed. For example $$A=[0,1]\times [0,1]\setminus \{(x,y)\in [0,1]\times [0,1]:x-y\in \mathbb{Q}\}.$$ AI: You already have an example of a set $A$ of positive measure such $A _1 \times A_2$ is not contained in $A$ for any $A_1$ and $A_2$ of positive measure. By regularity of Lebesgue measure on $\mathbb R^{2}$ $A$ contains a compact set $K$ of positive measure. It follows that $A _1 \times A_2$ is not contained in $K$ for any $A_1$ and $A_2$ of positive measure.
H: $R$ is symmetric if and only if it is equal to its converse Given a binary relation $R$ over set $A$ ,prove the following statement: $R$ is symmetric if and only if it is equal to its converse. $\implies$ $R$ symmetric iff $\forall a,b \in A$: $$(a,b) \in R \iff (b,a) \in R$$ But how to show that it is equal to its converse? $\Longleftarrow$ $R$ is equal to its converse iff $$R=R^T \iff \left\{\left(a,b\right)\mid aRb\right\}=\left\{\left(b,a\right)\mid aRb\right\} \iff {\left(a,b\right)}={\left(b,a\right)}$$ How does this imply the relation is symmetric? AI: Look at the very definition ov converse relation: $$(x,y)\in R^T\iff (y,x)\in R$$ So $$R\;\;\text{is symmetric}\;\iff \left[(a,b)\in R\iff(b,a)\in R\right]\iff R=R^T\;$$ Another, slower, way: 1) Suppose $\;R\;$ is a symmetric relation and let $\;(a,b) \in R\;$, then also $\;(b,a)\in R\implies (a,b)\in R^;$, and skipping intermediate steps, we actually got here$\;(a,b)\in R\implies (a,b)\in R^T\implies \color{red}{R\subset R^T}\;$ . But also $\;(x,y)\in R^T\implies (y,x)\in R\;$, and since $\;R\;$ is symmentric then also $\;(x,y)\in R\;$ , and skipping intermediate steps we got $$(x,y)\in R^T\implies (x,y)\in R\implies \color{red}{R^T\subset R}$$ Both opposite inclusions in red above yield $\;R=R^T\;$ (2) Suppose now $\;R=R^T\;$, and let $\;(a,b)\in R\;$. then also $\;(a,b)\in R^T\;$ and from here $\;(b,a)\in R\;$. Hoping over intermediate steps we actually got here that $\;(a,b)\in R\implies (b,a)\in R\implies R\;$ is symmetric. And we're done with both directions...
H: Is $\mbox{Rank}(A + A^2) \leq \mbox{Rank} (A)$? Here, $A$ is an $n \times n$ matrix. I am not able to find any counterexample but not able to prove this as well. The examples I have tried so far shows me that $\mbox{Rank} (A + A^2) = \mbox{Rank} (A)$. I don't know how to show the inequality. AI: $\DeclareMathOperator{\rank}{rank}$ $\DeclareMathOperator{\Img}{Im}$ To show the inequality, note that $\rank(M)=\dim(\Img(M))$. Since $\Img(A + A^2)$ is a subspace of $\Img(A)$, the dimension inequality $\dim(\Img(A + A^2))\leq\dim(A)$ holds. Note that equality doesn't always hold. Consider $A=-I$ where $I$ is the identity. Then $A + A^2 = 0$. Assuming $n > 0$, we have $\rank(A+A^2)=0<n=\rank(A)$.
H: A question about subring of Rational Numbers involving prime and maximal ideals Edited : I have this particular question in abstract algebra assignment given to me. I have been studying algebra from Thomas Hungerford as a textbook. Question : Let R be a subring of $\mathbb{Q}$ containing 1 . Then which 1 of following is nessesary true. A. R is Principal ideal Domain. B. R contains infinitely many prime ideals. C. R contains a prime ideal which is a maximal ideal. D. for every maximal ideal m in R, the residue field R/m is finite. Attempt : I don't think ring is PID as it need not have a single element which will generate it. Unfortunately , for B, C, D I am clueless on how can they be approached. I understand one should give his attempt but I am unable to think anything about B,C and D. Any hints please!! AI: B doesn't necessarily hold - take $R = \mathbb{Q}$. D also doesn't necessarily hold - again, take $R = \mathbb{Q}$. It seems like both A and C hold. A - let $w$ be an ideal of $R$. Let $k = \mathbb{Z} \cap w$. Then we see that $k$ is an ideal of $\mathbb{Z}$. Take $n \in k$ s.t. $k = (n)$ since $\mathbb{Z}$ is a PID. Then $n$ generates $w$. For if we have fully simplified $a/b$ in $w$, then clearly $a \in k$ and therefore $a$ is a multiple of $n$. And since $a$ and $b$ are relatively prime, we can take $x, y \in \mathbb{Z}$ such that $xa + yb = 1$. Then $x (a/b) + y = 1/b$. Then $1/b \in R$. Then $a (1/b)$ is a multiple of $n$. Alternately, since $n \in k$, we have $n \in w$ and thus every multiple of $n$ is in $w$. Then $R$ is a PID. C - every nonzero ring has a maximal ideal, and every maximal ideal is prime. So $R$ has a maximal, prime ideal.
H: $x_1 + x_2 + x_3 + x_4 + x_5=5$ . Determine the maximum value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$. Let $x_1 , x_2 , x_3 , x_4 , x_5$ be non-negative real numbers such that $x_1 + x_2 + x_3 + x_4 + x_5=5$ . Determine the maximum value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$. Normally in such questions I use the fact that the equation is symmetric and thus extremum is attained when all the variables are equal , but this can not be done here , and I have spent quite a long time on this but nothing worth-mentioning came to my mind . Could someone please help me find the maximum value ? Thanks ! AI: $x_1x_2+x_2x_3+x_3x_4+x_4x_5=(x_1+x_3+x_5)(x_2+x_4) - (x_2x_5+x_1x_4) $ Now we try to find maximum value of $(x_1+x_3+x_5)(x_2+x_4)$ when $(x_1+x_3+x_5)+(x_2+x_4)=5$ And try to minimize the value of $(x_2x_5+x_1x_4) $. Take, $a=(x_1+x_3+x_5)$ and $b=(x_2+x_4)$ By , A.M. $\ge $ G.M. $\implies$ $\sqrt(ab) \le \frac{a+b}{2} \implies (ab) \le (\frac{5}{2})^2 $ So, max value of $(x_1+x_3+x_5)(x_2+x_4)$ is $(\frac{5}{2})^2 $ And , clearly, minimum value of $(x_2x_5+x_1x_4) $ is $0$. So , max value of $x_1x_2+x_2x_3+x_3x_4+x_4x_5$ is $(\frac{5}{2})^2 $
H: Question on determinant Given $x \in \Bbb R$ and $$P = \begin {bmatrix}1&1&1\\0&2&2\\0&0&3\end {bmatrix}, \qquad Q=\begin {bmatrix}2&x&x\\0&4&0\\x&x&6\end {bmatrix}, \qquad R=PQP^{-1}$$ show that $$\det R = \det \begin {bmatrix}2&x&x\\0&4&0\\x&x&5\end {bmatrix}+8$$ for all $x \in \Bbb R$. My attempt: $\|R\|=\frac{\|P\|\|Q\|}{\|P\|}=\|Q\|$ $$\left\|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right\|=\left\|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right\|+\left\|\begin {array}&2&x&x\\0&4&0\\x&x&1\end {array}\right\|=\left\|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right\|+8-4x^2$$ What's my mistake? AI: You used multilinearity incorrectly. It should be $$\left\|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right\|=\left\|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right\|+\left\|\begin {array}&2&x&x\\0&4&0\\0&0&1\end {array}\right\|=\left\|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right\|+8.$$
H: Proof of a group (in the context of finite groups) my task is the following: Let $G$ be a finite set with an inner connection $\circ: G \times G \rightarrow G$, which is associative and for which a neutral element exists in $G$. In addition, for all $a,b,c \in G$, it applies that from $a \circ b = a \circ c$ also $b = c$ follows. Show that $ G $ is a group. My idea was: The following axioms must apply to a group: Closure Associativity Neutral element Existence inverse The first 3 axioms are already given according to the task definition (please correct me if I am wrong). So I only have to show the inverse. Here's what I'm thinking: Be $a \in G$ arbitrary. We want to show that there is a $a′\in G$ with $a \circ a′$ being the neutral element of $G$. Consider the left multiplication with a, i.e. the map $l_a : G \to G$ defined by $ x \rightarrow a \circ x$. Which property of $l_a$ can I deduce from the precondition $a \circ b = a \circ c \implies b = c$ and how can I continue? Could someone help me please. Thanks in advance. AI: You have to use that $l_a(x)=a \circ x$ is bijective since $G$ is finite, you know that it is injective since $a \circ x=a\circ y$ implies $x=y$. There exists $a^{-1}$ such that $a \circ a^{-1}=e$, use the associativity to show that $a^{-1}$ is a left inverse of $a$: $$\begin{align} (a^{-1} \circ a) \circ a^{-1}&=a^{-1} \circ(a \circ a^{-1})\\ &=a^{-1} \circ e\\ &=a^{-1}. \end{align}$$ Since $e \circ a^{-1}=(a^{-1} \circ a) \circ a^{-1}$, we deduce that $a^{-1}\circ a =e$.
H: Unimodularity: How are these notions related? 1. Definitions We call a Hopf algebra $H$ unimodular if the space of left integrals $I_l(H)$ is equal to the space of right integrals $I_r(H)$. We call a square integer matrix $M$ unimodular if $det(M)=\pm 1$. Apparently, there exists a notion of unimodular group: "a locally compact group whose left Haar measure equals its right Haar measure.” 2. Questions (How) are these three notions related? I haven't heard of unimodular groups, let alone locally compact groups or the Haar measure before. However, looking at this answer here, the unimodularity of a Hopf algebra seems to be somehow related to unimodular groups. How so? AI: Yes, they are related. A Lie group $G$ is unimodular if and only if, for all $g\in G$, $\operatorname{Ad}g\colon\mathfrak{g}\to\mathfrak{g}$ is a unimodular matrix when you choose a basis of $\mathfrak{g}$. For a locally compact group $G$, we know there exists, up to multiplicative constants, a unique left-invariant Haar measure and similarly right-invariant. We can also build a Hopf algebra of continuous functions $H\subseteq\mathbb{R}^G$. Now you can choose to integrate a function with respect to the left-invariant Haar measure or right-invariant Haar measure and these give you the two integrals for the Hopf algebra. If $G$ is unimodular, any left-invariant Haar measure is also right-invariant, so the two integrals agree, giving you the notion of unimodular Hopf algebra.
H: Normal distribution sample Since I'am beginner in statistics I'm stuck in simple exercise so will appreciate any help. I have mean, standard deviation and probability p(x) and need to get x. Here is the Exercise The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.3 days and a standard deviation of 2.1 days The 90th percentile for recovery times is? I know that it's possible to get x from probability formula but I was wondering if there is easier way to get it. AI: Look up a table for standard normal distribution. The random variable $X$ is distributed normally. Let $Z = \frac{X-\mu}{\sigma}$ be the standardisation. Then we're interested in $$ \mathbb P\left (Z\leq \frac{t_0-\mu}{\sigma}\right ) = 0.9 $$ The table can be used to find the closest desirable value and one can then solve for $t_0$. Also, verify quickly what happens when percentile increases or decreases. Or how affecting mean/st deviation changes result. Here is a flexible table and graph for standard normal distribution to try
H: Identity for sets Identity for a set X: The set X has an identity under the operation if there is an element j in set X such that j * a = a * j = a for all elements a in set X. According to my college book the counting numbers don' t have an identity. But for the operation * there is number 1 (= j ) such that a * 1 = 1 * a = a for all element a in set X. Why do natural numbers have no identity? AI: The operation is crucial here. $(\mathbb{N},\cdot)$ does have an identity, namely $1$ as you observed. For $(\mathbb{N},+)$ the identity would be $0$ and so it depends on your definition of $\mathbb{N}$ whether you include $0$ or not. I assume that your book uses the definition $\mathbb{N} = \{1,2,3,\ldots\}$ and so no number $j \in \mathbb{N}$ satifies $j + a = a$ for all $a \in \mathbb{N}$.
H: Is AC necessary to prove that whenever $S$ is a set of non-empty sets, there exist a set $A$ which for each $s\in S$, $s\cap A\neq\varnothing$? Is this theorem need Axiom of Choice ? or is it equal to that? Let $S$ be set of non-empty sets, there exist a set $A$ which for each $s\in S$, $s\cap A\neq\varnothing$ I think for making a set "$S$" such that satisfy $\forall s\in S, s\cap A\neq\varnothing$ we need choice function and we need AoC but apparently i was wrong. I don't know where i made a mistake. AI: No, the axiom of choice is not needed, since $A=\bigcup S$ works. The mistake is that knowing that $A\cap s\neq\varnothing$ is not enough to build a choice function: this intersection might contain many elements for every $s$, and then you need to choose one for every $s$ in order to get a choice function
H: Local representation of one map with respect to another map Let $M$ be a smooth manifold of dimension $m$ and $(U,\phi)$, $(V,\psi)$ be two maps on $M$ such that $U\cap V\neq\emptyset$. I will write $\phi=(\phi^1,\dots,\phi^m)$ and $\psi=(\psi^1,\dots,\psi^m)$. I want to prove that on $U\cap V$, we can write (using Einstein notation) $$\frac{\partial}{\partial \phi^k}=\frac{\partial \psi^i}{\partial\phi^k} \frac{\partial}{\partial \psi^i}$$ for all $k\in\mathbb N\cap[1,m]$. My attempt: See my answer below. My question: Is there a more elegant proof? AI: Let $p\in U\cap V$ and $f\in\mathcal C^\infty(M)$. Then, letting $E_k$ denote the $k$th basis vector of $\mathbb R^m$ and $\mathrm d$ the differential, \begin{split} \frac{\partial f}{\partial \phi^k}(p)&=\left(\frac{\partial (f\circ\phi^{-1})}{\partial x^k}\right)(\phi(p))\\ &=\mathrm d(f\circ\phi^{-1})\vert_{\phi(p)}(E_k)\\ &=\mathrm d(f\circ\psi^{-1}\circ\psi\circ\phi^{-1})\vert_{\phi(p)}(E_k)\\ &\overset{{\text{chain rule}}}=\mathrm d(f\circ\psi^{-1})\vert_{\psi(p)}\big(\mathrm d(\psi\circ\phi^{-1})\vert_{\phi(p)}(E_k)\big)\\ &= \mathrm d(f\circ\psi^{-1})\vert_{\psi(p)}\left(\frac{\partial(\psi\circ\phi^{-1})}{\partial x^k}(\phi(p))\right)\\ &=\frac{\partial(f\circ\psi^{-1})}{\partial x^i}(\psi(p)) \cdot \frac{\partial(\psi^i\circ\phi^{-1})}{\partial x^k}(\phi(p))\\ &=\left.\left(\frac{\partial \psi^i}{\partial\phi^k} \frac{\partial}{\partial \psi^i}\right)\right\vert_p f. \end{split}
H: Given $M\in\mathbb{R}^{m\times n}$ and $v_1,v_2\in\mathbb{R}^n,$ find $A\in\mathbb{R}^{m\times n}$ such that $Mv_1=Av_2$? Suppose that we are given a matrix $M\in\mathbb{R}^{m\times n}$ and two vectors $v_1,v_2\in\mathbb{R}^n$. Under which conditions there exists a matrix $A\in\mathbb{R}^{m\times n}$ such that $Mv_1=Av_2$? In general such matrix does not always exist (say, if $v_2$ is a zero vector and $Mv_1$ is not a zero vector, then, clearly, no such $A$ exists) and so the question is about sufficient conditions under which such matrix exists. AI: The linear map $v \mapsto \frac{\langle v, v_2 \rangle}{\langle v_2, v_2 \rangle} Mv_1$ maps $v_2$ to $Mv_1$ unless $v_2 = 0$. So the answer is: always as long as $v_2 \neq 0$. Of course, if $Mv_1 = 0$, then also $v_2 = 0$ works.
H: Complex integration around unit circle centresd at origin I am trying masters entrance exam question papers and I am unable to solve this particular question, so I am asking it here. Let C denote the unit circle centred at origin in $\mathbb{C} $ then find value of $$\frac{1}{2\pi i}\int_{C} \|1+z+z^2\|^2 dz$$. Attempt: Integration of any analytic function along a closed curve is 0 but the function is not analytic , also as ${\|z\|}^2 $ is analytic anywhere except 0 so I can't use residue theorem. So, I am struck and kindly give hint!! AI: Hint: The integral can be written as $$\frac 1 {2\pi i} \int_0^{2\pi} (1+e^{i\theta}+e^{2i\theta})( \overline {1+e^{i\theta}+e^{2i\theta}}) ie^{i\theta} d\theta.$$ Just expand the product and integrate term by term.
H: Series. Uniformly convergent on $\Bbb R$ vs. any interval $[-K,K]$ As the title indicates I am slightly unsure about a setup used for convergence of a series. I was wondering if there is/what the difference is between showing that a series is uniformly convergent on $\Bbb R$ vs. uniformly convergent in any interval $[-K,K]$, for fixed $K$ where $0<K<\infty$. If I want to show that the sum function is continuous in any $x$ in the real numbers I would just choose $K=\text{numerical($x$)}$ and obtain continuity on $[-x,x]$, right? AI: I will assume that your series is made up of terms which are continuous. The sum function being continuous is only a necessary condition for uniform convergence. It is not sufficient. The series $\sum \frac x {n^{2}}$ is uniformly convergent on $[-K,K]$ for any $K >0$ (by M-test) but it is not uniformly convergent on $\mathbb R$ because the general term $\frac x {n^{2}}$ does not tend to $0$ uniformly on $\mathbb R$. The sum is $\frac {\pi^{2}} 6 x$ which is continuous on the whole line but still the series is not uniformly convergent.
H: Elements of a subgroup and explanation. Find all of the elements in the subgroup K = $\langle (12)(34),(125)\rangle \leq S_5$. I understand that you need to take the powers of the given elements and products, but I don't understand how for example $\beta^2=(125)(125)=(152)$. Where do you get $(152)$ from? AI: $(1 \space 2 \space 5)$ signifies the permutation that sends $1$ to $2$, sends $2$ to $5$, sends $5$ to $1$, and fixes all other elements. We understand $(1 \space 5 \space 2)$ in a similar way. Both of these permuataations are examples of cycles because they consist of precisely one cycle involving some of the numbers, and leave the rest of the numbers fixed. Any permutation in $S_5$ (or any $S_n$) can be written as a product of disjoint (i.e., non-overlapping) cycles. So we want to figure out how to write the permutation $$ (1 \space 2 \space 5)(1 \space 2 \space 5) $$ in this way. Remember to read from right to left. What happens to $1$? Starting from the right, $1$ is sent to $2$ by the righthand copy of $(1 \space 2 \space 5)$. Then $2$ is sent to $5$ in the lefthand copy of $(1 \space 2 \space 5)$. So, altogether, moving through both copies, $1$ is sent to $5$. Now we want to continue making a cycle and ask what happens to $5$. If you follow the same steps you'll find that $5$ is sent to $2$. Then ask what happens to $2$ and you'll see it is sent to $1$. So this gives us the cycle $(1 \space 5 \space 2)$. To conclude that this is the only cycle we need in the product, we just have to observe that the rest of the numbers ($3$ and $4$) are fixed by $(1 \space 2 \space 5)(1 \space 2 \space 5)$. So there are no more cycles in the product.
H: The definition of the set of positive integers in "Topology 2nd Edition" by James R. Munkres. I am reading "Topology 2nd Edition" by James R. Munkres. In Ch. 1.4, Munkres defines the set of real numbers $\mathbb{R}$ with the field axioms (including completeness), and then defines $\mathbb{Z}_{+}$ as the smallest inductive set in $\mathbb{R}$, as follows: A subset $A$ of the real numbers is said to be inductive if it contains the number $1$, and if for every $x$ in $A$, the number $x+1$ is also in $A$. Let $\mathcal{A}$ be the collection of all inductive subsets of $\mathbb{R}$. Then the set $\mathbb{Z}_{+}$ of positive integers is defined by the equation $$\mathbb{Z}_{+} = \bigcap_{A\in \mathcal{A}} A.$$ Munkres didn't define $\mathbb{Z}_{+} := \{1, 1 + 1, 1 + 1 + 1, \dots\}$. Why? AI: In general, definitions containing "$\dots$" means that everybody can intuitively understand them, but they don't provide details for a formal and rigorous definition of the object. What is the problem? If you want to prove some properties about an object defined by means of "$\dots$", since you don't have a rigorous definition, you don't know exactly how to formally prove it, even though you intuitively understand what you have to prove: your proof will be inevitably hand-waved, and this could be a way to overlook some important and unexpected details (for instance, the fact that you need some further hypothesis to prove the desired property). So, your definition of the set of positive integers as $\mathbb{Z}_+= \{1, 1+ 1, 1+ 1 + 1, \dots\}$ is perfectly understandable but then if you want to prove something about $\mathbb{Z}_+$, what do you do? Thanks to the rigorous and formal definition of $\mathbb{Z}_+$ as the smallest inductive subset of $\mathbb{R}$, it is clear which are the elements of $\mathbb{Z}_+$ and how you can use and refer to them. By the way, there are many different but equivalent ways to define $\mathbb{Z}_+$ formally and rigorously. As @Théophile pointed out in his witty comment, topologists love unions and intersections so this is a possible reason why, in his topology handbook, Munkres defined $\mathbb{Z}_+$ as the intersection of some subsets of $\mathbb{R}$. This is not only a joke, but also due to the fact that this definition (among all the possible ones) is maybe the handiest one to deal with positive integers in a topological context, where you usually cope with intersections and unions.
H: A deduction in 1st course of complex analysis if a particular series is given absolutely I am trying assignment problems in complex analysis and I couldn't deduce the reasoning behind a particular Statement. Suppose f is holomorphic in an open neighborhood of $z_{0} $ $\epsilon $ $\mathbb{C} $. Given that the series $\sum_{n=0}^{\infty} f^{n} (z_{0}) $ converges absolutely, then how can we conclude that f can be extended to an entire function? I know of a result that uniform convergence on compact sets implies analyticity. By M-test (Uniform convergence ) I can deduce that f is uniformly convergent locallyand on compact intervel it is entire. But, can it be extended on whole of $\mathbb{C} $ ? I think it cannot be as convergence is not uniform everywhere. Am I right in my reasoning? AI: Convergence of $\sum f^{(n)}(z_0)$ implies that $f^{(n)}(z_0) \to 0$. In particular this implies boundedness of $(f^{(n)}(z_0))$ and this is enough to conclude that $\frac {\sum f^{(n)}(z_0)} {n!} (z-z_0)^{n}$ converges for every complex number $z$. This series has radius of convergence $\infty$ and its sum is an entire function.
H: Why did we call a row operation "elementary"? Why we called the three actions of row operation "elementary"? Is there a thing called "advanced" or "complicated" row operation? I've seen the word "non-elementary" row operation is used to describe things like $R_1-R_2$, which is not written in the conventional $-1R_2 + R_1$. Is this usage correct? In particular, what should $R_1-R_2$ be called? \begin{align} &\text{a) A non-elementary row operation} \\ &\text{b) An elementary row operation} \\ &\text{c) Just a row operation} \\ &\text{d) It is not a row operation} \end{align} AI: The sense of "elementary" here is that all the operations that preserve the row-space of a matrix can be be produced by combining various elementary row operations. Thus these are the elementary steps that can be taken to calculate a (reduced) row echelon form of a matrix, a basic tool for solving several kinds of problems involving the row-space of a matrix and the solutions (if any) of a linear system of equations. With regard to what $R_1 - R_2$ ought to be called, something is missing from the description. It would indeed be an elementary row operation if this "new row" immediately replaces $R_1$. If you wanted it to replace $R_2$, you would have to perform that row operation by combining two "elementary row operation" steps (first replace $R_2$ with $R_2 - R_1$, then multiply the resulting new second row by non-zero scalar $-1$). If you wanted to do something completely different with $R_1 - R_2$, then it would possibly either not be a row operation or not a row operation that preserved the row space of the matrix. For example, if you replaced $R_3$ with $R_1 - R_2$ (leaving $R_1,R_2$ as they are), you might well be making the row space of the matrix smaller.
H: Is weak connectivity sufficient for $0$ to be a simple eigenvalue of the weighted Laplacian? $L$ is the weighted Laplacian of a weakly connected directed graph $G$,$$L=D-A$$ with the $L_{ij} \leq 0$ when $i \neq j$, $L_{ii} \geq 0$ and $\sum_{i=1}^{n}L_{ij}=0$. My question is: Is $0$ a simple eigenvalue of $L$ ? AI: Counterexample 1: $$\begin{bmatrix} 2 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ which is to say $A \leftarrow B \rightarrow C$. Counterexample 2: In a previous version, you imposed a requirement that $D$ has a kernel with dimension at most $1$. You can make it zero, in fact, by doing this: $$\begin{bmatrix} 2 & -1 & -1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 0 & -1 \\ 0 & -1 & 0 & 1 & 0 \\ 0 & 0 & -1 & 0 & 1 \end{bmatrix}.$$ This is two cycles plus a single vertex that enters both cycles. Now there are two eigenvectors with eigenvalue zero (as can be seen by the fact that rows 2 and 4 are multiples of each other and rows 3 and 5 are as well).
H: Problem in Understanding the following steps in an integral While doing quartic integral,I was unable to understand the step that leads to the answer. Can somebody illustrates me how do we got from first integral to the next one? AI: Just looking at the antideriavtive $$I=\int\left(\frac{1-\cos (\theta )}{(1-a) \cos (\theta )+a+3}\right)^r\frac {d\theta}{1-\cos (\theta )}$$ making $$\Phi=\frac{1-\cos (\theta )}{(1-a)\cos (\theta ))+3} \implies \theta=\cos ^{-1}\left(\frac{(a+3) \Phi -1}{(a-1) \Phi -1}\right)$$ I let you the pleasure of calculating $d\theta$. Replacing $$I=\frac 1{2\sqrt 2}\int\frac{\Phi ^{r-\frac{3}{2}}}{ \sqrt{1-(a+1) \Phi}}\,d\Phi$$ Now $t=(a+1) \Phi$ to get $$I=\frac {(a+1)^{\frac{1}{2}-r} }{2\sqrt 2} \int \frac{t^{r-\frac{3}{2}}}{\sqrt{1-t}}\,dt$$
H: coordinates of focus of parabola Find the coordinates of focus of parabola $$\left(y-x\right)^{2}=16\left(x+y\right)$$ rewriting: $(\frac{x-y}{\sqrt{2}})^2=8\sqrt2(\frac{x+y}{\sqrt{2}})$ comparing with $Y^2=4aX$ $4a=8\sqrt2,a=2\sqrt2 $ $\Rightarrow$ coordinates of focus=2,2 Is this the correct approach? AI: Using Rotation of axes, let $$x=X\cos t- Y\sin t,y=X\sin t+Y\cos t$$ set $\cos2t=0$ to eliminate $XY$ If $t=\dfrac\pi4$ $$2Y^2=16(\sqrt2X)\iff Y^2=4(2\sqrt2)X$$
H: Is $\operatorname{div}(X)=\partial_i X^i$? We know that for a vector field $X$, $\operatorname{div}(X)$ is defined as $\nabla_I X^I$. This is not the same as $\partial_i X^i$ right? I would assume that $\nabla_I X^I=\partial_i X^i-X^j\Gamma_{ij}^i $. AI: Yes, you are correct. Another way to expand the divergence on a Riemannian manifold in local coordinates is to take the expression similar to what you find in the Laplace-Beltrami operator: $$ X^i \partial_i \mapsto \frac{1}{\sqrt{\|g\|}} \partial_i \left( \sqrt{\|g\|} X^i \right) $$ and you see that $\nabla_I X^I$ and $\partial_i X^i$ are equal only when either the volume element relative to the local coordinate chosen has a critical point, or that $X^i$ itself vanishes at that point.
H: Prove that $\frac{1}{2} (x-1) x + y$ is a bijection. (on p.45 Munkres Topology 2nd Edition) I am reading "Topology 2nd Edition" by James R. Munkres. On p.45, Munkres leaves it to the readers to show that $g$ is bijection: Show that $g(x, y) = \frac{1}{2} (x-1) x + y$ is a bijection from $\{(x, y) \in \mathbb{Z}_{+} \times \mathbb{Z}_{+} \mid y \leq x\}$ to $\mathbb{Z}_{+}$. I proved the above fact, but I am not sure my proof is right or not. And if my proof is right, please give me a better proof. My proof: $g$ is injective: Let $(x, y), (x^{'}, y^{'}) \in \mathbb{Z}_{+} \times \mathbb{Z}_{+}$ and $y \leq x$ and $y^{'} \leq x^{'}$. Let $(x, y) \neq (x^{'}, y^{'})$. If $x \neq x^{'}$, then $x < x^{'}$ or $x > x^{'}$. Without loss of generality, we can assume $x < x^{'}$. Then, $x+1 \leq x^{'}$, because if $x+1 > x^{'}$, then $0 < x^{'} - x < 1$ and $x^{'} - x \in \mathbb{Z}_{+}$. But there is no element $x \in \mathbb{Z}_{+}$ such that $0 < x < 1$. $$\frac{1}{2}(x-1)x+y \leq \frac{1}{2}(x-1)x+x =\frac{1}{2}x(x+1)\leq \frac{1}{2} (x^{'}-1)x^{'}<\frac{1}{2} (x^{'}-1)x^{'}+y^{'}.$$ If $x = x^{'}$ and $y \neq y^{'}$, then $y < y^{'}$ or $y > y^{'}$. Without loss of generality, we can assume $y < y^{'}$. $\frac{1}{2}(x-1)x+y = \frac{1}{2}(x^{'}-1)x^{'}+y < \frac{1}{2}(x^{'}-1)x^{'}+y^{'}$. So, $g$ is injective. $g$ is surjective: We prove by induction. $1 = \frac{1}{2} (1 - 1) 1 + 1$ and $1 \leq 1$. Assume that $n = \frac{1}{2} (x - 1) x + y$ and $y \leq x$. If $y < x$, then $n+1 = \frac{1}{2} (x - 1) x + (y+1)$ and $y+1 \leq x$. If $y = x$, then $n+1 = \frac{1}{2} (x - 1) x + y+1 = \frac{1}{2} (x - 1) x + x+1 = \frac{1}{2} x (x + 1) + 1$ and $1 < x+1$. So, $g$ is surjective. AI: Your proofs are fine. Here is a quicker proof of injectivity. Suppose $T_{x-1}+y=T_{x'-1}+y'$ for $y\le x$ and $y'\le x'$ where $T_x=x(x+1)/2$ is the $x$th triangular number. Without loss of generality let $x<x'$ and thus $$T_{x-1}+y\ge T_x+1\implies y\ge(T_x-T_{x-1})+1=x+1$$ which is a contradiction. Hence $x=x'$ from which it follows that $y=y'$. A quicker proof of surjectivity is that any positive integer can be written as a triangular number $T_{x-1}$ plus the remainder which is at most $x$, since $T_{x-1}+x=T_x$.
H: Consequence of inequality My question: Let $\Omega$ in $\mathbb{R}^n$ bounded. For all $\varepsilon>0$, there exists a constant $C(\varepsilon)>0$ such that $$ \label{lemma_gagliardo_nirenberg_2} \\|\varphi\\|_{L^2(\Omega)}^2 \le \varepsilon \\|\nabla \varphi\\|_{L^2(\Omega)}^2 + C(\varepsilon) \\|\varphi\\|_{L^1(\Omega)}^2 \quad\text{ for all }\varphi \in W^{1,2}(\Omega). $$ AI: From Gagliardo-Nirenberg it holds that $$ \\|u\\|^2_{L^2} \leq C_1 \\|\nabla u\\|_{L^2}^{2n/(n+2)} \\|u\\|_{L^1}^{4/(n+2)} + C_2 \\|u\\|^2_{L^1}. $$ Now you can apply Young's inequality $$ a b \leq \frac{\varepsilon}{C_1} a^p + C(\varepsilon) b^{p'}, \qquad a,b>0, $$ with $p = \frac{n+2}{n}$ (and $p' = \frac{n+2}{2}$).
H: Expected value with a die with 9 faces You have a die with $9$ faces, which are numbered $1, 2, 3, \dots,9$. All the numbers have an equal chance of appearing. You roll the die repeatedly, write the digits one after another, and you stop when you obtain a multiple of $3$. For example, you could roll a $4$, then a $1$, then a $7$. You would stop at this point, because $417$ is a multiple of $3$ but $4$ and $41$ are not. Find the expected number of times that you roll the die. I'm not sure how to find the expected value of something, can I get a solution to this problem? AI: Note that at any point, there are $3$ values from $1\to 9$ that would make the number a multiple of $3$ if rolled, since a number is a multiple of $3$ iff its sum of digits is. If the sum of the currently rolled digits mod $3$ is $1$, then $2,5,8$ would make the number a multiple of $3$. If the sum mod $3$ is $2$, then $1,4,7$ would make the number a multiple of $3$. If this is the first roll, then $3,6,9$ would make the number a multiple of $3$. So, the expected value is just $$\frac1{1/3}=3$$
H: A general summation of powers of roots of $x^2-x+q=0$ If $\alpha,\beta$ are the roots of the equation $x^2-x+q=0$ and $S_r=\alpha ^r + \beta ^r$, find $S_n$ in terms of $\sum a_iS_{n-i}$, where $a_i$ are constant terms for each $S_{n-1}$. I tried to observe a pattern in the $S_r$: $S_1=1, S_2=1-2q, S_3=1-3q,S_4=1-4q(1-2q)+6q^2$ I looked up for Vietta's theorem as well, did not work. AI: Note that the roots satisfy $x^2 = x - q$. In particular, they satisfy $$x^n = x^{n-1} - qx^{n-2}$$ for $n \ge 2$. In matrix form, we may write this as $$\begin{bmatrix}\alpha^n \\ \alpha^{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -q \\ 1 & 0 \end{bmatrix}\begin{bmatrix} \alpha^{n-1} \\ \alpha^{n-2}\end{bmatrix}; \quad n \ge 2.$$ Inductively, we get $$\begin{bmatrix}\alpha^n \\ \alpha^{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -q \\ 1 & 0 \end{bmatrix}^{n-1}\begin{bmatrix} \alpha \\ 1\end{bmatrix}; \quad n \ge 2.$$ And similarly, we have the same for $\beta$. Adding the similar equation for $\beta$ to the above equation gives us $$\begin{bmatrix}S_n \\ S_{n-1} \end{bmatrix} = \begin{bmatrix} 1 & -q \\ 1 & 0 \end{bmatrix}^{n-1}\begin{bmatrix} 1 \\ 2\end{bmatrix}; \quad n \ge 2.$$ Thus, we have $$[S_n] = \begin{bmatrix} 1 & 0\end{bmatrix}\begin{bmatrix} 1 & -q \\ 1 & 0 \end{bmatrix}^{n-1}\begin{bmatrix} 1 \\ 2\end{bmatrix}; \quad n \ge 2.$$ This is a "closed form" for a lenient enough definition. If you are not satisfied, you can do better by diagonalising (if $\alpha \neq \beta$) the above matrix and getting a better form for it. (However, that will not be particularly helpful since it would involve $\alpha^n$ and $\beta^n$.)
H: Interpolation of three positive values at 0, 1 and 2 by a polynomial with non-negative coefficients I am asking myself the question: Let $ y_0, y_1, y_2 $ be positive real numbers. Is there always a polynomial $ f $ with non-negative real numbers as coefficients which satisfies $ f( i ) = y_i $ for $ i = 0, 1, 2 $? Thanks for thinking about it. AI: Hint: Suppose $y_0>y_1$. What do you get when you evaluate $f(1)-f(0)$ ?
H: Showing the subset $\{(x_1,x_2) \in \mathbb{R}^2 : x_1 > x_2 \}$ is open The metric is the typical Euclidean metric, $ \sqrt{ (x_1 - x_2)^2 + (y_1 - y_2)^2 } $. I have solved this one, albeit with in my opinion quite excessive steps. I would love to know if there is a simpler way to do it. Below is my approach. Denote $A = \{(x_1,x_2) \in \mathbb{R}^2 : x_1 > x_2 \}$. Take $p \in A$ and write $p = (x_1, x_2)$, where $x_1 > x_2$. Want to find a ball around $p$ entirely contained in $A$. The smallest distance from $p$ to the line $x_2 = x_1$ is $\frac{x_1-x_2}{\sqrt 2}$. Take any smaller radius, such as $ r = \frac{x_1 - x_2}{10} > 0$. Want to show the open ball $B_r (x) \subset A$. Let $q \in B_r (x)$ and write $ q = (y_1, y_2)$. Want to show $y_1 > y_2$. Since $q \in B_r (x)$, we have $\sqrt{ (x_1 - y_1)^2 + (x_2 - y_2)^2 } < r$. This implies respectively $$\|x_1 - y_1\| < r \ \ \ \text{ and } \ \ \ \|x_2 - y_2\| < r,$$ owing to $\|a\| \leqslant \sqrt{a^2 + b^2} < r$. Adding these inequalities we get the condition $ \|x_1 - y_1\| + \|x_2 - y_2\| < 2r $. Using the symmetry of absolute value, write $\|x_1 - y_1\| = \|y_1 - x_1\|$. The triangle inequality in reverse gives us: $$\|y_1 - x_1 + x_2 - y_2\| \leqslant \|x_1 - y_1\| + \|x_2 - y_2\| < 2r$$ Exploiting the fact that $2r = \frac{x_1 - x_2}{5} < x_1 - x_2 $, we now write: $$ x_2 - x_1 < -2r < y_1 - x_1 + x_2 - y_2 < 2r < x_1 - x_2 $$ Adding and subtracting $x_1$ and $x_2$ respectively, we get: $$ 0 < y_1 - y_2 < 2(x_1 - x_2) $$ Since $x_1 > x_2$, this inequality is legitimate and shows $y_1 > y_2$. Done. Could there by a simpler way to show this? I found all the inequality juggling somewhat circuitous, which made me wonder whether there is a slightly more shorter and elegant way to show it. AI: You are off to a good start. Here is a shorter version of your argument: let $p \in A$ and write $p = (x_1,x_2)$ so of course $x_1 > x_2$. Define $r = x_1 - x_2$. Select $0 < \epsilon < \dfrac r2$. If you are given a second point $q = (y_1,y_2)$ with $\|p-q\| < \epsilon$ then $$\|x_1 - y_1\| \le \|p-q\| < \epsilon$$ and $$\|x_2 - y_2\| \le \|p-q\| < \epsilon$$ so that $$y_2 \le x_2 + \epsilon = x_1 - r + \epsilon \le y_1 - r + 2\epsilon < y_1$$ and in particular, $q \in A$. It follows that $B(p,\epsilon) \subset A$. This implies every $p \in A$ has a neighborhood contained in $A$ so that $A$ is open.
H: General proof for square–cube law Can someone present a general (and easy) proof for square-cube law? For similar objects 1 and 2, $$ \frac{A_1}{A_2}=k^2 \ \mathrm{and} \ \frac{V_1}{V_2}=k^3, $$ where k is the scale of objects 1 and 2. AI: For a cube we have that $A_1=l_1^2$ $V_1=l_1^3$ and $A_2=l_2^2$ $V_2=l_2^3$ then by $k=\frac{l_1}{l_2}$ $$\frac{A_1}{A_2}=\left(\frac{l_1}{l_2}\right)^2=k^2 \ \mathrm{and} \ \frac{V_1}{V_2}=\left(\frac{l_1}{l_2}\right)^3=k^3$$ For a complex boby we can think to divide it in many small cubes and apply the same reasoning to obtain the same result as a limit.
H: Evaluate $\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$ Evaluate:$$\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$ Using the property:$$r\binom{m}{r}=m\binom{m-1}{r-1}$$ It is same as $$\sum_{r=2}^{m} \frac{(r-1)m^{r-1}}{m\cdot\binom{m-1}{r-1}}$$ How I do now? AI: Let $$S=\sum_{r=1}^{m} \frac{(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$ Multiply both sides by $m+1$. $$S(m+1)=\sum_{r=1}^{m} \frac{(m+1)(r-1)m^{r-1}}{r\cdot\binom{m}{r}}$$ $$S(m+1)=\sum_{r=1}^{m} \frac{(mr-(m-r+1))m^{r-1}}{r\cdot\binom{m}{r}}$$ $$S(m+1)=\sum_{r=1}^{m} \frac{(rm^r-(m-r+1)m^{r-1})}{r\cdot\binom{m}{r}}$$ $$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{(m-r+1)m^{r-1}r!\cdot(m-r)!} {r\cdot m!} $$ $$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}(m-r+1)!(r-1)!}{m!}$$ $$S(m+1)=\sum_{r=1}^{m} \frac{m^r}{\binom{m}{r}}-\frac{m^{r-1}}{\binom{m}{r-1}}$$ Now this becomes a telescoping series. $$\boxed {S(m+1)=m^m-1}$$
H: Trouble in understanding the proof of the existense of the HOMFLY polynomial Following Page 252-254 of Christian Kassel's text "Quantum Groups" Chapter 10 very closely: $Definition$ $4.1$ A triple $(L^+,L_-,L_0)$ of oriented links in $\mathbb{R}^3$ is a Conway triple if they can be represented by link Diagrams $D_+,D_-,D_0$ which coincide outside a disk in $\mathbb{R}^2$ and which are respectively isotopic to $(X_+),(X_-),$ and $(\|\|)$, respectively, inside the disc: (Enter here a picture of the standard diagram of skein relations for a polynomial knot invariant) $Theorem$ $4.2$ $\exists !$ map $L \rightarrow P_L$ from the set of all oriented links in $\mathbb{R}^3$ to the ring $\Delta=\mathbb{Z}[x,x^{-1},y,y^{-1}]$ of two variable Laurent polynomials such that: $(i)$: If $L \cong L'$, then $P_L=P_{L'}$ $(ii)$: The value of $P$ on the trivial knot is $1$ $(iii)$: Whenever $(L_+,L_-,L_o)$ is a conway triple we have: $$xP_{L_x}-x^{-1}P_{L_-}=yP_{L_0} ()$$ The invariant $P_L$ is called the Jones-Conway polynomial, or the HOMFLY polynomial. Let $K$ be the set of isotopy equivalence classes of all oriented links in $\mathbb{R}^3$, and let $\Delta[K]$ be the free $\Delta$-module generated by $K$. Denote $\Gamma$ to be the quotient of $\Delta[K]$ by: $$x[L_+]-x^{-1}[L_-]-y[L_0]$$ where $(L_+,L_-,L_0)$ runs over all Conway triples of $K$. The $\Delta$-module $\Gamma$ is called the skein module of $\mathbb{R}^3$. $Proposition$ $4.4$: Let $Q: \Delta \rightarrow \Gamma$ be the $\Delta$-linear map sending $1$ to the class $[0]$ to the trivial knot. Then Q is an isomorphism. As a consequence of $Proposition$ $4.4$, the skein module $\Gamma$ is a free $\Delta$-module of rank one generated by $[0]$, the equivalence class of the trivial knot. Proposition $4.4$ implies Theorem $4.2$. To see this, let $[L] \in \Gamma$ and set $P_L=Q^{-1}([L])$. So we just need to prove Proposition $4.4$. We need to show that $Q$ is surjective and injective. To show that $Q$ is surjective, it suffices to show that the $\Delta$-module $\Gamma$ is generated by the class $[0]$ of the trivial knot. I can't understand the following proof of lemma $4.5$. $Lemma$ $4.5$: The $\Delta$-module $\Gamma$ is generated by the family $\{[O^{\otimes n}]\}_{n>0}$ of isotopy classes of all trivial links. $proof$: Let $\Gamma_m$ be the $\Delta$-submodule generated by the isotopy classes of links reprsentable by link diagrams with $\leq m$ crossing points. Clearly, $\Gamma_{m}$ maps to $\Gamma_{m+1}...$. (What the heck is the author talking about here? What map is he talking about that maps $\Gamma_{m}$ maps into $\Gamma_{m+1}$? Surely not $Q$!!! $Q$ is not an endomorphism of $\Gamma$!!) The proof then continues... .. and $\Gamma$ is the union of $\Gamma_m$ and so it suffices to prove lemma 4.5 for each $\Gamma_m$. We proceed by induction, the base case is trivial. Suppose lemma 4.5 is true when there for all integers $<m$. Let $[L] \in \Gamma_m$. It may be represented by a link diagram wtih $m$ crossing points. Consider one of them. Then there exists a conway triple $(L_+,L_-,L_0)$ such that $L=L_+$ or $L=L_-$ and the diagram $L_0$ has less than $m$ crossing points. It follows from $()$ that $[L_+]=x^{-2}[L_-]$ mod $\Gamma_{m-1}$. In other words, a change of crossings changes the class of $L$ mod $\Gamma_{m-1}$ by multiplication by an invertible element of $\Delta$. Lemma 3.3 (stated below) implies that the class of $L$ belongs to the submodule generated by the trivial links and $\Gamma_{m-1}$. The latter is also generated by trivial links in view of the induction hypothesis. (I also don't understand the bold part above, how does Lemma 3.3 imply this? Thanks!!) $Lemma$ $3.3$: Any link diagram may be turned after appropriate changes of crossings into a link diagram representing a trivial link in $\mathbb{R}^3$ AI: Any diagram of $\leq m$ crossings is also a diagram of $\leq m+1$ crossings, so $\Gamma_m\hookrightarrow\Gamma_{m+1}$ by sending generators over to generators and extend using $\Delta$-linearity. Start with a diagram of $m$ crossings. We know there is a sequence of crossing changes that result in the trivial link (of $\ell$ circles). We use $[L_+]=x^{-2}[L_-]\pmod{\Gamma_{m-1}}$ to change one crossings, then another would be mod $\Gamma_{m-2}$ by multiplying by $x^{\pm2}$, etc. until you get to the trivial link. But $\Gamma_i$ are nested, so you can take everything mod $\Gamma_{m-1}$ and hence $[L]\equiv x^{2k}[O^{\otimes \ell}]\pmod{\Gamma_{m-1}}$, some integer $k$.
H: How many angles can be drawn using only a ruler and a compass? So far I know that it’s possible to draw angles which are multiples of 15° (ex. 15°, 30°, 45° etc.). Could anybody please tell me if it's possible to draw other angles which are not multiples of 15° using only a compass and a ruler. AI: You can construct a regular $n$-gon with straightedge and compass if and only if $n$ is a power of $2$ times a product of Fermat primes - primes of the form $2^{2^j} +1$. That tells you what fractional angles you can construct. For example, the $17$-gon is constructible, so you can construct an angle of $360/17$ degrees.
H: Anti-canonical bundle of a bundle Let $ F \to E \stackrel{\pi}{\to} B$ be a smooth fibre bundle, so that $F$, $E$, and $B$ are smooth manifolds. I'm interested in what one can say about the anti-canonical bundle $K_E^$ of the total space $E$, given the anti-canonical bundles of the fibre $F$ and base $B$. In particular, I want to show that if $K_F^$ and $K_B^$ are free (i.e. if their linear systems have no base loci), then $K_E^$ is free too. According to this page, the tangent bundle of $E$ splits as $$ TE \cong \pi^(TB) \oplus T_\pi E \,, $$ where $T_\pi E$ consists of those tangent vectors tangent to the fibres of $\pi$. Taking the determinant bundle of both sides, and noting the determinant commutes with the pull-back, we have $$ K_E^ \cong \pi^(K_B^) \otimes \mathrm{det}(T_\pi E) \,. $$ Since the pull-back of a free bundle is free, I know the first factor is free. Also, the tensor product of free bundles is free, so if the second factor is free then $K_E^$ is too, so we would be finished. However I'm not quite sure how to deal with $\mathrm{det}(T_\pi E)$, and how it is related to $K_F^$. If the bundle were trivial, $E = F \times B$, with projection $p: E \to F$, then it must be that $T_\pi E \cong p^(TF)$, so that $K_E^ \cong \pi^(K_B^) \otimes p^(K_F^)$, which is free. But I don't know how to treat the general case. AI: What you want to show is not true. Maybe the simplest counterexample is the Hirzebruch surface $\mathbf F_3$, meaning the projective bundle $\mathbf P(O \oplus O(3))$ over $\mathbf P^1$. This is a bundle in which the base and fibres are both $\mathbf P^1$, so the anticanonical bundle of both is very ample, in particular basepoint free. However, this bundle contains a curve $C$ isomorphic to $\mathbf P^1$ with selfintersection $C^2=−3$. Adjunction then tells you that $K_{\mathbf F_3}^* \cdot C<0$, so $C$ is in the base locus of $K^∗$.
H: A question about whether an ideal is maximal or not Edited New question : my earlier asked question was assosiated with this question:Consider the ideal $I=(x^2+1,y)$ in the polynomial ring $\mathbb{C}[x,y]$.Then which of the following is true I already knew this question exists on MSE but here it's not answered by question asker clearly why I is not maximal and Answering username Gone didn't answered about maximal ideal. So, I should have asked in comments but user: Gone has left website and user which asked the question was last seen in Mid of 2017 . So, the chances that question asker will reply my comment are very bleak. So, I am asking again here. Due to above mentioned reasons please don't close this question. I am trying assignment question for my abstract algebra course and I am unable to solve this particular problem. I is an ideal defined by $I=(x^2+1 , y)$ in polynomial ring $\mathbb{C} [x,y] $ . Then is Prime? Is I maximal? I showed that I is not prime by proving that $x-i$ doesn't belongs to $I$. But I am unable to think how can I prove it maximal. Can someone please give hints. AI: To check whether $I$ is maximal or prime, simply consider the quotient as for a commutative ring with unity say $R$, an ideal $J$ is maximal / prime $\iff$ the quotient ring $R/J$ is a field / an integral domain. Now in our case, $$\frac{\Bbb C[x,y]}{I}\cong \frac{\Bbb C[x]}{(x^2+1)}$$ Since, $x^2+1 $ splits as $x^2+1=(x+i)(x-i)$ in $\Bbb C[x]$ and $x+i+x-i=2i$ thus, the principal ideals $(x+i)$ and $(x-i)$ are co-maximal, thus by CRT, $$\frac{\Bbb C[x]}{(x^2+1)} \cong \frac{\Bbb C[x]}{(x+i)} \times \frac{\Bbb C[x]}{(x-i)} \cong \Bbb C \times \Bbb C$$ which is not even an integral domain, thus also not a field. Hence by the fact we mentioned it follows that $I$ is neither a Prime ideal nor a Maximal ideal.
H: Total combinations from number of unique combination? How can you find count of the total combinations possible if you have the count of unique combinations? Suppose I need to form a sum of $3$ using $\{1,2\}$ Unique combn. $(1+1+1)$, $(1+2)$ hence '$2$' unique but total is '$3$' i.e, $(1+1+1)$, $(1+2)$, $(2+1)$ I know the elements used to make the combination Note: Number of unique combinations (ie. 2 here) is known. How can I find the total number of combination given the unique combinations? AI: To find the number of restricted compositions (use the correct terminology) with a desired sum $n$ and available numbers $\{1,2\}$, what I will label $f(n)$, recognize the following: $f(1)=1$ (as the only possibility is the sum with only one term: $(1)$) $f(2)=2$ (as there are two possibilities: $(1+1)$ and $(2)$) alternatively, and preferably, recognize that $f(0)=1$ as there is in fact a sum whose sum is zero using terms $1$ and $2$... the "empty sum"... $(~)$. This is somewhat abstract, but I strongly encourage coming to grips with this as using counts related to objects of size zero can greatly simplify arithmetic in many instances Next, recognize that for each $n\geq 3$ you have $f(n) = f(n-1)+f(n-2)$ which is seen by noting that every such restricted composition either begins with a "$1+\dots$" with a sum of $n-1$ following which can be arranged in $f(n-1)$ ways, or it will begin with $2+\dots$ with a sum of $n-2$ following which can be arranged in $f(n-2)$ ways. Now... you should recognize this... $f(1)=1, f(2)=2, f(n)=f(n-1)+f(n-2)$... this is precisely the fibonacci sequence. So... here we found that $f(n)$ is precisely equal to $F_n$. No knowledge of the related problem counting restricted partitions rather than restricted compositions was needed. The technique can obviously be adapted... the number of restricted compositions with sum $n$ using terms $\{1,2,3\}$ for instance would be the tribonacci numbers satisfying the recurrence $t(1)=1, t(2)=2, t(3)=4, t(n)=t(n-1)+t(n-2)+t(n-3)$ or the number of restricted compositions with sum $n$ using terms $\{2,5\}$ for instance would be $a(1)=a(3)=0, a(2)=a(4)=a(5)=1$ and $a(n)=a(n-2)+a(n-5)$ for each $n\geq 6$ and so on... Coming to terms with the zero object can simplify a lot of the above for finding initial conditions. We could instead have had $a(n)=0$ for all $n<0$ and $a(0)=1$, reducing the need to actually manually find initial conditions Coming up with a closed form for these types of linear recurrences is already covered in great detail in related chapters in books, and otherwise for now the current presentation of the answer in terms of a recurrence relation is generally considered sufficient.
H: sum of terms of series If $$F(t)=\displaystyle\sum_{n=1}^t\frac{4n+\sqrt{4n^2-1}}{\sqrt{2n+1}+\sqrt{2n-1}}$$ find $F(60)$. I tried manipulating the general term(of sequence) in the form $V(n)-V(n-1)$ to calculate the sum by cancellation but went nowhere. I also tried using the fact that $$2n+\sqrt{4n^2-1}=\frac{1}{2}{(\sqrt{2n-1}+\sqrt{2n+1})}^2$$ Could someone please give me a hint? AI: With more details, we have that $$\frac{2n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)$$ then $$\frac{4n+\sqrt{(4n^2-1)}}{\sqrt{2n+1}+\sqrt{2n-1}}=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+\frac{2n}{\sqrt{2n+1}+\sqrt{2n-1}}=\\=\frac12\left({\sqrt{2n+1}+\sqrt{2n-1}}\right)+n\left({\sqrt{2n+1}-\sqrt{2n-1}}\right)=\\=\frac12(2n+1)\sqrt{2n+1}-\frac12(2n-1)\sqrt{2n-1}=\\$$ $$=\frac12\left(\sqrt{(2n+1)^3}-\sqrt{(2n-1)^3}\right)$$
H: What is the integer part of the following fraction: $\frac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$ What is the integer part of the following fraction: $\dfrac{2012^{2013}+2013^{2014}}{2012^{2012}+2013^{2013}}$. This is a competition problem for 7th grade students. The answer to this question is $2012$. Is there any way to simplify/evaluate it so we can see the integer part clearly? Or can we just estimate? AI: Let $y=2012^{2013}+2013^{2014}$ Now $2012^{2013}+2013^{2014}=2012^{2012}(2012)+2013^{2013}(2013)=2012\Big(2012^{2012}+2013^{2013}\Big)+2013^{2013}$ Let $x=2012^{2012}+2013^{2013}$ Thus $y=2012x+2013^{2013}\Rightarrow \dfrac{y}{x}=2012+\dfrac{2013^{2013}}{x}$ Since $x>2013^{2013}$, $\dfrac{2013^{2013}}{x}<1$ $\therefore $ Integral part of $\dfrac{y}{x}$ is $2012$.
H: On hypergeometric square integral $\int_0^{\infty } \, _2F_1(a,b;c;-x){}^2 \, dx$ Preliminary. When $a,b>\frac12$, $c\not=0, -1,-2, ...$, one have (using Mellin transform) $I=I(a,b,c)=\int_0^{\infty } \, _2F_1(a,b;c;-x){}^2 \, dx=\frac{\Gamma (c)^2 G_{4,4}^{3,3}\left(1\left\| \begin{array}{c} 0,1-a,1-b,c-1 \\ 0,a-1,b-1,1-c \\ \end{array} \right.\right)}{\Gamma (a)^2 \Gamma (b)^2}$ When $a+b<c+\frac12$ it equals $I=\frac{(c-1) \, _4F_3(1,a,b,2-c;2-a,2-b,c;1)}{(a-1) (b-1)}-\frac{2 \pi ^3 \csc (2 \pi a) \csc (2 \pi b) \Gamma (c)^2 \cos (\pi (a+b)) \Gamma (a+b-1) \Gamma \left(-a-b+c+\frac{1}{2}\right)}{\Gamma \left(\frac{3}{2}-a\right) \Gamma (a)^2 \Gamma \left(\frac{3}{2}-b\right) \Gamma (b)^2 \Gamma \left(c-\frac{1}{2}\right) \Gamma (c-a) \Gamma (c-b)}$. Examples: Special cases of formula above. When $a+b<\frac32$: $\int_0^{\infty } \, _2F_1(a,b;1;-x){}^2 \, dx=-\frac{\pi ^{3/2} 2^{-2 a-2 b+3} \csc (2 \pi a) \csc (2 \pi b) \cos (\pi (a+b)) \Gamma \left(-a-b+\frac{3}{2}\right) \Gamma (a+b-1)}{\Gamma (2-2 a) \Gamma (a)^2 \Gamma (2-2 b) \Gamma (b)^2}$ When $c>3/2$: $\int_0^{\infty } \, _2F_1(a,2-a;c;-x){}^2 \, dx=-\frac{\frac{\sec (\pi a) \Gamma (c)^2}{\Gamma (c-a) \Gamma (a+c-2)}+(c-1)^2}{(a-1)^2 (2 c-3)}$ By using analytic continuation more results are found. Here are $2$ more examples: $\int_0^{\infty } \, _2F_1\left(\frac{11}{8},\frac{5}{8};\frac{5}{4};-x\right){}^2 \, dx=\frac{8}{9}-\frac{256 \sqrt[4]{2} \Gamma \left(\frac{9}{8}\right) \Gamma \left(\frac{5}{4}\right)}{9 \sqrt{\left(2-\sqrt{2}\right) \pi } \Gamma \left(-\frac{1}{8}\right)}$ $\int_0^{\infty } \, _2F_1\left(\frac{3}{4},\frac{5}{6};1;-x\right){}^2 \, dx=-\frac{2 \sqrt[6]{2} \left(\sqrt{3}-3\right) \pi \Gamma \left(\frac{7}{12}\right)}{\Gamma \left(\frac{5}{12}\right) \Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{3}{4}\right)^2}$ Question: What more can be found for this kind of integrals? This is rather an open question, and any suggestions will be appreciated. AI: If you see this as a Mellin transform with $s=1$, then by the Ramanujan master theorem you are talking about the integral being $$ \int_0^\infty x^{s-1} \;_2F_1(a,b;c;-x)^2 \; dx = \Gamma(s) C_{-s} $$ for the power series parameterised as $$ _2F_1(a,b;c;-x)^2 = \sum_{k=0}^\infty \frac{(-1)^k}{k!}C_k x^k $$ but it depends on the RMT still holding for this product of power series. I find this is why the negative argument in the hypergeometric function works well from $(-1)^k x^k = (-x)^k$. So perhaps think about the Cauchy product $$ \left(\sum_{i=0}^\infty \frac{(a)_i (b)_i}{(c)_i i!} (-x)^i\right)\left(\sum_{j=0}^\infty \frac{(a)_j (b)_j}{(c)_j j!} (-x)^j\right) $$ also you might want to rewrite the $\pi\csc(\pi s)$ terms in the form $\Gamma(s)\Gamma(1-s)$ to spot patterns. I have a note about spotting patterns in the gamma functions, I'll see if it can find it... Edit: The following might be useful if there is a way to consider a 'confluence' of sorts from integrals of the form $$ \int_0^\infty \int_0^\infty x_1^{s_1-1} x_2 ^{s_2-1} f_1(x_1) f_2(x_2) \; dx_1 dx_2 \to \int_0^\infty x^{s-1} f(x) f(x) \; dx $$ we could view this as a multidimensional Mellin transform, but I have found there can be different results from the order of integration. If some Fubini type condition is there, then: If functions $f_k(x)$ have Mellin transforms $g_k(s)$ and coefficients are included then this nested $D$ dimensional type Mellin transform of the product of the functions is given by $$ \mathcal{M}_D\left[\prod_{k=1}^n f_k\left(\alpha_k \prod_{l=1}^n x_l^{a_{kl}}\right) \right] = \frac{\prod_{k=1}^n \alpha_k^{-(A^\top)^{-1}_k \mathbf{s}}}{\|\det(A)\|}\prod_{k=1}^n g_k((A^\top)^{-1}_k \mathbf{s}) $$ where $A_{kl}=a_{kl}$. An example Solve $$ I = \int_0^\infty \int_0^\infty \int_0^\infty x_1^{s_1-1} x_2^{s_2-1} x_3^{s_3-1} e^{-\frac{\alpha x_1 x_2}{x_3}}J_n(\beta x_1^2 x_2)\mathrm{Ai}(\gamma x_3) \; dx_1 dx_2 dx_3 $$ with Bessel function $J_n(x)$, Airy function $\mathrm{Ai}(x)$. We have that $f_1(x) = e^{-x}$,$f_2(x) = J_n(x)$, $f_3(x) = \mathrm{Ai}(x)$. We look up that $$ g_1(s) = \Gamma(s) $$ $$ g_2(s) = \frac{2^{s-1} \Gamma \left(\frac{n}{2}+\frac{s}{2}\right)}{\Gamma \left(\frac{n}{2}-\frac{s}{2}+1\right)} $$ $$ g_3(s) = \frac{3^{\frac{2 s}{3}-\frac{7}{6}} \Gamma \left(\frac{s}{3}+\frac{1}{3}\right) \Gamma \left(\frac{s}{3}\right)}{2 \pi } $$ we inspect the integrand and find the coefficient matrix $$ A = \begin{bmatrix} 1 & 1 & -1 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \;\; (A^\top)^{-1} = \begin{bmatrix} -1 & 2 & 0 \\ 1 & -1 & 0 \\ -1 & 2 & 1 \end{bmatrix}, \;\; \det(A) = -1 $$ we have $$ I = \alpha^{s_1 - 2 s_2}\beta^{s_2-s_1}\gamma^{s_1-2 s_2-s_3} \Gamma(2s_2-s_1) \frac{2^{s_1-s_2-1} \Gamma \left(\frac{n}{2}+\frac{s_1-s_2}{2}\right)}{\Gamma \left(\frac{n}{2}-\frac{s_1-s_2}{2}+1\right)} \frac{3^{\frac{2 (2s_2-s_1+s_3)}{3}-\frac{7}{6}} \Gamma \left(\frac{(2s_2-s_1+s_3)}{3}+\frac{1}{3}\right) \Gamma \left(\frac{(2s_2-s_1+s_3)}{3}\right)}{2 \pi } $$ Implication When I see your example result, there are immediate patterns that hint at this linear combination of variables type approach $$ A = -\frac{\pi ^{3/2} 2^{-2 a-2 b+3} \csc (2 \pi a) \csc (2 \pi b) \cos (\pi (a+b)) \Gamma \left(-a-b+\frac{3}{2}\right) \Gamma (a+b-1)}{\Gamma (2-2 a) \Gamma (a)^2 \Gamma (2-2 b) \Gamma (b)^2} $$ for example $2^{-2 a - 2 b + 3}=4^{-a-b+3/2}$ and the linear combination is $-a-b+3/2$ as seen in the gamma function. One possible goal is to split your expression apart into the product of $N$ distinct Mellin transforms and reverse engineer the original integral as a product of simpler integrals? We can rewrite your first result using $$ \cos\left(\frac{\pi s}{2}\right) = \frac{\pi}{\Gamma\left(\frac{1}{2} + \frac{s}{2}\right)\Gamma\left(\frac{1}{2}-\frac{s}{2}\right)} $$ and $$ \pi \csc(\pi s) = \Gamma(s)\Gamma(1-s) $$ to $$ A = -\pi\csc (2 \pi a) \pi\csc (2 \pi b) \cos (\frac{\pi}{2} (2a+2b)) \frac{4^{-a-b+3/2}}{\pi^{1/2}}\frac{\Gamma \left(-a-b+\frac{3}{2}\right) \Gamma (a+b-1)}{\Gamma (2-2 a) \Gamma (a)^2 \Gamma (2-2 b) \Gamma (b)^2} $$ $$ A = - \pi^{1/2} 4^{-a-b+3/2}\frac{\Gamma(2a)\Gamma(1-2a) \Gamma(2b)\Gamma(1-2b)}{\Gamma\left(\frac{1}{2} + a+b\right)\Gamma\left(\frac{1}{2}-a-b\right)} \frac{\Gamma \left(-a-b+\frac{3}{2}\right) \Gamma (a+b-1)}{\Gamma (2-2 a) \Gamma (a)^2 \Gamma (2-2 b) \Gamma (b)^2} $$ As a base observation $$ \int_0^\infty \int_0^\infty x_1^{s_1-1} x_2 ^{s_2-1} \;_2F_1(a,b;c;-x_1)\;_2F_1(a,b;c;-x_2) \; dx_1 dx_2 = \frac{\Gamma (c)^2 \Gamma (\text{s1}) \Gamma (\text{s2}) \Gamma (a-\text{s1}) \Gamma (a-\text{s2}) \Gamma (b-\text{s1}) \Gamma (b-\text{s2})}{\Gamma (a)^2 \Gamma (b)^2 \Gamma (c-\text{s1}) \Gamma (c-\text{s2})} $$ I feel there might be a parameterisation with $s_1 =a+b-1$ and $s_2=3/2-a-b$, therefore $\Gamma(1/2+a+b) = \Gamma(3/2+s_1)$ and $\Gamma(1/2-a-b)=\Gamma(s_2-1)$. But I have fried my brain for the time being...
H: Checking whether $\mathbb C^n$ is semisimple or not Consider $\mathbb C^n$ as a Banach algebra over $\mathbb C$. Is it semisimple? A Banach algebra is semisimple if the intersection of all maximal ideals of it is ${0}$. I can't figure out what are the maximal ideals of $\mathbb C^n$. How do I solve it? AI: What about putting $I_k:=\{x\in\mathbb{C}^n\mid x_k=0\}$? They are surely ideals, they are maximal since their dimension is $n-1$, and their intersection is zero.
H: Does there exist some linear factor for every quadratic such that their product's $x^2$ and $x$ terms disappear Can any quadratic with integer coefficients be multiplied by some linear factor, also with integer coefficients, such that the coefficients of the product's $x^2$ and $x$ terms are both zero and the coefficient of the $x^3$ is nonzero and the constant term can be any integer? Alternatively, for readability, is there always some $px+q$ such that for any quadratic $ax^2+bx+c$ with $a,p \ne 0$ and $a,b,c,p,q \in \mathbb{Z}$, $(ax^2+bx+c)(px+q) = apx^3+cq$ ? I tried searching for the question online but it looks like it was too specific to already have been asked (unless I'm not looking hard enough). What I've tried so far is that if I have the quadratic $ax^2+bx+c$ and the linear factor $px+q$ with $a,p \ne 0$ and $a,b,c,p,q \in \mathbb{Z}$, multiplying these two expressions gives us $(ax^2+bx+c)(px+q)=apx^3+(aq+bp)x^2+(bq+cp)x+cq$ and now we have the following statements and equations: $$ap\text{ is a nonzero integer}$$ $$aq+bp = 0$$ $$bq+cp = 0$$ $$cq \text{ is an integer}$$ The first and last statements are redundant due to the fact that $a,b,c,p,$ and $q$ are all integers and if $a$ or $p$ were $0$, then $ax^2+bx+c$ would not be a quadratic and $px+q$ would be a constant. This means that we only need to solve the following system of equations: $$aq+bp = 0$$ $$bq+cp = 0$$ Multiplying both sides of the top equation by $q$ and both sides of the bottom equation by $p$, we get that $aq^2+bpq = 0 = bpq+cp^2$ and so $aq^2=cp^2$. Rearranging, we get that $\dfrac{a}{c}=\dfrac{p^2}{q^2} = \left(\dfrac{p}{q}\right)^2$. Does this mean that such a linear factor only exists when $\dfrac{a}{c}$ is the square of a rational number? I'm sorry if this was hard to read, English is my first language and I'm still bad at it. AI: Assuming $a,b,c$ are integers and $a \neq 0$, we want to find $p,q$ so that $p\neq 0$ and $$(ax^2 + bx + c)(px+q) = apx^3 + cq$$ where $ap$ is an integer and $cq$ is an integer. Claim: The necessary and sufficient condition is $b^2=ac$. Proof: Necessity: We need $aq+bp=0$ and $bq+cp=0$. Substituting $q=-bp/a$ into the second equation gives that either $p=0$ or $b^2=ac$. Since $p\neq 0$ we must have $b^2=ac$. Sufficiency: If $b^2=ac$ then we can choose $p=a$ and $q=-b$. $\Box$
H: Let the sum of the coefficients of the polynomial $(4x^2 - 4x + 3)^4(4 + 3x - 3x^2)^2$ be $S$ . Find $\frac{S}{16}$ . Let the sum of the coefficients of the polynomial $(4x^2 - 4x + 3)^4(4 + 3x - 3x^2)^2$ be $S$ . Find $\frac{S}{16}$ . Actually I have absolutely no other idea on how to find this without opening up the brackets, and that will seem to be a very tiring work. I don't know any other method of getting the sum of the coefficients of this type of problems. Wolfram Alpha gives the expansion of this expression :- https://www.wolframalpha.com/input/?i=%284x%5E2+-+4x+%2B+3%29%5E4%284+%2B+3x+-+3x%5E2%29%5E2 . That is ok, but here I want a short and an easy answer which can immediately tell me the value of $\frac{S}{16}$. Can anyone help me and give some hint or idea ? AI: Hint: What do you get when you substitute $x=1$ into a polynomial $p(x)$?
H: Ornstein - Uhlenbeck - Stochastic Differential Equation liminf and limsup I came across an exercise which deals with a stochastic differential equation of the form $$\mathrm dX(t)=-\theta[X(t)-\mu]\mathrm dt+\sigma\,\mathrm dW(t)$$ for $t>0$, where $\theta,\sigma>0$ and $\mu$ are fixed parameters. It requests to show that $$\lim \sup X(t)=\infty$$ almost certainly when $t$ approaches infinity. Also it requests to show that $$\lim \inf X(t)=-\infty$$ almost certaintly as $t$ approaches infinity. It says to use the Dambis–Dubins–Schwarz theorem and the law of iterated logarithms for Brownian motion: $$ \lim \sup \frac{W(t)}{\sqrt{2t}\log\log t}=1\quad\text{and}\quad\lim \inf \frac{W(t)}{\sqrt{2t}\log\log t}=-1 $$ almost certainly as $t$ approaches infinity. Any help on how I can solve/prove these? Thanks. AI: The quadratic variation of the Ornstein-Uhlenbeck process above is $$[X]_t = \sigma^2 [W]_t = \sigma^2 t,$$ which is unbounded. Moreover, the O.-U. process is a continuous local martingale, and we assume that $X_0 = 0$. We define the process $X^\tau = \{X_{\tau_t}, t \geq 0\}$ defined by the times $$\tau_t = \inf\{s \geq 0 : [X]_s > t\}.$$ By the theorem of Dambis-Dubins-Schwarz, the process $X^\tau$ is indistinguishable from a Brownian motion up until time $[X]_\infty$. Since the process is divergent, the result applies for all $t \geq 0$. Since $X^\tau$ is a Brownian motion, the Law of Iterated Logarithm implies that, almost surely, $$\limsup_{t\rightarrow\infty} X^\tau_t = \infty\tag{1}$$ and $$\liminf_{t\rightarrow\infty} X^\tau_t = -\infty.\tag{2}$$ We can translate these results about the process $X^\tau$ into ones about the original process $X$, since any time we have $X^\tau_t > N$ for some large $N$, we necessarily have $X_{\tau_t} > N$; ditto for $X^\tau_t < -N$. Thus, $(1)$ and $(2)$ hold for $X_t$ in place of $X^\tau_t$, as was to be shown.
H: List all the possible values for $\int_{\mathbb{R}}\sup_{k\in\mathbb{N}}f_k(x)dx$ under these conditions... Question: Let $\{f_k(x)\}_{n=1}^\infty$ be a sequence of nonnegative functions on $\mathbb{R}$ such that $\sup_{x\in\mathbb{R}}f_k(x)=\frac{1}{k}$, and $\int_{\mathbb{R}}f_k(x)dx=1$. List all the possible values for $\int_{\mathbb{R}}\sup_{k\in\mathbb{N}}f_k(x)dx$. My Thoughts: I am a bit confused on how we are going to relate the supremum over $x\in\mathbb{R}$ of $f_k(x)$, as in the hypothesis, and the supremum over $k\in\mathbb{N}$ of $f_k(x)$, as in what we are trying to prove... are they the same? Since the $f_k$'s are integrable, we can say they are measurable, so the Monotone Convergence Theorem would imply, I believe, that $\lim_{k\rightarrow\infty}\int_{\mathbb{R}}\sup_{x\in\mathbb{R}}f_k(x)dx=\int_\mathbb{R}\frac{1}{k}$ (maybe I can't put that supremum inside the integral like that....) But then I am not sure how to use $\int_{\mathbb{R}}f_k(x)dx=1$.... any help is greatly apprecaited! Thank you. AI: I think the answer is meant to be $\{\infty\}$. Suppose $$ \int_{\mathbb R} \sup_{k} f_k = I < \infty. $$ By the monotone convergence theorem, there exists a positive integer $n$ such that $$ \int_{[-n,n]} \sup_{k} f_k \ge I-\tfrac13. $$ Now consider $f_{6n}$. Since $\sup_{x\in\mathbb R} f_{6n}(x) \le \frac1{6n}$, we see that $\int_{[-n,n]} f_{6n} \le \frac13$. Therefore \begin{align} \int_{\mathbb R} \sup_{k} f_k &= \int_{[-n,n]} \sup_{k} f_k + \int_{\mathbb R \setminus [-n,n]} \sup_{k} f_k \\&\ge \int_{[-n,n]} \sup_{k} f_k + \int_{\mathbb R \setminus [-n,n]} f_{6n} \\&\ge I - \tfrac13 + \tfrac23 \\&= I + \tfrac13.\end{align}
H: Is there any other controller than PID Controller? I have been given a Project to search for a controller having better transient response than PID Controller. I searched but I didn't find any research paper on it. All are talking about improving PID's transient response but I am unable to find any other Controller with the better transient Response. Can anyone share comparative analysis between the aforementioned controllers Or a Research Paper? Or can someone list controllers? P.S: I'm new in Control Theory field. AI: Of course there are other controllers than PID... there's a reason for the large amount of ongoing work in Control Theory at research universities. Perhaps the next level up in terms of sophistication from PID is the Linear Quadratic Regulator (LQR). LQR control uses a linear state-space model of the system. The basic idea is that you define a convex quadratic cost function, which encodes which inputs/actuators and states you care about most. The cost function penalizes the use of the inputs and also penalizes the deviation of the states from your desired setpoint. For your example, since you care mostly about transient response, you would probably heavily weight the state variables associated with signal you care about (for instance, position or voltage), as well as the derivative of that signal (which is typically taken as a state variable for second- and higher-order systems). With your cost function defined, you then minimize the cost with respect to the control input. LQR is particularly remarkable since the optimal control input ends up being a time-invariant linear state feedback of the form $u(t) = Kx(t)$, where $K$ solves the algebraic Ricatti equation. LQR is also powerful because it naturally encodes various robustness properties into the closed-loop system. Another level up in sophistication is model predictive control. Model predictive control (MPC) can be thought of as online optimal control, i.e., solve LQR (or a finite-time optimal control problem) at each time step, then send the first time step of the optimal control input to the system, then repeat the process at the next time step. MPC performs particularly well on systems whose dynamics include slight unmodeled dynamics or unmodeled disturbances, since the act of re-solving the optimal control problem at each time step acts as a feedback mechanism to take the effects of these unmodeled behaviors into account. More recently, there has been a lot of work in the research community where data-driven methods from machine learning are being used in conjunction with model-based control strategies in order to achieve certain control goals. The bottom line is, yes, there is an entire literature on more sophisticated control methods than PID, and this body of work is growing by the day now that the fields of machine learning, control theory, and optimization are finding novel intersections. A great new conference in this area is L4DC (Learning for Dynamics and Control). I suggest you take a look at this year's papers to get an idea of what is going on in the field. Two other staple conferences for the control theory community are CDC (Conference on Decision and Control) and ACC (American Control Conference). I hope this helps give you some new directions!
H: A question based on analyticity of entire functions This particular question was asked in quiz yesterday and I was unable to solve it. Suppose f and g are entire functions and g(z) $\neq$0 for all z $\epsilon \mathbb{C} $ . If \|f(z) \|$\leq$ \|g(z) \| , the which one of them is true. 1.f is a constant function. 2.f(0) =0 . 3.for some C $\epsilon \mathbb{C} $ , f(z) = C g(z) . 4.f(z) $\neq$ 0 for all z $\epsilon \mathbb{C} $ . 4 th option can be removed by taking f(z) =0 and g(z) =2 . But I am unable to find any way to remove rest of options. AI: Simply consider the function $z \mapsto \frac{f(z)}{g(z)}$. Since, $g(z) \ne 0, \forall z \in \Bbb C$ we have that this map is entire and as $\|f(z)\| \le \|g(z)\|$ we have that our map $z \mapsto \frac{f(z)}{g(z)}$ has absolute value bounded by 1 $\forall z \in \Bbb C$, hence Liouville's theorem gives (3). (1) and (2) aren't true by taking $f(z)=g(z)=e^z$ (4) isn't true by taking $f \equiv 0$ and $g \equiv 1$
H: What's wrong with this method of evaluating an integral? I was trying to evalute the integral $$\int \frac{1}{x^2+1} \,dx$$ by partial fractions. $$\frac{1}{x^2+1} = \frac{1}{2i}\left(\frac{1}{x-i} - \frac{1}{x+i}\right)$$ Therefore, \begin{aligned} \int \frac{1}{x^2+1} \,dx &= \frac{1}{2i} \int \left(\frac{1}{x-i} - \frac{1}{x+i}\right) \,dx \\ &= \frac{1}{2i} (\ln(x-i) - \ln(x+i)) \\ &= \arctan\left(\frac{1}{x}\right) +C \end{aligned} Because $$ x - i = \exp\left(\sqrt{x^2+1}+i\arctan{\frac{1}{x}}\right)$$ and $$ x + i = \exp\left(\sqrt{x^2+1}-i\arctan{\frac{1}{x}}\right).$$ This differs from what you get from trigonometric substitution, which is where I'm having difficulties in finding my error. In the last step, I'm assuming that $\ln(z)$ works as you would expect for complex numbers. AI: In the your fifth equation you will actually get $$I=-\tan^{-1}(1/x)+C$$ and due to the identity: $$\tan^{-1} x+\tan^{-1} (1/x)= \pi/2$$ We get $$I=\pi/2+\tan^{-1} x+ D$$ Uther wise the standard formula gives $$I=\tan^{-1} x+E$$ These two results differ only by a constant, which is normal in indefinite integration.
H: Question 2 from Bredon's Topology and Geometry page 39 I got stuck on the following question from Bredon Topology and Geometry Chap 1 sec 12. Suppose $X$ is paracompact. For any open subset $U$ of $X \times [0,\infty)$ which contains $X \times \{0\}$ show that there is a map $f:X\rightarrow (0,\infty)$ such that $(x,y)\in U$ for all $y\leq f(x)$. The first thoughts on top of my mind is to use a partition of unity, but I have no clue how to proceed. Any hint will be appreciated.. AI: Each $\xi \in X$ admits an open neigbhorhood $U(\xi)$ in $X$ and $t(\xi) > 0$ such that $U(\xi) \times [0,t(\xi)] \subset U$. Now let $\{\phi_\xi\}_{\xi \in X}$ be partition of unity subordinate to the open cover $\{U(\xi)\}_{\xi \in X}$ of $X$. Define $$f : X \to (0,\infty), f(x) = \sum_{\xi \in X} t(\xi)\phi_\xi(x) .$$ This is well-defined because $\phi_\xi(x) > 0$ only for finitely many $\xi \in X$ and $\phi_\xi(x) > 0$ for at least one $\xi$. It is moreover continuous because each $x \in X$ admits an open neigborhood $V(x)$ such that all but a finite number of the $\phi_\xi$ are $0$. Now let $y \le f(x)$. Let $\xi_1,\ldots,\xi_n$ be the finitely many points such that $\phi_{\xi_i}(x) > 0$. Let $k$ be an index such that $t(\xi_k)$ is maximal among the $t(\xi_i)$. Then $$y \le f(x) = \sum_{i=1}^n t(\xi_i)\phi_{\xi_i}(x) \le \sum_{i=1}^n t(\xi_k)\phi_{\xi_i}(x) = t(\xi_k) \sum_{i=1}^n \phi_{\xi_i}(x) = t(\xi_k) .$$ But $x \in \operatorname{supp} \phi_{\xi_k} \subset U(\xi_k)$, thus $$(x,y) \in U(\xi_k) \times [0,t(\xi_k)] \subset U .$$
H: Books: homotopy groups of inverse limits Can anybody recommend a book, which contains comprehensive information about homotopy groups of inverse limits? AI: I doubt that there is a book as you desire. There is only a loose relation between homotopy groups of the inverse limit and the homotopy groups of the spaces in the inverse system which makes it very difficult to obtain useful results. More interesting are the homotopy pro-groups and the shape groups of spaces. See for eaxmple Mardešic, Sibe, and Jack Segal. Shape theory: the inverse system approach. Elsevier, 1982. Edited: In your comment you refer to a paper by Joel M. Cohen. This only considers a very special case: The inverse systems under consideration are inverse sequences $\mathbf X$ having fibrations as bonding maps. As Tyrone comments, the inverse limit of such a system agrees with its homotopy limit. The exact sequence $$0 \to \lim\nolimits^1 [SY, \mathbf X] \to [Y, \lim \mathbf X] \to \lim [Y, \mathbf X] \to 0$$ is therefore a special case of $$0 \to \lim\nolimits^1 [SY, \mathbf X] \to [Y, \operatorname{holim} \mathbf X] \to \lim [Y, \mathbf X] \to 0$$ which is valid for all inverse sequences $\mathbf X$. See for example Bousfield, Aldridge Knight, and Daniel Marinus Kan. Homotopy limits, completions and localizations. Vol. 304. Springer Science & Business Media, 1972- Chapter IX Edwards, David A., and Harold M. Hastings. Cech and Steenrod homotopy theories with applications to geometric topology. Vol. 542. Springer, 2006 - in particular (5.2.1) For $Y=S^n$ we obtain the exact sequence $$0 \to \lim\nolimits^1 \pi_{n+1}(\mathbf X) \to \pi_n(\operatorname{holim} \mathbf X) \to \lim \pi_n(\mathbf X) \to 0$$ The term $\lim \pi_n(\mathbf X)$ is nothing else than the $n$-th shape group of $\lim \mathbf X$ and $\pi_n(\operatorname{holim} \mathbf X)$ is the $n$-th strong shape shape group (or strong homotopy group) of $\lim \mathbf X$. This has in general nothing to do with $\pi_n(\lim \mathbf X)$. For the shape groups see 1. and for the strong shape groups see e.g. Mardesic, Sibe. Strong shape and homology. Springer Science & Business Media, 2013 - Remark 19.3 As an example for "non-nice behavior" take the Warsaw circle which can be written as the inverse limit of plane annuli $A_n$ (i.e. copies of $S^1 \times I$) bonded by inclusions (i.e. maps of degree $1$). The Warsaw circle is simply connected, but $\lim \pi_1(A_n) = \mathbb Z$ and $\lim\nolimits^1 \pi_2(A_n) = 0$.

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