Sunday, July 17, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


Question about estimating number of individuals in a population that belong to 2 sets...

Posted: 17 Jul 2022 08:33 AM PDT

Suppose we have a population containing $x$ number of individuals. Out of these individuals we know a number $x_{men}$ are men and $x_{women}$ are women. We also know that there are $x_{veg}$ vegetarians and $x_{non-veg}$. We also assume the following:

$$x_{men} + x_{women} = x$$ $$x_{veg} + x_{non-veg} = x$$

Can we deduce how many vegetarian men there are? If not can we at least make an estimate?

Will a curve having infinite length, contained into a set of $\mathbb R^2$ always intersect itself?

Posted: 17 Jul 2022 08:37 AM PDT

I have a question(for which I am sure there is already an answer) however not being able to spot the answer online I decided to ask my question in this forum. Suppose that we have a set $S$ of $\mathbb R^2$ as shown in the picture:enter image description here. My question is: Can there exist a curve of infinite length contained in $S$ so that the curve does not intersect itself? Thanks in advance for any answers.

CFG for L = $a^{i}$ $b^{j}$ $c^{k}$ ∈ $\{a, b, c\}^{*}$ | i ≠ j or j ≠ k}

Posted: 17 Jul 2022 08:20 AM PDT

im trying to figure out a CFG for the given Language:

L = $a^{i}$ $b^{j}$ $c^{k}$$\{a, b, c\}^{*}$ | i ≠ j or j ≠ k}

I've tried many variations, but I keep coming back to the problem that I find some words that are not in the language

one attempt i had was:

A -> ABB | AAB | a

...

further i didn't came because when i try to make my B with this grammar its the same problem, i get again the same amount of i's and j's

Convex function and curves

Posted: 17 Jul 2022 08:10 AM PDT

Is there diffrence between convex function and convex curve i.e Is it true that A function is convex iff its curve is convex .or they are not related with each other . It will better if there is mathematic proof if its true or counter example if its false

Relative dualising sheaf of cover of elliptic curves

Posted: 17 Jul 2022 08:06 AM PDT

I am trying to understand the relative dualising sheaf of a morphism of varieties in general but also in the special case of covers of the projective line by a elliptic curve, let us say of the form $Y=Spec(R) \to X=Spec(S)$ with $R=k[x,y]/(y^2=x(x-1)(x-\lambda))$ and $S=k[x]$ with the morphism coming from the inclusion.

In the literature I have found two different global descriptions of the dualising sheaf of a morphism by

  1. $\mathcal{Hom}_{f^{-1}\mathcal{O}_X}(f^{-1} \Omega_X, \Omega_Y)$
  2. $(f^*\Omega_X)' \otimes \Omega_Y$

and the local description as

  1. the coherent module associated to $\Omega_{R/S}$.

I would like to know how these concepts are related, so in the best case what an explicit isomorphism is. Especially, I am interested in what the image of the "canonical" element mapping $d \varphi \to d(\varphi \circ f)$ in the first case in concept $2$ and $3$ is.

This whole thing is quite confusing for me, as in my case I think $\Omega_X$ is just associated to the $k[x]$ module generated by $dx$ and $\Omega_Y$ is the $R$-module generated by $dx, dy$ modulo the relation $r=2dy-(3x^2-2(\lambda+1)x-\lambda)dx$.

But then description $2$ gives me the module associated to $R[dy,dx,(dx)^{-1}]/r$ while concept $3$ gives me (as $dx=0$) just $R[dy]/2dy=R$, and in none of these cases I am able to describe the "canonical element". So I would be grateful for some clarification.

Maximal abelian extension of finite abelian Galois extension of local fields of characteristic 0

Posted: 17 Jul 2022 08:11 AM PDT

Let $K$ be a non-archimedean local field of characteristic $0$. One might take a finite extension of $\mathbb{Q}_p$ for some prime $p$. We fix an algebraic closure of $K$, and, throughout, every extension of $K$ is in that fixed algebraic closure. Let $L$ be a finite abelian Galois extension of $K$. I am wondering if there is any known relation between the maximal abelian extensions $L^{ab}$ of $L$ and $K^{ab}$ of $K$. For example, do we have $K^{ab}\subset L^{ab}$? If then, can we expect $\operatorname{Gal}(L/K)\simeq\operatorname{Gal}(L^{ab}/K^{ab})$?

Proof of inequality $(1+X(1-A))*(1+X)^{A+B}>(1+X)$

Posted: 17 Jul 2022 08:10 AM PDT

I'm getting crazy over a proof that I do not manage to perform.

I want to show for my seminar group that

$$(1+X(1-A))*(1+X)^{A+B}>(1+X) \quad∀\quad X>0\quad and\quad A,B\in(0,1)$$

and playing with numbers shows that this is always true. But no matter what I try, i don't get closer. The problem is that $1+X(1-A)<(1+X)$ and $(1+X)^{A+B}<(1+X)$ but jointly they are $>1+X$.

Do you have any ideas how to show this?

Simple Matrix inequality

Posted: 17 Jul 2022 08:18 AM PDT

Let $A, B$ be two $n \times n$ matrices.

It is known that $(A-B)(A-B)^*\leq 2(AA^*+BB^*)$

How about the lower bound in terms of $AA^*$ and $BB^*$?

Can anyone please tell me why x must belong to natural number and must be greater than equal to 2? in these xth root type of log questions?

Posted: 17 Jul 2022 08:23 AM PDT

$$\log 4 \big(1+\frac1{2x}\big )\log 3=\log\big(\sqrt[x]{3}+27)$$

My teacher said that in xth root type of logarithmic questions x must be greater than equal to 2 and be a natural number but did not explained why Can someone please help me?

If $x^2 + y^2 =36$ and $x,y<0,$ what can we conclude?

Posted: 17 Jul 2022 08:36 AM PDT

I got this question on a mock exam:

If $x$ and $y$ are both negative numbers and $x^2 + y^2 =36,$ which of the following must be true?

a) $x + y > 36$

b) $x + y > 6$

c) $x + y = -6\quad$ (my original guess)

d) $x + y < -6$

e) $x + y < -36\quad$ (the "correct" answer)

My answer: $$x + y = -6,$$ because the square of that is $36$.

Answer according to the answer sheet: $$x + y < -36,$$ which makes no sense to me. How does this work?

If $lim_{n\to \infty} na_n \to 0$ and $\sum_{n=1}^{\infty} a_n x^n \to S$ $\forall$ $x\in (0,1)$ then $\sum_{n=1}^{\infty} a_n=S$. Is this false?

Posted: 17 Jul 2022 08:10 AM PDT

Question :If $\lim_{n\to \infty} n a_n \to 0$ and $\sum_{n=1}^{\infty} a_n x^n \to S$ for all $x\in (0,1)$ then $\sum_{n=1}^{\infty} a_n=S$. Is the statement true or false?

If I choose $a_n=\frac{1}{n!}$ (its limit is $0$ when multiplied by n) and $S=e^x$ (I can choose any of the $x\in (0,1)$, but consequently $\sum_{n=1}^{\infty}a_n=e$. So it should be false.

My problem is that the given answer is true, what is wrong with this counter? Thanks.

roll until lose game with changing probability

Posted: 17 Jul 2022 08:33 AM PDT

  • Start with probability p $= 1$

  • Keep rolling until you get a failure.

  • Modify p after each roll, multiplying it by k

  • Count the number of successes.

    event_count = 0      p = 1      while random() < p:          event_count += 1          p *= k  

$E$ = expected value of event_count at the end

I'm looking for a formula for k in terms of $E$.

(This would be the inversion of the function for $E$, given k.)


random() gives a continuous uniform distribution $[0, 1)$

The probability p changes at each iteration (multiplied by k).

event_count will always be at least $1$, since p starts at $1$, so the first random roll will always be a success.


If k $>= 1$, event_count is infinity.

If k $<= 0$, event_count is $1$.

So the interesting values of k are $(0, 1)$


Some samples give these numbers:

            E: k              1.0: 0.0,              1.01: 0.01,              1.1: 0.099,              1.25: 0.238,              1.5: 0.42,              2.0: 0.645,              3.0: 0.833,              4.0: 0.904,              5.0: 0.938,              6.0: 0.957,              7.0: 0.9685,  

Most of these digits are significant.

I found $0.968$ to be pretty reliably giving $E < 7$ and $0.969$ to reliably give $E > 7$


What's the formula to relate $E$ and k?

Find $\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $

Posted: 17 Jul 2022 08:38 AM PDT

Let $a \in (0,1)$, I would like to integrate

$$\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $$

Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $t^*$ such that $\sin(\pi t^*/2)^2 = a^2.$ Now $t^* <t$ such that the integral is well-defined.

Please let me know if you have any questions.

Equivalent definitions of oscillation

Posted: 17 Jul 2022 07:56 AM PDT

Yesterday I had to deal for the first time with the concept of oscillation: I found this wide general definition:

Let $X$ a topological space, $(Y,d)$ a metric space and $f:X\to Y$ a map. The value $$\omega_f(x_0):=\inf\Big\{\sup_{x,y\in U}d(f(x),f(y))\,|\,U\subseteq X\,\mathrm{neighbourhood\,of}\,x_0\Big\}\in[0,+\infty]$$ is called oscillation of $f$ in $x_0\in X$.

In particular, if we assume $f:(X,d)\to\mathbb R$ with its natural metric, its pretty obvious that $$\omega_f(x_0)=\inf_{\varepsilon>0}\,\sup_{x,y\in B_\varepsilon(x_0)} |f(x)-f(y)|=\inf_{\varepsilon>0}\left[\sup_{x\in B_\varepsilon(x_0)} f(x)-\inf_{x\in B_\varepsilon(x_0)} f(x)\right]\qquad(\ast)$$ where $B_\varepsilon(x_0)\subseteq X$ is the open ball centered $x_0$ with radius $\varepsilon>0$ in the metric space $(X,d)$. Now I ask if there's a relationship between oscillation, limsup and liminf of $f$ in $x_0$: I imagine that the following holds $$\omega_f(x_0)=\limsup_{x\to x_0} f(x)-\liminf_{x\to x_0} f(x).$$ On this way, I think to have proved one inequality: if we say $B_\varepsilon:=B_\varepsilon(x_0)$ and $B^*_\varepsilon:=B_\varepsilon\setminus\{x_0\}$, then we have $$\limsup_{x\to x_0} f(x)-\liminf_{x\to x_0} f(x)=\inf_{\varepsilon>0}\sup_{x\in B^*_\varepsilon} f(x)-\sup_{\varepsilon>0}\inf_{x\in B^*_\varepsilon} f(x)\leq\inf_{\varepsilon>0}\sup_{x\in B_\varepsilon} f(x)-\sup_{\varepsilon>0}\inf_{x\in B_\varepsilon} f(x)=\\=\inf_{\varepsilon>0}\sup_{x\in B_\varepsilon} f(x)+\inf_{\varepsilon>0}\sup_{x\in B_\varepsilon} (-f(x))\leq\inf_{\varepsilon>0}\left[\sup_{x\in B_\varepsilon} f(x)+\sup_{x\in B_\varepsilon} (-f(x))\right]=\omega_f(x_0)$$ where the last equality holds for $(\ast)$. Suppose these lines up here are correct, I can't help myself proving the opposite inequality and so I ask for your ideas -- any hint is appreciated :-)

Roll a pair of 6-sided dice. $S=$ total score. What is the expected number of pair-rolls to get the same value of $S$ consecutively?

Posted: 17 Jul 2022 08:41 AM PDT

Roll a pair of unbiased 6-sided dice.

Let $S=$ total score. So the possible values of $S$ are 2, 3, 4, ..., 12.

Let $X$ = number of pair-rolls up to and including the first time we get the same value of $S$ consecutively.

For example, if the values of $S$ are {9, 7, 12, 3, 8, 5, 5} then $X=7$.

Question: What is the value of $E(X)$?

What makes this question difficult for me is that the values of $S$ have different probabilities, so I cannot directly use a geometric distribution. And I have not been able to find a useful recurrence relation.

I made a simulation on excel with 1000 trials, and $E(X)$ seems to be approximately $10$.

(Context: No special context; I've been looking at various questions involving dice n MSE, and I came up with this question.)

Coin tossing game - probabilities of $(K,K,K)$ and $(Z,K,Z)$

Posted: 17 Jul 2022 08:09 AM PDT

Let's consider the following game of two players:

A fair coin is tossed as long as no one has won. Player $A$ wins if the sequence $(Z,K,Z)$ appears and player $B$ wins if $(K,K,K)$ appears. What is the probability that $A$ wins?

(Note: To solve this problem we are only allowed to use the notion of conditional probability, law of total probability and bayes theorem!)


My approach:

Let be $P(A)$ the probability that $A$ wins. I am not sure how to tackle this problem in a rigorous way because I am not able to find an appropriate probability space. What I could do is assume that we are at some stage of the game where nobody has won. Then, we define two sets \begin{align*} &B_{K}:=\{\omega_1=K,\omega_2\in\{Z,K\},\omega_3\in\{Z,K\},\dots\}\\ &B_{Z}:=\{\omega_1=Z,\omega_2\in\{Z,K\},\omega_3\in\{Z,K\},\dots\}. \end{align*}

Those two sets are obviously disjoint and all future sequences are either member of $B_{K}$ or $B_{Z}$. Further, we can easiliy see that there exists a bijection between both sets.

Is it correct if I simply define a sample space $\Omega$ which consists of all future sequences that either start with $K$ or $Z$? This would mean that $P(B_{K})+P(B_{Z})=1$ and $P(B_{K})=P(B_{Z})=0.5$ and allows us to apply the law of total probability $$ P(A)=P(A\mid B_{K}) P(B_{K})+P(A\mid B_{Z}) P(B_{Z})=0.5\cdot(P(A\mid B_{K}) +P(A\mid B_{Z})). $$


Is this the way to go? It doesn't feel rigorous because of two things.

1.) The sample space is only valid for some arbitrary stage of the game - not the whole game.

2.) How do I know that there exists a probability function which I can use? The sample space I have defined is not countable, so I don't know if I can assign a probability to each element of $\Omega$?

The congruence $f(x) \equiv 0 \mod p^2$ has either $p^2$ solutions or it has at most $p^2-p+1$ solutions in $\mathbb{Z}_{p^2}$.

Posted: 17 Jul 2022 08:12 AM PDT

Let $f(x)$ be a polynomial with integer coefficients and $p$ be a prime.

Prove that the congruence $$f(x) \equiv 0 \mod p^2$$ has either $p^2$ solutions or it has at most $p^2-p+1$ solutions in $\mathbb{Z}_{p^2}$.


If $f(x)$ has $p^2$ roots then we are done and $f(x) \equiv 0 \mod p^2$ has $p^2$ solutions.

But I am stuck with the other part to show that in other cases it has at most $p^2-p+1$ solutions in $\mathbb{Z}_{p^2}$.

One thing I have observed is that $p^2-p+1 = p^2 - (p-1)$ so we have to eliminate $p-1$ terms from $\mathbb{Z}_{p^2}$.


Any hints will be helpful.

How to avoid mis-reading? And why its happening? [closed]

Posted: 17 Jul 2022 08:20 AM PDT

Recently I'm observing that in many math questions I'm reading the question wrongly by skipping a sentence or seeing a value which is not that or reading additional words that aren't there. This results in sometimes making the problem unsolvable as the complexity increases too much or sometimes the opposite happens or in other cases the question ceases to make any sense. But in most of these cases even re-reading doesn't helps as you re-read the way you already did!

Did such a thing happened or happens to anyone else? Do you know why it happens so? And most importantly, do you know how to make this happen less often or completely overcome it?

Some of the examples where such a thing happened to me, one of them is this and the other one is this apart from the examples where I find such a mistake myself, after seeing the solution that is with me already or after asking a friend.

A question about operators on Hilbert spaces

Posted: 17 Jul 2022 08:08 AM PDT

Let $H$ be a Hilbert space and $T,S \in B(H).$ Suppose that $T,S$ satisfy the following condition: $$\exists c\geq 0:~~|\langle Tu,v\rangle | \leq c |\langle Su,v\rangle |, \text{ for all }u,v\in H. $$ Can we say that one of $T$ or $S$ is a multiple of the other; i.e., $T=aS$ or $S=aT$ for some scalar $a?$

Thanks

Is there a group action which is not free but induces covering map?

Posted: 17 Jul 2022 08:31 AM PDT

Let group G act on topological space X, i.e. there is a group homomorphism from G to the homeomorphism group of X.

As we all know, if this group action is properly discontinuous, then it is free and induces covering map. Now I wonder the relationship between being free and inducing covering map.

Either prove the following statement or give a counterexample: If the quotient map of the group action is a covering map, then the group action must be free.

Arguments of my classmates:If there were a fixed point of the group action, then the fiber number of the quotient map couldn't be constant.

But I guess there may be such an example: The isotropy group of each point of X is not trivial, but they are isometric to each other. Then the group action is still not free, but the fiber number of the quotient map is constant.

Unfortunately, I can't give such an example.

Edit:OK, Moishe and Ruy give good counterexamples for my original question.

Now either prove the following statement or give a counterexample: If the group action is effective and its quotient map is a covering map, then this group action must be free.

I'm sorry for my greed, but I guess this is a natural question for any student who majored in math when seeing such counterexamples.

Equilibrium problem with 3 unknown forces | Hanging mass

Posted: 17 Jul 2022 07:59 AM PDT

This problem is related to my previous question on the generalized Lami's theorem. I would like to see how you solve this problem and compare with my solution. My motivation for this problem is that I have not seen A SINGLE problem of this type on the internet that considers a 4-force system in static equilibrium. All the problems that I have seen consider 3 forces and those that consider 4 never ask for three unknowns, but offer more information in a way that can be solved by vector components. How do you solve this problem using vector components? I apologize for the ugly problem.

enter image description here

Note: The cable for $T_2$ only hangs from the vertical line, NOT the horizontal.

Let $G$ be a finite group. Suppose that there exists $n\in\mathbb N_0$ such that $G_n=G_{n+1}$ and $Z(G)=1$. Show that $C(G_n)\subseteq G_n$.

Posted: 17 Jul 2022 07:55 AM PDT

Let $G$ be a finite group. Recursively define the sequence $G_1=G$ and $G_{n+1}=[G_n,G]$. Suppose that there exists $n\in\mathbb N_0$ such that $G_n=G_{n+1}$ and $Z(G)=1$. Show that $C(G_n)\subseteq G_n$.

Here, $C(G_n)$ denotes the centralizer of the subgroup $G_n$.

My attempt: I first showed that $G/G_n$ is nilpotent and then that there exists a nilpotent subgroup $H$ of $G$ such that $G=HG_n$. Assume the result is false, then $C(G_n)G_n/G_n$ is a non-trivial normal subgroup of $G/G_n$. Nilpotency of the latter implies that $$[C(G_n)G_n/G_n,G/G_n]\neq C(G_n)G_n/G_n.$$ Thus, $$[C(G_n)G_n,G]/G_n\neq C(G_n)G_n/G_n$$ and so $[C(G_n)G_n,G]\neq C(G_n)G_n $.


I can't proceed from here.

Use Abel-Plana formula to calculate an integral $\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt $

Posted: 17 Jul 2022 08:19 AM PDT

I have asked a related problem here:

$$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt $$

But I want to use the Abel-Plana formula to calculate above integral. The Abel-Plana formula says: $$\sum_{n=0}^\infty f(n)=\int_0^\infty f(x)dx+\frac{1}{2}f(0)+i\int_0^\infty \frac{f(it)-f(-it)}{e^{2\pi t}-1}dt$$ This formula requires: $$|f(z)|\le \frac{C}{|z|^{1+\epsilon}}$$ to guarantee both $\sum_{n=0}^\infty f(n)$ and $\int_0^\infty f(x)dx$ converge.

Can I use it if $\sum_{n=0}^\infty f(n)$ and $\int_0^\infty f(x)dx$ diverge, but $\lim_{N\rightarrow \infty} \left( \sum_{n=0}^N f(n) -\int_0^N f(x)dx\right) $ exists?

Here is my attempt: $$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1} dt ~~~~~~\text{let}~~~f(z)=\frac{\ln(1+z)}{2}$$

$$f(it)-f(-it)=\frac{\ln(1+it)}{2}-\frac{\ln(1-it)}{2}=\mathrm{artanh}(it)=i\cdot \arctan(t)$$

plug into the Abel-Plana formula:

$$\sum_{n=0}^\infty \frac{\ln(1+n)}{2}=\int_0^\infty \frac{\ln(1+x)}{2}dx+0-\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1}dt$$

$$I=\int_0^\infty \frac{\arctan(t)}{e^{2\pi t}-1}dt=\lim_{N\rightarrow \infty} \left( \int_0^N \frac{\ln(1+x)}{2}dx-\sum_{n=0}^N \frac{\ln(1+n)}{2} \right) $$

$$I=\frac{1}{2}\lim_{N\rightarrow \infty} \left( ~~(N+1)\ln(N+1)-N- \ln[(N+1)!] ~~\right) $$

let $n=N+1$ and apply Stirling's formula:

$$I=\frac{1}{2}\lim_{n\rightarrow \infty} \left( ~~n\ln(n)-n+1- \ln\left[\sqrt{2\pi n} \left(\frac{n}{e}\right)^n \right] ~~\right) $$

expand and cancel the terms:

$$I=\frac{1}{2}\lim_{n\rightarrow \infty} \left( 1-\frac{1}{2}\ln(2\pi)-\frac{1}{2}\ln(n)\right)$$

Surely this limit diverges. The correct result is without the last term $-\frac{1}{2}\ln(n)$

I am wonder if there is a way to amend it? Thank you very much!

Can this property of certain pythagorean triples in relation to their inner circle be generalized for other values of $n$?

Posted: 17 Jul 2022 08:38 AM PDT

This question was raised in comments of

Is the $(3,4,5)$ triangle the only rectangular triangle with this property?

and I was suggested to ask it as a separate one.

First some notation, let's write $(a,b,c)$ where $a<b<c$ for a pythagorean triple, and let's write $(x,y,z)$ where $x<y<z$ for the distances from the vertices to the center of the inner circle of the corresponding right triangle.

a b c and x y z

($XYZ$ should be lowercase but geogebra did not allow these labels)

The answer to above question proved (partially left to reader) the property:

$x * y = z$ if and only if $c - b = 1$

In comments was asked if a similar property would exist for $c - b = 2$

and it was confirmed (and proof was left to reader) that:

$x * y = 2 * z$ if and only if $c - b = 2$

Somewhat natural question then was raised (by me) if one can generalize for other values of $n >= 1$, that is:

for which $n$ (perhaps all) holds:

$x * y = n * z$ if and only if $c - b = n$

?

Thanks to @heropup for the suggestion (and the answers for $n=1,2$)

update

A simple computer programmed enumeration seems to confirm equivalence. At least for all $(a,b,c)$ with maximum $c <= 10000$. Note that it is not known to me (but perhaps it is known to others) if all $c - b$ cover all $n >= 1$.

More in particular $n$ was never equal to $101$ (maybe no coincidence as $100*100=10000$) in the enumeration for primitive triples. It was often equal for non-primitive ones but even for non-primitive ones there were gaps. But much more near the maximum.

So asking for all $n$ is a bit ambiguous since some $n$ might never occur.

An alternative, perhaps better, question rephrase is:

Prove equality

$$x * y = (c - b) * z$$

for all pythagorean triples.

Question on: it holds "trivially".

Posted: 17 Jul 2022 07:54 AM PDT

I am not really interested in the proof per say, but only the red underlined part. I have some idea how to do it but I just want to make sure. It feels like it should be obvious to me by now but still:

The theorem

enter image description here

My Question

I am not really sure what they mean that result hold trivially

Here is what I attempted:

let $x = 0$ and $y > 0$

thus $|<x,y>| = |<\textbf{0},y>| = |<0 \cdot \textbf{0},y>| = 0 \cdot|<\textbf{0},y>| = 0$

and as $\sqrt{<x,x>}\cdot \sqrt{<y,y>} = 0\cdot \sqrt{<y,y>} = 0$

thus by equating the two equations above: $|<x,y>| =\sqrt{<x,x>}\cdot \sqrt{<y,y>}$

which implies: $|<x,y>| \leq \sqrt{<x,x>}\cdot \sqrt{<y,y>}$

Why Abel-Plana formula does not work for exponent?

Posted: 17 Jul 2022 08:26 AM PDT

Abel-Plana formula:

$$\sum _{n=0}^{\infty }f(n)=\int _{0}^{\infty }f(x)\,dx+{\frac {1}{2}}f(0)+i\int _{0}^{\infty }{\frac {f(it)-f(-it)}{e^{2\pi t}-1}}\,dt$$

If we take $f(x)=e^{-x}$, the right-hand side is $3/2$, the left-hand side is $\frac{e}{e-1}$.

How do I rotate 3 groups of 4 people into teams of 3 so that each person in one group works with each person in the other groups?

Posted: 17 Jul 2022 08:07 AM PDT

I have three teams of four people. I would like to create rotating groups of three, where each group has one person from each team, and those people rotate on a staggered schedule so that each person is in the group for 3 slots. For example, if team 1 is people a, b, c, and d; team 2 is people e, f, g, and h; team 3 is people i, j, k, l, the start of the schedule might look like this

  1. a, e, i
  2. b, e, i
  3. b, f, i
  4. b, f, j
  5. c, f, j
  6. c, g, j
  7. c, g, k
  8. d, g, k
  9. d, h, k
  10. d, h, l
  11. a, h, l
  12. a, i, l

How do I shift the order in future to make sure that the groups change and everyone gets to work with all of the people on the other teams?

Thanks!

Why do we have to take the real part in when solving $\int_{-\infty}^{\infty}\frac{\cos(x)}{x^2 + x + 1}\ dx$ (Contour integral)

Posted: 17 Jul 2022 08:17 AM PDT

Here is an example from Priestley's Complex Analysis textbook ($\Gamma_R$ is arc of semi-circle sitting on real axis centered at 0):

enter image description here

She is stating that to evaluate the given integral, we must equate real and imaginary parts in the following equation:

$$ \int_{-R}^R f(x) dx + \int_{\Gamma_R} f(z) dz = 2\pi i\ \mathrm{res}(f(z); \omega)$$

However, the LHS of the equation is equal to the RHS, and as $R \to \infty$, the second integral in the LHS goes to 0. The remaining integral on the LHS is real – so shouldn't the RHS automatically be real? Why is there a need to explicitly take the real part? Where have I gone wrong?

Conditional probabilities involving the exponential distribution

Posted: 17 Jul 2022 07:52 AM PDT

The number of years the laptop functions is exponentially distributed with mean = 5 years. If a customer purchased an old laptop which was used for last two years, what is the probability that it will function for at least 3 years?

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