Wednesday, June 15, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


Uniform convergence of $\{f_n\}$ satisfying $f_n\left(x + \frac{1}{n}\right) = f_n(x)$ implies that the limit is a constant function

Posted: 15 Jun 2022 05:57 PM PDT

I encountered this problem on a graduate school entrance test :

Let $\{f_n\}$ be a sequence of real-valued continuous functions on $\mathbb{R}$ such that $$f_n\left(x + \frac{1}{n}\right) = f_n(x) \hspace{2mm} \forall \hspace{2mm} x \in \mathbb{R} \text{ and } n \in \mathbb{N}.$$ Suppose $f:\mathbb{R} \to \mathbb{R}$ is such that $\{f_n\}$ converges uniformly to $f$ on $\mathbb{R}$, then show that $f$ is a constant function.

My attempt :

Let $x,y \in \mathbb{R}$ and $\epsilon >0$ be arbitrary. It suffices to show that $|f(x) - f(y)| < \epsilon$. By triangle inequality, given any $n \in \mathbb{N}$ : $$|f(x) - f(y)| \leq |f(x) - f_n(x)| + |f_n(x) - f_n(y)| + |f_n(y) - f(y)|$$ By uniform convergence, $\exists$ $N \in \mathbb{N}$ such that $|f(x) - f_n(x)| < \epsilon/3$ and $|f(y) - f_n(y)| < \epsilon/3 $ whenever $n > \mathbb{N}$.

So, now it suffices to show that $|f_n(x) - f_n(y)| \to 0$. Now comes the confusing part :

  • I fix an $n$.
  • Continuity of $f_n$ implies that existence of $\delta >0$ such that $|f_n(t)-f_n(y)| < \epsilon$ whenever $|t-y|<\delta$.
  • Find $n'$ such that $\frac{1}{n'} < \delta$. Then, $\exists$ $k \in \mathbb{N}$ such that $\left|\left(x+\frac{k}{n'}\right) - y\right| < \delta$.
  • But now I can't say that $|f_{n'}(x)-f_{n'}(y)| = |f_{n'}(x+\frac{k}{n'})-f_{n'}(y)| < \epsilon$ as $f_{n'}$ might require a smaller $\delta'$ than $f_n$.

I hope I have made my point clear. If not, feel free to ignore my attempt and post your own solution.

Any help/hints shall be highly appreciated.

Product of first n primes modulo 9 is always either 3 or 6?

Posted: 15 Jun 2022 05:59 PM PDT

Why is the product of n sequential primes seemingly always either 3 or 6?

ie: $2 \cdot 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 = 510510$ and $510510 \pmod{9} \equiv 3$

$510510 \cdot 19 \pmod{9} = 3$

$9699690 \cdot 23 \pmod{9} = 6$

$223092870 \cdot 29 \pmod{9} = 3$

$6469693230 \cdot 31 \pmod{9} = 3$

$200560491030 \cdot 37 \pmod{9} = 3$

etc

So if a number * the first n primes (mod 9) isn't 3 or 6 then that number isn't the next prime?

Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.

Posted: 15 Jun 2022 06:00 PM PDT

Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.

Answer:

$a+b = 7, ab = 2$

$$\begin{align} (a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt] a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\ &= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3) \end{align}$$

now, $$\begin{align} a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\ &= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\ &= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\ &= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\ &= 2017 \end{align}$$

so

$$\begin{align} &\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\ &= 7^6 - (6*2*(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\ &= 90585 \end{align}$$ correct?

If two tangents drawn from a point to the two circles and the ratio of length of tangents is constant then find the locus of that point

Posted: 15 Jun 2022 05:34 PM PDT

Given to circles find the locus of point P such that the ratio of the length of the tangents drawn from P to the given circles is a constant

Solving the differential equation $y' = x - y^2$

Posted: 15 Jun 2022 05:47 PM PDT

I know this differential equation is "unsolvable", this is my attempt at trying to solve it.

I will start by finding a formula for a general DE of the form $ y' = f(x, y) $, where $ f(x,y) $ can be written as $ g(x)+h(y) $

  1. Take the differential of both sides

$ d(y') = d(f(x, y)) $

$ y''(x)dx = \cfrac{\partial f}{\partial x}dx + \cfrac{\partial f}{\partial y}dy $

  1. Divide both sides by $ dx $

$ y'' = \cfrac{\partial f}{\partial x} + \cfrac{\partial f}{\partial y}\cfrac{dy}{dx} $

$ y'' - \cfrac{\partial f}{\partial y}y' = \cfrac{\partial f}{\partial x} $

  1. Using the definition of $ f $ we have $ \cfrac{\partial f}{\partial y} = \cfrac{\partial}{\partial y}(h(y)) = \cfrac{dh}{dy} $ and $ \cfrac{\partial f}{\partial x} = \cfrac{\partial}{\partial x}(g(x)) = \cfrac{dg}{dx} $

$ y'' - \cfrac{dh}{dy}\cfrac{dy}{dx} = \cfrac{dg}{dx} $

$ y'' - \cfrac{dh}{dx} = \cfrac{dg}{dx} $

$ y'' = \cfrac{dg}{dx} + \cfrac{dh}{dx} $

  1. Integrate both sides

$ \displaystyle \int y''dx = \int \cfrac{dg}{dx}dx + \int \cfrac{dh}{dx}dx $

$ y' = g(x)+h(x) + c_1 $

$ \displaystyle \int y'dx = \int (g(x)+h(x)+c_1)dx $

$ \displaystyle y(x)= \int (g(x)+h(x))dx + c_1x + c_2 $

  1. Now we can use this formula to solve $ y'= x - y^2 \implies g(x) = x $ and $ h(y)=y^2 $

$ \displaystyle y(x) = \int (x - x^2)dx + c_1x+c_2 = \cfrac{x^2}{2} - \cfrac{x^3}{3} + c_1x + c_2 $

  1. Apply the initial conditions

$ y(0) = 1 \implies (c_2 = 1) \wedge (y'(0) = 0 - y(0)^2 = -1) $

$ y'(0) = -1 \implies c_1 = -1 $

$ y(x) = \cfrac{x^2}{2} - \cfrac{x^3}{3} -x + 1 $

This function satisfies the initial conditions but doesn't solve the DE and I can't see where I went wrong. Which step above was invalid? I'm thinking it has something to do with $ \cfrac{\partial f}{\partial y} = \cfrac{\partial}{\partial y}(h(y)) = \cfrac{dh}{dy} $ and $ \cfrac{\partial f}{\partial x} = \cfrac{\partial}{\partial x}(g(x)) = \cfrac{dg}{dx} $ but I'm pretty sure that is valid.

Is the tower property true conditional on vector?

Posted: 15 Jun 2022 05:14 PM PDT

Let $X|\theta \sim F$ with $\theta$ a random vector and $X$ a random variable. Is it true that $$ E[X] = E[E[X|\theta]] $$ even though $\theta$ is a vector? If so, can one say that if $X|\mu,\sigma \sim N(\mu, \sigma^2)$ then $E[X] = E[E[X|\mu,\sigma]] = E[\mu]$?

Proving a closed form of an integral

Posted: 15 Jun 2022 05:36 PM PDT

Is there any proof for this integral, $$\int_0^1 \frac{1}{a^2x^2+1}[(1-\frac{x}{2}\ln\frac{1+x}{1-x})^2+\frac{\pi^2x^2}{4}]^{-1}dx\;=\frac{\arctan\,a}{a-\arctan\,a}-\frac{3}{a^2},\text{Re}(a)\gt0.$$ I tried substituting $x\,=\,\frac{1-x}{1+x}$ but the integral seems to be harder. Please pardon the nature of the question as I'm still new to LaTeX and MathJax.

Calculating Probability For 2 Exploding Dice

Posted: 15 Jun 2022 05:24 PM PDT

Background: For those who don't know, exploding dice refers to when you roll the maximum on a die and reroll, cumulatively adding the rolled values until you don't roll the maximum anymore.

Problem: I am trying to find the probability of rolling a value $x$ when rolling 2d10, where each d10 can explode. I have already come up with an equation to calculate the probability of rolling a value on a single n-sided die: $$P\left(x\right)=\left(\frac{\left(\frac{x}{n}-\lfloor(\frac{x}{n})\right\rfloor)}{\left\{x-\lfloor(\frac{x}{n})\rfloor\cdot n=0:1, x-\lfloor(\frac{x}{n})\rfloor\cdot n\right\}}\right)^{\left(1+\lfloor(\frac{x}{n})\rfloor\right)}$$ where $x$ is the expected value, and $n$ is the number of sides on the die.

Is there a way I can use the equation I already have that I'm unaware of to calculate the probability of value $x$ with 2 exploding dice?

Orthonormal basis consisting of simultaneous eigenvectors for $A$ and $A^*$

Posted: 15 Jun 2022 05:55 PM PDT

Suppose that $H$ is a finite-dimensional Hilbert space over $\mathbb{C}$ and $A$ is a normal linear operator on $H$. Let $V$ be a closed subspace of $H$ that is invariant under $A$. Show that there exists an orthonormal basis for $V$ consisting of simultaneous eigenvectors for $A$ and $A^*$.

HINT: Show that since A is normal, then there exists a simultaneous eigenvector of A and $A^*$. Furthermore, show that if V is invariant under A, then $V^\perp$ is invariant under $A^*$.

I have proven both of the hints. From here, it seems like we should take the simultaneous eigenvector we have form the hint, consider the orthogonal complement of its span, and the find a simultaneous eigenvector in the orthogonal complement using the second part of the hint. Then we could continue this process until we have all of our desired simultaneous eigenvectors. However, I'm not sure how to do this.

Any help would be greatly appreciated!

Why does $Ax+By+C+\lambda(A'x+B'y+C')=0$ represent the set of lines that pass through the intersection of $Ax+By+C=0$ and $A'x+B'y+C'=0$

Posted: 15 Jun 2022 05:54 PM PDT

For example, given $\color{green}{l_1:5x-2y-8=0}$ and $\color{blue}{l_2:3x+8y-8=0}$,

img1

We can compute the set of lines that pass through the intersection of $l_1$ and $l_2$

$$5x-2y-8+\lambda(3x+8y-8)=0$$ $$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$ for any $\lambda\in\Bbb{R}$

$$\lambda$$ $$(5+3\lambda)x+(-2+8\lambda)y+(-8-8\lambda)=0$$
$$-1$$ $$x-5y=0$$
$$1$$ $$4x+3y-8=0$$
$$2$$ $$11x+14y-24=0$$

img2

Why does this method work?


The closest thread I could find is, A general circle through the intersection points of line $l$ and circle $S_1$ has the form $S_1+\lambda L$. What is the significance of $\lambda$?.

For example if we want to find lines through the point of intersection of 3x+4y+5=0 and 2x+y+4=0 . The required lines would be obtained by substituting different values of λ in 3x+4y+5+λ(2x+y+4)=0

The accepted answer is,

Let us take up the case of lines first. Let $L_1(x,y)$ and $L_2(x,y)$ be two lines which intersect at $(a,b)\\$. Thus $$L_1(a,b)=0\\L_2(a,b)=0$$Now let $L_3(x,y)$ be another line such that $$L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$$Now, if we are able to show that $L_3$ passes through $(a,b)$,i.e. the intersection point of $L_1$ and $L_2$, our job will be complete.To do this we put $(a,b)$ in our expression for $L_3$ $$L_3(a,b)=L_1(a,b)+\lambda L_2(a,b)$$ $$\Rightarrow L_3(x,y)=0+\lambda .0$$ $$\Rightarrow L_3(x,y)=0$$So as you can see, for any value of $\lambda$, our line $L_3$ always passes through the intersection of lines $L_1$ and $L_2\\$. You can the same with any two curves(e.g. two circles) .

What I understood from this answer is,

  • Consider the lines $L_1(x, y)$ and $L_2(x, y)$ which intersept at $(a, b)$ such that $L_1(a, b)=0$ and $L_2(a, b)=0$
  • Assume that $L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$
  • Then this means that $L_3$ passes through $(a, b)$

I do not understand how this proves that $L_3(x,y)=L_1(x,y)+\lambda L_2(x,y)$ spans a set of distinct lines that pass through $(a, b)$.

In the case of circles of the form $x^2+y^2+Dx+Ey+F=0$ and $x^2+y^2+D'x+E'y+F'=0$ that intersect and are not concentric, we can't have $\lambda=-1$ because we would get their radical axis, not another circle.

I'm looking for a proof by deduction if possible (as opposed to assumption).

How do you create a novel numerical integration algorithm based on your own rules/axioms?

Posted: 15 Jun 2022 05:14 PM PDT

For example, if you defined the symbolic derivative with respect to $x$ of an equation of the general form $x^n$ to be $\sin(n)x^{\ln(n)},$ instead of $nx^{n-1}.$ How would you design the numerical derivative algorithm for an array of numbers using these arbitrary rules?

History of special quadratic reciprocity $(-3/p)_2$ and $(5/p)_2$

Posted: 15 Jun 2022 05:12 PM PDT

https://hsm.stackexchange.com/questions/14533/special-quadratic-reciprocity-3-p-2-and-5-p-2-in-addition-to-1-p-2

asks about the history of special cases of quadratic reciprocity that are understandable for "geometric" reasons. I'm curious whether Euler or Lagrange knew these cases before Gauss. I'd imagine that Franz Lemmermeyer @franzlemmermeyer knows this very well, but I only belatedly realized that he is not attending the history of science and math stack exchange...

Do we have to show that the integral is convergent

Posted: 15 Jun 2022 05:20 PM PDT

I have seen many proofs that calculate by using some substitutions and arriving at some simple equalities that show $$\int_0^{\pi/2}\ln(\cos x)dx = -\frac{\pi}{2}\ln2$$

They do the following: $$I=\int_0^{\pi/2}\ln(\cos x)dx = -\int_{\pi/2}^0\ln(\cos(\frac{\pi}{2}-x))dx = \int_0^{\pi/2}\ln(\sin x)dx$$

$$2I=I+I=\int_0^{\pi/2}\ln(\cos x)dx + \int_0^{\pi/2} \ln(\sin x)dx= \int_0^{\pi/2} \ln(\sin x \cos x)dx$$$$=\int_0^{\pi/2} \ln\left(\frac{\sin(2x)}{2}\right)dx$$$$= -\frac{\pi}{2}\ln2 + \int_0^{\pi/2} \ln(\sin(2x))dx=-\frac{\pi}{2}\ln2 + \frac{1}{2}\int_0^{\pi} \ln(\sin x)dx$$$$=-\frac{\pi}{2}\ln2 + \frac{1}{2}\left(\int_0^{\pi/2} \ln(\sin x)dx + \int_{\pi/2}^{\pi} \ln(\sin x)dx\right)$$$$=-\frac{\pi}{2}\ln2 + \frac{1}{2}\left(\int_0^{\pi/2} \ln(\sin x)dx + \int_0^{\pi/2} \ln(\cos x)dx\right)$$$$2I=-\frac{\pi}{2}\ln2 + \frac{1}{2}\cdot2I$$$$I=\int_0^{\pi/2}\ln(\cos x)dx = -\frac{\pi}{2}\ln2$$

My question is as follows:
In many calculus courses, to show that the result is the number we manipulate as in here and arrive at a more solvable equation, we had also to show that these things (integrals, limits, etc.) converge. Is the proof enough if we do not show that the integral converges? If it is not enough, how can we show that the integral converges?

A functional calculus question

Posted: 15 Jun 2022 05:25 PM PDT

Let $f(x)$ be a continuous function in the real number set, such that

$$f(x)=\begin{cases} \pi, \,\, \text{if} \,\, x> 1\\ g(x), \,\,\text{if}\,\,x≤ 1\end{cases}$$

Which of these statements are always correct?

1. $$\lim_{x\to 1} g(x)=\pi$$

2. $$g(1)=\pi$$

3. $$\lim_{x\to 1^-}\frac{f(x)}{g(x)}=1$$

My attempts.

Maybe I can not see the right connection about the continuity between $f(x)$ and $g(x)$ . That is my first problem.

We have $$f(1)=\lim_{x\to 1^+} f(x)=\pi=\lim_{x\to\ 1^-}f(x)=g(1)\implies \pi=g(1) $$

I think this doesn't imply,

$\lim_{x\to 1} g(x)=\pi$.

I want to say that about $3$, that is not always correct. Because, it can be

$$\lim_{x\to 1^-}g(x)=0$$

But, $$g(1)=\pi$$

So, my answer is $2$. It is possible that, I am completely wrong. I am not sure, what is going on here, exactly.

Is $S=\{\overline{0},\overline{5},\overline{10}\}\subset\mathbb{Z}_{15}$ isomorphic to$\mathbb{Z}_{3}$ as rings?

Posted: 15 Jun 2022 05:58 PM PDT

My book asked me to prove that it is in fact. I generated the tables for addition

\begin{array}{c|ccc} + & 0 & 5 & 10 \\ \hline 0 & 0 & 5 & 10 \\ 5 & 5 & 10 & 0 \\ 10 & 10 & 0 & 5 \end{array} \begin{array}{c|ccc} + & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \end{array}

and for the product

\begin{array}{c|ccc} \cdot & 0 & 5 & 10 \\ \hline 0 & 0 & 0 & 0 \\ 5 & 0 & 10 & 5 \\ 10 & 0 & 5 & 10 \end{array}

\begin{array}{l|lll} . & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 2 & 0 & 2 & 1 \end{array}

If I choose 0 $\mapsto 0$, $10 \mapsto 1$, because 10 and 1 are units of those rings, respectively and $5 \mapsto 2$, there is no problem to prove that $f(xy)=f(x)f(y)$, but $$f(5+5)=f(10)=1 \neq 2=1+1=f(5)+f(5)$$

Did I do anything wrong?

Continuity condition of the derivative for the function to be linear

Posted: 15 Jun 2022 05:10 PM PDT

We are given some properties of a function $f$ defined on $\mathbb{R}$:

$f'(x)$ is defined everywhere

The function satisfies $f'(x) = f'\left({\frac{x}{{2^n}}} \right)$ for all $n\in\mathbb{Z}^{+}$

The questions are:

Can we verify that $f(x)$ is linear if $f'(x)$ is continuous?
Do we have to have the condition of $f'(x)$ being continuous at all points to verify that $f(x)$ is a linear function?

How many turns are needed to get a length L of the steel blade roll?

Posted: 15 Jun 2022 05:08 PM PDT

I'm doing a college project and I came across the need to calculate how many turns it takes to get a length $L$ of a steel blade roll. I am not able to develop a formula for this.
The roll length formula I'm using is:$$L=\frac{\pi}{T}\cdot \left (\left (\frac{d_2}{2}\right )^2-\left (\frac{d_1}{2}\right )^2\right ).$$And here is a sketch of the roll:

Number of draws before you see all candies?

Posted: 15 Jun 2022 05:07 PM PDT

Recently, I thought of the following question: Suppose there are 5 candies in a bag - you choose two candies, and then put these two candies back in the bag (assume each candy has an equal probability of being selected). On average, how many times do you need to choose candies before you are guaranteed to have seen every candy at least once?

In a way, this problem kind of reminds me of the "Coupon Collector Problem" (https://en.wikipedia.org/wiki/Coupon_collector%27s_problem), but I am not sure how to solve this problem using the Coupon Collector framework. I thought of framing this problem as a Markov Chain:

  • State 2 : You have observed 2 unique candies
  • State 3: You have observed 3 unique candies
  • State 4: You have observed 4 unique candies
  • State 5: You have observed 5 unique candies (Absorbing State)

It took me a long time, but I think I was able to create a Transition Matrix for this problem :

 A = matrix(     c(0.1, 0.6, 0.3, 0, 0,0.3, 0.6, 0.1, 0,0, 0.6, 0.4, 0,0,0, 1), # the data elements     nrow=4,              # number of rows     ncol=4,              # number of columns    byrow = TRUE)              [,1] [,2] [,3] [,4]  [1,]  0.1  0.6  0.3  0.0  [2,]  0.0  0.3  0.6  0.1  [3,]  0.0  0.0  0.6  0.4  [4,]  0.0  0.0  0.0  1.0  

From here, I suppose I could use the Theory of Markov Chains and find out the expected number of transitions until you reach the Absorbing State - but it was quite difficult to correctly calculate the transition probabilities. I imagine that once the number of states (i.e. "candies") increase, it will become very difficult to calculate all these transition probabilities.

I was hoping for an easier way which would directly allow you to calculate the expected number of draws needed to observe "M" candies (at least once) with "N" draws and each draw of size "K" (e.g. M = 5, K = 2, N = ?) - provided you are given the probability of selecting any given candy (e.g. suppose the candies did not have equal probabilities of being selected).

Can someone please suggest another way of solving this problem?

Thanks!

  • "Food" for Thought: Suppose there were "M" candies" and you draw "K" candies "N" number of times. Suppose this time, you don't know the true value of "M" and you only have information on "K" and "N" - is there a way to estimate "M" based on the data you collect from "K" and "N"?

Prove that the set of all odd order elements in a group of order $2022$ form a normal subgroup [closed]

Posted: 15 Jun 2022 05:44 PM PDT

Q. Let $G$ be a group of order 2022. Let $H=\{g \in G \mid \operatorname{order}(g)$ is odd $\}$. Prove or disprove that $H$ is a normal subgroup of $G$.

My thoughts:

  1. The subgroup generated by $H$ will be normal (in any group irrespective of the order) considering the conjugate isomorphism.

  2. In the groups like $G=S_4 $,$S_5,...,$ the set of all odd order elements need not be a subgroup as well.

My doubt: Is there any relevance for the order $2022$?

How to say that $H$ need not be even a subgroup for this particular group $G$?

Joining $\infty$ and -$\infty$ in complex contour integral.

Posted: 15 Jun 2022 05:20 PM PDT

When solving the following integral $$\int_{-1}^1 \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$$ using complex contour integration, I decided to take the keyhole contour, looping around the branch point at $-1$. This however led me to the same integral up top, but with the limits between $-1$ and $\infty$.

To get around this, I decided to use the following reasoning, that I'm not quite sure is correct. Hear me out.

$\int_{-1}^\infty \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$ = -$\int_{1}^{-\infty} \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$ = $\int_{-\infty}^{1} \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$ by a simple swapping of signs.

thus, using the idea of the Riemann sphere, I said that $\int_{-1}^\infty \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$ + $\int_{-\infty}^{1} \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$ = $\int_{-1}^1 \frac{1}{(x^4+1)\sqrt{1-x^2}} \, dx$. The way I thought about this is that we can connect -$\infty$ and +$\infty$ on a Riemann sphere, so going around it amounts to the same thing as going from $-1$ to $1$. Again, I'm not a rigorous mathematician, more of a Physicist, so I only care if this reasoning would lead me to the correct answer.

Optimization problem on simplex

Posted: 15 Jun 2022 05:01 PM PDT

Let $ \Delta_2 = \left\lbrace \boldsymbol{\pi} \in \mathbb{R}^{3} | \; \forall i \in \{1, 2, 3\}: \, \pi_i \geq 0, \, \sum_{i = 1}^{3} \pi_i = 1 \right\rbrace $ denote the standard 2-Simplex. Now, I want to find all extreme points when I constraint my $\pi_i$ for all $i \in \{1,2,3\}$, e.g.: $\; 0.2 \leq \pi_1 \leq 0.8,\; 0.4 \leq \pi_2 \leq 1,\; 0.1 \leq \pi_3 \leq 0.9$.

Help how to approach such a problem is very much appreciated!

In how many ways can we partition $k$ (distinct) items into $n$ blocks so that we have $k_i$ blocks of size $i$ for each $i$?

Posted: 15 Jun 2022 05:11 PM PDT

This question originates from Problem 139 at https://bogart.openmathbooks.org/ctgd/s3-2-partitions.html, which I located a solution file at

https://github.com/OpenDiscreteMath/ibl-combinatorics/blob/update-bogart/ComboNoteswSolutions11-6-04.pdf

The solution, however, appears to contain some serious typos:

Solution to Problem 139

In particular, as I understand it, all "$n$" in the Solution section should be replaced by "$k$", resulting in what I think is the correct solution:

\begin{align*} \frac{k!}{\prod_{i=1}^k(i!)^{k_i}k_i!} \end{align*}

Is my understanding correct?

Proof that natural numbers with multiplication are cancellative

Posted: 15 Jun 2022 05:56 PM PDT

i want to show that the multiplicative semigroup of the natural numbers without zero are cancellative, that is: For all $h,x,y$, $hx=hy \implies x=y$

I tried proving this by induction over $h$ and letting $x,y$ fixed.

For $h=1$ it is trivial.

But now I'm stuck in the induction step, I gotta show that $(h+1)x=(h+1)y \implies x=y$ But I don't know how to do that, I tried to write it like this $hx+x=hy+y$ but that doesn't help :(

Can anyone help me?

Constructive proof of how many trees on [n] have exactly 3 vertices degree 3 and all other vertices degree $\leq 2$

Posted: 15 Jun 2022 05:52 PM PDT

How many trees on [n] have exactly 3 vertices degree 3 and all over vertices degree $\leq2$

I argued from prufer codes and got that there must be $\frac{(n-3)!}{5!}*{n-2\choose 2}*{n-4\choose 2}*{n-6\choose 2}$. I am unsure whether this result is correct (would err on the side of saying not) - however I am interested in seeing a constructive approach to this problem (i.e by constructing trees). Can't get very far with my attempt but I think I can picture the solution.

Computing mean and variance of a compound normal distribution with $\mu$ and $\sigma^2$ being random

Posted: 15 Jun 2022 05:40 PM PDT

Given the compound probability density $Y\sim\mathcal N(\mu,\sigma^2)$ with $\mu\sim F_\mu$ and $\sigma^2\sim F_{\sigma^2}$ how do we evaluate $\mathsf E(Y)$ and $\mathsf{Var}(Y)$ in terms of $\mathsf E(\mu)$, $\mathsf E(\sigma^2)$, $\mathsf{Var}(\mu)$, and $\mathsf{Var}(\sigma^2)$?

If we had only one random parameter this would be a direct application of the law of total expectation/variance. For example, suppose $\mu$ is fixed and only $\sigma^2$ is random. Then $$ \mathsf E(Y)=\mathsf E(\mathsf E(Y|\sigma^2))=\mathsf E(\mu)=\mu $$ and $$ \mathsf{Var}(Y)=\mathsf E(\mathsf{Var}(Y|\sigma^2))+\mathsf{Var}(\mathsf E(Y|\sigma^2))=\mathsf E(\sigma^2)+\mathsf{Var}(\mu)=\mathsf E(\sigma^2). $$ I am not clear on how to generalize these results to multiple random parameters ($\mu$ and $\sigma^2$ both random and possibly dependent). Can someone please explain?

Question about definition of normal bundle from the quotient space

Posted: 15 Jun 2022 05:26 PM PDT

According to Wikipedia, the definition of normal bundle is defined as,

Defintion $1$. [Normal bundle]

Let $(M,g)$ be a Riemannian manifold, and $S\subset M$ a Riemannian submanifold.

  • For a given $p \in S$ , a vector $n \in T_pM$ to be normal to $S$ if whenever $g(n,v)=0$ for all $v \in T_pS$. Then the set $N_pS$ of all such $n$ is called the normal space to $S$ at $p$.
  • $NS:= \coprod _{s \in S}N_pS$ is called the total space of normal bundle to $S$ at $p$. ............. $(*)$

From the above definition, I understand

$$N_p S := \left\{ n \in T_pM : g(n,v)=0~ for~ all~~ v ~\in T_pS \right\}$$

For example, let $M=\mathbb{R}^2$ and $S=S^1$ and pick a point $p\in S$. Then, since $S$ is embedded in $M$, naturally $p\in M$. and $T_pM$ is also well defined. then since the vector $n$ is easily constructed ( clearly the every vector $v$ who lives in $T_pM$ is perpendicular to $n$, literally, $g(n,v)=0$. ) and like the second bullet, $NS$ is also well defined for any point $p \in M$. (The following image is just visualizing my description. the pink vector is one of element of $N_pS $ )

enter image description here

Meanwhile, how about the formal definition of normal bundle? Even though such definition is slightly different from each source, essentially, the basic idea seems to use a quotient space, However, I think that the formal definition contradicts the first definition, $(*)$. To begin with, based on wikipedia description,

one can define a normal bundle of $N$ in $M$, by at each point of $N$, taking the quotient space of the tangent space on $M$ by the tangent space on $N$. For a Riemannian manifold one can identify this quotient with the orthogonal complement, but in general one cannot...

I will define for convenience,

Defintion2. [General definition of normal bundle] Let $(M,g)$ be a Riemannian manifold, and $S\subset M$ a Riemannian submanifold. Then the quotient space $TM/TS$ is called a normal bundle to $S$ at $p$.

For example, also consider $M=\mathbb{R}^2$ and $S=S^1$ and metric tensor still is Riemannian metric, $g$. if I pick two vector bundle $v,w \in TM$, the equivalence relation ~ is given,

$$v\sim w ~if~and~only~if~ v-w \in TS ...... (**)$$

Then we can viusalize both vector and $v$ and $w$ like the below picture,

enter image description here

and since $\sim$ is equivalent relation, we consider a representation $\nu \in TM/TS $. Then the representation $\nu$ would be a normal bundle. Of course, when considering $(**)$, the representation is descirbed

$$\nu=\left\{ w + \alpha u : w \in TM , \alpha \in \mathbb{R}, u \in TS \right\}$$

However, when comparing to the first definition, $\nu $ clearly does not represent normal vector. Obviously, for any point $p \in S$ , then $w+u$ is not perpendicular to the tangential vector $T_pS$

Therefore, I cannot understand why the formal definition of normal bundle , Defintion2, is a reasonable statement when considering Definition1.

Expressing any even natural number as a sum of primorials with coefficients

Posted: 15 Jun 2022 04:59 PM PDT

I'm having a hard time trying to solve the following problem:

Given any random even natural number, $x$, prove that it can or cannot be written as the product of some integer, $b$, times the primorial function with an argument being some other integer, $n$, s.t. $1< n \leq a$, where $a$ is the smallest integer s.t. $p_a\#\geq x$ and $b$ can be expressed as the sum of coprime partitions of other integers, but each partition must be coprime to $p_n\#$.

For example: if the natural number is $2^{448}-1$, can it be written as $b*p_{n}\#$ or not?

I've tried transforming the problem into another problem where instead I have multiple "coefficients" like $b$ and different arguments like $n$, then summing them. Sort of like a "primorial decomposition".

Perhaps this is relevant? Showing $\prod\limits_{p \leq x} p> e^{(1+\epsilon )x}$ and $\prod\limits_{p \leq x} p < e^{(1-\epsilon) x}$ are false for $x$ large enough.

Minimum of a sum over the simplex

Posted: 15 Jun 2022 05:09 PM PDT

Setup: I was trying to simplify the following minimization problem

$$\min\limits_{x\in\mathcal{X}}\left\{\mathbb{E}\left[\max\limits_{k\in[m]}\left\{s_k \sum\limits_{i=1}^n \xi_i x_i\right\}\right]\right\}$$ where $\xi_i\sim$ Exp$(i)$, $s_k\geq 0$ are known constants, and $\mathcal{X}$ denotes the simplex in $\mathbb{R}^n$. I was able to reduce the problem to the following $$\min\limits_{x\in\mathcal{X}}\left\{\left(\sum\limits_{i=1}^n ix_i\right)^{-1}\sum\limits_{\mathcal{K}\in\mathcal{P}[m]} (-1)^{|\mathcal{K}|-1}\left(\sum\limits_{k\in\mathcal{K}} s_k\right)^{-1}\right\}$$

where $\mathcal{P}[m]$ denotes the powerset of the set of the first $m$ natural numbers excluding the empty set. If the sum inside over the sets in the powerset were positive, then we would take $x=e_n$ to reach a minimum. But we don't know this as it depends on the sequence $\{s_k\}_{k=1}^m$.

Question: If $s_k=k$, then the sum is positive (by numerically checking) implying $x^*=e_n$, but is there a general, simple combinatorial argument to see this sum is positive or does it require more painful computation?

Source: I read this paper by Nemirovski and Juditsky on a similar stochastic utility problem, and thought that instead of i.i.d normals, independent exponentials with varying rates and no shifts would make for an interesting problem to optimize.

The convergence of the series $\sum\limits_{n=3}^\infty \frac {1} {n^2\ln(\ln(n))}$

Posted: 15 Jun 2022 05:12 PM PDT

Is the series $$\sum\limits_{n=3}^\infty \frac {1} {n^2\ln(\ln(n))}$$ convergent?

First order ODE with Dirac delta funtcion

Posted: 15 Jun 2022 05:01 PM PDT

I am looking for a direct method to solve this first order ODE with Dirac delta funtcion

$$\frac{dU(t)}{dt}+k^2U(t)=\frac{1}{\sqrt{2\pi}}\delta(t)$$

with the initial condition $U(0)=\frac{1}{\sqrt{2\pi}}$.

The solution to this problem is $$U(t)=\frac{1}{\sqrt{2\pi}}e^{-k^2t}$$

My try

The integrating factor for this ode is $$I=e^{\int k^2 dt }=e^{ k^2 t }$$

then multiplying both sides of the differential equation by $\,\,e^{ k^2 t }$, we get

$$\frac{d }{d t}\left(e^{ k^2 t } U(t)\right)=\frac{1}{\sqrt{2\pi}}e^{ k^2 t } \delta(t)$$ Integrating both sides, we have

$$e^{ k^2 t } U(t)=\frac{1}{\sqrt{2\pi}}\int{e^{ k^2 t } \delta(t)}dt+C$$

From here on, I am lost. Any suggestions?

Back ground of this problem

The above ode we got after applying Fourier transform to the following PDE $$u_{t}=u_{xx}+\delta{(x)}\delta{(t)}$$ with $u(x,0)=\delta(x)$.

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