Wednesday, December 29, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Ideals in Frobenius Lie Algebras

Posted: 29 Dec 2021 01:47 PM PST

Let $\mathfrak{g}$ be a Frobenius Lie algebra, that is, there exist $f \in \mathfrak{g}^{*}$ such that the bilinear form defined by $b(x,y)=f([x,y])$ is non-degenerate. Since $b$ is non-degenerate there exist a unique $x_p \in \mathfrak{g}$ called the principal element(asociated to $f$) which satisfies $$ f \circ ad(x_p)=f $$ and it can be proved that $$tr(ad(x_p))=\frac{\dim (\mathfrak{g})}{2}$$ which implies that $x_p \notin [\mathfrak{g},\mathfrak{g}]$. Here is my question: Why if $\lbrace x_p,x_1,\ldots, x_{2n+1} \rbrace$ is a basis of $\mathfrak{g}$ then $span\lbrace x_1,\ldots, x_{2n+1} \rbrace$ is an ideal of $\mathfrak{g}$?.

I read in an article, namely, Principal derivations and codimension one ideals in contact and Frobenius Lie algebras that this follows from the last equation but I don't agree with this. I mean, if we consider the $2$-dimensional Lie algebra $\mathfrak{g}$ whose only non-trivial Lie bracket is given by $[e_1,e_2]=e_1+e_2$, then $e_1 \notin [\mathfrak{g},\mathfrak{g}]$ but $span \lbrace e_2 \rbrace$ is not an ideal of $\mathfrak{g}$, am I wrong? If so, why?

In advance, thank you.

How do we globalize the relative cotangent sequence?

Posted: 29 Dec 2021 01:43 PM PST

I'm trying to show that if $X \stackrel{\pi}\to Y \stackrel{\rho}\to Z$ is a sequence of scheme morphisms, there is an exact sequence of quasicoherent $\mathcal{O}_X$ modules $$\pi^*\Omega_{Y/Z} \to \Omega_{X/Z} \to \Omega_{X/Y} \to 0.$$

If $X$, $Y$, and $Z$ are affine, we have the usual cotangent sequence of algebras, and I'd like to construct the exact sequence above from those maps. In the affine case, the map is given by $a \otimes db \mapsto a\;db$ and the second map is $db \mapsto db$. Since $d$ is defined globally, I would expect this second map to behave well on intersections, giving rise to a global map $\Omega_{X/Z} \to \Omega_{X/Y}$ which restricts to the usual affine one. I'm having trouble globalizing the first map, though.

I'm assuming the way to do this is defining a map $\Omega_{Y/Z} \to \pi_*\Omega_{X/Z}$ given by $db \mapsto db$, and then appealing to adjunction to get a map $\pi^* \Omega_{Y/Z} \to \Omega_{X,Z}$, which will be the map we want.

Am I overcomplicating this? Is there an easier way to see these maps glue to give the exact sequence above?

Thank you!

How to compute this exponential matrix

Posted: 29 Dec 2021 01:42 PM PST

Let $$ \begin{align} [\omega] &= \begin{bmatrix}0 &-\omega_3 &\omega_2 \\ \omega_3 & 0 & -\omega_1 \\ -\omega_2 & \omega_1 & 0\end{bmatrix} \in so(3) \\ e^{[\omega]\theta} &= I + \sin\theta[\omega] + (1-\cos\theta)[\omega]^2 \\ e^{[\mathcal{S}]\theta} &= \begin{bmatrix} e^{[\omega]\theta} & (I\theta + (1-\cos\theta)[\omega]+(\theta-\sin\theta)[\omega]^2)v \\ 0 & 1\end{bmatrix}, \quad v\in \mathbb{R}^3 \end{align} $$ In the book I'm reading, the authors define the following formula, $$ e^{[\mathcal{B}]\theta} = e^{M^{-1}[\mathcal{S}]M\theta}, \quad \quad M \in\mathbb{R}^{4\times 4} $$ The matrix $M$ is given. How can I compute $e^{[\mathcal{B}]\theta}$ using the equations at the top?

Prove a family of functions is a normal family

Posted: 29 Dec 2021 01:40 PM PST

I'm working on the following problem: Let $\{f_n\}$ be a sequence of analytic functions on $\mathbb{D}$ (the open unit disk) such that $\text{Re}(f_n) > 0$ and $\lvert f_n(0) - i \rvert < \frac{1}{2}$ for every $n$. Show that there is a subsequence of $\{f_n\}$ that converges uniformly on compact subsets of $\mathbb{D}$.

I want to apply Montel's theorem by showing that $\{f_n\}$ is uniformly bounded on each compact subset of $\mathbb{D}$. However, I'm not sure how to make use of the hypotheis.

Prove that exists $x\in[0,2]$ such that $f(x)=\frac{1}{x}$.

Posted: 29 Dec 2021 01:46 PM PST

Let $f$ be a continuous function in the interval $[0,2]$ such that $f(2)=3$. Prove that exists $x\in[0,2]$ such that $f(x)=1/x$.

So I've tried defining a new function $g$ such that $g(x)=f(x)-1/x$ although it didn't lead me anywhere, I tried getting to a situation where I can use the IVT although it seems my attempts haven't gotten me anywhere.

I haven't used any math formatting in my post because I'm not too familiar with the formatting syntax, due to my studies not being in English.

Hopefully if a mod sees my post, they will be able to update my post according to the right formats.

Thanks in advance for any advice!

Enumerate alle Combinations of Integer in different ranges

Posted: 29 Dec 2021 01:34 PM PST

I have m "slots" with indivual capacity c_m. Now I want to distribute a certain amount of integer k to those m slots without exceeding their capacity c_m. Such that every slot has a certain "contingent" between 0 and c_m and the sum of all contingents equal k.

I'm in search of every combination, without duplicates and in a most efficient way. Obviously I don't need to differntiate between coningents of same value. The slots have an identity though, so that their order matters.

I hope that describes my problem understandable (no homework btw).

My guess is that the best way would be to restrict every combination to have a difference to the last combination of at most 2, for subtracting 1 at one slot and adding 1 to another slot. But I'm open for other ideas.

If you have a constant velocity over some time interval, how can you say that the average velocity is equal to that constant velocity?

Posted: 29 Dec 2021 01:40 PM PST

My physics textbook claimed that if velocity was constant over an interval, then the average velocity was that constant velocity. The book used this to prove that the area under a curve with constant velocity is the displacement(since average velocity = displacement/time). My question is how could you (only using the definition of average velocity) initially say that average velocity is equal to the constant velocity? Wouldn't you need to first know that displacement was the area under the velocity curve?

Why in math 2*-2=-4 but also sqrt (2*-2) =2i? why both are not the same?

Posted: 29 Dec 2021 01:23 PM PST

The square exponent of number is the product of number by itself, and square roots cannot be negative this rules taught in math from the beginning of time.

But also this laws violate the fundamental law of algebra when adding anything in both sides the result of the equation will be the same.

Nevertheless, the "imaginary numbers" describe perfectly the natural world. So why we distinguish "Real numbers" from "imaginary numbers" when both are just the same but square root can be negative?

Can we describe what is a dimension but in terms of set theory and not with geometry?

Posted: 29 Dec 2021 01:22 PM PST

Imagine you have a square with vertices, points A =(0,0), B=(0,1), C=(1,0,), D=(1,1). Now get all parallel straight segments from K0 = (A,B) to Kn = (C,D) (where K = segment).

This set of line segments is aleph 1 in size.. greater than the set of straight segments that make up an infinite line ,for example I0=(0,1] , I1=(1,2] I2=(2,3] , I3=(3,4] .. In etc (where I = segment) ..same cardinality as the natural numbers aleph 0 * .

If just a square has more (the parallels*) segments from an infinite line then a plane has too.

This is my observation with Cantor's prove.

I do not know if it is true but it is interesting as we can say when a body is one dimensional or two, not in terms of geometry but through set theory.

How can a Cross-Product give a vector solutions is missing

Posted: 29 Dec 2021 01:25 PM PST

I am trying to understand the function of a cross-product

If you take the cross-product of V x U gives a another vector that is pendicular say call it c-vector?

V * c-vector = 0

U * c-vector = 0

But what i don't understand why can't this vector be written as linear combinations of

aV + bU = c-vector ?

And another question is the Cross-product used to find basis vectors?

Characterization of elements in non-algebraic, finitely generated field extensions

Posted: 29 Dec 2021 01:18 PM PST

I'm in the process of self-studying Galois theory from Dummit and Foote. I'm currently reading through section 13.2, on algebraic extensions, and I'm a little confused about something.

On page 525, the authors note that, for a field $F$, and algebraic elements $\alpha$ and $\beta$ over $F$, elements of $F(\alpha, \beta)$ are of the form $\sum_{\substack{i = 0, 1, \ldots, n - 1 \\ j = 0, 1, \ldots, d - 1}} a_{i, j} \alpha^i \beta^j$, where $n$ is the degree of the extension $F(\alpha)/F$, $d$ is the degree of the extension $F(\alpha, \beta)/F(\alpha)$, and each $a_{i, j}$ is an element of $F$.

Then, on the next page, they say that, if $\alpha$ and $\beta$ are not algebraic over $F$, one must use the same construction but also "close" under inversion to describe the elements of $F(\alpha, \beta)$. This is where I'm hung up -- I'm not quite sure what that means.

To simplify things, let's assume we're dealing with a simple, non-algebraic extension $F(\alpha)/F$ (since every finitely generated extension can be obtained recursively by a series of simple extensions). I understand this to mean that every element of $F(\alpha)$ is of the form $\frac{a_0 + a_1\alpha + a_2\alpha^2 + \cdots + a_n\alpha^n}{b_0 + b_1\alpha + b_2\alpha^2 + \cdots + b_m\alpha^m}$, where each $a_i$ and $b_j$ is an element of $F$; in other words, every element of $F(\alpha)$ is the ratio of two polynomials of arbitrarily large degree in $\alpha$ with coefficients in $F$. Is my understanding correct?

What is The highest number of information bits $N_b$ and the smallest number of bits?

Posted: 29 Dec 2021 01:18 PM PST

I want to find the highest number of information bits $N_b$ that can be reliably transmitted over a binary symmetric channel (BSC) with fixed channel parameters, error probability $ \sigma = 0.5$, and channel uses $N_c = 100$ bits.

The next question is to find the smallest number of bits $N_b$ that is required for representing $N_s $ source symbols $s$ for the source distribution and k-bit quantizer.

Is the highest number of information bits related to the entropy of the source symbols? and What are the formulas to calculate these quantities?

Taylor series at a singular point of an algebraic variety

Posted: 29 Dec 2021 01:17 PM PST

Shafarevich writes in his book Basic algebraic geometry 1 (exercise 6 of Chapter 2 section 2.4) that if $x$ is singular then any function $f\in\mathcal{O}_x$ has an infinite number of different Taylor series.

It is clearly enough to prove this for the function $0$, but the proof of uniqueness in the smooth case, Theorem 2.7, does not give hints to prove the exercise.

For example, for $x^2=y^3$ in $\mathbb{A}^N=\mathbb{A}^2$, the local parameters at $0$ are $x$ and $y$, so we can set $x^2-y^3$ as the Taylor series, as well as $\lambda x^2-\lambda y^3$. However, if the tangent space at the singular point is not the whole space, i.e., if $x_1,\dots,x_N$ are not local parameters, I do not know how to do this trick. In other words, I do not know how to do this with any kind of singularity, in particular if $u_1,\dots,u_{N-i}$ are local parameters ($0<i< codim X$).

The inclusion $N \hookrightarrow M$ defines a natural morphism $\mathcal{A}^*_{cpt}(N) \rightarrow \mathcal{A}^*_{cpt}(M)$ , what is it?

Posted: 29 Dec 2021 01:17 PM PST

In some lecture notes that I'm reading I found the following statement:

Let $M$ be a smooth manifold. The inclusion $N \hookrightarrow M$ of an open subset defines a natural morphism $\mathcal{A}^*_{cpt}(N) \rightarrow \mathcal{A}^*_{cpt}(M)$ between the differential forms with compact support.

I Know that in general if $f : M \rightarrow N$ is smooth map between two manifolds we can define a map (The pullback) $f^*:\mathcal{A}^*(N) \rightarrow \mathcal{A}^*(M) $, but how can we define a map from $\mathcal{A}^*(M)$ to $\mathcal{A}^*(N)$ ?

Hessian approximation from Jacobian

Posted: 29 Dec 2021 01:13 PM PST

I've seen two expressions for using the Jacobian to compute an approximation for the Hessian, and I don't see how they can give you the same answer in general. The papers are here(1) and here(2) (which actually have common authors). Both are approximations for the Hessian of a cost function, and are centred on some particular parameter combination $\theta^*$

Cost function: $$C(θ)=\frac{1}{2} ∑_μ(y_μ(θ)-y_μ(θ^* ))^2 $$

Jacobian matrix of y: $$J_{μ,i}=\frac{∂y_μ}{∂θ_i}$$

Hessian of cost, approximation 1:

$$H(C)=J^T J$$

Hessian of cost, approximation 2:

$$ H(C)_{i,j}=∑_μ \frac{∂y_μ}{∂θ_i} \frac{∂y_μ}{∂θ_j} $$

True Hessian:

$$ H(C)_{i,j} = \frac{\partial^2 C}{\partial \theta_i \partial \theta_j}$$

Do these actually give the same answer, even though they're different expressions?

Matrix exponential via Cayley-Hamilton

Posted: 29 Dec 2021 01:45 PM PST

Problem

For any $t\in\mathbb{R}$ compute $\exp(A_\omega t)$, where \begin{equation*}A_\omega\triangleq\left[\begin{array}{c|c} 0_2 & I_2 \\ \hline 0_2 & \Omega \end{array}\right]\end{equation*} and

  • $0_2$ is the $2\times2$ matrix whose entries are all zero;
  • $I_2$ is the $2\times2$ identity matrix;
  • $\omega$ is a given parameter and\begin{equation*}\Omega \triangleq \begin{bmatrix} 0 & -\omega\\ \omega & 0 \end{bmatrix}\end{equation*}

partial solution

Just to refresh my mind, I want to use the method (which I don't remember anymore) based on the Cayley-Hamilton theorem. First of all, the characteristic polynomial. Since $A_\omega$ is upper-triangular, holds \begin{equation*}\begin{aligned} \chi_{A_{\omega}}(s) &\triangleq \text{det}(sI_4-A_{\omega})\\ &=\text{det} (sI_2-0_2)\text{det} (sI_2-\Omega)\\ &=s^2(s^2+\omega^2)=s^4+\omega^2 s^2 \end{aligned}\end{equation*} Now the Cayley-Hamilton theorem says that \begin{equation*}\begin{aligned} \chi_{A_{\omega}}(A_{\omega}) = 0_4 \end{aligned}\end{equation*} or, more explicitly, \begin{equation*}\begin{aligned} A_{\omega}^4+\omega^2 A_{\omega}^2 = 0_4 \end{aligned}\end{equation*} so we know that \begin{equation}A_{\omega}^4=-\omega^2 A_{\omega}^2 \tag{1}\end{equation} but now how can we exploit this information to compute $\exp(A_{\omega}t)$? I don't remember very well.

Probably we can use $(1)$ to find a closed expression for $A_{\omega}^k$ that figures in the definition \begin{equation*}\exp(A_\omega t)\triangleq \sum_{k=0}^\infty \frac{(A_\omega t)^k}{k!}=\sum_{k=0}^\infty A_\omega^k \frac{t^k}{k!}\end{equation*} but honestly I don't remember how to do it.

questions

  1. How can we use CH to compute $\exp(A_{\omega}t)$?
  2. There is a more clever way to compute $\exp(A_{\omega}t)$? If yes, what is the procedure?

A prime ideal of an Integral Domain polynomial ring can be generated by two elements.

Posted: 29 Dec 2021 01:13 PM PST

I am studying for a qualifying exam, and have been working on this problem:

Let $D$ be a $PID$ and let $P$ be a prime ideal of the polynomial ring $D[x]$. Suppose that $P$ contains a non-zero constant polynomial. Prove that $P$ can be generated by two elements.

I saw this Every prime ideal in $\mathbb{Z}[x]$ is generated by at most two elements but I'm having trouble generalizing it for rings that aren't $\mathbb{Z}$. I know that $D[x]$ is also an integral domain (not a $PID$) so I don't get that $D[x]/P$ is a field, and I'm trying to understand the case where $P=(c,f(x),g(x))$ for some constant $c$ and two nonconstant polynomials $f(x),g(x)$. Does it work to take replace $f$ and $g$ with their gcd? If so, can someone please explain why?

Thanks for your help.

Does tangent line of inflection point always passes through the curve?

Posted: 29 Dec 2021 01:45 PM PST

I saw many functions on my book and all of the tangent line of inflection point always pass through the curve, Here are examples :

Example 1 :

$$f(x) = x^3 \quad (x=0)$$

Tangent line at $x = 0$, $l:y=0$

Then we know that $l$ passes through the $f(x)$.

Example 2 : $$f(x)=\sin x\quad (x=0)$$

Tangent line at $x = 0$, $l: y = x$

We know $l$ passes through the $f(x)$.

I got a curiosity about this, so my question is :

Let $f$ is differentiable function and $f$ is not a constant, linear function.

If a line $l$ is a tangent line of inflection point, $l$ passes through $f$ near of inflection point?

I think this is true, but I have no idea to prove this. Thanks for help.

Looking for help in understanding a proof of the fixed point lemma in mathematical logic.

Posted: 29 Dec 2021 01:30 PM PST

In the fixed point lemma on page 58 of https://www.math.ucla.edu/~dam/114L.18s/114Lnotes.pdf don't the implications in χ(v1) and σ need to be double implications? Is this an error in the proof?

If the proof in the above pdf is in fact correct, then I don't get the very last line of the proof asserting that you can deduce σ from φ.

I have the same issue with the Wikipedia proof: https://en.wikipedia.org/wiki/Diagonal_lemma It seems to me that {\displaystyle {\mathcal {B}}(z):=(\forall y)[{\mathcal {G}}_{f}(z,,y)\Rightarrow {\mathcal {F}}(y)]} needs to be a double implication so that the deduction can be made from {\mathcal {F}}(y)]}

Many thanks for your help.

Show that a cubic second order ODE only has periodic solutions

Posted: 29 Dec 2021 01:19 PM PST

Consider the following ODE $$x''+x+x^3=0$$ Show that it only has periodic solutions.

I tried to do the following, we consider the equivalent system

$$\begin{cases} x' = y\\ y' = -x(1+x^2) \end{cases}$$

Let $(a(t), b(t))$ be a solution of the last ODE and $I \subseteq \mathbb{R}$ the maximum interval in which it is defined, we can assume that $0\in I$. Then consider the IPV given by the ODE and the initial condition $x(0)=a(0); y(0)=b(0)$ since $(y,-x(1+x^2))$ is locally Lipschitz there is a unique solution $(x(t),y(t))$ therefore $x(t)=a(t);y(t)=b(t),\quad \forall t\in I$.

If $x(0)=y(0)=0$ the solution is trivial so we can assume that $x_0:=x(0)\neq 0 \neq y(0)=:y_0$. The function $F(x,y)=2(x^2+y^2)+x^4$ is a constant of motion, thus $(x(t),y(t))\in H:=\{(x,y):F(x,y)=F(x_0,y_0)\}$ $\forall t\in I$. It is easy to see that this is a compact set and a Jordan curve, and contains no fixed points since $F(x_0,y_0)>0$. We also that $\gamma^+(x_0,y_0)$ is bounded therefore the limit set is non-empty, compact and connected. Since $H$ is compact we also know that the limit set is contained in $H$.

I'm stuck at this point, I tried to use Poincaré-Bendixson's theorem from which we deduce that the limit set is periodic but I don't know how to continue from there to conclude that the orbit is periodic.

Getting width and height values ​that best fit a certain amount of pixels in an image

Posted: 29 Dec 2021 01:20 PM PST

If I have n number of pixels, what's the best way to get the factor of n that most resembles a square shape when using it as a dimension of an image?

Using n == 512 as an example, its factors are:

1, 2, 4, 8, 16, 32, 64, 128, 256, 512

The values that most resembles a square in this list of factors would be 16 or 32, precisely both. I don't know if this type of "middle" factor has a specific name, but is there a formula to get it, or is this more of a programming problem than a math problem?

And if n is prime, what would be the best way to fit all the pixels while keeping a square-like shape? (no problem if extra pixels need to be added).

Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$

Posted: 29 Dec 2021 01:20 PM PST

Find the smallest $n$ for which there are real $a_{1}, a_{2}, \ldots,a_{n}$ such that

$$\left\{\begin{array}{l} a_{1}+a_{2}+\ldots+a_{n}>0 \\a_{1}^{3}+a_{2}^{3}+\ldots+a_{n}^{3}<0 \\a_{1}^{5}+a_{2}^{5}+\ldots+a_{n}^{5}>0 \end{array}\right.$$

I don't have a solution, but I suspect $n_{\min}=5$ is optimal. I think so mainly because I could not find anything by brute force for either $4$ or $3$. There are quite a few solutions for $n=5$, for example:

$$a_{1}=1, a_{2}=-0.88, a_{3}=-0.88, a_{4}=0.5, a_{5}=0.5$$

$$a_{1}=-8, a_{2}=-7, a_{3}=3, a_{4}=4, a_{5}=9$$

It is also not difficult to prove by contradiction that $n_{\min}>2$. The case $n=2$ is trivial, and there is not much sense in describing it.

However, $n=3$ is more interesting. The impossibility of $n=3$ can be proved by setting one of the variables to $1$ (WLOG) and plotting the inequalities on the plane. The same logic should work for $n=4$ as well (I think). I can consider the case $n=3$ more rigorously if required of course.

Calculating closing speed between pair of aircraft

Posted: 29 Dec 2021 01:17 PM PST

Time(hh:mm:ss) Latitude(degrees) Longitude(degrees) Ground Speed(knot) Bearing(degree) Altitude(feet)
01:03:01 34.13 127.66 320 179 22000
01:03:05 34.17 127.68 322 178 21900
01:03:09 34.18 127.69 322 178 21800
... ... ... ... ... ...
Time(hh:mm:ss) Latitude(degrees) Longitude(degrees) Ground Speed(knot) Bearing(degree) Altitude(feet)
01:03:04 34.09 127.59 331 179 21600
01:03:08 34.11 127.60 330 178 21700
01:03:12 34.12 127.62 322 177 21800
... ... ... ... ... ...

Hi, I want to calculate the closing speed of all pairs of aircraft to determine the conflict rate.

Suppose that I have these trajectory data table for 2 aircraft, how can I calculate the closing speed (in knots) of those 2 aircraft?

I saw this post (https://gamedev.stackexchange.com/q/118162/) about calculating closing speed (the first answer), but it's quite confusing to calculate like that with my data on Matlab.

I understood that calculating distance between two aircraft at each timestamp and divide by the elapsed time (in this case, 4 sec) produces closing time.

Q1) But the recorded time is not exactly same per aircraft, so this 'time discrepancy' should be considered to calculate distance.

Q2) Also, the aircraft altitude changes every timestamp. So, how to calculate closing speed between different altitudes?

Actually, if the time discrepancy matter is solved, I'll calculate the horizontal distance between pairs of aircraft using the Haversine formula, and the vertical distance from the altitudes. Then I'll use Pythagoras' theorem to calculate the diagonal, and divide the diagonal distance by the elapsed time to produce closing time. Does this make sense?

Those are my questions.. any helps will be appreciated!

Note: Since it's not allowed to upload the original data publicly, I change those values manually.

Probability of receiving gifts on time

Posted: 29 Dec 2021 01:43 PM PST

Alex plans to order a birthday gift for his friend from an online retailer. However, the birthday coincides with the festival season during which there is a huge demand for buying online goods and hence deliveries are often delayed. He estimates that the probability of receiving the gift, in time, from the retailers A, B, C and D would be 0.6, 0.8, 0.9 and 0.5 respectively.

Playing safe, he orders from all four retailers simultaneously. What would be the probability that his friend would receive the gift in time?

My solution:-

If we even get 1 gift on time, the task would be done, so I thought of finding 1-P(No gift on time)

P(No gift on time)=P(A)*P(Not on time | A) + P(B)*P(Not on time | B) + P(C)*P(Not on time | C) + P(D)*P(Not on time | D)

Choosing any of gifts among A,B,C,D are equally likely therefore P(A)=P(B)=P(C)=P(D)=1/4

P(No gift on time) = 1/4 * (0.4 + 0.2 + 0.1 + 0.5) = 0.3

Therefore P(At least 1 gift on time)= 1-0.3 = 0.7

but the answer is given as 0.996 , what mistake am I making ?

Update :-

I have understood that P(A)=P(B)=P(C)=P(D)=1 as it is certain that alex has ordered the gifts from all these retailers, but my question now is why is the following notation wrong ?

P(No gift on time)=P(A)*P(Not on time | A) + P(B)*P(Not on time | B) + P(C)*P(Not on time | C) + P(D)*P(Not on time | D)

A= event of ordering gift from retailer A

P(Not on time | A) = Probability of gift not arriving on time given that it was ordered from retailer A

Is this Lauricella $\text F_\text D$ to hypergeometric R, from DLMF, conversion formula correct?

Posted: 29 Dec 2021 01:24 PM PST

Here is another post about uncommon standard special functions in the form of Elliptic Integrals which will be special cases of the Lauricella D function. Most elliptic integrals can be put in terms of a double hypergeometric functions like the Jacobi functions and others like the Arithmetic Geometric Mean, Elliptic Logarithm, and Legendre Ellipctic functions can all be put into closed form with a single or double hypergeometric series, but the Carlson Elliptic functions and Bulirsch's Elliptic Integrals. Let's start with some definitions which will allow integral and sum representations for the more complex elliptic integrals of a triple hypergeometric function with the Pochhammer Symbol. The Lauricella D function is also in Wolfram functions

$$\text F_\text D^{(n)}(a,b_1,…,b_n,c;x_1,…,x_n)=\sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(a)_{\sum_{j=1}^n m_j}}{(c)_{\sum_{j=1}^n m_j}}\prod_{j=1}^n \frac{(b_j)_{m_j}x_j^{m_j}}{m_j!}=\frac{\Gamma(c)}{\Gamma(a)\Gamma(c-a)}\int_0^1t^{a-1}(1-t)^{c-a-1}\prod_{j=1}^n (1-x_j t)^{-b_j}dt$$

When the Hyperelliptic hypergeometrc function from DLMF can give any elliptic integral as a special case:

$$\text R_a(b_1,..,b_n;z_1,…,z_n)=\frac 1{\text B(-a,(-a)')}\int_0^\infty t^{(-a)'-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt,(-a)'=a+\sum_{j=1}^n b_j$$

The problem here is that the integral bounds for the Lauricella function, so some substitutions help fix the bounds to put the Hyperelliptic function into Lauricella form: $$\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\text F^{(n)}_\text D(a,b_1,…,b_n,c;x_1,…,x_n)=\int_0^1t^{a-1}(1-t)^{c-a-1}\prod_{j=1}^n (1-x_j t)^{-b_j}dt \mathop=^{t\to\frac1t}-\int_\infty ^1t^{-a+1}(t-1)^{c-a-1}t^{-(c-a-1)}\prod_{j=1}^n (t-x_j )^{-b_j}t^{b_j}t^{-2}dt =\int_1^\infty t^{-c} (t-1)^{c-a+1}t^{b_1}\cdots t^{b_n}\prod_{j=1}^n (t-x_j)^{-b_j} dt =\int_1^\infty t^{-c+\sum\limits_{j=1}^n b_j}(t-1)^{c-a-1}\prod_{j=1}^n(t-x_j)^{-b_j}\mathop=^{t\to t+1}_{t-1\to t} \int_0^\infty (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j} $$

Now assume the two hypergemoetric functions are equal:

$$\frac{\Gamma(a)\Gamma(c-a)}{\Gamma(c)}\text F^{(n)}_\text D(a,b_1,…,b_n,c;x_1,…,x_n) = \int_0^\infty (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j}= \text B(-a,(-a)') \text R_a(b_1,..,b_n;z_1,…,z_n)=\int_0^\infty t^{(-a)'-1}\prod_{j=1}^n(t+z_j)^{-b_j}dt $$

Therefore the following is implied. Note that we had the same $a$ variable in both hypergeometric functions, so let's represent one variable for the hyperelliptic function as $a=\alpha$:

$$t^{\alpha+\sum\limits_{j=1}^n b_j-1}\prod_{j=1}^n(t+z_j)^{-b_j} = (t+1)^{-c+\sum\limits_{j=1}^n b_j}t^{c-a-1}\prod_{j=1}^n(t+1-x_j)^{-b_j} \implies z_j=1-x_j,c=\sum_{j=1}^n b_j=(-a)'-a,c-a-1= \alpha+\sum\limits_{j=1}^n b_j-1 =\alpha +c-1\implies -a\to\alpha,a=-\alpha$$

and:

$$\implies \frac{\Gamma(-\alpha)\Gamma\left(\sum\limits_{j=1}^n b_j+\alpha\right)}{\Gamma\left(\sum\limits_{j=1}^n b_j\right)}\text F^{(n)}_\text D\left(-\alpha,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right)= \text B\left(-\alpha,\alpha+\sum_{j=1}^n b_j\right) \text R_\alpha(b_1,..,b_n;z_1,…,z_n)\implies \text R_\alpha(b_1,..,b_n;z_1,…,z_n) = \frac{\Gamma(-\alpha)\Gamma\left(\sum\limits_{j=1}^n b_j+\alpha\right)}{\Gamma\left(\sum\limits_{j=1}^n b_j\right) \text B\left(-\alpha,\alpha+\sum\limits_{j=1}^n b_j\right)} \text F^{(n)}_\text D\left(-\alpha,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right) $$

The coefficients cancel. I have derived this simple looking formula before, and got the same result, but am not sure if it is correct: $$\boxed{\text F^{(n)}_\text D\left(-a,b_1,…,b_n, \sum\limits_{j=1}^n b_j;1-x_1,…,1-x_n \right)=\text R_a(b_1,…,b_n;x_1,…,x_n) =\frac{\int_0^\infty t^{a-1}\prod\limits_{j=1}^n(1+tx_j)^{-b_j} dt}{\text B\left(-a,a+\sum\limits_{j=1}^nb_j\right)}= \sum_{m_1\ge0}\cdots\sum_{m_n\ge0}\frac{(-a)_{\sum_{j=1}^n m_j}}{\left(\sum\limits_{j=1}^nb_j\right)_{\sum_{j=1}^n m_j}}\prod_{j=1}^n\frac{(b_j)_{m_j}(1-x_j)^{m_j}}{m_j!} }$$

This is the basic expression of the conversion formula, but is it correct? We can now put any elliptic integral into a Lauricella hypergeometric form and have many new decomposition formula for the Lauricalla D function which is a standard function.

Maybe there will be some examples of special cases of the formula if it is correct. Specifically, the Carlson and Bulirsch's elliptic functions, which also appear on Wolfram, many of which can only be expressed as a special case of the Lauricella D function. Please correct me and give me feedback!

Divide a ball of volume $\frac{e^2}{6}n$ into $n$ slices of equal height. What is the product of the volumes of the slices as $n\rightarrow\infty$?

Posted: 29 Dec 2021 01:23 PM PST

Divide a ball of volume $\frac{e^2}{6}n$ into $n$ slices of equal height, as shown below with example $n=8$. What is the limit of the product of the volumes of the slices as $n\rightarrow\infty$?

Ball

(If the image doesn't load for you, just imagine $n+1$ equally-spaced horizontal planes, and a ball that is tangent to the top and bottom planes. The planes, between the top and bottom planes, are where you cut the ball.)

I used volume of revolution, and after simplifying I got:

$$\lim_{n\to\infty}\exp{\left(2n-2n\ln{n+\sum_{k=1}^{n}}\ln{\left(k(n+1-k)-\frac{n}{2}-\frac{1}{3}\right)}\right)}$$

Wolfram does not evaluate this limit, but desmos tells me that when $n=10, 100, 1000, 10^6$, the product is approximately $1.847, 1.977, 1.997, 1.99999513$, respectively. So apparently the limit converges and equals $2$, but I do not know how to prove this.

(In case you're wondering how I got the number $\frac{e^2}{6}$: I used trial and error on desmos to hunt for the number that makes the limit converge, assuming such a number exists. I obtained a number like 1.231509. I entered this number into Wolfram and it suggested $\frac{e^2}{6}$.)

Summing $\sum_{i=k}^n 3\cdot 5^i$

Posted: 29 Dec 2021 01:47 PM PST

The problem is $\sum_{i=k}^n 3\cdot5^i$

I calculated it using the geometric sum formula as

$$\frac{(3 \cdot 5^k)-(3 \cdot 5^{n-k+1})}{1-5}$$

but the answer says

$$\frac{3\cdot5^k(5^{n-k+1}-1)}{4}$$

How?

Upper-bound on volume of polytope inscribed in the sphere

Posted: 29 Dec 2021 01:14 PM PST

To my great surprise, I was unable to find any general reference on the volume of (symmetric) polytopes inscribed in, say, the unit sphere.

Let $P = absconv(v_1,\dots,v_k) $ where the $v_i$'s are unit vectors. Then, I would like to have an upper bound on the quantity :

$$\frac{vol(P)}{vol(B^2)} $$ where $B^2$ is the euclidean unit ball. Intuitively the maximal volume polytope with k vertices should be made of vertices $v_i$'s forming an $\epsilon$-net of the sphere for some $\epsilon$ (that is being "uniformly" distributed) but that's beyond the point.

In particular, I'm expecting that if $k$ is sub-exponential in $n$ then this volume ratio should go to zero. That is if $k_n = o(e^{cn})$ for every $c>0$ and $P_n$ is a polytope with $k_n$ vertices on the unit sphere of $R^n$, then $$vol(P_n) = o(vol(B^2_n))$$

Is this true ? Is there any quantitative statement about it ?

What happens when apply vanishing viscosity method to $|u'| = 1$ with $u(0) = u(1) = 0$?

Posted: 29 Dec 2021 01:46 PM PST

I know the viscosity solution for $ |u'| = 1$ with $u(0) = u(1) = 0$ is $\frac{1}{2} - |x - \frac{1}{2}|$, but just wondering what will I get if I use the vanishing viscosity method?

For a similar equation $u' = 1$ I can solve the ode and get $u_{\epsilon} = \frac{e^{\frac{x}{\epsilon}} - 1}{1 - e^{\frac{1}{\epsilon}}} + x$, then as $\epsilon$ goes to $0$, $u_{\epsilon}$ will go to $x, x\in [0,1)$ and $0$ at $x = 1$ which is not a viscosity solution.

But how shall I deal with the absolute value in the vanishin viscosity ode?

Thank you!

Expected Value of Absolute Value of Difference between Two Independent Uniform Random Variables

Posted: 29 Dec 2021 01:32 PM PST

Let X be Uniform Random Variable on [0,1]. Y is Uniform Random Variable on [0,2]. Given that X and Y are independent, calculate E[|X-Y|].

I have tried calculating the answer using two different methods - both of which I think are correct approaches - but each produces differing results of 2/3 and 3/4.

Method 1) $$\int_{0}^2\int_{0}^1 (|x-y|/2)dxdy$$ I plug the following into Wolfram Alpha and get 2/3: Wolfram calculation

I'm pretty sure this a correct way of going about getting the expected value, since in general for independent x and y, E[g(x,y)] = $$\int_{a}^b\int_{c}^d g(x,y)f(x)f(y)dxdy$$ where of course f(x) and f(y) are the respective distributions of random variables X and Y, and in this particular instance is (1/2)*(1) = (1/2).

Method 2) E[X-Y|X>Y] + E[Y-X|Y>X]. This method yields me 3/4 when I calculate it. The first integral is $$\int_{0}^1\int_{0}^x (.5(x-y))dydx$$ This equals 1/12. The second integral is $$\int_{0}^2\int_{0}^y (.5(y-x))dxdy$$ This equals 2/3. (2/3) + (1/12) = (3/4).

I'm also fairly certain this is a correct procedure, since there are 3 cases: X>Y, X=Y, and X<Y. X=Y will always produce |X-Y|=0. If X>Y, then |X-Y| = X-Y. If X<Y, then |X-Y| = Y-X. Summing the expectations should yield me a correct result.

Which answer is correct (or neither)? Have I made a careless computational blunder in one of the methods? Or is one of the methods I used a fundamentally incorrect approach to solving the given problem?

EDIT: So, Method 2 is full of errors on my part, which can be noted in the comment section. While I'm certain the correct answer is 2/3, I have found a line of reasoning that yields 3/4 using conditional expectation that I am not able to disprove.

E[|X-Y|] = Pr(Y>1)(E[Y-X|Y>1]) + Pr(X>Y)(E[X-Y|X>Y]) + Pr(Y>X And 0<Y<1)*(E[Y-X|Y>X And 0<Y<1])

Pr(Y>1) = 1/2

Pr(X>Y) = 1/4

Pr(Y>X and 0<Y<1) = 1/4

E[Y-X|Y>1] = 1 (this is intuitive if you think about the symmetry)

E[Y-X|Y>X And 0<Y<1] = E[X-Y|X>Y] (once again intuitive by symmetry)

Now, by linearity of expectation, E[X|X>Y] - E[Y|X>Y] = E[X-Y|X>Y] = 0.5. Given what we know about Y, it is fair to treat Y as uniform on [0,1] in this instance. Therefore, if we expect X to be 0.5, then Y is uniform on [0,0.5], giving expectation of 0.25. The same logic applies in reverse to give us E[X|X>Y] = 0.75. You can see integral proof here: OSU Conditional Expectation Slides.

Finally, summing it all up, (1/2)(1) + (1/4)(1/2) + (1/4)*(1/2) = 3/4.

Can anyone see where I went wrong with my logic on the conditional expectation? In general, when you apply the above procedure, I think what is being spit back is actually the average of the means of X and Y. However, I'm unsure where the logic is breaking down.
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