Sunday, July 25, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


How to find the mean and median of x-(x^2 /4) function?

Posted: 25 Jul 2021 08:25 PM PDT

I am trying to find the mean and median for a function: x-(x^2 /4).

  1. For mean I have tried this using symbolab, the resulting answer was wrong.

  2. And for the median I had found that 1.11 for T in the integral resulted in an answer very close to 0.5 (Something like 0.5000032).

What is the actual solution?

Image of the problem

Can convolution be equivalent to stochastic differential equation?

Posted: 25 Jul 2021 08:19 PM PDT

Consider this equation (convolution of of two functions)

$dy(t)=f(\tau)g(t-\tau)d\tau$.

For the case when $g$ and $f$ are continuous and continuously differentiable functions, we can say that it is well-defined. However if one function say $f$ is a random variable then can we write the above equation in the form of a stochastic differential equation:

$dX_t=\mu(X_t,t)dt+\sigma(X_t,t)dB_t$.

In other words are we solving some stochastic differential equation when one of the function of a convolution is a random variable?

If alternate angles are equal, does this always mean the lines are parallel?

Posted: 25 Jul 2021 08:15 PM PDT

If alternate angles are equal, does this always mean the lines are parallel?

I want to know is this always true, or are there shapes that do not satisfy these property?

Given N arrows, A arrows and K bows find the maximum probability of winning

Posted: 25 Jul 2021 07:55 PM PDT

I am stuck with one of the questions involving probability distribution. The question mentions about an archery tournament where you have made it to the finals and knowing the rules of the tournament you have to score exactly N points to win, with A arrows and K bows. The possible points that can be scored are 0 to 5. Probability distribution of scoring points from 0 to 5 using each of K bows is given we have to find maximum probability of winning. At any point of time it is allowed to voluntarily discard the arrow and register a 0 score for that turn. Below is the sample question and answer -

int n, the exact number of points you must score to win int a, the number of arrows you have available int k, the number of bows from which you can choose for each arrow double p[][6], the probability distribution for each of the k bows: p[i][j] is the probability of scoring exactly i points using bow i The return value is the probability that you defeat your opponent.

Sample N = 3,
A = 4, K = 2, P[][] = 0.3 0.3 0.0 0.4 0.0 0.0 0.1 0.8 0.0 0.1 0.0 0.0

Ans - 0.8548

I thank you for your time and help!

Finding the probability that at least 1 student doesn't get admitted to any college.

Posted: 25 Jul 2021 07:52 PM PDT

enter image description here

enter image description here

I'm having quite a bit of trouble with this problem. I'd posted a similar question before, where all $k_i$ were 1, but now that restriction is gone, and I'm having a harder time with this.

So $\text{P(at least 1 student does not get admitted to any college)} = 1 - \text{P(all students get admitted to at least 1 college)}$. We are also given that $\frac{k_1 +... + k_n}{n} \ge 2 \ln(n)$.

Let the event A denote all students getting admitted to at least 1 college.

The problem is, I'm not sure how exactly to compute P(A), because while each college $i$ chooses $k_i$ students and accepts them, I need to make sure within the union of all of $k_i$ there is at least one of each student.

Without worrying about having everyone, I could just do :

Each of the n colleges has n! possible list of students (hence $(n!)^n$ total possibilities).

College 1 has n C $k_1$ students chosen, and has $(n - k_1)!$ permutations

College 2 has n C $k_2$ students chosen and has $(n - k2)!$ permutations

...

College n has n C $k_n$ students chosen and has $(n - k_n)!$ permutations.)

But since I have to have all students accepted into at least one college, I thought of doing for each College $i$, the college has $n - (i-1) C k_i$ students chosen and has $(n-k_i)!$ permutations. But then in this pattern, for College n, I'd be doing $1 C k_n$, which may be impossible.

How can I calculate P(A), or are there other "simpler" routes?

Has there been any attempt to classify the countable ordinals that correspond to models of set theory?

Posted: 25 Jul 2021 07:52 PM PDT

Let $S:=\{ \alpha\in \aleph_1 \mid \exists M, M \models ZFC \wedge M\cap \textbf{ORD} =\alpha \wedge |M|=\aleph_0 \}$. Has there been any research into the properties of S?

In particular, I'm interested in the connection to Shoenfield Absoluteness. Sense any two models of set theory containing the same ordinals agree on all $\Sigma_2^1$ and $\Pi_2^1$ arithmetical sentences, forcing is insufficient to prove independence of such statements. It seems to me that the logical first place to start is exploring this S. I've found a proof that this S is unbounded, but anything more seems incredibly difficult.

Finding global maximum of a function

Posted: 25 Jul 2021 07:49 PM PDT

Define $f(x) = (ax + b)e^{-x}$ where $a > 0$. I'm trying to prove that there exists a global maximum.

The only thing I could do first was differentiate. After doing so, I get $$ f'(x) = -e^{-x} (ax - a + b). $$ I have to check where $f'(x) = 0$. $-e^{-x} \neq 0$, of course, so $ax - a + b = 0$. So $ax = a - b$, so $x = \frac{a-b}{a}$. The corresponding function value is $$ f\left(\frac{a-b}{a}\right) = - e^{- \frac{a-b}{a}} \left(a \cdot \frac{a-b}{a} - a + b \right) = - e^{\frac{b-a}{a}} \left(a - b - a + b \right) = - e^{\frac{b-a}{a}} \cdot 0 = 0. $$ I don't believe this is right, because there are supposed to be $a,b$ for which the global maximum is $(2,2)$. So it seems I have done something wrong.

What is the quantifier complexity of the function sending an ordinal to the power set iterated that many times?

Posted: 25 Jul 2021 08:25 PM PDT

One use of the Axiom Schema of Replacement in ZFC is that it allows one to prove that for a set $X$ and a limit ordinal $\alpha$, the transfinitely iterated powerset $P^\alpha(X)$ exists (for a successor ordinal $\beta+1$, one only needs $P^\beta(X)$ to exist beforehand). Sometimes people consider restricted versions of Replacement with restricted complexity on the description of the function to which it is being applied. I was wondering what the quantifier complexity of the statement that Replacement is then applied to in order to get $P^\alpha(X)$. This would then put a floor on how low complexity Replacement instances can be before iterated powerset no longer becomes available.

Effective Interest Rate

Posted: 25 Jul 2021 07:47 PM PDT

appreciate any help to enlighten me on the below question.

Andrew borrowed a car loan of $80,000 with an interest rate of 1.9% per annum for 5 years. The bank offered zero interest for first 6 months, calculate the effective rate of interest of this loan?

Question about my proof of: $\displaystyle \lim_{h \to 0}f(ch)=\displaystyle \lim_{ch \to 0}f(ch)$ for $c\neq 0$

Posted: 25 Jul 2021 08:04 PM PDT

This post will be broken up into two sections: the first section will contain the proof, and the second section will contain the question. The proof will be written formally, as the question is more easily understood referencing the formal description. (Note, the context of this post is in Spivak's Calculus, which treats all functions, unless otherwise stated, as having a domain of $\mathbb R$).


Prove: $\displaystyle \lim_{h \to 0}f(ch)=\displaystyle \lim_{ch \to 0}f(ch)$ for $c\neq 0$, which is equivalent to:

If $c\neq 0$ and $\displaystyle \lim_{h \to 0}f(ch)=L$, then $\displaystyle \lim_{ch \to 0}f(ch)=L$ $\quad$ and $\quad$ If $c\neq 0$ and $\displaystyle \lim_{ch \to 0}f(ch)=L$, then $\displaystyle \lim_{h \to 0}f(ch)=L$

We will only prove the first implication (the converse is completed similarly):

By assumption: $\displaystyle \lim_{h \to 0}f(ch)=L \iff \forall \varepsilon \gt 0 \ \exists \delta \gt 0 \ \forall h \in \mathbb R \left [ 0 \lt |h| \lt \delta \rightarrow |f(ch)-L| \lt \varepsilon \right ]$

We want to show that for an arbitrary $\varepsilon$, we can construct a $\delta$ such that $\color{red}{\forall ch} \in \mathbb R \left [ 0 \lt |ch| \lt \delta \rightarrow |f(ch)-L| \lt \varepsilon \right]$

For $\varepsilon$, we know by assumption that there is a $\delta_{\varepsilon}$ such that:

$\forall h \in \mathbb R \left [ 0 \lt |h| \lt \delta_{\varepsilon} \rightarrow |f(ch)-L| \lt \varepsilon \right ]$

Now, consider a $\delta ^* = \min\left(\delta_{\varepsilon},\frac{\delta_{\varepsilon}}{|c|}\right)$.

If $0 \lt |h| \lt \delta^* \leq \delta_{\varepsilon}$, by assumption we have: $|f(ch)-L| \lt \varepsilon$.

Further, if $0 \lt |h| \lt \delta^* \leq \frac{\delta_{\varepsilon}}{|c|}$, then $0 \lt |ch| \lt |c|\delta^*$.

Therefore, let our desired $\delta$ be defined as $\delta = |c|\delta^*$. As long as $0\lt|ch| \lt \delta$, all of our criteria is met.


In the above proof, I made use of the following statement:

$\color{red}{\forall ch} \in \mathbb R \left [ 0 \lt |ch| \lt \delta \rightarrow |f(ch)-L| \lt \varepsilon \right]$

Through my brief experience in maths, the universally quantified object $ch$ is atypical. I suspect the proper way to denote this is by establishing a function of the form: $g(h)=ch$ and then defining a single symbol as representing its output. i.e. something like $s_h :=g(h)$. More specifically, we should write $g$ formally as: $g: \mathbb R \to \mathbb R$ where $h \mapsto ch$.

We would then rewrite the statement as:

$\color{red}{\forall s_h} \in \mathbb R \left [ 0 \lt |s_h| \lt \delta \rightarrow |f(s_h)-L| \lt \varepsilon \right]$

This seems to emulate the more familiar notation of a universal quantifier, where only one symbol follows the the quantifier.

I do not really know the deep theory behind first order logic, but I suspect the reason this proof "works out" is because my new symbol $s_h$ has the capacity of sweeping through all objects within $\mathbb R$. Said differently, the previously defined function $g$ can be shown to be surjective with respect to $\mathbb R$.

If the above is true, are there times where the change of variable function is not surjective, and this causes the equality between two limits to fail?

intersection of ker [closed]

Posted: 25 Jul 2021 08:19 PM PDT

Let $E$ be a normed space in $\mathbb R$ with infinite dimension Let $f_0,f_1,f_2,\ldots, f_n \in E^*$ Show that if $\bigcap_{i=1}^n \ker f_i \subset \ker f_0$ then $f_0 \in \operatorname{span}\{f_1,f_2,\dots,f_n\}$.

I would be very much appreciated if you answered.

High school exercise about polinomics.

Posted: 25 Jul 2021 08:00 PM PDT

I'm getting troubles with this exercise so, hope you can help me.

Find a and b, for each item separately, so that the following conditions are met:

a) P(x) is divisible by Q(x) and P(-1) = 10, where P(x) = $ax^4$ + bx ^ 2 + 4 and Q(x) = x + 2

b) P and Q have common roots, where P(x) = $ax^3$ + $bx^2$ + x and Q(x) = $x^2$-x

Thank you!

probability of a single student not being accepted into any college

Posted: 25 Jul 2021 07:53 PM PDT

enter image description here (Just to clarify, it says that the average of all $k_i$ is at least $2\ln n,$ NOT $21nn$.)

I'm having some problem trying to figure out part b.

First I tried picking an arbitrary college $i$, which accepts $k_i$ students. The college has $n!$ ways of ranking all the students. For Alice to note get admitted, we can select the $k_i$ students $(n-1) P k_i$ ways.

Then the probability simplifies down to $\frac{1}{(n-1-k_i)!}$

Now I want to extrapolate this, and the problem also gives that $\frac{k1 + \cdots + kn}{n} = 2 \ln n,$ but I'm not sure how to use that to give the result the problem wants. Any help would really be appreciated!

Solving the system $ax + by = 1$, $cx + dy = 2$

Posted: 25 Jul 2021 08:17 PM PDT

From Serge Lang's book:

Let $a, $b, $c, $d be numbers such that $ ad - bc \neq 0 $. Solve the following systems of equations for $x$ and $y$ in terms of $a, $b, $c, $d.

$$ \begin{cases} ax + by = 1 \\ cx + dy = 2 \end{cases} $$

What I did so far step by step

\begin{cases} ax + by = 1 \\ cx + dy = 2 \end{cases}

\begin{cases} cax + cby = c \\ cax + ady = 2a \end{cases}

\begin{cases} cax = c - cby \\ cax + ady = 2a \end{cases}

\begin{cases} cax = c - cby \\ c - cby + ady = 2a \end{cases}

\begin{cases} cax = c - cby \\ y(ad - cb) = 2a - c \end{cases}

\begin{cases} cax = c - cby \\ y = \dfrac{2a - c}{ad-cb} \end{cases}

\begin{cases} x = \dfrac{c - cby}{ca} \\ y = \dfrac{2a - c}{ad-cb} \end{cases}

\begin{cases} x = \dfrac{c(1 - by)}{ca} \\ y = \dfrac{2a - c}{ad-cb} \end{cases}

\begin{cases} x = \dfrac{1 - by}{a} \\ y = \dfrac{2a - c}{ad-cb} \end{cases}

\begin{cases} x = \dfrac{1 - b * \dfrac{2a - c}{ad-cb}}{a} \\ y = \dfrac{2a - c}{ad-cb} \end{cases}

Book's solution says that

but after multiple tries I can't get my equation to a point where $x$ looks near like the desired answer. Could someone hint me where I did a mistake in my thought process or what I am doing wrong in general?

$f(x) = 1 - \int_0^x [f(t)]^2 \ dt$ for all $x \ge 0.$. Calculus Homework Problem

Posted: 25 Jul 2021 08:25 PM PDT

The Problem: Find the function $f(x)$ that satisfies $f(x) = 1 - \int_0^x [f(t)]^2 \ dt$ for all $x \ge 0.$

What I've done: I tried solving the integral, and arrived at $\frac{F(x)^3}{3}.$ So from here, would I just be solving for the derivative of this "ultimate function" so to speak? Just lost on the integral part, I can take it from there once I know that part.

How is it possible to prove this theorem?

Posted: 25 Jul 2021 07:57 PM PDT

Could someone give me tips on how to resolve this issue? I don't even know where to start.

Let $f:X\longrightarrow \mathbb{R}$ be differentiable at point $a \in X.$ If $x_n<a<y_n$, for all $n$ and $\lim x_n=\lim y_n=a$. Prove that $\lim \left(\frac{f\left(y_n\right)-f\left(x_n\right)}{\left(y_n-x_n\right)}\right)=f'\left(a\right).$ Interpret this fact geometrically.

Thanks!

Average Value Of a Random Grid That Contains Multipliers

Posted: 25 Jul 2021 08:19 PM PDT

I am a software developer and recently came across this problem when trying to make a piece of software and it has since put that on hold.

I have a bag inside the bag there are 5 zeros, 15 ones, 5 twos, 5 threes. The threes also multiply all adjacent (all 8) squares by 2. The tiles are set randomly on a 5 by 5 grid with 5 tiles left in the bag. What is the best way to calculate the average value of grid that will be made by any number and value of numbers?

Eg. Bag ( 2 1 3 3 3 ), Grid Value: 39

$$\begin{array}{c|c|c|c|c} 3& 2&1&1&1\\ \hline 1&1&0&0&1\\ \hline 1&0&3&2&2\\ \hline 1&0&1&2&1\\ \hline 1&1&1&1&0 \end{array}$$

Without the multiplier, the solution is quite simple:

The number of values in the grid (25) × The sum of all of the numbers in the bag / the number of numbers in the bag. Or $(n * s) / a$

Since they do not interact, the fact they are in a grid does not even matter. For the same reason, it's not that useful for solving this problem.

Proof Verification of squeeze theorem

Posted: 25 Jul 2021 08:19 PM PDT

Theorem:If $X=(x_n)$, $Z=(z_n)$ and $Y=(y_n)$ be three given sequences such that
$$x_n\leq y_n\leq z_n\space for\space all \space n\in \mathbb N$$ and $lim(x_n)=lim(z_n)$.Show that $Y$ is convergent and has limit $lim(x_n)$.
Proof: We see that $0 \leq y_n-x_n \leq z_n-x_n$ and thus $|(y_n-x_n)-0|\leq |z_n-x_n|$. $lim(z_n-x_n)=lim(z_n)-lim(x_n)=0$ and thus,for every $\varepsilon>0$ we can find $K\in \mathbb N$ such that for all $n \geq K$ such that $|(y_n-x_n)-0|\leq |(z_n-x_n)-0|<\varepsilon$ and thus,$lim(y_n-x_n)=0$. Let $A=(y_n-x_n)$.We see that the sequence $A+X$ is convergent and further by algebra of limits,we get $$lim \space A+X=lim(y_n-x_n+x_n)=lim(y_n)=lim(y_n-x_n)+lim(x_n)=0+lim(x_n)=lim(x_n)$$.
I doubt there is an glitch in the last part of my proof.Please let me know if it is correct or not.

Examples of modules that $M_\mathfrak{p} \subset N_\mathfrak{p}$ that does not imply $M \subset N$?

Posted: 25 Jul 2021 08:13 PM PDT

I came across a theorem in algebraic number theory:

Theorem Let $A$ be a Dedekind ring and $M, N$ two modules over $A$. If $M_\mathfrak{p} \subset N_\mathfrak{p}$ for all prime ideals $\mathfrak{p} \subset A$, then $M \subset N$.

Proof. Let $a \in M$. For each $\mathfrak{p}$ we can find $x_\mathfrak{p} \in N$ and $s_\mathfrak{p} \in A \setminus \mathfrak{p}$ such that $a/1=x_\mathfrak{p}/s_\mathfrak{p}$. Let $\mathfrak{b}$ be the prime ideal generated by the $s_\mathfrak{p}$. Then $\mathfrak{b}$ is the unit ideal, and we can write \[ 1=\sum y_\mathfrak{p}s_\mathfrak{p} \] with elements $y_\mathfrak{p} \in A$ all but a finite number of which are $0$. This yields \[ a = \sum y_\mathfrak{p}s_\mathfrak{p}a = \sum y_\mathfrak{p}x_\mathfrak{p} \in N \] as desired. $\square$


Here $M_\mathfrak{p}$ means the localisation of $M$ at $\mathfrak{p}$.

My question is not about the theorem per se, but I guess there should be some counterexamples. Chances are there are modules $M,N$ such that

$$ M_\mathfrak{p} \subset N_\mathfrak{p} \text{ does not imply } M \subset N. $$

But I find it pretty hard to construct an example. I try to breakdown this question into pieces:

  1. Does this coincide with local property? It's pretty close:

Let $\phi:M \to N$ be an $A$-modue homomorphism (no futher property assumed on $A$ at this point), then the followings are equivalent:

  1. $\phi$ is injective.

  2. $\phi_\mathfrak{p}:M_\mathfrak{p} \to N_\mathfrak{p}$ is injective for each prime ideal $\mathfrak{p}$.

  3. $\phi_\mathfrak{m}:M_\mathfrak{m} \to N_\mathfrak{m}$ is injective for each maximal ideal $\mathfrak{m}$.

It looks like we are talking about 2 implies 1. But the inclusion map $M \to N$ is not given in this theorem. Also in the proof of the theorem, Serge Lang used properties of Dedekind ring (fractional ideals form a group) explicitly. If local property is the case, why would he ignore it?

  1. Is it possible to get a desired counterexample given by $M_\mathfrak{p} \not\subset N_\mathfrak{p}$ for at least one prime ideal?

  2. Is it possible to get a desired counterexample given by $A$ not being Dedekind, or even wilder, $A$ is not Noetherian? Noetherian counterexamples will certainly be more interesting.

Thanks in advance!


Edit 1: I added proof of the theorem to avoid unnecessary confusion.

What is the homotopy type of $S^2 - p - q\ $?

Posted: 25 Jul 2021 08:16 PM PDT

Let $p,q$ be two distinct points of $S^2.$ Then what would be the homotopy type of $S^2 - p - q\ $?

If we join $p,q$ by two $1$-cells, one cell (say $A$) lying entirely on $S^2$ and one (say $B$) lying outside $S^2$ then both the cells are contractible and $S^2$ along with those two cells gives us a CW complex structure in which both the cells are included as subcomplexes. Let $X$ be that CW complex. Then $X / A \simeq X \simeq X/B.$ But $X/A \approx S^2/\{p,q\}$ and $X/B \approx S^2 \vee S^1.$ So we have $S^2/\{p,q\} \simeq S^2 \vee S^1.$

Here instead of making that construction if we say that $S^2 - p \approx \mathbb R^2$ then the point $q$ is mapped homeomorphically to a point of $\mathbb R^2.$ By translating the point suitably we can think of that the point is the origin. Then doesn't deleting the points $p$ and $q$ from $S^2$ amount to deleting the origin from $\mathbb R^2\ $?

Any help in this regard would be much appreciated. Thanks for your time.

Conjecture about divergent periodic summations of odd functions

Posted: 25 Jul 2021 07:51 PM PDT

$$f(x)=\lim_{m\to\infty}\frac{\sum_{n=-m}^m(x-n)^{1/3}}{\sum_{n=-m}^m(1-n)^{1/3}}\stackrel{?}{=}x$$

I had a 'proof' but I made the simple mistake of assuming 2 limits could be swapped, so I only really proved that $f(x)=x$ for $x\in[-1,0,1]$, which is pretty trivial. Here's the 'proof' that I had. This was a first draft so it may be confusing to read.

The $(\text{stuff})^{1/3}$ in the numerator and denominator is really just $g(\text{stuff})$. Where $g$ just needs to be any odd function and relatively smooth and have a few other properties that I don't remember.

Guessing number of colors of beads in an urn

Posted: 25 Jul 2021 08:02 PM PDT

Motivation from cocktail bar

Every time when I order the cocktail "Latex and Prejudice" ("Латекс и предубеждение") in the Tesla bar in Saint Petersburg (Russia) the barkeeper selects by random a small interesting photo$^1$ and attaches it with a clamp to the cocktail glass. At the beginning I got always different pictures and started to collect them. The more cocktails I ordered the more often the motifs repeated. Finally, I had drunken so much that almost every time I had to ask for a different photo. I was wondering how many different pictures there are but due to drunkenness couldn't solve the problem by myself.

Mathematical form using urn model

Given is an urn with an unknown number of balls that have an unknown number of colors. It is assumed that every color has equal probability. From this urn in total $n$ balls with $m$ different colors were drawn, $k_i$ balls for color $i$ were sampled. How many different colors are in the urn? Sampling with and without replacement is of interest.

Questions for bounty

  1. If we do not know if the balls were drawn with or without replacement, shall we always assume drawing with replacement?

  2. Does this answer direct to the right direction?

  3. If the previous answer is yes: Can we consider these likelihoods as discrete probability distributions if they were be normalized (with support on $(m,\infty)$)?

Related problems

In this SE post the number of colors in the urn is also unknown but in the problem given here it is assumed that the colors have equal probability. Another SE post deals with lending books from a library that were already lent at an earlier time.


$\small{^1 \text{Because the site is accessible to minors,$\\$ the content of the photos is not discussed here.}}$

Finding another form of $\mathrm{\int (a^t)^{(a^t)}}dt$

Posted: 25 Jul 2021 08:20 PM PDT

I know there exist functions like this one for simplifying tetration based sums. There may be a way to simplify this type of sum at least using a lesser known and widely accepted function. Here are some results as proof: proof of results and calculation of special case and integral version of special case.

It was a nice idea to find the sophomore's dream, but I thought that it would be more interesting if I could find a "generalized exponential sophomore's dream". This uses the regularized lower gamma function which is quite rare, the exponential integral function, and tetration. There was an annoying discontinuity at n=0 hence the constant term:

$$\mathrm{\int_0^b \, ^2\left(a^t\right)dt=\int_0^b a^{ta^t}dt=b+\sum_{n=1}^\infty\frac{ln^n(a)}{n!}\int_0^b t^n a^{tn}dt=\boxed{\mathrm{b+\frac1{ln(a)}\sum_{n=1}^\infty\frac{(-1)^nP\big(n+1,-n\,b\,ln(a)\big)}{n^{n+1}}}}\implies \int_{-\frac1e}^0 e^{{te}^t}dt=1+\sum_{n=1}^\infty \frac{(-1)^nP(n+1,n)}{n^{n+1}}=0.77215…}$$

$$\mathrm{\implies A(a,t)\mathop=^\text{def}\int \,^2\left(a^t\right)dt=C+t+\frac1{ln(a)}\sum_{n=0}^\infty \frac{(-1)^n Q(n+1,-nt\,ln(a))}{n^{n+1}}=\quad C+t-t\sum_{n=0}^\infty \frac{(t\,ln(a))^n E_{-n}(-nt\,ln(a))}{n!}}$$

This series reminds me of the Marcum Q function for non negative integers: $$\mathrm{Q_m(a,b)=1-e^{-\frac{a^2}2}\sum_{n=0}^\infty\left(\frac{a^2}{2}\right)^n\frac{P\left(m+n,\frac{b^2}2\right)}{n!}}$$

This representation is good, but is quite tedious to use as the formula requires summing an infinite amount of regularized gamma functions. I would like to find a way to get rid of the summation .I see the gamma function with powers which reminds me of the summation definition of a hypergeometric function. I would even like to see the use of a generalized hypergeometric function like Meijer G or Kampé de Fériet functions seen in the link. Please correct me and give me feedback!

Flipping Fractions

Posted: 25 Jul 2021 08:18 PM PDT

enter image description here

I'm currently reading Differential Equations for Dummies, and this is what it says on pg 60.

I wasn't too sure whether "flipping fractions" still satisfy the equation, and I searched online (including this site), but it seems that most people say flipping fraction is not correct. At the same time, there does seem to be a specific method that allows you to "flip fractions" correctly. Could someone help me understand how? (i.e. how to get from the first eq shown to the second eq.)

Does the Hadamard product of two vectors have a geometric interpretation?

Posted: 25 Jul 2021 07:46 PM PDT

Pointwise multiplication (or element-wise) is also known as the Hadamard product. Is there a geometric interpretation similar to addition or subtraction in vector spaces?

Intuition & Proof of rank(AB) $\le$ min{rank(A), rank(B)} (without inverses or maps) [Poole P217 3.6.59, 60]

Posted: 25 Jul 2021 08:07 PM PDT

I'm aware of analogous threads; I hope that mine is specific enough not to be esteemed one.

$\mathbf{a^i}$ is a row vector. $A, B$ are matrices. Prove: $1$. $\mathbf{a^i}B$ is a linear combination of the rows of $B$.
$2.$ Row space of $AB \subseteq$ row space of $B$. $\qquad$ $3.$ Column space of $AB \subseteq$ Column space of $A$.
$4.$ If $\mathbf{a_i}$ is a column vector, then $A\mathbf{a_i}$ is a linear combination of the columns of $A$.
$5. \operatorname{rank}(A\color{#B8860B}{B}) \color{#B8860B}{\le} \operatorname{rank}\color{#B8860B}{B} \qquad \qquad$ $6.\operatorname{rank}(AB) \leq \operatorname{rank} A$.
In general, $x \leq a \text{ & } x \le b \implies x \le \min\{a, b\}$.
So by $5 \, \& \, 6$, $\operatorname{rank}(AB) \leq \min\{\operatorname{rank}A,\operatorname{rank} B\}$.

$\bbox[2px,border:2px solid grey]{\text{ Proof of #5 :}} \;$ The rank of a matrix is the dimension of its row space. Need to show :
If $\operatorname{rowsp}(AB) \subseteq\operatorname{rowsp}(B)$, then $\operatorname{dim rowspace}(AB) \le \operatorname{dim rowspace}(B). $
Pick a basis for $\operatorname{rowsp}(AB)$. Say there are $p$ vectors in this basis.
By $\#2$, row space of $AB \subseteq$ row space of $B$, $\color{green}{\text{so all of these $p$ vectors also $\in \operatorname{rowsp}(B)$}}$. Moreover, they must be linearly independent (hereafter dubbed l-ind).
${\Large{\color{red}{[}}} \;$ Since the dimension of a space $=$ the maximum number of l-ind vectors in that space, $\; {\Large{{\color{red}{]}}}}$
and $\color{green}{\text{$\operatorname{rowsp}(B)$ has $\ge p$ l-ind vectors}}$, thus $ \operatorname{dim rowspace}(B) \; \ge \; \operatorname{dim rowspace}(AB) = p. $

$\bbox[2px,border:2px solid grey]{\text{ Proof of #6 :}} \;$ Apply $ \operatorname{rank}M = \operatorname{rank}M^T$ and $\#5$: $ \operatorname{rank}(AB)^T = \operatorname{rank}(B^T\color{#B8860B}{A^T}) \quad \color{#B8860B}{\le} \quad \operatorname{rank}\color{#B8860B}{A^T} = \operatorname{rank}(A)$.

$Q1.$ Please elucidate the above proof of $5$? I'm bewildered. What's the strategy?

$Q2.$ On P209, Poole defines dimension as the number of vectors in a basis.
So shouldn't the red bracket refer to a basis? If so, why doesn't the proof simply declare:
By $2$, the basis for $\operatorname{rowsp}(AB)$ can be reused as a basis for $\operatorname{rowsp}(B).$ ?

$Q3.$ How'd one previse to invert $AB$ and apply $\#5$ (the key strategem) for #6?

$Q4.$ What's the intuition behind results $5$ and $6$? I'd be grateful for pictures.

Sources: P147, 4.48, Schaum's Outline to Lin Alg, web.mit.edu/18.06/www/Spring01/Sol-S01-5.ps

Dense subset of Cantor set homeomorphic to the Baire space

Posted: 25 Jul 2021 07:56 PM PDT

Does anyone know a proof that the Cantor set, $\{0,1\}^{\mathbb{N}}$, has a dense subset homeomorphic to the Baire space, $\mathbb{N}^{\mathbb{N}}$? Thank you.

Representation of a linear functional in vector space

Posted: 25 Jul 2021 08:24 PM PDT

In the book Functional Analysis, Sobolev Spaces and Partial Differential Equations of Haim Brezis we have the following lemma:

Lemma. Let $X$ be a vector space and let $\varphi, \varphi_1, \varphi_2, \ldots, \varphi_k$ be $(k + 1)$ linear functionals on $X$ such that $$ [\varphi_i(v) = 0 \quad \forall\; i \in \{1, 2, \ldots , k\}] \Rightarrow [\varphi(v) = 0]. $$

Then there exist constants $\lambda_1, \lambda_2, \ldots, \lambda_k\in\mathbb{R}$ such that $\varphi=\lambda_1\varphi_1+\lambda_2\varphi_2+\ldots+\lambda_k\varphi_k$.

In this book, the author used separation theorem to prove this lemma. I would like ask whether we can use only knowledge of linear algebra to prove this lemma.

Thank you for all helping.

How do I calculate the equation of a circle given 3 complex numbers?

Posted: 25 Jul 2021 08:18 PM PDT

Given three complex values (for example, $2i, 4, i+3$), how would you calculate the equation of the circle that contains those three points? I know it has something to do with the cross ratio of the three points and $z$ and the fact that the cross ratio is a real number, but I don't know what to do with that information.

Why is the volume of a cone one third of the volume of a cylinder?

Posted: 25 Jul 2021 08:24 PM PDT

The volume of a cone with height $h$ and radius $r$ is $\frac{1}{3} \pi r^2 h$, which is exactly one third the volume of the smallest cylinder that it fits inside.

This can be proved easily by considering a cone as a solid of revolution, but I would like to know if it can be proved or at least visual demonstrated without using calculus.

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