Saturday, December 25, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange


properties of lebesgue integral in Apostol

Posted: 25 Dec 2021 12:52 AM PST

there is this theorem in Apostol mathematical analysis :

if $ f,g \in L(I) $ which is the set of lebesgue integrable functions on interval I then $ \int_I f \geq \int_I g $ if $f(x)\geq g(x)\quad a.e \quad on\quad I$.

in proof he says if we write $f=u-v$ which $u,v\in U(I)$ ( the set of upper functions on I) then $ u(x)\geq v(x)$ a.e on $I$ . my question is what if $f$ is a.e negative on $I$ , the how should I reach to $ u(x)\geq v(x)$ a.e on $I$?

Solve the ODE $y'=1-\frac{y}{x}$

Posted: 25 Dec 2021 12:53 AM PST

I substituted $u=\frac{y}{x}$ then tried to solve the ODE $$\frac{u'}{2u-1}= -\frac{1}{x}$$ and I came this far $$\frac{1}{2}\ln |{2u-1}|=- \ln |{x}| + c_1$$ but then in the solution there was the step $$c_1=\ln c_2 \in \mathbb{R}, c_2 > 0$$ to get $$\ln |2u-1|=\ln{(\frac{c_2}{x})^2}$$ but why do we have this additional step? Couldn't we just calculate the solution without this step?

How to prove bijective continuous functions(Rn -> Rn) map open-in-Rn sets to open-in-Rn sets ?

Posted: 25 Dec 2021 12:48 AM PST

How to prove bijective continuous functions(Rn -> Rn) map open-in-Rn sets to open-in-Rn sets?

For example, an bijective continuous functions F:Rn -> Rn, it map closed-in-Rn set A to closed-in-Rn set B. A1 is an open-in-Rn set included in A. A1's inverse image is B1. I know that an open set's inverse image is an open set,but the openness is relative, that means B1 is open in B, but how to prove B1 is also open in Rn?

Maximal number of intersections within a bipartite graph

Posted: 25 Dec 2021 12:41 AM PST

Consider $n$ line segments in the Cartesian plane. For $1\leq k\leq n$, the $k$-th line segment is drawn from $(k,0)$ to $(x_k,1)$, where $\{x_1,x_2,...,x_k\}$ is a permutation of $\{1,2,...,n\}$. Define $a(n)$ to be the maximal number of distinct points at which two or more line segments intersect.

For example, take $n=3$. A maximal construction for $a(n)$ would be a segment from $(1,0)$ to $(3,1)$, $(2,0)$ to $(1,1)$, and $(3,0)$ to $(2,1)$, giving $2$ distinct intersection points.

I have simulated the maximal values up to $n=13$ (and my friend has posted it on OEIS already here: https://oeis.org/A332774), but I am limited by the extreme time complexity of my program (around $n*n!*\log n$). The values from $n=1$ to $13$ are:

$0,1,2,5,8,13,17,23,30,39,47,57,67$

I was wondering if a formula for $a(n)$ could exist, either recursive or explicit. Could it have something to do with the number of intersections in certain types of bipartite graphs? If no formula can be found, can there be any good estimates for upper / lower bounds?

Existence and unicity of solution for a Cauchy problem

Posted: 25 Dec 2021 12:34 AM PST

I have this problem $$\begin{cases} y'=-x^2 y^2+\exp(-y^2)\\ y(0)=0\end{cases}$$ the question is: study the existence and unicity of solution for this problem on $R=\{(x,y)\in\mathbb{R}^2, |x|\leq \frac12, |y|\leq 1\}$ using two methods.

can someone give me a hint on what is the two methods ?

Thank you

Write the event of drawn cards

Posted: 25 Dec 2021 12:33 AM PST

From a deck of 52 cards take out 26 cards. The event $A_{ij}$ means that there are at least $i$ clubs and at least $j$ spades. I need to write the event using set-theoretic operations and $A_{ij}$ that all of this 26 cards are red.
if I understand correctly, I should wrute the event when only clubs and spades were drawn and take the inverse.
My answer is: $$\Biggl(\Bigl(\bigcup_{i=0}^{26}\bigcup_{j=0}^{26-i}A_{ij}\Big)\setminus A_{00}\Bigg)^C$$ Am I right?

Probability that sums of $N(0,1)$ variables exceeds 5

Posted: 25 Dec 2021 12:24 AM PST

I'm trying to work out what is the probability that a process $W(t) = \sum_{i = 1}^t N(0,1)$ ever exceeds 5 before reaching a certain number of steps (say 100). So my thinking is that the probability that $W(1)$ is less than 5 is:

  • $P(W(1) < 5) = \Phi_{0,1}(5)$ where $\Phi_{0,1}$ is the CDF of a normal distribution with $\mu = 0, \sigma^2 = 1$

the probability that $W(2)$ is less than 5 is:

  • $P(W(2) < 5) = \Phi_{0,2}(5)$

and so on until the probability that $W(100)$ is less than 5 is $\Phi_{0,100}(5)$. For the process to never exceed 5, all the $W(t)$ from $t=1$ to $100$ need to be less than five, so the probability that $W(t)$ never exceeds 5 in the first 100 steps should be

$P(W(1) < 5) \times P(W(2) < 5) \times ... P(W(100) < 5) \approx \Phi_{0,1}(5) \times \Phi_{0,2} (5) \times ... \Phi_{0,100}(5) \approx 0.9999 \times 0.9997 \times ... 0.691 \approx 0 $

So it says that the process is guaranteed to reach 5, which is not correct based on some simulations. What am I missing here?

Does $\mathbb{N}$ have fewer primes than any arithmetic progression?

Posted: 25 Dec 2021 12:31 AM PST

Let $f(x):=ax+b$ for naturals $x,a,b$, such that $f(x)$ will take infinitely many primes as $x\to\infty$.

Is it the case that for any choice of $a,b$, there exists some $N$ such that for all $n>N$, $\pi(n)<\pi_{a,b}(n)$?

In other words, in the limit, do the naturals yield fewer primes than all admissible arithmetic progressions? This seems likely to be true heuristically, but it's non-obvious to me past that.

Is there an Integral domain that is a GCD domain but NOT a UFD?

Posted: 25 Dec 2021 12:18 AM PST

For most Integral domains I found that are Not UFD, they just does not have the uniqueness of factorization, while they still possess the existence of factorization. ($\mathbb{Z}[\sqrt{-5}]$ for example)

I searched on Wiki and found that the Uniqueness corresponds to being in a GCD domain, and Existence corresponds to Accending Chain Condition of Principal ideal(ACCP)

So, I am wondering is there any integral domain between the GCD domain & UFD?

Resolution of Curry's paradox in natural language?

Posted: 25 Dec 2021 12:02 AM PST

There are a lot of resources on Curry's paradox and its resolution regarding formal systems, but what about the "top-level" natural language.

My take is that there are two ways to introduce a new symbol during natural discourse.

  • a) the algorithmic shorthand kind, where whenever we observe said symbol we immediately replace that with its definition
  • b) the other one is labeling, labeling parts of the discourse, like done here with a) and b)

(note that we don't have to worry about definitional extension since the a) kinds are not "real" symbols they are eliminated, while b) kinds obviously exist since the document fragment they label exist)

In this spirit an introduction like

Let A be the sentence 1+1=2

can be of the first kind, while

Let A be the sentence if A is true then the sky is red

can only be of the second kind, since trying to "normalize it" (eliminate all symbols of the first kind) is futile.

If we accept that in the second example A is a simple label then either we cannot speak about the truth/provability of the sentence behind such a label, or A is true does not entail that we can use the sentence behind A (which then kind of has the same result as before)

What resolutions exist to Curry's paradox without reference to formal systems?

Prove a specific function is not the Fourier transform of any function in $L^1$

Posted: 25 Dec 2021 12:02 AM PST

For $\xi \geq 0$, define $ g(\xi)=\left\{\begin{array}{ll}\frac{1}{\ln \xi}, & \xi>e \\ \frac{\xi}{e}, & 0 \leq \xi \leq e\end{array} \right.$, when $ \xi<0 $ , define $g(\xi)=-g(-\xi)$. Prove that There is no $f(x) \in L^{1}(\mathbb{R})$, s.t. $\hat{f}(\xi)=g(\xi)$, where $\hat{f}(\xi)=\int_{\mathbb{R}} f(x) e^{-i x \xi} \mathrm{d} x, \xi \in \mathbb{R}, i=\sqrt{-1}$.

I know that $g(\xi) $ is not in $L^1(\mathbb{R})$ , and $\|\hat{f}(\xi)\|_{L^\infty}\leq C\|f\|_{L^1}$. But these seem to be useless for this problem , can anyone help me?

How to solve this problem on alligations using successive technique?

Posted: 24 Dec 2021 11:58 PM PST

Question :

"The diluted wine contains only 8 litres of wine and the rest is water. A new mixture whose concentration is 30 % is to be formed by replacing wine. How many litres of mixture shall be replaced with pure wine if there was initially 32 litres of water?"

Doubt:

I tried solving by finding out the new wine concentration that is 30 % of 40 litre which comes out to be 12 litre while the initial concentration is 8 litre for the wine. So the increased concentration is 4 litre of wine while the answer reads 5 litre. Where did I go wrong also I feel that this problem can be solved by using the alligation technique by using successive. Please give me a detailed outlook about where I went wrong along with the successive technique.

Cardinality Of The Continuum(s?)

Posted: 25 Dec 2021 12:43 AM PST

considering The Continuum Hypothesis in it's most simplified form States that $|\Re| = 2^{\aleph_0} = \aleph_1$.

  1. is $3^{\aleph_0} = |\Re| = 2^{\aleph_0} = \aleph_1$? what about other $n\epsilon N: n^{\aleph_0}?$ for $n=1$ looks like $\aleph_0$ to me :) so what would $n=0$ be ?!
  1. what if $|\Re| = \aleph_0^{\aleph_0} = \aleph_{\aleph_0} > ... > 4^{\aleph_0} > 3^{\aleph_0} > 2^{\aleph_0} > 1^{\aleph_0} = \aleph_0$ and let there be countably infinite number of continuums and call $\Re$ something like The Absolute Continuum?

ps: just learned a bit of latex to write this post plz don't judge my language. i'm more used to turing-complete universes :)

send bacon and huggies

H

Determining whether $ \sum_{m=0}^{\infty}(-1)^{\sin(5\pi/m)}\sin\frac{5\pi}{m}$ converges

Posted: 25 Dec 2021 12:41 AM PST

Consider the function $$ \sum_{m=0}^{\infty}(-1)^{\sin(5\pi/m)}\sin\frac{5\pi}{m}$$ How do we determine the convergence or divergence the series? Can we plot the graph of the partial sums?If the partial sum is a complex function ,how do we plot it?

For which values of $\alpha$ is $\frac{3}{2}(\frac{1}{\alpha}-1)$ an integer?

Posted: 25 Dec 2021 12:01 AM PST

How can I prove that for which values of $\alpha$ the expression $$\frac{3}{2}\left(\frac{1}{\alpha}-1\right)$$ where $0\leqslant\alpha< 1$ is an integer, and for which of them it is non-integer?

Clearly for $\alpha=\frac{1}{3}$, the given expression is integer.

Any hint to prove this will be of great help.

What is the mean of the largest exponent among the prime factor of natural numbers?

Posted: 25 Dec 2021 12:52 AM PST

Let $f(n)$ the be largest exponent among exponents of the prime factor of $n$. E.g. $f(80) = 4$ since $80 = 2^4.5$ and the prime factor of $80$ which has the largest exponent is $2$ which occurs with the exponent $4$. Trivially for all square-free numbers $n, f(n) = 1$. Experimental data for $n \le 3.5 \times 10^9$ shows that.

$$ \mu = \lim_{n \to \infty}\frac{1}{n}\sum_{k = 1}^nf(n) \approx 1.784744 $$

Q 1: Does this limit exist and does it have a closed form?

If all positive integers were square-free, the above mean would have been exactly $1$. But this is not the case because there are non-squarefree numbers. Thus, we can say that the square-free numbers contribute exactly $1$ to the above mean value while all numbers which have prime factors with an exponent $> 1$ contribute the remaining $0.784744$.

Using the natural density of $k$-th power free numbers, I can show that

$$ \mu \ge \sum_{k = 1}^{\infty}k \Big(\frac{1}{\zeta(k+1)} - \frac{1}{\zeta(k)}\Big) \approx 1.70521 $$

Q 2: Is it possible to show a weaker result like $\mu < 2$?

Transformation of random variable using CDF not making sense?

Posted: 25 Dec 2021 12:50 AM PST

I have two exponential random variables $X,Y$ that are both identical (i.e. they have the same $\lambda$). I am attempting to find their sum using the method of CDFs (please note, I understand how to use the convolution but this is for completeness).

We have $f_X(x) = f_Y(y) = \begin{cases} \lambda e^{-\lambda x} & x \ge 0 \\ 0 & elsewhere \end{cases}$

The transformation we are given is $U = X+Y$.

Because $f_X(x) = f_Y(y)$, I propose we may treat this transformation as $U=2X$.

$F_U(u) = P(U\le u) = P(2X \le u) = P(X \le u/2) = F_X(u/2)$.

We know that the CDF of an exponential random variable is given by $F_X(x)= \begin{cases} 1-e^{-\lambda x} & x \ge 0 \\ 0 & elsewhere \end{cases}$

So $F_X(u/2) =\begin{cases} 1-e^{-\lambda u/2} & u \ge 0 \\ 0 & elsewhere \end{cases} $

Taking the derivative, we reach $f_U(u) = \begin{cases} (\lambda /2)e^{-\lambda u/2} & u \ge 0 \\ 0 & elsewhere \end{cases}$

But, when I solve by convolution I get $f_U(u) = \begin{cases} \lambda ^2ue^{-\lambda u} & u \ge 0 \\ 0 & elsewhere \end{cases}$

So I'm really unsure what's going on, any suggestions?

How to demonstrate that $\frac{\sqrt{n}}{1+\sqrt{nx}}$ converges in $L^{1}( (0,1),\mu)?$

Posted: 24 Dec 2021 11:54 PM PST

Let, $\mu$ be the Leb. measure. Let, $g_n =\frac{\sqrt{n}}{1+\sqrt{nx}} $ for $n \ge 1$. I know that,

$$\int^{1}_{0}|g_n| ~ d\mu \le \int^{1}_{0}\frac{\sqrt{n}}{\sqrt{nx}} ~ d\mu \le \int^{1}_{0}\frac{1}{\sqrt{x}} ~d\mu < \infty.$$ Thus, we can use the Lesbegue Dominated Convergence Theorem in this context. I know that,

$$\lim_{n\rightarrow \infty}g_n = \frac{1}{\sqrt{x}}.$$

I don't know where to go from here, any help would be appreciated. Thanks!

Find the remainder when $1111^{22} + 2222^{33} + 3333^{44} + 4444^{55} + 5555^{66}$ is divided by $13$. [duplicate]

Posted: 24 Dec 2021 11:56 PM PST

I have to find the remainder when $1111^{22}+2222^{33}+3333^{44}+4444^{55}+5555^{66}$ is divided by $13$. I tried to use Fermat's little theorem and wrote the numbers using 1111 so I could use $1111^{12}$ in theorem but I didn't really get close to the solution. Am I doing something wrong? Is there any other way to solve this? Thank you.

Exercise 5, Section 13 of Munkres’ Topology

Posted: 25 Dec 2021 12:48 AM PST

Show that if $\mathscr{A}$ is a basis for a topology on $X$, then the topology generated by $\mathscr{A}$ equals the intersection of all topologies on $X$ that contain $\mathscr{A}$. Prove the same if $\mathscr{A}$ is a subbasis.

I would first rephrase the problem:

If $\mathscr{A}$ is a basis for a topology on $X$, then $\mathscr{I}_\mathscr{A}=\cap_{\alpha \in A} \mathscr{T}_{\alpha}$, where each $\mathscr{T}_{\alpha}$ is topology on $X$, $\mathscr{A} \subseteq \mathscr{T}_{\alpha}$

My attempt: let $\mathscr{I}_\mathscr{A}$ denotes topology generated by basis $\mathscr{A}$. To show $\cap_ {\alpha \in A} \mathscr{T}_{\alpha} \subseteq \mathscr{I}_{\mathscr{A}}$.Proof: If $U\in \cap_ {\alpha \in A} \mathscr{T}_{\alpha}$, then $U\in \mathscr{T}_{\alpha}, \forall \alpha \in A$. Since $\mathscr{A}$ is basis, $\forall x\in X, \exists \mathcal{A} \in \mathscr{A}$ such that $x \in \mathcal{A}$. So, $\forall x \in U, \exists \mathcal{A}_{x}\in \mathscr{A}$ such that $x\in \mathcal{A}_{x}$. Thus, $U \subseteq \cup_{x \in U} \mathcal{A}_{x}$. If $y \in \cup_{x \in U} \mathcal{A}_{x}$, then $y\in \mathcal{A}_{j}$, for some $j\in U$. Since $\mathcal{A}_{j}\in \mathscr{A}$, $\mathcal{A}_{j} \subseteq X$. So, $y\in \mathcal{A}_{j} \subseteq X$. Thus, $\cup_{x \in U} \mathcal{A}_{x} \subseteq U$. Hence $U = \cup_{x \in U} \mathcal{A}_{x}$. By lemma 13.1, $U\in \mathscr{T}_{\mathscr{A}}$. Is this proof correct?

The way I proved this inclusion, I don't think it's correct, because it shows any subset of $X$ can be written as unions of elements of $\mathscr{A}$. Maybe somewhere in the proof I did something wrong.

The ideal solution to this problem is here: Prob. 5, Sec. 13 in Munkres' TOPOLOGY, 2nd ed: How to distinguish between a basis and a subbasis?

How do I prove this statement about cyclic groups? [duplicate]

Posted: 25 Dec 2021 12:02 AM PST

I have the following problem:

Let $C_m,C_n$ be two finite cyclic groups of card. $m,n$ respectivly. Let $d=\gcd(m,n)$ be the greatest common divisor of $m$ and $n$. And consider $f:C_m \rightarrow C_n$ a morphism of groups. Show that $\forall x\in C_m$, $ord(f(x))$ is a divisor of $d$.

My Idea was the following.

Proof

We can choos $g_m, g_n$ to be the generator of $C_m,C_n$, i.e. $$\langle g_m\rangle=C_m, \langle g_n \rangle =Cn$$Now take $x\in C_m$ then $x$ is of the form $g_m^i$ for some $1\leq i \leq m-1$ furthermore $f(x)=g_n^{ij}$ for some $1\leq j\leq n-1$. From the Bézout relation we get that $d=um+vn$ for some $u,v \in \mathbb{Z}$. But we can assume $m,n \neq 0 \Rightarrow d\neq 0$ otherwise $d$ would be zero and everthing divides zero. Thus $u,v\in \mathbb{Z}\setminus 0$. Now let us compute $f(x)^d$:$$f(x)^d=f(x)^{um+vn}=f(x)^{um}\cdot f(x)^{vn}=f(g_m^i)^{um}\cdot (g_n^{ji})^{vn}\stackrel{\text{f is a morphism}}{=}f(g_m^m)^{iu}\cdot (g_n^{jiv})^{n}\stackrel{\text{using of orders}}{=}f(1)^{iu}\cdot 1\stackrel{\text{f is a morphism}}{=}1^{iu}\cdot 1=1 $$ We know that since $f(x)^d=1$ we have $ord(f(x))|d$ and we are done.

Could someone please correct this version if it is totally wrong? Since it is my idea it would be helpful to work with it.

Thanks a lot.

Finding asymptotic variance of MLE using Fisher information - comprehension of steps

Posted: 25 Dec 2021 12:36 AM PST

I'm working on finding the asymptotic variance of an MLE using Fisher's information.

The distribution is a Pareto distribution with density function $f(x|x_0, \theta) = \theta \cdot x^{\theta}_0 \cdot x^{-\theta - 1}$.

There are two steps I don't get, namely step 3 and 5.

(step 1) We have that

$1 = \int^{\infty}_{-\infty} f(x|x_0, \theta)$

(Step 2) We take derrivative wrt $\theta$:

$0 = \int^{\infty}_{-\infty} \frac{\partial f(x|x_0, \theta)}{\partial \theta} dx$

(Step 3) We can (according to my textbook) write the above as:

$0 = \int^{\infty}_{-\infty} \frac{\partial f(x|x_0, \theta)/\partial \theta }{f(x|x_0,\theta)} f(x|x_0,\theta) dx$

(Step 4) This can again be written as an expectation:

$E[\frac{\partial \log f(x|x_0,\theta)}{\partial \theta}] = 0$

(Step 5) According to my textbook we can differentiate again and get:

$0 = \int^{\infty}_{-\infty} \frac{\partial^2 \log f(x|x_0, \theta)}{\partial \theta^2} f(x|x_0, \theta) dx + \int^{\infty}_{- \infty} \frac{\partial \log f(x|x_0,\theta)}{\partial \theta} \frac{\partial \log f(x|x_0,\theta)}{\partial \theta} f(x|x_0, \theta) dx$

I can't figure if I have just stared at these steps for too long, or if there is something fundamental about differentiation and integration I have missed.

I would be very grateful if someone could explain the steps 3 and 5 to me in layman's manner?

(Textbook is Hogg & Craig "introduction to mathematical statistics").

Localization, saturation and Nilradical (help needed and proof verification)

Posted: 25 Dec 2021 12:32 AM PST

This question is from my commutative algebra assignment and I am struck with it. So, I am posting here for guidance.

Question: Let A be a commutative ring.

(a) Let $S\subseteq A$ be a multicatively closed subset. Then prove that $S^{-1}$ commutes with the nilradical i.e. $nil (S^{-1} A) = S^{-1} (nilA)$.

Attempt: I think that both sets are equal because if $x\in nil (S^{-1} A) $ => $x^n =(a/s)^n=0=(a)^n / (s.s... n times)= a^n /s=0$ which implies that $a^n=0$ for some $n \in \mathbb{N}$. So, $x\in S^{-1} (nil A)$.Conversaly, If $x\in S^{-1} (nil A)$ then there exists $a\in A$ and $n \in \mathbb{N}$ such that $\frac{a^n} {s}=0$ . Now s must be non zero and hence I can write $\frac{a^n} {s^n} =0$ which means that $x\in nil (S^{-1} A)$.

Is the proof fine?

(b) A prime ideal $p\in (Spec A, \subseteq ) $ iff $Spec(A_p)$ is singleton.

I think p always lie in Spec A if p is prime so there is nothing to prove in the case when $SpecA_p$ is singleton is given. Am i right?

Converse, I am not able to prove. Can you please give some hints on how to prove that $SpecA_p$ must be singleton if p is a prime ideal.

Kindly help!

(c) If A is reduced and $p\in (Spec A, \subseteq )$ is minimal, then $A_p$ is a field.

A is reduced means that A has no non-zero nilpotent elements and p is minimal means that p is not a proper subset of another member set of Spec A is called a minimal set. Let $x\in A_p$, I have to prove that x is a unit. But I don't have any ideas and I am struck.

Unfortunately, I am not so good in solving problems in Localization of rings and would very much appreciate help.

Thanks!

Asymptotically Unbiased [closed]

Posted: 24 Dec 2021 11:58 PM PST

Let $X_1, \ldots, X_n$ be $U[0, θ]$ distributed random variables. Show that is asymptotically unbiased.

How to calculate $\int_{0}^{\infty} x^{-x} \,dx$?

Posted: 25 Dec 2021 12:31 AM PST

I am trying to solve this improper integral: $\int_{0}^{\infty} x^{-x} \,dx$.

First I replace de infinity by a another variable $y$, so: $\int_{0}^{y} x^{-x} \,dx = \int_{0}^{y} e^{\ln(x^{-x})}\,dx = \int_{0}^{y} e^{-x\ln(x)}\,dx = \int_{0}^{y} \sum_{n=0}^{\infty} \frac{(-x\ln(x))^{n}}{n!}\,dx = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \int_{0}^{y} (x\ln(x))^n\, dx$.

Solving $\int (x\ln(x))^n\,dx$:

  1. Substitute $u=\ln(x) \implies \int u^ne^{(n+1)n}\,du $
  2. Substitute $v=u^{n+1} \implies \frac{1}{n+1}\int e^{(n+1)v^{\frac{1}{n+1}}}\,dv$
  3. Substitute $w=(n+1)^{(n+1)}v \implies \int e^{(n+1)v^{\frac{1}{n+1}}}\,dv = (n+1)^{-n-1}\int e^{w^{\frac{1}{n+1}}}\,dw$
  4. $\int e^{w^{\frac{1}{n+1}}}\,dw = -(n+1)(-1)^n \operatorname{\Gamma}(n+1,-w^{\frac{1}{n+1}})$
  5. Undo substitutions, $\int (x\ln(x))^n\,dx = \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(x\right)\right)}{\left(-1\right)^n}$
  6. $\int_{0}^{y} (x\ln(x))^n\, dx \implies \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(x\right)\right)}{\left(-1\right)^n} \Big|_0^y$, when $x=0$, the incomplete gamma function will be evalueate between plus infinity and plus infinity so, $\int_{0}^{y} (x\ln(x))^n\, dx = \dfrac{\left(n+1\right)^{-n-1}\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(y\right)\right)}{\left(-1\right)^n}$
  7. $\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \int_{0}^{y} (x\ln(x))^n\, dx = \sum_{n=0}^{\infty} \frac{\operatorname{\Gamma}\left(n+1,-\left(n+1\right)\ln\left(y\right)\right)}{n!(n+1)^{n+1}} = \sum_{n=1}^{\infty}\frac{Q\left(n,\ \ln\left(\frac{1}{y^n}\right)\right)}{n^n}$.

Where $Q$ is the normalized or regularized incomplete gamma function

But what's happen when $y \to \infty$? The incomplete gamma function will be evaluated between zero and minus infinity, is it valid? Is there another way to find this value? Because de the improper integral converges.

Why are tetrations not useful?

Posted: 25 Dec 2021 12:40 AM PST

I've always wondered after learning addition, multiplication, and power facts (and their inverse operations) what the next higher level of facts I would need to memorize would be. However, instead of learning about the next higher operator, math instead took an entirely different turn and started going into all sorts of other things that never seemed to need any higher operator than just the three and their inverses. From my perspective, operators went from addition to multiplication (repeated addition) , to exponentiation (repeated multiplication), but I've never heard of any higher-order operators than exponentiation. So recently I tried to learn about this operator myself. I started like this: repeated addition = multiplication, repeated multiplication = exponentiation, repeated exponentiation = ?.

However, I see that because of exponentials being non-commutative,
$$k^{\left( k^{\left( k^{\left( ... \right)} \right)} \right)} \neq \left( \left( \left( \left( k \right)^k \right)^k \right)^{...} \right)$$ there's two ways to repeat exponentiation, leading to two new operators, with one being tetration, with right-associative exponentiation. If I represent the other left-associative operator as $?$, then:

$$k+k+k+k = k * 4\text{,}$$

$$k * k * k * k = k ^ 4\text{, and}$$

$$\left(\left(\left(k\right)^k\right)^k\right)^k = k ? 4\text{.}$$

It seems that something close to what I want is $\left(\left(\left(k\right)^k\right)^k\right)^k = k^{(k^3)}$, and while this could be shorthand, it doesn't follow the convention of $k$ (operator) $4$. I think that left-associative exponentiation is still the consistent way with repeated addition and repeated multiplication. It's what I usually think of when I think of repeated exponentiation.

Then, according to the interwebs, it seems that tetration is usually considered the "next" operation after exponentiation. However, it seems that it isn't very common or useful. Not much can actually be represented with tetration. It seems that Tetration basically only gives you big numbers. For example, 2^(2^(2^2)) is already a huge number, and that's with really small inputs.

My question then is: Why is addition, multiplication, exponentiation (plus their 3 inverses) incredibly useful whereas tetration is the sudden cutoff for usefulness?

I mean, surely the reason for repeated operators is not only to represent bigger numbers? There are a myriad of useful things for just addition, multiplication, and exponentiation, and their inverses, but suddenly nothing useful from tetration? That doesn't make sense. It seems multiplication gives you more advanced math than just addition, and exponentiation gives more advanced math than just multiplication, so why doesn't tetration give you even more advanced math? Or are there actually advanced uses for Tetration at least as diverse and meaningful as the previous 3 that I don't know about?

EDIT: So far, it seems the general consensus is that tetrations are in fact not useful. Most of the answers/comments seem to reiterating this. But my central confusion is WHY this is. Multiplication seems to model more advanced real-world applications than addition. For example, multiplication finds area or volume, vs addition for finding number of apples or sheep. Going by this pattern, exponentials seem to model even more advanced real-world situations than multiplication, like expressing gravity as g = 9.8 m s^-2, or expressing polynomials which have a myriad of advanced applications. So does tetration have even more real-world applications than exponentiation? It appears no, thus far. In fact, tetration seems to have even less applications than plain old addition. This is mind boggling for me. (Does this mean that humanity simply hasn't reached that level of math yet?)

If $N = q^k n^2$ is an odd perfect number with special prime $q$, then can $N$ be of the form $q^k \cdot (\sigma(q^k)/2) \cdot {n}$?

Posted: 25 Dec 2021 12:30 AM PST

Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$.

A number $M$ is said to be perfect if $\sigma(M)=2M$. For example, $6$ and $28$ are perfect since $$\sigma(6) = 1 + 2 + 3 + 6 = 2\cdot{6}$$ and $$\sigma(28) = 1 + 2 + 4 + 7 + 14 + 28 = 2\cdot{28}.$$

It is currently unknown if there are infinitely many even perfect numbers. It is also an open problem whether any odd perfect numbers exist. It is widely believed that there are no odd perfect numbers.

Euler proved that an odd perfect number $N$, if one exists, must necessarily have the so-called Eulerian form $$N = q^k n^2$$ where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.

QUESTION

Here is my question:

Can the odd perfect number $N$ be of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{n}?$$

I would certainly appreciate it if anybody could point me to papers/articles/publications in the literature where this particular inquiry is covered.

CONTEXT

Slowak (1999) proved that the odd perfect number $N$ must be of the form $$\frac{q^k \sigma(q^k)}{2}\cdot{d},$$ where $d > 1$.

Dris (2017) showed further that $d$ must have the form $$\frac{D(n^2)}{\sigma(q^{k-1})}=\gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{n^2}{\sigma(q^k)/2},$$ where $D(x)=2x-\sigma(x)$ is the deficiency of $x$.

Metric Spaces: Convergent Sequences and Limit Points

Posted: 25 Dec 2021 12:05 AM PST

I have the following question regarding Metric Spaces from Fred H. Croom's book Principles of Topology

Show that the limit of a convergent sequence of distinct points in a metric space is a limit point of the range of the sequence. Give an example to show that this is not true if the word "distinct" is omitted.

How would I approach this question properly? My attempt revolved around the following understanding.

If $(X,d)$ is a metric space and $\{x_n\}_{n=1}^{\infty}$ a sequence of points of $X$. Then $\{x_n\}_{n=1}^{\infty}$ converges to the point $x\in X$. Now if we assume $A\subset X$. A point $x \in X$ is a limit point of $A$ provided that there are infinitely many elements of the subset inside the neighborhood (open ball) $B_d(x,r)=\{a\in A: d(x,a)<\epsilon\}$. Thus, $x$ is a limit point of the sequence $\{x_n\}$ if given $\epsilon >0$, $x_n \approx x$ for infinitely many $n$.

Would I just have to simply show that both limit of the sequence $\{x_n\}$ and the limit point of $A$ are the same? Am I on the right track, and how would I properly state it?

What would then be an example to shows that this is not true if the word "distinct" is omiited?


I want to thank you for taking the time to read this question. I greatly appreciate any assistance you provide.

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