Sunday, June 13, 2021

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

Is $\{x|x-1=3\}=4$ True or False?

Posted: 13 Jun 2021 08:04 PM PDT

Is it true that $\{x|x-1=3\}=4$ ? I'm not sure if this is true, I think $\{x|x-1=3\}=\{4\}$ is correct. Am I right?

Convergence of numerical method for Burgers' equation

Posted: 13 Jun 2021 07:59 PM PDT

This is a classic question from Leveque's book Numerical Methods for Conservation Laws.

Consider the 1-dimensional Burgers equation $u_{t}+(\frac{u^{2}}{2})_{x}=0$ with the inital data:

\begin{cases} u(x,0)=-1 & x < 1 \\ u(x,0)=+1 & x>1 \\ \end{cases}

Let $k_{l}=\frac{1}{2l},h_{l}=\frac{1}{l}$,where l is a positive interger,k is the step length in t-direction,l is the step length in x-direction. We can discretized the inital data as:

\begin{cases} u_{j}^{0}=-1 & j<l \\ u_{j}^{0}=0 & j=l \\ u_{j}^{0}=1 & j>l \end{cases} And we notice that $u_{l}^{0}$ is the point $x=1,t=0$.

Consider the upwind method with flux:

\begin{cases} F(v,w)=f(v) & \frac{f(v)-f(w)}{v-w}\geq 0 \\ F(v,w)=f(w) & \frac{f(v)-f(w)}{v-w}< 0 \\ \end{cases}

Consider $u_{j}^{n+1}=u_{j}^{n}-\frac{k}{h}(F(u_{j}^{n},u_{j+1}^{n})-F(u_{j-1}^{n},u_{j}^{n}))$, we need to prove the sequence $u_{j}^{n}$ converges to the rarefaction wave solution as $l \rightarrow \infty$.

The rarefaction wave solution is: \begin{cases} u(x,t)=-1 & x<1-t \\ u(x,t)=\frac{x-1}{t} & 1-t<x<t-1 \\ u(x,t)=1 & x>t-1 \end{cases}

I have spent a lots of days in this question but still I don't know how to do. Can anyone help me?

Examples of interesting cones in infinite-dimensional Hilbert spaces

Posted: 13 Jun 2021 07:58 PM PDT

Let $X$ be a real Hilbert space, and let $K$ be a closed convex cone.

The dual cone is defined by $K^*=\{x^*\in X \mid (\forall k\in K)\, \langle x^*,k\rangle \leq 0 \}$.

I am looking for some interesting examples of $K$ and $K^*$, especially when $X$ is infinite-dimensional.

For example,

  • $X=\ell_2$ and $K=\ell_2^+ = -K^*$
  • $X=L_2[0,1]$ and $K=L_2^+[0,1]=-K^*$

Note that these examples satisfy $K-K=X$. I wonder if there are some concrete examples where

$$K-K\neq X\quad \text{but}\quad \overline{K-K}=X$$

and $K^*$ is actually known?

Any examples/comments/references would be greatly appreciated.

A congruence lemma, do I need the chinese reminder theorem?

Posted: 13 Jun 2021 07:49 PM PDT

I am trying to prove the next statement.

For $r$ different primes such that the product $p_{1}\cdot p_{2}\cdots p_{k}>n^{2}$ and $a,b\in \{-n^{2},...,n^{2}\}$ $a=b \iff \forall r : a \mod p_{r}=b \mod p_{r}$

I made a table to build a numbering system applying mod on the naturals for each prime, I get sequences of numbers that I can put in my table, in such a way that the columns are coordinates and the new elements of my counting system, but I think that this is not a proof at all. I am stuck with this problem and I can't see how to proceed, especially the return.

Is there a connection between functions mapped from a double cover and spinor fields when there are ramification points?

Posted: 13 Jun 2021 07:48 PM PDT

Apologies, I'm very new and naive to topology.

I have been learning about different covering maps between manifolds and recently was learning about the map $T^2 \rightarrow S^2$ which is a two-to-one covering map with four ramifications points of ramification index two.

my understanding is that a function on the torus when mapped to the two-sphere takes on a multi-function behavior, which fails at the ramification points. If we take said function and rotate it $2\pi$ around one of our four points it returns to the negative of it's value, only returning to its original value after two full revolutions around our ramification points.

The behavior of a function in this case appears precisely like spin $1/2$ spinor fields in quantum mechanics (physics is my background).

Can someone explain to me how these two concepts are (or aren't) related. I find myself picturing the spinors of quantum mechanics as maps of functions from some double covering space. I know that the $Spin(n)$ groups are double covers of the $SO(n)$ groups from which arise spinor representations for connections via covariant derivatives in physics.

Doubt regarding Dirichlet's proof for infinitely many primes in an arithmetic progression

Posted: 13 Jun 2021 07:45 PM PDT

I have been studying the proof given by Dirichlet for infinitely many primes in an arithmetic progression and I cannot help but wonder this :

Why did Dirichlet need to first show $$\sum_{p\leq x, \hspace{2mm} p \hspace{1mm} \equiv \hspace{1mm} h (mod \hspace{1mm} k)} \frac{log (p)}{p} = \frac{1}{\phi(k)}log(x) +O(1)$$ in order to prove his point? I understand how this implies the infinitude primes in an arithmetic progression but why use $\frac{log (p)}{p}$ in the first place ?

Isn't there a better way ? What motivated Dirichlet to do this ?

Example on impulse

Posted: 13 Jun 2021 07:37 PM PDT

the question goes like this: A rubber ball of mass 50g falls from a height of 1 m and rebounds to a height of 0.5 m. Find the impulse and the average force between the ball and the ground if the time for which they are in contact was 0.1 s The way it was solved in my book, they worked out the final velocity for each case (upward and downward motion of the ball) and finally determined the change in momentum. Once the impulse is out avg force was resolved. My question is, the ball was acting under the force of gravity so cann't we just multiply g and given mass of the body to find F

Non-existence of supremum norm bound in terms of $L^2$ norm of gradient

Posted: 13 Jun 2021 07:35 PM PDT

Show there does not exist any constant $C>0$ such that, for any $\phi\in C^{\infty}_c(\mathbb{R}^2)$, the inequality $$\lVert\phi\rVert_\infty\leq C\lVert\nabla\phi\rVert_2$$ holds.

Obviously, we can scale to take $\lVert\phi\rVert_\infty=1$. I am thus trying to construct a sequence $\phi_n$ which reaches $1$ at $0$ but for which the $L^2$ norm of the gradient becomes arbitrarily small. Surely some type of bump function will work, but I can't seem to find an explicit form. Or is there a slicker way to approach this question?

(Why just $\mathbb{R}^2$ and not $\mathbb{R}^n$? One can prove that any inequality of the form $\lVert\phi\rVert_q\leq C\lVert\nabla\phi\rVert_p$ for $\phi\in C^{\infty}_c(\mathbb{R}^n)$ must satisfy $q^{-1}=p^{-1}-n^{-1}$)

counting the number of ways

Posted: 13 Jun 2021 07:34 PM PDT

Three colored boxes, i.e., Red (R), Blue (B), Green (G), contains 3 balls each, of the same colors as that of the box. A ball is chosen from the R box and put in one of other two boxes. Again a ball is chosen randomly from the box B and put in one of other boxes. Same process is repeated with box G. If given that each box have equal number of balls after this process and P is the probability that composition of balls is not exactly same as in the beginning of random experiment, then value of 8P is __

Strong law for additive functionals ( Durrett : Probability theory and examples )

Posted: 13 Jun 2021 07:33 PM PDT

This is excercise 5.6.5 from durrett's book.

Suppose $p$ is irreducible and has stationary distribution $\pi$. Let $f$ be a function that has $\sum\limits |f(y)|\pi(y)<\infty$. Let $T^k_{x}$ be the time of $k$ th return to $x$. Let $K_n=\inf \{k:T^k_x \geq n\}$ and $V^f_k = f(X(T^k_x))+\dots + f(X(T^{k+1}_x-1))$

Then $$\frac{1}{n}\sum\limits^{K_n}_{m=1}V^f_m \rightarrow \frac{EV_1^f}{E_xT_x^1}=\sum\limits f(y)\pi (y)\quad P_{\mu}-\text{a.s.}$$

Using Strong Law of large number and property of recurrence, I managed to show that $\frac{1}{n}\sum\limits^{K_n}_{m=1}V^f_m \rightarrow \frac{EV_1^f}{E_xT_x^1}$.

However I am having difficulty proving $\frac{EV_1^f}{E_xT_x^1}=\sum\limits f(y)\pi (y)$.

My try was $$EV^f_1 =E\sum\limits_{n=T^1_x}^{T^2_x-1} f(X(n))=E\sum\limits_y\sum\limits_{n=T^1_x}^{T^2_x-1} f(y)1\{x=y\} $$

How to prove the equality?

Matlab code to solve maximum likelihood problem.

Posted: 13 Jun 2021 07:53 PM PDT

There are 25 columns in a dataset each having 10000 points of the form - x(n) = A + w(n) where w(n) ~ N(0,A) AWGN.

  1. Estimate A' using maximum likelihood estimation.
  2. Using Monte carlo simulations find E(A') and var (A').
  3. Plot (E(A')-A)^2 and var(A') as function of N.
  4. Find pdf of A'. Plot for N= 5,10,20. Take any of N columns.

Does irreducibility and aperiodicity of transition matrix imply all eigenvalues are non-negative?

Posted: 13 Jun 2021 07:24 PM PDT

If $P$ is the transition matrix of a Markov chain that is irreducible and aperiodic, then is it true that all of its eigenvalues are non-negative?

Maximizing magnitude given a complex valued function

Posted: 13 Jun 2021 07:24 PM PDT

I am presented with a problem, which asks you to maximize $|z|$ with the constraint of $$\left|z+\frac 2z \right| = 2.$$

I tried to approach this with components, however this degraded into a messy cubic, letting $z=a+bi$ only to receive $$\frac{a^3+a^2b+ab^2+b^3+2a-2b}{a^2+b^2},$$ trying to maximize $\sqrt{a^2+b^2}.$

How does one go through solving this, without calculus?

How can you use logic to find the indicated term of a sequence(arithmetic and geometric sequence)?

Posted: 13 Jun 2021 07:19 PM PDT

Pre-Calc(Sequences and Series) When only given 1st term, the common difference/ratio, and number term, how do you find it logically without making the entire list of terms? Examples plz.

Confusion on when to use CDF and Poisson process

Posted: 13 Jun 2021 07:20 PM PDT

I'm going through the MIT OCW probability course (6.041sc), but I'm having trouble on when to use CDF and the Poisson process. Here's the problem (Recitation 15, problem 1).

Problem Statement:

Beginning at time $t=0$, we begin using bulbs, one at a time, to illuminate a room. Bulbs are replaced immediately upon failure. Each new bulb is selected independently by an equally likely choice between a a type-A bulb and a type-B bulb. The lifetime, $X$, of any particular bulb of a particular type is a random variable, independent of everything else, with the following PDF: \begin{aligned}\text{for type-A bulbs: }f_X(x) &= \begin{cases}e^{-x}, x\geq0,\\0, \text{ otherwise}\end{cases}\\\text{for type-B bulbs: }f_X(x) &= \begin{cases}3e^{-3x}, x\geq0,\\0, \text{ otherwise}\end{cases}\end{aligned}

Find the probability that there are no bulb failures before time $t$.

My Attempt:

I used the total probability theorem and then computed the CDF, $F_X(t)=P(X\leq t)$ : \begin{aligned}P(\text{no bulb failure before time }t)&=P(A)P(X\leq t|A)+P(B)P(X\leq t|B)\\&=\frac{1}{2}\int_0^t{e^{-x}}{dx}+\frac{1}{2}\int_0^t{3e^{-3x}}{dx}\\&=\frac{1}{2}\left(1-e^{-t}\right)+\frac{1}{2}\left(1-e^{-3t}\right)\end{aligned}

Solution:

\begin{aligned}P(\text{no bulb failure before time }t)=\frac{1}{2}e^{-t}+\frac{1}{2}e^{-3t}\end{aligned}

I was able to reproduce this result using the PMF for the number of arrivals $N_t$ in a Poisson process with rate $\lambda$, over an interval of length $t$. \begin{aligned}P_{N_t}(k)=e^{-\lambda t}\frac{(\lambda t)^k}{k!}, \text{ }k=0,1,\dots\end{aligned} In this context, we're looking at no arrivals, so $k=0$. And I figured that the arrival rate would be $\lambda=1,3$ for type-A(and type-B respectively) but I'm not sure why. Plugging in the appropriate numbers and using the total probability theorem we get the answer above.

My questions:

  1. Why did the CDF give me a different result? I'm sure that computing $P(X\leq t)$ was a valid approach, because that's the probability of the lifetime being at most $t$, but I must have some sort of conceptual misunderstanding on this.
  2. How would I know that the arrival rate for type-A is $1$(and $3$ for type-B)? The only way I'd think of that is the fact that both type A and B are exponentially distributed with parameter $\lambda=1,3$.

Determine the rate of decay

Posted: 13 Jun 2021 08:03 PM PDT

A radioactive element decays so that after $t$ years, the amount remaining, expressed in milligrams, of the original amount, is $A(t) = 100 (0.5)^{t/3.8}$. Determine the rate of decay when $25\%$ of the element is gone. Round to two decimal places.

Proving that the sum of a series tends to 0 as $n\to \infty$

Posted: 13 Jun 2021 07:51 PM PDT

Let $f(n)=\sum_{k=2}^{n-1} \binom{n}{k}\left(1-\frac{k}{n}\right)^{2n-2k+2} \left(\frac{k-1}{n}\right)^{2k}$

I'm trying to show $f(n)\to 0$ as $n\to \infty$

My attempt: Firstly, numerically I observe that the above statement is true. Analytically, I have a loose understanding of the sum and am wondering if someone can help me prove it more formally?

\begin{align*} \text{For $k=2$}&: \binom{n}{2}\left(1-\frac{2}{n}\right)^{2n-2} \left(\frac{1}{n}\right)^{4} \approx \frac{1}{n^2} \\ \text{For $k=3$}&: \binom{n}{3}\left(1-\frac{3}{n}\right)^{2n-4} \left(\frac{1}{n}\right)^{6} \approx \frac{1}{n^3} \\ &\vdots \\ \text{For $k=n/2$}&: \binom{n}{n/2}\left(\frac{1}{2}\right)^{n+2} \left(\frac{n/2-1}{n}\right)^{n} \approx \frac{2^n}{\sqrt{n}}\frac{1}{2^n}\frac{1}{2^n} \quad\text{By Stirlings's approx} \\ &\vdots \\ \text{For $k=n-1$}&: \binom{n}{n-1}\left(\frac{1}{n}\right)^{4} \left(\frac{n-1}{n}\right)^{2n-2} \approx \frac{1}{n^3} \\ \\ \implies f(n)&\leq (n-2)\times \frac{1}{n^2}\to 0 \quad\text{as}\quad n\to \infty \end{align*}

Thanks!

Finding a $3 \times 3$ matrix that only has 1 non-trivial invariant subspace of $\mathbb{R}^3$.

Posted: 13 Jun 2021 07:59 PM PDT

Does such a matrix exist, and if so how can I find it? My intuition was the matrix $$M = \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$$ as the only eigenspace is $\text{span} \{\mathbf{e}_1\}$, but I'd also need to prove that there's no other 2-dimensional invariant subspace.

If we let $\mathbf{u}$ and $\mathbf{w}$ be arbitrary linearly independent nonzero vectors in $\mathbb{R}$, we want $M\mathbf{u} = \alpha \mathbf{u} + \beta \mathbf{w}$ and $M\mathbf{w} = \gamma \mathbf{u} + \delta \mathbf{w}$.

We then get the system of equations $$\begin{align*} u_1 + u_2 &= \alpha u_1 + \beta w_1 \\ u_2 + u_3 &= \alpha u_2 + \beta w_3 \\ u_3 &= \alpha u_3 + \beta w_3 \\ w_1 + w_2 &= \gamma u_1 + \delta w_1 \\ w_2 + w_3 &= \gamma u_2 + \delta w_3 \\ w_3 &= \gamma u_3 + \delta w_3 \end{align*}$$

I'm sure we could solve this by turning it into a matrix and RREF and finding values of the constants to make it nonzero, but I don't feel like this is the way.

With some research I found the theorem:

Let $T: V \to V$ be a linear operator on a finite dimensional vector space. If $T$ is diagonalizable, and $W$ is a $T$-invariant subspace of $W$, then the restriction of $T$ to $W$, $T_W$, is also diagonalizable.

Does this mean, because the only eigenspace is the above and is one-dimensional, any 2-dimensional subspace cannot be invariant because the eigenspaces of $T_W$ must be eigenspaces of $T$? Also, does this imply that any $n \times n$ Jordan matrix only has a single non-trivial invariant subspace? And does this all generalise to $\mathbb{C}$ as well?

Thanks for any help!

Can you help me with this multiplier problem of lagrange multipliers?

Posted: 13 Jun 2021 08:03 PM PDT

$f\left(x,y\right) = x^4 + 3y^{\frac{4}{3}}$ with restriction $xy = c$ and the constant $c > 0$. I can't find a way to find the critical points with Lagrange multipliers.

I put together the following system of equations:

$$ L_x = 4x^3 - y \cdot \lambda = 0 \\ L_y = 4\sqrt[3]{y} - x \cdot \lambda = 0 \\ L_\lambda = x \cdot y - c=0 $$

How $\int_{0}^{\infty}\left(\frac{1}{x}\right)^{\log x}dx=e^{\frac14} \sqrt{\pi}$?

Posted: 13 Jun 2021 07:40 PM PDT

I have tried to evaluate $$\displaystyle\int_{0}^{\infty}\left(\dfrac{1}{x}\right)^{\log x}\,dx$$ using variable change $z=\frac{1}{x}$ yield to integrate $\frac{-z^{-\log z}}{z^2}$ from $0 \to \infty $ but the obtained quantite is not standard for me for integration ,Probably this is well known integral ? Any simple way for integration?.

Note:The result using WA is $e^{\frac14} \sqrt{\pi}$

If $K$ is an algebraic semantics for a deductive system $S$, what is the signature of the algebras in $K$ and what is $\models_K$?

Posted: 13 Jun 2021 07:48 PM PDT

A deductive system $S$ my be algebraizable over a language $L$ if there's an a class of algebras $K$ satisfying some conditions.

I'd like a better sense of what the class $K$ looks like before moving on to the conditions placed on it. What is the signature of the individual algebras that make up $K$ and what exactly is the relation $\models_K$?

I'm going to give a language $L$ a little bit more structure than it is given in BP and say that a language is a set of well-formed formulas and a family of sets of connectives indexed by arity.

A deductive system is defined by Blok and Pigozzi in Algebraizable Logics on page 5 as a pair $(L, \vdash_S)$ where $\vdash_S$ is a consequence relation. A consequence relation must satisfy some axioms to be a consequence relation, which I am omitting for brevity.

With that out of the way, on page 14 as well as in the abstract, Blok and Pigozzi describe what exactly an algebraic semantics is, paraphrased below.

Let $S$ be a deductive system. Let $K$ be a class of algebras. $K$ is an algebraic semantics for $S$ if and only if $\vdash_S$ can be interpreted in $\models_K$.

At this point, I'm having a lot of trouble following the text. I'd like to nail down what $\models_K$ and the class $K$ itself are before moving on to understanding the interpretability condition.

  • What is $\models_K$ exactly? The text suggests that it is an equivalence relation, but I'm not sure whether it is an equivalence relation within individual alegbras or an equivalence relation between algebras in my class.

  • I'm assuming that a class of algebras means a class of algebraic structures. I'm not sure what the signature is for these structures, but I'm assuming it's the connectives in $L$ with their respective arities.

For completeness, here is the definition of algebraizability. $K$ is the class of algebras that witnesses the algebraizability of $S$. This definition appears on page 14 in BP. In addition, $\text{Fm}$ refers to the set of well-formed formulas.

$K$ is called an algebraic semantics for $S$ if $\vdash_S$ can be interpreted in $\models_K$ in the following sense: there exist a finite system $\delta_i(p) \approx \epsilon_i(p)$, for $i<n$, of equations with a single variable $p$ such that, for all $\Gamma \cup \{ \varphi \} \subset \text{Fm}$ and each $j < n$

$$ \Gamma \vdash_S \varphi \iff \{ \delta_i[\psi/p] \approx \epsilon_i[\psi/p] : i < n , \psi \in \Gamma \} \models_K \delta_j[\varphi/p] \approx \epsilon_j[\varphi/p] \tag{2.1i}$$

Proving a point does not belong to a set in $\mathbb{R}^2$

Posted: 13 Jun 2021 07:54 PM PDT

I have a set $\mathcal{A}$ in $\mathbb{R}^2$ which is a closed and simply connected set. I need to find conditions under which the point $(x^*,y^*)$ does not belong to the set $\mathcal{A}$. I think the point $(x^*,y^*)$ does not belong to the set $\mathcal{A}$ if we at least have one of the following cases,

  1. $x^*>x^\ddagger$,
  2. $y^*<y^\ddagger$,
  3. $x^\dagger<x^*\leq x^\ddagger$,
  4. $y^\ddagger\leq y^*<y^\dagger$,

where $$ x^\ddagger\triangleq\max\{x:(x,y)\in\mathcal{A}\},\\ y^\ddagger\triangleq\min\{y:(x,y)\in\mathcal{A}\},\\ x^\dagger\triangleq\max\{x:(x,y^*)\in\mathcal{A}\},\\ y^\dagger\triangleq\min\{y:(x^*,y)\in\mathcal{A}\}.\\ $$ For otherwise $(x^*,y^*)\in\mathcal{A}.$

Is this a right idea and any idea how can I prove that?

As an example, the set $\mathcal{A}$ is the orange region in the following picture. The lower boundary of set $\mathcal{A}$ is monotonically increasing.

enter image description here

If $\displaystyle \int_\gamma f(z) \, dz = i$ and $\displaystyle\int_\gamma g(z)\,dz = 3-i,$ find $\displaystyle\int_{-\gamma} (2f(z) + ig(z))\,dz$?

Posted: 13 Jun 2021 07:41 PM PDT

Can't figure our this reverse path integral at all, can anyone help?

Natural Numbers as $0$ and its Successors

Posted: 13 Jun 2021 07:38 PM PDT

I've been thinking about the propositions one can prove using the Peano axioms lately and there's this one question that crossed my mind.

I understand that the axiom of induction was introduced to remove 'unwanted' elements from $\mathbb{N}$. These 'unwanted' elements would be objects that obey the first four axioms, but are detached from the list of objects obtained through applying the successor function repeatedly on $0$ (i.e. detached from the $0$, $S(0)$, $S(S(0)) ...$ list).

I believe I understand the logic behind how the axiom of induction achieves this. My question is whether we can rigorously prove the statement $\forall n \in \mathbb{N} (\text{n can be reached by applying the successor function sufficiently many times to 0})$.

I know that this statement that I just wrote down is informal, but I'm not sure how I should restate such a statement in a more rigorous way. However, if I stick to this informal statement, I can use induction to heuristically justify my claim:

Proof: Let $P(n)$ be the statement 'n can be reached by applying the successor function sufficiently many times to 0'.

$P(0)$ is automatically true, since $0$ is already equal to $0$.

Now we suppose $P(n)$ is true for some arbitrary $n \in \mathbb{N}$. This means $n$ can be expressed as something like $S(...(0)...)$.

Then this implies $P(S(n))$ is also true, since $S(n) = S(S(...(0)...))$.

So this closes the induction, and $\forall n \in \mathbb{N} (P(n))$ is true.

I was wondering whether it's possible to reframe this claim and proof into a more logically rigorous form.

Thank you very much in advance for taking the time to read through my question.

Fermat's Last Theorem for $n=4$ revisited

Posted: 13 Jun 2021 07:26 PM PDT

Fermat's Last Theorem for $n=4$ states that
If $$b^4+c^4=a^4\tag{1}$$ Then, no positive integers $a, b, c$ exist simultaneously.
We can re-arrange (1) as $$a^4-b^4=c^4\tag{2}$$ But, by direct expansion, we see that $$a^4-b^4 = (a^2+b^2)^2-(b(a+b))^2-(b(a-b))^2\tag{3}$$ So, equating (2) and (3), we get $$c^4 = (a^2+b^2)^2-(b(a+b))^2-(b(a-b))^2$$ $$(c^2)^2 = (a^2+b^2)^2-(b(a+b))^2-(b(a-b))^2$$ $$(c^2)^2+(b(a+b))^2+(b(a-b))^2 = (a^2+b^2)^2\tag{4}$$ We can see that (4) is a Pythagoras' quadruple.
Pythagoras' quadruple states that
If $$w=m^2+n^2-p^2-q^2$$ $$x=2(mq+np)$$ $$y=2(nq-mp)$$ $$z=m^2+n^2+p^2+q^2$$ Then $$w^2+x^2+y^2=z^2\tag{5}$$ Where $gcd(w,x,y,z)=1$
Since $w,x,y,z$ are co-prime, then $w$ and $z$ are odd while $x$ and $y$ are even.
This shows that two of $b(a+b)$, $b(a-b)$ and $a^2+b^2$ must be even while the remaining one is odd.
Now, comparing (4) and (5), we can see that $$(c^2)^2=m^2+n^2-p^2-q^2$$ Thus, $c$ is odd. And if $c$ is odd, then we see from (2) that $a$ and $b$ must be of opposite parity. But $b$ can't be odd because then, $b(a+b)$, $b(a-b)$ and $a^2+b^2$ will all be odd which does not conform with Pythagoras' quadruple.
If $a$ is odd and $b$ is even, we see that $b(a+b)$ is even, $b(a-b)$ is even and $a^2+b^2$ is odd which corresponds with Pythagoras' quadruple. So, we are forced to conclude that $a$ is odd while $b$ is even.
Now, we can see that $$(c^2)^2=m^2+n^2-p^2-q^2\tag{6}$$ $$a^2+b^2=m^2+n^2+p^2+q^2\tag{7}$$ $$b(a+b)=2(mq+np)\tag{8}$$ $$b(a-b)=2(nq-mp)\tag{9}$$ Adding (8) and (9), we have $$b(a+b) + b(a-b)=2(mq+np)+ 2(nq-mp)$$ $$b(a+b) + b(a-b)=2q(n+m)+ 2p(n-m)\tag{10}$$ We can see that to make both expressions of (10) obviously equal, we only need to set $p=q=\frac{m}{2}$. By doing this, we have $$b(a+b) + b(a-b)=2\frac{m}{2}(n+m)+ 2\frac{m}{2}(n-m)$$ $$b(a+b) + b(a-b)=m(n+m)+ m(n-m)\tag{11}$$ Comparing both sides of (11), we can see that $b=m$ and $a=n$. Specifically, $m$ is even while $n$ is odd.
Putting $p=q=\frac{m}{2}$ in (6) gives $$(c^2)^2=m^2+n^2-(\frac{m}{2})^2-(\frac{m}{2})^2$$ $$(c^2)^2=m^2+n^2-\frac{m^2}{2}$$ $$(c^2)^2=\frac{m^2}{2}+n^2$$ Since $m$ is even, say $m=2m_1$, then $$(c^2)^2=2{m_1}^2+n^2$$ $$c^4=2{m_1}^2+n^2\tag{12}$$ Also, putting $p=q=\frac{m}{2}$ in (7) gives $$a^2+b^2=m^2+n^2+(\frac{m}{2})^2+(\frac{m}{2})^2$$ $$a^2+b^2=m^2+n^2+\frac{m^2}{2}$$ $$a^2+b^2=\frac{3m^2}{2}+n^2$$ Since $m$ is even, say $m=2m_1$, then $$a^2+b^2=6{m_1}^2+n^2\tag{13}$$ The final jigsaw is now to prove that at least one of (12) and (13) does not have integers solution.
Putting $a=n$, $b=m=2m_1$ in (13), we have;
$$n^2+(2m_1)^2=6{m_1}^2+n^2$$ $$4m_1^2=6{m_1}^2$$ $$2m_1^2= 0$$ Thus we have that $m_1=m=p=q=b=0$ which means (1) becomes $$0^4+c^4=a^4$$ which is trivial. This makes us conclude that no three natural numbers satisfy (1)
Question: Is my proof correct? If not, where is/are the flaw(s)?

Prove, using the definition of a closed set, that $S = [0, \Omega)$ is not a closed subset of $X = [0, \Omega]$ with order topology

Posted: 13 Jun 2021 07:27 PM PDT

I am using the book A First Course in Topology by Robert Conover. I can assume everything up to this point and info on ordinals.

Prove using the definition of a closed set that S = [0,$\Omega$) is not a closed subset of $X = [0, \Omega]$ with order topology (prob 4.a, pg 68).

Assumptions and definitions

$\Omega$ is the the initial ordinal of $\text{card}(\aleph_1)$.

https://en.wikipedia.org/wiki/Ordinal_number

5.1 Theorem Any countable subset of $[0, \Omega)$ is bounded above (i.e., if $S \subset[0, \Omega]$ with $\lvert S \rvert \leq \aleph_0$, then there exists a number x $\in$ [0,$\Omega$] such that $x \geq s$ for every $s \in S$).

My attempt(edit)

Let X,S=[0,$\Omega$) be with the order topology If $S$ is closed $X \setminus S$ is open in $X$.Then X$\setminus$ S =($\Omega$ )But $\Omega$ has finite countable infinite sets and infinite uncountable set and is open in the order topology and each set is uncountably infinite. Since X\S is uncountably infinite by 5.1 is not bounded above, and since $\Omega$ is a limit ordinal, there is no x$\in$ S for a positive r>0 such that x$\in$(x-r,x+r) Thus Sis not closed in X

I hope I don't get downvoted. I put a lot of thought into it. I know ordinals are Hausdorff,Conover mentions it, but can't use it,cause l have to prove it later. Any help would be appreciated

Making $x$ the subject of $y=x\sqrt{2-x}$

Posted: 13 Jun 2021 07:58 PM PDT

I find it hard to change the subject of equation from $y$ to $x$.

Here is the example: $$y=x\sqrt{2-x}$$

Could you help me how to change it into a function $x(y)$?

Integral over domain of infinite tetration of x over its real domain from 0 to $\sqrt[e]e$. Partially found a non-integral exact form.

Posted: 13 Jun 2021 07:34 PM PDT

I have been trying to find an interesting constant over the domain of the infinite tetration of x and have just almost figured out the area with a non integral infinite sum representation. Just one constant is in my way. D denotes the domain. This question is different from this one as it has the full domain such that the imaginary part is $0$ and not for a single point. Here is a demo of my expansion. Here is my source for the product logarithm/W-Lambert function series. My inspiration for the series is here.

Finally here is a graph of the constant. My work is as follows. I used a bit of software to help with the evaluation at the end. Here is data about the generalized incomplete gamma function used here.

enter image description here

$$ \mathrm{G=\int_D x^{x^{x^…}}dx=\int_D {^\infty x}\, dx=\int_D-\frac{W(-ln(x))}{ln(x)}dx= 1.265188689361227081430914184615901039501069191363542653701819999950085943915822836313002058708863484…\implies G+\int_0^{\sqrt[-e]e }\frac {W(-ln(x))}{ln(x)}dx= \int_{\sqrt[-e]e}^{\sqrt[e]e}\frac {W(-ln(x))}{ln(x)}dx= \sum_{n=1}^\infty\frac{n^{n-1}}{n!} \int_{\sqrt[-e]e}^{\sqrt[e]e} ln^{n-1}(x)dx= \sum_{n=1}^\infty\frac{(-n)^{n-1}}{n!}Γ\left(n,-\frac 1e,\frac 1e\right)=\sum_{n=1}^\infty\frac{(-n)^{n-1}}{n}Q\left(n,-\frac 1e, \frac 1e\right)= 0.886369135921835965080748…=G-0.378819553439391116350165…} $$

I have also found the amazing result of being able to integrate the $\mathrm{x^{th}}$ root of x using a theorem on the integral of an inverse function. Here is my work.

$$\mathrm{\int_{eW\left(\frac 1e\right)}^e \left(x^\frac 1x=\sqrt[x]x\right)dx+\int_{e^{-\frac 1e}}^{e^\frac 1e} {^\infty x}\,dx=e^{1+\frac1e}-e^{1-\frac1e} W\left(\frac 1e\right)=e^{1-\frac1e}\left(e^\frac2e-W\left(\frac 1e\right)\right)=e^{1+\frac1e}\left(1-e^{-\frac2e}W\left(\frac 1e\right)\right)\implies \int_{eW\left(\frac 1e\right)}^e x^\frac 1xdx= e^{1+\frac1e}-e^{1-\frac1e} W\left(\frac 1e\right)-\sum_{n=1}^\infty\frac{(-n)^{n-1}}{n}Q\left(n,-\frac 1e, \frac 1e\right)}$$

In order to find an exact form of G, I need to find the following. The other form uses the following identity here: $$\mathrm{I= \int_0^ {e^{-\frac 1e}} x^{x^{x^…}}dx=\int_0^{e^{-\frac 1e}} {^\infty x}\, dx=\int_0^ {e^{-\frac 1e}} -\frac{W(-ln(x))}{ln(x)}dx=e^{1-\frac1e}W\left(\frac1e\right)-\int_0^ {eW\left(\frac1e\right)} \sqrt[x]x dx=0.378819553439391116350165…}$$

The previous result is proof of the following. I guess this link here is not accurate anymore. I will give an example if wanted of this result. Using another Wikipedia theorem proves that: $$\mathrm{\int x^{\frac1x}dx=x^{\frac1x+1}+\sum_{n=1}^\infty (-1)^nn^{n-2} Q\left(n,-\frac{ln(x)}{x}\right)+C,eW\left(\frac1e\right)\le x\le e}$$

How do I evaluate this integral? A closed form is wanted, but optional. Please give me any hints as the already used series expansion for this other integral is not in the interval of convergence. This is the main constant that I need to find. Also, please correct me and give me feedback!

Laguerre's theorem on power of a point w.r.t. an algebraic curve

Posted: 13 Jun 2021 07:46 PM PDT

So on Wikipedia article for a power of a point there is a short section about Laguerre's theorem. The problem is, the article has no references, and whenever I'm trying to Google it the only things I get are either power diagrams or some different definitions of power of a point w.r.t. algebraic curve, but none of these seem to trace back to Laguerre's work.

So my question is, as expected

Can anyone give a reference to Laguerre's original work, or at least provide a place where properties of his definition (such as independence of the choice of the circle) are proven?

G as a graph without self loops and parallel edges with n vertices and m edges

Posted: 13 Jun 2021 08:03 PM PDT

EDITED to include c.

Could someone help me understand this problem? I haven't been able to comprehend what I am supposed to do here.

1) Let G be a graph without self loops and parallel edges with n vertices and m edges. Consider the following procedure:
a) $G' \leftarrow G$
b) While there is a vertex of degree stricly less than $m/n$ do:
c) Remove this vertex and all its edges from the graph, and assign to G` the new graph. d) Recompute the degrees (note that $M$ and $n$ are fixed)

2) Return $G'$

Honestly, I feel like I should be doing derivatives in Calculus after reading this. (If that's what I'm actually supposed to do I may break my computer).

Thank you to anyone for your help.
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