Sunday, April 24, 2022

Recent Questions - Mathematics Stack Exchange

Recent Questions - Mathematics Stack Exchange

How to show that if  is hermitian then U= exp(iÂ) is an unitary operator using the fact that A is diagonalizable

Posted: 24 Apr 2022 06:11 AM PDT

It is given that  can be expressed as Â=S $\Lambda$ $S^\dagger$

My progress so far is:

$ÛÛ^\dagger$= $exp(iÂ)exp(iÂ)^\dagger$ = (S exp(iÂ)$S^\dagger$)(S exp(-iÂ)$S^\dagger$) but I do not know how to progress further to show that this equals the identity matrix.

Complex Numbers, Euler polar coordinates form

Posted: 24 Apr 2022 06:10 AM PDT

enter image description here  

Hello everyone can you please help in this question's solution of complex numbers based on euler polar coordinates summation..

enter image description here

ANY HELP WILL BE APPRECIATED

THANK YOU!!

Show that the Range of the multiplication operator $M_a$ is dense in $L_p (\Omega)$ provided that $a \neq 0$ a.e. in $\Omega \subset \mathbb{R}^n$

Posted: 24 Apr 2022 05:56 AM PDT

Here, $1 \le p < \infty$

I thought that maybe - for $a \neq 0 $ a.e. in $\Omega$ - it is true, that

$$\mathcal{C}^{\infty}_0 \subset \mathcal{R}(M_a)$$

And because $\mathcal{C}^{\infty}_0$ is dense in $L_p (\Omega)$, then this would mean that the range of $M_a$ is dense too in $L_p (\Omega)$

But I don't know if we can argue like that here.

The next step in that exercise would be to show that $\mathcal{R}(M_a) = L_p (\Omega)$, if there exists a constant $c>0$ such that $|a(x)| \ge c$ a.e. in $\Omega$

That's why I doubt my solution would be the right one here

Is the following set simply connected?

Posted: 24 Apr 2022 05:55 AM PDT

I have the following problem:

$\Omega=\{x+iy:x,y\geq 0\}\cup \{x+iy:x,y\leq 0\}$

I need to check if $\Omega$ is simply connected, if not give an example of a closed curve on $\Omega$ which is not homologous to $0$ in $\Omega$. We had the following definition:

Let $\Omega\subset \Bbb{C}$ be open and connected, we say $\Omega$ is simply connected if $(\Bbb{C}\cup \{\infty\})\setminus \Omega$ is connected.

I'm somehow a bit confused since $\Omega$ is not open. Hence I think $\Omega$ is not simply connected. But using this definition I can't finde a curve which is not homologous to $0$ in $\Omega$

Could maybe someone help me?

$E[(d W_t)^2] =?$

Posted: 24 Apr 2022 05:55 AM PDT

I confused myself here. Let $W_t$ be Brownian motion. It is obvious that the variance of the Brownian motion increment $dW_t$ is $dt$, i.e. $E[(d W_t)^2] = dt$. However, if I write this like so, I get a different result:

\begin{align} E[(dW_t)^2] &= E[(W_{t+dt} - W_t)^2] \\ &= E[W_{t+dt}^2] - 2E[W_{t+dt}]E[W_t] + E[W_t^2] \\ &= (t+dt) - 0 + t \\ &= 2t + dt \end{align}

Where is my mistake in the expansion above?

Integral $\int_{-\infty}^\infty \frac{\tan^{-1}(x) - \tan^{-1}(x+a)}{(x-i)(x+a-i)} dx$

Posted: 24 Apr 2022 05:54 AM PDT

For $a\in\mathbb R$, I want to compute the integral $$I = \int_{-\infty}^\infty \frac{\tan^{-1}(x) - \tan^{-1}(x+a)}{(x-i)(x+a-i)} dx.$$ Previously, I asked a question on a similar integral:

Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$

In that question, several ideas, such as (1) taking derivative with respect to $a$ or (2) contour integration, are suggested. However, both method seems to be not working for this integral.

This integral is motivated from physics calculation.

How can I evaluate $I$?

doubling metric measure spaces and Lebesgue differentiation theorem

Posted: 24 Apr 2022 05:52 AM PDT

If $(X,d,\mu)$ is a doubling metric measure space then it known that if $f \in L^1(X)$ is a positive function with $\|f\|_{L^1(X)}=1$ then

$$ \lim_{n \to \infty} \frac{1}{\mu(B(x,\frac{1}{n}))}\int_{B(x,\frac{1}{n})} \phi\, f\, d \mu=\phi(x) $$ for almost all $x \in X$. See the book "Sobolev Spaces on Metric Measure Spaces" page 77.

Suppose in addition that $X$ is locally compact and that $f_n$ is a positive function with compact support such that $\mathrm{supp} f_n \subset B(x,\frac{1}{n})$ and $\|f_n\|_{L^1(X)}=1$. Do we have $$ \lim_{n \to \infty} \int_{B(x,\frac{1}{n})} \phi\, f_n d \mu_G =\phi(x) $$ for almost all $x \in X$ ?

Derivative of a matrix of euclidean distances

Posted: 24 Apr 2022 05:48 AM PDT

I'm trying to compute the gradient of a cost function wrt. to some matrices. During the computations, I reached an expression for which I wasn't sure about its derivative. Here's my problem :

Let $X = (x_1,x_2,\cdots,x_n) \in \mathbb(R)_+^{M\times N}$, $F = (f_1, f_2,\cdots,f_k)\in \mathbb(R)_+^{M\times K}$ is the matrix of cluster centroids, $G = (g_1, g_2,\cdots,g_k)\in \mathbb(R)_+^{K\times N}$ indicates the matrix clustering indicators, and $D \in \mathbb(R)_+^{K\times N}$ is the matrix of the euclidean distances between each data point $X$ and the set of centroids $F$, more precisely $d_{kn} = \Vert x_n - k \Vert$.

What is the derivative of $Tr(G\circ D.(G\circ D)^T)$ with respect to $F$ ?
$st. Tr$ is the trace, and $\circ$ is the Hadamard Product.

The solution I found, using the matrix cookbook formulas, is :
$\frac{\partial}{\partial F} Tr(G\circ D.(G\circ D)^T) = 2(G\cdot F\circ G - G\cdot X\circ G)$

But I'm not sure about it. In case my solution is wrong, can you please provide a step by step explanation?
Thanks

Question about discontinouties of complex valued functions and isolated singularities.

Posted: 24 Apr 2022 05:45 AM PDT

Our definitions for an isolated singularity is the following: $z_0$ is called a isolated singularity of $f: U \rightarrow U$ iff: $z_0$ is an isolated point of $U$.

Now we classified 3 types of such singularities, removable, poles and essential singularities. It is clear to me what those are, we called $z_0:$

(i) removable, if one cand define $f(z_0)$ s.t. $f$ is holo on $U \cup {z_0}$

(ii) a Pole, if (i) applies to $f \cdot (z-z_0)^m$

(iii) essential, if (i),(ii) do not apply.

Now my question is the following, in what of those categories does a discontinouity fall, where the limit does not exist (therefore it cant be removable) and where f is bounded in a region of $z_0$.

So I am thinking of a point where $f(x_n)$ converges to some complex value for alle sequences $x_n \rightarrow z_0$ but does not yield the same result for them. Something like the signum function in $0$, but in a complex way. Does something like this even exist?

My intuition tells me, that this must either be a pole or be impossbile to exist, but i feel like poles are points where f tends to infinity. It is very hard to think of complex functions for me, i would be happy for some thoughts on this. Thank you!

an integral inequality having function and its derivatives

Posted: 24 Apr 2022 05:44 AM PDT

If $f\in C^1[0,1]$ and $f(0)=0$ , show that $\int_0^1\frac{|f(x)|^2}{x^2}dx\leq4\int_0^1|f^{'}(x)|^2dx$

I want to seek for an elementary solution for the problem.

SOSTOOLS in MATLAB: findbound gives wrong output

Posted: 24 Apr 2022 05:43 AM PDT

Here is the MATLAB code that uses SOSTOOLS toolbox:

clear;  syms a12 a13 C;  deg = 6;  vartable = [a12, a13, C];  prog = sosprogram(vartable);  solver_opt.solver = 'mosek';  [sol_C, vars, xopt] = findbound(C,[0],[-1 -54*a12^2 - 81*a12^4 - 54*a13^2 - 486*a12^2*a13^2 - 81*a13^4 + C*(1 + 6*a12^2 + a12^4 + 6*a13^2 + 6*a12^2*a13^2 + a13^4)],deg,solver_opt);

Output says:

No lower bound could be computed. Unbounded below or infeasible?

Whereas the Mathematica code:

Minimize[{c, -1 - 54*a12^2 - 81*a12^4 - 54*a13^2 - 486*a12^2*a13^2 -       81*a13^4 +       c*(1 + 6*a12^2 + a12^4 + 6*a13^2 + 6*a12^2*a13^2 + a13^4) ==      0}, {a12, a13, c}]

gives the expected (correct) output:

{1, {a12 -> 0, a13 -> 0, c -> 1}}

What went wrong in the MATLAB code?

Prove that we can express $n^k$ as a sum of $n$ consecutive odd natural numbers

Posted: 24 Apr 2022 05:59 AM PDT

Prove that $\forall n \in \mathbb{N},$ we can express $n^k$ where $k \geq2$ is an integer, as a sum of $n$ consecutive odd numbers.

My Solution -

Let $2m+1$ be the first odd number , $m \geq 0 , m \in \mathbb{N_0.}$

$\Rightarrow 2m+1 , 2m+3 , 2m+5,...,2m+2n-1 $ are $n$ consecutive odd numbers.

$⇒$ Sum of $n$ odd numbers ,

$⇒\sum_{k=1}^{n} 2m+2k-1 = 2mn+n^2=n(2m+n)$

Now notice that $2m+n$ should be a power of n.

$⇒2m+n=n^p ; $ For some $p\geq2, p \in \mathbb{N}.$

$⇒2m = n^p-n=n(n^{p-1}-1)$

Now $2$ cases :

$\text{Case 1:}$ $n$ is odd $n=2a+1 ; a \geq 0 , a \in \mathbb{N_0}$

$⇒(2a+1)[(2a+1)^{p-1}-1]$

Now note that $(2a+1)^{p-1}-1 \equiv 0 \pmod {2}⇒$ Above expression is divsible 2 $⇒$ Can be expressed in terms of $m$ as $2m$ is even.

$\text{Case 2 : }$ $n$ is even $⇒n=2a ; a \geq 0 , a \in \mathbb{N_0}$

$⇒2a[(2a)^{p-1}-1]$

Now note that $2a[(2a)^{p-1}-1] \equiv 0 \pmod{2}⇒ $ Divisible by $2⇒$ Can be expressed in terms of $m$.

Hence there exists $m$ such that it makes $n^k$.

Hence Proved.

Can someone tell if this proof is mathematically correct?

How does the degree of a polynomial affect its domain and range?

Posted: 24 Apr 2022 05:42 AM PDT

So, my knowledge is that odd degree polynomials have a range of all real numbers and that the range of even degree polynomials need to be derived from global minimum and maximum points.

I want to know about the degree's effect on the domain and the logic behind the validity of the above statement.

Thank you.

Finding an efficient algorithm that computes the polynomial $p \in \mathbb{P}_n$ that interpolates an even function

Posted: 24 Apr 2022 05:34 AM PDT

Let $f: [-a, a] \rightarrow \mathbb{R}, x \mapsto x \cdot arctan(x)$ for $a > 0$, $n \in \mathbb{N}$ and $p \in \mathbb{P}_n$ the polynomial that interpolates $f$ at the interpolation nodes $x_j = -a +2a \frac{j}{n}$ for $j = 0, ..., n$.

Since $f$ is an even function I have already deduced that $c_i = 0$ (the coefficients of the polynomial p) for all odd $i$, and I am asked to exploit this fact in order to simplify the computation necessary for the coefficients $c_i$.

The professor wants our interpolation to only use the function values $f(x_{\lceil n/2\rceil}), ..., f(x_n)$ and to define a suitable change of variables $\phi$ and interpolate $g(y) := f(\phi^{-1}(y))$.

My idea was to simplify the system of linear equations $p(x_0) = f(x_0)), ..., p(x_n) = f(x_n)$ to $p(x_{\lceil n/2\rceil}) = f(x_{\lceil n/2\rceil}), ..., p(x_n) = f(x_n)$, since we already know that $c_i = 0$ for all odd $i$ and thus only need $n/2$ equations, but I don't know how to define $\phi$. Any help would be appreciated.

How to find minimal distance (and coordinates where is it) between two conic sections?

Posted: 24 Apr 2022 05:31 AM PDT

I have equations of two conic sections in general form. Is it possible to find minimal distance between them (if they are not intercross)?

I need it to calculate is two spacecrafts on two orbits (2d case) can collide or not. If minimal distance bigger than sum of radiuses of bounding circles I don't need to care about collision.

Tensor fields defining $G$-structure are parallel

Posted: 24 Apr 2022 05:38 AM PDT

Suppose $G \leq GL_n(\mathbb{R})$ is the stabilizer of some tensors $T^0_1, ..., T^0_k$, let $P$ be a $G$-structure on a manifold, i.e. a principal $G$ subbundle of the frame bundle of $M$ and let $T_1, ..., T_k$ be tensor fields that are pointwise the image of $T^0_1, ..., T^0_k$ through the frames of $P$, as in this question. Is it true that the tensor fields $T_1, ..., T_k$ are parallel with respect to a (torsion-free) connection on $P$? Alternatively, is it true that if the holonomy of a Riemannian manifold is contained in $G$ then the tensors defining (in the sense of the linked question) the $G$ structure are parallel? By looking at Berger's classification it looks true.

Cauchy schwarz in vectors || meaning

Posted: 24 Apr 2022 05:48 AM PDT

What does "|| .. ||" mean in the cauchy schwarz inequality in vectors ? I was interpreting that modulus of two vectors is always greater than or equal to scalar product magnitude . That is |x.y| <= |x| |y| . But given was |x.y| <= ||x|| ||y||

Could someone validate this logic proof for me?

Posted: 24 Apr 2022 05:57 AM PDT

Prove that if the sentence (∀x P (x)) ∨ (∀x Q(x)) holds, then ∀x (P (x) ∨ Q(x)) holds too

Assume $\forall x$ P(x) = True

$\forall x$ (P(x)$\lor$Q(x)) must be true by assumption.

By symmetry this must be hold if Q(x) is true for all x

Norm of a map in Banach space

Posted: 24 Apr 2022 05:37 AM PDT

Let $X$ be a banach space and $f:(\Omega, \mathcal{F}, \mathbb{P}) \to (X,\mathcal{B})$ measurable function. I want to define $||f||_{L_1(X)}$. I saw this definition in a paper: $$||f||_{L_1(X)} = \int|f|d\mathbb{P}$$ but I don't understand what is $|f|$.

Show set of functions forms a group: $x, x+1, x+2, 2x, 2x+1, 2x+2$.

Posted: 24 Apr 2022 05:48 AM PDT

Prove that the set of transformations forms a group with order six, under function composition.

$f_1(x) = x$,
$f_2(x) = x+1$,
$f_3(x) = x+2$,
$f_4(x) = 2x$,
$f_5(x) = 2x+1$,
$f_6(x) = 2x+2$

Function composition is finding all possibilities of $f_n(f_m(x))$, and then testing for the group properties.

But, being function, the composition is to be associative. But, am not clear on this part. Say, taking up a few examples:

  1. $$f_6(f_5(f_3(x)))= f_6(2(x+2)+1)=f_6(2x+5)=2(2x+5)+2= 4x+12.$$ $$(f_6(f_5(x))f_3(x)= (2(2x+1)+2)f_3(x)= (4x+4)f_3(x)=(4(x+2)+4)= 4x+12.$$

  2. $$f_5(f_3(f_2(x)))= f_5(x+3)= 2(x+3)+1= 2x+7.$$ $$(f_5(f_3(x))f_2(x)= (2(x+2)+1)f_3(x)= (2x+5)f_3(x) = 2(x+2)+5= 2x+9.$$

Q.1. Here, associative property is not seen?

The table is drawn below with $n$ being row, column being $m$.

\begin{array}{|c|c|c|c|} \hline & x & x+1 & x+2 & 2x & 2x+1 & 2x+2 \\ \hline x & x & x+1& x+2& 2x & 2x+1& 2x+2\\ \hline x+1 & x+1& x+2& x+3& 2x+1& 2x+2 & 2x+3\\ \hline x+2 & x+2& x+3& x+4& 2x+2& 2x+3& 2x+4\\ \hline 2x & 2x& 2x+1& 2x+2& 4x& 4x+1& 4x+2\\ \hline 2x+1 & 2x+1& 2x+3& 2x+5& 4x+1& 4x+3 & 4x+5\\ \hline 2x+2 & 2x+2& 2x+4& 2x+6& 4x+2& 4x+4 & 4x+6\\ \hline \end{array}

But, if take with $n$ being column, row being $m$.

\begin{array}{|c|c|c|c|} \hline & x & x+1 & x+2 & 2x & 2x+1 & 2x+2 \\ \hline x & x & x+1& x+2& 2x & 2x+1& 2x+2\\ \hline x+1 & x+1& x+2& x+3& 2x+2& 2x+3 & 2x+4\\ \hline x+2 & x+2& x+3& x+4& 2x+4& 2x+5& 2x+6\\ \hline 2x & 2x& 2x+1& 2x+2& 4x& 4x+1& 4x+2\\ \hline 2x+1 & 2x+1& 2x+2& 2x+3& 4x+2& 4x+3 & 4x+4\\ \hline 2x+2 & 2x+2& 2x+3& 2x+4& 4x+4& 4x+5 & 4x+6\\ \hline \end{array}

Q.2. Are the two approaches equivalent, or not?

enter image description here

Integral $ \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx$

Posted: 24 Apr 2022 05:28 AM PDT

For $a\in\mathbb R$, I want to evaluate the integral $$ I = \int_{-\infty}^\infty \frac{1}{x^2+1} \left( \tan^{-1} x + \tan^{-1}(a-x) \right) dx.$$ I tried to integration by parts by considering $\left( \tan^{-1}(x) \right)' = \frac{1}{x^2+1}$, so that $$I = - \int_{-\infty}^\infty \tan^{-1}(x) \left(\frac{1}{1+x^2} - \frac{1}{1+(x-a)^2}\right).$$ However, I cannot proceed further. Mathematica cannot solve both integrals.

How to evaluate $I$?

Why subtracting positive same as adding negative

Posted: 24 Apr 2022 05:34 AM PDT

Here I didn't get this why subtracting a positive same as adding negative Ex 2-4=2+(-4) please explain me. And is 2+-4 is correct or 2+(-4).

Area of the region traced by a point on a variable line segment

Posted: 24 Apr 2022 06:09 AM PDT

Let $C$ be a variable point in the first quadrant on line $x+y=1$, which intersects the axes at $A$ and $B$ respectively. Let $D$ and $E$ be the foot of perpendiculars on the axes from the point $C$, And $P(h,k)$ be a point on the line segment $DE$.

We need to find

1.the maximum possible value of product $hk$.

2.the area traced by point $P$.

I assumed a point $C(\cos^2\theta,\sin^2\theta)$ ,giving us $$ \frac{h}{\cos^2\theta} + \frac{k}{\sin^2\theta}=1 $$ from here we can apply AM-GM inequality , yielding $$hk\leq \sin^22\theta/16$$

Making the maximum required value to be $1/16$, This solves first part of the question.

But now because $P$ only lies on curve $xy=1/16$ at one point $(1/4,1/4)$, We cannot use this function ($y=x/16$) to find the area of the locus of $P$ ( what we need here is the area that the line segment $DE$ is 'sweeping')

So my question is how should we visualise and find the area traced by point P?

Is the "parallel axiom" sentence, a proposition in absolute geometry?

Posted: 24 Apr 2022 05:50 AM PDT

I will try to explain in other words. Given a formal theory of absolute geometry, we are able to write the expression whose meaning is the paralel axiom.

The question is, that expression is a "proposition"? AFAIK a "proposition" in propositional logic is a "thing" that is true or false. It cannot be undecided.

On the other side, we can add this expression as an axiom and we will have euclidean geometry.

Now I guess there is no issue to consider that these expression is a proposition.

Probably I need some precise definitions of what a proposition is, what a predicate is, what an axiom is, how a language is defined, ...

Lagrange multipliers to find the max/min of the $x_1x_2\cdots x_n$ Where $1/x_1+\dots+1/x_n=1$ and all $x_i>0$ [closed]

Posted: 24 Apr 2022 06:13 AM PDT

Question: $f(x_1,x_2,\dots,x_n)=x_1x_2\cdots x_n$ subject to the constraint $$\frac1{x_1}+\frac1{x_2}+\cdots+\frac1{x_n}=1.$$ Use the Lagrange multipliers to find the max or min.

Original Question Photo

Hi all. I worked until the sub-question of "explain why the maximum and minimum exist or not" I guess is to use EVT to show it has global min/max. Also I think the function only has minimum, as it seems to be biggest when x=1, and start to shrink afterwards. Is it true?

My Answer

How to calculate $\prod _{n=1}^{\infty}\frac{2^n-1}{2^n}$?

Posted: 24 Apr 2022 05:28 AM PDT

Suppose we have a kind of lottery as follow:

$1.$ You have a $\frac{1}{2}$ possibility of getting a prize on the first try.
$2.$ You have a $\frac{1}{4}$ possibility of getting a prize on the second try.
$\quad\vdots$
$n.$ The probability is $\frac{1}{2^{n}}$ on the $n$th try.
$\quad\vdots$

What is the probability of getting at least one prize?

I know that it is

$$p = 1- \prod _{n=1}^{\infty}\frac{2^n-1}{2^n},$$

But how to calculate it? I don't know. I want to find the answer.

Thanks.

Deformations of finite schemes

Posted: 24 Apr 2022 05:42 AM PDT

I am reading some texts about the tangent space to the Hilbert scheme. Apparently, $T_{[Z]}(Hilb (X)) = H^0(Z, N_{Z/X})$ for any $Z$ is a consensus which I agree with, but then regarding the hilbert scheme of points, people generally use that $T_{[Z]}(Hilb_r (X)) = \text{Hom}_{O_X}(I_Z, i_*O_Z)$, and there are a lot of proofs of this last equality but I don't really see where they use that $Z$ is finite.

I know that $ H^0(Z, N_{Z/X}) = \text{Hom}_{O_Z}(I_Z/I_Z^2, O_Z)$, so maybe I am missing something and the equality $\text{Hom}_{O_Z}(I_Z/I_Z^2, O_Z) = \text{Hom}_{O_X}(I_Z, i_*O_Z)$ is obvious, or is obvious when $Z$ is a finite scheme.

The usual proofs of $T_{[Z]}(Hilb_r (X)) = \text{Hom}_{O_X}(I_Z, i_*O_Z)$ basically try to characterize all quotients $Q$ of $O_X [ \varepsilon ]$ that are flat over $k[ \varepsilon ]$ and such that, $Q/\varepsilon Q = O_Z$. Chasing a bit in some diagram they show that this is equivalent as giving a $O_X$-morphism from $I_Z$ to $i_*O_Z$. For some links, see for example page 22 here. The proof convinces me but it doesn't say where thye use that $Z$ is a finite scheme.

Can someone shed some light on this topic? Thank you

Notes: Here $X$ is a projective variety, and $F[\varepsilon] = F\otimes k(t)/(t^2)$. When I say a finite scheme a mean a scheme of finite type over a field whose topological space is finite. but it can be non-reduced

Possibility that all lights $\mathbf{X}=(X_1,X_2,\cdots)$ turn off again with every time turn a light with its number $n\sim\text{geom}(\frac{1}{2})$.

Posted: 24 Apr 2022 05:47 AM PDT

Problem: Let $\mathbf{X} = (\mathbb{Z}_2)^\mathbb N$, i.e., $\mathbf{X} = (X_1,X_2,\cdots,X_N,\cdots)$, $X_i\in \{0,1\} $. It can be considered as countable lightbulbs. $0$ means off, $1$ means on. We start with $\mathbf{X}_0 = 0$. Keep generating independent geometric random variables, whose distribution are $geom(1/2)$. Denote them as $K_1, K_2,\cdots$. Now let $\mathbf{X}_m$ (for $m \ge 1$) be as follows $$(\mathbf{X}_m-\mathbf{X}_{m-1})_k = \mathbf{1}(k = K_m), $$ i.e, in the $m$- th turn, we only change the status of the $K_m$-th light bulb. Then what is the probability of all lights being off again, i.e., $$\mathbb P(\exists m>1, \mathbf{X}_m =0)$$

My first intuition below was wrong I'm afraid.

New intuition: We may define a function $f(x_1,x_2,\cdots)$ as the probability that status $\mathbf{X}=(x_1,x_2,\cdots)$ will eventually become all $0$, then our goal is to calculate $\sum_{n=1}^{\infty}\frac{1}{2^n}f(0,\cdots,0,1,0,\cdots)$ where $1$ only appears on $n$-th term, and we can achieve all the transporting equations of $f$, which is an uncountable dimensional equation system. We can see immediately that the equation system has a solution $f(x_1,x_2,\cdots)\equiv 1$. Can we conclude that this solution is unique (Then the answer to this problem is $1$)?
Continue the intuition above, we define $$g(x_1,x_2,\cdots)=f(x_1,x_2,\cdots)-1,$$ moreover we define $g(0,0,\cdots,0,\cdots)=0$.
Obviously $f\le1$, hence $g$ can achieve a maximum at $(0,0,0,\cdots,0,\cdots)$. Note that after changing all the equations involving $f$ into $g$, the constant term will be cancelled and the coefficients at the righthand side is strictly larger than $0$ and the sum of all coefficients are always $1$. For example, the equation $$f(1,0,0,\cdots,0,\cdots)=\frac{1}{2}+\frac{1}{4}f(1,1,0,\cdots,0,\cdots)+\frac{1}{8}f(1,0,1,\cdots,0,\cdots)+\cdots$$ is changed into $$g(1,0,0,\cdots,0,\cdots)=\frac{1}{2}g(0,0,0,\cdots,0,\cdots)+\frac{1}{4}g(1,1,0,\cdots,0,\cdots)+\frac{1}{8}g(1,0,1,\cdots,0,\cdots)+\cdots$$ We want to use the maximum principle, but we lack an equation with $\text{LHS}=g(0,0,0,\cdots,0,\cdots)$, are there any ways to supple this equation?

Edit: Another possible intuition from here: ignore $g(0,0,0,\cdots,0,\cdots)$, we only focus on the remaining term. We wish to discard the term that has infinity $1$ since the probability of getting this term is $0$. If the maximum happens on one remaining element we're done, what we don't want is if the maximum happens on the limiting term, and we wish to prove that this won't happen because the limit is (somewhat?) decaying. (I don't know how to proceed here, but I'll post my intuitions these days anyway)

Solve the pde $z=pq$?

Posted: 24 Apr 2022 06:02 AM PDT

i tried tried it using charpit method $$f=z-pq$$ $$\frac{\partial f}{\partial x}=0,\frac{\partial f}{\partial y}=0,\frac{\partial f}{\partial p}=-q,\frac{\partial f}{\partial q}=-p,\frac{\partial f}{\partial z}=1$$ $$\frac{dx}{f_p}=\frac{dy}{f_q}=\frac{dz}{pf_p + qf_q}=\frac{dp}{−(f_x + pf_z)}= \frac{dq}{−(f_y + qf_z)}$$ $$\frac{dx}{q}=\frac{dy}{p}=\frac{dz}{2pq }=\frac{dp}{p}= \frac{dq}{q}$$ $\log p=\log qC$ ,$p=qC$ now i can put the this in $z=pq,z=p^2C$ $p$ in terms of $(x,y,z,C)$ $$p=\sqrt{\frac{z}{C}},q=\sqrt{Cz}$$

writing $dz = p(x, y, z,C)dx + q(x, y, z, C)dy$ $$dz=\sqrt{\frac{z}{C}}dx+\sqrt{Cz}dy$$ $$\frac{dz}{\sqrt{z}}=adx+\frac{dy}{a}$$ $${2\sqrt{z}}=ax+\frac{y}{a}+b$$ $$4z=(ax+\frac{y}{a}+b)^2$$ Is my solution right ?

Example of an infinite group where every element except identity has order 2

Posted: 24 Apr 2022 05:40 AM PDT

Find an infinite group, in which every element g not equal identity (e) has order 2

Does this question mean this:

the group that fail condition (2) which is no inverse and also that group must have the size 2

My answer:

Z*
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