问:
Prove the $\lim_{x \to -1} x^2 + 4x + 2 = 0$ is wrong by negating the $\epsilon-\delta$ definition of limit
I'm struggling to disprove the following limit is wrong by negating the definition

$$
\lim_{x\to-1} x^2 + 4x + 2 = 0
$$

The negation of the definition applied to my case should be

$$
\exists \epsilon > 0 \forall \delta > 0 : 0 < |x+1| < \delta \text{ and } |x^2 + 4x + 2| \geq \epsilon
$$

This is to me equivalent to find some $\epsilon > 0$ such that the following system has solution for each $\delta > 0$

$$
\left\{
\begin{array}{l}
|x^2 + 4x + 2| \geq \epsilon \\
0 < |x+1| < \delta
\end{array}
\right.
$$

I started by picking $\epsilon = 1$, and assumed that $\delta \geq 1$ in such a case since
$$
x^2 + 4x + 2 = (x + 1)^2 + 2(x + 1) - 1
$$

it follows that if $x = 0$ solves the system, I'm having trouble however when $0 < \delta < 1$. I cannot manage to construct a solution of that system... any suggestion?

Thank you

答: Take $\varepsilon=\frac34$. If $x\in\Bbb R$, then$$x^2+4x+2=(x+1)^2+2(x+1)-1.\tag1$$Therefore, for any $\delta>0$, if $-\delta<x+1<0$, and if $x+1\geqslant-\frac12$, you have $(x+1)^2\leqslant\frac14$ and $2(x+1)\leqslant0$. So, it follows from $(1)$ that $x^2+2x+2\leqslant-\frac34$, and therefore$$|x^2+4x+2|\geqslant\frac34=\varepsilon.$$


答:
There is a missing quantifier: The definition of limit actually has an interior $\forall x$, so the negation should really be formalized as

$$\exists\epsilon\gt0\forall\delta\gt0\exists x\in\mathbb{R}:0\lt|x+1|\lt\delta\text{ and }|x^2+4x+2|\ge\epsilon$$

Added later (in response to comments from Rob Arthan and the OP):  At this point, the issue becomes one of making a judicious choice of $\epsilon$ and then finding a "nice" relationship of $x$ to $\delta$ so that two inequalities are satisfied. The key idea is to keep $x$ away from either value for which $x^2+4x+2=0$ (i.e., $x=-2\pm\sqrt6$) even when $\delta$ is so large that the inequality $|x+1|\lt\delta$ allows it.  There are many ways to do this, but one I like is to let $x=-1-{\delta\over1+\delta^2}$. It's clear that $|x+1|=\delta/(1+\delta^2)\lt\delta$ (when $\delta\gt0$), so the first inequality is satisfied. As for the second, we have

$$\begin{align}
x^2+4x+2
&=(x+1)^2+4(x+1)-1\\
&={\delta^2\over(1+\delta^2)^2}-{4\delta\over1+\delta^2}-1\\
&={\delta^2-4\delta(1+\delta^2)-(1+\delta^2)^2\over(1+\delta^2)}\\
&=-{\delta^4+2\delta^3+\delta^2+2\delta+1\over\delta^4+2\delta^2+1}
\end{align}$$

at which point it's easy to see, for example, that $|x^2+4x+2|\ge{1\over1000}$. (In fact, $\epsilon=1$ works, since showing that $\delta^4+2\delta^3+\delta^2+2\delta+1\ge\delta^4+2\delta^2+1$ for all $\delta\gt0$ reduces to verifying that $2\delta^2-\delta+2\ge0$, which is straightforward to check.)

Remark: The choice of $x=-1-\delta/(1+\delta^2)$ comes mainly from experience at finagling with limit problems of this type. Sometimes the first thing you try doesn't work, in which case you wad that scratch paper up and try try again. It can be fun to make choices that just barely work, but it's hardly imperative to do so.

ContentLicense: CC BY-SA 4.0

Title: Prove the $\lim_{x \to -1} x^2 + 4x + 2 = 0$ is wrong by negating the $\epsilon-\delta$ definition of limit

Body:
I'm struggling to disprove the following limit is wrong by negating the definition

$$
\lim_{x\to-1} x^2 + 4x + 2 = 0
$$

The negation of the definition applied to my case should be

$$
\exists \epsilon > 0 \forall \delta > 0 : 0 < |x+1| < \delta \text{ and } |x^2 + 4x + 2| \geq \epsilon
$$

This is to me equivalent to find some $\epsilon > 0$ such that the following system has solution for each $\delta > 0$

$$
\left\{
\begin{array}{l}
|x^2 + 4x + 2| \geq \epsilon \\
0 < |x+1| < \delta
\end{array}
\right.
$$

I started by picking $\epsilon = 1$, and assumed that $\delta \geq 1$ in such a case since
$$
x^2 + 4x + 2 = (x + 1)^2 + 2(x + 1) - 1
$$

it follows that if $x = 0$ solves the system, I'm having trouble however when $0 < \delta < 1$. I cannot manage to construct a solution of that system... any suggestion?

Thank you


问:
Is it legal to define a function that gives different results for 1.0 and for 1?
In programming languages I can define such function, because in most programming languages 1.0 is not 1, because 1.0 has type "float", and 1 has type "integer". In math I don't see axioms that forbid sets $\mathbb{R}$ and $\mathbb{N}$ to be disjoint. So can I define function in such a way, that f(1) = 1, and f(1.0) = 2, motivating it by saying that 1.0 is taken from $\mathbb{R}$, 1 is taken from $\mathbb{N}$, and these sets are defined in the completely separate ways?

答:
Usually (emphasis on "usually") we regard $\Bbb N \subset \Bbb Z \subset \Bbb Q \subset \Bbb R \subset \Bbb C$ as subsets.

To do this, after you constructed $\Bbb Z$ from $\Bbb N$ etc., you retroactively redefine $\Bbb N, \Bbb Z, \Bbb Q,$ and $\Bbb R$ to be subsets of $\Bbb C$.

I'm only talking about what happens in fields of mathematics that aren't focused on foundations of mathematics.


问:
Isomorphism of affine algebraic variety
https://math.stackexchange.com/a/2886099/884739
Can Someone please explain why over the ground field $\mathbb{C}$ the above affine variety is isomorphic to $\{(z_1,z_2):z_1z_2\ne0\}$? What is the change of variable that you do?

if you write $x^2+y^2=(x+iy)(x-iy)$ then can you simply replace $(x+iy)$ by $z_1$ and $(x-iy)$ by $z_2$? because $x-iy$ is conjugate of $x+iy$, so they are not really independent variables.

答:
There is (as indicated in the comments) a ring isomorphism (an isomorphism of $\mathbb{C}$-algebras):

$$ \phi: A_1:=\mathbb{C}[u,v]/(uv) \cong \mathbb{C}[x,y]/(x^2+y^2):=A_2$$

defined by $\phi(u):=x+iy, \phi(v):=x-iy$. The inverse ring isomorphism is

$$\phi^{-1}(x)=1/2(u+v), \phi^{-1}(y)=1/2i(u-v).$$

Hence we get an ismorphism of affine schemes (over $\mathbb{C}$):

$$f:Spec(A_1) \cong Spec(A_2).$$

Question: "Can Someone please explain why over the ground field C the above affine variety is isomorphic to ${(z1,z2):z1z2≠0}$? What is the change of variable that you do?"

Answer: The variable change given above induce an isomorphism $f$ of affine schemes. You may check that the above definition gives the inverse.
There are structure maps $\pi_i: Spec(A_i) \rightarrow Spec(\mathbb{C})$
and the isomorphism $f$ commute with these maps: $\pi_2 \circ f =\pi_1$.

Title: Isomorphism of affine algebraic variety


问:
Example of non-commutative ring such that every ideal is two sided
Can you give an example of non-commutative ring $R$ such that every ideal is two sided?

答: The quaternions. Every ideal is either $(0)$ or the whole ring.


答: Let $R$ be a noncommutative division ring - for example, take $R = \mathbb{H}$. Since it's a division ring, the only (right/left) ideals are the trivial ideal. Thus every ideal is two sided.

Title: Example of non-commutative ring such that every ideal is two sided

Body: Can you give an example of non-commutative ring $R$ such that every ideal is two sided?


问:
Prove or disprove that there exists a family of entire functions F that is not normal in any neighborhood of any z∈C.
A family of holomorphic function is normal if and only if it is uniformly bounded on any compact subset of any region.Typically being locally bounded is equivalent to being normal.</p>

答: Take the family of entire functions $F$ that are not normal in any neighbourhood of any $z \in \Bbb C$. That is the family required.


问:
$x \in \partial E \iff \exists x_k \subset E$ and $\exists x_j \subset E^C$ such that lim $x_k$= lim $x_j$ = x
Let $E$ a set in $\mathbb R^n$, I want to show that $x \in \partial E \iff \exists$ a sequence $x_k \subset E$ and $\exists$ a sequence $x_j \subset E^C$ such that lim $x_k$ = lim $x_j$ = $x$.

I did the $\Longleftarrow$ implication, but I struggle with the $\Longrightarrow$ way, I think I need to build two explicit sequences of limit $x$ by myself but I don't know which one.

答:
Every ball $B(x,\frac1n)$ around $x$ intersects $E$ and $E^\complement$, so we can pic $x_n \in B(x,\frac1n) \cap E$ and $y_n \in B(x,\frac1n) \cap E^\complement$ for each $n \in \Bbb N$.

Then show that $x_n \to x$ and $y_n \to y$, and these are as required.

It's notationally not very handy to have two sequences $x$ with different index symbols, I find it confusing. The indices $k,j$ are both in $\Bbb N$ presumably, so $x_1$ would be ambiguous: is it from one or the other sequence?


问:
How to solve $x \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+x y=0$
How to solve $$x \frac{d^{2} y}{d x^{2}}+4 \frac{d y}{d x}+x y=0$$
First of all i could not find the two linearly independent solutions $y_1(x),y_2(x)$.

I tried polynomials, but in vain.

答:
$$\frac{d^2y}{dx^2} + \frac{4}{x} \frac{dy}{dx} + y = 0$$
has $0$ as a regular singular point.  ($\frac{4}{x}$ is not analytic, but $\frac{4}{x}\cdot (x)$ is.)

So we try a solution of the form $ y = x^r \cdot \sum_{0}^{\infty} a_n x^n$
Substituting in we get an indicial equation for $r$, $$r(r-1) + p_0 r + q_0 = 0$$ where $p_0 = 4$ and $q_0 = 0$ giving $r = 0$ or $r = 3$. Since these solutions differ by an integer, plugging in the larger root, $r = 3$ means that we have an ordinary power series for the first solution.  The second solution may be Frobenius as well or it may require multiplying the series by $\ln (x)$ and finding new coefficients.  In that case the formula is $$y_2(x) = Cy_1(x) \ln x + x^{r_2}\sum_{n = 0}^{\infty} b_n x^n$$
Then we get a recurrence relation for $a_n$ in the usual way.


问:
Probability Example Question - Understanding Independence

An insurance company pays hospital claims.  The number of claims that include emergency room or operating room charges is 85% of the total number of claims.  The number of claims that do not include emergency room charges is 25% of the total number of claims.  The occurrence of emergency room charges is independent of the occurrence of operating room charges or hospital claims.  Calculate the probability that a claim to the insurance company includes operating room charges.

From the question stem, I interpreted the given data as follows:

Let E = Emergency room charges, and O= operating room charges

$\Pr(E \cup O) = 0.85$

$\Pr(E') = 0.25$

$\Pr(E) = 0.75$

The part of the question stem that confused me is this statement:

The occurrence of emergency room charges is independent of the occurrence of operating room charges or hospital claims.

Why is it necessary to say or hospital claims?  At first, I understood that there were two sets, namely Emergency room charges and Operating room charges, but when they state or hospital claims, it makes it seem as if there is a third set when in fact it is just the main space encompassing everything.

So you really should get:

$\Pr(E \cap O) = \Pr(E) * \Pr(O)\tag{1}$

When they say or hospital claims, it makes me want to write:  $\Pr(E \cap O \cap H) = \Pr(E) * \Pr(O) * \Pr(H)$.

Is there something about independence I don't understand?  Is it really necessary to state this in the problem?

Once you got (1), I understand you simply plug it in the inclusion exclusion equation and solve for Pr(O):

$\Pr(E \cup O) = \Pr(E) + \Pr(O) - \Pr(E \cap O)\tag{2}$

substitute (1) into (2).

$\Pr(O) = \cfrac{\Pr(E \cup O) - \Pr(E)}{\Pr(E')}= .4$

I appreciate any help.  Thank you!

答:
The "or hospital" is unfortunate wording, and should be disregarded. </p>

<p>For calculations, there are many ways. I would probably note that $\Pr(O' \cap E')=0.15$. Then from $\Pr(O')\Pr(E')=0.15$ we find that $\Pr(O')=0.6$, and hence $\Pr(O)=0.4$. Your calculation may be a little more efficient. </p>

ContentLicense: CC BY-SA 3.0

Body:
The "or hospital" is unfortunate wording, and should be disregarded.

For calculations, there are many ways. I would probably note that $\Pr(O' \cap E')=0.15$. Then from $\Pr(O')\Pr(E')=0.15$ we find that $\Pr(O')=0.6$, and hence $\Pr(O)=0.4$. Your calculation may be a little more efficient.


问:
Relationship between discriminants and smoothness of curves
My understanding of the use of the discriminant in elliptic curve theory is to test whether an elliptic curve in Weierstrass normal form over a field not of characteristic either 2 or 3, $y^{2} = x^{3}+Ax+B$, is smooth, i.e., non-singular. What I am confused about is the relationship between the discriminant and smooth curves. For instance, $y=x^{2}$ over $\mathbb{R}$ has a zero discriminant, yet is a smooth curve.

Why does a zero discriminant of an elliptic curve tell us that the curve is not smooth, and why does this  not seem to apply likewise to other types of curves, such as, quadratics?

Is it perhaps because smoothness implies non-singularity for elliptic curves, but in general this is not the case?

答:
Question: "Why does a zero discriminant of an elliptic curve tell us that the curve is not smooth, and why does this not seem to apply likewise to other types of curves, such as, quadratics? Is it perhaps because smoothness implies non-singularity for elliptic curves, but in general this is not the case?"

Answer: If $f(x,y):=y-x^2$ it follows the jacobian ideal $J(f)$ equals

$$J(f)=(\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y})=(-2x, 1)=(1)$$

Hence the singular scheme $C_{sing}$ of the curve $C$ (which is the zero scheme of the jacobian ideal) is empty.

If $k$ is any field and $F(x,y):=y^n-f(x)$ with $n\geq 2$ an integer and $deg(f)=d$, you may consider the algebraic curve $C:=Spec(A)$ where $A:=k[x,y]/(F(x,y))$. The jacobian ideal $J$ given by

$$J:=(ny^{n-1},-f'(x))$$

defines the singular subscheme $C_{sing}:=V(J) \subset C$. Assume for simplicity $char(k) \neq n$. Let $(a,b)\in C_{sing}(K)$ be a $K$ rational point of $C_{sing}$, where $k\subseteq K$ is a field. It follows $a,b\in K$ and $f(a)=f'(a)=0$.
Hence the curve $C$ has a $K$-rational singular point $(a,0)\in K^2$ iff

$$f(a)=f'(a)=0.$$

If $f(x):=a_0+a_1x+\cdots +a_dx^d$ we define the discriminant $\Delta(f)$ to be the resultant $\Delta(f):=Res(f(x),f'(x))$ of the polynomial $f(x)$ and it's derivative $f'(x)$. It follows $\Delta(f)=0$ iff $f(x)$ has a multiple root in $K$, where $K$ is a finite extension of $k$. Hence the discriminant $\Delta(f)=0$ iff the curve $C$ has a $K$-rational singular point, where $k\subseteq K$ is a finite extension.

Note: The discriminant $\Delta(f):=Res(f(x),f'(x))$ is a polynomial in the coefficients of $f(x)$, hence you can calculate  $\Delta(f)$ without knowing anything on the roots of $f(x)$ or if the roots of $f(x)$ live in a finite extension $k \subseteq K$. There is a well known determinantal formula for the resultant $Res(f(x),g(x))$ of any pair of polynomials $f(x),g(x)$ - the "Sylvester determinant".

As remarked in the comments: If $k$ is not algebraically closed you may get singular points defined over non-trivial extensions $k \subseteq K$. The curve $C$ is finite type over $k$ and closed points $x\in C$ have residue field $k \subseteq \kappa(x)$ which is a finite extension in general. Hence points of $C_{sing}$ are not in general defined over the base field $k$.

Example: If $k=\mathbb{Q}$ is the field of rational numbers and $f(t):=g(t)^2$ where $g(t):=t^2-2$. Let $F(x,y):=y^n-f(t)$. It follows

$$f'(t)=2g(t)g'(t)=4t(t^2-2) \in k[t].$$

Hence $f(t)=0$ and $f'(t)=0$ does not have a solution in $k$. It has a solution in $k(\sqrt{2})$. This phenomenon also happens in characteristic $p>0$.

Note: If the polynomial $f(x)\in k[x^p]$ is a polynomial in $x^p$ it follows $f'(x)=0$ hence $\Delta(f)=Res(f(x), f'(x))$ is identically zero in characteristic $p>0$. Over the algebraic closure $\overline{k}$ it follows $f(x^p)=g(x)^p$, hence $f(x)$ has multiple roots and the curve is singular.


问:
Integrating the Pochhammer symbol. $\int_{-1}^{1} \prod_{j=0}^{\infty} (1-x^{j+1})$
I had a question regarding the Pochhammer symbol, specifically integrating $[x;x]_{\infty} = \prod_{j=0}^{\infty} (1-x^{j+1})$. Firstly, is there a closed form for the $n^{th}$ partial product of this, or $[x;x]_{n+1}$? I tried computing the first 3 partial products and couldn't find a pattern.

Secondly, is there a closed form for the area under this from $0$ to $1$?  I was thinking about trying to find a closed form for $[x,x]_n$ as a series, then you can integrate the series and get a series representation of the integral, but if I don't know a closed form for $[x,x]_n$, this isn't useful. Can someone help?

This is just for personal interest, as I just learned about the Pochhammer symbol.

答:
I do not know a closed form for
$$I_n=\int_{-1}^{+1} (x;x)_{n+1}\,dx$$

For even values of $n$, this generates the sequence
$$\left\{2,\frac{152}{105},\frac{4672}{3465},\frac{333280768}{253161909},\frac{141841
   28217088}{10883356008525},\frac{257964528813440303104}{198899418641149766925}\right\}$$ For odd values of $n$
$$\left\{\frac{4}{3},\frac{128}{99},\frac{328448}{255255},\frac{34963456}{27188525},
   \frac{530068713377923072}{412106981947605945}\right\}$$ which both decrese when $n$ increase. As you wrote, I do not see any clear pattern.
I did not find these terms in $OEIS$.

Concerning

$$I_\infty=\int_{-1}^{+1} (x;x)_\infty\,dx$$ it is just a number $\color{red}{\text{(this is WRONG})}$
$$I_\infty=1.28830088867392123018090\cdots$$ which is not recognized by inverse symbolic calculators.

Edit (after @Brian Constantinescu's comment)

In Quora, using residues, @Brian Constantinescu provided a very elegant closed form of the result.

Simplifying his last result
$$I_\infty=\int_{-1}^{+1} (x;x)_\infty\,dx=4\pi \sqrt{\frac{6}{23}}\,\, \frac{\sinh \left(\frac{\sqrt{23}
   \pi }{4}\right)+\sqrt{2} \sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi }{3}\right)-1}$$

Using the same steps and notations as @Brian Constantinescu, computing the sums in terms of polylogarithms and simplifying, we have
$$A=\sum_{-\infty}^{\infty}\frac{(-1)^n}{3 n^2-n+2}$$ $$A=-\frac{i \pi  \left(\tan \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi \right)+\cot
   \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi \right)-2 \csc \left(\frac{1}{6}
   \left(1+i \sqrt{23}\right) \pi \right)\right)}{2 \sqrt{23}}$$ Expanding the complex numbers
$$\color{blue}{A=4 \pi \sqrt{\frac{3}{23}}\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi}{3}\right)-1}} $$
$$B_x=\sum_{-\infty}^{\infty}\frac{(-1)^x}{12 x^2-2 x+2}$$
$$B_x=\frac{i \pi  \left(\csc \left(\frac{1}{12} \left(1+i \sqrt{23}\right) \pi
   \right)-\text{sech}\left(\frac{1}{12} \left(\sqrt{23}-5 i\right) \pi \right)\right)}{2\sqrt{23}}$$
$$\color{blue}{B_x=-\pi\sqrt{\frac{2}{23}} \left(\sqrt{3}+1\right)\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{3}-2 \cosh
   \left(\frac{\sqrt{23} \pi }{6}\right)}}$$
$$B_y=\sum_{-\infty}^{\infty}\frac{(-1)^y}{12 y^2+10 y+4}$$
$$B_y=\frac{i \pi  \left(\csc \left(\frac{1}{12} \left(5+i \sqrt{23}\right) \pi
   \right)-\text{sech}\left(\frac{1}{12} \left(\sqrt{23}-i\right) \pi \right)\right)}{2 \sqrt{23}}$$
$$\color{blue}{B_y=\pi\sqrt{\frac{2}{23}} \left(\sqrt{3}-1\right)\,\,\frac{\sinh \left(\frac{\sqrt{23} \pi }{12}\right)}{\sqrt{3}+2 \cosh
   \left(\frac{\sqrt{23} \pi }{6}\right)}}$$
$$\color{red}{I_\infty=2(A+B_x+B_y)=4\pi \sqrt{\frac{6}{23}}\,\, \frac{\sinh \left(\frac{\sqrt{23}
   \pi }{4}\right)+\sqrt{2} \sinh \left(\frac{\sqrt{23} \pi }{6}\right)}{2 \cosh \left(\frac{\sqrt{23} \pi }{3}\right)-1}}$$


问:
Probability that a Particle which moves Unit distance in a Random direction on each step will be inside the Unit Sphere after $n$ steps
The following integral equation arises while calculating the probability that, a particle which starts at the origin and moves a unit distance in a random direction on each ‘move’, will be within the unit sphere after $n$ moves:
$$
f_n(x) = \begin{cases}
\int_{1-x}^{1+x} \frac{df_{n-1}(t)}{dt}  \left( \frac{x^2-(t-1)^2}{4t} \right) dt, & 0\le x\lt 1
 \\ f_{n-1} (x-1) + \int_{x-1}^{x+1} \frac{df_{n-1}(t)}{dt}  \left( \frac{x^2 -(t-1)^2}{4t} \right) dt, & 1 \le x \le n-2
\\ f_{n-1} (x-1) + \int_{x-1}^{n-1} \frac{df_{n-1}(t)}{dt}  \left( \frac{x^2 -(t-1)^2}{4t} \right) dt, & n-2\lt x\lt n
\\ 1, & x\ge n
\end{cases}
$$
Here, $n \ge 3$. At first glance, this looks quite unsolvable, as it is a mixture of a recurrence relation and an integral equation, that too with differing arguments in $x$. But just to make sure, is there a way to solve for $f_n(x)$? I’m ultimately looking for $f_n(1)$ so it’s also fine if that can be obtained without actually solving the equation.

Note: $f_n(x)$ is defined to be the probability that the particle is inside the sphere of radius $x$ centered at the origin after $n$ moves. As the ‘base case’, $$f_2(x) =\begin{cases} \frac{x^2}{4}, & 0\le x\le 2 \\ 1, & x\gt 2 \end{cases}$$

Here are the graphs of $\color{blue}{f_2(x)}, \color{green}{f_3(x)}, \color{red}{f_4(x)} $,

and some initial values:

$$f_0(1) = 1 \\ f_1(1) = 0 \\ f_2(1) = \frac 14 \\ f_3(1)= \frac 16 \\ f_4(1) = \frac{23}{192} \\ f_5(1) =\frac{11}{120} $$

答:
NOT AN ANSWER.

I know that Borwein studied that problem in various papers. The following one seems to be relevant to you:

Jonathan M. Borwein, Armin Straub, Christophe Vignat. Densities of short uniform random walks in higher dimensions. Journal of Mathematical Analysis and Applications, Elsevier, 2016, 437 (1), pp.668-707. 10.1016/j.jmaa.2016.01.01

I hope this link works; https://hal-centralesupelec.archives-ouvertes.fr/hal-01261938/document


问:
Thomae's function
I am self-learning Real Analysis from Stephen Abbott's Understanding Analysis. I'd like to ask, if my below proof is technically correct and rigorous.

Problem 4.2.3. Consider the Thomae's function $t(x)$ defined as follows:

\begin{align*}
t(x) = \begin{cases}
1 & \text{ if } x = 0\\
\frac{1}{n} & \text { if } x = \frac{m}{n} \in \mathbf{Q}\setminus \{0\}\\
0 & \text { if } x \notin \mathbf{Q}
\end{cases}
\end{align*}

(a) Construct three different sequences $(x_n)$, $(y_n)$ and $(z_n)$ each of which converges to $1$ without using the number $1$ as a term in the sequence.

(b) Now, compute $\lim t(x_n)$, $\lim t(y_n)$ and $\lim t(z_n)$.

(c) Make an educated conjecture for $\lim_{x \to 1} t(x)$ and use the definition of functional limits to verify the claim. Given $\epsilon > 0$, consider the set of points $\{x \in \mathbf{R}, t(x)\ge \epsilon\}$. Argue that all the points in this set are isolated.

Proof.

Consider $x_n:= \frac{n}{n+1}$. The first few terms of the sequence are,

\begin{align*}
(x_n) = \frac{1}{2}, \frac{2}{3}, \frac{3}{4},\frac{4}{5},\ldots
\end{align*}

Consider $y_n:= 1$.

The first few terms of the sequence are,

\begin{align*}
(y_n) = 1,1,1,1,1, \ldots
\end{align*}

Consider $(z_n):= \frac{n+\sqrt{2}}{n}$.

The first few terms of the sequence are,

\begin{align*}
(z_n) = \frac{1+\sqrt{2}}{1}, \frac{2+\sqrt{2}}{2}, \frac{3+\sqrt{2}}{3}, \frac{4+\sqrt{2}}{4}, \ldots
\end{align*}

(b) The corresponding image sequences are:

\begin{align*}
t(x_n) &= \frac{1}{2}, \frac{1}{3},\frac{1}{4},\frac{1}{5},\ldots\\
t(y_n) &= 1,1,1,1,\ldots\\
t(z_n) &= 0,0,0,0,\ldots
\end{align*}

Thus, $\lim_{x_n \to 1}t(x_n) = \lim_{z_n \to 1}t(z_n) \ne 1 = t(1)$

(c) We conjecture that $\lim_{x \to 1} t(x)$ does not exist. Clearly, from the sequential characterization of functional limits, for all sequences $(x_n)$ in the domain of $t$, whenever $(x_n) \to 1$ and $x_n \ne 1$, we have $t(x_n) \to L$, then $\lim_{x_n \to 1}t(x_n) = L$.

Clearly, this condition is not satisfied. $\lim_{x \to 1}t(x)$ is not well-defined.

答:
It depends on the definition of limit you take. If you consider \begin{equation} \lim_{x \to c}f(x) = L \iff \forall \varepsilon > 0 \ \exists \delta > 0 : 0 < \vert x\vert < \delta \Rightarrow \vert f(x) \vert < L \end{equation}
then the limit exists and its value is $0$ as you can conjecture from the sequences $(x_{n})$ and $(z_{n})$ you have defined. Indeed, exercise $(a)$ tells you that you cannot consider $1$ as a term in the sequences so $(y_{n})$ is not properly defined according to the exercise. Define another sequence $(y_{n})$ without using $1$ as a term and you will get $f(y_{n}) \to 0$

Title: Thomae's function


问:
Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

Find all triples of natural numbers $x$, $y$, $z$ such that $x+2y+2z=xyz$.

I found a solution, but it involves a lot of case work. Can someone help me find a solution which doesn't involve a lot of case work?

Body:
"OP" equation is:

$x+2y+2z=xyz$  ---(1)

Eqn. (1) has parameteric solution's shown below.

$(x,y,z)=(3k^2+1)/(k),(3k),(1/k)$

For, $k=1$, we get: $(x,y,z)=(4,3,1)$

& for, $k=1/3$, we get: $(x,y,z)=(4,1,3)$

Another parametrization is:

$(x,y,z)=(4k^2+2)/(k),(2k),(1/k)$

For, $k=1$, we get: $(x,y,z)=(6,2,1)$

& for, $k=1/2$, we get: $(x,y,z)=(6,1,2)$


问:
Let $X$ and $Y$ normed spaces and $T:X\rightarrow Y$ one bounded surjective linear transformation. Show that they are equivalent:
$(I)$ $T$ is reversible and $T^{-1}:Y\rightarrow X$ is bounded.

$(II)$ Exists $b>0$ such that $||T(x)||\geq b||x||,\forall x\in X.$

$(III)$ All linear inverse on right of $T$ is bounded, and $$\sup \{||S||: S\text{ is linear inverse on right of }T\}.$$

I proved that $(I)\Leftrightarrow(II)\Rightarrow (III)$. Now, I triyng prove that $(III)\Rightarrow (II)$ or $(III)\Rightarrow (I)$.

答: Suppose that $T$ is not injective, and let $U$ be a bounded right inverse, and $e\in Ker(T), \|e\|=1$. Let $\alpha$ be a bounded non zero $1$-form defined on $Y$, defined $U_n(x)=U(x)+n\alpha(x)e$, $U_n$ is a right inverse and the sequence $U_n$ is not bounded since $\|U_n(x)|\geq |n\alpha(x)\|-\|U(x)\|$. If $\|x\|=1$ and $\alpha(x)\neq 0$, $lim_n\|U_n(x)\|=+\infty$. Contradiction.


问:
Limit of non-archimedean absolute values is eventually constant
Let $(x_n)$ be a Cauchy sequence in a field $K$ with respect to a non-archimedean absolute value $|\cdot|$ arising from a discrete valuation $v$ (meaning $|x|:=c^{v(x)}$ for some $c\in(0,1)$. Suppose $\operatorname{lim}|x_n|\neq 0$. I'd like to understand why the sequence $(v(x_n))$ becomes evetually constant.

I found this statement at the top of page three in these notes by Andrew Sutherland.

The only tool at hand seems to be the non-archimedean triangle inequality, but I'm not sure how to use it. What am I missing?

答:
Your questions on $p$-adic numbers are strange, as if to you it was something completely abstract, but $p$-adic numbers are very concrete. If $x\ne 0$ and $x_n\to x$ then $v(x_n-x)>v(x)$ for $n$ large enough so that $v(x_n)=v(x),|x_n|=|x|$. When $K$ is not complete and $(x_n)$ is Cauchy then $x=\lim x_n$ is in its completion instead of $K$ but it doesn't change anything.

$v(x_n-x)>v(x)\implies v(x)=v(x_n)$ is part of the definition:

If $v(x_n) > v(x)$ then $v(x)=v(x_n+(x-x_n))\ge \min(v(x_n),v(x-x_n))$ a contradiction.


问:
Subobjects and quotient objects in category of topological spaces
Are the following definitions are equivalent?

An embedding of topological spaces is just a monic in Top, i.e. it is a continuous $e: A \rightarrow X$ s.t. for any space $Z$ the function $f \mapsto e \circ f:$ Top $\left(Z,A\right)\rightarrow$ Top $\left(Z,X\right)$ is injective i.e. $e_\circ f_1=e\circ f_2 \implies f_1 =f_2$.

and

$e:A \rightarrow X$ is an embedding iff

$U\left(e\right):U\left(A\right)\rightarrow U\left(X\right)$ is injective in Set, where $U$ is the underling set functor.

for any $Z \in$ Top a set map  $U\left(f\right):U\left(Z\right) \rightarrow U\left(A\right)$, $f$ is continues iff $e \circ f : Z \rightarrow X$ is continuous.

Is the analogue definitions for epic and quotient space are also equivalent? Many thanks.

答: The first definition actually characterises injective continuous maps, but the second definition does indeed characterise topological embeddings. More generally there is a notion of initial topology and final topology for a family of maps, and an embedding is precisely an injective map that induces the initial topology on its domain, and a quotient is precisely a surjective map that induces the final topology on its codomain. The universal property you have written down is a translation of that into the language of category theory.


问:
Prime Ideals: $ab\:\in \:I$ such that $a\:\in \:I\:\:\vee \:\:b\in I$ vs. either $a\:\in \:I\:$ or $b\:\in \:I\:$ but not both?
Everywhere I read it's "a is element of I or b is element of I", but I couldn't find an explanation if it's a mathematical or (so that it can be the case that a is element of I and b is element of I), or if it's a language or (so that it's impossible that both belong to I)</p>

答:
It's the former - inclusive not exclusive or. In math, we always use the inclusive or unless explicitly stated otherwise.

As a sanity check, your other definition doesn't really even make sense. Take an ideal $I \subseteq R$ that is "prime" per your second definition, and take $a, b \in I$. Then by definition of an ideal, $ab \in I$. But by this exclusive or definition, precisely one of $a, b$ can be in $I$. In other words, no ideals are "prime" per your second definition.


问:
Bijection between $S^1$ and $S^n$
I have read in this post Clarify: "$S^0$, $S^1$ and $S^3$ are the only spheres which are also groups" that there is a bijection between  $S^1$ and the sphere $S^n$, $n \geq 1$ and this was confirmed in the answer!

What is an explicit bijection $f : S^1 \rightarrow S^n$ ?

Thanks!

答:
There is no point seeking an explicit bijection. It will not have any special property. It will not be close to continuous.

$S^1$ and $S^n$ are in bijection because they have the same cardinality, and here is why:

If you remove a point from each of them, then you end up with $\Bbb R$ and $\Bbb R^n$ because of the stereographic projection.

Now, $\Bbb R$ clearly injects into $\Bbb R^n$, so by the Schröder–Bernstein theorem, we only need to inject $\Bbb R^n$ back into $\Bbb R$: given $(x_1, \cdots, x_n) \in \Bbb R^n$, let $0.a_{i1}a_{i2} \cdots$ be the binary expansion of $\frac1{1+\exp(x_i)}$
(this is the sigmoid function, an injection $\Bbb R \to (0, 1)$). Then say that the result is $0.a_{11} a_{21} \cdots a_{n1} a_{12} a_{22} \cdots a_{n2} \cdots$ as ternary. This different base is taken to avoid things like $0.012022222\ldots_3 = 0.121_3$. Then this gives an injection $\Bbb R^n \to \Bbb R$, so we are done.

Title: Bijection between $S^1$ and $S^n$


问:
Choice set between a single element and nothing
I'm trying to represent a choice set between an element, $a$ and an alternative, nothing, $\emptyset$. Can this set by represented by

S={a,$\emptyset$} or does the empty set have to be a subset of my set S? S={a,{$\emptyset$}}?

答:
It's actually the other way around for your brackets.

$S = \{a\}$, and $\mathcal{P}(S) =\{ \emptyset, \{a\}\}$. You're looking for this powerset.


问:
Looking for the formulas of the Cox-F test to compare to survival functions
In survival analysis, several tests are available to compare two independent survival curves, e.g. Logrank test with different weights, Gehan's test, Cox-F test, Peto's test, etc.

I found in my daily work as a statistician that the Cox-F test has more power than the log-rank tests.

I think the Cox-F test is not the test within the Cox-regression model. It is a separate test.

Now, I want to make Monte Carlo simulations to compare the power of the Cox-F test with the family of log-rank tests.

In order to do that, I need the underlying formulas for the Cox-F test. I ask you to give me a hint where I can find such a formula, e.g. a pdf file, weblink or other reference.

I think this is important with a far reaching impact, e.g. in medical statistics most of the time logrank test is used, but I strongly suspect that the Cox-F test has much higher power.

Thank you.

答: Guess the paper "The Relationship Between Two Statistical Tests for the Comparison of Two Exponential Distributions" by Le (Biometrical Journal, 33)  contains the answer. The very last formula is the one proposed by Cox in his 1953 paper "Some simple approximate tests for Poisson variates" (Biometrika, 40, 354-360).</p>


答:
When $K$ is a $p$-adic field (rather than just a nonarchimedean field), what you are asking is true, and you have much more tools on hand than just the nonarchimedean inequality.  Let $f: K \rightarrow [0,\infty)$ be the function $f(x) = |x|$.

Let $\omega \in K$ be a uniformizer (if $K = \mathbb Q_p$ you can take $\omega = p$), so that $f(\omega) = \rho$ for some $0 < \rho < 1$.  Since every $0 \neq x \in K$ is equal to $u \omega^n$ for a unique integer $n$ and unique $u \in K$ with $|u| = 1$, it is clear that

$$\operatorname{Image}(f) = \{0\} \cup \{..., \rho^2, \rho, 1, \rho^{-1}, \rho^{-2}, ... \}$$

Now if $\{x_n\} \in K$ is a Cauchy sequence, then $\{f(x_n)\}$ is a convergent sequence in $\operatorname{Image}(f)$.  If we assume that $\{x_n\}$ does not converge to $0$, then $\{f(x_n)\}$ is a convergent sequence in

$$\{..., \rho^2, \rho, 1, \rho^{-1}, \rho^{-2}, ... \}.$$

This last set being discrete (every singleton set is open), the only convergent sequences therein are those which are eventually constant.

Title: Limit of non-archimedean absolute values is eventually constant

Body:
Let $(x_n)$ be a Cauchy sequence in a field $K$ with respect to a non-archimedean absolute value $|\cdot|$ arising from a discrete valuation $v$ (meaning $|x|:=c^{v(x)}$ for some $c\in(0,1)$. Suppose $\operatorname{lim}|x_n|\neq 0$. I'd like to understand why the sequence $(v(x_n))$ becomes evetually constant.

I found this statement at the top of page three in these notes by Andrew Sutherland.

The only tool at hand seems to be the non-archimedean triangle inequality, but I'm not sure how to use it. What am I missing?


问:
Proof regarding convergence
I am supposed to show that the sequence converges and to find the limit.
We've just started going over sequences and induction so I'm very limited in what I know/can use and I was wondering if the general pattern of the proof is okay and if it's correct.. Also it took me quiet a bit of time to prove it this way and I was wondering if there are any shortcuts I can do in the future?

$$ \{a_n\}=\left\{
\begin{array}{lcl}
a_1 = 3\\
a_{n+1} = 1+\sqrt{6+a_n}\\
\end{array}
\right.$$

First we show by induction $ \{a_n\} $ is increasing, that is, to prove that the following hold for every $n∈N$

$1+\sqrt{6+a_n}\ge a_n$

$\sqrt{6+a_n}\ge a_n-1$

$6+a_{n}\ge\left(a_{n}+1\right)^{2}$

$6+a_{n}\ge a_{n}^{2}-2a_{n}+1$

$0\ge a_{n}^{2}-3a_{n}-5$

Base case: $n=1$

$1+\sqrt{6+3}\ge 3$

$4\ge 3$

Inductive step: Assume $0\ge a_{n}^{2}-3a_{n}-5$, we show $0\ge a_{n+1}^{2}-3a_{n+1}-5$

$0\ge a_{n}^{2}-3a_{n}-5$

$6+a_{n}\ge a_{n}^{2}-2a_{n}+1$

$\sqrt{6+a_{n}}\ge a_{n}-1$

$0\ge a_{n}-1-\sqrt{6+a_{n}}$

$0\ge1+2\sqrt{6+a_{n}}+6+a_{n}-3-3\sqrt{6+a_{n}}-5$

$0\ge\left(1+\sqrt{6+a_{n}}\right)^{2}-3\left(1+\sqrt{6+a_{n}}\right)-5$

$0\ge a_{n+1}^{2}-3a_{n+1}-5$

Therefore by induction, $\{a_n\}$ is increasing for every $n∈N$.

Now we show by induction $\{a_n\}$ is bounded, that is, $a_n ≤ 4.2$ for every $n∈N$.

Base case: $n=1$, which is trivially true.

Inductive step: Assume $a_n ≤ 4.2$, we show $a_{n+1} ≤ 4.2$.

$a_n ≤ 4.2$

$6+a_{n}\le10.2$

$\sqrt{6+a_{n}}\le\sqrt{10.2}$

$\sqrt{6+a_{n}}\le3.2$

$1+\sqrt{6+a_{n}}\le4.2$

Therefore by induction $\{a_n\}$ is bounded from above.

As $\{a_n\}$ is both bounded and incresing for every $n∈N$, $\{a_n\}$ converges and $\lim _{n\to \infty }\left(a_n\right)=L$ for some $L∈R$

As $\lim _{n\to \infty }\left(a_n\right)=L$, then $\lim _{n\to \infty }\left(a_{n+1}\right)=L$, therefore

$L\ =\ 1+\sqrt{6+L}$

$L\ -1=\ \sqrt{6+L}$

$\left(L\ -1\right)^{2}=\ 6+L$

$L^{2}-2L+1=\ 6+L$

$L^{2}-3L-5=\ 0$

$L = \dfrac{ 3 \pm \sqrt{(-3)^2 - 4*(-5)}}{ 2 }$

$L_1 = 4.19258$

$L_2 = −1.19258$

As $a_n ≥ 0$ for every $n∈N$, $L = 4.19258$

答: Your proof is honestly very tedious and should be replaced ! You must make use of the recursive property of the sequence directly. Here is how induction works for your problem. Base case: $a_2 = 4 \ge 3 = a_1$ is true. Assume $a_n  \ge a_{n-1}$ for $n \ge 2$, then using the step $a_n \ge a_{n-1} \implies a_{n+1} = 1+\sqrt{6+a_n} \ge 1 +\sqrt{6+a_{n-1}} = a_n$. Hence by induction $a_n \ge a_{n-1}$ for all $n$ and the sequence is monotonically increasing. Next you can show too by induction again that the sequence is bounded above by $5$ and below by $4$. Here again, base case $a_1 = 3 < 5$, assume $a_n < 5$, then $a_{n+1} = 1 +\sqrt{6+a_n} < 1 +\sqrt{6+5} = 1+\sqrt{11} < 1+\sqrt{16} = 1+4 = 5$. Thus $a_n < 5$ for all $n$. Thus the $\{a_n\}$ has a limit $L$ which is the solution of the equation: $L = 1+\sqrt{6+L}$. It can be solve by squaring both sides and note that $4 \le L \le 5$. Can you solve for $L$ ?

Title: Proof regarding convergence


答:
For one-dimensional case see Theorem 5 on page 2: https://www.mimuw.edu.pl/~nayar/wdk_2020/notatki_wdk_2020.pdf
Maybe You will find a way to generalise it to more dimensions.

I suck at probability so I cannot help more.

Title: Probability that a Particle which moves Unit distance in a Random direction on each step will be inside the Unit Sphere after $n$ steps

Body:
The following integral equation arises while calculating the probability that, a particle which starts at the origin and moves a unit distance in a random direction on each ‘move’, will be within the unit sphere after $n$ moves:
$$
f_n(x) = \begin{cases}
\int_{1-x}^{1+x} \frac{df_{n-1}(t)}{dt}  \left( \frac{x^2-(t-1)^2}{4t} \right) dt, & 0\le x\lt 1
 \\ f_{n-1} (x-1) + \int_{x-1}^{x+1} \frac{df_{n-1}(t)}{dt}  \left( \frac{x^2 -(t-1)^2}{4t} \right) dt, & 1 \le x \le n-2
\\ f_{n-1} (x-1) + \int_{x-1}^{n-1} \frac{df_{n-1}(t)}{dt}  \left( \frac{x^2 -(t-1)^2}{4t} \right) dt, & n-2\lt x\lt n
\\ 1, & x\ge n
\end{cases}
$$
Here, $n \ge 3$. At first glance, this looks quite unsolvable, as it is a mixture of a recurrence relation and an integral equation, that too with differing arguments in $x$. But just to make sure, is there a way to solve for $f_n(x)$? I’m ultimately looking for $f_n(1)$ so it’s also fine if that can be obtained without actually solving the equation.

Note: $f_n(x)$ is defined to be the probability that the particle is inside the sphere of radius $x$ centered at the origin after $n$ moves. As the ‘base case’, $$f_2(x) =\begin{cases} \frac{x^2}{4}, & 0\le x\le 2 \\ 1, & x\gt 2 \end{cases}$$

Here are the graphs of $\color{blue}{f_2(x)}, \color{green}{f_3(x)}, \color{red}{f_4(x)} $,

and some initial values:

$$f_0(1) = 1 \\ f_1(1) = 0 \\ f_2(1) = \frac 14 \\ f_3(1)= \frac 16 \\ f_4(1) = \frac{23}{192} \\ f_5(1) =\frac{11}{120} $$


问:
PDF of volume of tetrahedron with random coordinates

Question

What is the probability distribution function (PDF) of the absolute volume of a tetrahedron with random coordinates?

The 4 random tetrahedron vertices in $\mathbb{R}^3$ are
$$ \mathbf{\mathrm{X}_1} =(x_1^1,x_1^2,x_1^3),\;\;   \mathbf{\mathrm{X}_2}=(x_2^1,x_2^2,x_2^3),\;\;    \mathbf{\mathrm{X}_3}=(x_3^1,x_3^2,x_3^3),\;\;    \mathbf{\mathrm{X}_4} =(x_4^1,x_4^2,x_4^3)$$
where $x_i^j$ are independent standard normal distributed variables
$$x_i^j\sim\mathcal{N}(0,1)$$
The non-oriented volume of a random tetrahedron instance is
$$V=\frac{1}{6}\left| ( \mathbf{\mathrm{X}_1}-  \mathbf{\mathrm{X}_4})\cdot \left(( \mathbf{\mathrm{X}_2}-  \mathbf{\mathrm{X}_4} ) \times (  \mathbf{\mathrm{X}_3}-  \mathbf{\mathrm{X}_4} )\right) \right|  \tag{1} $$ $$=\frac{1}{6}\left|x_1^2 x_2^3 x_3^1 - x_1^1 x_2^3 x_3^2
+ x_1^3 x_2^1 x_3^2- x_1^2 x_2^1 x_3^3
+ x_1^1 x_2^2 x_3^3- x_1^3 x_2^2 x_3^1
+ x_1^3 x_2^2 x_4^1- x_1^1 x_2^2 x_4^3
+ x_1^1 x_2^3 x_4^2- x_1^2 x_2^3 x_4^1
+ x_1^2 x_2^1 x_4^3- x_1^3 x_2^1 x_4^2
+ x_2^3 x_3^2 x_4^1 -x_2^1 x_3^2 x_4^3
+ x_2^1 x_3^3 x_4^2- x_2^2 x_3^3 x_4^1
+ x_2^2 x_3^1 x_4^3- x_2^3 x_3^1 x_4^2
+ x_1^2 x_3^3 x_4^1- x_1^1 x_3^3 x_4^2
+ x_1^3 x_3^1 x_4^2- x_1^2 x_3^1 x_4^3
+ x_1^1 x_3^2 x_4^3- x_1^3 x_3^2 x_4^1
\right|$$

Known relations

The expectation value of $V$  is
$$\mathbb{E}[V]=\frac{2}{3}\sqrt{\frac{2}{\pi}}\tag{2}$$
A proof can be found in a Math Stack Exchange post.

The variance of $V$ is
$$\mathbb{Var}[V]=\mathbb{E}[V^2]-(\mathbb{E}[V])^2=\frac{2}{3}-\frac{8}{9\pi}\tag{3}$$
where $\mathbb{E}[V^2]$ can be calculated by multiple integration.

Approximate relations based on empirical data

The remaining part contains only unproven statements that could give indications of the true solution.

The probability distribution of empirical data of $V$ can be fitted quite well with a function of the form
$$f(V)=\text{exp}\left(-\left(\frac{V}{c_2}\right)^{c_1}\right)c_3\tag{4}$$
where $c_1,c_2,c_3$ are fit parameters. As a PDF must fulfill the conditions
$$\int_0^\infty f(V) \mathrm{d}V=1\ \,\, \text{and}\ \int_0^\infty V f(V)\mathrm{d}V=\mathbb{E}[V]$$
the fit parameters $c_2$ and $c_3$ in eq.(4) can be expressed in dependence of $c_1$
$$c_2=\mathbb{E}[V]\frac{\Gamma(1/c_1)}{\Gamma(2/c_1)}\ ,\;\; c_3=\frac{c_1}{c_2\Gamma(1/c_1)}$$
with $\Gamma$ being the Gamma function. Only $c_1$ remains to be fitted. The best fit is for $c_1\approx\pi/4$, i.e. $c_1\approx 0.7854, c_2\approx 0.3491, c_3\approx 2.4944$. However it is not known whether eq.(4) is the true form of the PDF at all. It just models well experimental data.

答:
We'll try to solve the following similar problem. It's easy to see how the solution of such problem answers to your question.

Given
$$ \mathbf{\mathrm{X}_1} =(x_1^1,x_1^2,x_1^3),\;\;   \mathbf{\mathrm{X}_2}=(x_2^1,x_2^2,x_2^3),\;\;    \mathbf{\mathrm{X}_3}=(x_3^1,x_3^2,x_3^3)$$
where $x_i^j$ are independent standard normal distributed variables
$$x_i^j\sim\mathcal{N}(0,1)$$

What is the PDF of the volume of a parallelepiped whith vertices on $(0,0,0), \mathbf{\mathrm{X}_1}, \mathbf{\mathrm{X}_2}, \mathbf{\mathrm{X}_3}, \mathbf{\mathrm{X}_1}+\mathbf{\mathrm{X}_2}, \mathbf{\mathrm{X}_1}+\mathbf{\mathrm{X}_3}, \mathbf{\mathrm{X}_2}+\mathbf{\mathrm{X}_3}, \mathbf{\mathrm{X}_1}+\mathbf{\mathrm{X}_2}+\mathbf{\mathrm{X}_3}$?

The idea to solve this problem is to decompose each random vector in two random variables that correspond to their unit vector and their (eucidean) norm.
Luckily, the variables obtained can be shown to be independent.

So, we define
$$\mathbf{\mathrm{U}_i}=\mathbf{\mathrm{\hat{X}}_i}$$
$$\mathbf{\mathrm{N}_i}=||\mathbf{\mathrm{X}_i}||_2$$

We observe that the volume (with sign) $V$ of the parallelepiped can be expressed as $$V=N_1N_2N_3(U_3\cdot(U_1\times U_2))$$
The distribution of the $N_i$'s is known to be the chi distribution with parameter $k=3$.

Let's then focus on $U_3\cdot(U_1\times U_2)$. It can be shown that:

1).the angle $\theta$ between $U_1$ and $U_2$ has a uniform probability distribution on the interval $[-\pi,\pi]$.

2).the angle $\phi$ between $U_1\times U_2$ and $U_3$ has a uniform probability distribution on the interval $[-\pi,\pi]$.

3).The random variables $\phi$ and $\theta$ are independent.

Moreover, $U_3\cdot(U_1\times U_2)=\text{sin}(\theta)\text{cos}(\phi)$. It's not too hard to find that the probability distribution of $\text{sin}(\theta)$ (which is equal to that of $\text{cos}(\phi)$) is $\frac{1}{\pi \sqrt{1-x^2}}$.

I don't think we can go further than that. If we want the PDF explicity we should apply some convolution on all those density functions, and the result (if you can get one) wouldn't be pretty at all.


问:
Can a machine decide the statement 0>0?
My question may seem silly, but I've been having some difficulties with it... I saw that order relation in the set of computable numbers applies only if $a \neq b$, that is, we can say that $a > b$ or $b > a$ if $a \neq b$. But then I was thinking, I know $0$ is computable - because it is a natural number - then what output a computer would give if it was to decide whether $0>0$ or not...

答: In Python, print(0>0) will print False. Just about any programming language can do the same. The trichotomy law states exactly one of $a>b, \, a=b, \, a<b$ is true.

Title: Can a machine decide the statement 0>0?


问:
Prove the $\lim_{x \to -1} x^2 + 4x + 2 = 0$ is wrong by negating the $\epsilon-\delta$ definition of limit
I'm struggling to disprove the following limit is wrong by negating the definition

$$
\lim_{x\to-1} x^2 + 4x + 2 = 0
$$

The negation of the definition applied to my case should be

$$
\exists \epsilon > 0 \forall \delta > 0 : 0 < |x+1| < \delta \text{ and } |x^2 + 4x + 2| \geq \epsilon
$$

This is to me equivalent to find some $\epsilon > 0$ such that the following system has solution for each $\delta > 0$

$$
\left\{
\begin{array}{l}
|x^2 + 4x + 2| \geq \epsilon \\
0 < |x+1| < \delta
\end{array}
\right.
$$

I started by picking $\epsilon = 1$, and assumed that $\delta \geq 1$ in such a case since
$$
x^2 + 4x + 2 = (x + 1)^2 + 2(x + 1) - 1
$$

it follows that if $x = 0$ solves the system, I'm having trouble however when $0 < \delta < 1$. I cannot manage to construct a solution of that system... any suggestion?

Thank you

答:
Set $f(x) = x^2 + 4x + 2.$

There are actually two ways of proving that the stated limit is wrong.  One way is the direct approach:

You can simply evaluate $f(x = -1)$ directly.

$f(-1) = (-1)^2 + 4(-1) + 2 = -1 \neq 0.$

Therefore the $\lim_{x \to -1} f(x) \neq 0.$

When $f(x)$ is well defined at the value that $x$ is going towards, (in this case, $[x = -1]$), as in this problem, this approach is viable.

The alternative approach, which seems to be a requirement of the problem is to demonstrate that the limit is wrong through the negation of the $\epsilon,\delta$ approach.

Note: the comment of lone student makes a serious point.  If you know that a function can't converge to two distinct points, then you could simply use the $\epsilon,\delta$ approach to show that $f(x)$ does in fact converge to $(-1)$ rather than $0$.

I will take the more pedantic approach:

Your negation has a flaw.

The approach should be:

$\exists ~\epsilon > 0~$ such that $\forall \delta > 0$ at least one value for $x$ can be found such that

$$0 < |x - (-1)| < \delta ~~\text{and}~~ |f(x) - 0| > \epsilon.\tag{Constraint-1}$$

I will simply set $\epsilon = (1/2).$

It is desired, given any specific value of $\delta > 0$, no matter
how small $\delta$ is, to provide a formula for finding $x$ such that (Constraint-1) above is satisfied.

This means, that with $\epsilon$ fixed at $[\epsilon = (1/2)]$, you want to demonstrate that for any $\delta > 0$, some value of $x$ can be found such that:

$$-\delta < (x+1) < \delta ~~\text{and}~~ |(x^2 + 4x + 1)| > (1/2). \tag{Constraint-2}$$

Specify $x$ as follows: $x = \min[(\delta/2) -1, (1/10) - 1].$

This specification serves the following purposes:

It guarantees that $x$ satisfies the LHS of Constraint-2.

It is a general specification for $x$ in terms of $\delta$, as required.

It guarantees that $-1 < x \leq -0.9$.

There are two possibilities, given how $\delta > 0$ is chosen.

If $(\delta/2) \leq (1/10)$, then $x = (\delta/2 - 1) \implies$

$f(x) ~=~ [\frac{\delta^2}{4} + 1 - \delta] + 4[(\delta/2) - 1] + 2$

$= ~ \frac{\delta^2}{4} + \delta - 1$.

Here, since it is assumed that $(\delta/2) \leq (1/10)$, you clearly have that $|f(x)| > (1/2)$, so the RHS of Constraint-2 is satisfied.

If $(\delta/2) > (1/10)$, then $x = (1/10 - 1) \implies$

$f(x) ~=~ [(1/100) + 1 - (2/10)] + 4[(1/10) - 1] + 2$

$= ~ (1/100) + (2/10) - 1$.

Again, you clearly have that $|f(x)| > (1/2)$, so the RHS of Constraint-2 is satisfied.


问:
Success count of Bernoulli rv.
We have a sequence of iid Bernoulli variables, $X_n \sim \text{Bernoulli}(\pi)$, with $n \in \mathbb{N}$.  Let $S_t$ and $F_t$ be the number of successes and failures by time $t$, respectively.

How would $S_t$ be distributed? I know this is a count process with $\mathbb{E}[S_t] = \pi t$. I don't know what this count process is called  though.

More generally, consider the infinitesimal limit, $X_n = \pm \sqrt{dt}$ (as done in the derivation of Brownian motion). $S_t$ would be the time spent increasing up to time $t$. What's the distribution of $S_t$?

Edit: if we let $dt = t/n$, then by the CLT, $\frac{1}{n}\text{Binom}(n,\pi) \simeq \mathcal{N}(\pi, \pi(1-\pi)/n)$. With a step size of $dt$, we have $\frac{1}{n} \sum_{i=1}^n dt X_i \sim \mathcal{N}(t \pi , t^2/n \pi(1-\pi)) $.

答:
How would $S_t$ be distributed?

It is a Binomial $Bin(t;\pi)$. If you want to prove it consider the MGF (or CF) of $S_t$

Title: Success count of Bernoulli rv.


问:
What is an endomorphism?
This question is really basic, but I have essentially no background on algebra so I really do not know basic things. On page 4 of these notes, the author defines $\text{End}S$ to be the set of all endomorphisms of a given finite-dimensional vector space $S$. What is the definition of an endomorphism?

答:
An endomorphism is a homomorphism $\varphi : A \to A$ with $A$ being an mathematical structure. So its a mapping from a structure to itself with respect to the structure ("not losing it through the mapping").

To be more precise and for example: In terms of $(A, \circ), (B, \ast)$ being groups or monoids with $\varphi : A \to B$:

$\varphi(1_A) = 1_B$ for any $1_A, 1_B$ being neutral elements in $A$ and $B$

$\varphi(a \circ_A b) = \varphi(a) \ast_B \varphi(b)$ for all $a \in A$ and $b \in B$

Title: What is an endomorphism?


答:
(It's not actually totally clear to me what you're asking, but I think that you're running into one of the standard confusions around equality of computable real numbers - namely, how do we reconcile the computational difficulty of comparing numbers with the obviousness of comparisons like $0\not<0$? If this doesn't address your question, let me know and I'll delete it.)

What's the input?

When we say that equality of real numbers is undecidable, or something similar, we're referring to reals as infinite bit strings. In this context, even something as $0$ is presented as an infinite object. Crucially, we are not given any additional "contextual information" about the bit string in question; so in order to tell apart the bitstrings corresponding to $0$ and to $1\over 10^{10^{10}}$ we basically have to spend $\log_2(10^{10^{10}})$-many computational steps.

By contrast, when you ask "Is $0<0$?" you've given the number $0$ as an input in a much nicer way: the input now tells me that I'm dealing with an (exact) integer, and so I can decide - without difficulty - that in fact $0\not<0$.

Title: Can a machine decide the statement 0>0?

Body: My question may seem silly, but I've been having some difficulties with it... I saw that order relation in the set of computable numbers applies only if $a \neq b$, that is, we can say that $a > b$ or $b > a$ if $a \neq b$. But then I was thinking, I know $0$ is computable - because it is a natural number - then what output a computer would give if it was to decide whether $0>0$ or not...


问:
Books and resources on internet for coloring and domination in graphs
I am currently undergraduate student and I want to explore about coloring and domination of graphs. There are lot of materials on web related to this but asking advice like which resource or book to follow might help me in progressing.

I know the concept of coloring and I do not have any idea regarding domination in graphs. I will be thankful if someone provides links of resources or books which will be easy for me to follow as a beginner in the field.

答:
The problem with studying any specific "sub-subdiscipline" in math, is that each one relates to many other topics within the broader topic. Coloring for example, relates closely to combinatorics or computational algorithms.

That being said, I used A First Course in Graph Theory by Chartrand and Zhang (2012) in my undergraduate days to learn about domination numbers and coloring. These have two good sections on these topics. Beyond that, finding easy-to-read texts specific to coloring and domination numbers gets tricky. Graph Theory by Diestel gives an up-to-date graduate level overview of graph theory in general, with a small section on coloring, and casual mentions of the topic in other chapters. However, discussion of domination is missing, probably because it is a settled NP-complete problem, so Chartrand might be the end of the story, beyond individual papers.


问:
How to prove this particular set is a countable set?
Suppose $f(x)$ is defined on $[0,1]$, and $\exists M$, s.t. for any finite $x_1, \cdots, x_n \in [0,1]$, we have $|f(x_1)+\cdots+f(x_n)|\le M$. Prove E = $\{x \in [0,1]:f(x)\neq 0\}$ is a countable set.

I've got a simple idea to change to $\cup\{x:|f(x)|\ge \frac{1}{n}\}$, but don't know what to do next.

答: Hint : Each of the $\{x : |f(x)| \geq {1 \over n}\}$ has to be a finite set due to Archimedean property of $\mathbb R$.


问:
How do you divide a complex number with an exponent term?
Ok, so basically I have this:

$$ \frac{3+4i}{5e^{-3i}} $$

So basically, I converted the numerator into polar form and then converted it to exponent form using Euler's formula, but I can have two possible solutions.

I can have $5e^{0.972i}$ (radian) and $5e^{53.13i}$ (angle).

So my real question is, the exponent, does it have to be in radian or angle?

I ended up with $e^{3.927i}$.

答:
Radians are indeed conventional, and using degrees will lead to trouble once you start messing around with derivatives and integrals. Whenever an angle appears "bare" in a mathematical formula (without, e.g., a degree symbol, indication of gradians, turns, etc.), it is assumed to be in radians.

You can either get both numbers into rectangular form or both into polar form.

With both in polar form, $$\frac{re^{i\alpha}}{qe^{i\beta}} = \frac r q e^{i\alpha} e^{-i\beta} = \frac r q e^{i(\alpha - \beta)}.$$

With both in rectangular form, $$\frac{a+bi}{c+di} = \frac{a+bi}{c+di}\frac{c-di}{c-di} = \frac{(a+bi)(c-di)}{c^2+d^2} = \frac{ac+bd}{c^2+d^2}+\frac{bc-ad}{c^2+d^2}i.$$


问:
How to prove that $\sum(-1)^n\sin \frac{1}{n^\alpha \ln n}$ is divergent for any $\alpha\leq 0$?

How to prove that $\sum(-1)^n\sin \frac{1}{n^\alpha \ln n}$ is divergent for any $\alpha\leq 0$?

It is well-known that $\lim \sin n$ does not exist. But that procedure could not be easily adapted to those that $\lim \sin \frac{1}{n^\alpha\ln n}$ does not exists. How to do then for this above problem?

答:
Let $a_n=\frac{1}{n^\alpha \ln n}, n \ge 2$

For $\alpha=0$ we have that $a_n$ decreases to $0$ as $n \to \infty$ hence $\sin a_n$ does so even from $n=2$ on ($a_2=1/\ln 2 < \pi/2$) so the series converges by the alternating test.

If $0 < \beta=-\alpha \le 1$ we let $f_\beta(x)=f(x)=\frac{x^{\beta}}{\pi \ln x}, x>3/2$

Then $f'(x)=(\beta-\frac{1}{\ln x})\frac{f(x)}{x}$ so $f'(x)$ (eventually) decreases to zero as $x \to \infty$ (for monotonicity we use that the leading term of $f''$ which dominates is negative as it easily seen, so $f'$ eventually starts decreasing) and also $xf'(x) \to \infty, x \to \infty$ since $f(x) \to \infty, x \to \infty$.

Hence by Fejer's theorem, we get that $f(n)$ is equidistributed modulo $1$ so $\sin a_n=\sin \pi f(n)$  cannot have limit zero as $n \to \infty$, so the series doesn't converge.

If now $\beta>1$ and $k\ge 1$ the unique positive integer for which $0< \beta -k \le 1$, one applies the same ud modulo $1$ criterion as above but for the $k+1$ derivative of $f$ - namely, $f^{(k+1)}(x)$ decreses eventually to zero at infinity and $xf^{(k+1)}(x) \to \infty$ which is an easy inductive exercise since the leading term of $f^{(k)}$ will be $cf_{\beta-k}$ and which is as in the previous case.

This shows that $f(n)$ is still ud modulo $1$ by the easy inductive generalization of Fejer's theorem (using Van der Corput difference criterion) in this case hence $\sin \pi f(n)$ cannot converge to zero.


问:
Expectation of a game where player doubles the bet every time he wins
Player P plays a game whereby he bets on an event E occuring.

Event E occurs with probability q. So ¬E occurs with probability (1 - q)

The payoff per round is 1:1 -> For every 1 unit P bets on E , if E occurs, E collects 1 unit as the payoff, and keeps the 1 unit he bet; and if ¬E occurs, P loses the 1 unit he bet.

He adopts a strategy of playing agressive when he is winning, and playing safe when he is losing:

P plays exactly 10 rounds of the game

P always bets on E in every round

P will start the first round with a bet of 1 unit

After each and every loss, P will bet 1 unit on the next round, if there is a next round

If P wins the i^th round, he will increase his bet for the next round ( (i + 1)^th round ) by 2 times of the amount he bet in the i^th round.

ie, P bets 1 unit in i^th round and won. He will bet 2 units in the (i+1)^th round. If he wins again, he will bet 4 units in the (i+2)^th round.

After a triple consecutive win, P returns to betting 1 unit on the following round.

Rules 1 to 6 applies to every round of the game.

An example of a game,

Outcomes
E
E
E
E
¬E
¬E
E
E
¬E
¬E

Bets
1
2
4
1
2
1
1
2
4
1

Question: What is the expectation of the payoff if P adopts this strategy?

Note: No use of markov chain

##################################################

I need help with this question. I was thinking of using the usual way of calculating expectation, until I realised that the payoffs for games with the same number of wins and losses differs with the sequence of occurances of the wins and losses.

I tried to calculate the expectation of playing 1 round, 2 rounds, 3 rounds ... individually to try to find a relationship between them, but to no avail. I could not find any perceivable relation. In fact, the expectation became too tedious to calculate beyond 3 rounds.

Is there any intuitive way of solving this problem?

答:
This solution was written before the poster modified the question and added rule 6.  The following is only valid if P continues to double indefinitely after each win.

Let $E(n)$ be the expected payoff on the next $n$ rounds, assuming P stakes 1 unit on the first of those rounds.  So we want $E(10)$.

The first thing to note is that with n rounds to play, if P stakes $x$ units on the first of those rounds, the expected payoff will be $xE(n)$ (the expectation scales up).

With $n+1$ rounds left to play, with P staking 1 unit on the first of those:

with probability $q$, P will gain 1 and (because they will then stake 2) expect to gain $2E(n)$ on their remaining rounds

with probability $1-q$, P will lose 1 and then expect to gain $E(n)$ on their remaining rounds.

So

$E(n+1) = q(1+2E(n)) + (1-q)(-1+E(n))$

$E(n+1) = 2q-1+(q+1)E(n)$

This is a recurrence relation that we can solve, using the initial condition that:

$E(1) = q - (1-q) = 2q - 1$

By considering a solution of the form $E(n) = A\alpha^n+B$ the recurrence relation leads to $\alpha=(q+1)$ and $B=\frac{1-2q}{q}$.  The initial condition leads to $A = \frac{2q-1}{q}$.  So

$E(n) = \frac{(2q-1)((q+1)^n-1)}{q}$

Substitute $n=10$ to answer the question.


问:
Prove that an infinte ring with identity must have an infinite number of non-invertible elements
The problem is to prove that an infinte ring $K$ with identity can not contain a finite number $n \ge 1$ of non-invertible elements.

The hint advices to use a proof by contradiction. Assume that $N = \{a_1, \ldots, a_n\}$ is the set of all non-invertible elements of $K$ without $0$. A map $p_x: a_i \to xa_i$ is a bijection $N \to N$, $\forall x \in K \setminus(N \cup{0})$. And the kernel of the map $p: x \to p_x$ is infinite.

I wanted to show that in this case the map $p$ is an isomorphism from $K \setminus(N \cup{0})$ to a subset $H \subset S_n$ of all permutations on $N$ and therefore $K \setminus(N \cup{0})$ must contain a finite number of elements, but the kernel of $p$ which is a susbset of $K \setminus(N \cup{0})$ has an infinite number of elements by contradition. I could make a mistake, but I see that $p$
is indeed the isomorphism. I'm not able to prove that its kernel has an infinite number of elements. By its kernel I assume $\ker p = \{ x \in K \setminus(N \cup{0})| p_x = e \}$, where $e$ is the identity permutation on $N$, but the author could think of something different. I can only think of $p_1$ where $1$ is an identity element in $K$ .

答:
Challenge: "Prove that an infinte ring with identity must have an infinite number of non-invertible elements."

Answer: If $K$ is an infinite field it follows every element in $K-0$ is invertible. The only non-invertible element is $0$, hence this is not true in general.


问:
Question of proving the limit of $\lim\limits_{x\to0}x\ln(x)$
I know that there are already numerous questions on $\lim\limits_{x\to0}x\ln(x)$. However, I am interested in a specific trait when it comes to prove the existence of the limit by the use of rule of L'Hopital.

Why is it a dead end if we start off like this:

$$
\lim\limits_{x\to0}x\ln(x)=\frac{\lim\limits_{x\to0}x}{\lim\limits_{x\to0}\frac{1}{\ln(x)}}\underset{\text{L'Hopital}}{=}\frac{\lim\limits_{x\to0}1}{\lim\limits_{x\to0}\frac{(-1)}{x\ln(x)^2}}=\lim\limits_{x\to0}(-x)\ln(x)^2=\cdots?
$$

If we start with the right expression then we see that the limit exists and if not we get lost. This seems a bit unsatisfying as it involves a kind of "luck". Maybe there is a deeper sense in this point which I don't see. Or did I simply made a mistake?

Hope this question is not too vague. If it is then let me know and I will delete it.

答:
The equality$$\lim_{x\to0}x\log(x)=\frac{\lim_{x\to0}x}{\lim_{x\to0}\frac1{\log x}}$$is false, since the LHS is $0$, whereas the RHS doesn't make sense. And the equality$$\frac{\lim_{x\to0}x}{\lim_{x\to0}\frac1{\log x}}=\frac{\lim_{x\to0}1}{\lim_{x\to0}-\frac1{x\log^2(x)}}$$simply says that a thing that doesn't make sense is equal to another thing which also doesn't make sense.

On the other hand, it is indeed true that if you try to apply L'Hopital's rule to compute the limit$$\lim_{x\to0}\frac{x}{\frac1{\log x}}$$you get another indeterminate forme. So what? That happens a lot.


问:
How to find the angle Theta from the Figure?
I have the following figure and want to find angle $\theta$ from the figure

I know the slope of the two lines can be obtained from :

\begin{align}
m1 = \frac{Y2 - Y1} {X2 - X1}
\end{align}
Similarly,
\begin{align}
m2 = \frac{Y4 - Y3}{X4 - X3}
\end{align}

Now i know that to find an angle between the given two lines is:

\begin{align}
\theta = \tan^{-1}(\frac{m2 - m1}{1 + m1*m2})
\end{align}
But i am not sure that the $\theta$ obtained is certainly the $\theta$ from my drawing.

Please help me out how I can resolve this.

答:
From dot product between 2 vectors:

$$a \cdot b = |a||b|cos\theta \\
cos\theta = \frac{a \cdot b}{|a||b|}
$$

Building vectors from coordinates:

$$a = (X1-X2, Y1-Y2) \\
b = (X3-X4, Y3-Y4)
$$

To sum up it all together:

$$cos\theta = \frac{(X1 - X2)(X3-X4) + (Y1-Y2)(Y3-Y4)}{\sqrt{(X1-X2)^2 + (Y1-Y2)^2}\sqrt{(X3-X4)^2 + (Y3-Y4)^2}}$$


答:
Draw a horizontal line that intersects the vertex of $\theta,$ and let $\theta=\theta_1+\theta_2$ such that
$\theta_1$ is below $\theta_2$ and both together are vertically opposite $\theta.$

By definition of gradient, $$\theta_1=-\arctan(m_1)$$
and $$\theta_2=\arctan(m_2).$$

Thus, $$\theta=\arctan(-m_1)+\arctan(m_2).$$

Since $\theta$ is acute, the arctangent addition formula applies: $$\theta=\arctan\frac{(-m_1)+(m_2)}{1-(-m_1)(m_2)}\\
=\arctan\frac{m_2-m_1}{1+m_1m_2}.$$

Title: How to find the angle Theta from the Figure?

Body:
I have the following figure and want to find angle $\theta$ from the figure

I know the slope of the two lines can be obtained from :

\begin{align}
m1 = \frac{Y2 - Y1} {X2 - X1}
\end{align}
Similarly,
\begin{align}
m2 = \frac{Y4 - Y3}{X4 - X3}
\end{align}

Now i know that to find an angle between the given two lines is:

\begin{align}
\theta = \tan^{-1}(\frac{m2 - m1}{1 + m1*m2})
\end{align}
But i am not sure that the $\theta$ obtained is certainly the $\theta$ from my drawing.

Please help me out how I can resolve this.


问:
For which $\alpha$ this sum converges? $\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$
Given:

$$\sum_{n=3}^{\infty} {\frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$$
I am asked: For what values of $\alpha$ does this summation converge?

So I said, $f(n) = \frac{1}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}$. $f(n)$ is obviously monotonically decreasing function. Then, by using the integral test, this summation converges if and only if $I = \int_{3}^{\infty} {\frac{dn}{n \cdot \ln(n) \cdot \ln(\ln(n))^{\alpha}}}$ converges. (has a value)

But I am finding this very hard to go on with. Any direction will be appreciated!

答:
Note that $$\frac{1}{1- \alpha}\frac{d}{dx}(\log\log(x))^{1-\alpha}=(\log\log(x))^{-\alpha}\frac{1}{x\log x}$$

since $$(\log\log x)^\prime =\frac{1}{x\log x}$$


问:
What will be the probability density function of a function of the random variable?
This question just came to my mind while thinking in the shower, and it's answer too will if I think hard enough but I would like to get the answer lot quicker by sharing it with you.

So, let's say X is some random variable, whose range is from say 0 to 1.
And let's say it follows some, probability distribution function, let us assume that it is a rectangular one for the sake of simplicity. So the probability of obtaining any number will be equal and exhaustive and mutually exclusive. i.e. 1/100.

Now let's say we introduced a function of X, f(X) = x*100 to increase the range from 0 to 1 to 0 to 100
This will make no change to the probability density function of f(X). i.e., f(X) will also have a rectangular distribution.

But now, if we want to increase the range of the random variable from 0 to 100 to 0 to 200 but I want to do it by defining the third random variable g(X) as
g(X) = x*100 + 100

instead of doing x*200, what will happen to the probability distribution of g(X)?

EDIT: I just read that a thing called skewness also exists so is it related to something like that?

答:
Moving comments to answer:

It is not related to skewness, which is scale and location invariant.

Your $g(x)=100x+100$ will produce a distribution in the interval $[100,200]$ while $200x$ would produce a distribution in the interval $[0,200]$.

So $g(x)$ would have a higher expectation and lower dispersion than the alternative.


问:
What is the equation to make a concentric spiral that isn't an Archimedean Spiral?
I've googled spirals, and the only ones I get are Archimedean Spirals. I'm looking for an equation that would create the spiral below:

答:
$r = k \theta$, where k is a constant.

Edit - As saulspatz pointed out, this describes the Archimedian spiral. But that is what the picture shows.

There are other spirals. Wikipedia lists some.


问:
Is every finite subset of $\mathbb{C}^2$ is a zero set of a polynomial in $\mathbb{C}[x,y]$?
I am trying to do the following exercise:

Is any finite subset of $\mathbb{C}^2$ the zero set of a polynomial $\mathbb{C}[x,y]$? Prove or find a counterexample.

Here I interpreted the "any" to mean "every". Then my thought was that this is indeed true since suppose the finite subset is given by
$$A= \{(a,b),(c,d),(e,f),...\}$$

Then take the polynomial

$$P(x,y)= ((x-a)+(y-b))((x-c)+(y-d))((x-e)+(y-f))...$$

Then the zero set of this polynomial would the set $A$ right? Is my solution correct or am I misinterpreting the question?

答:
Question: "Is any finite subset of C2 the zero set of a polynomial C[x,y]? Prove or find a counterexample."

Answer: If $f \in \mathbb{C}[x,y]:=A$ with $B:=A/(f), V(f):=Spec(B)$, and $f$ is a non-constant polynomial, it follows $dim(Spec(B))=1$. Hence $Spec(B)$ cannot be a finite set of points: A finite set of points is a set of dimension $0$.


问:
Value of the following limit
I'm just a high school math student who was writing a paper on a new type of probability but I got stuck here because even the teachers who taught me were lost as to what the value of this would be. As long as the calculator could allow me, I got the answer to be 0.288788. I actually wanted to know what was the derivation of this rather than just a numeric value, for example could it be something like $\frac{1}{e^2}$ or any number which was rather peculiar.
$$
=\frac{(2^1-1)(2^2-1)(2^3-1)(2^4-1)...(2^n-1)}{2^{\frac{n(n+1)}{2}}}
$$
What will be the limit as n tends to $\infty$

In retrospect the question can also be written as
$$
\lim_{i \to \infty}\Pi_{n=1}^i \frac{2^{n}-1}{2^n}
$$
Your help would be very much appreciated, thank you.

Title: Value of the following limit

Body:
$$\prod_{n=1}^\infty \frac{2^{n}-1}{2^n}=\prod_{n=1}^p \frac{2^{n}-1}{2^n} \prod_{n=p+1}^\infty \frac{2^{n}-1}{2^n}$$
$$y=\prod_{n=p+1}^\infty \frac{2^{n}-1}{2^n}\implies \log(y)=\sum_{n=p+1}^\infty \log \left(1-2^{-n}\right)$$ Using Taylor expansion we have
$$\log(1-2^{-n})=-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\frac{x^5}{5}-\frac{x^6}{6}+O\left(x^7\right)$$ with $x=2^{-n}$.

This gives
$$\log(1-2^{-n})=-\frac{1}{60} a^{-6 n} \left(10+12a^n+15 a^{2 n}+20 a^{3 n}+30 a^{4 n}+60 a^{5 n}\right)$$ with $a=2$. So we have geometric series. Using $p=5$ and summing, this will give
$$\log(y)=-\frac{1976285950427}{62910533468160}$$ and then
$$\frac{9765}{32768}\times e^{\log(y)} \sim 0.288788095086612$$

The "exact" value is
$$0.28878809508660242127889972192923078008891190484069\cdots$$

Edit

Making the problem more general for
$$z_a=\prod_{n=1}^\infty \frac{a^{n}-1}{a^n}\implies \log(z_a)=\sum_{n=1}^\infty \log \left(1-a^{-n}\right)$$ and switching the order of summation, we have
$$\log(z_a)=-\sum _{k=1}^{\infty } \frac{a^{-k}}{k \left(1-a^{-k}\right)}$$ For $a=2$,this will give
$$\log(z_2)=-\Big[1+\frac{1}{6}+\frac{1}{21}+\frac{1}{60}+\frac{1}{155}+\frac{1}{378}+\frac{1}
   {889}+\frac{1}{2040}+\frac{1}{4599}+\frac{1}{10230}+\frac{1}{22517}+\frac{1}{49140}+\frac{1}{106483}+\frac{1}{229362}+\frac{1}{4
   91505}+\frac{1}{1048560}+\frac{1}{2228207}+\frac{1}{4718574},\frac{1}{9961453}+\frac
   {1}{20971500}+\frac{1}{44040171}+\frac{1}{92274666}+\frac{1}{192937961}+\frac{1
   }{402653160}+\frac{1}{838860775}+\frac{1}{1744830438}+\cdots   \Big]$$ Limited to these only terms, this would give
$$\log(z_2)=-1.24206209427900$$ Taking the exponential and computing, this  would give as an approximation
$$z_2=0.28878809524064630741$$

As a function of $a$, we then generate the following constants
$$\left(
\begin{array}{cc}
a & z_a \\
 2 & 0.2887880950866024212788997 \\
 3 & 0.5601260779279489449697922 \\
 4 & 0.6885375371203397154565144 \\
 5 & 0.7603327958712324201014883 \\
 6 & 0.8056877281621649409237502 \\
 7 & 0.8367954070890378710267298 \\
 8 & 0.8594059944007028662007586 \\
 9 & 0.8765603540359642058360198 \\
 10 & 0.8900100999989990000001000
\end{array}
\right)$$


问:
Is it possible to find function that contains every given point?
Let say we have a arbitrary number of given points and there is at least one function, for which every point lies on its graph. Is it possible to find that function using only X and Y coordinates of every given point?

Example:

We are given points $A(0,1)$; $B(0.27;0)$ and $C (3.73;0)$. All of these points lie on the graph of the function $f(x) = x^2 - 4x + 1$. Is it possible to find this function using only X and Y coordinates of the point A,B and C.

Note: The number of point is can be really big and also the exponent of the function.

答:
Suppose you have $n$ given points. Assume a polynomial function of order $n-1$ with unknown constant coefficients. You'll now have to solve simultaneously $n$ number of equations for coefficients that make the polynomial curve pass through the points.

For example in your case, there are $3$ given points $A(0,1)$, $B(0.27,0)$ and  $C(3.73,0)$.

Now assume a polynomial of order $2$ with unknown constant coefficients.

$$y=A+Bx+Cx^2$$

Plug in the $3$ points to get $3$ equations in $A$, $B$ and $C$ and simultaneously solve them.

And indeed  $A=1$, $B=-4$ and $C=1$.

This can be employed for any arbitrary set of points, though the process gets increasingly tedious with large number of points.


问:
Proof regarding basic properties of sequences

We've just started learning sequences and I was faced with this question. I have a strong feeling that I'm missing something and that I'm over simplifying things as these questions almost seem trivial to the point I'm not sure how to correctly prove them..

A. True

Assume $(a_n)_{n=1}^{\infty }$ converges to some $L∈R$.

Then as $\{a_n | n∈N\}$ = $\{b_n | n∈N\}$ for every $n∈N$

$(a_n)_{n=1}^{\infty }=(b_n)_{n=1}^{\infty }=L$

Similarly, if $(b_n)_{n=1}^{\infty }$ converges

$(b_n)_{n=1}^{\infty }=(a_n)_{n=1}^{\infty }=L$.

B. True

Assume $(a_n)_{n=1}^{\infty }$ converges to some $L∈R$.

As $\{n∈N | a_n ≠ b_n\}$ is bounded from above, there exists $M∈N$ such that for every $n≥M$ we have

$a_n = b_n$

Therefore $(a_n)_{n=1}^{\infty }=(b_n)_{n=1}^{\infty }=L$.

Similarly if $(b_n)_{n=1}^{\infty }$ converges, there exists $M∈N$ such that for every $n≥M$ we have

$b_n = a_n$

Therefore $(b_n)_{n=1}^{\infty }=(a_n)_{n=1}^{\infty }=L$.

C. False.

Counter example:

$\{a_n\} = \frac{1}{n}$ for every $n∈N$

$\{b_n\} = Z^{+}$

We've seen in recitation that $(a_n)_{n=1}^{\infty } = 0$. But $(b_n)_{n=1}^{\infty }=∞$, and therefore diverges.

答:
$A)$ is false : take $a_1=-1$, and $a_n=1$ for all $n\geq 2$ and $b_n=(-1)^n$ for all $n\geq 1$ for a counter example.

$B)$ is indeed true, but your justification is strange (what means $(a_n)=(b_n)=L$?) Indeed, $a_n=b_n$ for all $n\geq M$ for some $M\in \mathbb N$. From here, obviously $(a_n)$ converges if and only if $(b_n)$ converges.

$C)$ is true (I even don't understand your counter-example). If $(a_n)$ converges, since $(b_n)$ is increasing, then $(b_n)$ converges as well. If $(b_n)$ converges, then, by squeeze theorem, $(a_n)$ converges as well.


问:
Matrix gradient descent method with power constraints
I'm solving an optimization problem

\begin{equation}
  \begin{aligned}\label{}
    & \underset{A\in \mathbb{R}_{m,d_1} B\in \mathbb{R}_{n,d_2}}{\text{minimize}}
    & & \phi({A},B) \\
    & \text{subject to}
    & & \|Ha_j\|_2^2=1 , \ j = 1, \ldots, d_1.\\
    &&& \|Vb_j\|_2^2=1 , \ j = 1, \ldots, d_2.
  \end{aligned}
\end{equation}
where $H$ and $V$ are known matrices, $a_j$'s and $b_j$'s are columns of $A$ and $B$

I can't solve the optimal solution, so I try to use numerical methods like gradient descent

But I don't know how to use gradient descent with such constraint

I already have derivative of  $\phi$ w.r.t. to  $A$ and $B$
My thought is to seek the matrix $A'$ in the constraint set which has the smallest distance (like Frobenius norm) to $A_{k+1}^{temp} = A_k- \alpha_k\frac{\partial\phi}{\partial A}(A_k,B_k)$ to be $A_{k+1}$and follow the similar way to update $B$.

That is, solve
\begin{equation}
  \begin{aligned}\label{}
    & \underset{A\in \mathbb{R}_{m,d_1} }{\text{minimize}}
    & & \|A-A_{k+1}^{temp}\|_F^2 \\
    & \text{subject to}
    & & \|Ha_j\|_2^2=1 , \ j = 1, \ldots, d_1.\\
  \end{aligned}
\end{equation}
Am I on the right way to constrainted gradient descent?

Any help would be appreciated!

Refer to Rem's answer, BFGS may work better. But how BFGS work on matrix variable and under some constraints?

答:
You probably want to look at the dual problem. Introduce a penalty in your cost function and solve a bigger but easier problem instead. See Duality, Lagrange multiplier.

The idea is solve a new, unconstrained problem, namely find a saddle point of:
$$L(A,B,\lambda,\mu) = \phi(A,B) - \sum_{j=1}^{d1}\lambda_j(\lVert Ha_j\rVert^2-1) - \sum_{j=1}^{d2}\mu_j(\lVert Vb_j\rVert^2-1)$$


问:
Is it legal to define a function that gives different results for 1.0 and for 1?
In programming languages I can define such function, because in most programming languages 1.0 is not 1, because 1.0 has type "float", and 1 has type "integer". In math I don't see axioms that forbid sets $\mathbb{R}$ and $\mathbb{N}$ to be disjoint. So can I define function in such a way, that f(1) = 1, and f(1.0) = 2, motivating it by saying that 1.0 is taken from $\mathbb{R}$, 1 is taken from $\mathbb{N}$, and these sets are defined in the completely separate ways?

答: Yes. You can use the set $\mathbb{R}\sqcup\{1\}$, which is the disjoint union of the reals with the integer $1$ and define a function on that set which distinguishes between the $1.0$ in the reals and the $1$ in the set $\{1\}$. You can extend this idea to $\mathbb{R} \sqcup\mathbb{Z}$ and give distinct values depending on their membership in the $\mathbb{R}$ elements or the $\mathbb{Z}$ elements. The dijoint union is essentially creating labeled versions of each number with $1.0=1_{\mathbb{R}}$ and the other being $1_{\mathbb{Z}}$. While you lose many of the useful properties of the reals and integers by constructing this disjoint union it's still a perfectly valid set and you can use it to define the domain of a function.


问:
How to express this computer science problem mathematically?
I am writing my dissertation for a software degree and want to provide a mathematical formula for a function I have.

The code I want to describe consists of two parts. First to build a collection of values by doing a calulation:

The input to this code is three lists of factors labled High, Medium and Low. What the factors are arn't important but how many there are is. Say there may be 5 high factors 7 meduim and 4 low. The number of each catagory of factors is timesed by a weighting. Say 100 for high 50 for meduim and 10 for low.

This looks like:
Total Weight $= $(weighting of a high factor $\times$ Count of high factors) $+$ (weighting of a medium factor $\times$ Count of medium factors ) $+$ (weighting of a low factor $\times$ Count of low factors )

Each time it does that sum it adds the outcome to the collection of values.

Secondly afterwards the code selects the top 5 values from this collection and returns them

How would I describe this in a mathematical formula?

答:
Let be $S = \{a_i \; | \; i \in [[0, n[[\}$ a collection (array) of $n$ elements. We define a map $f: S \to V$, which to each to each element of $S$ associates a value; with $V = \{1, 2, 3, 4, 5\}$.

We denote $argmax_5(S, f)$ the subset of $S$ (ie, $argmax_5(S, f) \in \mathcal{P}(S)$) containing the $5$ elements of $S$ which have the highest values for the function $f$.

You can then use $\#(E)$, the cardinal of a set $E$, for your "count". For example, if the high factors are the set $H$, then you can define the weight of the high factors as $w_H$ and multiply that by $\#(H)$. If you have another question, comment below.


答:
Your error was stated in the other answer.  You could do it this way.

$$\lim_{x\to 0}x\ln(x)=\lim_{x\to 0} \frac{\ln x}{\frac 1x}=\lim_{x\to 0}\frac{\frac 1x}{- \frac {1}{x^2}}=\lim_{x\to 0}(-x)=0.$$

Title: Question of proving the limit of $\lim\limits_{x\to0}x\ln(x)$

Body:
I know that there are already numerous questions on $\lim\limits_{x\to0}x\ln(x)$. However, I am interested in a specific trait when it comes to prove the existence of the limit by the use of rule of L'Hopital.

Why is it a dead end if we start off like this:

$$
\lim\limits_{x\to0}x\ln(x)=\frac{\lim\limits_{x\to0}x}{\lim\limits_{x\to0}\frac{1}{\ln(x)}}\underset{\text{L'Hopital}}{=}\frac{\lim\limits_{x\to0}1}{\lim\limits_{x\to0}\frac{(-1)}{x\ln(x)^2}}=\lim\limits_{x\to0}(-x)\ln(x)^2=\cdots?
$$

If we start with the right expression then we see that the limit exists and if not we get lost. This seems a bit unsatisfying as it involves a kind of "luck". Maybe there is a deeper sense in this point which I don't see. Or did I simply made a mistake?

Hope this question is not too vague. If it is then let me know and I will delete it.


问:
Hilbert Schmidt norm
Suppose $Q \in L(U)$ is non negative operator with $Tr Q < \infty$ where $(U, \langle,\rangle)$ is Hilbert space. Suppose that $e_k, k=0...$ is an orthonormal base of $U$ consisting of eigenvectors of $Q$ with $\lambda_k$ eigenvalues.
Suppose that $U_0=Q^{1/2}(U)$ is the Hilbert space with scalar product
$$\langle u,v\rangle_0=\langle Q^{-1/2}u,Q^{-1/2}v\rangle$$
Call $(L_2^0=L_2(U_0,H),||\cdot||_{L_2^0})$ the Hilbert Schmidt operators from $U_0$ to $H$ where $H$ is another Hilbert space.

Then by the Parseval identity
$$||\Psi||_{L_2^0}^2=\sum_{k,h}^\infty |\langle\Psi g_h,f_k\rangle|^2=\sum_{k,h}^\infty \lambda_h |\langle\Psi e_h,f_k\rangle|^2=\sum_{k,h}^\infty  |\langle\Psi Q^{1/2} e_h,f_k\rangle|^2=||\Psi Q^{1/2} ||_{L_2(U,H)}^2$$

Now why does the following hold [Da prato - Stochastic equations in infinite dimensions 1999] ?

$$||\Psi||_{L_2^0}^2=Tr (\Psi Q \Psi^*)$$

where $g_k=\sqrt{\lambda_k}e_k, k=0...$ is an orthonormal base of $U_0$ and $f_k,k=0...$ is an orthonormal base of $H$

答:
Using the definition:
\begin{align}
\|\Psi\|^2_{L_2^0}&=\|\Psi Q^{1/2}\|^2_{L_2(U,H)}=\operatorname{Tr}((\Psi Q^{1/2})^*\Psi Q^{1/2})\\[0.3cm]&=\operatorname{Tr}(Q^{1/2}\Psi^*\Psi Q^{1/2})=\operatorname{Tr}(\Psi Q\Psi^*).
\end{align}
In case the problem is the apperance of the trace,
$$
\sum_{h,k}|\langle \Phi g_h,f_k\rangle|^2=\sum_h\|\Phi g_h\|^2=\sum_h\langle \Phi^*\Phi g_h,g_h\rangle=\operatorname{Tr}(\Phi^*\Phi).
$$

Title: Hilbert Schmidt norm


问:
Prove the $\lim_{x \to -1} x^2 + 4x + 2 = 0$ is wrong by negating the $\epsilon-\delta$ definition of limit
I'm struggling to disprove the following limit is wrong by negating the definition

$$
\lim_{x\to-1} x^2 + 4x + 2 = 0
$$

The negation of the definition applied to my case should be

$$
\exists \epsilon > 0 \forall \delta > 0 : 0 < |x+1| < \delta \text{ and } |x^2 + 4x + 2| \geq \epsilon
$$

This is to me equivalent to find some $\epsilon > 0$ such that the following system has solution for each $\delta > 0$

$$
\left\{
\begin{array}{l}
|x^2 + 4x + 2| \geq \epsilon \\
0 < |x+1| < \delta
\end{array}
\right.
$$

I started by picking $\epsilon = 1$, and assumed that $\delta \geq 1$ in such a case since
$$
x^2 + 4x + 2 = (x + 1)^2 + 2(x + 1) - 1
$$

it follows that if $x = 0$ solves the system, I'm having trouble however when $0 < \delta < 1$. I cannot manage to construct a solution of that system... any suggestion?

Thank you

答:
As indicated in my comment to question, let us take $\epsilon=1/2$. And let $\delta>0$ be given. We need to show that there is some $x$ with $0<|x+1|<\delta$ and $|x^2+4x+2|\geq 1/2$.

Let us observe that $$|x^2+4x+2|=|(x^2+4x+3)-1|\geq ||x+1||x+3|-1|$$ and the last expression is greater than or equal to $1/2$ if (not iff) $|x+1||x+3|<1/2$.

Let us assume that $|x+1|<1$ and then $|x+3|<3$ and the inequality $|x+1||x+3|<1/2$ is satisfied if $|x+1|<1/6$. Thus it should be clear that if $|x+1|<1/6$ then our job is done.

Thus we can take any value of $x$ which satisfies $0<|x+1|<\min(\delta, 1/6)$ and it will ensure that $|x^2+4x+2|\geq 1/2$.


问:
Proving whether ideals are prime in $\mathbb{Z}[\sqrt{-5}]$
I am working with the ideals $\mathfrak{p}=\left<2,1+\sqrt{-5}\right>, \mathfrak{q}=\left<3,1+\sqrt{-5}\right>, \mathfrak{t}=\left<3,1-\sqrt{-5}\right>$ and I am trying to prove that they are prime in $\mathbb{Z}[\sqrt{-5}]$.

I understand a good method to do this involves taking norms of each element that generates the ideals. This gives, for instance, the norms of the generators of $\mathfrak{p}=4,6$. These are both divisible by $2$. I have also calculated $\mathfrak{p}^2=\left<2\right>$. I get the feeling that since both norms are divisible by the generator of the ideal squared this proves it is prime, but I am not quite sure why.

In a similar vein, the norms of the generators of $\mathfrak{q,t}=9,6$ which are both divisible by $3$. I know $\mathfrak{qt}=\left<3\right>$, but I am not sure if this is relevant, since it is not either ideal squared. I think the norm of the ideal itself being prime implies the ideal is prime, but I am not sure how to find the norm of the ideal from the norm of its generating elements.

This question is similar to: Prove that ideals are prime, but I don't quite understand the reasoning behind the chosen answer.
I am not sure how the answerer deduces that the example there is prime either. Following their reasoning as far as I can, the  fact that $2$ divides both norms means that $\left<2\right> \subset \mathfrak{p} \subset \mathbb{Z}[\sqrt{-5}]$ but I'm not sure how, and then I don't know how this proves it is prime.

答: Here is how I see it. $\mathbb{Z}[\sqrt{-5}] = \mathbb{Z}[1+\sqrt{-5}]$. Let $1+\sqrt{-5} = \alpha$. Then $(\mathbb{Z}.1 \oplus \mathbb{Z}.\alpha)/(2\mathbb{Z} \oplus \mathbb{Z}.\alpha) \equiv \mathbb{Z}/\mathbb{2Z}$, since $\mathbb{Z}/\mathbb{2Z}$ is a field, our ideal is maximal and hence prime.


问:
How do you divide a complex number with an exponent term?
Ok, so basically I have this:

$$ \frac{3+4i}{5e^{-3i}} $$

So basically, I converted the numerator into polar form and then converted it to exponent form using Euler's formula, but I can have two possible solutions.

I can have $5e^{0.972i}$ (radian) and $5e^{53.13i}$ (angle).

So my real question is, the exponent, does it have to be in radian or angle?

I ended up with $e^{3.927i}$.

答:
the exponent is always is radians. converting into polar form would give you $$r e^{i \theta}$$ where $$r=\sqrt{3^2 + 4^2}=5$$ and $$\theta=\tan^{-1}(\frac43)$$ or $\arctan (\frac43)$ which gives $$\theta=0.927$$ so your fraction becomes $$\frac{5e^{0.927i}}{5e^{-3i}}$$
$$=e^{3.927i}$$</p>


问:
Fair die - probability - geometric
Throw a fair die.
X is the number of throws until the first appearance of 6, Y is the number of throws until the first appearance of 5.

I need to calculate E(x), so I used geometric expectency (1/1/6) = 6

Now, I need to calculate -

答:
Hint:

$$E(X|Y=1)$$

means that the first die roll is 5

thus

$$E(X|Y=1)=1+6=7$$

Title: Fair die - probability - geometric


问:
In a PID, the intersection of descending chain of ideals is trivial

Suppose $R$ is PID and $I_1 \supseteq I_2 \supseteq \cdots$ is a descending chain of ideals in $R$. I would like to prove that $\bigcap^\infty_{n=1} I_n=(0)$.

Now, since $R$ is a PID, every ideal is principal, so each $I_n=(a_n)$ for some $a_n \in R$. So I need to show that
$$\bigcap^\infty_{n=1} (a_n)=(0).$$
We have $(a_1) \supseteq (a_2) \supseteq \cdots$, so $a_i \mid a_{i+1}$ for all $i$.

 I am not sure what to do next.

Also, I don't think this is true if $R$ is just a UFD. What would an example of that be?

答: Assuming strict descent (as otherwise any constant sequence is a counter-example). Let $a$ be the generator of the intersection. Then, $(a) \subset (a_i) \, \forall i$ and hence $a_i | a \, \forall i$. But since the descent is strict, none of the $a_i$'s are associates. Also $a_i | a_{i+1}$ so $\exists p_i$, a prime that divides $a_{i+1}$ but not $a_i$ for each $i$. That gives us infinitely many prime divisors for $a$, forcing $a=0$.


问:
prove that the order of the elements of restricted direct product is finite .
if $G_i = \Bbb Z/p_i \Bbb Z$

($\Bbb Z$ means integers)

where $p_i$ is the ith  integer prime , I=the positive integers

show that , every element of the restricted direct product of the $G_i$'s has finite order

my trial to solve it ,

i supposed that $S\subset I$  is the indes of the restricted direct product .

let $S=${$i,i+1,i+2,...,m-1,m$}

$ G = G_i \times G_{i+1} \times G_{i+2} \times ... \times G_m \times G_1 \times G_2 \times ... \times G_{i-1} \times G_{m+1} \times ... $

let  $g=(1_i ,1_{i+1} , 1_{i+1} , ... ,1_m , g_1 ,g_2 ,...,g_{i-1} , g_{m+1} , ... ) \in G$

where , $1_i$ is the identity of the ith group , $ g_1 \in G_i$

now the problem is that to compute the order of this element g , i have to calculate l.c.m of the orders of the elements $g_1 , g_2 , ... ,g_{i-1} , g_{m+1} , ... $
and those orders are infinte numbers of primes so the l.c.m can't be finite !
so the order must be infinte .

but i'm asked to prove it's finite !

so i think i made a error in my definitions " may be i understand the definition of restricted wrongly , i'm not sure "

any help plz , thanx

答:
A "tuple" in the restricted direct product only differs from the identity on finitely many coordinates. So, you will be able to find the LCM of orders of coordinates to find the order of an element, as you began to do.

I think a good way to understand the restricted direct product is as a subgroup of the full direct product. The direct product, as you know, is just all "vectors" of elements where the $i$'th entry is from $G_i$, and you make this into a group with pointwise operations.

The restricted direct product is a subgroup of that consisting of elements which are the identity except possibly on finitely many coordinates. You can easily see that the two products share the identity, and that the elements I described are closed under multiplication and inversion.

In fact, I think you will have a good time showing that the restricted direct product is a normal subgroup of the direct product :)


答:
You are correct that UFD does not suffice: you can have an infinite strictly decreasing chain of ideals whose intersection is not trivial. In $\mathbb{Z}[x]$, the ideals $I_n=(2,x^n)$ satisfy $I_{n+1}\subsetneq I_n$, but $\cap I_n = (2)$. But you can still leverage the UFD property to get what you want in the PID.

Note that in a UFD, if the principal ideals $(a)$ and $(b)$ satisfy $(a)\subseteq (b)$, then $b|a$; and if $(a)\subsetneq (b)$, then $b$ is a proper divisor of $a$.

Proposition. Let $R$ be a UFD, and let $(a_1)\supseteq (a_2)\supseteq\cdots\supseteq (a_n)\supseteq\cdots$ be a chain of principal ideals in $R$. If $a\neq 0$ lies in $\cap (a_k)$, then there exists $k$ such that $(a_k)=(a_{k+r})$ for all $r\geq 0$; that is, the chain stabilizes.

Proof. If $a$ is a unit, then the intersection is $R$, so each ideal is $R$ and they are all equal. So we may assume $a$ is not a unit. Since $a\neq 0$, then it has a factorization into irreducibles,
$$a = p_1\cdots p_r.$$
Since $(a)\subseteq (a_n)$, then $a_n$ is a divisor of $a$. But $a$ has only finitely many proper divisors up to associates, so there are finitely many principal ideals $I$ such that $(a)\subsetneq I$. Thus, at some point, the chain $(a_i)\supseteq (a_{i+1})\supseteq\cdots$ must stabilize. $\Box$

Thus, in your PID, either your descending chain stabilizes, or else the intersection does not contain nonzero elements.

(Intuitively, the intersection is a least common multiple; but if at each step you are adding an irreducible factor, then the least common multiple would have an “infinite” factorization into irreducibles, which is impossible.)

Title: In a PID, the intersection of descending chain of ideals is trivial

Body:
Suppose $R$ is PID and $I_1 \supseteq I_2 \supseteq \cdots$ is a descending chain of ideals in $R$. I would like to prove that $\bigcap^\infty_{n=1} I_n=(0)$.

Now, since $R$ is a PID, every ideal is principal, so each $I_n=(a_n)$ for some $a_n \in R$. So I need to show that
$$\bigcap^\infty_{n=1} (a_n)=(0).$$
We have $(a_1) \supseteq (a_2) \supseteq \cdots$, so $a_i \mid a_{i+1}$ for all $i$.

 I am not sure what to do next.

Also, I don't think this is true if $R$ is just a UFD. What would an example of that be?


问:
Which permutation matrix is this diagonal matrix similar to?
I was trying to solve this question from GATE exam 2015.

Q43 2015 EC GATE PAPER SET 1

Q 4) Two sequences $[a,b,c]$ and $[A,B,C]$ are related as,
$$\left [ \begin{matrix}A \\B \\C  \\\end{matrix} \right ] = \left [ \begin{matrix}1 & 1 & 1 \\1 & W_3^{-1} & W_3^{-2} \\1 & W_3^{-2} & W_3^{-4}\\\end{matrix} \right ] \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]  \ \ \ \ \ \ \ \ \ \ \ (1)$$
Where $W_3 = e^{j\frac{2\pi}{3}}$. If another sequence  $[p,q,r]$ is derived as,
$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \left [ \begin{matrix}1 & 1 & 1 \\1 & W_3^{1} & W_3^{2} \\1 & W_3^{2} & W_3^{4}\\\end{matrix} \right ] \left [ \begin{matrix}1 & 0 & 0 \\0 & W_3^{2} & 0 \\0 & 0 & W_3^{4}\\\end{matrix} \right ] \left [ \begin{matrix}\frac{A}{3} \\\frac{B}{3}  \\\frac{C}{3}   \\\end{matrix} \right ]   \ \ \ \ \ \ \ \ \ \ \ (2)$$

The the relationship between the sequences $[p,q,r]$ and $[a,b,c]$ is,

(a) $[p,q,r] = [b,a,c]$
(b) $[p,q,r] = [b,c,a]$
(c) $[p,q,r] = [c,a,b]$
(d) $[p,q,r] = [c,b,a]$

I think the answer is (b), the official answer key says it is (c). (page 19 of the pdf) Can you find the mistake in my solution or confirm that I am right?

My solution

First we notice $W_3$ is the 3rd root of unity. Next if we take $\frac{1}{3}$ out of the vector on the right of (2) we can substitute (1) into (2).
$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \left [ \begin{matrix}1 & 1 & 1 \\1 & W_3^{1} & W_3^{2} \\1 & W_3^{2} & W_3^{4}\\\end{matrix} \right ] \left [ \begin{matrix}\frac{1}{3} & 0 & 0 \\0 & \frac{W_3^{2} }{3}& 0 \\0 & 0 & \frac{W_3^{4}}{3}\\\end{matrix} \right ] \left [ \begin{matrix}1 & 1 & 1 \\1 & W_3^{-1} & W_3^{-2} \\1 & W_3^{-2} & W_3^{-4}\\\end{matrix} \right ] \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]   \ \ \ \ \ \ \ \ \ \ \ (3)$$
At this point it should be clear that what we are seeing is a similarity transform on a diagonal matrix, which would turn it into a permutation matrix (based on  the options).
Since its a diagonalisation we can see how the eigenvector matrix used to diagonalize the permutation matrix is actually the 3 point DFT matrix. Since this matrix is unitary, the inverse is just the conjugate transpose. So now we know the eigenvalues of that permutation matrix are the diagonal elements and the eigenvectors are the column vectors.
$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \frac{1}{\sqrt{3}}\left [ \begin{matrix}1 & 1 & 1 \\1 & \omega^{-1} & \omega ^{-2} \\1 & \omega ^{-2} & \omega ^{-4}\\\end{matrix} \right ] \left [ \begin{matrix}1 & 0 & 0 \\0 & \omega ^{-2}  & 0 \\0 & 0 & \omega ^{-4}\\\end{matrix} \right ] \frac{1}{\sqrt{3}} \left [ \begin{matrix}1 & 1 & 1 \\1 & \omega & \omega ^{2} \\1 & \omega ^{2} & \omega ^{4}\\\end{matrix} \right ] \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]   \ \ \ \ \ \ \ \ \ \ \ (3)$$

Now we can use the fact that the roots of unity are n-periodic.
$$\omega^{-5}, \omega ^{-4}, \omega^{-3}, \omega ^{-2}, \omega^{-1}, \omega ^{0}, \omega^{1}, \omega ^{2}, \omega^{3}, \omega ^{4}, \omega^{5}, \omega ^{6}$$
Since we are dealing with the 3rd root of unity the above sequence is the same as,
$$\omega^{1}, \omega ^{2}, 1, \omega ^{1}, \omega^{2}, 1, \omega, \omega ^{2}, 1, \omega, \omega^{2}, 1$$
Now the crucial observation is that $\omega^{-2}$ or $\omega$ is the eigenvalue for the 2nd column in the eigenvector matrix. So multiplying the second column with $\omega$ is the same as applying that permutation matrix $P_\pi$.
$$P_\pi \left [ \begin{matrix} 1  \\ \omega  \\ \omega ^{2}\\\end{matrix} \right ]  = \omega \left [ \begin{matrix} 1  \\ \omega  \\ \omega ^{2}\\\end{matrix} \right ] = \left [ \begin{matrix}\omega \\ \omega^{2}  \\ 1 \\\end{matrix} \right ] $$
We can say the same for the 3rd eigenvector too.
$$P_\pi \left [ \begin{matrix} 1 \\\omega ^{2} \\ \omega ^{4}\\\end{matrix} \right ]   = \omega^2 \left [ \begin{matrix} 1 \\\omega ^{2} \\ \omega ^{4}\\\end{matrix} \right ]   = \left [ \begin{matrix}\omega ^{2} \\ \omega  \\ 1 \\\end{matrix} \right ] $$
So we can conclude the permutation matrix is,

$$P_\pi = \left [ \begin{matrix}0 & 1 & 0 \\0 & 0 & 1 \\1 & 0 & 0 \\\end{matrix} \right ]$$
So the answer is (b) $[p,q,r] = [b,c,a]$.
I write so much for the sake of clarity. If you want to solve this quickly we can just quickly check what the eigenvalue $W_3^{2} = W_3^{2} W_3^{-3} = W_3^{-1}  $ does to the second eigenvector.
$$P_\pi \left [ \begin{matrix} 1 \\W_3^{-1}\\W_3^{-2}\\\end{matrix} \right ]   = W_3^{-1} \left [ \begin{matrix} 1 \\W_3^{-1}\\W_3^{-2}\\\end{matrix} \right ]    = \left [ \begin{matrix} W_3^{-1} \\W_3^{-2}\\1\\\end{matrix} \right ] $$
So we immediately see how the second element becomes the first, 3rd becomes 2nd and 1st becomes last.

答:
We can confirm the answer is indeed (c) by multiplying the 3 matrices in equation (3). We will need to use the fact that the sum of the roots of unity is zero.

$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \frac{1}{\sqrt{3}}\left [ \begin{matrix}1 & 1 & 1 \\1 & \omega & \omega ^{2} \\1 & \omega ^{2} & \omega ^{4}\\\end{matrix} \right ]  \left [ \begin{matrix}1 & 0 & 0 \\0 & \omega ^{2}  & 0 \\0 & 0 & \omega ^{4}\\\end{matrix} \right ] \frac{1}{\sqrt{3}} \left [ \begin{matrix}1 & 1 & 1 \\1 & \omega^{-1} & \omega ^{-2} \\1 & \omega ^{-2} & \omega ^{-4}\\\end{matrix} \right ]\left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]$$

The diagonal elements scale the rows when you right multiply,

$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \frac{1}{3}\left [ \begin{matrix}1 & 1 & 1 \\1 & \omega & \omega ^{2} \\1 & \omega ^{2} & \omega ^{4}\\\end{matrix} \right ]  \left [ \begin{matrix}1 & 1 & 1 \\  \omega ^{2} & \omega ^{1}  &  \omega ^{0}\\  \omega ^{4}  & \omega ^{2} & \omega ^{0}\\\end{matrix} \right ] \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]$$

$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \frac{1}{3}\left [ \begin{matrix}1+\omega^{2} +\omega^{4}  & 1+\omega+\omega^{2}   & 3 \\1+\omega^{3}+\omega^{6} & 1+\omega^{2}+\omega^{4} & 1+\omega+\omega^{2} \\ 1+\omega^{4}+\omega^{8} & 1+\omega^{3}+\omega^{6} & 1+\omega^{2}+\omega^{4} \end{matrix} \right ]  \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]$$

$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \frac{1}{3}\left [ \begin{matrix}1+\omega^{2} +\omega  & 1+\omega+\omega^{2}   & 3 \\3 & 1+\omega^{2}+\omega & 1+\omega+\omega^{2} \\ 1+\omega+\omega^{2} & 3 & 1+\omega^{2}+\omega \end{matrix} \right ]  \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ]$$

Since we know $1,\omega,\omega^2$ are 3 roots of unity, their sum is zero.

$$\left [ \begin{matrix}p \\q \\r  \\\end{matrix} \right ] = \left [ \begin{matrix} 0  & 0  & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{matrix} \right ]  \left [ \begin{matrix}a \\b \\c  \\\end{matrix} \right ] = \left [ \begin{matrix}c \\a \\b  \\\end{matrix} \right ] $$

Your answer was wrong because you took the wrong matrix as the eigenvector matrix. The first matrix on the left is the eigen vector matrix. When you diagonalize a matrix then the matrix is on the right.

$$P_\pi = W\Lambda W^{-1} $$

$$P_\pi W = W\Lambda $$

$$\Lambda = W^{-1}P_\pi W $$

The final solution has been updated.