Title: Is there a formula for $1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$?

Body:
Given
$$f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$$
I am wondering if there is a straightforward formula to compute $f(n)$ and how it may be derived.

The only reduction I thought about so far would be:

$$n\cdot 1 + (n - 1)\cdot 2 + (n - 3)\cdot 3 +\cdots $$

which seems symmetrical; for example, an odd and even n

```<code> 5 4 3 2 1        4 3 2 1
*                *
 1 2 3 4 5        1 2 3 4
</code>```

but I'm not sure if and how it may help derive a formula.

Body:
The hockey stick identity states that
$$\sum_{k=r}^{n} \binom{k}{r}=\binom{n+1}{r+1}$$

Note that $$1+2+...+n = \frac{n(n+1)}{2}=\binom{n+1}{2}.$$

Hence

$$f(n)=\binom{2}{2}+\binom{3}{2}+...+\binom{n+1}{2}=\binom{n+2}{3}=\frac{(n+2)(n+1)n}{3!}$$


问:
Knowing $f(x) \to 0$ for $x \to 0$, why is $f(x)/x \to 0$ for $x \to 0$?
Let's say, I've found out, that $\lim\limits_{x\rightarrow 0} f(x)= 0$, why is then $\lim\limits_{x \to 0} f(x)/x = 0$? I want to prove, that the function is differentiable at $x=0$, for what I have to show that this limit exists.
$f:\mathbb{R} \to \mathbb{R}$ is defined as $0$ for $x \leq 0$ und $f(x) = e^{-1/x}/x^n$ for x>0, where I've proven, that in case $x<0$ and x>0 it is differentiable.

答:
Not true.

Take $f(x) = \sqrt{x}$.
Then,
as $x \to 0$,
$f(x) \to 0$
but
$\dfrac{f(x)}{x}
=\dfrac1{\sqrt{x}}
\to \infty
$.

ContentLicense: CC BY-SA 4.0


问:
Is it possible to separate a triple integral into a difference of two triple integral?
Is it possible to do this subtraction ?

$\int\limits_{z = 0}^{z = \frac{b}{a}} {\int\limits_{x = 0}^{x = \infty } {\int\limits_{y = \frac{a}{x}}^{y = \frac{b}{{xz}}} {f\left( y \right)f\left( x \right)f\left( z \right)dydxdz = } } } \int\limits_{z = 0}^{z = \frac{b}{a}} {\int\limits_{x = 0}^{x = \infty } {\int\limits_{y = 0}^{y = \frac{b}{{xz}}} {f\left( y \right)f\left( x \right)f\left( z \right)dydxdz} } }  - \int\limits_{z = 0}^{z = \frac{b}{a}} {\int\limits_{x = 0}^{x = \infty } {\int\limits_{y = 0}^{y = \frac{a}{x}} {f\left( y \right)f\left( x \right)f\left( z \right)dydxdz} } } $

where $a,b,f\left( y \right),f\left( x \right),f\left( z \right) \geqslant 0$ and ${f\left( y \right),f\left( x \right),f\left( z \right)}$ are all decaying exponential function.

If it wrong then why ?
Thank you very much ?

答:
I don't know what does "decaying exponential function" exactly mean, but one can claim that you are right if all the integrals converges. From the Linearity of integrals, we have:
\begin{aligned}
&\quad \int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\int\limits_{\frac{a}{x}}^{\frac{b}{xz}}{f\left( y \right) f\left( x \right) f\left( z \right) dydxdz=}}}\int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\left( \int\limits_{\frac{a}{x}}^{\frac{b}{xz}}{f\left( y \right) dy} \right) f\left( x \right) f\left( z \right) dxdz}}
\\
&=\int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\left( \int\limits_0^{\frac{b}{xz}}{f\left( y \right) dy-\int\limits_0^{\frac{a}{x}}{f\left( y \right) dy}} \right) f\left( x \right) f\left( z \right) dxdz}}
\\
&=\int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\left( \int\limits_0^{\frac{b}{xz}}{f\left( y \right) dy} \right) f\left( x \right) f\left( z \right) dxdz}}-\int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\left( +\int\limits_0^{\frac{a}{x}}{f\left( y \right) dy} \right) f\left( x \right) f\left( z \right) dxdz}}
\\
&=\int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\int\limits_0^{\frac{b}{xz}}{f\left( y \right) f\left( x \right) f\left( z \right) dydxdz}}}-\int\limits_0^{\frac{b}{a}}{\int\limits_0^{\infty}{\int\limits_0^{\frac{a}{x}}{f\left( y \right) f\left( x \right) f\left( z \right) dydxdz}}}
\end{aligned}
As desired.


问:
How should I get the minimum of a sum of root square and a linear part
$\frac {\sqrt{x^2 + 100}}  a + \frac {b - x}  c$

where $a$, $b$, $c$ are positive constants.

I can draw the curve in a graph paper and get the minimum, but don't know how to calculate it in a mathematical way?

答:
The most straightforward way to do this is by taking the derivative and setting it to 0, as Ross Millikan commented.

Let

$$ y = \frac{\sqrt{x^2+100}}{a} + \frac{b-x}{c} $$

Then, differentiating, we have

$$ \frac{dy}{dx} = \frac{2x}{2a\sqrt{x^2+100}} - \frac{1}{c} $$

Setting $ y' = 0 $, we have

$$ \frac{2x}{2a\sqrt{x^2+100}} - \frac{1}{c} = 0 $$

After a bit of algebra, we get to

$$ x = \frac{10a}{\sqrt{c^2-a^2}} $$

You can use the second derivative test to confirm that it is a minimum, and not a maximum.

Given that the parameters $ a, b, c > 0$, we have:

$ c > a $ : The minimum is at the unique local minimum we solved for, $ x = \frac{10a}{\sqrt{c^2 - a^2}} $

$ c \le a$ : $ y' < 0 $ everywhere, so there are no local minima, but on any bounded interval, the minimum will be at the greatest value of $x$ on the interval.


问:
Non-Noetherian ring with a single prime ideal

What are the most simple examples of a commutative ring $R$ satisfying both of the following two properties:

  1. $R$ is not Noetherian.

  2. $R$ has exactly one prime ideal.

答: Take a localization $O:=R_M$ of the ring $R$ of all algebraic integers at a maximal ideal $M$. Then $O/xO$ has exactly one prime for every non-zero $x\in MO$.</p>

ContentLicense: CC BY-SA 3.0


问:
Distribution of sum of dependent Gaussians
If we know $X \sim \mathcal N(\mu, \Sigma)$, is there a closed-form expression for the distribution of $(X-\mu)^\top \Lambda^{-1} (X - \mu)$? I know there is one if $\Lambda = \Sigma$, but if we just assume $\Lambda$ to be any arbitrary positive definite matrix, can we still obtain some distribution? Maybe a gamma or some chi-squared distribution? Any help would be appreciated!

答:
$X - \mu = \Sigma^{1/2} Z$ where $Z \sim N(0, I_n)$. So
$$(X-\mu)^\top \Lambda^{-1} (X-\mu) = Z^\top \Sigma^{1/2} \Lambda^{-1} \Sigma^{1/2} Z = \|\Lambda^{-1/2} \Sigma^{1/2} Z\|^2.$$

So it suffices to understand the distribution of $\|AZ\|^2$ where $A$ is positive definite and $Z \sim N(0, I_n)$, since you can take $A=\Lambda^{-1/2} \Sigma^{1/2}$ above for your specific question.

Because $A$ is positive definite, it can be written as $A=UDU^\top$ where $U$ is orthogonal and $D$ is diagonal with positive entries. Then $\|AZ\|^2 = \|UDU^\top Z\|^2 = \|D \tilde{Z}\|^2$ where $\tilde{Z} := U^\top Z$ is also $N(0, I_n)$ due to rotational invariance of the standard multivariate Gaussian distribution. So your random variable is a weighted sum of squares of independent Gaussians, i.e. $\sum_{i=1}^n d_{ii}^2 \tilde{Z}_i^2$. This is a generalization of the chi-squared distribution, but I don't think it has a name.


问:
are all the terms of this chaotic map 0s?
I'm trying to implement the algorithm proposed on this paper: A Simple Method for Image Encryption Using
Chaotic Logistic Map
but i cannot figure out the equation (7), i'm pretty sure that $X_0=0$, because is the sum of integer products (5) and (6). So, if that is the case, it means: the terms  $X_1,X_2, X_3, X_4..$ of (1) are also 0, which make no sense to me. Is not supposed that the terms of the map (1) will be pseudo-random numbers?

I searched on other journals to check if it was a typo error, but is not the case. So the only thing that come to my mind is that i don't undertand the notation used in this paper. Thanks in advance.

答: Apparently the square brackets in $(5)$ and $(6)$ indicate division by $2^{40}$; compare $(6)$ and $(7)$ in this PDF. That reduces $X_{01}$ and $X_{02}$ to the range $(0,1)$, so that $(8)$ puts $X_0\in(0,1)$, where it should be.


问:
How many eigenvalues does an $n\times n$ matrix have, and how does this relate to the algebraic multiplicity of the dominant eigenvalue?
I'm having to reevaluate my understanding of eigenvalues and how many eigenvalues an $n\times n$ matrix possesses. Previously, I had thought that such a matrix $A$ possessed $d\leq n$ complex eigenvalues, and that this number $d$ was determined by the number of distinct roots of the matrix's characteristic polynomial. Furthermore, the sum of the complex eigenvalues' algebraic multiplicities equals $n$, since the characteristic polynomial of $A$ necessarily has degree $n$ and therefore has $n$ complex roots (where a root of multiplicity $m$ is counted $m$ times). However, I came across a problem involving dominant eigenvalues that prompted me to question this interpretation:

If $A$ has a dominant eigenvalue $\lambda_1$, prove that the eigenspace $E_{\lambda_1}$ is one-dimensional.

Solution: If $\lambda$ is dominant, then $|\lambda|>|\gamma|$ for all other eigenvalues $\gamma$. But this means that the algebraic multiplicity of $\lambda$ is $1$, since it appears only once in the list of eigenvalues listed with multiplicity, so its geometric multiplicity is $1$ and thus its eigenspace is one-dimensional.

So if I'm reading this explanation correctly, a list of all of the eigenvalues of $A$ should include $i$ instances of an eigenvalue with algebraic multiplicity $i$. In other words, every $n \times n$ matrix has exactly $n$ complex eigenvalues, and there is a distinction between the number of eigenvalues that a matrix possesses and the number of distinct eigenvalues that a matrix possesses. This subtle distinction seemed arbitrary until I considered the solution to this problem, which seems to require that all eigenvalues be treated as separate entities, even if they possess the same scalar values. For example, an eigenvalue $\lambda =2$ with algebraic multiplicity $2$ should actually be thought of as two eigenvalues $\lambda _1 = \lambda _2 = 2$. With this understanding, it is clear that neither $\lambda _1$ nor $\lambda _2$ can be a dominant eigenvalue, since it is not true that $|\lambda _1|>|\lambda _2|$, nor is it true that $|\lambda _2|>|\lambda _1|$. In fact, it is impossible for any eigenvalue with algebraic multiplicity greater than $1$ to be dominant. From here, I am comfortable with the fact that a dominant eigenvalue (if it exists) must also have geometric multiplicity $1$, since the geometric multiplicity of an eigenvalue is always less than or equal to the corresponding algebraic multiplicity.

Is this the correct way to interpret the preceding proof/eigenvalues in general? Hopefully I've articulated my thought process clearly, and thank you for taking the time to help!

答:
I don't think a convention is well-established: in some contexts, I see "different eigenvalues" refer to a set of distinct values with associated algebraic multiplicities, while in other contexts, I see "different eigenvalues" refer to the set of $n$ eigenvalues, possibly with repetitions due to multiplicity. Typically one can either discern which convention is being used, or the author should take care to clarify what is meant.

In your case, I think you just have to read the definition of "dominant eigenvalue" carefully. Based on the problem writing "dominant eigenvalue $\lambda_1$," I suspect the definition is written as

if $\lambda_1, \ldots, \lambda_n$ are the eigenvalues of $A$, then $\lambda_1$ is considered dominant if $|\lambda_1| > |\lambda_i|$ for all $i \ne 1$

or something like that, which is unambiguous compared to

$|\lambda| > |\gamma|$ for all other eigenvalues $\gamma$

which is very ambiguous for the reasons you raise.

Now that we know that the context of your question is the power method, then my above guess on what "dominant eigenvalue" means is incorrect.

Let $\lambda_1, \ldots, \lambda_m$ be the distinct eigenvalues of $A$ with multiplicities $n_1, \ldots, n_m$. If $|\lambda_1| > |\lambda_i|$ for all $i \ne 1$, then $\lambda_1$ is said to be  the dominant eigenvalue. The power method will converge to something in the eigenspace corresponding to $\lambda_1$. To ensure that it does not converge to zero, the initial vector must not be orthogonal to the eigenspace.


问:
All elements of factor ring $M_2(\mathbb{Z})/ M_2(2\mathbb{Z})$
I came across this example in the book Abstract Algebra by Gregory T. Lee.  I found it hard to understand.

Let $R= M_2(\mathbb{Z})$ and let $I$ be the ideal consisting of all matrices whose entries are even. Then notice that for any $a_{ij} \in \mathbb{Z}$, we have
$$\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}+I=\begin{pmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{pmatrix}+I$$ where $b_{ij}$ is if $a_{ij}$ is even and 1 if $a_{ij}$ is odd. Thus, $R/I$ consists of the sixteen different elements $ \begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix} + I, \, b_{ij} \in \{0,1\}$.

$M_2(\mathbb{Z})$ is the set of 2 x 2 matrices over $\mathbb{Z}$.

My thoughts:

For $a_{ij} \in \mathbb{Z}$ and $ r, s, t, u \in 2 \mathbb{Z}$, we have:

$$\underbrace{a_{ij}}_{\text{even}} + r = \overbrace{0 + s}^{\text{even}}$$

and

$$\underbrace{a_{ij}}_{\text{odd}} + t = \overbrace{1 + u}^{\text {odd}}. $$

The elements of the $ R / I $ quotient ring are of the type $ \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} + I $ with $ b_{ij} \in \{0, 1\}. $ Since there are $ 2 ^ 4 = 16$ arrangements with repetition of elements 0 and 1, then the set $ R / I $ has 16 elements.

I really need a more detailed explanation for this example.

Question:

How to prove that by double inclusion? Thanks in advance!

答:
Let $$S = \bigg\{ \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} + I : b_{ij} \in \{0,1\} \bigg\}$$ and we claim that $R/I = S$.

For every $a_{ij} \in \mathbb Z$ define $b_{ij}=0$ if $a_{ij}$ is even and $b_{ij}=1$ if $a_{ij}$ is odd. Thus $a_{ij}-b_{ij}$ is always even, and then $$\begin{pmatrix} a_{11} & a_ {12} \\ a_ {21} & a_ {22} \end{pmatrix} + I = \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} + I$$ because $$\begin{pmatrix} a_{11} & a_ {12} \\ a_ {21} & a_ {22} \end{pmatrix} - \begin{pmatrix} b_ {11} & b_ {12} \\ b_ {21} & b_ {22} \end{pmatrix} = \begin{pmatrix} a_{11}-b_{11} & a_ {12}-b_ {12} \\ a_ {21}-b_ {21} & a_ {22}-b_ {22}  \end{pmatrix} \in I.$$ This proves that every element of $R/I$ is equal to an element of $S$, and then $R/I \subseteq S$.

Clearly every element of $S$ is also an element of $R/I$, so $S \subseteq R/I$.


问:
Sequence whose limit is given set
Let $(X,T)$ be a topological space with infinite elements. Consider $\{x_n\}$ the sequence of infinite distinct points in $X$. We say this sequence is convergent if there is a $x$ in $X$ such that for each nbd $G$ of $x$ a positive integer $N$ can be found such that $x_n \in G$ for all $n \geq N$. This point $x$ is said to be a limit of the sequence $\{x_n\}$.

If I consider topology $T = \{\, \varnothing, X \, \}$ then any sequence converges every point of the space. This demonstrates that in an arbitrary space limit need not be unique.

If I consider the discrete topology and let $x$ be a point, suppose a sequence $\{x_n\}$ with infinitely many distinct points and consider $G = \{x, x'\}$. Clearly $G$ is an open nbd of $x$. Then it is easy to see that $\{x_n\}$ doesn't converge to $x$ since there are infinitly many distinct $x_n$ s.

However if the space is Hausdorff then we can easily observe that every convergent sequence has unique limit.

Let $A$ be a subset of $X$. Now my aim is to find a topology in which there is a sequence whose limit points are precisely the set $A$. It is clear that if $A$ is full space and empty then trivial and discrete topologies would work. If $A$ is singleton then Hausdorff space will work. So for a proper subset $A$ of $X$ does there exist topology in which the limit points of a sequence is $A$?

答: Take $\tau = \{\varnothing,A,X \setminus A,X\}$ and $(x_n)_{n \in \mathbb N}$ a sequence of elements in $A$. Observe that for any $x \in A$, $G$ only can be $A$ or $X$.

Title: Sequence whose limit is given set


问:
Are terminating decimals dense in the reals?
I have read that rationals are dense in the reals, but I've been curious if the terminating decimals are as well. Since, in applied math, all effectively "real" numbers are approximated with terminating decimals, I would expect the terminating decimals to have this property, and if not I'm curious why. Here I'm defining the terminating decimals like this:

$$D = \{\frac{m}{10^n} \in \mathbb{Q} : m \in \mathbb{Z} \wedge n \in \mathbb{N}_0\}$$.

答:
Suppose $a$ and $b$ are terminating decimals. We will find a terminating decimal that lies between them, namely  $\frac{a+b}2$.  For example, between $1.7$ and $0.0001$ is $\frac{1.7+0.0001}2 = 0.850005$.

First, as you said, write $$a=\frac A{10^i}, b=\frac B{10^j}.$$

Without loss of generality, $i\le j$. Then write $$a=\frac{A\cdot 10^{j-i}}{10^j}.$$

Now $$\frac{a+b}2=   \frac{A\cdot10^{j-i} + B}{2\cdot 10^j} =
 \frac{5A\cdot 10^{j-i} + 5B}{10^{j+1}} $$

is a terminating decimal between $a$ and $b$.


答:
Combine the two extreme cases that you already know: let

$$\tau=\{X\}\cup\wp(X\setminus A)\,.$$

Let $\sigma=\langle x_n:n\in\Bbb N\rangle$ be a sequence in $X$ with infinitely many distinct points. If $x\notin A$, $\{x\}$ is an open nbhd of $x$ that does not contain a tail of $\sigma$, so $\sigma$ does not converge to $x$. If $x\in A$, then $X$ is the only open nbhd of $x$, so $\sigma$ converges to $x$.

In fact, the only sequences in $X$ that do not have $A$ as their set of limit points are those that are eventually constant.

Title: Sequence whose limit is given set

Body:
Let $(X,T)$ be a topological space with infinite elements. Consider $\{x_n\}$ the sequence of infinite distinct points in $X$. We say this sequence is convergent if there is a $x$ in $X$ such that for each nbd $G$ of $x$ a positive integer $N$ can be found such that $x_n \in G$ for all $n \geq N$. This point $x$ is said to be a limit of the sequence $\{x_n\}$.

If I consider topology $T = \{\, \varnothing, X \, \}$ then any sequence converges every point of the space. This demonstrates that in an arbitrary space limit need not be unique.

If I consider the discrete topology and let $x$ be a point, suppose a sequence $\{x_n\}$ with infinitely many distinct points and consider $G = \{x, x'\}$. Clearly $G$ is an open nbd of $x$. Then it is easy to see that $\{x_n\}$ doesn't converge to $x$ since there are infinitly many distinct $x_n$ s.

However if the space is Hausdorff then we can easily observe that every convergent sequence has unique limit.

Let $A$ be a subset of $X$. Now my aim is to find a topology in which there is a sequence whose limit points are precisely the set $A$. It is clear that if $A$ is full space and empty then trivial and discrete topologies would work. If $A$ is singleton then Hausdorff space will work. So for a proper subset $A$ of $X$ does there exist topology in which the limit points of a sequence is $A$?


问:
Find the maximum between the minima

For every $n\in\mathbb Z^+ = \{1, 2,\dots\}$ set of positive integers, let $r_n$ be the minimum value of $|c−d\sqrt3|$ for all nonnegative integers $c$ and $d$ with $c+d=n$.

Find, with proof, the smallest positive real number $g$ with $rn\leq g$ for all $n\in\mathbb Z^+$.

What I understood is that we need to find the maximum between the minima $r_n$, that is, the greatest $r_n$.
So I reduced $|c−d\sqrt3|$ to $|n−d(1+\sqrt3)|$.

Now note that $1 \leq n-d$.
If we suppose $n−d(1+\sqrt3)>0$, then $$|n−d(1+\sqrt3)| = n−d(1+\sqrt3),$$
so that
$$2n − d(2+\sqrt3) > n-d > 1 \geq n−d(1+\sqrt3) \geq 1 + n + d \geq\dots$$

I stopped here because I saw that it will lead us to nothing basically, so I would appreciate any help.

答:
Writing $|c- d \sqrt 3| = |n - d(1+\sqrt 3)|$ is a nice observation. We can go a bit further:

$$|n-d (1+\sqrt 3)| = (1+\sqrt 3) \left|\frac n{1+\sqrt 3} - d\right|$$

Since $\dfrac n{1+\sqrt 3}$ is just some real number and $d \in \mathbb Z^+$, the value inside the absolute value is minimized when $d$ is the closest integer to $\dfrac n{1+\sqrt 3}$.

This shows that, for the optimal $d$,  $\left|\dfrac n{1+\sqrt 3} - d\right| \le \dfrac12$ (Side note: this inequality can be strict as the fraction is always irrational.) Hence: $$r_n \le \dfrac {1+\sqrt 3}2$$

To show that this is the smallest real number that can bound this absolute value, we need to show that $\dfrac n{1+\sqrt 3}$ can be arbitrarily close to some $d \pm \dfrac12$. This will require a density argument.

Title: Find the maximum between the minima


问:
On the opposite group
When the group $G$ acts on the set $S$, e.g. $G \colon= {\mathrm{GL}}_2({\Bbb C})$, $S \colon= {\Bbb C}^{\oplus 2}$, $G$ have the right to act on the left or on the right. Both actions can be naturally identified as follows$\colon$

$\sigma \circ \tau (s) = (s)\tau \circ \sigma$,

where both sides means $\tau$ acts on $s \in S$ first and then $\sigma$ acts on $\tau(s) ~(= \!(s)\tau)$ next.

That is, the equality

\begin{equation*}
\sigma \circ \tau = \tau \circ \sigma \quad \cdots\cdots \quad (\lozenge)
\end{equation*}
automatically makes sense under the sequence of actions (or time).

Or when one starts from the left action, the set in the R.H.S. in $(\lozenge)$ can be put, by the more sophisticated terminology, the opposite group $G^{\mathrm{op}}$, for which the unique isomorphism $G \cong G^{\mathrm{op}}$ is given by $g \mapsto g^{-1}$.

However, the identification in $(\lozenge)$ is accomplished by switching the multiplication between two elements $\sigma$ and $\tau$ according to the sequence of actions on the left or right (on $S$).

Q. Is it correct that I never care about $G^{\mathrm{op}}$ but always automatically identify $G$ and $G^{\mathrm{op}}$ as in $(\lozenge)$ ignoring the group isomorphism between $G$ and $G^{\mathrm{op}}$?

答:
It sounds like you are saying if you start with a left action of $G$ on $S$ (with $g$ acting on $s$ written "$gs$"), that you want to define a "right action" by just declaring that "$sg$" is an alternate notation for $gs$. This is what I assumed when you said $\sigma \tau (s) = (s) \tau \sigma$ both mean do $\tau$ first, then $\sigma$.

Here is the problem with this: the right-hand side, $s \tau \sigma$ is not really consistent. You get different results depending on if you intepret it as meaning $(s\tau)\sigma$ or $s(\tau\sigma)$ (but they need to agree to be a right action).

I will show by example. Consider the example you give, of $G = \mathrm{GL}_2(\Bbb{C})$ acting on $S=\Bbb{C}^2$. Consider
$$ g = \begin{pmatrix} 0&1\\1&0 \end{pmatrix} \hspace{0.5cm} \text{and} \hspace{0.5cm} v=\begin{pmatrix}x\\y\end{pmatrix}$$
Then of course $gv = \begin{pmatrix}y\\x\end{pmatrix}$. So if we define our right action like you suggest, then $vg = \begin{pmatrix}y\\x\end{pmatrix}$.

But $g$ can be factored as $g=g_2g_1$, where
$$ g_2 = \begin{pmatrix}0&-1\\1&0\end{pmatrix} \hspace{0.5cm} \text{and} \hspace{0.5cm} g_1 = \begin{pmatrix} 1&0\\0&-1 \end{pmatrix} $$
So in the usual left action, multiplication by $g$ is the same as first multiplying by $g_1$, and then by $g_2$. So your suggestion is that if we write things on the right, we should have $g_2g_1v = vg_1g_2$.

This is where the problem occurs: do we interpret the expression $vg_1g_2$ as meaning $(vg_1)g_2$ or as $v(g_1g_2)$? To be a valid right action, it shouldn't matter and both should be the same. The first interpretation is $(vg_1)g_2$, meaning act first by $g_1$, and then by $g_2$. This agrees with the calculation above (and the left action), and we get $\begin{pmatrix}y\\x\end{pmatrix}$. The other intrepretation is $v(g_1g_2)$, which means first multiply $g_1g_2$ in the group, and then act by that element. But $g_1g_2 = \begin{pmatrix}0&-1\\-1&0\end{pmatrix}$, which is not the same as $g$, and we get $\begin{pmatrix}-y\\-x\end{pmatrix}$ as the result.

Title: On the opposite group


问:
Roots of $|x|^{\tan(x)}-x$
For $f(x)=|x|^{\tan(x)}-x$, I am curious about the relation between the roots of $f(x)$. On plotting it in Desmos, I got the roots as $\frac{\pi}{4}$, 1, $\frac{5\pi}{4}$, $\frac{9\pi}{4}$, $\frac{13\pi}{4}$, $\frac{17\pi}{4}$, and so on. But I'm not able to establish a relation between them like the $k^{th}$ root is equal to $\frac{f(k)\pi}{4}$. I tried to simplify, $$f(x)=|x|^{\tan(x)}-x=0$$ $$|x|^{\tan(x)}=x$$ $$\log_x(|x|^{\tan(x)})=1$$ $$\tan(x)=1$$ That is fine till here. But then why are the roots not at $\frac{n\pi}{4}$?

答:
HINT. What are the roots of $\tan(x)$? Check your solution. (Really $\dfrac{n\pi}{4}$?)

Meanwhile, we can't take $\log_x$ when $x=1$. The case of $x=1$ should be separated to consider.

Title: Roots of $|x|^{\tan(x)}-x$


问:
Spherical Trigonometry: Spherical triangle
ABC is an equilateral spherical triangle in which small displacements are made, in the sides and angles, of such a nature that the triangle remains equilateral. Prove that
$$
\frac{da}{dA} = \cos\left({\frac{A}{2}}\right) \cot\left({\frac{a}{2}}\right)
$$

I would have mentioned my progress. But all I have been able to done is differentiate different trigonomeric formulae to get an expression which is of no use to simplify, since it never matches with answer on a numerical verification.

Please help. A hint on how to start might be enough.

NOTE: This question is Q9 of Execrcise 1 from W.M. Smart's Spherical Astronomy.

答:
I don’t believe the formula is correct as quoted.

We’re dealing with an equilateral (and equiangular!) spherical triangle, and to avoid (my own) confusion, I’ll call the side-length $s$ and the angle $\theta$. Both are angles, of course. The most obvious thing is to bisect the triangle into two right triangles, each with hypotenuse $s$ and one side $s/2$, this latter side being opposite the angle $\theta/2$, while the other angle of the triangle is still $\theta$.

Now, one of the standard formulas for right spherical triangles is
$$
\cos c=\cot A\cot B\,,
$$
and for us, $c=s$, $A=\theta$, $B=\theta/2$. So we differentiate our relation
$\cos s=\cot\theta\cot(\theta/2)$, and get
\begin{align}
-\sin s\frac{ds}{d\theta}&=-\cot\theta\csc^2(\theta/2)\cdot\frac12
-\csc^2\theta\cot(\theta/2)\\
\frac{ds}{d\theta}&=\frac{\frac12\cot\theta\csc^2(\theta/2)+\csc^2\theta\cot(\theta/2)}{\sin s}\,.
\end{align}
Now let’s test this out on the nicest possible equilateral triangle, the one whose sides and angles are both $90^\circ$. Here, the first term in the numerator drops out, ’cause $\cot90^\circ=0$, but everything else evaluates to $1$, and so even $ds/d\theta=1$ at this test-point. Yet, the formula evaluates to $\sqrt2/2$, unequal to $1$.


问:
What is the difference between "expectation", "variance" for statistics versus probability textbooks?
It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $X$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like,
$$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$. The variance is,
$$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide  (def 2)? How do I know which mean the person is talking about?/

答:
You ask a very insightful question that I wish were emphasized more often.

EDIT: It appears you are seeking reputable sources to justify the above. Sources and relevant quotes have been provided.

Here's how I would explain this:

In probability, the emphasis is on population models. You have assumptions that are built-in for random variables, and can do things like saying that "in this population following such distribution, the probability of this value is given by the probability mass function."

In statistics, the emphasis is on sampling models. With most real-world data, you do not have access to the data-generating process governed by the population model. Probability provides tools to make guesses on what the data-generating process might be. But there is always some uncertainty behind it. We therefore attempt to estimate characteristics about the population given data.

From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 1.6:

The objective of statistics is to make an inference about a population based on information contained in a sample taken from that population...

A necessary prelude to making inferences about a population is the ability to describe a set of numbers...

The mechanism for making inferences is provided by the theory of probability. The probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer this from the characteristics of an unknown population. Thus, probability is the foundation of the theory of statistics.

From Shao's Mathematical Statistics, 2nd edition, section 2.1.1:

In statistical inference... the data set is viewed as a realization or observation of a random element defined on a probability space $(\Omega, \mathcal{F}, P)$ related to the random experiment. The probability measure $P$ is called the population. The data set or random element that produces the data is called a sample from $P$... In a statistical problem, the population $P$ is at least partially unknown and we would like to deduce some properties of $P$ based on the available sample.

So, the probability formulas of the mean and variance assume you have sufficient information about the population to calculate them.

The statistics formulas for the mean and variance are attempts to estimate the population mean and variance, given a sample of data. You could estimate the mean and variance in any number of ways, but the formulas you've provided are some standard ways of estimating the population mean and variance.

Now, one logical question is: why do we choose those formulas to estimate the population mean and variance?

For the mean formula you have there, one can observe that if you assume that your $n$ observations can be represented as observed values of independent and identically distributed random variables $X_1, \dots, X_n$ with mean $\mu$,
$$\mathbb{E}\left[\dfrac{1}{n}\sum_{i=1}^{n}X_i \right] = \mu$$
which is the population mean. We say then that $\dfrac{1}{n}\sum_{i=1}^{n}X_i$ is an "unbiased estimator" of the population mean.

From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.1:

For example, suppose we want to estimate a population mean $\mu$. If we obtain a random sample of $n$ observations $y_1, y_2, \dots, y_n$, it seems reasonable to estimate $\mu$ with the sample mean $$\bar{y} = \dfrac{1}{n}\sum_{i=1}^{n}y_i$$

The goodness of this estimate depends on the behavior of the random variables $Y_1, Y_2, \dots, Y_n$ and the effect this has on $\bar{Y} = (1/n)\sum_{i=1}^{n}Y_i$.

Note. In statistics, it is customary to use lowercase $x_i$ to represent observed values of random variables; we then call $\frac{1}{n}\sum_{i=1}^{n}x_i$ an "estimate" of the population mean (notice the difference between "estimator" and "estimate").

For the variance estimator, it is customary to use $n-1$ in the denominator, because if we assume the random variables have finite variance $\sigma^2$, it can be shown that
$$\mathbb{E}\left[\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2 \right] = \sigma^2\text{.}$$
Thus $\dfrac{1}{n-1}\sum_{i=1}^{n}\left(X_i - \dfrac{1}{n}\sum_{j=1}^{n}X_j \right)^2$ is an unbiased estimator of $\sigma^2$, the population variance.

It is also worth noting that the formula you have there has expected value
$$\dfrac{n-1}{n}\sigma^2$$
and $$\dfrac{n-1}{n} < 1$$
so on average, it will tend to underestimate the population variance.

From Wackerly et al.'s Mathematical Statistics with Applications, 7th edition, chapter 7.2:

For example, suppose that we wish to make an inference about the population variance $\sigma^2$ based on a  random sample $Y_1, Y_2, \dots, Y_n$ from a normal population... a good estimator of $\sigma^2$ is the sample variance
$$S^2 = \dfrac{1}{n-1}\sum_{i=1}^{n}(Y_i - \bar{Y})^2\text{.}$$

The estimators for the mean and variance above are examples of point estimators. From Casella and Berger's Statistical Inference, Chapter 7.1:

The rationale behind point estimation is quite simple. When sampling is from a population described by a pdf or pmf $f(x \mid \theta)$, knowledge of $\theta$ yields knowledge of the entire population. Hence, it is natural to seek a method of finding a good estimator of the point $\theta$, that is, a good point estimator. It is also the case that the parameter $\theta$ has a meaningful physical interpretation (as in the case of a population) so there is direct interest in obtaining a good point estimate of $\theta$. It may also be the case that some function of $\theta$, say $\tau(\theta)$ is of interest.

There is, of course, a lot more that I'm ignoring for now (and one could write an entire textbook, honestly, on this topic), but I hope this clarifies things.

Note. I know that many textbooks use the terms "sample mean" and "sample variance" to describe the estimators above. While "sample mean" tends to be very standard terminology, I disagree with the use of "sample variance" to describe an estimator of the variance; some use $n - 1$ in the denominator, and some use $n$ in the denominator. Also, as I mentioned above, there are a multitude of ways that one could estimate the mean and variance; I personally think the use of the word "sample" used to describe such estimators makes it seem like other estimators don't exist, and is thus misleading in that way.

In Common Parlance

This answer is informed primarily by my practical experience in statistics and data analytics, having worked in the fields for about 6 years as a professional. (As an aside, I find one serious deficiency with statistics and data analysis books is providing mathematical theory and how to approach problems in practice.)

You ask:

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide (def 2)? How do I know which mean the person is talking about?

In most cases, you want to just stick with the statistical definitions. Most people do not think of statistics as attempting to estimate quantities relevant to a population, and thus are not thinking "I am trying to estimate a population quantity using an estimate driven by data." In such situations, people are just looking for summaries of the data they've provided you, known as descriptive statistics.

The whole idea of estimating quantities relevant to a population using a sample is known as inferential statistics. While (from my perspective) most of statistics tends to focus on statistical inference, in practice, most people - especially if they've not had substantial statistical training - do not approach statistics with this mindset. Most people whom I've worked with think "statistics" is just descriptive statistics.

Shao's Mathematical Statistics, 2nd edition, Example 2.1 talks a little bit about this difference:

In descriptive data analysis, a few summary measures may be calculated, for example, the sample mean... and the sample variance... However, what is the relationship between $\bar{x}$ and $\theta$ [a population quantity]? Are they close (if not equal) in some sense? The sample variance $s^2$ is clearly an average of squared deviations of $x_i$'s from their mean. But, what kind of information does $s^2$ provide?... These questions cannot be answered in descriptive data analysis.

Other remarks about the sample mean and sample variance formulas

Let $\bar{X}_n$ and $S^2_n$ denote the sample mean and sample variance formulas provided earlier. The following are properties of these estimators:

They are unbiased for $\mu$ and $\sigma^2$, as explained earlier. This is a relatively simple probability exercise.

They are consistent for $\mu$ and $\sigma^2$. Since you know measure theory, assume all random variables are defined over a probability space $(\Omega, \mathcal{F}, P)$. It follows that $\bar{X}_n \overset{P}{\to} \mu$ and and $S^2_n \overset{P}{\to} \sigma^2$, where $\overset{P}{\to}$ denotes convergence in probability, also known as convergence with respect to the measure $P$. See https://math.stackexchange.com/a/1655827/81560 for the sample variance (observe that the estimator with the $n$ in the denominator is used here; simply multiply by $\dfrac{n-1}{n}$ and apply a result by Slutsky) and Proving a sample mean converges in probability to the true mean for the sample mean. As a stronger result, convergence is almost sure with respect to $P$ in both cases (Sample variance converge almost surely).

If one assumes $X_1, \dots, X_n$ are independent and identically distributed based on a normal distribution with mean $\mu$ and variance $\sigma^2$, one has that $\dfrac{\sqrt{n}(\bar{X}_n - \mu)}{\sqrt{S_n^2}}$ follows a $t$-distribution with $n-1$ degrees of freedom, which converges in distribution to a normally-distributed random variable with mean $0$ and variance $1$. This is a modification of the central limit theorem.

If one assumes $X_1, \dots, X_n$ are independent and identically distributed based on a normal distribution with mean $\mu$ and variance $\sigma^2$, $\bar{X}_n$ and $S^2_n$ are uniformly minimum-variance unbiased estimators (UMVUEs) for $\mu$ and $\sigma^2$ respectively. It also follows that $\bar{X}_n$ and $S^2_n$ are independent, through - as mentioned by Michael Hardy - showing that $\text{Cov}(\bar{X}_n, X_i - \bar{X}_n) = 0$ for each $i = 1, \dots, n$, or as one can learn from more advanced statistical inference courses, an application of Basu's Theorem (see, e.g., Casella and Berger's Statistical Inference).


问:
Can I convert $\min_x\max_y: f(x,y)$ to $\max_x\min_y: -f(x,y)$?
Can I convert $\min_x\max_y: f(x,y)$ to $\max_x\min_y: -f(x,y)$? There are lots of methods for minmax problem. So I think there should be a way to convert maxmin problem to minmax problem.

答:
Use the fact that $\min(A)=-\max(-A)$ for any set $A$ of real numbers (where, obviously, $-A=\left\{-a:a\in A\right\}$. [If you don't know that already, try to prove it yourself.]

\begin{align*}\min_x\max_y f(x,y)&=-\max_x(-\max_y f(x,y))\\&=-\max_x(\min_y-f(x,y))\end{align*}

so the process of "maximizing $f(x,y)$ for $y$ and then minimizing for $x$" is the same as "minizing $-f(x,y)$ for $y$ and then maximizing for $x$".

The only deal is that the results are opposites.

More formally, a point $(x_0,y_0)$ such that $f(x_0,y_0)=\min_x\max_y f(x,y)$ will also yield $-f(x_0,y_0)=\max_x\min_y-f(x,y)$ and vice-versa.


问:
$(X,\tau)$ a topological space with $X\neq \emptyset$ and $D = \{D \subseteq X : D$ is dense in $(X, \tau) \}$.
Be $(X,\tau)$ a topological space with $X\neq \emptyset$ and $D = \{D \subseteq
X : D$ is dense in $(X, \tau) \}$.

(a)Prove that it is a discrete topology over X if, and only if, $D = \{X\}$.

Definition. The discrete topology is formed by the parts of X, i.e, ℘(X).

From the imprint that only set X is dense in itself, and no other subset of $X$ is dense in $X$.

(b) Prove that τ is the chaotic topology on X if, and only if, $D$ =
℘(X) - {∅}.

Definition. The caotica topology consists of: $\{\emptyset, X\}$

Can you help me, please?

答:
Adding to Brian M. Scott Answer:
Claim: In (a) part, Converse part:
Suppose $D=\{X\}$ and Let there is some subset of $X$ not open in $X$ , let's say $\{x\}$
then,
Claiming $X\setminus\{x\}$ is dense in $X$ :
Then for every $U$ subset of $\tau$ such that
if $x \in U$ then $U\cap(X\setminus\{x\})$ not equal to $\phi$ as $\{x\}$ is not open in $\tau$.
So, clearly $X\setminus\{x\}$ is dense in $X$.


答:
The first definitions you gave are correct and standard, and statisticians and data scientists will agree with this. (These definitions are given in statistics textbooks.) The second set of quantities you described are called the "sample mean" and the "sample variance", not mean and variance.

Given a random sample from a random variable $X$, the sample mean and sample variance are natural ways to estimate the expected value and variance of $X$.

Title: What is the difference between "expectation", "variance" for statistics versus probability textbooks?

Body:
It seems that there are two ideas of expectation, variance, etc. going on in our world.

In any probability textbook:

I have a random variable $X$, which is a function from the sample space to the real line. Ok, now I define the expectation operator, which is a function that maps this random variable to a real number, and this function looks like,
$$\mathbb{E}[X] = \sum\limits_{i = 1}^n x_i p(x_i)$$ where $p$ is the probability mass function, $p: x_i \mapsto [0,1], \sum_{i = 1}^n p(x_i) = 1$ and $x_i \in \text{range}(X)$. The variance is,
$$\mathbb{E}[(X - \mathbb{E}[X])^2]$$

The definition is similar for a continuous RV.

However, in statistics, data science, finance, bioinformatics (and I guess everyday language when talking to your mother)

I have a multi-set of data $D = \{x_i\}_{i = 1}^n$ (weight of onions, height of school children). The mean of this dataset is

$$\dfrac{1}{n}\sum\limits_{i= 1}^n x_i$$

The variance of this dataset (according to "science buddy" and "mathisfun dot com" and government of Canada) is,

$$\dfrac{1}{n}\sum\limits_{i= 1}^n(x_i - \sum\limits_{j= 1}^n \dfrac{1}{n} x_j)^2$$

I mean, I can already see what's going on here (one is assuming uniform distribution), however, I want an authoritative explanation on the following:

Is the distinction real? Meaning, is there a universe where expectation/mean/variance... are defined for functions/random variables and another universe where expectation/mean/variance... are defined for raw data? Or are they essentially the same thing (with hidden/implicit assumption)

Why is it no probabilistic assumption is made when talking about mean or variance when it comes to dealing with data in statistics or data science (or other areas of real life)?

Is there some consistent language for distinguishing these two seemingly different mean and variance terminologies? For example, if my cashier asks me about the "mean weight" of two items, do I ask him/her for the probabilistic distribution of the random variable whose realization are the weights of these two items (def 1), or do I just add up the value and divide  (def 2)? How do I know which mean the person is talking about?/


问:
Redefining a set of transformations as a single linear transformation using tensor product?
I learned today that a multilinear map $A \times A \rightarrow B$ can be re-defined as a single linear map $A \otimes  A \rightarrow B$, where $\otimes$ is the tensor product.

I have a set of transformations in an iterative algorithm, and I would like to determine if they collectively count as a multilinear map, in order to "reduce" the multiple transformations to one linear map. Here are the set of transformations I want to represent as one transformation:

$\mathbf{y}^{t}_d = \mathbf{y}^{t-1}_d \odot (W \cdot \left(\mathbf{y}^{t}_{d-1} \oslash (W^{\text{T}} \cdot \mathbf{y}^{t-1}_d)\right) + \left( (V \cdot \mathbf{y}^{t-1}_{d+1}) \oslash \mathbf{y}^{t-1}_d\right))$

This can be rewritten as a function $\mathbf{y}^{t}_d = \mathbf{f}(\mathbf{y}^{t-1}_d, \mathbf{y}^{t}_{d-1}, \mathbf{y}^{t-1}_{d+1}, W, W^{\text{T}}, V)$. This function involves three linear transformations, $W, W^\text{T}, V$. My goal is to determine whether I can use a "tensor product trick" to rewrite $\mathbf{f}$ as a function of a single linear transformation, $U$. That is, I want to find $U$ such that:

$\mathbf{y}^{t}_d = \mathbf{f}(\mathbf{y}^{t-1}_d, \mathbf{y}^{t}_{d-1}, \mathbf{y}^{t-1}_{d+1}, W, W^{\text{T}}, V) = \mathbf{g}(\mathbf{y}^{t-1}_d, \mathbf{y}^{t}_{d-1}, \mathbf{y}^{t-1}_{d+1}, U)$

Is this possible? Am I even on the right track with this sort of idea?

EDIT: $\odot$ and $\oslash$ represent element-wise multiplication and division respectively. $\mathbf{y}^{t}_{d-1}, \mathbf{y}^{t}_{d}, \mathbf{y}^{t}_{d+1}$ are vectors in three different vector spaces. $t$ represents the iteration during which the vector was computed.

答:
I think the answer is "no", it is not multi-linear for a couple reasons.

(1) If you look at the last term in your equation, the vector $V \cdot y^{t-1}_{d+1}$ is being both multiplied and divided (component-wise) by $y^{t-1}_d$. The multiplication and division cancel, so that term actually does not depend at all on $y^{t-1}_d$. Thus it is not multilinear.

For a simpler example, think of $f(x,y) = xy + x$. Then$f(x,y_1+y_2) = xy_1 + xy_2 + x$, but on the other hand, $f(x,y_1)+f(x,y_2) = xy_1+xy_2 + 2x$, which is different. They are not the same because the second term does not depend on $y$. This is what is happening in your function.

(2) Also, the component-wise division is not a linear operation, so you shouldn't expect your function to be multi-linear because of that. Consider just $v,w \in \Bbb{R}^2$ given by $v = \binom{x}{y}$ and $w = \binom{a}{b}$. Then
$$ v \oslash w = \begin{pmatrix} x/a \\ y/b \end{pmatrix} $$
It is linear in $v$, but not in $w$:
$$ v \oslash (w_1+w_2) = \begin{pmatrix}\frac{x}{a_1+a_2} \\ \frac{y}{b_1+b_2} \end{pmatrix} $$
This is definitely not the same as $v \oslash w_1 + v \oslash w_2$.


问:
Hausdorff property is not homotopy invariant
In Topology: A Categorical Approach it's stated that the Hausdorff property is topological but not homotopy invariant. The only example that I know of a "naturally occurring" non-Hausdorff space is the line with two origins, and the only thing I can think to do is use the quotient procedure that created it, but I don't think that produces a homotopy equivalence. Is there an elementary example of a homotopy equivalence between a Hausdorff and non-Hausdorff space? Or am I wrong in thinking that the quotient procedure $(x,a) \sim (x,b)$ if $x \neq 0$ does not give a homotopy equivalence?

答:
Here's an example of a non-Hausdorff space:

Let $X$ be your favourite nontrivial Hausdorff space. Let $Y=X\times\left\{1,2\right\}$ be the product space of $X$ with the chaotic topological space $\left\{1,2\right\}$, where the only open subsets of $\left\{1,2\right\}$ are $\varnothing$ and $\left\{1,2\right\}$ itself.

More informally, $Y$ consists of two copies of $X$, namely $X_1=X\times\left\{1\right\}$ and $X_2=X\times\left\{2\right\}$. Obviously, $Y$ is not Hausdorff.

But $X$ and $Y$ are homotopy invariant: Take the maps $f\colon X\to Y$ which maps $X$ to the first copy of it inside $Y$: $f(x)=(x,1)$; and let $g\colon Y\to X$ maps whatever element of any copy of $X$ to the same element of $X$: $g(x,i)=x$.

Then $gf=\operatorname{id}_X$, and the function
$$H\colon Y\times[0,1]\to Y,\qquad H((x,i),t)=\begin{cases}(x,1)&\text{ if }t=0\\(x,i)&\text{ if }t>0\end{cases}$$
is a homotopy from $fg$ to $\operatorname{id}_Y$. [The only nontrivial part is continuity of $H$: Note that $H^{-1}(U)=U\times[0,1]$ for all $U\subseteq Y$ open.]


问:
Does the notion of average rate of change make sense for discontinuous functions?
The "slope of the secant line" interpretation makes perfect sense for continuous functions and anyone familiar with the derivative knows its applications. For example, if $f(x)=x^2$, then its average rate of change (AROC) over the interval $[1, 2]$ is $f_{\rm AROC}=\frac{f(2)-f(1)}{2-1}=3$, and this can be used to roughly approximate the derivative of $f$.

But what about a discontinuous function, such as $g(x)=\begin{cases}0&\text{for }x<\frac12\\1&\text{for }x\ge\frac12\end{cases}$ over $[0, 1]$, or $h(x)=\frac1x$ over $[-1, 1]$? Does the notion of AROC make sense, so that $g_{\rm AROC}=\frac{1-0}{1-0}=1$ and $h_{\rm AROC}=\frac{1-(-1)}{1-(-1)}=1$?

I cannot seem to find a formal definition for the AROC, although some textbooks and calculators like Wolfram|Alpha express the AROC of $f(x)$ in terms of the average value of $f'(x)$, e.g.

$$f(x)=x^2,x\in[1,2]\\
\implies f_{\rm AROC}=\frac1{2-1}\int_1^2\frac{\mathrm d(x^2)}{\mathrm dx}\,\mathrm dx=2\int_1^2x\,\mathrm dx=3$$

and this discussion usually accompanies the mean value theorem, which does not apply to discontinuous functions.

答:
The best I can come up with is defining $\lambda(x,y)=\frac{f(y)-f(x)}{y-x}$ and averaging it, using $$f_{AROC}=\frac{1}{\#U}\iint_U\lambda(x,y) dy dx$$

where I use $\#U$ to represent the length, area, volume, size etc of $U$.

For example, when $f: [1,2]\to [2,4] ; f(x)=x^2$,
then $\lambda : [1,2]\times [1,2] \mapsto [2,4]; \lambda(x,y)=x+y$, and $$f_{AROC}=\frac{1}{1\cdot 1}\int_1^2\int_1^2 (x+y) dy dx =3$$

Applying this to $g$, we observe that: $$\lambda_g(x,y)=\begin{cases}\frac{-1}{y-x} & x\geq \frac 12, y< \frac 12 \\ \frac{1}{y-x} & x<\frac12, y\geq \frac12 \\ 0& \text{otherwise} \end{cases}$$

Again the area is $1\cdot 1=1$, and $$g_{AROC}=\bigg[\int_0^\frac 12 \int_\frac 12^1 \frac{1}{y-x}dy dx+\int_\frac12^1 \int_0^\frac 12 \frac{-1}{y-x} dy dx\bigg]=2\ln2$$

As seen here, note both double integrals are equal (but nasty!)

As for $h$, that integral does not converge, as is to be expected, due to the graph's asymptote at $x=0$


问:
How do I find the series expansion of the meromorphic function $\frac{1}{e^z+1}$?
in a theoretical physics book, the author makes the following claim:

$$\frac{1}{e^z + 1} = \frac{1}{2} + \sum_{n=-\infty}^\infty \frac{1}{(2n+1) i\pi - z}$$
and justifies this as

These series can be derived from a theorem which states that any meromorphic function may be expanded as a summation over its poles and residues at those poles

What's the name of that theorem? It's not really a Laurent series, since the Laurent series is for an expansion around one particular point only. I can see that the poles occur whenever $z = (2n+1)i\pi$ for $n \in \mathbb{N}$, but then where does that constant $1/2$ come from?

EDIT: Well, it appears that the general claim isn't valid, so now I'd be interested in a justification for the expansion in my particular example...

答:
This seems to be some kind of application of the Residue theorem: https://en.wikipedia.org/wiki/Residue_theorem. Here, given $z_0$ at which $f$ is holomorphic, define $$g(z) = \frac{f(z)}{z - z_0}.$$ Then $g$ has poles where $f$ does, and of the same residue, but also has one extra pole, at $z_0$, whose residue we can compute via Cauchy's integral theorem to be exactly $f(z_0)$. The residue at the pole $p$ can be computed by taking $$\lim_{z\rightarrow p} zg(z) = \frac{\mbox{Res}(f,p)}{p - z_0}.$$

Now, choose a finite set of poles and your favorite point $z_0$ and a contour $\gamma$ around them so that the winding numbers are all 1 (a large box will do). Integrating around the contour will give, by the residue theorem and the above calculations:

$$f(z_0) = \frac{1}{2\pi i}\int_\gamma g(z) dz - \sum_{p \mbox{ inside } \gamma} \frac{Res(f,p)}{p - z_0}.$$

Now, the catch is to "let $\gamma$ go to infinity", to get an infinite sum. As already pointed out, there are some issues with that, so more careful estimates are needed, and these answer the question of "where did that constant come from"? I unfortunately don't have time to carry out those right now, but I will update this answer with those estimates once I have them.


问:
What is the error of FEM-like spline discretizations?
Let's have $f(x)$ a nice*, scalar valued function. It is approximated by
$$
f(x)\approx g(x)=\sum_i f(x_i) N_i(x),
$$
where

$x_i = x_0+i \Delta x$ are uniformly spaced points,

$N_i(x) = N(x-x_i)$, and

$N$ is a linear, quadratic, or cubic spline that is a partition of unity, and has finite support.

Based only this information, what can be said about bounds for $|f(x)-g(x)|$,  $|f'(x)-g'(x)|$, and $|f''(x)-g''(x)|$?

What if $f$ is a degree $n$ polynomial? Can these approximations be exact?

Note:

Based on $g$'s definition
$$
f'(x)\approx g'(x)=\sum_i f(x_i) N_i'(x)\\
f''(x)\approx g''(x)=\sum_i f(x_i) N_i''(x)
$$

*let's say at a minimum twice continuously differentiable, but it's fine if more strict requirements are needed of the analysis.

答: Look in Carl deBoor’s book “A Practical Guide to Splines”. The error bounds are typically obtained by truncating Taylor series expansions. This means that the error for cubic spline approximation (for example) is related to the maximum value of the fourth derivative of $f$.


问:
The shape of the Rational Basis Functions and the Basis Functions for generating NURBS
Here is the definition of the basis functions:

\begin{align}
        N_{i,0}(u) = &
        \begin{cases}
            1 & \text{se } u_i \leq u < u_{i+1} \\
            0 & \text{caso contrário}
        \end{cases}\label{eq:b-spline-inicial}\\ %\notag\\
        N_{i,k}(u) = & \, \frac{u - u_i}{u_{i+k} - u_i}N_{i,k-1}(u) + \frac{u_{i + k + 1} - u}{u_{i + k + 1} - u_{i + 1}}N_{i+1, k-1}(u) \label{eq:b-spline-recursiva}
\end{align}

wich generates the following Basis functions of degree $2$ for the Knot Vector $U = \{0, 1, 2, 3, 4, 5, 6, 7 \}$:

And here is the definition of the rational basis functions:

$$
R_{i,k}(u) = \frac{N_{i,k}(u)w_i}{\sum\limits_{j=0}^n N_{j,k}(u)w_j}
$$

wich generates the following Rational Basis functions of degree $2$ for the same Knot Vector $U = \{0, 1, 2, 3, 4, 5, 6, 7 \}$ and weigth vector $W = \{1, 1, 1, 1, 1\}$:

Now my question:

The "NURBS Book" and "An introduction to NURBS" say that NURBS curves (rational) are a generalization of the B-Spline curves (polynomial)... but for periodic knot vectors the basis functions aren't the same... they are the same only for open knot vectors. My question is: have I implemented something wrong?

答:
Yes, you have implemented something wrong.

Since your weights are all equal to $1$, the rational basis functions will be the same as the polynomial ones (just set all $w_i=1$ in your second formula).


问:
1, 333, 55555, .... find an expresion for the general term
$$1, 333, 55555, ....$$

My work:

If $n\in [u^2+1, (u+1)^2]$,

then $a_n = 2u+1$ for $u=0,1,2,\ldots$

This seems to work but is it possible to have a nice expression for the general term without conditions?

答:
Here are some ideas:

If you're allowed to use the sum:

$$(2n + 1 - 10\lfloor\frac{2n + 1}{10}\rfloor)\sum_{i=0}^{2n + 1}10^i$$

The sum can be rewritten as:

$$(2n + 1 - 10\lfloor\frac{2n + 1}{10}\rfloor)(10^{2(n + 1)} - 1)$$

The first term can also be rewritten:

$$(10\left\{\frac{2n + 1}{10}\right\})(10^{2(n + 1)} - 1)$$

In each form the first term extracts consecutive odd numbers in a circular way: $1, 3, 5, 7, 9, 1, 3, \dots$ whereas the second term is of the form $11\dots1$.

You can mix them as you want.


答: If I am interpreting the query correctly, you want $\left(2 \times \lceil\sqrt{n}\rceil\right) - 1,$ where $\lceil x\rceil$ is the ceiling function (smallest integer $\geq x$).

Title: 1, 333, 55555, .... find an expresion for the general term

Body:
$$1, 333, 55555, ....$$

My work:

If $n\in [u^2+1, (u+1)^2]$,

then $a_n = 2u+1$ for $u=0,1,2,\ldots$

This seems to work but is it possible to have a nice expression for the general term without conditions?


问:
Representations of signed areas
I am studying differential forms and have difficulty with the following question：Suppose $v_1,v_2$ are two vectors in $\mathbb{R}^n$, $S$ is the triangle spanned by $v_1,v_2$ and $S_1$ is the projection of $S$ onto the plane $x_1Ox_2$.Then how to use $v_1,v_2$ to represent the signed area of $S_1$? And this induces a 2-form.

答: (Signed) area of $S_1$ $=dx_1\wedge dx_2(v_1,v_2)$

Title: Representations of signed areas


问:
A Computation in Polar Coordinates
Let $B_t$ be a ball of radius t. There's a function $u$ that's defined everywhere. I'm reading a calculation that makes the following equality:
$$\int_o^t\int_{\partial B_r}\chi_{\left\{u\left(t \frac{x}{\lvert x \rvert} \right)> 0 \right\}} \, d\mathcal{H}^{n-1}(x)\,dr = \frac t n \int_{\partial B_t}\chi_{\left\{u> 0 \right\}} \, d\mathcal{H}^{n-1}$$
I don't exactly follow the computation, I thought $\frac t n$ should be $\frac{t^{n}}n$ but I read this equality elsewhere in the paper so I don't think its a typo. Perhaps its the change of variables $x \rightarrow \frac x t$ but I'm having trouble getting this to work because this conversion is cartesian, not polar.

答:
Transform $x\in\partial B_r\mapsto \frac{t}rx=t\frac{x}{\lvert x\vert}\in\partial B_t$.  So $$\int_{\partial B_r}f\left(t\frac{x}{\lvert x\rvert}\right)\,\mathrm{d}\mathcal{H}^{n-1}(x)=\int_{\partial B_t}f(y)\,\left(\frac{r}t\right)^{n-1}\,\mathrm{d}\mathcal{H}^{n-1}(y)$$
and hence
$$
\begin{aligned}
\int_0^t\int_{\partial B_r}f\left(t\frac{x}{\lvert x\rvert}\right)\,\mathrm{d}\mathcal{H}^{n-1}(x)
&=\int_0^t\int_{\partial B_t}f(y)\,\left(\frac{r}t\right)^{n-1}\,\mathrm{d}\mathcal{H}^{n-1}(y)\,\mathrm{d}r\\
&=\int_{\partial B_t} f(y)\,\mathrm{d}\mathcal{H}^{n-1}(y)
\underbrace{\int_0^t\left(\frac{r}t\right)^{n-1}\,\mathrm{d}r}_{t/n}.
\end{aligned}
$$

Title: A Computation in Polar Coordinates


问:
Proving the Variance of the standard normal distribution
I am trying to prove the variance of the standard normal distribution $\phi(z)=\frac{e^{-\frac{1}{2}z^2}}{\sqrt{2\pi}}$ using high school level mathematics only. The proof given in my textbook seems wrong to me. Here is what it says:

Because the mean is zero, the variance is given by the integral

$\operatorname{Var}(Z)=\int_{-\infty}^{\infty} z^{2} \phi(z) d z$

We showed above while finding the second derivative of $\phi(z)$ that

$\phi^{\prime \prime}(z)=\phi(z)\left(z^{2}-1\right)$

and rearranging, $z^{2} \phi(z)=\phi^{\prime \prime}(z)+\phi(z)$

Hence $\operatorname{Var}(Z)=\int_{-\infty}^{\infty} z^{2} \phi(z) d z$

$=\int_{-\infty}^{\infty} \phi^{\prime \prime}(z) d z+\int_{-\infty}^{\infty} \phi(z) d z$

$=\left[\phi^{\prime}(z)\right]_{-\infty}^{\infty}+\int_{-\infty}^{\infty} \phi(z) d z$

$=0+1$

$=1$

The first integral above is zero, because the integrand $\phi^{\prime}(z)=-z \phi(z)$ is odd, as we saw before. The second integral above is 1 because $\phi(z)$ is a probability density function.

The only part that I don't understand is the 3rd last line, where they say that $\left[\phi^{\prime}(z)\right]_{-\infty}^{\infty}=0$

Because indeed $\phi^{\prime}(z)=-z \phi(z)$, so

$\left[\phi^{\prime}(z)\right]_{-\infty}^{\infty}$

$=\left[-z \phi(z)\right]_{-\infty}^{\infty}$

$=-\{\infty \phi(\infty)+\infty  \phi(-\infty)\}$

$=-(\infty \times 0+\infty \times 0)$ which we can't say is equal to zero, right?

I think what the textbook is trying to say is that since the integrand $\phi^{\prime}(z)=-z \phi(z)$ is odd, $\int_{-\infty}^{\infty} \phi^{\prime}(z) d z=0$, but clearly the integrand is $\phi^{\prime \prime}(z)$ not $\phi^{ \prime}(z)$, and $\phi^{\prime \prime}(z)$ which is equal to $(z^2-1)\phi(z)$ is not odd.

Please let me know if I am missing something here.. And if the textbook is wrong, is there another way to prove the variance of the standard normal distribution using high school level mathematics only? (I had a look around here and saw the use of Gamma function, which I don't know what it is..)

答: Actually, $\lim_{z\to\pm\infty}z\phi(z)=0$ because $\phi(z)$ decays exponentially at infinity. You can use L'Hopital's rule to see that the limit is zero. Therefore, $[\phi'(z)]|_{-\infty}^\infty=0$ and the argument in your book is correct.


问:
Compute lower bound for standard normal tail
Let $X\sim N(0,1)$. I want to prove the next innequality holds for $x\geq 0$:

$$\left(\dfrac{x}{x^2+1}\right)f(x)\ \leq \ \mathbb{P}[X\geq x]$$

where $f(x)$ is the pdf of $X$. I've already read a proof here, but I'm having trouble with the calculation.

The idea is to proof $g(x) = \mathbb{P}[X\geq x] - \left(\frac{x}{x^2+1}\right)f(x) > 0$ for $x\geq 0$. The proof for $x=0$ it's easy, but the problem comes when assuming $x>0$. At one step, the author takes the derivative of $g(x)$ to conclude $g(x)$ is positive, which results:

$$g^{\prime}(x) = -2 \dfrac{e^{-x^2/2}}{(x^2 + 1)^2}.$$

My problem comes when I compute the derivative trying to reach that result: computing directly the derivative my problem comes in the term

$$\dfrac{d}{dx} \mathbb{P}[X\geq x] = \dfrac{d}{dx}\int_{x}^{\infty}\dfrac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}}dt,$$

I try to apply Leibniz integral rule but the upper limit of integral was a problem, and sincerely, although that derivative wasn't a problem, I have no idea how the author reach that expression since

$$\dfrac{d}{dx}\left[\left(\frac{x}{x^2+1}\right)f(x)\right] = -\dfrac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} \dfrac{x^4 + 2 x^2 - 1}{(x^2 + 1)^2}.$$

Having this trouble with this approach to the proof, I found that another way to prove the bound was answered here where the next innequality holds since $x\leq t$:

$$\dfrac{1}{x^2}\int_{x}^{\infty} e^{-\frac{t^2}{2}}dt \geq \int_{x}^{\infty}\dfrac{1}{t^2}e^{-\frac{t^2}{2}}dt,$$

the suggestion here was applying integration by parts to the right side of the innequality, so, I consider:

$$u = e^{-\frac{t^2}{2}}, \quad dv = \dfrac{1}{t^2}dt$$

but then again I'm having trouble trying to compute the integral since:

$$\int \dfrac{1}{t^2}e^{-\frac{t^2}{2}}dt = -\dfrac{e^{-\frac{t^2}{2}}}{t} - \int e^{-\frac{t^2}{2}}dt$$

and I'm stuck trying to understand how can I reach that this equality hold

$$-\dfrac{e^{-\frac{t^2}{2}}}{t}\bigg\vert_{t=x}^{\infty} - \int_{t=x}^{\infty} e^{-\frac{t^2}{2}}dt = \dfrac{1}{x(x^2 + 1)}e^{-\frac{x^2}{2}}$$

since that will finish the alternative proof. Any help to conclude any of the calculations here would be appreciated.

答: $\frac d {dt} [\int_t^{\infty} e^{-x^{2}/2}-\frac t {1+t^{2}} e^{-t^{2}/2}]=-e^{-t^{2}/2}+\frac {t^{2}} {1+t^{2}} e^{-t^{2}/2}-e^{-t^{2}/2} \frac {(1+t^{2})-t(2t)} {(1+t^{2})^{2}}$ by product rule and quotient rule. A little simplification gives exactly the derivative quoted in the link.


问:
Question Regarding $\int_0^1\int_0^∞ e^{-xy}-xye^{-xy}dxdy$
Specifically for $\int_0^{\infty}e^{-xy}-xye^{-xy}dx$,
I see that this inner integral evaluates to $xe^{-xy}$ from x = zero to x = infinity, which after plugging in, comes out to be zero (according to my textbook). However, I stuck on how to evaluate $xe^{-xy}$  for x = infinity. I tried to solve it by rephrasing it to $$\lim \limits_{x \to {\infty}}\frac{x}{e^{xy}}$$
But I realized there is no l'hopital's rule for multiple variables.

Any hints or answers would be greatly appreciated. Thanks.

Body:
$$I=\int_{0}^{1} \int_{0}^{\infty} (1-xy)^{-xy}dx~dy$$
Let $xy=t$,then
$$I=\int_{0}^{1}\frac{dy}{y}\int_{0}^{\infty} (1-t)e^{-t} dt=0$$


问:
Cannot find the inverse of Z-transform.
Find the invers of $Z$-transform of
$X(z)=\dfrac{z}{(z-2)(z^2+6z+9)}$.

I try as below.

Let
\begin{align*}
  X(z)&=\dfrac{z}{(z-2)(z^2+6z+9)}
  =\dfrac{z}{(z-2)(z+3)^2}\\
  &=\dfrac{A}{z-2}+\dfrac{Bz+C}{(z+3)^2}\\
  &=\dfrac{A(z+3)^2+(Bz+C)(z-2)}{(z-2)(z+3)^2}\\
  &=\dfrac{A(z^2+6z+9)+Bz^2-2Bz+Cz-2C}{(z-2)(z+3)^2}\\
  &=\dfrac{(A+B)z^2+(6A-2B+C)z+(9A-2C)}{(z-2)(z+3)^2}
 \end{align*}

Now we have system of linear equation
\begin{align*}
 A+B&=0\\
 6A-2B+C&=1\\
 9A-2C&=0
\end{align*}
and now we have $A= \dfrac{2}{25}$, $B= -\dfrac{2}{25}$, and $C= \dfrac{9}{25}$.

We have
\begin{align*}
 X(z)&=\dfrac{2}{25}\cdot \dfrac{1}{z-2}+\dfrac{-\dfrac{2}{25}z+\dfrac{9}{25}}{(z+3)^2}\\
 &= \dfrac{2}{25}\cdot \dfrac{1}{z-2}-\dfrac{2}{25} \dfrac{z}{(z+3)^2}+\dfrac{9}{25} \dfrac{1}{(z+3)^2}.
\end{align*}

Now I don't know what the inverse Z-transform of $\dfrac{1}{z-2}$, $\dfrac{z}{(z+3)^2}$, and $\dfrac{1}{(z+3)^2}$.

I just have this Z-transform table

Any one can give me hint for this problem?

Title: Cannot find the inverse of Z-transform.

Body:
Hints:
$$\frac{1}{z-2} = z^{-1}\cdot\frac{z}{z-2}$$

$$\frac{z}{(z+3)^2} = \frac{-1}3\cdot\frac{(-3z)}{(z-(-3))^2}$$

$$\frac{1}{(z+3)^2} = \frac{-z^{-1}}{3}\cdot\frac{(-3z)}{(z-(-3))^2}$$

Then use shifting property

$$\mathcal{Z}\{x(n-m)\} = z^{-m}X(z) $$


问:
Determine if a transformation is self-adjoint with respect to a non-orthonormal basis
I have matrix that represent $T:V \to V$ (linear map over $\mathbb{R}$) according to basis $E$.

$E$ is not an orthonormal set.

how can I know if this $T$ is self-adjoint.

I know that if $E$ was orthonormal basis we would take the transpose matrix.

but what about here ?

the matrix is :
$$\begin{bmatrix}
1 & 2 \\
2 & 1 \\
\end{bmatrix}$$

答:
@SandraWest, so we need to compute. (Or use the matrix short-cut in my comment.) Remember that $T(v_1)=v_1+2v_2$ and $T(v_2)=2v_1+v_2$. It suffices (by linearity) to check whether $\langle Tx,y\rangle=\langle x,Ty\rangle$ when $x$ and $y$ are basis vectors. You can see the only thing that really requires checking is whether $\langle Tv_1,v_2\rangle = \langle v_1,Tv_2\rangle$.

Well,
\begin{align*}
\langle Tv_1,v_2\rangle & =\langle v_1+2v_2,v_2\rangle=−1+2=1, \quad\text{and} \\
\langle v_1,Tv_2\rangle &= \langle v_1,2v_1+v_2\rangle = 4−1=3 \,,
\end{align*}

so T is not self-adjoint.


问:
Domain of function $x^{1/2}$
What different of 2 function: $\sqrt{x}$ and $x^{1/2}$? Why domain of $\sqrt{x}$ is $x \ge 0$ but domain of $x^{1/2}$ is $x>0$?

答: The two functions are actually the same. Maybe you're talking about $x^{-1/2}$, whose domain is $x>0$, instead of $x^{1/2}$.

Title: Domain of function $x^{1/2}$


问:
Checking convergence of an increasing sequence: hint
It is known (see this link ) that $(1\sqrt{1}) + (1/\sqrt{2})+\cdots + (1/\sqrt{n})\ge \sqrt{n}$ for integers $n\ge 1$.

I took the sequence $x_n=(1\sqrt{1}) + (1/\sqrt{2})+\cdots + (1/\sqrt{n}) -\sqrt{n}$.

First, we can see that this is increasing sequence
$$
x_{n+1}-x_n = \frac{1}{\sqrt{n+1}}-\sqrt{n+1}+\sqrt{n}
$$
and $\frac{1}{\sqrt{n+1}} \ge \sqrt{n+1}-\sqrt{n}$.

I do not know whether sequence is bounded above or not. Can anyone suggest for it, how to proceed for boundedness and so for convergence.

Body:
Hint - notice that:

$$
x_{n+1}-x_n = \frac{1}{\sqrt{n+1}}-\sqrt{n+1}+\sqrt{n} = \frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+1}+\sqrt{n}}
$$


问:
Quick doubt about chinese theorem: Giving the wrong answer
i was doing an exercise, where I ended up with this equation:

$p = 1 mod (5)$

$p = 1 mod (2)$

I was trying to use Chinese theorem to see what number p takes form, but it is giving a wrong answer:
$p = \sum a^{i}x^{i}M^{i}$

$a^{i} = x^{i} = 1$

$p = \sum M^{i}$

$ p = 5 + 2 = 7$
:q

答:
Recall the procedure of CRT:

Given positive integers $n_1, \dots, n_k>1$ which are coprime to each other，define $N=\prod_{i=1}^k n_i$ and $N_i=\frac{N}{n_i}$. Given integers $a_1, \dots, a_k$, then the following system of coresidual equations
$$
x\equiv a_1 \pmod{n_1}\\ x\equiv a_2 \pmod{n_2}\\ \vdots\\ x\equiv a_k\pmod{n_k}
$$
have one unique solution when $0\le x < N$.
Since $\gcd(n_i, N_i)=1\quad \forall i\in\{1, 2, \dots, k\}$, from Bézout's identity, we have for each $i\in\{1, \dots, k\}$, there exists integer $M_i, m_i$ s.t.
$$
M_iN_i+m_in_i=1
$$
solve the equations above for $M_1, \dots, M_k$, then one special solution of this system of coresidual equation can be represented as:
$$
x_0=\sum_{i=1}^k a_i M_iN_i
$$
And general solutions are
$$
x=Nq+x_0,q\in\mathbb Z
$$

In your case, it seems that you have the wrong $x^i$.

What we are to solve is
$$\begin{cases}
2M_1+5m_1=1
\\
5M_2+2m_2=1\end{cases}
$$
Apply Extended Euclidean algorithm, we have
$$
\begin{cases}
 M_1=3,m_1=-1\\
 M_2=-1,m_2=3
\end{cases}
$$
Thus a special solution can be calculated:
$$
x_0=\sum a_iM_iN_i=1\times 3\times 2+1\times (-1)\times 5=1
$$
And all solutions are of the form
$$
x = Nq+x_0,q\in\mathbb Z
$$


答: $x_n \to \infty$. For this note that $\int_k^{k+1} \frac 1 {\sqrt x} dx  <\frac 1 {\sqrt k}$ Sum this from $k=1$ to $n-1$ to see that $x_n >\frac 1 {\sqrt n} +\sqrt n-2$. [Note that $\int_1^{n} \frac 1 {\sqrt x} dx=2\sqrt n -2$]

Title: Checking convergence of an increasing sequence: hint

Body:
It is known (see this link ) that $(1\sqrt{1}) + (1/\sqrt{2})+\cdots + (1/\sqrt{n})\ge \sqrt{n}$ for integers $n\ge 1$.

I took the sequence $x_n=(1\sqrt{1}) + (1/\sqrt{2})+\cdots + (1/\sqrt{n}) -\sqrt{n}$.

First, we can see that this is increasing sequence
$$
x_{n+1}-x_n = \frac{1}{\sqrt{n+1}}-\sqrt{n+1}+\sqrt{n}
$$
and $\frac{1}{\sqrt{n+1}} \ge \sqrt{n+1}-\sqrt{n}$.

I do not know whether sequence is bounded above or not. Can anyone suggest for it, how to proceed for boundedness and so for convergence.


问:
How to prove that $\frac{\pi}{2}=\left[1, 1, \tfrac{1}{2},\tfrac{1}{3},\tfrac{1}{4},\ldots\right]$?
I don't know how to prove the generalized continued fraction
$\pi/2=[1;1/1,1/2,1/3,1/4,...]=1+\cfrac{1}{1/1+\cfrac{1}{1/2+\cfrac{1}{1/3+\cfrac{1}{1/4+\ddots}}}}$

It appears on wikipedia without proof, it contains the terms of the harmonic series but it seems not to be very well known.

I would appreciate any help, thanks.

答:
Matrix Representation of Convergents

One method to compute the convervgents of continued fractions uses matrices:
if $[c_0;c_1,c_2,\dots,c_n]=\frac{p_n}{q_n}$, then
$$
\begin{bmatrix}0&1\\1&0\end{bmatrix}
\prod_{k=0}^n\begin{bmatrix}0&1\\1&c_k\end{bmatrix}
=\begin{bmatrix}p_{n-1}&p_n\\q_{n-1}&q_n\end{bmatrix}\tag1
$$
Since $c_0=1$, and for $k\ge1$, $c_k=\frac1k$ we get
$$
\begin{bmatrix}1&1\\0&1\end{bmatrix}
\prod_{k=1}^n\begin{bmatrix}0&1\\1&\frac1k\end{bmatrix}
=\begin{bmatrix}p_{n-1}&p_n\\q_{n-1}&q_n\end{bmatrix}\tag2
$$

Closed Form

We will show by induction that
$$
\begin{bmatrix}1&1\\0&1\end{bmatrix}
\prod_{k=1}^{2n}\begin{bmatrix}0&1\\1&\frac1k\end{bmatrix}
=\begin{bmatrix}
\frac{4^n}{\binom{2n}{n}}&\frac{4^n}{\binom{2n}{n}}\\
\frac{\binom{2n}{n}2n}{4^n}&\frac{\binom{2n}{n}(2n+1)}{4^n}
\end{bmatrix}\tag3
$$
Note that $(3)$ is true for $n=0$. Assume $(3)$ is true for some $n$, then
$$
\begin{bmatrix}
\color{#C00}{\frac{4^n}{\binom{2n}{n}}}&\color{#C00}{\frac{4^n}{\binom{2n}{n}}}\\
\color{#090}{\frac{\binom{2n}{n}2n}{4^n}}&\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}
\end{bmatrix}
\begin{bmatrix}
0&1\\1&\frac1{2n+1}
\end{bmatrix}
=\begin{bmatrix}
\frac{4^n}{\binom{2n}{n}}&\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\\
\frac{\binom{2n}{n}(2n+1)}{4^n}&\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}
\end{bmatrix}\tag4
$$
The left column of the result is easy, the right column top element is
$$
\begin{align}
\color{#C00}{\frac{4^n}{\binom{2n}{n}}}\cdot1+\color{#C00}{\frac{4^n}{\binom{2n}{n}}}\cdot\frac1{2n+1}
&=\frac{4^n}{\binom{2n}{n}}\frac{2n+2}{2n+1}\tag{5a}\\
&=\frac{4^n}{\binom{2n}{n}}\frac{(2n+2)^2}{(2n+1)(2n+2)}\tag{5b}\\
&=\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\tag{5c}
\end{align}
$$
the right column bottom element is
$$
\begin{align}
\color{#090}{\frac{\binom{2n}{n}2n}{4^n}}\cdot1+\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\cdot\frac1{2n+1}
&=\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\tag6
\end{align}
$$
We can continue
$$
\begin{bmatrix}
\color{#C00}{\frac{4^n}{\binom{2n}{n}}}&\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\\
\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}&\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}
\end{bmatrix}
\begin{bmatrix}
0&1\\1&\frac1{2n+2}
\end{bmatrix}
=\begin{bmatrix}
\frac{4^{n+1}}{\binom{2n+2}{n+1}}&\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\\
\frac{\binom{2n}{n}(2n+1)}{4^n}&\color{#090}{\frac{\binom{2n+2}{n+1}(2n+3)}{4^{n+1}}}
\end{bmatrix}\tag7
$$
The left column of the result is easy, the right column top element is
$$
\begin{align}
\color{#C00}{\frac{4^n}{\binom{2n}{n}}}\cdot1+\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\cdot\frac1{2n+2}
&=\frac{4^n}{\binom{2n}{n}}+\frac{4^n}{\binom{2n}{n}}\frac4{\frac{(2n+1)(2n+2)^2}{(n+1)^2}}\tag{8a}\\
&=\frac{4^n}{\binom{2n}{n}}+\frac{4^n}{\binom{2n}{n}}\frac1{2n+1}\tag{8b}\\
&=\frac{4^n}{\binom{2n}{n}}\frac1{\frac{(2n+1)(2n+2)}{(2n+2)^2}}\tag{8c}\\
&=\color{#C00}{\frac{4^{n+1}}{\binom{2n+2}{n+1}}}\tag{8d}
\end{align}
$$
the right column bottom element is
$$
\begin{align}
\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\cdot1+\color{#090}{\frac{\binom{2n}{n}(2n+1)}{4^n}}\cdot\frac1{2n+2}
&=\frac{\binom{2n}{n}(2n+1)}{4^n}\frac{(2n+2)(2n+3)}{(2n+2)^2}\tag{9a}\\
&=\color{#090}{\frac{\binom{2n+2}{n+1}(2n+3)}{4^{n+1}}}\tag{9b}
\end{align}
$$
$(4)$ and $(7)$ show that $(3)$ is true for $n+1$.

Estimating the Convergents

$(1)$ and $(3)$ show that
$$
\frac{p_{2n-1}}{q_{2n-1}}=\frac1{2n}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\tag{10}
$$
$$
\frac{p_{2n}}{q_{2n}}=\frac1{2n+1}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\tag{11}
$$
$(10)$ and $(11)$ can be unified as
$$
\frac{p_n}{q_n}=\frac1{n+1}\left(\frac{4^{\lceil n/2\rceil}}{\binom{2\lceil n/2\rceil}{\lceil n/2\rceil}}\right)^2\tag{12}
$$
$(9)$ from this answer says that
$$
\pi\left(n+\tfrac14\right)\le\left(\frac{4^n}{\binom{2n}{n}}\right)^2\le\pi\left(n+\tfrac13\right)\tag{13}
$$
Since $\frac n2\le\lceil n/2\rceil\le\frac{n+1}2$, $(12)$ and $(13)$ show that
$$
\frac\pi2\left(1-\frac1{2n+2}\right)\le\frac{p_n}{q_n}\le\frac\pi2\left(1+\frac2{3n+3}\right)\tag{14}
$$
which, by the Squeeze Theorem, says that
$$
\bbox[5px,border:2px solid #C0A000]{\lim_{n\to\infty}\left[1;1,\tfrac12,\tfrac13,\dots,\tfrac1n\right]=\frac\pi2}\tag{15}
$$

Wallis Product

The Wallis Product for $\pi$ is
$$
\frac\pi2=\prod_{k=1}^\infty\frac{4k^2}{4k^2-1}\tag{16}
$$
Consider the partial products
$$
\begin{align}
\prod_{k=1}^n\frac{4k^2}{4k^2-1}
&=\prod_{k=1}^n\frac{2k}{2k+1}\frac{2k}{2k-1}\tag{17a}\\
&=\prod_{k=1}^n\frac{2k-1}{2k+1}\left(\frac{2k}{2k-1}\right)^2\tag{17b}\\
&=\frac1{2n+1}\left(\prod_{k=1}^n\frac{(2k)^2}{(2k-1)2k}\right)^2\tag{17c}\\
&=\frac1{2n+1}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\tag{17d}
\end{align}
$$
Note the similarity between $(11)$ and $\text{(17d)}$.


问:
Find % difference between a positive and a negative number
I am trying to work out the percentage difference between two numbers (one of which is negative), NOT PERCENTAGE CHANGE. I have done quite a bit of reading, but there doesn't appear to be a definitive answer. Where both numbers are positive, I have been using the formula listed on the following site (https://www.calculatorsoup.com/calculators/algebra/percent-difference-calculator.php). If both numbers are negative, I simply input the absolute values, making them positive. I have issues arise when one of the numbers is negative - for instance:

The percentage difference between 1.6 and -8.8. If I use the formula in the link above, it provides me a result of -293%, but I'm not sure if this is correct - from a methodological stand point. Is this something that would be frowned upon in the mathematics world?

Cheers, Bosshard

答:
The percentage difference is defined as the difference of the values, divided by the average and written in percentage. If you have two values, the formula is
$\frac{|x-y|}{\frac{|x+y|}{2}} \times 100$. This also holds for negative numbers.

Remark:

(1) It is better known as relative change in percentage, for example here https://en.m.wikipedia.org/wiki/Relative_change_and_difference.


问:
In Grassmann algebra a la Browne, why are vectors dependent if their wedge product vanishes?
I'm reading John Browne's Grassmann Algebra, Vol 1 : Foundations. Early on, he asserts without proof that if $x$ and $y$ are any two vectors in the underlying (real) vector space such that $x \wedge y = 0$, then $x$ and $y$ are linearly dependent. Take the vector space to be $R^3$, say. The result is equivalent to proving that if $e_i, e_j$ are two of the standard basis vectors, then $e_i \wedge e_j \neq 0$.

In the framework of axioms and or constructions that Browne provides, how does one prove that simple fact?

答:
The point is that the determinant gives you a non-zero linear function from $\wedge^n V$ to the ground field (which can be arbitrary) when $n=\mathrm{dim}(V)$. So this space is non-zero. Now if you have a basis $e_1,\dots,e_n$ of $V$, then multilinearity and skew-commutativity together imply $e_1 \wedge \cdots \wedge e_n$ spans the top exterior power, and must therefore be non-zero. The result you want follows.

On the other hand, as you point out in the comments below and I confirm, it seems that Browne's axioms are not enough to imply this fact about determinant, and indeed the result you want might fail: any quotient of the Grassmann algebra also satisfies his axioms.


问:
Show that two topologies coincide on a subset
Let $E$ be a set, $\tau_i$ be a topology on $E$ and $B\subseteq E$.

I wonder how we would need to show that $$\left.\tau_1\right|_B:=\{\Omega\cap B:\Omega\in\tau_1\}=\{\Omega\cap B:\Omega\in\tau_2\}=\left.\tau_2\right|_B\tag1.$$

I already know the following:

Lemma 1: The following are equivalent:

$\tau_1$ is coarser than $\tau_2$;

$\operatorname{id}_E$ is $(\tau_2,\tau_1)$-continuous;

If $x\in E$ is a $\tau_2$-limit point of a net $(x_t)_{t\in I}\subseteq E$, then $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$.

So, from this we immediately obtain that the following are equivalent:

$\tau_1=\tau_2$;

$\operatorname{id}_E$ is a homeomorphism from $(E,\tau_1)$ onto $(E,\tau_2)$;

If $x\in E$ and $(x_t)_{t\in I}\subseteq E$ is a net, then $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$ if and only if $x$ is a $\tau_2$-limit point of $(x_t)_{t\in I}$.

So, I guess we need to find a characterization of the net convergence in a subspace topology. Let $\tau$ be any topology on $E$, $(x_t)_{t\in I}\subseteq B$ be a net and $x\in B$.

Are we able to show that
7. $x$ is a $\left.\tau\right|_B$-limit point of $(x_t)_{t\in I}$;
8. $x$ is a $\tau$-limit point of $(x_t)_{t\in I}$

are equivalent?

This should be correct: Assume (7.) and let $N$ be a $\tau$-neighborhood of $x$. Then there is a $\Omega\in\tau$ with $x\in\Omega\subseteq N$ and, since $x\in B$, $O:=\Omega\cap B$ is a $\left.\tau\right|_B$-neighborhood of $x$. So, by (7.)$, $$\exists t_0\in I:\forall t\ge t_0:x_t\in O\subseteq\Omega\subseteq N;$$ i.e. (8.).

Now assume (8.) and let $N$ be a $\left.\tau\right|_B$-neighborhood of $x$. Then there is a $\Omega\in\tau$ with $x\in O:=\Omega\cap B\subseteq N$ and obviously $\Omega$ is a $\tau$-neighborhood of $x$. So, by (8.), $$\exists t_0\in I:\forall t\ge t_0:x_t\in\Omega.$$ Since $(x_t)_{t\in I}\subseteq B$, we conclude $$\forall t\ge t_0:x_t\in\Omega\cap B=O\subseteq N;$$ i.e. (7.).

So, $(1)$ should be equivalent to

If $(x_t)_{t\in I}\subseteq B$ is a net and $x\in B$, then $x$ is a $\tau_1$-limit point of $(x_t)_{t\in I}$ if and only if $x$ is a $\tau_2$-limit point of $(x_t)_{t\in I}$,

right?

答:
It’s correct, but you can shorten it considerably. For $i\in\{1,2\}$ let $\tau_i^B$ be the relative topology on $B$ induced by $\tau_i$. It’s trivial that if $\tau_1^B=\tau_2^B$, $\nu$ is a net in $B$, and $p\in B$, then $\nu$ converges to $p$ in $\tau_1^B$ iff if $\nu$ converges to $p$ in $\tau_2^B$.

For the converse, suppose that $\tau_1^B\ne\tau_2^B$. Without loss of generality there is some $U\in\tau_1^B\setminus\tau_2^B$. Let $F=B\setminus U$; $F$ is not closed in $\tau_2^B$. Since $F$ is not closed in $\tau_2^B$, there is a $p\in(\operatorname{cl}_{\tau_2^B}F)\setminus F$, and there is a net $\nu$ in $F$ converging to $p$ in $\tau_2^B$. In $\tau_1^B$, however, $U$ is an open nbhd of $p$ that contains no term of the net $\nu$, so $\nu$ does not converge to $p$ in $\tau_1^B$.

Thus, $\tau_1^B=\tau_2^B$ iff $\left\langle B,\tau_1^B\right\rangle$ and $\left\langle B,\tau_2^B\right\rangle$ have the same convergent nets.


问:
Calculus Books, preferably Soviet.
I would like some book recommendations on calculus books for novices. I love old Russian texts and if you you suggest such books(no issue if they are rigorous), I would be very grateful. If the text is quite challenging, I have no problem. But it should be interesting with proper exposition of topics.

The book should be good for self-study, preferably a textbook(again, for beginners). And it's not necessary for it to be Soviet. Just anything good. I merely like Soviet/Slavic books and so, the title.

Thanks to you all.

Edit again:Please try to abstain from suggesting analysis books. All I want is a nice, rigorous Book on Integral and Differential Calculus. Doesn't matter which origin(at this point), but should be translated into English. Thanks.

答:
I like to recommend the both parts of the calculus books by Balmohan Limaye here is the link
https://g.co/kgs/iexgM8</p>

Title: Calculus Books, preferably Soviet.


问:
Build graph with exactly n automorphisms
Construct graph with exactly distinct n automorphisms. For n $\geq$ 2.

I wonder if we can just take an asymmetric graph, such as this one as building block.

答:
This graph has exactly five automorphisms:

And it's easy to see how to construct a similar graph with $n$ automorphisms for any $n\ge 3$.


问:
How to find critical point of t distribution when 1-α can't be determine directly from the table.
I was following a tutorial on Bonferroni's method. As a challenge I was ask to
Compute the Bonferroni simultaneous confidence interval For a 95 % overall confidence coefficient using the Bonferroni method, the t value is $t_{1-0.05/(2\cdot2), \, 16} = t_{0.9875, \, 16}$

I Can solve these problem but what's really getting be confuse is how they managed to derive 2.473 as the t-value and how I can go about finding a column for 0.9875 (not in my statistic table).
My knowledge about finding critical value from a t distribution
For example, if i want a t-value for a 90% confidence interval when i have 9 degrees of freedom, go to the bottom of the table, find the column for 90%, and intersect it with the row for df = 9. This gives me a t–value of 1.833 (rounded).
But how about
$t_{0.9875, \, 16}$ how is it 2.473. I would appreciate your clarification. thank you in advance.

答: In cases where the value you're looking for isn't in a t-table, you can use several different programs to the value you want. For example, in Excel, the formula you're looking for is T.INV(0.9875,16). You could also use WolframAlpha to calculate the same thing.


问:
If $\alpha,\beta$ are two real numbers s.t. $\alpha \sqrt{P(A|B)}+\beta\sqrt{P(\bar{A}|B)}=2/3$, what's the value of $\frac{27}{8}(\alpha^2+\beta^2)$?
Let $A$ and $B$ be two independent events. If $\alpha$ and $\beta$ are any two real numbers such that $$\alpha \sqrt{P(A|B)}+ \beta\sqrt{P(\bar{A}|B)}=2/3,$$ then the value of $\frac{27}{8}(\alpha^2+\beta^2)$ lies between.

MY ATTEMPT:

By using AM - GM mean we can figure out the lower boundary of the value is $4$  but I am facing difficulty in finding the upper boundary. I have tried to expand the equation which simplifies to
$$\alpha\sqrt{P(A)}+ \beta\sqrt{P(\bar{A})}=2/3 $$
assuming $P(A) = x$ and $P(\bar{A}) = 1-x$ and simplify and tried to isolate the $x$ and use the fact that $x$ lies on the interval $[0,1]$, but I failed.

Please, help me.

答: There is no upper bound. Take $\alpha=n$ (a positive integer) and solve the given equation for $\beta$. Note that the hypothesis is now satisfied but $\alpha^{2}+\beta^{2} \geq \alpha^{2} \geq n^{2}$. Hence there is no upper bound for $\alpha^{2}+\beta^{2}$


问:
Find the number of positive divisors of $(2008^3 + (3 * 2008 * 2009) + 1)^2$

Find the number of positive divisors of $(2008^3 + (3 * 2008 * 2009) + 1)^2$

What I Tried: I have no idea of an elegant solution to this. I cannot seem to guess or figure out the factors as there is a $(+1)$ added to it, which seems to bother me. As the number is a perfect square I can guess it will have even number of positive divisors or so.

I decided to solve it by hand. I calculated the expression to be $(8108486729)^2$ , and turns out , the prime factorization of $8108486729$ is amazingly $7^6 * 41^3$ .

My first question is, how is that factorization coming? It looks a bit like magic to me, can someone explain that?

So from here $(8108486729)^2$ will have prime factorization as $7^{12} * 41^6$ , so total number of factors will be $(13)(7) = 91$ , problem solved. This is however, not an elegant solution, and I can't seem to find any.

Can anyone help?

Body: Hint: $${(2008)}^3+3\cdot 2008\cdot 2009+1={(2008+1)}^3={(2009)}^3$$


Body:
Find the number of positive divisors of $(2008^3 + (3 * 2008 * 2009) + 1)^2$

What I Tried: I have no idea of an elegant solution to this. I cannot seem to guess or figure out the factors as there is a $(+1)$ added to it, which seems to bother me. As the number is a perfect square I can guess it will have even number of positive divisors or so.

I decided to solve it by hand. I calculated the expression to be $(8108486729)^2$ , and turns out , the prime factorization of $8108486729$ is amazingly $7^6 * 41^3$ .

My first question is, how is that factorization coming? It looks a bit like magic to me, can someone explain that?

So from here $(8108486729)^2$ will have prime factorization as $7^{12} * 41^6$ , so total number of factors will be $(13)(7) = 91$ , problem solved. This is however, not an elegant solution, and I can't seem to find any.

Can anyone help?

Body:
$z^3+3z^2+3z+1=(z+1)^3$

$z=2008$

$z+1=2009$

$2009 = 7^2 * 41$

$2009^3 = 7^6 * 41^3$


问:
Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$

Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.

Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any  textbooks with similar problems?
Thank you so much!

答:
You need to show the sequence
$$
x_n = \sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$ converges. Only then you can take $x = \sqrt{12 + \sqrt{12 +\cdots}} $ and continue to other steps.
It is easy to see that $x_n = \sqrt{12 + x_{n-1}}$ with $x_1 = \sqrt {12}$.
As answered by @user2661923 and @QC_QAOA the sequence $x_n$ is bounded. Here is an easy way of showing $x_n$ is strictly increasing. Observe that $x_1 < x_2$ and we assume that $x_{n-1} < x_n$. Then
$$x_{n+1} = \sqrt{12 +x_n} > \sqrt{12 + x_{n-1}} = x_n$$
which completes our inductive argument and we conclude that $x_n$ converges.

Observing that $x_n > 0$ for any $n$ we can conclude that $x \ge0$ by limit theorems. So we must have $x = 4$


问:
How solve an Increase function problem
I was trying to figure out to solve this question. I understand getting increases and decrease after getting the derivatives based on what I've learned, I am stuck and don't understand how to solve this math problem below.

If $f(t)=t^2 + t$, Between $t = 10$ and $t = 12$, what is the increase in f divided by the increase in t?

Title: How solve an Increase function problem

Body: $\frac{f(12)-f(10)}{12-10}=\frac{156-110}{2}=\frac{46}{2}=23$


答:
Hint:

The question is asking you to calculate $$\frac{f(12)-f(10)}{12-10}$$

Where the numerator shows the difference in $f$, and the denominator the difference in $t$.

Title: How solve an Increase function problem

Body:
I was trying to figure out to solve this question. I understand getting increases and decrease after getting the derivatives based on what I've learned, I am stuck and don't understand how to solve this math problem below.

If $f(t)=t^2 + t$, Between $t = 10$ and $t = 12$, what is the increase in f divided by the increase in t?


问:
How is it possible that a cyclic group like $\mathbb Z_{2p}$ is isomorphic to a non-cyclic group of order $2p$?
We know that a non-cyclic group can not be isomorphic to a cyclic group. But in Gallian there is a theorem which states that all non-cyclic groups of order $2p$ (where $p$ is prime) are isomorphic to $\mathbb Z_{2p}$ or $D_p$.

Since $\mathbb Z_n$ is a cyclic group, how can it be isomorphic to a non-cyclic group?

答: It is not possible. Let $f$ be a homomorphism from $\mathbb{Z}_n$ to a group $G$. If it is an isomorphism then the image of the generator of $\mathbb{Z}_n$, for example, $f(1)$ is a generator of $G$ as well.


答:
Edit

First see the comment of NinadMunshi immediately following this answer.  I have edited the answer accordingly.

Continuing Infinity_hunter's answer, and letting

$$x_n~ \text{denote}~
\sqrt{12 + \sqrt{12 +  \cdots  \text{n times} }}
$$

To show convergence, all that is necessary is to show that

the sequence is bounded

the sequence is strictly increasing.

Clearly, $0 < x_1 < 4.$  Assume that $0 < x_n < 4.$  Then

$$x_{(n+1)} = \sqrt{12 + x_n} < \sqrt{12 + 4} < 4$$

and

$$x_{(n+1)} = \sqrt{12 + x_n} > \sqrt{12 + 0} > 0.$$

Therefore,

$$0 < x_n < 4 \implies 0 < x_{(n+1)} < 4.$$

To show that the sequence is (therefore) strictly increasing:

Since

$$ 0 < x_n < 4$$

and

$$[(x_n)^2 - x_n - 12] = (x_n - 4)(x_n + 3)$$

I conclude that

$$[(x_n)^2 - x_n - 12] < 0 \implies (x_n)^2 < x_n + 12.$$

However, by the definition of the sequence

$$\left[x_{(n+1)}\right]^2 = x_n + 12.$$

Therefore

$$\left[x_{(n+1)}\right]^2 > (x_n)^2.$$

Therefore, since each element in the sequence is positive,

$$x_{(n+1)} > (x_n).$$

Therefore, the sequence is strictly increasing.

Title: Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$

Body:
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3)(x-4)=0$$
So, the answer is $\boxed{4}$.

Is correct my solution?
Can you show other ways for to solve this problem?
Can you suggest me any  textbooks with similar problems?
Thank you so much!


答:
Let $a_1=\sqrt{12}$ and $a_n=\sqrt{12+a_{n-1}}$. We must show that $a_n\to 4$. In order to do this, it is sufficient to show that

$1)\ \lim_{n\to\infty}a_n\text{ exists}$

$2)\ \text{This limit is }4$

You have basically skipped step $1)$ and gone straight to step $2)$. To fill in the first step, first note that $0<a_n<4$ for all $n$. We proceed by induction:

$$0<a_1=\sqrt{12}<\sqrt{16}=4$$

$$0<\sqrt{12}\leq a_n=\sqrt{12+a_{n-1}}<\sqrt{12+4}=\sqrt{16}=4$$

Second, we may now show that $a_n$ is increasing. This is obvious as

$$\frac{a_{n+1}}{a_n}=\frac{\sqrt{12+a_n}}{a_n}=\sqrt{\frac{12}{a_n^2}+\frac{1}{a_n}}$$

Since $a_n<4$, this becomes

$$\sqrt{\frac{12}{a_n}^2+\frac{1}{a_n}}>\sqrt{\frac{12}{16}+\frac{1}{4}}=\sqrt{\frac{3}{4}+\frac{1}{4}}=\sqrt{1}=1$$

Thus, $a_{n+1}>a_n$. Since we have already established that $a_n<4$ we may conclude that

$$\lim_{n\to\infty}a_n=L$$

exists. This concludes step $1)$. As you have already proved step $2)$, we may conclude that

$$\lim_{n\to\infty}a_n=4$$

The reason we may discard the $-3$ answer is that $0<a_n<4$ so $L\in[0,4]$.

EDIT: Edited as OP requested more resources for these types of infinite radicals. The best place to start (at least in my mind) is Herschfeld's Convergence Theorem. The theorem states that

$$\lim_{k\to\infty} \left[x_0+(x_1+(x_2+(\cdots +(x_k)^p)^p)^p)^p\right]$$

exists if and only if $(x_n)^{p^n}$ is bounded. We see that at each step, $x_n=12$. Since $12^{2^{-n}}$ is bounded, we may conclude that the series converges.


问:
Is it true that Hasse diagram of ToSET(totally ordered set) is vertical line?
I am learning hasse diagram concept. I read that toset means poset + every pair of elements in set must be comparable(ie.., related). If any pair of elements are comparable then there will not be branching right(pls correct me if wrong). If there is branching (say V shape) then 2 top elements on top of V-shape will not be comparable. THen i think is it true that toset will always be a vertical line? pls correct me if wrong.</p>

答:
You are correct that a finite toset can be represented by a entirely vertical Hasse diagram as you describe. For infinite sets it becomes debatable there even is a Hasse diagram, although the vertical intuition is still a good one since it works "locally" (i.e. for finite subtosets). Indeed, this is the reason another term for totally ordered is "linearly" ordered - because our intuition is that it can exist on a line. The Hasse diagram of a toset of size $n$ is the corresponding "line graph."

For infinite tosets, we can have vertices that are not connected by edges and edges necessarily of arbitrarily small length - for instance the ordinal $\omega+1$, which you can think of as $\mathbb{N}\sqcup\{\infty\}$, or more topologically as $\{-1/n\mid n>0\}\sqcup\{0\}$. This also shows the cardinality does not determine the diagram, since $\omega$ itself ought to have a different diagram. It may not even make sense to say there are edges, for instance what would the Hasse diagram of $\mathbb{Q}$ be? And the toset may not even embed in $\mathbb{R}$ at all, for instance if its cardinality is greater than $\mathfrak{c}=|\mathbb{R}|$.


问:
Closed unbounded operator with domain not closed
I am looking for an example for further understanding of the Closed Graph Theorem:

Let $X,Y$ be Banach spaces and $T:X\to Y$ closed (i.e. the graph of $T$ is closed in $X\times Y$). Then if $\mathcal{D}(T)$ is closed in $X$, $T$ is bounded.

I am looking for an unbounded operator whose graph $\mathcal{G}(T)$ is closed in $X\times Y$ and whose domain $\mathcal{D}(T)$ is not closed in $X$, to clearify the necessity of $\mathcal{D}(T)$ being closed.

This question arose due to the definition of the norm of a graph $\lVert (x,Tx)\rVert:=\lVert x\rVert+\lVert Tx\rVert$, where I thought the following statement would be true:  [$\mathcal{G}(T)$ closed $\Rightarrow\mathcal{D}(T)$ closed] which in general is false.

答:
If there is a Cauchy sequence $\{x_n\} \subset \mathcal D(T)\subset X$ whose $\{T x_n\}$ is not Cauchy in $Y$, then the definition doesn't cover this particular sequence.

In order to make $\mathcal D(T)$ closed, you need to guarantee ALL Cauchy sequences in $\mathcal D(T)$ to accumulate inside $\mathcal D(T)$. The definition of a closed operator, however, can only guarantee those Cauchy sequences $\{x_n\}$, which BOTH the $\{x_n\}$ and $\{T x_n\}$ are Cauchy, has a limit inside $D(T)$.


问:
Is the Dirac delta a function?
Is Dirac delta a function? What is its contribution to analysis?

What I know about it:
It is infinite at 0 and 0 everywhere else. Its integration is 1 and I know how does it come.

答:
$ \delta(x) = \begin{cases} +\infty, & x = 0 \\ 0, & x \ne 0
> \end{cases}$

and which is also constrained to satisfy the identity $
> \int_{-\infty}^\infty \delta(x) \, dx = 1$

For example, the objects f(x) = δ(x) and g(x) = 0 are equal everywhere except at x = 0 yet have integrals that are different. According to Lebesgue integration theory, if f and g are functions such that f = g almost everywhere, then f is integrable if and only if g is integrable and the integrals of f and g are identical.

This is merely a heuristic characterization. The Dirac delta is not a
  function in the traditional sense as no function defined on the real
  numbers has these properties. The Dirac delta function can be
  rigorously defined either as a distribution or as a measure.

Source: Wikipedia

Title: Is the Dirac delta a function?


问:
determinant of matrix in polynomial
Suppose $X=\sqrt{k^2 - k_1^2}, Y= \sqrt{k^2 - k_2^2},Z=\sqrt{k^2 - k_3^2}, S=\sqrt{k^2 - k_4^2}, T= \sqrt{k^2 - k_5^2}$. \
And there is $5 \times 5 $ matrix as given below.
\begin{equation}
\begin{vmatrix}
a_{11} & a_{12}  & a_{13} & a_{14}*S & a_{15}*T\\
a_{21}*X & a_{22}*Y  & a_{23}*Z & a_{24} & a_{25}\\
a_{31}*X & a_{32}*Y  & a_{33}*Z & 0 & 0\\
a_{41} & a_{42}  & a_{43} & a_{44}*S & a_{45}*T\\
a_{51}*X & a_{52}*Y  & a_{53}*Z & a_{54}& a_{55}\\
\end{vmatrix} = 0,
\end{equation}

I have to find the determinant in terms of k with inter powers. By usual exapnsion method (squaring the terms), finding the expression for k is getting complex. Is there any other approach to find the expression for above determinant. And that expression should not contain fraction powers of k, means expression should have $X^2, Y^2,Z^2,S^2,T^2$ instead of $X, Y,Z,S,T$.

答:
Here is a special case of your matrix
$$
\begin{vmatrix}
1 & 0  & 0 & 0 & 0\\
0& 0  & 0 & 1 & 0\\
0 & 0  &  Z & 0 & 0\\
0 & 1  & 0 & 0 & 0\\
0 & 0  & 0 & 0& 1\\
\end{vmatrix}.
$$

It evaluates to $Z=\sqrt{k^2 - k_3^2}$. (Or is it $-Z$? No matter.) You can't achieve what you are asking for.

Title: determinant of matrix in polynomial


答:
Your solution is correct. The other answers have addressed formality and rigour as a concern, but from the sound of your comments, you're more interested in the kind of problem more than the right answer, so I hope the following helps.

This kind of problem is an infinitely nested radical. As far as I know, there's no other way as straightforward as recognizing the infinitely nested square root is just a quadratic in disguise. In fact, an infinitely nested cube root would be a cubic in disguise.

Solution of Polynomial Equations with Nested Radicals on Arxiv actually derives a generic nested radical as a solution to a generalized quadratic equation, and dives deeper into different kinds of polynomials.

On Infinitely Nested Radicals is a less crazier article from a magazine that discusses their convergence and what type of numbers could be represented by them.

A Chronology of Continued Square Roots and Other Continued Compositions [...] on Arxiv is exactly what it sounds like. At a glance it's written pretty well.


问:
Algebraic multiplicity of Complex solutions
Just wondering how to know the Algebraic multiplicity of Complex solutions..

For example : $(x^2+2)^3$

The solutions are: $i\sqrt{2}$ , $i\sqrt{-2}$

How can I see what is algebraic multiplicity of these solutions?

is it $3$ for both of them or less?

Thank you in advance

答:
Note that

$$x^2 + 2 = (x-i\sqrt 2)(x + i \sqrt 2)$$

In this expression, each have multiplicity $1$. However, since you seek their results in the cube, you have

$$(x^2 + 2)^3 = (x-i\sqrt 2)^3(x + i \sqrt 2)^3$$

which clearly shows each has multiplicity $3$ for the polynomial you care about.