问:
GCD of $n^2-3n-1$ and $n-4$
$n$ is a natural number and after trying the division algorithm

$$\gcd(n^2-3n-1,n-4)=\gcd(n-4,n-1)=\gcd(n-1,-3)=\gcd(n-1,3)$$

For the last part I'm not sure whether it does equal to $\gcd(n-1,3)$.
If yes, then should I take the cases where $n$ is $3k+i, 0\le i\le2$?

答:
Using this implementation of the Extended Euclidean Algorithm, rotated $90^{\large\circ}$ and applied to polynomials, gives
$$
\begin{array}{c|c|c}
\color{#00F}{n^2-3n-1}&1&0\\
\color{#00F}{n-4}&0&1\\
\color{#090}{n-1}&\color{#C00}{1}&\color{#C00}{-n}&n\\
\color{#090}{-3}&\color{#C00}{-1}&\color{#C00}{n+1}&1
\end{array}
$$
which says that
$$
\overbrace{\begin{bmatrix}
\color{#C00}{1}&\color{#C00}{-n}\\\color{#C00}{-1}&\color{#C00}{n+1}
\end{bmatrix}}^{\det=1}
\begin{bmatrix}
\color{#00F}{n^2-3n-1}\\\color{#00F}{n-4}
\end{bmatrix}
=\begin{bmatrix}
\color{#090}{n-1}\\\color{#090}{-3}
\end{bmatrix}
$$
which means any integer linear combination of $n-1$ and $3$ is an integer linear combination of $n^2-3n-1$ and $n-4$ and vice-versa (since the determinant of the matrix is $1$). This means that
$$
\gcd\left(n^2-3n-1,n-4\right)=\gcd(n-1,3)
$$

Title: GCD of $n^2-3n-1$ and $n-4$


问:
Find $x$ such that $ 2^{16^x} = 16^{2^x} $.
Find $x$ such that $ 2^{16^x} = 16^{2^x}.$

I am a bit confused, what will happen when we expand $16^{2^x}$, will we get  $4^{2^{2^x}}$ or $4^{2^{x+1}}$ ?

答:
Alternative approach

$$16^{2^x} = [2^4]^{2^x} = 2^{4 \times 2^x}.$$

Since this is equal to $2^{16^x}$ you have

$$16^x = 4 \times 2^x \implies 8^x = 4 \implies x = (2/3).$$

ContentLicense: CC BY-SA 4.0


Title: Find $x$ such that $ 2^{16^x} = 16^{2^x} $.

Body:
Find $x$ such that $ 2^{16^x} = 16^{2^x}.$

I am a bit confused, what will happen when we expand $16^{2^x}$, will we get  $4^{2^{2^x}}$ or $4^{2^{x+1}}$ ?

Body:
See that $2^{16^x} = 16^{2^x}$ is $2^{2^{4x}} = 2^{2^{x+2}}$

then $4x = x + 2$ and get $x = 2/3$.


问:
Kernel of a Second derivative linear map

Hi I was wondering if anyone could help me tackle this question.

I have recently been introduced to the idea of a Kernel and have predominantly been looking at it in terms of matrices and not linear maps. So I'm a bit shaky on this area.

However I am aware that the kernel is the set of vectors in the domain of the mapping which are mapped to the zero vector.
Hence I belive it must be the domain of the first derivative here which is mapped to the zero vector.

So my thinking is that the kernel of D_2 should represent any constants given from the first derivative as when derived again these will disappear.

Hence I think the option should be b). But I have no clue whether this is correct or not.

Also could someone explain what the subscripts mean next to the polynomial symbol.

Please can someone help me consolidate this. Thanks in advance.

Body:
Let $p=\sum_{i=0}^na_ix^i$ be a polynomial then

$p''=\sum_{i=2}^ni(i-1)a_ix^{i-2}=2a_2+6a_3x+...+n(n-1)a_nx^{n-2}$

$Ker(D_2)=\{p \in P(R):p''=0\}$ $\Rightarrow$  $Ker(D_2)=\{p \in P(R):2a_2+6a_3x+...+n(n-1)a_nx^{n-2}\}=0$

$2a_2+6a_3x+...+n(n-1)a_nx^{n-2}=0$ $\Rightarrow$ $a_2=a_3=...=a_n=0$ which means

$Ker(D)=\{p \in P(R):p=a_0+a_1x+a_2x^2\}=P_2(R)$


问:
Integration over an area bounded between two curves
Determine $\int_Axy\space\space d(x,y)$, when $A$ is a closed set, bounded by the curve $y=x-1$, and the parabel $y^2 = 2x+6.$

I believe we only need to know the range, where $A$ is actually bounded by these two curves. I think this will be $-3\leq x \leq 5$. But to be specific, isn't $A\subset Int ( G_h \cap G_g)$, s.t $g(x)=x-1$, and $h(x)=y^2 = 2x+6$? Where $-3\leq x \leq 5$. Can we determine the integral as $$\int_{-3}^5\Bigg(\int_{x-1}^{2x+6}xy\space\space dy\Biggr)dx\space\space?$$

Also, I'm assuming that $(x,y)\in \Bbb{R^2}$.

答: The curves intersect at $(-1,-2),(5,4)$. For $x\in(-3,-1),y$ goes from $-\sqrt{2x+6}$ to $\sqrt{2x+6}$ but for $-1<x<5,y$ ranges from $x-1$ to $\sqrt{2x+6}$. The required integral is$$\int_{-3}^{-1}\int_{-\sqrt{2x+6}}^\sqrt{2x+6}xy~dy~dx+\int_{-1}^5\int_{x-1}^\sqrt{2x+6}xy~dy~dx$$


问:
Recursive sequence is increasing
I have to show, that the sequence $g(n+1) = \frac{2ag(n)}{a+g(n)^2}$
is increasing for $a>0$ and $g(0)>0$. I know that it holds for $g(n)^2<a$
since $2a/(a+g(n)^2)>1$. But for $g(n)^2>a$ the opposite seems to be true.

I assume my professor forgot some initial restrictions but maybe I am overseeing something. I’d appreciate your thoughts on the matter.

答: It's very easy to see that such a recurrence cannot be strictly increasing for all such initial choices.  For instance, $a = 1$ and $g(0) = 2$ yields $g(1) = \frac{4}{5}$.  In fact, we can see that $\sqrt{a}$ is a fixed point of the sequence by solving $$g = \frac{2ag}{a+g^2}$$ for $g$.  If $g(0) > \sqrt{a}$, then we have $$\frac{g(1)}{g(0)} = \frac{2a}{a+g(0)^2} < \frac{2a}{a+a} = 1,$$ hence $g(1) < g(0)$.

Title: Recursive sequence is increasing


答:
Hint:

Let $f(x)=\dfrac{2ax}{a+x^2}$ be the function that defines the recursive sequence $\bigl(g(n)\bigr)_{n\ge 0}$.

This sequence is increasing if all $g(n)$ live in $A\subset\mathbf R$ such that, for all $x\in A$, $f (x)\ge x\:$ and $\:f(A)\subset A$.

Now $\quad f(x)\ge x \iff 2ax \ge x(a+x^2)\iff ax\ge x^3\iff\begin{cases}x^2\le a&\text{if }x\ge 0,\\x^2\ge a&\text{if }x<0.\end{cases}$

It is clear  from the hypotheses that we're in the first case ($g(n)>0$). So there remains to prove by induction that all $g(n)$ live in the interval $I=\bigl[0,\sqrt a\bigr]$, i.e. $f(I)\subset I$.

Title: Recursive sequence is increasing

Body:
I have to show, that the sequence $g(n+1) = \frac{2ag(n)}{a+g(n)^2}$
is increasing for $a>0$ and $g(0)>0$. I know that it holds for $g(n)^2<a$
since $2a/(a+g(n)^2)>1$. But for $g(n)^2>a$ the opposite seems to be true.

I assume my professor forgot some initial restrictions but maybe I am overseeing something. I’d appreciate your thoughts on the matter.


问:
Poisson limit process with divergence condition
Please help me with problem 3.26 of 'Probability Theory' by Varadhan:

For each $n\in \mathbb{N}$, $\ X_{n,j}:(\Omega,P) \to \mathbb{R}$ with $1\leq j\leq k_n$ are $k_n$ independent random variables with $$P[X_{n,j}=1]=p_{n,j} \text{ and } P[X_{n,j}=0]=1-p_{n,j}.$$

Let $S_n:=\sum_{j} X_{n,j}$. Let $\lambda_n:=\sum_{j} p_{n,j}$ be the mean of the $S_n$.

Question: If $\lambda_n \to \infty$, show that the distribution of $\frac{S_n-\lambda_n}{\sqrt{\lambda_n}}$ converges to the standard normal distirbution.

Though not explicitly mentioned, I think the assumption that the $X_{n,j}$ be uniformly infinitesimal is required. That is, we assume
$$\lim\limits_{n\to \infty} \sup\limits_{1\leq j\leq k_n} p_{n,j} = 0.
$$

Attempt: This looks similar to the central limit theorem without the variables being identically distributed. So anyhow, I try to compute characteristic function -

\begin{equation}
\begin{split}
\widehat{\frac{S_n-\lambda_n}{\sqrt{\lambda_n}}}(t) &= \int\exp \left[it\frac{S_n(\omega)-\lambda_n}{\sqrt{\lambda_n}}\right]dP(\omega) \\
&= \int\exp\left[ \frac{itS_n(\omega)}{\sqrt{\lambda_n}}\right] \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right]dP(\omega)\\
&= \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right]\int\exp\left[\frac{it}{\sqrt{\lambda_n}}\sum x_j\right]d((X_{n,1},\dots,X_{n,k_n})_*P)(x_1,\dots,x_{k_n})\\
&= \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right]\int\dots\int\exp\left[\frac{it}{\sqrt{\lambda_n}}\sum x_j\right]d((X_{n,1})_*P)(x_1)\dots d((X_{n,k_n})_*P)(x_{k_n})\\
&= \exp\left[\frac{-it\lambda_n}{\sqrt{\lambda_n}}\right] \prod \int \exp\left[\frac{itx}{\sqrt{\lambda_n}}\right]d((X_{n,j})_*P)(x)\\
&= e^{\frac{-it}{\sqrt{\lambda_n}}\sum p_{n,j}} \prod \left[(1-p_{n,j}) + p_{n,j}e^{\frac{it}{\sqrt{\lambda_n}}}\right]\\
&= \prod \left[(1-p_{n,j})e^{\frac{-itp_{n,j}}{\sqrt{\lambda_n}}} + p_{n,j}e^{\frac{it(1-p_{n,j})}{\sqrt{\lambda_n}}}\right].
\end{split}
\end{equation}

Not sure what to do now. Maybe take log, but that doesn't really look too promising. Or should I be looking at the Levy-Khintchine representation?

Question: Why are people interested in such a sequence of $S_n$? I understand that the iid case with $p_{n,j}=1/2$ is talking about someone tossing a bunch of coins.

Edit: I see that this question is very similar although I'm not sure if it's exactly the same since there is some different assumption involving some variables following a Poisson distribution.

答:
Ok, Maybe I was being lazy and should have used Levy-Khintchine representation + Accompanying laws theorem as indicated in the book:

For each $n$, we have $k_n$ independent variables $Y_{n,j}:=\frac{X_{n,j}-p_{n,j}}{\sqrt{\lambda_n}}$ as above and we want to show that the distribution $\mu_n$ of $\sum Y_{n,j}$ converges to the standard normal distribution.

Let $\alpha_{n,j} = (Y_{n,j})_*P$, the distribution of $Y_{n,j}$. Note that the $Y_{n,j}$ are uniformly infinitesimal and have truncated means $\int_{|x|\leq 1}x d\alpha_{n,j}(x)=0$. The Accompanying laws theorem then says $\mu_n$ converges to the standard normal if and only if the distributions $\lambda_n$ (defined below) also converge to the same limit. We have
$$\lambda_n := (\beta_{n,j})^{*k_n}
$$
where $\beta_{n,j}$ is defined by the property $\widehat{\beta_{n,j}}(t)= \exp(\widehat{\alpha_{n,j}}(t)-1)$.

If we compute the characteristic function of $\lambda_n$, we see
\begin{equation}
\begin{split}
\widehat{\lambda_n}(t)= \prod\widehat{\beta_{n,j}}(t) &= \prod \exp\left(\widehat{\alpha_{n,j}}-1\right) = \exp\left(\sum\int(e^{itx}-1)d\alpha_{n,j}(x)\right)\\
&=\exp\left(\int\left(e^{itx}-1-it\theta(x)\right)dM_n(x) + ita_n\right)
\end{split}
\end{equation}
where $M_n$ is the $\textit{admissible Levy measure}$ defined by $\sum d\alpha_{n,j}$ and $\theta(x)$ is the bounded continuous function defined by $x$ for $|x|\leq 1$ and $\text{sgn}(x)$ otherwise. Note, the only requirement on $\theta$ is that it be bounded continuous and have $|\theta(x)-x| < C|x|^3$ near $0$. The term $a_n$ is, of course, $\int \theta dM_n$.

The expression for $\widehat{\lambda_n}$ above is written as  e$(M_n,0,a_n)$, the $\textit{Levy-Khintchine representation}$ for the infinitely divisible distribution $\lambda_n$. (The $0$ indicates that the variance term $\frac{-t^2\sigma^2}{2}$ within the exponential is absent.) For e$(M_n,0,a_n)$ to converge to the standard normal distribution e$(0,1,0)$, it is necessary and sufficient that

For every bounded continuous $f$ with $0\notin \text{supp}(f)$ $$\int f dM_n \to 0.$$

For every $l > 0$, we have $$\int_{|x|\leq l} x^2 dM_n(x) \to 1.$$

$a_n \to 0$.

Proof of (1):
\begin{equation}
\int f dM_n = \sum \int f(x) d\alpha_{n,j}(x) = \sum \int f\left(\frac{X_{n,j}(\omega)-p_{n,j}}{\sqrt\lambda_n}\right)dP(\omega).
\end{equation}
Since $0$ is not in the support of $f$, since $p_{n,j}, X_{n,j}$ are uniformly bounded, and since $\lambda_n \to \infty$, one can observe that $(1)$ does indeed hold.

Proof of (3):
\begin{equation}
\begin{split}
a_n = \int\theta dM_n &= \sum \int \theta(x) d\alpha_{n,j}(x)\\
&= \sum \left(\int_{|x|\leq 1} x d\alpha_{n,j}(x) + \int_{|x| > 1} \text{sgn}(x) d\alpha_{n,j}(x)\right)
\end{split}
\end{equation}
Again, as we reasoned in $(1)$, the second summands must all be $0$ for $n$ large enough. More precisely, the support of the integrals in the second summands are zero, so we are left with computing (assuming $n$ large)
\begin{equation}
\begin{split}
\sum \int_{|x|\leq 1} x d\alpha_{n,j}(x) &= \sum \int x d\alpha_{n,j}(x) \\
&= \sum \int \left(\frac{X_{n,j}(\omega)-p_{n,j}}{\sqrt\lambda_n}\right) dP(\omega) =0
\end{split}
\end{equation}
since these variables clearly have zero means.

Proof of (2): We compute
\begin{equation}
\begin{split}
\int_{|x|\leq l} x^2 dM_n(x) &= \sum \int {\bf 1}_{\{|x|\leq l\}}\left(\frac{X_{n,j}(\omega) - p_{n,j}}{\sqrt\lambda_n}\right) \cdot\left(\frac{X_{n,j}(\omega) - p_{n,j}}{\sqrt\lambda_n}\right)^2dP(\omega)\\
&= \sum \int \left(\frac{X_{n,j}(\omega) - p_{n,j}}{\sqrt\lambda_n}\right)^2dP(\omega)
\end{split}
\end{equation}
for $n$ large enough, since $\lambda_n\to \infty$.
Thus we have, for large $n$,
\begin{equation}\begin{split}
\int_{|x|\leq l} x^2 dM_n(x) &= \sum \int \left(\frac{X_{n,j}(\omega) - p_{n,j}}{\sqrt\lambda_n}\right)^2dP(\omega)\\
&= \frac{1}{\lambda_n}\sum \int \left(X_{n,j}^2(\omega) + p_{n,j}^2 - 2p_{n,j}X_{n,j}(\omega)\right)dP(\omega)\\
&= \frac{1}{\lambda_n}\sum p_{n,j}(1-p_{n,j})
\end{split}\end{equation}
which converges to $1$ by the infinitesimality assumption.


问:
Bernoulli differential equation help?
We have the equation $$3xy' -2y=\frac{x^3}{y^2}$$ It is a type of Bernouli differential equation. So, since B. diff equation type is
$$y'+P(x)y=Q(x)y^n$$
I modify it a little to:

$$y'- \frac{2y}{3x} = \frac{x^2}{3y^2}$$

$$y'-\frac{2y}{3x}=\frac{1}{3}x^2y^{-2}$$

Now I divide both sides by $y^{-2}$. What should I do now?

答:
$$\text{We have $3xy^2 y'-2y^3 = x^3 \implies x (y^3)' - 2y^3 = x^3 \implies \dfrac{(y^3)'}{x^2} + y^3 \times \left(-\dfrac2{x^3}\right) = 1$}$$
$$\text{Now note that }\left(\dfrac1{x^2}\right)' = -\dfrac2{x^3}. \text{ Hence, we have }\dfrac{d}{dx}\left(\dfrac{y^3}{x^2}\right) = 1\implies \dfrac{y^3}{x^2} = x + c$$
$$\text{Hence, the solution to the differential equation is }\boxed{\color{blue}{y^3 = x^3 + cx^2}}$$</p>

ContentLicense: CC BY-SA 3.0

Title: Bernoulli differential equation help?


答:
Instead of dividing integral into two parts, for simplification, you can set it up as below (integrating over $x$ first between the two curves) -

$\displaystyle \int_{-2}^4\int_{(y^2-6)/2}^{y+1} \, xy \, dx \, dy$

For the region, $-2 \leq y \leq 4$.

Title: Integration over an area bounded between two curves

Body:
Determine $\int_Axy\space\space d(x,y)$, when $A$ is a closed set, bounded by the curve $y=x-1$, and the parabel $y^2 = 2x+6.$

I believe we only need to know the range, where $A$ is actually bounded by these two curves. I think this will be $-3\leq x \leq 5$. But to be specific, isn't $A\subset Int ( G_h \cap G_g)$, s.t $g(x)=x-1$, and $h(x)=y^2 = 2x+6$? Where $-3\leq x \leq 5$. Can we determine the integral as $$\int_{-3}^5\Bigg(\int_{x-1}^{2x+6}xy\space\space dy\Biggr)dx\space\space?$$

Also, I'm assuming that $(x,y)\in \Bbb{R^2}$.


问:
Integration of the product of two polynomials
Consider this integration:

$$\int (x-\mu_x)^2  \cdot (mx+q) dx$$

I can square the first element, do the products and then integrate but I was wondering, is there a smarter way to perform this integration?

答:
$$I=\int (x-\mu_x)^2  \cdot (mx+q) dx$$
Substitute $z=x-\mu_x$:
$$I=\int z^2  (m(z+\mu_x)+q) dz$$
$$I=\int   mz^3+(m\mu_x+q)z^2)dz$$
It's a polynomial.


答:
By simplification,

$$g(n+1)=\frac{2ag(n)}{a+g^2(n)}\lessgtr g(n)$$

is equivalent to

$$a\lessgtr g^2(n).$$

But note that after a single iteration,

$$g(n)\le\sqrt a.$$

Title: Recursive sequence is increasing


问:
Prove that there is no simple group $G$ of order $3^3\cdot7\cdot13\cdot409$
This is from Dummit and Foote Chapter 6.2 Exercise 29 and there is a hint to work in the permutation representation of degree $819$.

In Exercise 28 we reduced to:

$n_3=2863$

$n_7=47853$

$n_{13}=25767$

$n_{409}=819$

I found a proof here: https://www.slader.com/textbook/9780471433347-abstract-algebra-third-edition/214/exercises/29/

What I don't understand is, why can't we have besides an orbit of $1$ and an orbit of order $409$ a second orbit of order $409$. This would neatly sum up to $819$, the number of Sylow $409$-subgroups.

答:
It's easy to see that NG(P) is not an abelian group (otherwise the normalizer of sylow 3 subgroup would have prime factor 409). Then n3 of NG(P) is 409.

Suppose Q is another sylow 409 subgroup, denote NG(P)=U, NG(Q)=V, then U intersects with V is a group of order 1 or 3 (it can't be 409 since otherwise they would share the same sylow 409 group).

That means there exists a sylow 3 subgroup in U\V, acting on Q as an orbit of length 3 by conjugation. That is, U acts on Q by conjugation results in an orbit of length 3*409, overflowing n409.


问:
Prove that every simple planar which is **connected** can be coloured with 6 or less colour.
Prove that every simple planar which is connected can be coloured with 6 or less colour. I heard a very similar theorem which doesn't have the connected condition.

(Hints in the book:using the fact that simple planar which is connected has a vertex which is degree smaller than or equal to 5 and apply mathematical induction on the number of vertices of the graph.)

Attempts: I have tried to use Mathematical induction by assumimg the graph of r vertice is true. But when considering r+1, a problem come. I first want to, say remove the vertix which has degree smaller than or equal to 5 (there exist at least one)and the edges connected to it, but it will return to a subgraph with vertice r, however i found that the reduced graph may not be connected, and once it is disconnected, we cannot apply the induction assumption. So how should I prove the theorem?

答:
Assuming you are asking if a simple planar graph can be colored with six or more colors NOT less.

Notation : $m$ - number of edges, $n$ - number of vertices.

A graph is planar implies it has a vertex $v$ such that $degree(v) \leq 5$. This result is unconnected to whether the graph is connected or disconnected but is related to the fact that planar graphs has an upper bound on the number of edges $m \leq 3(n-2)$. Euler's formula does only apply for connected planar but the inequality $m \leq 3(n-2)$ holds for connected and disconnected graphs. This can be seen by adding edges to disconnected planar graph till it is connected, such that new edge count $m^+ > m$. Then since the new graph is now a connected planar graph $m^+ \leq 3(n-2)$ therefore the original edge count $m < 3(n-2)$.

The fact that a vertex exits in disconnected or connected planar graphs of $degree(v) \leq 5$ be proven by contradiction.
For contradiction assume there exists a planar graph with $degree(v) \geq 6$ for all vertices. Then $2m = \sum{degree(v)} \geq 6n$ or $m \geq 3n$. which is inconsistent with $m \leq 3n-6$.

The proof of the six coloring follows the lines you stated:
By induction you can assume graphs of order $n$ can be colored with 6 colors. Then consider a graph $G$ of order $n+1$. Since $G$ is planar it contains a vertex with degree less than 5. Then consider the graph $G-v$. Since $G-v$ is planar and of order $n$ it must have a 6 coloring. Since $degree(v) \leq 5$ we can use the sixth color to color $v$ in $G$ thus producing a 6 coloring.

But notice we never assumed that the graph is connected. Since we proved a more general result for all planar graph it applies equally to connected and disconnected graphs.


问:
Double summation index notation: $\Sigma_{i<j}$ versus $\Sigma_{i\neq j}$?
What is the difference between the summations using  $i<j$  and $i\neq j$ in the formula below:
$$\sigma^{2}(\boldsymbol{w})=\sum_{i} \tilde{w}_{i}^{2}+2 \sum_{i<j} \tilde{w}_{i} \tilde{w}_{j} \rho_{i, j}=\sum_{i} \tilde{w}_{i}^{2}+\rho(\boldsymbol{w}) \sum_{i \neq j} \tilde{w}_{i} \tilde{w}_{j}$$
Screenshot here.

Are both summations operationally equivalent?

If so, why break consistency and have two competing representations?

Which summation is more correct, or which to use for which situations?

One of the non-best answers here seem to apply, but not sure how in my case.

答:
$\sum_{i<j}$ sums over all the possible pairs $(i,j)$ for which $i<j$ holds. Similarly, $\sum_{i\neq j}$ sums over all the possible pairs $(i,j)$ for which $i\neq j$ holds.

For example, if $i$ and $j$ can take values in $\{1,2,3\}$, then

$$\sum_{i<j}a_{i,j}=a_{1,2}+a_{1,3}+a_{2,3},$$

whereas

$$\sum_{i\neq j}a_{i,j}=a_{1,2}+a_{1,3}+a_{2,1}+a_{2,3}+a_{3,1}+a_{3,2}.$$

If the summand is symmetric, i.e., $a_{i,j}=a_{j,i}$ holds for all $i$ and $j$, these two quantities are related by

$$\sum_{i\neq j}a_{i,j}=2\sum_{i<j}a_{i,j}.$$


问:
Property of partial traces
Consider the Kronecker product of $A \in M_m, B \in M_n$:
$A \otimes B = \left( \begin{matrix} a_{11}B&...&a_{1m}B\\ \vdots&\ddots\\a_{m1}B&...&a_{mm}B \end{matrix} \right)$

$A \otimes B$ can also be thought of as an $n \times n$ block matrix. This representation can be seen as $\left( \begin{matrix} T_{11}&...&T_{1n}\\ \vdots&\ddots\\T_{n1}&...&T_{nn} \end{matrix} \right)$, where each $T$ entry in the block is an $m \times m$ matrix.

If we define the two partial traces of a Kronecker product between two matrices as follows:

$Tr_1(A \otimes B) = \sum_{i=1}^n T_{ii}$. We have $Tr_1 \in M_m$.

$Tr_2(A \otimes B) = \left( \begin{matrix} Tr(T_{11})&...&Tr(T_{1n})\\ \vdots&\ddots\\Tr(T_{n1})&...&Tr(T_{nn}) \end{matrix} \right)$. We have $Tr_2 \in M_n$.

How does one show that for any $X \geq 0$, $Im(Tr_2(X) \otimes Tr_1(X)) \supset Im(X)$ (where $X$ is Hermitian and has the form $A \otimes B$ for some $A,B$)?

答:
For clarification, following the question, let us emphasize some notations first.

Let $M_n$ be the collection of $n\times n$ complex valued matrices, let $H_n$ be the collection of $n\times n$ complex valued Hermitian matrices, and given $X,Y\in H_n$, denote $X\ge 0\Leftrightarrow$ $X$ is positive semidefinite, and $X\ge Y\Leftrightarrow X-Y\ge 0$. Given $m,n$, for $X\in M_{mn}$, write $X=(X_{ij})_{n\times n}$, where $X_{ij}\in M_m$. Define
$${\rm tr}_1 (X)=\sum_{i=1}^n X_{ii}\in M_m,\quad {\rm tr}_2 (X)=({\rm tr}(X_{ij}))_{n\times n}\in M_n\tag{1}$$
and
$$\mathcal P(X)={\rm tr}_2(X)\otimes {\rm tr}_1(X)\in M_{mn}.\tag{2}$$
Note that  the definitions $(1)$ and $(2)$ for $X\in M_{mn}(X)$ not only depend on the the product $mn$, but also depend on $m$ and $n$ themselves. Since there is no much ambiguity in the following, for simplicity, we will not specify the dependence on $m,n$.

Claim: Given $X\in H_{mn}$,  $$X\ge 0\Rightarrow {\rm Im}(\mathcal P(X))\supset {\rm Im}(X).\tag{3}$$

The proof is based on the following facts. Since the argument is very long, let us omit the proof of the facts.

Fact 1. Given  $X,Y\in H_n$ with $X,Y\ge 0$,  $${\rm Im}(X)\supset {\rm Im}(Y)\iff {\rm Ker}(X)\subset {\rm Ker}(Y)\iff X\ge c Y \quad\text{for some} \quad c>0.\tag{4}$$

Fact 2. Given $A\in H_n$ and $B\in H_m$, we have $A\otimes B\in H_{mn}$ and
$$A\ge 0, B\ge 0 \Rightarrow A\otimes B\ge 0.\tag{5}$$

Fact 3. Given $X\in H_{mn}$,
$$X\ge 0 \Rightarrow {\rm tr}_1(X)\ge 0,\ {\rm tr}_2(X)\ge 0.\tag{6}$$

From the definitions $(1)$, $(2)$ and the facts 2 and 3, we have

Fact 4.
Given $Y,Z\in H_n$, if $Y,Z\ge 0$, then for $X=Y+Z$,
$$\mathcal P(X)\ge \mathcal P(Y)+\mathcal P(Z).\tag{7}$$

Fact 5.(special case of Sylvester's law of inertia)
If $X\in H_n$, $X\ge 0$ and $X$ has rank $1\le r\le n$, then there exist $r$ nonzero vectors $v_1,\cdots,v_r\in\mathbb C^n$, such that
$${\rm Im}(X)={\rm span}(v_1,\dots,v_r),\text{ and } X=\sum_{i=1}^rv_iv_i^*.\tag{8}$$

Here $v_i$ is denoted as column vector and $v_i^*$ denotes its conjugate transpose.

Proof of Claim:
Write $X=\sum_{i=1}^rv_iv_i^*$ as in $(8)$. Note that $v_iv_i^*\ge 0$. Then apply $(7)$ inductively we can obtain that
$$\mathcal P(X)\ge\sum_{i=1}^r\mathcal P(v_iv_i^*).$$
Now let us assume that the claim is true when the rank of $X$ is $1$. Under this assumption, from $(4)$ we know that for every $1\le i\le r$, there exists $c_i>0$, such that
$$\mathcal P(v_iv_i^*)\ge c_i v_iv_i^*.$$
It follows that for $c:=\min_{1\le i\le r}c_i>0$,
$$\mathcal P(X)\ge c\sum_{i=1}^r v_iv_i^*=cX\Rightarrow {\rm Im}(\mathcal P(X))\supset  {\rm Im}(X).$$
Therefore, it suffices to show that the claim is true when the rank of $X$ is $1$, i.e. $X=vv^*$ for some $v\in\mathbb C^{mn}$.

Denote $v=(w_1,\cdots,w_n)^t$, where $w_1,\cdots,w_n\in\mathbb C^m$ and $t$ means transpose. By definition,
$X=(w_iw_j^*)_{n\times n}$, so by $(1)$,
$${\rm tr_1}(X)=\sum_{i=1}^nw_iw_i^*, \quad{\rm tr_2}(X)=(x_{ij})_{n\times n}, \text{ where } x_{ij}= w_j^*w_i.\tag{9}$$

Note that by $(6)$, for $i=1,2$, ${\rm tr}_i(X)\ge 0$, and hence ${\rm Ker}({\rm tr}_i(X))$ and and ${\rm Im}({\rm tr}_i(X))$ are orthogonal to each other. Then it is easy to see that
$${\rm Ker}(\mathcal P(X))= {\rm Ker}({\rm tr_2}(X))\otimes\mathbb C^m + \mathbb C^n\otimes {\rm Ker}({\rm tr_1}(X)).\tag{10}$$

Therefore, due to $(4)$ and $(10)$, it suffices to show that for $u=u_2\otimes u_1\in \mathbb C^n\otimes\mathbb C^m$, if ${\rm tr}_i(X)u_i=0$ for $i=1$ or $2$, then $Xu=0$.

Denote $u_2=(a_1,\dots,a_n)^t$. Then $u=(a_1u_1,\dots, a_nu_1)^t$, and
$$Xu=\big((\sum_{j=1}^n a_jw_j^*u_1)w_1,\cdots,(\sum_{j=1}^n a_jw_j^*u_1)w_n\big)^t.\tag{11} $$
From $(9)$ we know that

${\rm Ker}({\rm tr}_1(X))$ is the orthogonal complement of ${\rm span}(w_1,\dots,w_n)$, so if ${\rm tr}_1(X)u_1=0$, then $w_j^*u_1=0$ for $j=1,\dots,n$, and hence $(11)$ is $0$.

${\rm tr}_2(X)u_2=\big((\sum_{j=1}^n a_jw_j^*)w_1,\cdots,(\sum_{j=1}^n a_jw_j^*)w_n\big)^t$, so if ${\rm tr}_2(X)u_2=0$, $(11)$ is $0$.

The proof of claim is completed. $\quad\square$

Title: Property of partial traces


问:
The Novikov field is a field
I have a questions on why the "Novikov field" defined in below is indeed a field:

Specifically, why every element $\sum_{g\in \Gamma}a_gT^g$ has an inverse? If

$$T^0=\sum_{g\in \Gamma}a_gT^g * \sum_{h\in \Gamma}b_hT^h:=\sum_{l\in \Gamma}(\sum_{g\in \Gamma}a_gb_{l-g})T^l$$

Basically, this means we need to solve the systems:
$\sum_{g\in \Gamma}a_gb_{l-g}=\delta_{0l}$, where $0$ is the additive identity in $\Gamma$. I don't know how to solve this for $b_h$'s.

Although this has a background in symplectic geometry, I hope the solution could be purely algebraic since the definition of this "field" is.

Source

答:
Let $\Gamma_0:= \Gamma\cap \mathbb{R}_{\geq 0}$ be the non-negative subgroup of $\Gamma$. Then $\Lambda$ is the localisation of the ring
$$\Lambda_0 := \big\{\sum_{g\in\Gamma_0 }a_gT^g:a_g\in \mathcal{K} \big\}  $$
at the unique maximal idea of $\Lambda_0$. Since $\mathcal{K}$ is a field the ring $\Lambda_0$ is an integral domain. And hence the localisation is a field.

Title: The Novikov field is a field


问:
Find a transformation such that the Fisher Information is constant in terms of the parameter.
Say you have a Gamma distribution with parameters $\alpha$ known and $\theta$ unknown. Find a transformation of $\theta$, $\eta=g(\theta)$ such that ${\cal I} (\eta)$, the Fisher Information, is constant in terms of $\eta$.

My problem is that the only way I can do this is by trial and error. At the moment I haven't been able to come up with any transformations that make this work. Is there a method that this can be done?

答:
In fact, it is a trivial task ! Denoting by prime the derivative with respect to $\theta$, the differential equation writes
\begin{equation}
\frac{\mathcal{I}^\prime(\theta)}{2 \mathcal{I}(\theta)} = \frac{\eta^{\prime\prime}(\theta)}{\eta^{\prime}(\theta)}.
\end{equation}
Integration is straightforward
\begin{equation}
\log(\eta^{\prime}(\theta))= \frac{1}{2}\log(\mathcal{I}(\theta)) + \text{cst}
\end{equation}
which leads to
\begin{equation}
\eta(\theta) \propto \int_0^\theta \sqrt{\mathcal{I}(t)} \mathrm{d}t + \text{cst}.
\end{equation}


问:
How to derive the simplest forms of $\sum{\binom{n}{i} \times 2^{n-i}}$ and $\sum{i \times \binom{n}{i} \times 2^{n-i}}$?
How can we derive these two equations
 $\sum{\binom{n}{i} \times 2^{n-i}}=3^{n}$
 $\sum{i \times \binom{n}{i} \times 2^{n-i}} = n \times 3^{n-1}$

The second equation can be related to the following problem:

You have $n$ balls and you can color each ball red, green, or blue. There are $3^n$ possible ways to color the balls. Among all possible ways, how many times we can get a ball of the color red.
 If we select $i$ balls from $n$ balls and color them red and color the remaining balls either blue or green, we can get ${i \times \binom{n}{i} \times 2^{n-i}}$ red balls for all possible ways selecting $i$ red balls each time.

How can we derive the equations?

答:
The first equation is immediate, by binomial expansion of $(2+1)^n$.

For the second, use

$$i\binom ni=n\binom{n-1}{i-1}.$$


问:
$P\left( n,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ Disaggregating Tail of Poisson
I have a Poisson tail $P\left( x,\left( {{\lambda }_{1}}+{{\lambda }_{2}} \right)T \right)$ which is sum of two independent Poisson distribution with rate $\lambda_1$ and $\lambda_2$. I am trying to write the distribution tail  (probability of $x$ events or more) in terms of two Poisson tails. So I am trying to disaggregate it.

where: $P\left( x,\lambda T \right)=\sum\limits_{i=x}^{\infty }{p\left( i,\lambda T \right)}$

Thanks!

答:
If $X$ is Poisson with mean $a+b$, a way to exhibit $X_a$ and $X_b$ independent and Poisson with respective means $a$ and $b$ such that $X=X_a+X_b$ is to consider an i.i.d. sequence $(U_n)_{n\geqslant1}$ of Bernoulli random variables with $P[U_n=1]=a/(a+b)$ and $P[U_n=0]=b/(a+b)$ for every $n\geqslant1$, independent of $X$, and to define
$$
X_a=\sum_{n=1}^XU_n,\qquad X_b=\sum_{n=1}^X(1-U_n).
$$
This is called thinning, as explained in about every textbook on Poisson processes, and can be checked by elementary computations: define $X_a$ and $X_b$ as above, compute the distribution of $(X_a,X_b)$, note that obviously, $X=X_a+X_b$ (for starters, show that the distribution of $X_a$ is Poisson with mean $a$).


问:
Is it true the following theorem: If $\lim_{x\to a} g(x) = A $ and $\lim_{x\to A} f(x) = B $ then $\lim_{x\to a} f(g(x)) = B $ ? Justify your answer
Is it true the following theorem: If $\lim_{x\to a} g(x) = A $ and $\lim_{x\to A} f(x) = B $ then $\lim_{x\to a} f(g(x)) = B $ ? Justify your answer

I think that I should show a function which denies the theorem but I can't find which one...

答:
You have observed that if $f,g$ are continuous, the statement holds. Hence to find a counterexample, we should find a function that is not continuous at $a$. For example, let

$$f(x)=g(x)=\mathbb1_{\{0\}} = \begin{cases}1 &\text{for } x = 0\\0 &\text{for } x \ne 0\end{cases}$$

Then

$$f(g(x)) = \begin{cases}f(1)=0 &\text{for } x = 0\\f(0)=1 &\text{for } x \ne 0\end{cases}$$

We have $\displaystyle\lim_{x \to 0}f(x) = 0 = \lim_{x\to0}g(x)$ but $\displaystyle\lim_{x \to 0}f(g(x))=1\ne 0$.


问:
Is there anyway to do this with eigen vectors and eigen basis?
So, say we have a square matrix A, and lets say that $A^2, A^3....$ so on are defined. Now If I want to find a general term $A^n$ is there any way to do that with eigen vectors. Take for example $A = \begin{bmatrix} 2 & 0 \\ 1 & -1\end{bmatrix}$. Like I would like to know if its possible to do it with eigen vectors instead of just trying to guess a pattern ? Like can we do something by converting the matrix into an eigen basis ?

答:
In general:

$$A = TDT^{-1},$$

where $D$ is a diagonal matrix such that its diagonal entries correspond to the eigenvalues of $A$, and the columns of $T$ are the eigenvectors of $A$.

Starting from this, we get:

$$A^n = \underbrace{TDT^{-1} \cdot  TDT^{-1} \cdot \ldots \cdot TDT^{-1}}_{n~\text{times}}= \\
= T D (TT^-1) D (TT^-1) \ldots (TT^-1) D T^{-1}= \\
= T D I D I \ldots I D T^{-1} = \\
= T D^n T^{-1}.
$$

Notice that $D^n$ is a diagonal matrix such that its diagonal entries are the $n$-th power of the eigenvalues of $A$.


问:
How to compute integrals using Riemann sums

Compute the integral using Riemann sums $$\int_0^s x^2 \,dx$$
Find the sum $ U_n $ of all rectangles below the function $y = x^3$

Find the sum $O_n$ of all rectangles above the function $y = x^3$

Take the limits to show that
$$\lim \limits_{x \to ∞}U_n = \lim \limits_{x \to ∞}O_n = \frac{s^3}3$$

I am vaguely familiar with Riemann sums, however, given this problem I really am not sure how to approach it.

I know that $\Delta$x = $\frac{s}n$ and I know that $x_i$ = $\frac{si}n$.

The next thing I believe I am supposed to do is compute $f(x_i) = (\frac{si}n)^2$ and then use that to compute $$\lim \limits_{x \to ∞} \sum_{i=0}^n [\frac{s^2i^2}{n^2}]·\frac{s}n $$
At this point it feels like things start getting convoluted and I am not sure how I would use this to find the sum of the rectangles above and below the function $y=x^3$.

Guidance to solving this question would be much appreciated.

Body:
Hint :

$$\lim_{n \to \infty} \frac{\sum_{i=0}^{n} i^2}{n^3} = \lim_{n \to \infty}\frac{n(n+1)(2n+1)}{6n^3} = \frac{1}{3}  $$


答:
If we have function where $f'>0$ in the domain such as this, if we take the area of the rectangle in a given area $x\in[a,b]$ then it is clear that $(b-a)f(b)>A$ and $(b-a)f(a)<A$ so we end up with a limit we can squeeze:
$$(b-a)f(a)<A<(b-a)f(b)$$
now to cover the whole domain we are interested in just take the sum

Title: How to compute integrals using Riemann sums

Body:
Compute the integral using Riemann sums $$\int_0^s x^2 \,dx$$
Find the sum $ U_n $ of all rectangles below the function $y = x^3$

Find the sum $O_n$ of all rectangles above the function $y = x^3$

Take the limits to show that
$$\lim \limits_{x \to ∞}U_n = \lim \limits_{x \to ∞}O_n = \frac{s^3}3$$

I am vaguely familiar with Riemann sums, however, given this problem I really am not sure how to approach it.

I know that $\Delta$x = $\frac{s}n$ and I know that $x_i$ = $\frac{si}n$.

The next thing I believe I am supposed to do is compute $f(x_i) = (\frac{si}n)^2$ and then use that to compute $$\lim \limits_{x \to ∞} \sum_{i=0}^n [\frac{s^2i^2}{n^2}]·\frac{s}n $$
At this point it feels like things start getting convoluted and I am not sure how I would use this to find the sum of the rectangles above and below the function $y=x^3$.

Guidance to solving this question would be much appreciated.


问:
Find all intervals $\mathbb X$ of real numbers for which the function $f(x)$ is invertible
We have the function $f:\mathbb{R}\longrightarrow \mathbb{R}$ and $f(x)=x^2-4x+2$.
I have to prove that the function is not invertible in $\mathbb{R}$ (I have done that) and
I have to find all intervals $\mathbb{X}$ of real numbers, for which the function is invertible in $\mathbb{X}$ and non-invertible in $\mathbb{R} - \mathbb{X}$.
It is obvious that we can define an inverse function of $f(x)$ for two intervals:

$\mathbb {X} = (-\infty; 2] \implies f^{-1}(x)=2-\sqrt{x+2}$ and $\mathbb{R} - \mathbb{X}=(2; +\infty)$

and

$\mathbb {X} = [2; +\infty)\implies f^{-1}(x)=2+\sqrt{x+2}$ and $\mathbb{R} - \mathbb{X}=(-\infty; 2)$

However those two intervals cannot be merged into one answer. Morevover, I have the feeling I am missing something. For example, is $\mathbb {X} = \{2\}$ a valid interval, too? Then I have a third case and $\mathbb{R} - \mathbb{X}=(-\infty;2) \cup  (2; +\infty)$.
Have I understood the question correctly and should I unclude that last third interval? How should I write down my answer when I have all these cases?

答:
You do not explain what it means that $f$ is invertible on a subset $M \subset \mathbb R$. Probably you mean that $f$ maps $M$ bijectively onto $f(M)$. Then you get an inverse $f_M^{-1} : f(M) \to M$ for $f_M : M \to f(M)$.

If you agree with this interpretation, then clearly $f$ is invertible on $M$ of and onyl if $f \mid_M : M \to \mathbb R$ is injective.

If you admit degenerate intervals $[a,a]$, then each single point subspace $\mathbb X =\{a\}$ is an interval.

Let us first show that $f$ is injective on an interval $\mathbb X$ (in which case we call $\mathbb X$ admissible) if and only if either $\mathbb X \subset [2,\infty)$ or $\mathbb X \subset (-\infty,2]$.
One half has been shown by you. For the converse let $\mathbb X$ be any interval not contained in $[2,\infty)$ or $(-\infty,2]$. This means that $\mathbb X$ contains a point $a$ of $\mathbb R \setminus [2,\infty) = (-\infty,2)$ and a point $b$ of $\mathbb R \setminus (-\infty,2] = (2,\infty)$. Thus $a < 2 < b$ and $[a,b] \subset \mathbb X$. Let $r = \min(2-a,b-2)$. Then $2-r,2+r \in \mathbb X$ and $f(2-r) = f(2+r)$.

Now we are looking for admissible intervals $\mathbb X$ such that $f$ is not injective on $\mathbb Y  = \mathbb R \setminus \mathbb X$.

If $\mathbb X$ is bounded, then $f$ is not injective on $\mathbb Y$. In fact, $\mathbb X \subset (2-R,2+R)$ for some $R > 0$, thus $2-R,2+R \in \mathbb Y$ and $f(2-R) = f(2+R)$.

Now let $\mathbb X$ be unbounded. Then $\mathbb Y$ is also an unbounded interval and it must not be admissible, i.e. it must not be contained in $[2,\infty)$ or $(-\infty,2]$. Thus, if $\mathbb X \subset [2,\infty)$ has left boundary point $a$, we must have $a > 2$, and if $\mathbb X \subset (-\infty,2]$ has right boundary point $b$, we must have $b < 2$

This gives a complete answer: The desired intervals $\mathbb X$ are the precisely the  subintervals of $[2,\infty)$ which are bounded or have left boundary point $> 2$ and the  subintervals of $(-\infty,2]$ which are bounded or have right boundary point $< 2$.


答:
Factor out what can be factored:
$$\sum_{i=1}^n \frac{s^2i^2}{n^2}·\frac{s}n=\frac{s^3}{n^3}\sum_{i=1}^{n}i^2=\frac{s^3}{n^3}\frac{n(n+1)(2n+1)}6$$
for the upper Riemann sum, and almost the same for the lower one ($i=0$ to $n-1$).


问:
$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $
Let be a circle inscribed in the triangle $\triangle ABC$ wiht the center $I$. The intersection of the circle with $AI$ is $A'$, with $BI$ is $B'$ and with $CI$ is $C'$.

Prove that:
  $$\frac{AB}{A'B'}+\frac{BC}{B'C'}+\frac{CA}{C'A'} \geq 4 \left(\sin{\frac{A}{2}}+\sin{\frac{B}{2}}+\sin{\frac{C}{2}}\right). $$

thanks. seems to be to hard for me.

答:
$AB=r\left(\cot\dfrac{A}{2}+\cot\dfrac{B}{2}\right)=r\dfrac{\sin \left(\dfrac{A+B}{2}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}=r\dfrac{\cos \dfrac{C}{2}}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}=r\dfrac{\sin \dfrac{π+C}{2}}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}},$

$A'B'=r\sqrt{2\left(1-\cos \left(\dfrac{π+C}{2}\right)\right)}=2r\sin \left(\dfrac{π+C}{4}\right)$

$\dfrac{AB}{A'B'}=\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}$ then the inequality become:

$\dfrac{\cos \left(\dfrac{π+C}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{B}{2}}+\dfrac{\cos \left(\dfrac{π+B}{4}\right)}{\sin \dfrac{A}{2}\sin \dfrac{C}{2}}+\dfrac{\cos \left(\dfrac{π+A}{4}\right)}{\sin \dfrac{C}{2}\sin \dfrac{B}{2}} \ge 4\left(\sin {\dfrac{A}{2}}+\sin {\dfrac{B}{2}}+\sin {\dfrac{C}{2}}\right) $ $\iff$

$\cos \left(\dfrac{π+C}{4}\right)\sin \dfrac{C}{2}+\cos \left(\dfrac{π+B}{4}\right)\sin \dfrac{B}{2}+\cos \left(\dfrac{π+A}{4}\right)\sin \dfrac{A}{2} \ge 4\sin {\dfrac{A}{2}}\sin {\dfrac{B}{2}}\sin {\dfrac{C}{2}}\left(\sin {\dfrac{A}{2}}+\sin {\dfrac{B}{2}}+\sin {\dfrac{C}{2}}\right)$

$x=\dfrac{A}{2},y=\dfrac{B}{2},z=\dfrac{C}{2} →$

$f=\cos {\dfrac{π+2z}{4}}\sin {z}+\cos {\dfrac{π+2y}{4}}\sin {y}+\cos {\dfrac{π+2x}{4}}\sin {x} - 4\sin {x}\sin {y}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)$

$g=x+y+z-\dfrac{π}{2}$ , $F=f-\lambda g$

$F_{x}=\cos {\dfrac{π+2x}{4}}\cos {x}-\dfrac{1}{2}\sin {\dfrac{π+2x}{4}}\sin {x}-4\cos {x}\sin {y}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {x})=\lambda$ <1>

$F_{y}=\cos {\dfrac{π+2y}{4}}\cos {y}-\dfrac{1}{2}\sin {\dfrac{π+2y}{4}}\sin {y}-4\cos {y}\sin {x}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {y})=\lambda$ <2>

$F_{z}=\cos {\dfrac{π+2z}{4}}\cos {z}-\dfrac{1}{2}\sin {\dfrac{π+2z}{4}}\sin {z}-4\cos {z}\sin {y}\sin {x}\left(\sin {x}+\sin {y}+\sin {z}\right)-4\sin {x}\sin {y}\sin {z}\cos {z})=\lambda$ <3>

<1>- <2>  :
$ \sin {\dfrac{x-y}{4}}\left(\sin {\dfrac{π+2z}{8}}-3\sin {\dfrac{3z}{4}} \left(3-4\left(\sin {\dfrac{x-y}{4}}\right)^2 \right)+32\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)-16\sin {x}\sin {y}\sin {z}\cos {\dfrac{x-y}{4}}\sin {\dfrac{x+y}{2}} \right)=0 $ $→ \sin {\dfrac{x-y}{4}}P_{1}\left(x,y,z\right)=0$ <4>

<2>-<3>: $\sin {\dfrac{y-z}{4}}P_{2}\left(x,y,z\right)=0$, $P_{2}\left(x,y,z\right)$ is similar as above.

$|x-y|<\dfrac{π}{2} → \cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}} > 0.65 $ and $ \sin x+\sin y+\sin z \ge \sin \left(x+y+z\right)=1 → 16\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)> 9\sin {\dfrac{3z}{4}}$

it is trivial that:$16\cos {\dfrac{x-y}{4}}\cos {\dfrac{x-y}{2}}\sin {z}\left(\sin {x}+\sin {y}+\sin {z}\right)>16\sin {x}\sin {y}\sin {z}\cos {\dfrac{x-y}{4}}\sin {\dfrac{x+y}{2}}$

$→ P_{1}\left(x,y,z\right)>0$ \sin ce it is symmetric, we have $P_{2}\left(x,y,z\right)>0$ also.

so the only solution is $x=y=z=\dfrac{π}{6}$ ,we put in $ f, f=0$,put $x=y=\dfrac{π}{2},z=0, → f>0$ so $f_{min}=0$ QED.


问:
Hilbert spaces - linear and bounded operator
$X$ and $Y$ are two Hilbert spaces. $A: X\longrightarrow Y$ is linear and bounded operator.
How to Show that $ImA$ the image of A is not closed in general ( I'd like an example)?

答:
Let $X=Y$ have orthonormal basis $e_1, e_2, \dots$.  Then let $A(e_k) = \frac{1}{k} e_k$.

Every $e_k$ is in the image, so the image is dense.  Also
$$
x = \sum_{k=1}^\infty \frac{1}{k}e_k
$$
is in $X$ because $\sum\frac{1}{k^2}$ converges.  But this $x$ is not in the image of $A$, since the only choice $y$ with $A(y) = x$ would be
$$
y = \sum_{k=1}^\infty e_k
$$
which is not in $X$ because $\sum 1^2$ diverges.


问:
Show that this sequence of functions converges uniformly to $e^{-x}$
Define

$$
f_n(x)=
\begin{cases}
\left(1-\frac xn\right)^n & 0\le x<n \\
0 & x\ge n
\end{cases}
$$

I am able to show that $f_n(x)\to e^{-x}$ pointwise on $[0,+\infty)$, and I wonder if my justification for uniform convergence is valid:

By properties of logarithm, I transformed the sequence into

$$
\left(1-\frac xn\right)^n=\exp\left\{n\log\left(1-\frac xn\right)\right\}
$$

Employing the fact that for sufficiently large $n$

$$
\left|n\log\left(1-\frac xn\right)+x\right|=\left|n\left[-\frac xn+\mathcal O\left(1\over n^2\right)\right]+x\right|=\mathcal O\left(\frac1n\right)
$$

we have $n\log\left(1-\frac xn\right)\to-x$ uniformly, hence for each $\delta>0$ there exists an $N$ such that for all $n>N$:

$$
|\log f_n(x)+x|<\delta
$$

By the uniform continuity of the exponential function, for all $\varepsilon>0,x>0,y>0$, we have $|e^{-x}-e^{-y}|<\varepsilon$ whenever $|x-y|<\delta$. As a result let $y$ be $-\log f_n(x)$, so  for all $\varepsilon>0,x>0$ there exists an $N$ such that for all $n>N$

$$
|f_n(x)-e^{-x}|<\varepsilon
$$

I wonder if this proof is correct or please point out my mistake. Thank you all!

答:
I think you need to look closer at what's going on in the Taylor series. You can clean up the argument that $f_n(x):=n\log\left(1-\frac{x}{n}\right) \to -x$ uniformly, but this only works if $x\leq M$ for some $M>0$. To see this:

By Taylor's theorem, we know
$$
f_n(x/n)=n\underbrace{f_n(0)}_{=0}+\underbrace{f_n'(0)}_{=-1}x+\frac{f_n''(c)}{2}\left(\frac{x^2}{n}\right)
$$
for some $c$ between $0$ and $x/n$. Hence,
\begin{align}
|f_n(x/n)+x|\leq \left|\frac{f_n''(c)}{2}\right|\left(\frac{|x|^2}{n}\right)\leq \left|\frac{f_n''(c)}{2}\right|\left(\frac{M^2}{n}\right).
\end{align}
An easy calculation shows $f_n'(x)=\frac{-n}{1-\frac{x}{n}},\, f_n''(x)=\frac{-1}{(1-\frac{x}{n})^2}$. Now if $x$ is not demanded to be bounded above by some $M>0$ independent of $n$,  then clearly $f_n''(c)$ is unbounded. However, with this demand, we have
$$
|f_n''(c)|\leq \frac{1}{(1-\frac{M}{n})^2}\leq C<\infty
$$
From here, your argument about the exponential being uniformly continuous allows you to assert that $e^{f_n(x)} \to e^{-x}$ uniformly on $[0,M]$.

Finally, given $\epsilon>0$, pick $M$, such that $e^{-x}<\epsilon$ for all $x>M$. Then for sufficiently large $n$, it is obvious that we have uniform convergence.


问:
Solving for $d$ from $a=r^2\arccos{\frac{d}{2r}}-\frac{d}{2}\sqrt{r^2-\left(\frac{d}{2}\right)^2}$
I would like to solve this equation for d with a given r value. Is there a way to write in explicit form? $a$ is of course a known value, I am a little rusty on trigonometry.

$$a  = r^2\arccos{\frac{d}{2r}}-\frac{d}{2}\sqrt{r^2-\left(\frac{d}{2}\right)^2}
$$

Thank you very much.

EDIT: corrected the equation

Body:
The equation bing highly transcendental, there is no analytical solution (remember that this is already the case for $x=\cos(x)$) and you will need some numerical method.

However, we can make approximations. Let $d=2rx$ to make
$$\cos ^{-1}(x)-x \sqrt{1-x^2}=k \qquad \text{with} \qquad k=\frac a{r^2}$$
Using Taylor expansions built around $x=0$, the lhs write
$$\frac{\pi }{2}-2 x+\frac{x^3}{3}+\frac{x^5}{20}+\frac{x^7}{56}+\frac{5
   x^9}{576}+\frac{7 x^{11}}{1408}+\frac{21 x^{13}}{6656}+\frac{11
   x^{15}}{5120}+O\left(x^{17}\right)$$ which is quite good (draw the two functions).

Now, using series reversion, we should have (I could have added more terms)
$$x=t+\frac{t^3}{6}+\frac{13 t^5}{120}+\frac{493 t^7}{5040}+\frac{37369
   t^9}{362880}+O\left(t^{11}\right)$$ where $t=\frac {\pi-2k}4$.

Trying for a few values of $k$
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{exact} \\
 0.00 & 0.928261 &  1.000000 \\
 0.05 & 0.884346 &  0.910264 \\
 0.10 & 0.842841 &  0.856758 \\
 0.15 & 0.803428 &  0.811397 \\
 0.20 & 0.765836 &  0.770519 \\
 0.25 & 0.729828 &  0.732608 \\
 0.30 & 0.695202 &  0.696855 \\
 0.35 & 0.661782 &  0.662761 \\
 0.40 & 0.629417 &  0.629992 \\
 0.45 & 0.597977 &  0.598311 \\
 0.50 & 0.567349 &  0.567540 \\
 0.55 & 0.537436 &  0.537543 \\
 0.60 & 0.508153 &  0.508212 \\
 0.65 & 0.479429 &  0.479460 \\
 0.70 & 0.451198 &  0.451214 \\
 0.75 & 0.423406 &  0.423414 \\
 0.80 & 0.396003 &  0.396007 \\
 0.85 & 0.368946 &  0.368948 \\
 0.90 & 0.342198 &  0.342199 \\
 0.95 & 0.315725 &  0.315725 \\
 1.00 & 0.289494 &  0.289494
\end{array}
\right)$$ This seems to be quite decent except for small values of $k$ that is to say for $x$ close to $1$.

To improve it, doing the same kind of work around $x=1$, the lhs write
$$\frac{4\sqrt{2}}{3}  (1-x)^{3/2}+\frac{\sqrt{2}}{5}  (1-x)^{5/2}-\frac{\sqrt{2}}{56}(1-x)^{7/2}+\frac{\sqrt{2}}{288 }(1-x)^{9/2}+\cdots$$ and series reversion will give
$$\sqrt{1-x}=u-\frac{u^3}{20}+\frac{81 u^5}{5600}-\frac{1171 u^7}{201600}+\cdots$$ where $u=\sqrt[3]{\frac{3 k}{4 \sqrt{2}}}$.
which gives  better estimates for small values of $k$.
$$\left(
\begin{array}{ccc}
k & \text{estimate} & \text{exact} \\
 0.00 & 1.000000 &  1.000000 \\
 0.05 & 0.911847 &  0.910264 \\
 0.10 & 0.860754 &  0.856758 \\
 0.15 & 0.818272 &  0.811397
\end{array}
\right)$$


答:
Let $X,Y$ be the Hilbert space $\ell^2(\mathbb R)$ and $A$ defined by $(A \xi)_i = \frac{\xi_i}{i}$ for all $i \in \mathbb N$.

The sequence $x_1=(1, 0, \dots, ), x_2=(1, 1, 0, \dots, ), x_2=(1, 1, 1, 0, \dots, ), \dots$ is in $\ell^2(\mathbb R)$.

$\{A.x_n\}$ converges in $\ell^2(\mathbb R)$ to the harmonic sequence. However the harmonic sequence is not in $A[\ell^2(\mathbb R)]$ as the constant sequence equal to $1$ is not in $\ell^2(\mathbb R)$.

Title: Hilbert spaces - linear and bounded operator

Body:
$X$ and $Y$ are two Hilbert spaces. $A: X\longrightarrow Y$ is linear and bounded operator.
How to Show that $ImA$ the image of A is not closed in general ( I'd like an example)?


问:
Question 36 of chapter 16 from Contemporary Abstract Algebra by Gallian

Let $\phi$ be the ring homomorphism from $\mathbb{Z}[x]$ to $\mathbb{Z}$ given by $\phi(f(x)) = f(1)$. Find a polynomial $g(x)\in\mathbb{Z}[x]$ such that  $\ker \phi=\left<g(x)\right>$. To what ring is  $\mathbb{Z}[x]/\ker\phi$ isomorphic?

I tried in the following way

$g(x) = x-1$ satisfies $\ker \phi$

I know that first theorem of ring homomorphism says that $\mathbb{Z}[x]/\ker \phi$ is isomorphic to $\phi(\mathbb{Z}[x])$ but I don't know how do I proceed further.

答:
It seems that you know that $\ker \phi = \langle x-1 \rangle$ and that $\mathbb{Z}[x]/\ker \phi$ is isomorphic to $\phi(\mathbb{Z}[x])$. But $\phi$ is surjective since the constant polynomials $p_n(x) = n$ are mapped by $\phi$ to $n \in \mathbb Z$. Thus  $\phi(\mathbb{Z}[x]) = \mathbb Z$.

Edited:

To show that $\ker \phi = \langle x-1 \rangle$ observe that $f(x) \in \ker \phi$ iff $f(1) = 0$, i.e. $f$ has a zero at $x = 1$. It is well-known that then $f(x) = g(x) \cdot (x-1)$ for some polynomial $g(x)$ (use polynomial long division). Thus $\ker \phi = \langle x-1 \rangle$.


问:
Proving that $\mathbb{Z}[x]/\langle x^2 + 1\rangle$  is isomorphic to $\mathbb{Z}[i]$
I have some general idea of going about this. I'm trying to come up with a suitable mapping. I've come up with $\phi\colon a + \langle x^2 + 1\rangle\longmapsto a$ where $a$ is an irreducible element in $\mathbb{Z}[i]$ but this mapping has some issues. For example take
$5x + x^2 + 1$, which is $5x \pmod{x^2+1}$. $5x$ does not belong to $\mathbb{Z}[i]$. So my question is, is there a better mapping?

I know for a fact that if you add any integer call it $b$ to $\langle x^2 + 1\rangle$, you will get $b$ back which belongs to $\mathbb{Z}[i]$. I'm stuck when $b$ is not an integer.

答:
Hint: Let $\varphi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$ be the evaluation homomorphism and use the first isomorphism theorem.

Edit: Added for confusion on both evaluation homomorphism and first isomorphism theorem.

Let $R$ be a commutative ring. Then $R[x]$ is defined to be ring consisting of all polynomials in $x$ with coefficients in $R$. So for $\mathbb{Z}[x]$, we have elements such as $5, x^2+1,$ and $3x^7-9x^3+4$. We can naturally extend this definition to adjoin arbitrary elements instead of a placeholder variable. In particular, we can think of $\mathbb{Z}[i]$ as the ring of all polynomials in $i$ with coefficients in $\mathbb{Z}$. With some thought we realize that this is precisely equivalent to the normal definition of the Gaussian integers since $i^2=-1$ (i.e. polynomials in $i$ of degree greater than 1 can be reduced to the form $a+bi$).

Going back to polynomial rings, let $\psi: R \rightarrow S$ be a homomorphism of commutative rings. Then there is a conceptually natural way to extend this homomorphism to a homomorphism $\varphi: R[x] \rightarrow S$. By our definition of $R[x]$, every element can be written as $\sum r_i x^i$, with the $r_i \in R$. Then, since $\varphi$ is to be a homomorphism,
$$\varphi(\sum r_i x^i) =  \sum \varphi(r_i) \varphi(x)^i$$
So $\varphi$ is completely determined by the image of $R$ and $x$. Since our desire is to extend $\psi$ naturally, it makes sense to send all elements of $R$ to their image under $\psi$. So we only need choose what $x$ should be mapped to.

From this we can define the evaluation homomorphism between $R[x]$ and $R[\alpha]$ to be the evaluation of any polynomial in $R[x]$ at $x=\alpha$. So $\sum r_i x^i \mapsto \sum r_i \alpha^i$. Take time to convince yourself that it is a completely natural definition and that it is, in fact, unique. In addition to that, it is not hard to show that the homomorphism must be surjective.

To return to your example, let $\varphi: \mathbb{Z}[x] \rightarrow \mathbb{Z}[i]$ be the evaluation homomorphism. Then we have a natural, well-defined ring homomorphism and can look at its image. For any $a + bi \in \mathbb{Z}[x]$, $\varphi(a+bx) = a+bi$. So we see, as mentioned above, that $\varphi$ is surjective. Then we can use the first isomorphism theorem to say that $\mathbb{Z}[i] \simeq \mathbb{Z}[x]/\ker \varphi$.

Finally, the first isomorphism theorem states for any ring homomorphism $\varphi: R \rightarrow S$, $\text{im } \varphi \simeq R / \ker \varphi$. So if $\varphi$ is surjective, as is the case here, then $S \simeq R / \ker \varphi$.


问:
Linear transformation defined by a basis
I want to be sure that the follwing statement is true.

Let $V, W$ be $\mathbb K$-vector spaces. Let $\{v_1, \ldots, v_n\}$ be
a basis of $V$ and $\{w_1,\ldots,w_m\}$ ($m$ can be different from $n$)
be an arbitrary set of vectors of $W$. Then there exists a unique
linear map $F:V\longrightarrow W$ such that $F(v_i)=w_i$ for every
$i=1,\ldots,n$.

Is it true? I am not sure that $m$ can be different from $n$...

Thank you

答: If you let $n=m$, the statement is true, and this is a standard argument. If $n<m$, then you're essentially ignoring the last few vectors $w_{n+1},\dots, w_m$, so again it reduces to the case $n=m$. The final case $n>m$ doesn't make sense because if $n>m$, then the vector $w_n$ isn't even defined.


问:
if $f(x,y,y')$ does not depend on $x$ then the Euler-Lagrange equation becomes in:
if $f(x,y,y')$ does not depend on $x$ then the Euler-Lagrange equation becomes in:

$$f(x,u,u')-u'f_{y'}(x,u,u')=c  $$ in $[x_0,x_1]$ for some $c \in \mathbb{R}$

Ok I need prove this, I know that if $u$ is a local minimum then

$$f_{y}(x,u,u')-\frac{d}{dx}(f_{y}(x,u,u'))=0 $$ in $[x_0,x_1]$

答:
Given $f(x,y,y')$ we have

$$
\frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}y'+\frac{\partial f}{\partial y'}y''\ \ \ \ \ \ (1)
$$

and also

$$
\frac{d}{dx}\left(\frac{\partial f}{\partial y'}y'\right)=\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)y'+\frac{\partial f}{\partial y'}y''\ \ \ \ \ \ \ (2)
$$

and according the Euler-Lagrange equation we have

$$
\frac{\partial f}{\partial y}-\frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right)=0\ \ \ \ \ \ \ \ \ \ \ (3)
$$

then using $(3)$ into $(2)$ we obtain

$$
\frac{d}{dx}\left(\frac{\partial f}{\partial y'}y'\right)=\frac{\partial f}{\partial y}y'+\frac{\partial f}{\partial y'}y''\ \ \ \ \ \ \ (4)
$$

now from $(1),(4)$

$$
\frac{d}{dx}\left(\frac{\partial f}{\partial y'}y'-f\right)=-\frac{\partial f}{\partial x}
$$

but $\frac{\partial f}{\partial x} = 0$ because $f=f(y,y')$ hence

$$
\frac{\partial f}{\partial y'}y'-f = C
$$


问:
Every homeomorphism of the torus is homotopic to a homeomorphism with a fixed point
Let $f : T^2 \to T^2$ be a self-homeomorphism of the torus. I'm trying to prove that there exist a homeomorphism $g: T^2 \to T^2$ with a fixed point, such that $f$ and $g$ are homotopic.

My initial attempt was to lift $f$ to a homeomorphism $\tilde{f}$ of the universal cover $\mathbb{R}^2$ of $T^2$, and restrict $\tilde{f}$ to a fundamental domain of the covering $\mathbb{R}^2 \to T^2$ given by the unit square $U$. Since the unit square is homeomorphic to the closed disk, the domain and image of $\tilde{f}|_U$ are homeomorphic to a closed disk. One can then apply the Brouwer fixed point theorem to the composition of $\tilde{f}|_U$ with the homeomorphisms. However, I'm not sure how I can use this to find a map homotopic to $\tilde{f}$ (and such that the homotopy is equivariant under the action of the deck group on $\mathbb{R}^2$) with a fixed point.

答:
You don't need the Brouwer fixed point theorem, you just need a very concrete construction.

The key fact is that for every two points $P,Q \in T^2$ there exists a homeomorphism $g : T^2 \to T^2$ such that $g(P)=Q$ and such that $g$ is homotopic to the identity. Once you know that, choosing any $Q \in T^2$ and letting $P = f(Q)$, it follows that $g \circ f$ is homotopic to $f$ and $g \circ f(Q) = Q$.

The function $g$, and the homotopy, can be defined first up in the universal cover. Choose lifts $\tilde P, \tilde Q \in \mathbb R^2$ of $P,Q$. Now just do a straight line homotopy, sliding each point along the displacement vector $\tilde Q - \tilde P$:
$$\tilde H(a,t) = a + (1-t)(\tilde Q-\tilde P)
$$
and so $\tilde g(\tilde P)=\tilde H(\tilde P,0)=\tilde Q$, and $\tilde H(\tilde P,1)=\tilde P$.

Now you have to show that this straight line homotopy upstairs induces a homotopy downstairs. More precisely, letting $[a]=a+\mathbb Z^2 \in T^2$ denote the orbit of $a$ under the action of the deck group, you have to check that the formula
$$H([a],t) = [a+(1-t)(\tilde Q- \tilde P)]
$$
gives a well-defined homotopy on $T^2$, and that the function $g([a]) = H([a],0)$ satisfies $g(P)=Q$, and $g([a],1)=[a]$ for all $[a] \in T^2$.


问:
Functional equation $f(x)f(f(x))=x^2$
Find all real functions such that
$f(x)f(f(x))=x^2$ and $f(x)=x$ for some $x$.

Obviously $f(x)=x$ is a solution, but I have no clue how to find other solutions.

答:
Or solutions of the form : $f(x)=ax^b$

Substitutin gives $f(x)=x,\frac{1}{x^2}$

Title: Functional equation $f(x)f(f(x))=x^2$


问:
The weight of the extended ternary Golay code
I need to show that the extended ternary Golay code with parameters $[12,6]$ is a self dual code and has minimum weight $6$ so i will obtain a $[12,6,6]$-code.
I showed the self duality but how can I show that the minimum weight is $6$?

Thanks for your help...

答:
Hint: Show that the code has no words of weight three. Also use the fact that if
$\vec{c}=(c_1,c_2,\ldots,c_{12})$ is a codeword, then, by self-duality, its weight
$$
w=\vec{c}\cdot\vec{c}\equiv0\pmod3.
$$
The first equality follows from the fact $c_i^2=1$ for all non-zero values of $c_i$.

The most convenient way of excluding the possibility of weight three words may be to use the description of the (unextended) ternary Golay code as a QR-code (=quadratic residue code), and the general results on the minimum distance of QR-codes.


问:
What is the Op Category?
By what is, I do not mean the construction, the construction is easy. What I mean is what does the Op category really represent to us in terms of the original? Oh sure, it represents the dual... but that is circular reasoning.

For every category $\textbf{C}$ there is a mirror image category $\textbf{C}^{op}$. When we "walk" through $\textbf{C}$ we are walking through $\textbf{C}^{op}$ "contravariantly" BUT $f^{op}$ does something different for us.

So the question really is, what is $f^{op}$ actually doing to the internal structure? If I have some category $\textbf{C}$ and follow some path of arrows from a to b then in $\textbf{C}^{op}$ I have to take the path of arrows from b to a and this somehow tells me something relevant. In some sense it seems like an inverse. It is almost as if $f^{op}$ is some "generalized" inverse that takes us back from where we came. If that sorta is true then $\textbf{C}^{op}$ is sort of an "inverse" category that tracks all changes, a sort of "undo" class that allows us, in theory, to go back.

If that is true then it would not be much different than just saying that for every arrow in a category there is an op-arrow that sort of acts like a generalized inverse that in theory lets us reverse any path of arrows.

I guess this would make total since if a category that has inverse for all arrows is self-dual(hence the generalized inverses are just normal inverses and).

Anyways, I'm trying to get some type of intuitive understanding of the opposite category so that when it is talked about I can actually reason from it in a concrete way. This inverse analogy seems like it would work but it also seems that $f^{op}$ doesn't necessarily, by definition, have to have any relation to $f$, it just has to be any arrow that reverses direction.

This leads me to another problem I have, in the definition of the opposite category it seems that we can actually have many opposite categories. Only the arrow has to be reverses, but the definition doesn't say anything about it having to have any other relationship to $f$ other than exchanging the domain and codomain. Of course, it is clear that this is not true, I haven't been able to figure out way from the book.

I guess this is really ultimately asking how does $f^{op}$ really relate to $f$ rather than just saying it is the opposite?

答:
This leads me to another problem I have, in the definition of the opposite category it seems that we can actually have many opposite categories. Only the arrow has to be reverses, but the definition doesn't say anything about it having to have any other relationship to f other than exchanging the domain and codomain. Of course, it is clear that this is not true, I haven't been able to figure out way from the book.

I will try to clear up your confusion here.

Suppose we have a category $\mathcal C$. This is how we form the opposite category $\mathcal C^\text{op}$.

For every object $A$ in $\mathcal C$ we have an object $A^\text{op}$ in $\mathcal C^\text{op}$. There are no other objects.

For every morphism $f : A \to B$ in $\mathcal C$, we have a morphism $f^\text{op} : B^\text{op} \to A^\text{op}$. There are no other morphisms.

We also have to specify how composition works (we will denote composition in $\mathcal C$ by $\circ$, and in $\mathcal C^\text{op}$ by $\circ ^\text{op}$).

If we have two morphisms $A^\text{op} \to B^\text{op}$ and $B^\text{op} \to C^\text{op}$ we know they are of the form $f^\text{op}$ and $g^\text{op}$ for some $f : B \to A$ and $g : C \to B$ in $\mathcal C$. Then we will compose them according to this rule:

$g^\text{op} \circ^\text{op} f^\text{op} = (f \circ g)^\text{op}$

This is the full definition of the opposite category. It is not too hard to verify that it is indeed a category: we need to check that $\circ ^\text{op}$ is associative, and that there are identities ($id_A ^\text{op}$ is the identity on $A^\text{op}$). It is common to not write the $``{^\text{op}}''$, but for the sake of clarity---so that we do not get confused about which category we are in---I will leave it.

So, what is the relation between $f$ and $f^\text{op}$? In some sense, they are the same morphism. This is because $f^\text{op}$ composes in the same way as $f$ does, except in the other direction.

In particular, $f^\text{op}$ is not merely "any arrow [in $\mathcal C$, I assume you meant] that reverses the direction", because it's not an arrow of $\mathcal C$. It's also not in any sense an inverse to $f$, because they are in different categories, and cannot be composed.

So, the arrows of $\mathcal C ^\text{op}$ are precisely the arrows of $\mathcal C$, but we pretend they go in reverse. We denote this by writing $f^\text{op} : B ^\text{op} \to A ^\text{op}$ for the reverse of the arrow $f : A \to B$. To calculate a composition of these "reversed arrows", we unreverse them, compose them in the way that $\mathcal C$ tells us to, and reverse the result.

Title: What is the Op Category?


问:
Reference on different types of integral domains
I am looking for a good reference (book or otherwise) which has a comprehensive study of the different types of integral domains, including:

Euclidean domains

UFDs

GCD domains

PIDs

Dedekind domains

Bezout domains

etc...

and especially the relationships between them (e.g. all Euclidean domains are PIDs, a Bezout domain is a PID iff it has the accp, etc). Thanks!

答:
While I've never used a book that had all these topics at once, it does look like Bourbaki's commutative algebra volume might cover them:

Bourbaki, Nicolas. Commutative algebra. Vol. 8. Hermann, 1972.

And Kaplansky may have a good deal of them:

Kaplansky, Irving. Commutative rings. Allyn and Bacon, 1970.

Any good beginning ring theory book covers PIDs EDs and UFDs, namely

Isaacs, I. Martin. Algebra: a graduate course. Vol. 100. American Mathematical Soc., 2009.

Bhattacharya, Phani Bhushan, Surender Kumar Jain, and S. R. Nagpaul. Basic abstract algebra. Cambridge University Press, 1994.

Grillet, Pierre Antoine. Abstract algebra. Vol. 242. Springer Science & Business Media, 2007.

I think Jacobson's Basic Algebra books contain at least UFDs and Dedekind domains.

For other more special types you'll probably do fine just looking at the references to articles given in the wikipedia articles.

As for information about their interrelationships there is a (not necessarily totally comprehensive) graph at the Database of Ring Theory for domains.


问:
Relations between $K_3$ complex surface and $\mathbb{CP}^2$

What are the relations between $K_3$ complex surface and  $\mathbb{CP}^2$?

For one thing, I know that $K_3$ generates 4-dimensional bordism group $\Omega_4^{Spin}=\mathbb{Z},$
such that it has a signature $\sigma $ of $K_3$ is -16.

For another thing, I know that $\mathbb{CP}^2$ generates 4-dimensional bordism group $\Omega_4^{SO}=\mathbb{Z},$
such that it has a signature $\sigma $ of $\mathbb{CP}^2$ is 1.

Am I correct to claim that
$$
\bar{K_3} =\mathbb{CP}^2 \# ... \#  \mathbb{CP}^2 \text{ for 16 times of connected sums?}
$$
or
$$
{K_3} =\bar{\mathbb{CP}^2} \# ... \#  \bar{\mathbb{CP}^2} \text{ for 16 times of connected sums?}
$$
Here I consider $\sigma $ of $\bar K_3$ is +16. The $\sigma $ of $\bar {\mathbb{CP}^2}$ is -1.

If yes, how should we make the statements more rigorous? (or sketch a proof?)

答:
$K^3$ has a neat construction, it is somewhat easier to understand the analogous construction on the torus, though.

Consider $T^2\subset \mathbb{C}^2$. The action $(z,w)\mapsto (\overline{z},\overline{w})$ of conjugation has 4 fixed points. If we remove those fixed points, the action on the boundary circles is the antipodal map. $S^1$ quotiented out by the antipodal map is still $S^1$, so the quotient of the 4 punctured torus is the 4 punctured sphere. Then glue in 4 möbius bands.

We could have done this differently, though. We can remove the fixed points and glue the möbius bands in before we quotient out by conjugation. The antipodal map on the boundary circle extends over the möbius band and the quotient can be taken from there with the same resulting manifold.

The same thing is happening in 4 dimensions, except for $T^4$ you'll have 16 fixed points. So if you blow up 16 times before taking the quotient you'll get a manifold called $K^3$.

So, to stick with your notation,

$K = \left(T^4 \#\underbrace{\mathbb{CP}^2\#\dots\#\mathbb{CP}^2}_{\text{16 times}}\right)/(z \mapsto \overline{z})$

Viewing it this way, the remaining non-trivial homology classes come from tori that survive the quotient, although the basis can be formed entirely from spheres too.


问:
Prove $\lim_{x \rightarrow \infty} (1-e^{-x})^{x} = 1$ without l'Hôpital
A bit rusty on my calculus, trying to figure out whether this is correct.

I tried proving $\underset{x \rightarrow \infty}{\lim} (1-e^{-x})^{x} = 1$ without using L'hôpital.

$$\lim_{x \rightarrow \infty} \left(1-e^{-x}\right)^{x}=\lim_{x \rightarrow \infty}\left(1- \frac{x\cdot e^{-x}}{x}\right)^{x}\overset{(1)}{=}\lim_{x \rightarrow \infty} e^{e^{-x}\cdot x}=e^{\lim_{x \rightarrow \infty} e^{\frac{x}{e^{x}}}}=e^{0}=1$$

I'm a little unsure about the transition I marked as $(1)$.

答:
We have that as $x \to \infty$

$$\left(1+\frac a{x}\right)^{x} \to e^a$$

is true only for $a\in \mathbb R$ fixed, here we can use that

$$\left(1-\frac1{e^{x}}\right)^{x}=e^{x\log\left(1-\frac1{e^{x}}\right) }=e^{\frac{x}{e^x}\frac{\log\left(1-\frac1{e^{x}}\right)}{\frac1{e^x}} }$$

and use standard limits to get the result indeed

$\frac{x}{e^x} \to 0$

$\frac{\log\left(1-\frac1{e^{x}}\right)}{\frac1{e^x}} \to -1$

Refer also to

About $\lim \left(1+\frac {x}{n}\right)^n$

How to prove that exponential grows faster than polynomial?

Determine the following limit as x approaches 0: $\frac{\ln(1+x)}x$


问:
Monic Generator and Ideals of Polynomials

Is the set of all polynomials $P \in \mathbb{Q}[x]$ s.t. $P(1)=P(3)=0$ an ideal? If so find the monic generator.

I do not think it is because if we have $P(x)=x^2$, then $x*x^2=x^3$ which means $P(3)$ is not $0$, am i right?

I also do not know what monic generator is and have a hard time finding a good explanation so I was wondering if one could offer a simple explanation on what monic generator is and give an example on it, e.g. all polynomials of degree $\ge n$ (since i think that is ideal.)

答:
By definition, a subset $I$ of a commutative ring $R$ is an ideal if and only if for all $r \in R$ and $x, y \in I$, we have $rx \in I$ and $x + y \in I$. So, in your example, $R = \mathbb{Q}[x]$ and $I = \{P \in \mathbb{Q}[x] : P(1) = P(3) = 0\}$, and you have to determine whether, for all polynomials $P \in I$ and all polynomials $f \in \mathbb{Q}[x]$ (note: $f$ may or may not be in $I$, it's an arbitrary polynomial), the polynomial $f \cdot P$ is also in $I$.

Your example with $x^2$ and $x$ is not a counterexample, because neither of these polynomials evaluate to zero at $x = 1$ or $x = 3$. (Of course, that doesn't mean by itself that the set isn't an ideal, just that you haven't proven it is or it isn't yet.) Also, the set of all polynomials of degree at least $n$ is not an ideal, because the sum of two polynomials of degree $n$ can have degree less than $n$, so the set isn't closed under addition.

About "monic generators": This relies on a theorem that states that $\mathbb{Q}[x]$ is a principal ideal domain, meaning that every ideal is a principal ideal, that is, an ideal of the form $(g) = \{f \cdot g : f \in \mathbb{Q}[x]\}$ for some $g \in \mathbb{Q}[x]$. In fact, this is still true if $\mathbb{Q}$ is replaced with any field. The key idea of the proof is the Euclidean algorithm for polynomials, which lets us show that, if an ideal contains two polynomials, then it also contains their GCD.

It's an easy exercise to prove that, for any nonzero $c \in \mathbb{Q}$ and $g \in \mathbb{Q}[x]$, we have $(g) = (cg)$. So, by choosing $c$ to be the reciprocal of the leading coefficient of $g$, we see that each ideal of $\mathbb{Q}[x]$ has a generator that's a monic polynomial. Also, this is unique, since if there were two distinct such monic generators, we could take their GCD and find a new generator of smaller degree, which is a contradiction. (For example: the unique monic generator of the ideal $(2x^3 + 4)$ is $x^3 + 2$.)

So, if $I = \{P \in \mathbb{Q}[x] : P(1) = P(3) = 0\}$ is an ideal, the second part of the problem is asking you to find the unique monic polynomial $g \in I$ such that $I = (g)$.


问:
Geometric interpretation of the greatest common divisor
In a Cartesian coordinate system,  gcd(a,b) is equal to the number of vertices (points with integral coordinates) which lie on the segment with extremities $(0,0)$ and $(a,b)$ (excluding $(0,0)$).

How could I prove it? I have an idea why this works, but it is not clear to me how I could prove it. If you could give me a proof, suggest a book or article where this geometric interpretation appears, I would greatly appreciate it.I've been searching but still can't get anything

答: The segment has the equation $(x,y)=\lambda(a,b)=\lambda\gcd(a,b)\,\left(\dfrac a{\gcd(a,b)},\dfrac b{\gcd(a,b)}\right)$ with $\lambda\in [0,1]$. So $x,y$ are integer every time $\lambda\gcd(a,b)$ is an integer.


问:
Prove the set ext(S)$=\{x\in\mathbb{R}^n:x_i^2=1,i=1\ldots n\}$ where $S=\{x\in\mathbb{R}^n:||x||_{\infty}\leq 1\}$
Prove the set ext(S)$=\{x\in\mathbb{R}^n:x_i^2=1,i=1\ldots n\}$ where $S=\{x\in\mathbb{R}^n:||x||_{\infty}\leq 1\}$    I've got this exercise and I've never seen before how to prove that a group is an extreme points. Any guide will be helpful

答:
For $x\in\mathbb{R}^n$ with $|x_i|<1$ for some $i\leq n$. Then $x\notin\{x\in\mathbb{R}^n:x_i^2=1, i=1,...,n\}$. We choose $z_i=x_i+0,5(1-|x_i|)$ and $y_i=x_i-0,5(1-|x_i|)$ and $y_j,z_j=x_j$ for the other indices. Clearly $y_i, z_i\leq 1$ so $y,z\in S$ as well as $0,5y + 0,5z=x$ so $x\notin ext(S)$. This means $ext(S)\subset\{x\in\mathbb{R}^n:x_i^2=1, i=1,...,n\}$.

To prove the equality, take $x\in\{x\in\mathbb{R}^n:x_i^2=1, i=1,...,n\}$ and assume that there exist $y,z\in\S$ with $z_j,y_j\neq x_j$ for some $j\leq n$ as well as there exists $t\in(0,1)$ such that $ty+(1-t)z=x$. Since $x_i^2=1$, for all $i\leq n$, this means that $ty_j+(1-t)z_j<1$ with the index where $x$ differs from $y,z$, as $y,z\in S$ and thus $y_i,z_i\leq 1$ for all $i\leq n$.

I think this should do it.


问:
Trace of an endomorphism as a wedge product
Let $E$  be a finite dimensional vector space and $F=E^*$ its dual. We have that $Tr\in End(E)^*=(E\otimes F)^* = E^*\otimes F^* = \Lambda^1E^*\otimes \Lambda^1F^*\hookrightarrow \Lambda^2(E^*\oplus F^*)=\Lambda^2(E\oplus F)^*.$

I wonder what is the image of the trace under the last arrow, that is as a wedge product.

答:
Let $(e_i)_{i=1}^n$ be a basis for $E$ and $e^i$ the corresponding dual basis. In what follows I will use the natural isomorphism $F^{*} = \left( E^{*} \right)^{*} \cong E$ and Einstein's summation convention.

As an element in $\left( E \otimes E^{*} \right)^{*}$, the trace acts on elementary tensors by $\operatorname{Tr} \left( e \otimes \varphi \right) = \varphi(e)$.

Under the isomorphism $\left( E \otimes E^{*} \right)^{*} \cong \left( E^{*} \otimes \left( E^{*} \right)^{*} \right) \cong E^{*} \otimes E$, the trace becomes the element $e^i \otimes e_i$. To see this, note that
$$ \left( e^i \otimes e_i \right) \left( e \otimes \varphi \right) = e^i(e) \cdot e_i(\varphi) = e^i(e) \cdot \varphi(e_i) = \varphi \left( e^i(e) e_i \right) = \varphi(e). $$

Under the wedge product map $E^{*} \otimes E \hookrightarrow \Lambda^2 \left( E^{*} \oplus E \right)$, the trace corresponds to $(e^i,0) \wedge (0, e_i)$.

The natural isomorphism $E^{*} \oplus E \cong \left( E \oplus E^{*} \right)^{*}$ is given by
$$(\varphi,e)(e',\varphi') = \varphi(e') + e(\varphi') = \varphi(e') + \varphi'(e).$$

Finally, inder the isomorphism $\Lambda^2 \left( E^{*} \oplus E \right) \cong \Lambda^2 \left( \left( E \oplus E^{*} \right)^{*} \right) \cong \Lambda^2 \left( E \oplus E^{*} \right)^{*}$, using the two previous items, we get that the trace acts as
$$ \left( (e^i,0) \wedge (0,e_i) \right) \left( (e,\varphi) \wedge (e',\varphi') \right) = \left( (e^i,0)(e,\varphi) \right) \cdot \left( (0,e_i)(e',\varphi') \right) - \left( (e^i,0)(e',\varphi') \right) \cdot \left( (0,e_i)(e,\varphi) \right) = e^i(e) \cdot \varphi'(e_i) - e^i(e') \cdot \varphi(e_i) = \varphi' \left( e^i(e) e_i \right) - \varphi \left( e^i(e') e_i \right) = \varphi'(e) - \varphi(e').$$
In particular, if we identify elements of $E,E^{*}$ as elements in $E \oplus E^{*}$ we have
$$ \operatorname{Tr}(e \wedge \varphi) = -\operatorname{Tr}(\varphi \wedge e) = \varphi(e), \\
\operatorname{Tr}(e \wedge e') = \operatorname{Tr}(\varphi \wedge \varphi') = 0.$$


问:
Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $
Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$

Thanks!

Body:
$$\text{Note that, we have }\dfrac{x^4}{x^2+1} = \dfrac{x^4-1}{x^2+1}+\dfrac1{x^2+1} = x^2-1 + \dfrac1{x^2+1}$$
$$\text{Hence, }\int \dfrac{x^4}{x^2+1} dx = \int (x^2-1) + \int \dfrac1{x^2+1} = \dfrac{x^3}3 - x + \arctan(x) + \text{ constant}$$


问:
How to prove if a nxn matrix A is symmetric
If A is an nxn matrix such that rank(A) = 1 and the null space of A is the orthogonal complement of its column space, then how do I show that A is a symmetric matrix?
I have proven that the null space of A and A-transpose are equal. How do I proceed now?</p>

答:
Take an orthogonal basis $\{e_1, e_2, \dots, e_n\}$ of the linear space $V$ such that $\{e_1, e_2, \dots, e_{n-1}\}$ is a basis of $\ker A$. In this basis the matrix of $A$ is the zero matrix except the last column.

As the null space of $A$ is the orthogonal complement of its column space and the dot product is invariant under orthogonal transformations, the null space of $A$ is also the orthogonal complement of its column space in the basis $\{e_1, e_2, \dots, e_n\}$. This implies that the first $n-1$ coefficients of the last column vanish. Finally, $A$ is similar to the always vanishing matrix $B$ except the $(n,n)$ coefficient. We got the desired result as $B$ is symmetric.


问:
Recursive sequence is increasing
I have to show, that the sequence $g(n+1) = \frac{2ag(n)}{a+g(n)^2}$
is increasing for $a>0$ and $g(0)>0$. I know that it holds for $g(n)^2<a$
since $2a/(a+g(n)^2)>1$. But for $g(n)^2>a$ the opposite seems to be true.

I assume my professor forgot some initial restrictions but maybe I am overseeing something. I’d appreciate your thoughts on the matter.

答:
If you let $g_n=\sqrt a\, h_n$, you face the problem of
$$h_{n+1}=\frac {2h_n}{1+h_n^2}\implies h_n=\tanh \left(2^n\,\,k\right)\implies g_n=\sqrt a\,\tanh \left(2^n\,\,k\right)$$

So, no problem.

Title: Recursive sequence is increasing


问:
What is the geometrical meaning of matrix with the first row $(1,0,\cdots,0)$
What is the geometrical meaning of matrix $\mathbf{A}\in\mathbb{R}^{n\times n}$ in the form
\begin{align*}
\begin{bmatrix}
1 & 0 & 0 & \cdots & 0 \\
a_{21} & a_{22} & a_{23} & \cdots & a_{2n} \\
\vdots & & & \ddots  & \\
a_{n1} & & \cdots & & a_{nn}
\end{bmatrix}
\end{align*}
$a_{ij}\in\mathbb{R}$.

Is it some type of an affine transform?

答: Well it means that the first unit vector $e_1=(1,0,...,0)$ is an eigenvector of the matrix to the eigenvalue 1. In other words, when you consider the action of the matrix on the vector space, the subspace spanned by $e_1$ remains precisely the same. If it would be a 2 instead, this subspace would be scaled by a factor of 2...


问:
extended Euclidean (xgcd) in quadratic integer rings
Given a discriminant $D < 0$, I have the quadratic imaginary field $\mathbb{K} := \mathbb{Q}(\sqrt{D})$. And the quadratic integer ring is given by
$\mathcal{O} = \mathbb{Z} + \mathbb{Z} \frac{D + \sqrt{D}}{2}$.

For $a,b \in \mathcal{O}$, is it possible to calculate some numbers $p,q \in \mathcal{O}$ such that $ap + bq = d$ where $d$ is a divisor of $a$ and $b$?

$d$ must not be the greatest common divisor for my application, although I guess it would be? For some norm? (I'm not that much into this area.)

I was trying to generalize the extended Euclidean algorithm but I wasn't sure how to do division with remainder.

A bit more generic, where I need that: Given a matrix $M \in \operatorname{Sp}_2(\mathbb{K})$, I want to find $\gamma \in \operatorname{Sp}_2(\mathcal{O})$, $R \in \operatorname{Sp}_2(\mathbb{K})$ such that $M = \gamma \cdot R$ and I want the left lower $2 \times 2$ matrix of $R$ to be zero. In my construction, I heavily depend on the extended Euclidean algorithm.

答:
Given $a,b \in \mathcal{O}$, suppose that we can always find $x,y \in \mathcal{O}$ such that
$$ax+by=d,$$
with $d$ is a common divisor to $a$ and $b$.  If $d'$ is another common divisor, then $d' \mid (ax+by)$, hence $d'$ divides $d$.  Thus $d$ is a greatest common divisor (GCD) for $a$ and $b$.  In particular, $\mathcal{O}$ is a GCD-domain.

A GCD-domain is a (unique factorization domain) UFD if and only if it satisfies the ascending chain condition on principal ideals.  This chain condition holds for rings of integers, because such rings are Noetherian.  Thus it follows that $\mathcal{O}$ is a UFD.

For $\mathcal{O}$ the ring of integers in an imaginary quadratic field, unique factorization has been widely studied.  As a result of the Class Number Problem, we know that
$$-1,-2,-3,-7,-11,-19,-43,-67,-163$$
are the unique negative integers $d$ such that $\mathbb{Q}(\sqrt{d})$ generates a UFD, which answers your question negatively.

In general, if you wanted to classify $a,b \in \mathcal{O}$ such that we can find a GCD for $a,b$, I imagine that you would want to measure the degree to which unique factorization fails (i.e. over which primes).  This is measured, in some sense, by the class number, which is where I would look next.

Edit: You mentioned that this question arose in an attempt to generalize the Euclidean algorithm on $\mathbb{Z}$.  It should be noted that there exist GCD-domains that are not Euclidean domains (i.e. domains for which there exist analogues of the Euclidean algorithm).  These are not necessarily exotic, either: the integers $−19, −43, −67, −163$ from before give rise to PIDs (hence GCD-domains) that are not Euclidean domains.


问:
Prove that a Lipschitz function is continuous
Define $f:[a.b] \rightarrow \mathbb{R}$ as Lipschitz

By definition, then, $\exists M>0, M \in \mathbb{R}:|f(x)-f(y)\leq M|x-y|$  $ $ $\forall x,y\in[a,b]$

Rearranging this, we have $\frac{|f(x)-f(y)|}{x-y} \leq M$, positing that the slope must be finite on the interval $[a,b]$

I understand conceptually that Lipschitz functions must be continuous but I'm having trouble showing it.

Should I go about this by showing assuming the negation of Lipschitz and showing that for some $\epsilon,  \exists \delta>0: |f(x)-L|>\epsilon$ ?

答: It is not necessary to proceed by contradiction. Use the $\varepsilon-\delta$ definition of continuity and choose $\delta=\varepsilon/M$.


问:
How to merge two Gaussians
I have two multivariate Gaussians each defined by mean vectors and Covariance matrices (diagonal matrices). I want to merge them to have a single Gaussian i.e. I assume there is only one Gaussian but I separated observations randomly into two groups to get two different Gaussians which are not too different than each other.

Since I know the number of observations in each of two Gaussians, combined mean estimation is straight forward : $\frac{n_1\mu_1 + n_2\mu_2}{n_1+n_2}$

But, what about the Covariance matrix?

Thanks

EDIT:

The question was confusing in the original post, especially the "merging Gaussians" part. Maybe the following paragraph would be a better choice.

I have two sets of observations drawn from two multivariate Gaussians each defined by mean vectors and Covariance matrices (diagonal matrices). I want to merge the observations to have a single sample, and I assume to have another Gaussian (i.e. I assume initially there was only a single Gaussian, and observations were separated into two groups to get two different Gaussians).

答:
As the answer of JulienD is good. But I would like to rewrite the final formulas in a better way and extend them to many random variables, and want to clarify that the approved answer is likely not correct.

First, I rewrite the answer of JulienD with few additional steps to give it a better form as follows:

Single variate:
\begin{align}
\mu &= E[x] = \frac{n_1\mu_1 + n_2\mu_2}{n_1 + n_2}
\end{align}

\begin{align}
\sigma^2 & = E[x^2] - E[x]^2 \\
& = \frac{\sigma_1^2 n_1^2 + \sigma_2^2 n_2^2 + (n_1\mu_1 + n_2\mu_2)^2}{(n_1+n_2)^2} - \mu^2 \\
& = \frac{\sigma_1^2 n_1^2 + \sigma_2^2 n_2^2}{(n_1+n_2)^2} + \frac{(n_1\mu_1 + n_2\mu_2)^2}{(n_1+n_2)^2} - \mu^2 \\
& = \frac{\sigma_1^2 n_1^2 + \sigma_2^2 n_2^2}{(n_1+n_2)^2} + \mu^2 - \mu^2 \\
& = \frac{n_1^2}{(n_1+n_2)^2}\sigma_1^2  + \frac{n_2^2}{(n_1+n_2)^2}\sigma_2^2
\end{align}

Multivariate:
\begin{align}
\Sigma &= E[XX^T] - E[X]E[X]^T \\
& = \frac{n_1^2 \Sigma_1 + n_2^2 \Sigma_2 + (n_1\mu_1 + n_2\mu_2)(n_1\mu_1 + n_2\mu_2)^T}{(n_1 + n_2)^2} - \mu\mu^T \\
&= \frac{n_1^2 \Sigma_1 + n_2^2 \Sigma_2}{(n_1 + n_2)^2} + \frac{(n_1\mu_1 + n_2\mu_2)(n_1\mu_1 + n_2\mu_2)^T}{(n_1 + n_2)^2} - \mu\mu^T \\
& = \frac{n_1^2 \Sigma_1 + n_2^2 \Sigma_2}{(n_1 + n_2)^2} + \mu\mu^T - \mu\mu^T \\
&= \frac{n_1^2}{(n_1 + n_2)^2} \Sigma_1 + \frac{n_2^2}{(n_1 + n_2)^2} \Sigma_2
\end{align}

Another approach to get a formula of variance or covariance matrix (for multivariate) is that we use the theorem of variance (which can find in many textbooks of statistics, e.g., Larry Wasserman, All of Statistics, theorem 3.20, page 52, or prove via definition as a similar way as JulienD did):
$$
\mathbb{V}(X+Y) = \mathbb{V}(X) + \mathbb{V}(Y) + 2 \textbf{Cov}(X,Y)
$$
More generally, for random variables $X_1, X_2,..., X_N$:
$$
\mathbb{V}(\sum_{i=1}^{N}a_i X_i) = \sum_{i=1}^{N}a_i^2\mathbb{V}(X_i) + 2 \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} a_i a_j \textbf{Cov}(X_i,X_j)
$$

Consider $X_i \sim \mathcal{N}(\mu_i, \Sigma_i), i=1 \cdots N,$; and they are independent of each other. Hence, the covariances are zero, i.e., $\textbf{Cov}(X_i,X_j)=0.$
A newly aggregated these Gaussian distributions is defined as a weighted sum:
$$X=\sum_{i=1}^{N} a_i X_i=\sum_{i=1}^{N} \frac{n_i}{\sum_{l=1}^{N}n_l} X_i,$$
where $a_i=\frac{n_i}{\sum_{l=1}^{N}n_l}$ and $\sum_{i=1}^N a_i = 1$.

So, we get the mean and covariance of the aggregated distribution as follows:
$$
\mu = \sum_{i=1}^{N} a_i \mu_i
$$
\begin{align}
\Sigma &=\mathbb{V}(X)= \mathbb{V}(\sum_{i=1}^{N} a_i X_i)\\
&=\sum_{i=1}^{N}a_i^2\mathbb{V}(X_i) + 2 \sum_{i=1}^{N-1}\sum_{j=i+1}^{N} a_i a_j \textbf{Cov}(X_i,X_j) \\
& = \sum_{i=1}^{N}a_i^2 \Sigma_i + 0 \\
& = \sum_{i=1}^{N}a_i^2 \Sigma_i
\end{align}

Now, look at the answer of ahmethungari, it seems the answer is not correct.

Title: How to merge two Gaussians


问:
What are the most broadly applicable or general areas of mathematics?
Background:

I am in the final year of a bachelor's degree program in Data Science with the University of London. I am interested in exploring the possiblity of going to graduate school for pure mathematics, most likely at the master's level in Canada. (I live in Canada.)

All of my studies with the University of London are done remotely and with almost no support from my institution, so I don't really have anyone to ask for any kind of advice. I have recently had some correspondence with the math department chair at a top Canadian university and was told that students must have supervisors in order to be admitted, and that I would significantly improve my chances of admission if I had some idea of my research interests and who my supervisor would be.

To provide some context, I am currently about 130 pages into Analysis I by Amann and Escher, 20 pages into Algebra by Lang, and 20 pages into Topology: A Categorical Approach by Bradley et al. In other words, I am currently working on what I consider to be relatively foundational mathematics, so the idea of mathematical research seems very far off to me.

I do not know precisely what kind of research I would be interested in. As mentioned, I have no one to ask for advice about this at my university, which is why I am asking about this here.

I am not sure exactly what I will end up doing for a career---I think academia is probably quite unlikely for me---so I would like to try to keep my options open as much as possible. To my mind, part of keeping my options open is to try to learn math that is broadly useful as opposed to niche. If you permit the metaphor of mathematics knowledge as being tools in a toolbag, I am more interested in tools that are more commonly used and/or more versatile as opposed to highly specialized tools that are inapplicable outside of a narrow set of circumstances.

End of background

To that end, I am trying to figure out what the most "general" areas of math are, as opposed to niche areas of math. I do not quite know what I am looking for, but I will try to give some examples:

-Without citing any sources, I would say that calculus and linear algebra are probably tremendously versatile, although perhaps not advanced enough to be considered research interests (?)

-If you look at the list of applications of differential geometry on Wikipedia (https://en.wikipedia.org/wiki/Differential_geometry#Applications) you will see it's quite long and varied.

-Wikipedia says that representation theory is "pervasive across fields of mathematics"; also, p. xi of A Tour of Representation Theory by Lorenz says that "representation [theory] enjoys the additional benefit of having applications in myriad contexts other than algebra, ranging from number theory, geometry and combinatorics to probability and statistics [58], general physics [198], quantum field theory [210], the study of molecules in chemistry [49] and, more recently, machine learning [126]."

-At 12:30 in the video https://www.youtube.com/watch?v=vOboYNh_5is, Ravi Vakil seems to suggest that algebraic geometry is in some sense the "unification" of number theory, geometry, algebra and topology.

-Edward Frenkel has said that the Langlands program is "a kind of grand unified theory of mathematics."

Something that, as far as I know, is relatively niche and thus an example of something I'm not looking for would be algebraic number theory. (See for example: https://mathoverflow.net/questions/24971/practical-applications-of-algebraic-number-theory).

I appreciate any help.

TL;DR I am looking into grad school and have been told that I should try to figure out what my research interests are. I don't have any particular interests. I'm trying to figure out which areas of math are the most broadly useful.

答:
Too long for a comment:

I find it quite strange that a university would expect a b.a. in mathematics to already know what their "research interests" are. A student finishing their b.a. who is thinking about graduate work probably knows one thing: s/he is interested in mathematics . The purpose of MA studies (at least the first year), in my opinion is exactly supply a deep enough background in advanced math to decide what really sparks her up. Any decision made before that is probably based on misconceptions and might lead to surprises and disappointment further down the road. Find a university that will allow you to study at least one year of MA level courses before deciding.


问:
Does this sigma function correspond to this series?
I'm kind of self-taught in sigmas, and I'm not very proficient in calculus nor precalculus. Therefore, I'd like to see if the series I've thought of does correspond to this sigma function

$$\sum_{k=1}^n k\times\frac k2$$

Corresponds to this series:

$$1\times\frac 12+2\times \frac 22+3\times \frac 32 + 4\times\frac 42 ... n\times\frac n2$$

I'm wondering because I want to do a little recreational study of this series, and see if what kind of relations n has to the sums. To do that, it would be nice to compress that series to a sigma function, and I'm wondering if I did it correctly here.

答:
Adding this as an answer to avoid excessive comments.

First point:

Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=1\times\frac{1}{2}+2\times\frac{2}{2}+...+n\times\frac{n}{2}$$
PROOF: Um... plug the numbers in. The way Sigma notation works is that
$$\sum_{k=m}^n f(k)=f(m)+f(m+1)+...+f(n-1)+f(n)$$
But we can extend this to non-indexed sets (as long as they are countable), e.g
$$\sum_{x\in\mathcal{X}}f(x)=f(x_1)+f(x_2)+f(x_3)+...~~\forall x_i\in\mathcal{X}.$$

Second Point:

Notice that I am calling this "Sigma notation" not a "Sigma function" as it really is just a way of writing things to make it shorter. You could, very formally, think of this instead as a map ("function") that takes in two inputs - another function, and a set:
$$\Sigma(f,S)=f(s_1)+f(s_2)+f(s_3)+...~~\forall s_i\in S$$
E.g,
$$\Sigma(\text{id},\mathbb{N}_{\leq m})=\sum_{k=1}^m k=\frac{m(m+1)}{2}$$
But this is frankly overkill and unnecessarily complicates a simple idea.

Third point:

Yes, it is true that
$$\sum_{k=1}^n k\times\frac{k}{2}=\sum_{k=1}^n \frac{k\times k}{2}$$
There is nothing special about this! In a very crude manner, you can just "cancel out" the summation signs on both sides to get
$$k\times \frac{k}{2}=\frac{k\times k}{2}$$
Which is obviously true, e.g $(6\times 6)/2=36/2=18$ and $6\times (6/2)=6\times 3=18.$


问:
How to prove that $\frac{d}{dt}x(t)=-a x(t),\ x(0)=1,\ a>0$ will never go under 0?
So let $\frac{d}{dt}x(t)=-a\times x(t),\ a>1$ and $x(0)=1$, is it possible to prove that $\lim_{t\to \infty} x(t)\geq0$ hence no matter how long $x(t)$ won't be lesser than 0. It seems that this statement to be true, as I've run a simulation (where $dt=10^{-7},\ a=72$) for more than 20 million frames and it seems to still be greater than 0. But still I can't think of a way to prove it mathematically. The most I could say is that because the lesser you are, the lesser you'll decrease, so you're not likely to get below 0.

答: As said in the comments the solution of this ODE is $x(t)=e^{-at}$ as ${d\over dt}e^{-at}=-ae^{-at}$. So you need to prove that $\underset{t\rightarrow \infty}{\lim}e^{-at}=0$ if $a>0$.


问:
Isolate 3 variables in the compound interest formula
I'm trying to create three new formulas from a compound interest formula: I'm trying to respectively isolate M, r, and P. The original formula comes from https://www.thecalculatorsite.com/articles/finance/compound-interest-formula.php#regular-contributions, and it's the standard formula for finding future compound interest with an initial investment.

$$ A = P\left(1 + \frac{r}{n}\right)^ {n t} + M\biggl(\frac{\left(1 + \frac{r}{n}\right)^ {n t} - 1}{\frac{r}{n}}\biggr)$$

These are what the variables stand for:

A = the future value of the investment/loan, including interest

P = the principal investment amount

M = the monthly payment

r = the annual interest rate

n = the number of times that interest is compounded per unit t

t = the number of years the money is invested or borrowed for

I contacted the site owner and he wasn't able to help with this, since he's not on the mathematical side of things, he just builds calculators on the back of formulas that already exist. I've also been trying to achieve this in Wolfram Alphra with no luck:

```<code>solve P * (1 + r / n) ^ (n * t) + PMT * (((1 + r / n) ^ (n * t) - 1) / (r * n)) = A for r
</code>```

Any help here would be appreciated! Thanks. It's been too many years since my college math.

答:
There is no problem isolating $P$ or $M$, the equation is linear in these ($a=bP+c\to P=\dfrac{a-c}b$).

On the other hand, there is no analytical solution for $r$ and you have to resort to a numerical approximation approach. Newton's iterations will work fine.

As $(1+\frac rn)^{nt}\approx 1+rt$, a starting approximation is otained by solving

$$ A = P(1+rt) + Mnr$$
for $r$.


问:
Equivalence relation and class
Let X = {1, 2, 4, 7, 8, 10, 12, 14, 15, 18, 22, 23}. Consider the relation R on X
given by (a, b) ∈ R if a − b is divisible by 7.

(a) Show that R is an equivalence relation on X.

(b) Determine the equivalence class [1].

(c) Determine the partition of X into equivalence classes.

edit 1) I need help on C, I get a) is congruent modulo and therefore b) is "the quotient of X modulo R" by my understanding but I don't quite understand how to write out the notation of the partition in this regard. Thank you for the quick answers, sorry I was not immediately specific.

edit 2) Thank for the explanations and help in the comments, especially JMoravitz and Lee Mosher. Much appreciated!

答:
To answer C, you can do things one step at a time.

Starting with $1 \in X$, to determine the equivalence class $[1]$, ask yourself this question:

What elements $a \in X$ are equivalent to $1$?

Then translate:

For what elements $a \in X$ is it true that $(a,1) \in R$?

Then translate again:

For what elements $a \in X$ is it true that $a-1$ is divisible by $7$?

Now go through the elements $a \in X$ one-by-one, answering this question for each in turn:

For $a=1$, is $a-1=1-1=0$ divisible by $7$? Yes it is, so $1 \in [a]$.

For $a=2$, is $a-1=2-1=1$ divisible by $7$? No it is not, so $2 \not\in [a]$.

For $a=4$, is $a-1=4-1=3$ divisible by $7$? No it is not, so $4 \not\in [a]$.

And so on.

Along the way, you may have a leap of understanding which let's you cut off all of this one-by-one stuff. But until you do, keep doing the one-by-one stuff.

Title: Equivalence relation and class


答: If $\delta=\frac{\epsilon}{M}$. The definition of continuity says that we can suppose that $$0<|x-x_0|<\delta$$ from here: $$M|x-x_0|<\epsilon$$ but we now that $$|f(x)-f(x_0)|<M|x-x_0|$$ because $f\in Lip(\mathbb{R})$, then $|f(x)-f(x_0)|<\epsilon$ and therefore $f$ is continuous.

Title: Prove that a Lipschitz function is continuous

Body:
Define $f:[a.b] \rightarrow \mathbb{R}$ as Lipschitz

By definition, then, $\exists M>0, M \in \mathbb{R}:|f(x)-f(y)\leq M|x-y|$  $ $ $\forall x,y\in[a,b]$

Rearranging this, we have $\frac{|f(x)-f(y)|}{x-y} \leq M$, positing that the slope must be finite on the interval $[a,b]$

I understand conceptually that Lipschitz functions must be continuous but I'm having trouble showing it.

Should I go about this by showing assuming the negation of Lipschitz and showing that for some $\epsilon,  \exists \delta>0: |f(x)-L|>\epsilon$ ?


问:
is this function injective?
Is this function injective?

$f:\mathbb{N}\to \mathbb{N\times N}$ defined as

$f:n\to (n, n+1)$

$f(n_{1})=f(n_{2})\Rightarrow (n_{1},n_{1}+1)=(n_{2},n_{2}+1)\Rightarrow$

$n_{1}=n_{2} \wedge n_{1}+1=n_{2}+1\Rightarrow n_{1}=n_{2}$

Is this correct?

答:
Yes.

In general $f:\, A\to B$ is injective if $$\forall a_{1},a_{2}\in A:\, f(a_{1})=f(a_{2})\implies a_{1}=a_{2}$$

By definition, you need to show that for all $n_1,n_2\in \mathbb{N}$ $$f(n_{1})=f(n_{2})\implies n_{1}=n_{2}$$

and this exactly what you have shown.

Title: is this function injective?


问:
Discuss the continuity of $f_1$ in the following problem.
Define $f_1:[0,1]\rightarrow\mathbb{R}$ by $f_1(x)=\sum\limits_{n=1}^{\infty}x^2(1-x^2)^{n-1}$. Discuss the continuity of $f_1$.

please give a hint for the problem. I am trying to use the idea of uniform convergence but couldn't approach

答: It is clear that $f_1(0)=0$. And, if $x\in(0,1]$, $1-x^2\in[0,1)$ and therefore$$f_1(x)=\sum_{n=1}^\infty x^2\left(1-x^2\right)^{n-1}=\frac{x^2}{1-\left(1-x^2\right)}=\frac{x^2}{x^2}=1.$$So, $(\forall x\in[0,1]):f_1(x)=1$ and therefore $f_1$ is discontinuous.


问:
Calculate the surface area bounded by the curves $3x^2−1$ and its tangents that pass through point (0,1/4).
Ok here i have a calculus problem to solve to find the area of the curve:

Calculate the surface area bounded by the curves $ 3x^2-1 $ and its tangents that pass through point $(0,1/4)$.

My question is how to find the equation of tangents

答:
There are no tangents to the curve $y=3x^2-1$ passing through the point $A=\left(0,\frac{1}{4}\right)$

Any line passing through $A$ has equation $y=mx+1/4$ intersects the curve in two distinct points

Substitute in the curve equation $3x^2-1=mx+1/4$ and solve

$$x = \frac{m\pm\sqrt{m^2+15}}{6}$$
As $m^2+15>0$ for any $m$ the equation has always two real solutions.


答: For $r\in(0,1)$ you can use Weierstrass $M$-test with geometric series. At $0$, use $x^2$ to bound. At $1$ use the series to bound.

Title: Discuss the continuity of $f_1$ in the following problem.

Body:
Define $f_1:[0,1]\rightarrow\mathbb{R}$ by $f_1(x)=\sum\limits_{n=1}^{\infty}x^2(1-x^2)^{n-1}$. Discuss the continuity of $f_1$.

please give a hint for the problem. I am trying to use the idea of uniform convergence but couldn't approach


问:
Interpretation of a certain integral
What does $\int d(\sin x)$ actually mean?I mean to say when we have $\int f(x)dx$,we interpret it as the values $f(x)$ s are being multiplied by tiny $dx$ and are summed together.But how do we interpret the above from this point of view.Because there is $d(\sin x)$,not $dx$ and we take $\sin x$ in the Y axis.But in integration we take dx to be in the X axis.

答:
as

$$\frac{d}{dx}\sin x=\cos x$$

$$d(\sin x)=\cos x~ dx$$

Thus

$$\int d(\sin x)=\int \cos x ~dx=\sin x+C$$

It is used when you do a variable's change.

Example

$$\int \sin^{10}(x) \cos(x) ~dx=\int \sin^{10}(x) d(\sin(x))=\frac{\sin^{11}(x)}{11}+C$$

Title: Interpretation of a certain integral


Title: Integral of $ \int_{-1}^{1} \frac{x^4}{x^2+1}\,dx $

Body:
Any suggestions how to solve it? by parts?
$$ \int_{-1}^{1} \frac{x^4}{x^2+1}dx$$

Thanks!

Body:
$$\dfrac{x^4}{x^2+1}=\dfrac{(x^4+2x^2+1)-2x^2-1}{x^2+1}=\\
=\dfrac{(x^2+1)^2-2x^2-2+1}{x^2+1}=x^2+1-2+\dfrac{1}{x^2+1}=x^2-1+\dfrac{1}{x^2+1}.$$