问:
convex optimization (problem)
Let us have the following task:
$$\min f(x)$$
$$g_{i}(x)\le 0,\quad\forall i$$
$$h_{j}(x)= 0,\quad\forall j$$

What is the correct form of rewriting as a maximize problem?
$$\max - f(x)$$
$$g_{i}(x)\le 0,\quad\forall i$$
$$h_{j}(x)= 0,\quad\forall j$$

or $$-\max - f(x)$$
$$g_{i}(x)\le 0,\quad\forall i$$
$$h_{j}(x)= 0,\quad\forall j$$
Thanks for your answer.

答:
If $f$ is a convex function, you can write the maximization problem as:

$$ max -f(x) $$

Title: convex optimization (problem)


问:
Local analysis two-variable functions with absolute value
I have a problem with the local analysis of two-variable functions which have absolute value. For example, how can I study nature of point $(0,0)$ in $f:\mathbb{R}^2 \to \mathbb{R}, f(x,y)=sin(xy+|xy|)$? Or in the function $f:\mathbb{R}^2 \to \mathbb{R}, f(x,y)=e^{sin(|xy|)}$?

I think that in both cases $(0,0)$ is a local minimum, but I don't know how to prove it. Typically, I use the Hessian matrix to do the analysis, however in this case I don't think it will help me...

答:
This answer is example-specific. I don't think that there is one good method for investigating the extrema of non-differentiable functions.

For the first example note that $$xy + |xy| = \begin{cases}2xy,\quad &xy> 0,\\0,\quad &xy \leq 0.\end{cases}$$
In particular, the argument of the sine function is always non-negative, therefore $\sin(xy + |xy|)$ has a local minimum at $(0,0)$ because on the neighborhood of the origin $xy$ cannot be too large (and we will not obtain e.g. $\sin \frac 32\pi$). The minimum will not be strict though, as the function vanishes at all points $(x,y)$ for which $xy\leq 0$.

The other example is quite similar. The argument in the sine function is positive and small near the origin and thus the exponent $\sin |xy|$ will attain only non-negative values around $(0,0)$. Once again the minimum will not be strict.

ContentLicense: CC BY-SA 4.0


答:
$$\max f(x) = - \min - f(x)$$

If you have a black box algorithm that knows how to find minima, to find maximum:

Multiply the objective by $-1$.

Plug the modified problem into the black box

Multiply the resulting objective value by $-1$, to reverse the first step. No need to adjust the $x$ of the solution.

Title: convex optimization (problem)

Body:
Let us have the following task:
$$\min f(x)$$
$$g_{i}(x)\le 0,\quad\forall i$$
$$h_{j}(x)= 0,\quad\forall j$$

What is the correct form of rewriting as a maximize problem?
$$\max - f(x)$$
$$g_{i}(x)\le 0,\quad\forall i$$
$$h_{j}(x)= 0,\quad\forall j$$

or $$-\max - f(x)$$
$$g_{i}(x)\le 0,\quad\forall i$$
$$h_{j}(x)= 0,\quad\forall j$$
Thanks for your answer.


问:
Does an unbounded straight line have infinitely many axes of symmetry?
In a circle, any diameter is an axis of symmetry, so technically a cirle should have infinitely many axes of symmetry.

This got me thinking about the axes of symmetry of straight lines.
A line segment obviously has only one axis of symmetry, but what about an unbounded infinite straight line?

Does it have one axis of symmetry (seems unlikely), infinitely many, none at all?
Does it depend on the kind of Geometry we're talking about?

答:
A bound line has 2 symmetries. The line itself and a perpendicular line intersecting the midpoint.

An unbound line with a centre point has 2 symmetries. The line itself and a perpendicular line intersecting the centre.

An unbound line without a centre point has 1 symmetry. The line itself. The perpendicular symmetry is undefined.

If the line represents an infinitely large circle, circles have infinite symmetries.

ContentLicense: CC BY-SA 3.0


问:
Find the limit of the following sequence
$$\lim_{n\to\infty} (\frac{1^{10} + 2^{10} + 3^{10} + ... + n^{10}}{n^{10}} - \frac{n}{11})$$

So I am aware that Stolz theorem may be used here but I don't know how. What I tried doing is I combined the two fractions:

$$ (\frac{11\dot(1^{10} + 2^{10} + 3^{10} + ... + n^{10}) - n^{11}}{11n^{10}})$$

But I can't apply Stolz theorem to this because we have two sequences in the numerator. Moreover the bigger problem is the $-$ sign next to $n^{11}$. Any help would be appreciated!

答:
Let $ n\in\mathbb{N}^{*} $, $ f \in\mathscr{C}^{1}\left(\left[0,1\right],\mathbb{R}\right) $ be an increasing function, and $ F:\left[0,1\right]\rightarrow\mathbb{R},\ x\mapsto\int_{0}^{x}{f\left(y\right)\mathrm{d}y} $.

Denoting $ \left(\forall k\in\mathbb{N}\right),\ x_{k}=\frac{k}{n} \cdot $

\begin{aligned}\frac{1}{n}\sum_{k=1}^{n}{f\left(\frac{k}{n}\right)}-\int_{0}^{1}{f\left(x\right)\mathrm{d}x}&=\sum_{k=1}^{n}{\int_{x_{k}}^{x_{k-1}}{\left(f\left(x\right)-f\left(x_{k}\right)\right)\mathrm{d}x}}\\ &=\sum_{k=1}^{n}{\left(F\left(x_{k-1}\right)-F\left(x_{k}\right)-\left(x_{k-1}-x_{k}\right)F'\left(x_{k}\right)\right)}\end{aligned}

Applying the generalized main value theorem (Taylor-Lagrange's formula) to $ F $ on $ \left[x_{k-1},x_{k}\right], \ k\in\left[\!\left[1,n\right]\!\right] $, there exists some $ \gamma_{k} $ such that : \begin{aligned} F\left(x_{k-1}\right)-F\left(x_{k}\right)-\left(x_{k-1}-x_{k}\right)F'\left(x_{k}\right)&=\frac{\left(x_{k-1}-x_{k}\right)^{2}}{2}F''\left(\gamma_{k}\right)\\ &=\frac{\left(x_{k}-x_{k-1}\right)f'\left(\gamma_{k}\right)}{2n} \end{aligned}

Thus :\begin{aligned} \frac{1}{n}\sum_{k=1}^{n}{f\left(\frac{k}{n}\right)}-\int_{0}^{1}{f\left(x\right)\mathrm{d}x}&=\frac{1}{2n}\sum_{k=1}^{n}{\left(x_{k}-x_{k-1}\right)f'\left(\gamma_{k}\right)}\\ &=\frac{1}{2n}\int_{0}^{1}{f'\left(x\right)\mathrm{d}x}+\underset{\overset{n\to +\infty}{}}{\mathcal{o}}\left(\frac{1}{n}\right)\end{aligned}

(Riemann sums theorem)

Thus : $$\fbox{$\begin{array}{rcl}\displaystyle\sum_{k=1}^{n}{f\left(\frac{k}{n}\right)}-n\int_{0}^{1}{f\left(x\right)\mathrm{d}x}\underset{n\to +\infty}{\longrightarrow}\frac{f\left(1\right)-f\left(0\right)}{2}\end{array}$}$$

Now apply this to $ f_{p}:x\mapsto x^{p} $, $ p\in\mathbb{N} \cdot $


问:
Real numbers such that $|x^2 + ax + b| \le {1\over2}$?
What are real numbers $a$, $b$ such that, for every $x$ satisfying $0 \le x \le 2$, it holds that $|x^2 + ax + b| \le {1\over2}$?

答:
Hint:

Denote $f(x)=x^2+ax+b$. It achieves its minimum at $x=-\frac a2$ and $f(-\frac a2)=b-\frac{a^2}{4}$.

If $-\frac a2 < 0$ or $-\frac a2 > 2$, $f(x)$ is monotone increasing or decreasing in $[0,2]$, then $|f(x)|\le \frac 12 \iff |f(0)| \le \frac 12 \text{ and } |f(2)| \le \frac 12$.

If $0 \le -\frac a2 \le 2$, then $|f(x)|\le \frac 12 \iff |f(0)| \le \frac 12 \text{ and } |f(2)| \le \frac 12 \text{ and } |f(-\frac a2)|\le \frac 12$.

Can you start from here? It would be helpful if you plot all these inequalities on an $a,b$-plane.


问:
Proof that $\int_0^{\infty} e^{-st} (\frac{1-\cos t}{t^2})dt = \frac{\pi}{2} - s\ln(\frac{s^2}{s^2+1}) + 2\tan^{-1}(s)$
I'm trying to solve this Laplace problem from Laplace Transform by Schaum's Outline, chapter 1, math 167(a). But no matter what approach I follow, I end up with this line:

$L(\frac{1-\cos t}{t^2}) = \frac{\pi}{2} + \frac{1}{2}s\ln(\frac{s^2}{s^2+1}) - \tan^{-1}(s)$

And by definition, $L(\frac{1-\cos t}{t^2}) = \int_0^{\infty} e^{-st} (\frac{1-\cos t}{t^2})dt$

So I've no idea why and how the problem expects me to end up from $\frac{\pi}{2} + \frac{1}{2}s\ln(\frac{s^2}{s^2+1}) - \tan^{-1}(s)$ to $\frac{\pi}{2} - s\ln(\frac{s^2}{s^2+1}) + 2\tan^{-1}(s)$

Looking for your kind help.

答:
You have to use
$$ L\bigg\{\frac{f(t)}{t}\bigg\}=\int_s^\infty F(\tau)d\tau $$
twice, where $F(s)=L\{f(t)\}$.
Let $f(t)=\frac{1-\cos t}{t}$ and then
\begin{eqnarray}
F(s)=L\{f(t)\}&=&\int_s^\infty L\{1-\cos t\}(\tau)d\tau=\int_s^\infty\bigg(\frac1\tau-\frac{\tau}{\tau^2+1}\bigg)d\tau\\
&=&\frac12\ln\bigg(\frac{\tau^2}{\tau^2+1}\bigg)\bigg|_s^\infty=-\frac12\ln\bigg(\frac{s^2}{s^2+1}\bigg).
\end{eqnarray}
So
\begin{eqnarray}
L\bigg\{\frac{1-\cos t}{t^2}\bigg\}&=&L\bigg\{\frac{f(t)}{t}\bigg\}=\int_s^\infty F(\tau)d\tau\\
&=&-\frac12\int_s^\infty \ln\bigg(\frac{\tau^2}{\tau^2+1}\bigg)d\tau\\
&=&\cdots
\end{eqnarray}
You finish the rest.


问:
What does it mean a function to belong to a space in the sense of distributions?
In my book Sobolev's spaces are defined as follows

$$W^{m,p}(\Omega) = \{v \in L^p(\Omega): \,\, D^{\alpha}v \in L^p(\Omega) \,\, \forall \, |\alpha|\leq m \,\mbox{ in the sense of
distributions} \}$$

I don't understand what he means by distributional sense. To be more specific in an example, what does mean $u \in L^2 (\mathbb{R}) $ in the distributional sense?

答:
The phrase "in the sense of distributions" here refers to the differential operator $D^\alpha$. For example, $u \in W^{1,1}(\mathbb{R})$ if $u, Du \in L^1 (\mathbb{R})$, where $Du$ is the weak derivative, or derivative in the sense of distributions. This derivative is computed by "integrating" against an arbitrary smooth function with compact support $\phi$.

\begin{aligned}
\langle Du, \phi \rangle &= \int u^{'}  \phi \; dx \\
&= -\int u  \phi ^{'} \; dx
\end{aligned}
It's a way of extending the notion of differentiability to functions that are not necessarily continuous.


问:
Prove that power of radical is contained in ideal
$A \textrm{ is a commutative ring with multiplicative identity }.\\
\alpha \textrm{ is a finitely generated  ideal of }A.\: \: \sqrt{\alpha } =\left \{ x\in A\mid  x^{m}\in \alpha \mid m\in \mathbb{Z}^{+} \right \}.\\
\textrm{Prove that }\exists n\in \mathbb{Z}^{+}\textrm{ such that }(\sqrt{\alpha })^{n}\subseteq \alpha $

Not getting any idea on how to proceed. Will it help if I prove that the radical is finitely generated?

答:
Hint:

You're seeking an $N$ such that the product of every collection of $N$ elements of $\sqrt{\alpha}$ is in $\alpha$.  What you are given is a finite subset $\{a_i\mid i=1\ldots k\}$of $\sqrt{\alpha}$, each of which has its own power lying in $\alpha$.

Consider this subproblem: Let $\{x_i\mid i\in 1\ldots N\}$ be a family of elements from the set $\{a_1,\ldots, a_k\}$ ($N$ may be much larger than $k$, with repeated selections.) How big must $N$ be to guarantee $\prod_{i=1}^N x_i\in \alpha$?

Now, given elements of the form $b_j=\sum a_ir_{ij}$, you know that after the expansion of the product of $N$ such sums, each term will contain (a multiple of) something of the form described in the subproblem above.   If each of those terms is in $\alpha$, then their sum will be also in $\alpha$...


问:
Joint density of functions of two random variables
If $X$ and $Y$ are iid with $U(0,2)$, $Z=2X+Y$ and $V=e^X$

What is the joint density of $(Z,V)$?

$\begin{align}
V=e^X &\Rightarrow X=lnV\\
Z=2X+Y &\Rightarrow Y=Z-2lnV
\end{align}$

$
J=
\begin{vmatrix}
\frac{\partial x}{\partial z} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial z} & \frac{\partial y}{\partial v} \\
\end{vmatrix}
=
\begin{vmatrix}
0 & \frac{1}{v} \\
1 & -\frac{2}{v}
\end{vmatrix}
=-\frac{1}{v}
$

$\begin{align}
f_{Z,V}(z,v)&=f_{X,Y}(lnV,Z-2lnV)*|J| \\
&=f_X(lnV)f_Y(Z-2lnV)*|-\frac{1}{v}|
\end{align}$

$\begin{align}
0 \le lnV \le 2 &\Rightarrow 1 \le V \le e^x \\
0 \le Z-2lnV \le 2 &\Rightarrow 2lnV \le Z \le 2+2lnV \\
&\Rightarrow 0 \le Z \le 6
\end{align}$

For $1 \le v \le e^x$ and $0 \le z \le 6$,

$\begin{align}
f_{Z,V}(z,v)&=\frac{1}{2}*\frac{1}{2}*\frac{1}{v} \\
&=\frac{1}{4v}
\end{align}$

However, when I check the answer by using double integration

$\begin{align}
\int_{1}^{e^2} \int_{0}^{6}\frac{1}{4v} \,dz\,dv &= \frac{3}{2}\int_{1}^{e^2}\frac{1}{v} \,dv\\
&=\frac{3}{2}*(2-0) \\
&=3
\end{align}$

The integral of density by the support should instead be 1, that means my answer is wrong, what have I done wrong?

答: Indeed $Z\in [0,6]$, but if $\varphi $ is s.t. $\varphi (X,Y)=(U,V)$, then it's not true that $$\varphi(\{(X,Y)\in (0,2)^2\})= [1,e^2]\times [0,6].$$ What you nevertheless have is $$\varphi (\{(X,Y)\in (0,2)^2\}=\{(V,Z)\mid V\in (1,e^2), 2\ln(V)\leq Z\leq 2\ln(V)+2\}).$$


问:
Is the following matrix diagonalizable over $\mathbb{R}$?
I am getting conflicting answers depending on how I approach the problem and I don't understand why. Let $A=\begin{pmatrix}2&-5&7\\1&-2&0\\0&0&3
\end{pmatrix}$. If I go directly to the characteristic polynomial I get $\det(A-I\lambda)=\det\begin{pmatrix}2-\lambda&-5&7\\1&-2-\lambda&0\\0&0&3-\lambda
\end{pmatrix}$ and then calculating the determinant I get that $\lambda_1=3, \lambda_{2,3}=\pm i$.

I then conclude that $A$ is not diagonalizable over $\mathbb{R}$ as we have non-real eigenvalues.

However if I put $A$ into an upper diagonal matrix I have that $A=\begin{pmatrix}2&-5&7\\0&0.5&-3.5\\0&0&3
\end{pmatrix}$

I then see the eigenvalues as the entries of our upper diagonal matrix, i.e. distinct and real. Then isn't this representation of $A$ diagonalizable? But $A$ hasn't changed. What am I misunderstanding?

Thanks

答:
The operation you are performing on the matrix $A$ is equivalent to pre-multiplying by another elementary matrix $E$. Just because $EA$ is diagonalizable, does not mean that $A$ is. In particular, as you just demonstrated, the eigenvalues of $A$ and $EA$ are not necessarily the same.

Perhaps a more elementary example of such "innocent" alterations leading to changing eigenvalues are flipping the rows of
$$M = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix},$$
which results in the identity matrix. The original matrix $M$ has eigenvalues $-1,1$ and the resulting identity has only $1$ instead (with multiplicity $2$)...


问:
Supremum of $A = \{ \frac{5^{3n}*5^{2m}}{3*5^{5n}+5^n + 2*5^{5m}} , n, m \in \mathbb{N}\}$.
How do I find the supremum and infimum of set A?

$$A = \{ \frac{5^{3n}*5^{2m}}{3*5^{5n}+5^n + 2*5^{5m}} , n, m \in \mathbb{N}\}$$

I know that this expression is always greater than 0 since it is made of natural numbers only.

for $n = 1$, $m \implies \infty$, $A \implies 0$

for $m = 1$, $n = \implies \infty$, $A \implies 0$

Therefore, I know that the infimum is equal to $0$. However, how do I find the supremum?

From a graph I made I know that for $n = m = 1$ that expression is equal to $\frac{1}{5}$ and that might be the supremum. However, how do I deduct that from this expression? I tried dividing nominator and denominator by $5^n$ and $5^m$. I tried comparing nominator to denominator. Got nothing.

Any hint would be much appreciated.

Body:
Hint : Let $x=5^n$ and $y=5^m$. Notice that

$$
\frac{1}{5}-\frac{x^3y^2}{3x^5+x+2y^5} =
\frac{1}{5(3x^5+x+2y^5)}\bigg(
x+(y-x)^2\big(3x^3 + 6yx^2 + 4y^2x + 2y^3\big)\bigg)
$$


问:
The fastest way to find an obtuse triangle based on several sets of lengths
A question from ACT math:

I'm wondering except using the law of cosine to get the answer K, is there any faster way to figure it out?

答:
If $\angle A$ is an obtuse angle in a triangle.

$$90<\angle A<180, -1<\cos\angle A<0$$

$$\cos\angle A= \frac{b^2+c^2-a^2}{2bc} <0$$

using only numerator since denominator is always positive

$$a^2>b^2+c^2$$

where $a$ is the longest side

$$16^2>11^2+8^2$$


答: That means every matrix of order $n\times n$  with rank $n$ is diagonalizable. But this is certainly not true. The matrix you got after you row reduced your $A$ is not simlar to $A$ since there does not exist any invertible $P$ such that $A=PBP^{-1}$ where $B$ is the row reduced version of $A$

Title: Is the following matrix diagonalizable over $\mathbb{R}$?

Body:
I am getting conflicting answers depending on how I approach the problem and I don't understand why. Let $A=\begin{pmatrix}2&-5&7\\1&-2&0\\0&0&3
\end{pmatrix}$. If I go directly to the characteristic polynomial I get $\det(A-I\lambda)=\det\begin{pmatrix}2-\lambda&-5&7\\1&-2-\lambda&0\\0&0&3-\lambda
\end{pmatrix}$ and then calculating the determinant I get that $\lambda_1=3, \lambda_{2,3}=\pm i$.

I then conclude that $A$ is not diagonalizable over $\mathbb{R}$ as we have non-real eigenvalues.

However if I put $A$ into an upper diagonal matrix I have that $A=\begin{pmatrix}2&-5&7\\0&0.5&-3.5\\0&0&3
\end{pmatrix}$

I then see the eigenvalues as the entries of our upper diagonal matrix, i.e. distinct and real. Then isn't this representation of $A$ diagonalizable? But $A$ hasn't changed. What am I misunderstanding?

Thanks


问:
Proof that an embedding into $\ell^1$ is compact

Prove that any sequence $(x^{(n)})_{n\in\mathbb{N}}\subseteq\ell^1$ such that $\sum_{k=1}^\infty k\lvert x_k^{(n)}\lvert\leq1$ for all $n\in\mathbb{N}$ has a convergent subsequence.

My thoughts on this: Clearly $\lvert x^{(n)}_k\lvert\leq\frac{1}{k}$ uniform in $n$. Therefore the sequence $x_1^{(n)}$ has a convergent subsequence $x_1^{(\tilde{n}_k)}$. Further extracting subsequences for any fixed $N\in\mathbb{N}$ I find $({n_l})_{l\in\mathbb{N}}\subseteq\mathbb{N}$ such that $x_k^{(n_l)}$ converges to some $x_k$ for all $1\leq k\leq N$ as $n\rightarrow\infty$.

If for every $\epsilon>0$ I could find some $N\in\mathbb{N}$ such that $\sum_{k=N+1}^\infty \lvert x_k^{(n)}\lvert<\epsilon$ uniform in $n$ the proof would be complete. But I don't see why this should be true.

Can you give me some hint?

答:
I think you may want a proof like this:

By the given condition we know $|x_k^{(n)}|\le1$ for all $n$ and $k$. So for $k=1$, we may pick up a convergent subsequence $\{x_1^{n^1_j}\}_{j\in\mathbb{N}}$. Then based on this subsequence we pick up for $k=2$ a convergent subsequence $\{x_2^{n^2_j}\}_{j\in\mathbb{N}}$. Keep doing in this way and get a sequence for each $k$, and then take the diagonal elements $n^j_j$. It can be verified that $\{x^{n^j_j}\}_{j\in\mathbb{N}}$ converge pointwisely to some $\tilde{x}$. For simplicity we denote this new sequence $\{x^{(m)}\}_{m\in\mathbb{N}}$. It can be shown that $\sum_{k=1}^{\infty}k|\tilde{x}_k|\le1$, hence $\tilde{x}\in\mathcal{l}^1$. In fact, since $\{x^{(m)}\}_{m\in\mathbb{N}}$ converge pointwisely to $\tilde{x}$, for finite $N$ we have $\sum_{k=1}^{N}k|\tilde{x}_k|\le1$, then passing $N$ to infinity we know $\sum_{k=1}^{\infty}k|\tilde{x}_k|\le1$.

Then we show that for every $\epsilon>0$ we can find some $N\in\mathbb{N}$ such that $\underset{m}{\sup}\sum_{k=N+1}^{\infty}|x_k^{(m)}|<\epsilon$. In fact, only need to observe that
$$\sum_{k=N+1}^{\infty}|x_k^{(m)}|\le\frac{1}{N+1}\sum_{k=N+1}^{\infty}k|x_k^{(m)}|\le\frac{1}{N+1},\qquad\forall m.$$

Now we're already to prove your claim. For any $\epsilon>0$ take $N_1$ such that $\underset{m}{\sup}\sum_{k=N_1+1}^{\infty}|x_k^{(m)}|<\epsilon/4$. Note that $\tilde{x}\in\mathcal{l}^1$, so we can find $N_2$ such that $\sum_{k=N_2+1}^{\infty}|\tilde{x}_k|<\epsilon/4$. Therefore, if taking $N_3=\max\{N_1,N_2\}$, then
$$\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}-\tilde{x}_k|\le\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}|+\sum_{k=N_3+1}^{\infty}|\tilde{x}_k|<\epsilon/2.$$
On the other hand, since $\{x^{m}\}_{m\in\mathbb{N}}$ converge pointwisely to $\tilde{x}$, we can find $N_4$ such that
$$\sum_{k=1}^{N_3}|x_k^{(m)}-\tilde{x}_k|<\epsilon/2,\qquad\forall m>N_4.$$
Combining the above two we know
$$\|x^{(m)}-\tilde{x}\|_{\mathcal{l}^1}\le\sum_{k=1}^{N_3}|x_k^{(m)}-\tilde{x}_k|+\underset{m}{\sup}\sum_{k=N_3+1}^{\infty}|x_k^{(m)}-\tilde{x}_k|<\epsilon,\qquad\forall m>N_4.$$
This means $\{x^{(m)}\}$ converge to $\tilde{x}$ in $\mathcal{l}^1$. Proof is finished.


答: Your new matrix is equivalent to $A$, but not similar. The latter condition is stronger and required to preserve the Eigenvalues.


问:
$\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}$ equals to $1$ or $e^2$?
$$\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}$$ equals to $1$ or $e^2$?

My Attempt

\begin{align}
\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}
    &= e^{ \displaystyle \lim_{x\to\infty} 2x(x-\sqrt{x^2-2})}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{2x(x^2-(x^2-2))}{x+\sqrt{x^2-2}}}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{4x}{x+\sqrt{x^2-2}}}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{4}{1+\sqrt{1-\frac{2}{x^2}}}}\\
    &=e^2
\end{align}

However, Mathematica Alpha gives a different result which is $1$. Which is the correct one?
If the Mathematica version is correct, how is its answer derived?

Edit

The rule that I am using here is

$$
\lim_{x\to a} (1+f(x))^{g(x)} = e^{\lim_{x\to a} f(x)g(x)}
$$

I am not sure, the constraint for this rule is

both $f(x)$ and $g(x)$ must approach zero or

both $f(x)$ and $g(x)$ must approach infinity

答:
Using equivalents when $x$ is large
$$(2x+1)\sim 2x \qquad \text{and} \qquad x-\sqrt{x^2-2} \sim \frac 1x$$

Now what is the limit of $(2x)^{\frac 1x}$ ?


问:
How can I compute rotation angle of an image?
An image is translated with $x_0$ and $y_0$ (vertical and horizontal displacements), rotated with an angle $θ$, and then is scaled on both vertical and horizontal directions with the scaling parameters $α$ and $β$. The resulting composed geometric transform in homogeneous matrix form is
$$\begin{pmatrix}\sqrt 3 & 1 & 20 \\ -1 & \sqrt 3 & 10 \\ 0 & 0 & 1\end{pmatrix}$$

I need to compute the rotation angle $θ$, the scaling parameters $α$ and $β$, and the vertical and horizontal displacements $x_0$ and $y_0$.

答:
Let $F\colon \mathbb P^2\to\mathbb P^2$ be your transformation on the projective plane. It can be decomposed as $F = S_{(\alpha,\beta)}\, R_{\theta}\, T_{(x_0,y_0)}$ as explained by AugSB. To find the angle $\theta$ you can take any two points and see how the line segment between them gets rotated. The line segment from $(0,0)$ to $(1,0)$ get's mapped to the line segment from $(20,10)$ to $(\sqrt 3+20,-1+10)$, where I use $(x,y)$ to denote the point $(x:y:1)$ in $\mathbb P^2$. Hence
$$
\theta = \measuredangle (1,0), (\sqrt 3,-1) = \cos^{-1}\left(\frac{\sqrt 3}{2}\right) = \frac{\pi}{6} = 30^\circ.
$$
Now consider the line segment of length $1$ from $(0,0)$ to $(\sqrt{3}/2,-1/2)$, this has an angle of $-30^\circ$ to the $x$-axis, so after translation and rotation it will be parallel to the $x$-axis. Hence the length of its image under $F$ determines $\alpha$. We have $F(0,0)=(20,10)$ and $F(\sqrt{3}/2,-1/2)=(21, -\sqrt 3+10)$ so that
$$
\alpha = \|(1, -\sqrt 3)\| = 2.
$$
The same reasoning applied to the line segment from $(0,0)$ to $(1/2,\sqrt{3}/2)$ yields
$$
\beta = \|(\sqrt 3,1)\| = 2.
$$
Finally, we can find $(x_0,y_0)$ by applying the inverses of $S_{(\alpha,\beta)}$ and $R_\theta$ to $F(0,0)=(20,10)$:
\begin{align*}
S_{(2,2)}\, R_{30^\circ}\, T_{(x_0,y_0)}(0,0) &= (20,10) \\
R_{30^\circ}\, T_{(x_0,y_0)}(0,0) &= (10,5) \\
\end{align*}
Note that $R_\theta = \begin{pmatrix} \sqrt 3/2 & -1/2 & 0 \\ 1/2 & \sqrt 3/2 & 0 \\ 0 & 0 & 1\end{pmatrix}$, so that $R_\theta^{-1} = \begin{pmatrix} \sqrt 3/2 & 1/2 & 0 \\ -1/2 & \sqrt 3/2 & 0 \\ 0 & 0 & 1\end{pmatrix}$ and we get
$$
(x_0, y_0) = T_{(x_0,y_0)}(0,0) = R_\theta^{-1} (10,5) = \left(\frac 5 2(2\sqrt 3 - 1), \frac 5 2(\sqrt 3 + 2)\right).
$$


问:
Norm subscript notation in optimization
I'm reading a paper with the expression $H_k = \underset{H}{arg \hspace{3pt}min } \hspace{6pt} ||H-H_{k-1}||_{F,W}^2 \hspace{10pt} s.t. \hspace{5pt} H=H^T, Hy_k = s_k$

I've read that the subscript means that $F$ and $W$ are positive definite, and that makes sense for W since the article later goes on to state that $||X||_{F,W}^2 = ||XW||_F^2$ but $F$ is never mentioned again anywhere in the paper except in the subscript of the norm, so why is it there and what does it mean?
It's clear from the paper that the norm is the L2 norm, so the $F$ can't mean which norm it is.

答: Based on your hint that there seems to be a relation with the L2 norm, I guess that $\lVert \cdot \rVert_F$ indicates that the Frobenius Norm is used. That would also make sense as we need a matrix norm. My only concern is the notation $\lVert \cdot \rVert_{F,W}$ which I have not seen before. But as stated in your question, there seems to be an explanation for that in the paper.

Title: Norm subscript notation in optimization


问:
Relationship between covariant/contravariant basis vectors
I'm starting to learn some of the basics of covariant and contravariant vectors. </p>

<p>I'm a little confused about the difference between a covariant and a contravariant basis vector. I know that the vector components transform using the metric tensor: $V^i = M^{ij} V_j$. How do I transform between the covariant basis vectors and contravariant basis vectors? </p>

答:
Covariant and contravariant bases are dual to one another and are physics nomenclature for constructs that arise in differential geometry. The problem here is that physicists often need to use differential geometry (for example, for relativity) long before they have seen a proper course on differential geometry.

I will try to provide you with an extremely quick view of what contra/covariant bases are, and how you move between them.

First, let $M$ be a [smooth manifold][1]. For our purposes, the most important part of being a smooth manifold is the Local Euclidean property, which says that for each point $p \in M$ there are open neighbourhoods $U \subseteq M, \tilde U \subseteq \mathbb R^n$ and a homeomorphism $\phi: U \to \mathbb R^n$ for some $n$; that is, if we look closely at the manifold, it just looks like $\mathbb R^n$. If we denote $\phi$ component-wise as $\phi = (x^1,\ldots,x^n)$ then we can describe points $q \in U \subseteq M$ in terms of $(x^1(q),\ldots,x^n(q))$ in $\mathbb R^n$.

The next step is to consider the [tangent space][2] to $M$ at the point $p$, denoted by $T_pM$. There are quite a few ways of defining this guy, my favourite being as derivations on $M$ through $p$, but for intuition-sake, think of $T_pM$ as the collection of all vectors through $p$ which are tangent to $M$. This is a vector space, and so has a dual space $T_p^*M$ called the cotangent space.

The coordinates $(x^1,\ldots,x^n)$ we defined at $p$ induce a basis on the tangent space, and that basis is typically written as $ \left\{ \left.\frac{\partial}{\partial x^1}\right|_p, \ldots, \left.\frac{\partial}{\partial x^n}\right|_p\right\}$, but for our purpose let's call it $\{ v_1,\ldots,v_n\}$. Similarly, there is an induced basis on the cotangent space (which turns out to be the dual basis), often denoted $\{dx^1,\ldots,dx^n\}$ but again, for our purposes, we will write $\{v^1,\ldots,v^n\}$.

These are your covariant and contravariant bases, respectively. But you are now likely confused as covariant vectors have subscripts and contravariant vectors have superscripts. This is because the Einstein summation convention says that that an arbitrary vector $v= \sum_{i=1}^n a_i v^i$ should be written by omitting the summation sign, so that $v = a_iv^i$. But the $v^i$ are in a sense assumed to be given, so the physicists shorten it further and just write $v = a_i$, and now you get the subscripts that you wanted. The same argument works for contravariant vectors.

The next question is how to move between them. A priori, there is no reason to suspect this should even be possible: They live in completely different spaces! However, we can define a Riemannian metric on $M$ which, intuitively speaking, is an non-degenerate inner product on each tangent space. This is what the physicists call a metric (tensor). Non-degenerate inner products on finite dimensional vector spaces give a natural identification between the vector space and its dual. More precisely, if $\langle\cdot,\cdot\rangle: V \times V \to \mathbb R$ then we can define an isomorphism $V \to V^*$ by $v \mapsto \langle v, \cdot \rangle$ (You can actually more generally define what are called [musical isomorphisms][3]). In local coordinates, our inner product can be written (in Einstein convention) as $g^{ij}v_i \otimes v_j$ (or just $g^{ij}$ if we omit the vectors), and hence the isomorphism above corresponds to
$$ v \mapsto \langle v, \cdot \rangle , \qquad a_i \mapsto g^{ij} a_i.$$
Again by laziness, we avoid carrying around the metric tensor and just write $a^j = g^{ij}a_i$, which is the concept of contraction (better known to mathematicians as the interior product).

You can of course move the other direction as well. In local coordinates, the metric tensor $g^{ij}$ looks like an invertible matrix. Let $g_{ij}$ be its inverse, in which case $a_j = g_{ij} a^i$.

It is definitely a lot to take in, but if you understand the rigorous mathematics behind these things, the physics actually becomes significantly easier.

[1]: Differentiable Manifold at Wikipedia

[2]: Tangent Space at WP

[3] :  Musical Isomorphism at WP


问:
How to show that Span $(x) \subseteq W_1 + W_2 ?$
If  $W_1 + W_2$ are finite dimensional subspaces of a vector space , then $W_1 + W_2$ is finite  -dimensional and $\dim(W_1 + W_2)= \dim(W_1) + \dim(W_2) - \dim (W_1 \cap W_2)$

Proof :

let $\{a_1,a_2,...,a_k\}$  be  basis for  $W_1 \cap W_2$

let $a= b+c \in W_1 +W_2$  where $b \in W_1$  and  $c \in W_2$

let  $x$  is  a basis for  $W_1 +  W_2$ where $ x= \{a_1,a_2 ,...., a_k ,b_1,b_2 ,.....,b_m,c_1,c_2,.....,c_p\}$

Now to prove this theorem two condition must be satisfied given below

$1.$ $x$ is linearly independent

$2$. Span $(x)= W_1 + W_2$

I can easily proved the first condition

But im stuck in condition $2$
For condition $(2)$ we can write  $x= \underbrace{(x_1a_1+ ....+x_ka_k + y_1b_1+.....+y_mb_m)}_{W_1} + \underbrace{(z_1a_1+ ...+z_ka_k+ t_1c_1+...+t_pc_p)}_{W_2}$

$x= (( x_1 +z_1)a_1 +....+(x_k+z_k)a_k + y_1b_1 +....+y_mb_m + t_1c_1 +....+t_pc_p)$

This implies that $W_1 + W_2 \subseteq $Span $(  x)$

But here im confusing that

How to show that Span $(x) \subseteq W_1 + W_2 ?$

答:
Hint

Things are a bit confused. Let $\{u_1,...,u_n\}$ be a basis of $W_1\cap W_2$ and prolonge it in a basis of $W_1$ and a basis of $W_2$, i.e. take $\{u_1,...,u_n,v_1,...,v_k\}$ being a basis of $W_1$ and $\{u_1,...,u_n,w_1,...,w_p\}$ being a basis of $W_2$.

Now, prove that $\{u_1,...,u_n,v_1,...,v_k,w_1,...,w_p\}$ is a basis of $W_1+W_2$, and thus $$\dim(W_1+W_2)=n+k+p=\dim(W_1)+\dim(W_2)-\dim(W_1\cap W_2).$$


答:
You have

$$0<\log V<2$$

$$0<Z-2\log V <2$$

That is

$$2\log V<Z<2\log V+2$$

Graphically, $(V,Z)$ domain is the following

Concluding:

$$f_{VZ}(v,z)=\frac{1}{4v}\cdot\mathbb{1}_{(1;e^2)}(v)\cdot\mathbb{1}_{(2\log v;2(1+\log v))}(z)$$

Title: Joint density of functions of two random variables

Body:
If $X$ and $Y$ are iid with $U(0,2)$, $Z=2X+Y$ and $V=e^X$

What is the joint density of $(Z,V)$?

$\begin{align}
V=e^X &\Rightarrow X=lnV\\
Z=2X+Y &\Rightarrow Y=Z-2lnV
\end{align}$

$
J=
\begin{vmatrix}
\frac{\partial x}{\partial z} & \frac{\partial x}{\partial v} \\
\frac{\partial y}{\partial z} & \frac{\partial y}{\partial v} \\
\end{vmatrix}
=
\begin{vmatrix}
0 & \frac{1}{v} \\
1 & -\frac{2}{v}
\end{vmatrix}
=-\frac{1}{v}
$

$\begin{align}
f_{Z,V}(z,v)&=f_{X,Y}(lnV,Z-2lnV)*|J| \\
&=f_X(lnV)f_Y(Z-2lnV)*|-\frac{1}{v}|
\end{align}$

$\begin{align}
0 \le lnV \le 2 &\Rightarrow 1 \le V \le e^x \\
0 \le Z-2lnV \le 2 &\Rightarrow 2lnV \le Z \le 2+2lnV \\
&\Rightarrow 0 \le Z \le 6
\end{align}$

For $1 \le v \le e^x$ and $0 \le z \le 6$,

$\begin{align}
f_{Z,V}(z,v)&=\frac{1}{2}*\frac{1}{2}*\frac{1}{v} \\
&=\frac{1}{4v}
\end{align}$

However, when I check the answer by using double integration

$\begin{align}
\int_{1}^{e^2} \int_{0}^{6}\frac{1}{4v} \,dz\,dv &= \frac{3}{2}\int_{1}^{e^2}\frac{1}{v} \,dv\\
&=\frac{3}{2}*(2-0) \\
&=3
\end{align}$

The integral of density by the support should instead be 1, that means my answer is wrong, what have I done wrong?


问:
Geometric tiling puzzle
yesterday my sister send me a YouTube link about a tiling puzzle game: https://youtu.be/A9KU_gPOaGU.

Can one deduce from the geometry of the pieces alone whether there is something special about this puzzle in a mathematical sense? Can it really be the hardest tiling puzzle that can be designed with 66 parts? I am a little skeptical about that.

Best regards.

答:
Yep, that it's based on one of the classic kites-and-darts or rhombs Penrose Tilings is also my guess. I'm trying out a few things right now. When I find a substitution for the tiles I will post it.

I cut out some pieces from paper, and they can be put together well. So probably the frame can be filled very quickly, but only almost. That probably makes it difficult.

Title: Geometric tiling puzzle


问:
Tangent space exploration of function
I have a function $f, R^n \rightarrow R$ , which  I evaluate at a vector $x \in R^n$. So, $f(x) = y$.

Now, I calculate $f'(x) = g, g \in R^n$ . I want to "move" x in the tangent space of $f$. The part I'm stuck at, is rather silly.

I compute a vector projected to the tangent space as below:

$g \in R^n$ is the gradient of $f$ at $x$. Compute $g_u$, which is $g$ unit normalized.

Sample a random vector $v \in R^n$ from a Unit Gaussian.

Compute $ w = v - <v,g_u> g_u$ , where $<.>$ is the inner product. So, $w \in R^n$ is the vector $v$ projected to tangent space of $f$.

I have two questions:

Is the above procedure of finding a direction in the tangent space correct? If not, how do I find a random vector in the tangent space of $f$ for given $x$ and $f'(x)$ ?

If it is correct, now that I have $w$ which is a vector in the tangent space of $f$, how do I use it to "move" $x$ in the tangent space? Right now, I do $x = x + \alpha*w$, where $\alpha$ is simply the step size.

Thank you for the help.

答:
What is sure:

$w$ is a vector of the $f$ tangent space at $x$.

$x + \alpha w$ belongs to $f$ tangent plane at $x$.

What is less sure is:

What is the distribution of $w$ in the tangent space knowing that $v$ is distributed according to a Unit Gaussian in $\mathbb R^n$?

But why don't you use a Unit Gaussian distribution in $\mathbb R^{n-1}$ and a basis in the tangent plane?

At the end, the most important question is to understand what kind of distribution do you expect in the tangent plane?

Title: Tangent space exploration of function


答:
Well, first thing I thaught of was to use the fact that : $$ \left(\forall x\in\mathbb{R}^{*}\right),\ \frac{1-\cos{x}}{x^{2}}=\int_{0}^{1}{\left(1-y\right)\cos{\left(xy\right)}\,\mathrm{d}y} $$

Then, with Fubini's theorem will, I shall find the result.

Let $ s>0 $, we have : \begin{aligned}\int_{0}^{+\infty}{\mathrm{e}^{-sx}\left(\frac{1-\cos{x}}{x^{2}}\right)\mathrm{d}x}&=\int_{0}^{+\infty}{\mathrm{e}^{-sx}\int_{0}^{1}{\left(1-y\right)\cos{\left(xy\right)}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\left(1-y\right)\int_{0}^{+\infty}{\mathrm{e}^{-sx}\cos{\left(xy\right)}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=s\int_{0}^{1}{\frac{1-y}{s^{2}+y^{2}}\,\mathrm{d}y}\\ &=\int_{0}^{1}{\frac{\frac{\mathrm{d}y}{s}}{1+\left(\frac{y}{s}\right)^{2}}}-\frac{s}{2}\int_{0}^{1}{\frac{\frac{2y\,\mathrm{d}y}{s^{2}}}{1+\left(\frac{y}{s}\right)^{2}}}\\ &=\int_{0}^{\frac{1}{s}}{\frac{\mathrm{d}x}{1+x^{2}}}-\frac{s}{2}\int_{0}^{\frac{1}{s^{2}}}{\frac{\mathrm{d}x}{1+x}}\\ \int_{0}^{+\infty}{\mathrm{e}^{-sx}\left(\frac{1-\cos{x}}{x^{2}}\right)\mathrm{d}x}&=\arctan{\left(\frac{1}{s}\right)}-\frac{s}{2}\ln{\left(1+\frac{1}{s^{2}}\right)}\end{aligned}

Title: Proof that $\int_0^{\infty} e^{-st} (\frac{1-\cos t}{t^2})dt = \frac{\pi}{2} - s\ln(\frac{s^2}{s^2+1}) + 2\tan^{-1}(s)$

Body:
I'm trying to solve this Laplace problem from Laplace Transform by Schaum's Outline, chapter 1, math 167(a). But no matter what approach I follow, I end up with this line:

$L(\frac{1-\cos t}{t^2}) = \frac{\pi}{2} + \frac{1}{2}s\ln(\frac{s^2}{s^2+1}) - \tan^{-1}(s)$

And by definition, $L(\frac{1-\cos t}{t^2}) = \int_0^{\infty} e^{-st} (\frac{1-\cos t}{t^2})dt$

So I've no idea why and how the problem expects me to end up from $\frac{\pi}{2} + \frac{1}{2}s\ln(\frac{s^2}{s^2+1}) - \tan^{-1}(s)$ to $\frac{\pi}{2} - s\ln(\frac{s^2}{s^2+1}) + 2\tan^{-1}(s)$

Looking for your kind help.


问:
Sum of reciprocals of numbers with certain terms omitted
I know that the harmonic series $1 + \frac12 + \frac13 + \frac14 + \cdots$ diverges. I also know that the sum of the inverse of prime numbers $\frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \cdots$ diverges too, even if really slowly since it's $O(\log \log n)$.

But I think I read that if we consider the numbers whose decimal representation does not have a certain digit (say, 7) and sum the inverse of these numbers, the sum is finite (usually between 19 and 20, it depends from the missing digit). Does anybody know the result, and some way to prove that the sum is finite?

答:
It is not very surprising that the sum is finite, since numbers without a 7 (or any other digit) get rarer and rarer as the number of digits increases.

Here's a proof.

Let $S$ be the harmonic series with all terms whose denominator contains the digit $k$ removed. We can write $S =S_1 + S_2 + S_3 + \ldots$, where $S_i$ is the sum of all terms whose denominator contains exactly $i$ digits, all different from $k$.

Now, the number of $i$-digit numbers that do not contain the digit $k$ is $8\cdot9^{i-1}$ (there are $8$ choices for the first digit, excluding $0$ and $k$, and $9$ choices for the other digits). [Well, if $k=0$ there are $9$ choices for the first digit, but the proof still works.] So there are $8\cdot9^{i-1}$ numbers in the sum $S_i$.

Now each number in $S_i$ is of the form $\frac1a$, where $a$ is an $i$-digit number. So $a \geq 10^{i-1}$, which implies $\frac1a \leq \frac1{10^{i-1}}$.

Therefore $S_i \leq 8\cdot\dfrac{9^{i-1} }{10^{i-1}} = 8\cdot\left(\frac9{10}\right)^{i-1}$.

So $S= \sum S_i \leq \sum 8\cdot\left(\frac9{10}\right)^{i-1}$

which is a  geometric series of ratio $\frac9{10} < 1$, which converges. Since $S$ is a positive series bounded above by a converging series, $S$ converges.


问:
The orthogonal complement of the space of row-null and column-null matrices
I propose the following lemma and its proof. It is related to row-null and column-null matrices - i.e. matrices whose rows and columns both sum to zero. Could you please give your opinion on the plausibility of the lemma, and the validity of the proof?

Lemma: Let $Z\in\text{GL}(n,\mathbb{R})$ be a general $n\times n$ real matrix, and let $Y\in\mathcal{S}(n,\mathbb{R})$, where $\mathcal{S}(n,\mathbb{R})$ is the space of row-null column-null $n\times n$ real matrices. Then $\text{Tr}(ZY)=0$ for all $Y$ in $\mathcal{S}(n,\mathbb{R})$ if and only if $Z$ has the form $$Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)$$.

Proof:
Consider the space of row-null and column-null matrices

$$\mathcal{S}(n,\mathbb{R})= \left\{ Y_{ij}\in GL(n,\mathbb{R}):\sum_{i}Y_{ij}=0,\sum_{j}Y_{ij}=0 \right\} $$

Its dimension is
$$\text{dim}(S(n,\mathbb{R}))=N^{2}-2N+1$$
since the row-nullness and column-nullness are defined by $2N$ equations, only $2N-1$ of which are linearly independent.
Consider the following space

$$\mathcal{G}(n,\mathbb{R})=\left\{ Z_{ij}\in GL(n,\mathbb{R}):Z_{ij}=\left(p_{j}-p_{i}\right)+\left(q_{j}+q_{i}\right)\right\}$$

Its dimension is

$$\text{dim}(\mathcal{G}(n,\mathbb{R}))=2N-1$$
where $N-1$ is the contribution from the antisymmetric part and $N$ is from the symmetric part.

Assume $Y\in\mathcal{S}$ and $Z\in\mathcal{G}$, then the Frobenius inner product of two such elements is
$$
\text{Tr}(ZY) =\sum_{ij}\left[\left(p_{j}-p_{i}\right)Y_{ji}+\left(q_{j}+q_{i}\right)Y_{ji}\right]
$$
$$
=\sum_{j}(q_{j}+p_{j})\sum_{i}Y_{ji}+\sum_{i}(q_{i}-p_{i})\sum_{j}Y_{ji}=0
$$
Since $\text{dim}(\mathcal{G})+\text{dim}(\mathcal{S})=\text{dim}(GL)$ and $\mathcal{G}\perp\mathcal{S}$, then $\mathcal{G}$ and $\mathcal{S}$ must be complementary in $GL$. Therfore, if $Y$ is orthogonal to all the matrices in $\mathcal{S}$, it must lie in $\mathcal{G}$.

PS: How can I get the curly brackets {} to render in latex mode?

答:
Here is an alternate way of proving your Lemma. I'm not sure if its any simpler than your proof -- but it's different, and hopefully interesting to some.

Let $S$ be the set of $n\times n$ matrices which are row-null and column-null. We can write this set as:
$$
S = \left\{ Y\in \mathbb{R}^{n\times n} \,\mid\, Y1 = 0 \text{  and  }1^TY=0\right\}
$$
where $1$ is the $n\times 1$ vector of all-ones. The objective is the characterize the set $S^\perp$ of matrices orthogonal to every matrix in $S$, using the Frobenius inner product.

One approach is to vectorize. If $Y$ is any matrix in $S$, we can turn it into a vector by taking all of its columns and stacking them into one long vector, which is now in $\mathbb{R}^{n^2\times 1}$. Then $\mathop{\mathrm{vec}}(S)$ is also a subspace, satisfying:
$$
\mathop{\mathrm{vec}}(S) = \left\{ y \in \mathbb{R}^{n^2\times 1} \,\mid\, (\mathbf{1}^T\otimes I)y = 0 \text{  and  } (I \otimes \mathbf{1}^T)y = 0 \right\}
$$
where $\otimes$ denotes the Kronecker product. In other words,
$$
\mathop{\mathrm{vec}}(S) = \mathop{\mathrm{Null}}(A),\qquad\text{where: }
A = \left[ \begin{array}{c} \mathbf{1}^T\otimes I \\ I \otimes \mathbf{1}^T \end{array}\right]
$$
Note that vectorization turns the Frobenius inner product into the standard Euclidean inner product. Namely: $\mathop{\mathrm{Trace}}(A^T B) = \mathop{\mathrm{vec}}(A)^T \mathop{\mathrm{vec}}(B)$. Therefore, we can apply the range-nullspace duality and obtain:
$$
\mathop{\mathrm{vec}}(S^\perp) =
\mathop{\mathrm{vec}}(S)^\perp =
\mathop{\mathrm{Null}}(A)^\perp =
\mathop{\mathrm{Range}}(A^T)
$$
So every vector in $vec(S^\perp)$ is of the form $(\mathbf{1}\otimes I)a + (I\otimes \mathbf{1})b$ for some vectors $a$ and $b$ in $R^{n\times 1}$. It follows that every matrix in $S^\perp$ is of the form $a1^T + 1b^T$. This parametrization is equivalent to the one you presented if you set $a_i = q_i-p_i$ and $b_j = q_j + p_j$.

ContentLicense: CC BY-SA 2.5


问:
Derived category and so on
I am looking for an introductive reference to the theory of derived categories. Especially I need to start from the very beginning and I need to know how to use this in examples which comes from algebraic geometry. I don't want a too rigorous approach, made of a lot of definition and propositions but instead I would like to find an introduction which gives the main ideas and many examples.
Thank you</p>

答: Gelfand and Manin's "Methods of homological algebra" explains the subject quite nicely, though it takes them some pages to develop the theory (and there are many typos, at least in the first edition). Personally, I also use the first three chapters of Huybrecht's "Fourier-Mukai transforms in algebraic geometry", which is a bit more condensed but has a nice overview of the situation in the algebro-geometric setting. For more advanced material you could have a look at the notes on http://therisingsea.org.</p>

Title: Derived category and so on


问:
What is the $n$th term of the sequence $1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1 ...$
So there exist a lot of similar questions which ask the $n$th term of

$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider each number as a single term. I tried making a pyramid of terms and got this

$1^2\\
1^2~2^2\\
1^2~2^2~3^2\\
1^2~2^2~3^2~4^2\\
...$

I got the $n$th term as$$T_n= (n\mod(m(m-1)/2)^2$$
Where $m$ is the row number.

But now I am stuck, is there a way to eliminate $m$ and write the general term only in terms of $n$?

答:
OEIS has an entry for this sequence. And they give a formula:

a(n) = A000290(m+1), where m = n-t(t+1)/2, t = floor((-1+sqrt(8*n-7))/2)

The sequence named A000290 is just the perfect squares, so A000290(m+1) means $(m+1)^2$. And $m$ is given as $n-\frac{t(t+1)}2$, where $t=\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor$. Nesting out all of this, the finished formula becomes
$$
\left(n-\frac{\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor\left(\left\lfloor\frac{\sqrt{8n-7}-1}2\right\rfloor+1\right)}{2}+1\right)^2
$$
where the brackets indicate the floor function.


问:
Determining the conjugacy class of the non-abelian group of order $pq$
I'm trying to find the conjugacy classes of non-abelian group of order $pq$ $(p>q)$.
It is known that this group is uniquely expressed as

$$\begin{align*}
G&=F_{p,q}\\
&=\langle a,b:a^p=b^q=1,b^{-1}ab=a^u\rangle,
\end{align*} $$

where $u$ has order $q$ modulo $p$ in $\mathbb{Z}^*_p$.

Here, the choice of $u$ doesn't matter. I have a question in finding the conjugacy class of $b^n$ specifically. I was convinced that it has the conjugacy class of size $p$. Then the book I'm reading states the following argument, which I'm not sure of.

On the other hand, as $G/\langle a\rangle$ is abelian, every conjugate of $b^n$ has the form $a^mb^n$ for some $m$. Hence $(b^n)^G=\{a^mb^n:0\leq
m\leq p-1\}$.

Here $(b^n)^G$ is just the conjugacy class of $b^n$. Could anyone give some detailed explanation of this? Thanks!

答:
$\newcommand{\Span}[1]{\left\langle #1 \right\rangle}$Take $0 < n < q$. Note first of all that $b^{n}$ is not in the centre of $G$, as it does not commute with $a$. Hence its centraliser is $\Span{b^{n}}$, of order $q$, so that its conjugacy class has indeed order $p$, as you correctly stated.

Now consider, as suggested, $K = \Span{a}$, and the factor group $G / K$, which has prime order $q$, and is thus abelian. If you take any $x \in G$, you have then in $G / K$
$$
b^{n} K
=
(x K)^{-1} (b^{n} K) (x K)
=
x^{-1} b^{n} x K.
$$
This shows that
$$
x^{-1} b^{n} x = b^{n} a^{i}
$$
for some $i$. There are $p$ possibilities for $a^{i}$. Since you already know that $b^{n}$ has $p$ conjugates, they must be exactly the $b^{n} a^{i}$, for $0 \le i < p$.


问:
What type of equation is this called?
I'm trying to solve it. Whatever it is, it's been out of my (very basic) understanding level thus far.  Hahaha I have sheets and sheets of paper trying to work it out. I dunno if it helps, but it's an optimization problem, I'm trying to work out for a piece of software I'm writing. The 'original' equation looks like this...

$\frac{2y}{2 + x} = n$

And I want to see if I'm able to feed in any particular number to N, and solve for the x and y. Like this...

$\frac{2y}{2 + x} = 4148$

At first I tried solving for X and Y like this.(I don't know how to show the steps, so I'm just going to show the 'results'.

$y = 2074 + 1037x$

$x = \frac{8296}{y -1}$

Then I thought I could replace one side with the other.

$x = \frac{8296}{2073 + 1037x}$

But I think that led down the road to madness. And I came here straight away. I think I'm okay. But could someone perhaps shine a bit of light upon my situation?

What I mean by this...is I keep getting different answers, which makes me think I'm really doing it wrong. Also, and maybe this doesn't matter, but x and y are a constant product formula themselves. So, x * y will always equal k(the invariant). I super appreciate any thoughts! Thanks =)

答:
There are many $(x,y)$ pairs that will solve your equation:

$x=0, y=4148$

$x=1, y=6222$

$x=2, y=8296$

In fact, there is a matching $y$ for any real number $x$ as long as $x \ne  -2$.

Title: What type of equation is this called?


答:
The number of terms in $x$ rows is $1+2+...+x=x(x+1)/2$. The row number $m$ of $n^{\text{th}}$ term is thus the smallest positive integer solution of $x(x+1)/2\ge n$. The roots of $x^2+x-2n=0$ are $\frac{-1\pm\sqrt{8n+1}}2$. First, we reject the negative root. Next, we take the ceiling value of the positive root as it may not be a whole number.

Thus,$$m=\left\lceil\frac{-1+\sqrt{8n+1}}2\right\rceil$$

As a side note, you can ditch the mod and use regular subtraction in your formula.

Title: What is the $n$th term of the sequence $1, 1, 4, 1, 4, 9, 1, 4, 9, 16, 1 ...$

Body:
So there exist a lot of similar questions which ask the $n$th term of

$$1^2 , (1^2 + 2^2) , (1^2 + 2^2 + 3^2) ... $$
Which have a simple answer as
$$T_n = n^3/3 + n^2/2 + n/6$$
But I want to know what will be the $n$th term if we consider each number as a single term. I tried making a pyramid of terms and got this

$1^2\\
1^2~2^2\\
1^2~2^2~3^2\\
1^2~2^2~3^2~4^2\\
...$

I got the $n$th term as$$T_n= (n\mod(m(m-1)/2)^2$$
Where $m$ is the row number.

But now I am stuck, is there a way to eliminate $m$ and write the general term only in terms of $n$?


问:
Dynamic system show that it has only one equilibrium point
I have the following dynamic system $A\cdot\ddot{x}+B\cdot\dot{x}+ C\cdot x=0 $ and A,B,C are positive definite matrices.
I have to show that it has only one equilibrium point and that this point is asymptotically stable. Any help will be appreciated!

UPDATE
Following the comment about treating A,B,C as scalars
$\dot{x_1}=x_2 = f_1(x_2)\\
\dot{x_2}=(-B\cdot x_2-C\cdot x_1)/A = f_2(x_1,x_2)$

$\longrightarrow \dot{x}=A(x)|_{x=0} \cdot x \\ $

$ A(x)=\begin{bmatrix}
\frac{\partial f_1}{\partial x_1} & \frac{\partial f_1}{\partial x_2} \\
\frac{\partial f_2}{\partial x_1} & \frac{\partial f_2}{\partial x_2} \\
\end{bmatrix}=\begin{bmatrix}
0 & 1\\
-C/A&-B/A\\
\end{bmatrix}$
$det(A-\lambda\cdot I)= 0 \longrightarrow det(\begin{bmatrix}
-\lambda & 1\\
-C/A& -\lambda -B/A\\
\end{bmatrix}) =0 \longrightarrow \lambda^2+\lambda B/A+C/A=0$\

But I don't seem to get any idea on how to generalize it.

答:
At an equilibrium point you get $\dot x=0$ and $\ddot x=0$, so the remaining condition is $Cx=0$, now apply that $C$ is s.p.d.

For the normalization use $\tilde B=A^{-1/2}BA^{-1/2}$ and $\tilde C=A^{-1/2}CA^{-1/2}$ so that with $y=A^{1/2}x$ you get a system
$$
\ddot y+\tilde B\dot y+\tilde Cy=0
$$
with still s.p.d. coefficient matrices. Then try to conclude that any exponential basis solution $ve^{\lambda t}$ needs to have $\lambda$ real and negative, or possibly complex with negative real part.


问:
Rank of Laplacian matrix
Let $ L $ be a Laplacian matrix of a balanced and strongly connected digraph having $n $ nodes.
$ L[r]$ is a submatrix of $L$ which is obtained  by deleting $rth$ row and $rth$ column of Laplacian matrix $L$. It is observed for any $r$ the rank of $ L[r]$ is full$(n-1)$. How can I prove this fact?
 Thanks in advance for your suggestions.</p>

答:
The proof can start with the acknowledgment that the rank of the incidence matrix for a graph G with a number of connected components c(G) is n-c(G), where n is the number of vertices in that graph.

Then, we can show that the rank of the Laplacian Matrix L is equal to the rank of the Incidence Matrix M for G, i.e.$rank(L)=rank(M)$.

It is because: $L = MM^{T}$ and for a given vector v, if  $Lv=MM^{T}v=0$, we have $v^{T}MM^{T}v=0$, i.e. $||M^{T}v||=0$, hence $M^{T}v=0$;  On the other hand, if $M^{T}v=0$, we have $MM^{T}v$ for sure. That is to say, $L$ and $M^T$ has same Kernel, given that they share the same number of columns, i.e. both equal to n, we know $rank(L)=rank(M)$ by Rank–nullity theorem

Title: Rank of Laplacian matrix


问:
What is the relation between $\operatorname{tr}(\sigma \log(\sigma))$ and its projection?
Let $\sigma \in \mathbb R^{n\times n}$ be a positive semi-definite matrix ($n$ is an even number), and let the projections
\begin{equation*}
V_1 =
\begin{pmatrix}
I_{\frac{n}{2},\frac{n}{2}} & 0_{\frac{n}{2},\frac{n}{2}} \\
\end{pmatrix},
\end{equation*}
\begin{equation*}
V_2 =
\begin{pmatrix}
0_{\frac{n}{2},\frac{n}{2}} &I_{\frac{n}{2},\frac{n}{2}}
\end{pmatrix}
\end{equation*}

where $I_{n,n}$ and $0_{n,n}$ are the identity matrix and the zero matrix of size $n\times n$, respectively.

Now what is the relation between the two quantities:

$\operatorname{tr}(\sigma \log(\sigma))$, and

$\operatorname{tr}(V_1 \sigma V_1^T \log(V_1 \sigma V_1^T)) + \operatorname{tr}(V_2 \sigma V_2^T \log(V_2 \sigma V_2^T)) $

Is there any generalization for the relation bound when the projection involves more than $V_1$ and $V_2$?

The $\log$ here is the matrix logarithm.

答:
$\def\bb{\mathbb}$
Instead of the two partitions in your example, let's generalize to $m$ partitions.

Use the Kronecker product $(\otimes)$ to define matrix analogs of the standard base vectors $\,e_i\in{\bb R}^{m}$
$$\eqalign{
&E_1 &= e_1\otimes I_k,\quad &E_2 &= e_2\otimes I_k,
\quad\ldots\quad &E_m &= e_m\otimes I_k \\
}$$
where $k=\left(\frac nm\right),\,$ assuming that $n$ is divisible by $m$.

Use the $\{E_i\}$ matrices to extract $k\times k$ blocks along the diagonal of $\sigma$
$$\sigma_1 = E_1^T\sigma E_1,\quad \sigma_2 = E_2^T\sigma E_2,\quad\ldots\quad \sigma_m = E_m^T\sigma E_m$$
This gives rise to the generalized functions
$$\eqalign{
F_m = \sum_{i=1}^m\,{\rm Tr}\Big(\sigma_i\log(\sigma_i)\Big) \\
}$$
Using this notation, your first function is $F_{\tt1},\;$ your second is $F_{\tt2},\,$ etc.

and Rammus's answer has demonstrated that
$$F_{\tt1}\ge F_{\tt2}\ge F_{\tt3}\ge \ldots$$

Another approach avoids the summation by using
the all-ones matrix $J_k\in{\bb R}^{k\times k}$,
the identity matrix $I_m\in{\bb R}^{m\times m}$
and the Kronecker/Hadamard $(\otimes/\odot)$
products to create a block-diagonal matrix
$$\eqalign{
H_{m} &= I_m\otimes J_k \,\in\, {\bb R}^{n\times n} \\
B_{m} &= H_m\odot \sigma \\
}$$
Now the generalized functions can be written as
$$F_m = {\rm Tr}\Big(B_m\;\log(B_m)\Big)$$
The advantage of this form is that it only requires a single evaluation of the log function, and the gradient is easy to calculate
$$\frac{\partial F_m}{\partial \sigma}
 = I_n + H_m\odot\log(H_m\odot \sigma)$$


问:
Precompactness of subset of function space with pointwise convergence topology
I need help with proving following statement. Let $M[a, b]$ - space of real (complex) value functions with topology of pointwise convergence.  Then $$A \subseteq  M[a, b] \,\text{ is precompact } \iff  \exists g \in M[a,b], |f(t)| \leq g(t)  ,\forall t \ \in \ [a,b], \ f \in \ A$$
I know, that if functions are continuous equicontinuity and uniform boundness are needed for precompactness, but don't really know what to do in this case.

答:
Assume that $\exists g \in M[a,b]$ such that $|f| \le g$ for all $f \in A$. For $x \in [a,b]$ let $\pi_x : [a,b] \to \Bbb{C}$ be the evaluation $f \mapsto f(x)$.

For every $x \in [a,b]$ we have that $\pi_x(A)$ is contained in the closed ball $\overline{K}(0,g(x))$ of radius $g(x)$ around the origin in $\Bbb{C}$.

Taking the closure yields $\overline{\pi_x(A)} \subseteq \overline{K}(0,g(x))$ and hence $\overline{\pi_x(A)}$ is compact in $\Bbb{C}$. Now we have
$$\overline{A} \subseteq \prod_{x \in [a,b]} \pi_x(\overline{A}) \subseteq  \prod_{x \in [a,b]} \overline{\pi_x(A)}$$
and the latter is a product of compact sets and hence compact. Therefore $\overline{A}$ is also compact.

Conversely, assume that $A$ is precompact. Then for every $x \in X$ since $\pi_x$ is continuous, we have that $\pi_x(\overline{A})$ is compact in $\Bbb{C}$ as an image of a compact set. In particular, $\pi_x(\overline{A})$ is bounded and then from
$$\pi_x(A) \subseteq \pi_x(\overline{A})$$
we see that is  $\pi_x(A)$ is also bounded. Hence there exists a positive real number which we can denote by $g(x)$ such that $\pi_x(A) \subseteq \overline{K}(0,g(x))$.

Hence for any $f \in A$ and $x \in [a,b]$ we have
$$f(x) = \pi_x(f) \in \pi_x(A) \subseteq \overline{K}(0,g(x)) \implies |f(x)| \le g(x).$$


问:
Find the value of $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$
I want to know an objective approach to solve these type of expression in a quick time
Which of the expression equals to
$$\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$$

a)$1-\tan\theta-\cot\theta$

b)$1+\tan\theta-\cot\theta$

c)$1-\tan\theta+\cot\theta$

d)$1+\tan\theta+\cot\theta$

I've tried it several ways like taking LCM,change whole into $\sin\theta$ and $\cos\theta$.but  I've stuck.

答: I'm not sure where you got stuck after writing the expression in terms of $ \sin \theta $ and $ \cos \theta $ but here's how I would do it: $$ \begin{align*}\frac{\tan \theta}{1 - \cot \theta} + \frac{\cot \theta}{1 - \tan \theta} &= \frac{\sin \theta \tan \theta}{\sin \theta - \cos \theta} + \frac{\cos\theta \cot \theta}{\cos \theta - \sin \theta} \\ &= \frac{\sin\theta\tan\theta - \cos\theta \cot \theta}{\sin \theta - \cos \theta} \\&= \frac{\sin^3 \theta - \cos^3 \theta}{\sin \theta \cos \theta (\sin\theta - \cos\theta)} \\&= \frac{\sin^2 \theta + \sin \theta \cos \theta + \cos^2\theta }{\sin \theta \cos \theta}\\&= 1 + \tan \theta + \cot \theta\\&=\boxed{\text{D}} \end{align*} $$</p>


问:
If all injections from a set E onto itself $j:E ->E$ are surjective, then E is finite
For my Mathematical Logic class I need to prove that a set E is finite iff all injections $j:E ->E$ are surjective. I managed to prove the "forward" direction of the proposition, but I'm having a lot of trouble proving the "backward" implication (i.e. all injections are surjective $=>$ E finite).

We're given the following definition of finiteness of a set: A set E is finite if there exists a $n \in \Bbb{N}$ and a bijection $f:E-> \{k \in \Bbb{N} : k<n\}$

And a proposition we can simply assume: If a set E is not finite, then there exists an injection $i:\Bbb{N}->E$

I've tried to find a contradiction assuming that the proposition is true for our set E and I also tried to find a function f so that we have finiteness according to the definition but I don't get anywhere.

Thanks in advance for your help!

答:
Contraposition. If $E$ is infinite, there exists an injection $f:E\rightarrow W$ which is not surjective.

Take $f:\Bbb N_0\rightarrow\Bbb N_0: x \mapsto 2x$.


问:
Proving the existence of a unique set constructed from two given sets
Let $M$ and $N$ be sets. Show that there exists a unique set $Q$ such that $x \in Q$ if and only if either $x \in M $ and $ x \notin N $ or $ x \in N $ and $ x \notin M $.

Here is my attempt. First, we will prove the existence of $Q$. Here we will use axiom schema of specification

$$\forall w\,\forall A\, \exists B\,\forall x \big(x \in B \Leftrightarrow [x \in A \wedge \varphi(x, w)] \big) $$

Now, if we let $A = M$ and $w = N$ and $\varphi(x, w) = x \notin w = x \notin N$ then we get the following using universal instantiation

$$ \exists B\,\forall x \big(x \in B \Leftrightarrow [x \in M \wedge x \notin N] \big) $$

Now, if we let $P_1 = B$, then using existential instantiation, we get a set $P_1$ such that

$$\forall x \big(x \in P_1 \Leftrightarrow [x \in M \wedge x \notin N] \big) \cdots\cdots (1)$$

In a similar way, if we let $A = N$ and $w = M$ and $\varphi(x, w) = x \notin w = x \notin M$ then we get a set $P_2$ such that

$$\forall x \big(x \in P_2 \Leftrightarrow [x \in N \wedge x \notin M] \big) \cdots\cdots (2)$$

Now, that we have $P_1$ and $P_2$, we can use axiom of pairing

$$\forall A\, \forall B \, \exists C\,\forall D\, [ D \in C \Longleftrightarrow (D = A \vee D = B)] $$

Letting, $A = P_1$ and $B = P_2$ and then using universal instantiation, we get

$$ \exists C\,\forall D\, [ D \in C \Longleftrightarrow (D = P_1 \vee D = P_2)]  $$

and further using existential instantiation, we get a set $C$ such that

$$ \forall D\, [ D \in C \Longleftrightarrow (D = P_1 \vee D = P_2)]  $$

By caling $C = P_3$, we have the a set $P_3$ such that

$$ \forall D\, [ D \in P_3 \Longleftrightarrow (D = P_1 \vee D = P_2)]\cdots\cdots (3)  $$

Now, we will prove that $P_3$ is same set as $\{P_1, P_2 \}$. Let $x \in P_3$ be some arbitrary element. Then using eq (3), we conclude that $x = P_1 \vee x = P_2$. In either case, $x \in \{P_1, P_2 \} $. In other direction, if $x \in \{P_1, P_2 \} $, then $x = P_1 \vee x = P_2$. So, using equation (3) again, we can see that $x \in P_3$. So, invoking axiom of extensionality, we can see that $P_3 = \{P_1, P_2 \} $. So, finally, we will use axiom of union

$$\forall A\, \exists B\, \forall c\, (c \in B \Longleftrightarrow \exists D(c \in D \wedge D \in A)) $$

If we let $A = P_3$, then using universal instantiation, we get

$$ \exists B\, \forall c\, (c \in B \Longleftrightarrow \exists D(c \in D \wedge D \in P_3)) $$

Lets call $B = P_4$. So, using existential instantiation, we get a set $P_4$ such that

$$ \forall c\, (c \in P_4 \Longleftrightarrow \exists D(c \in D \wedge D \in P_3))\cdots\cdots(4) $$

Now, we will prove that $P_4 = P_1 \cup P_2$. Let $x \in P_4$ be some arbitrary element in $P_4$. Then using equation (4), we get that

$$ \exists D(x \in D \wedge D \in P_3) $$

By calling $D = Q$ and using existential instantiation, we get a  $Q$ such that $x \in Q$ and $Q \in P_3$. This means that $Q \in \{P_1, P_2 \} $. So, either $Q = P_1$ or $Q = P_2$. Which means that $x \in P_1 \vee x \in P_2$. Which shows that $x \in P_1 \cup P_2$. In other direction, if $x \in P_1 \cup P_2$. Here we have two cases. If $x \in P_1$ and since $P_1 \in \{P_1, P_2 \}$, then we have some $y$ such that $x \in y$ and $y \in \{P_1, P_2 \}$. This means that we have $y$ such that $x \in y$ and $y \in P_3$. To put in symbols,

$$ \exists y (x \in y \wedge y \in P_3) $$

Using, equation (4), this means that $x \in P_4$. The second case, where $x \in P_2$ also leads to $x \in P_4$ with similar reasoning. So, in any case, if $x \in P_1 \cup P_2$ then $x \in P_4$. This proves that $P_4 = P_1 \cup P_2$. So, using set builder notation and using equations 1 and 2, we can say that

$$ P_4 = \{ x \,| \,(x \in M \wedge x \notin N) \vee (x \in N \wedge x \notin M) \} $$

So here we have proven the existence of set $Q = P_4$. Next stage is to prove the uniqueness of the set $Q$. So assume that there two distinct sets $Q_1$ and $Q_2$ satisfying equation 4.

$$ \forall c\, (c \in Q_1 \Longleftrightarrow \exists D(c \in D \wedge D \in P_3))\cdots\cdots(5) $$

$$ \forall c\, (c \in Q_2 \Longleftrightarrow \exists D(c \in D \wedge D \in P_3))\cdots\cdots(6) $$

Now , we let some arbitrary $x \in Q_1$. Equation 5 leads us to

$$ \exists D(x \in D \wedge D \in P_3) $$

And using equation 6, we can conclude that $ x \in Q_2$. With similar reasoning, we can go in opposite direction and prove that $x \in Q_2$ leads to $x \in Q_1$. Invoking axiom of extensionality this proves $Q_1 = Q_2$ contradicting initial assumption that $Q_1 \ne Q_2$. So, we must have $Q_1 = Q_2$. This proves the uniqueness part. So, we proved both existence and uniqueness parts here.

I am new to learning set theory and wanted to check my proof.

答:
If you use the general form of axiom schema of specification, i.e.
$$\forall w_1 \forall w_2\,\forall A\, \exists B\,\forall x \big(x \in B \Leftrightarrow [x \in A \wedge \varphi(x, w_1,w_2)] \big) $$

then you can just chose

$$\varphi(x, w_1,w_2) \equiv (x \in w_1 \wedge \neg(x \in w_2)) \vee (\neg (x \in w_1) \wedge x \in w_2)$$ with $w_1 = M$, $w_2 = N$ and $A = M \cup N$.


问:
Weak topology = topology of pointwise convergence?
In a book, I found the following theorem (slightly simplified here):

Consider $\mathbb{R}$ with the Borel $\sigma$-field.

Let $B$ be the set of all measurable, bounded, real-valued functions on $\mathbb{R}$.

Let $M$ be the set of all finite, signed measures on $\mathbb{R}$.

Let the weak topology on $B$ be defined to be the coarsest topology such that all functions

$B \to \mathbb{R},\quad f \mapsto \int f \mathrm{d} \mu,\quad \mu \in M$,

are continuous.

Theorem: On any $\| \cdot \|_\infty$-bounded subset of $B$, the weak topology and the topology of pointwise convergence coincide.

Proof. Since $M$ contains all one-point probability measures, convergence in the weak topology implies pointwise convergence. Conversely, since sets bounded in $\| \cdot \|_\infty$-norm are also pointwise bounded, we can deduce from Lebesgue's dominated convergence theorem that pointwise convergence implies convergence in the weak topology. $\square$

Now, I do not understand why having the same convergent sequences in this case should imply that the topologies are the same. One direction seems clear: The topology of pointwise convergence is coarser than the weak topology, since the topology of pointwise convergence is the initial topology w.r.t. the family of all functionals $f \mapsto \int f \mathrm{d} \delta_x$.

Yet I have not been able to get rid of the sequences argument in the other direction, and, thinking about how open sets in $B$ look like in the two topologies, I am not even sure anymore that the theorem is correct.

Questions:

(a) Is the theorem correct?
(b) Is the proof correct, and, if yes, why?

答:
Neither the theorem nor its proof are correct. You can use that the dominated convergence theorem does not apply to nets. To quote t.b.'s answer here:

Here's a cautionary example on the unit interval which you can modify to concoct all sorts of counterexamples:

Let $\Lambda$ be the set of finite subsets of $[0,1]$ ordered by inclusion. For each $\lambda \in \Lambda$ choose a continuous function $f_{\lambda}: [0,1] \to [0,1]$ such that $f_{\lambda}(x) = 1$ for all $x \in \lambda$ and $\int f_{\lambda} \leq \frac{1}{2}$. Then $f_{\lambda} \to 1$ everywhere on $[0,1]$, while $\frac{1}{2} \leq \int (1 - f_{\lambda}) \nrightarrow 0$…

By a slight modification, we get a net $\{f_\lambda\}_{\lambda \in \Lambda}$ in the unit ball of your space $B$ that converges pointwise to $\chi_{[0,1]}$, but with $\int f_\lambda \, d\mu_0 \leq \tfrac{1}{2}$ for all $\lambda \in \Lambda$, where $\mu_0 \in M$ denotes Lebesgue measure on $[0,1]$. (The only modification we need for this, is adding the requirement that $f_\lambda$ is zero outside the interval $[-\tfrac{1}{|\lambda|} , 1 + \tfrac{1}{|\lambda|}]$, in order to guarantee that $\{f_\lambda\}_{\lambda\in\Lambda}$ converges to $0$ outside the interval $[0,1]$.)

To complete the proof, note that the net $\{f_\lambda\}_{\lambda\in\Lambda}$ converges pointwise to $\chi_{[0,1]}$, but not weakly, since
$$ \lim_{\lambda \in \Lambda} \int f_\lambda \: d\mu_0 \neq \int \chi_{[0,1]} \: d\mu_0. $$


答:
$$
\begin{align}
\frac{\tan(\theta)}{1-\cot(\theta)}+\frac{\cot(\theta)}{1-\tan(\theta)}
&=\frac{\cot(\theta)-\tan^2{\theta}}{1-\tan(\theta)}\\
&=\cot(\theta)\frac{1-\tan^3(\theta)}{1-\tan(\theta)}\\
&=\cot(\theta)+1+\tan(\theta)
\end{align}
$$</p>

Title: Find the value of $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$

Body:
I want to know an objective approach to solve these type of expression in a quick time
Which of the expression equals to
$$\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$$

a)$1-\tan\theta-\cot\theta$

b)$1+\tan\theta-\cot\theta$

c)$1-\tan\theta+\cot\theta$

d)$1+\tan\theta+\cot\theta$

I've tried it several ways like taking LCM,change whole into $\sin\theta$ and $\cos\theta$.but  I've stuck.


问:
$\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}$ equals to $1$ or $e^2$?
$$\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}$$ equals to $1$ or $e^2$?

My Attempt

\begin{align}
\lim_{x\to\infty} (2x+1)^{x-\sqrt{x^2-2}}
    &= e^{ \displaystyle \lim_{x\to\infty} 2x(x-\sqrt{x^2-2})}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{2x(x^2-(x^2-2))}{x+\sqrt{x^2-2}}}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{4x}{x+\sqrt{x^2-2}}}\\
    &= e^{ \displaystyle \lim_{x\to\infty} \frac{4}{1+\sqrt{1-\frac{2}{x^2}}}}\\
    &=e^2
\end{align}

However, Mathematica Alpha gives a different result which is $1$. Which is the correct one?
If the Mathematica version is correct, how is its answer derived?

Edit

The rule that I am using here is

$$
\lim_{x\to a} (1+f(x))^{g(x)} = e^{\lim_{x\to a} f(x)g(x)}
$$

I am not sure, the constraint for this rule is

both $f(x)$ and $g(x)$ must approach zero or

both $f(x)$ and $g(x)$ must approach infinity

答:
Here is a possible way using that $\lim_{t\to+\infty}t^{\frac 1t}=1$.

Rewrite your expression as follows:
\begin{eqnarray*} (2x+1)^{x-\sqrt{x^2-2}}
& = & \left((2x+1)^{\frac{1}{2x+1}}\right)^{(2x+1)(x-\sqrt{x^2-2})}\\
& = & \left(\underbrace{(2x+1)^{\frac{1}{2x+1}}}_{\stackrel{x\to+\infty}{\longrightarrow}1}\right)^{\underbrace{\frac{2(2x+1)}{x+\sqrt{x^2-2}}}_{\stackrel{x\to+\infty}{\longrightarrow}2}}\\
& \stackrel{x\to +\infty}{\longrightarrow} & 1^2 = 1
\end{eqnarray*}


问:
Finding the derivative- is my answer correct? Is the method correct?
Question - find the derivative of $y$ with respect to $z$
$$y=  2^{z^3} $$

So what I’d do is
Differentiate boths sides of the equation so
$$\frac{dy}{dz}= \frac{d(2^{z^3 })}{dz}$$
Then I applied the chain rule by assuming,
$$ u=z^3$$

which resulted in
$$ 2^{z^3}\ln2\frac{dz^3}{dz}$$

Using the power rule I finally got the answer to be
$$ 3z^2\cdot2^{z^3}\ln2$$

Is my answer correct?

答:
The chain rule in this case gives: $$\frac{dy}{dz}=\frac{dy}{du}\cdot\frac{du}{dz}$$

You set $u=z^3$, so $y=2^u$, then $\frac{du}{dz}=3z^2$ and $\frac{dy}{du}=2^u\cdot \ln (2)$

Combining the two and substituting out $u$ we get $$\frac{du}{dz}\cdot\frac{dy}{du}=\frac{dy}{dz}=3\ln (2) \cdot 2^{z^3}\cdot z^2$$
which you got.


问:
Is distributivity of a lattice needed if we want its collection of prime filters to be a Stone space?
Let $L$ be a bounded distributive lattice and let $PF(L)$ denote its set of prime filters.

It is well known that $PF(L)$ is a Stone space if it is equipped with the topology that has sets of the form $\phi(a):=\{P\in PF(L)|a\in P\}$ and their complements as subbase.

A proof that I encountered looked quite complicated especially on the part where compactness of $PF(L)$ is shown. For instance there it was used that for disjoint filter $F$ and ideal $I$ there always exist a prime filter containing $F$ and disjoint with $I$.

I found an alternative proof of my own but do not trust myself. This also because distributivity is not used in it. Please be so kind to check me on errors.

At first I observe that compact Hausdorff space $2^{L}$ equipped with product topology is a Stone space, and consequently every closed (hence compact) subset of it is a Stone space. For completeness let me say that here $2=\{0,1\}$ is equipped with discrete topology.

Then I identify $PF(L)$ with the set of indicator sets of prime filters in lattice $L$. Actually you can think of them as the lattice-morphisms $L\to2$. Then $PF(L)$ is a subset of $2^L$ and it is enough now to prove that it is closed.

Then I observe that function $f$ is an element of $PF(L)$ iff it satisfies:

$f(0)=0$ and $f(1)=1$

$f(a\wedge b)=\min\{f(a),f(b)\}$ for every $a,b\in L$

$f(a\vee b)=\max\{f(a),f(b)\}$ for every $a,b\in L$

But it seems to me that for every mentioned condition we can find a closed set in $2^L$ such that demanding that $f$ satisfies the condition is the same thing as demanding that $f$ is an element of that closed set.

This becomes more clear if we rewrite e.g. the second condition by means of (continuous) projections as:

$|\pi_{a\wedge b}(f)-\min\{\pi_a(f),\pi_b(f)\}|=0$

This shows that $f$ must be an element of the preimage of closed set $\{0\}$ with respect to a continuous function.

If that is correct then $PF(L)$ is the intersection of closed sets hence is closed itself (and we are ready).

Any mistake in this?

I am curious.

答: Yes, your proof is correct.  To explain why anyone would ever use the more complicated proof that you encountered, let me point out two things.  First, your proof relies on Tychonoff's theorem to conclude that $2^L$ is compact, and so is essentially hiding much of the complexity of the argument inside the proof of Tychonoff's theorem.  Second, the more complicated proof actually shows something stronger: it can also be used to show that the canonical map $L\to 2^{PF(L)}$ is a lattice embedding.  This stronger result really does require distributivity.  Indeed, since $2^{PF(L)}$ is distributive, any lattice that can embed in it must be distributive as well.


问:
Example of a matrix satisfying $\rho(A) < \|A\|$

Give an example of a matrix $A \in \mathbf{C}^{n \times n}$ such that for all operator norms $\| \cdot \|$, one has $\rho(A) < \|A\|$.

$$\rho(A) := \max \left\{\left|\lambda_{1}\right|, \ldots,\left|\lambda_{n}\right|\right\}$$

where $\lambda_1, \dots , \lambda_n$ are the eigenvalues of $A$. I thought about it for a long time, but unfortunately I could not find an example.

答: You can take, say, $A=\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]$. Then $\rho(A)=1$, but $A.(0,1)=(1,1)$ and therefore $\|A\|\geqslant\sqrt2>\rho(A)$.


问:
Prove the sequence $a_n=\sum_{k=1}^n \frac {\sin(k)}{k^2}$ converges
I'm stuck at this question.

I tried proving that the sequence converges through proving the sequence is Cauchy but I'm stuck there too.

My try : Let $\epsilon>0$ and $n\ge m$.

$$|a_n-a_m|=\left|\sum_{k=1}^n\frac{\sin(k)}{k^2}-\sum_{k=1}^m\frac{\sin(k)}{k^2}\right|\\
=\left|\sum_{k=m+1}^{n}\frac{\sin(k)}{k^2}\right|$$

And exactly there I'm stuck.

How to prove it ?
Thanks.

答: Note that$$\left|\sum_{k=m+1}^n\frac{\sin(k)}{k^2}\right|\leqslant\sum_{k=m+1}^n\frac1{k^2}$$and that the sequence $\left(\sum_{k=1}^n\frac1{k^2}\right)_{n\in\Bbb N}$ is a Cauchy sequence, since the series $\sum_{k=1}^\infty\frac1{k^2}$ converges.


问:
Coercive bilinear form on Hilbert space
I need to show the two following results. If true, it must be a simple proof but I do not seem to be able to make it work. Thank you in advance.

Consider a continuous symmetric bilinear form $B$ on a real Hilbert space $H$. Let $D$ be a closed subspace of that Hilbert space over which the form is coercive, i.e, there exists $\alpha >0 \in \mathbb{R}$ such that $B(d,d)\geq \alpha \|d\|^2$ for all $d\in D$. Take the new closed subspace of $H$ generated by $D$ and a one dimensional subspace. Assume $B$ is definite positive on the new subspace. Then I need to show that $B$ is also coercive on the new subspace.

It seems like the logical thing to do is to use the Lax-Milgram theorem but I am not exactly sure how to do this.

答:
Here is a proof from scratch, in particular, without the Lax-Milgram.

Lemma. Let $\widetilde H$ be a real vector space with positive definite inner product (completeness not assumed). Suppose that $K$ is a subspace of $\widetilde H$ which is complete. Then any element $v\in \widetilde H$ can be written as $v_1+v_2$ where $v_1\in K$ and $v_2\in K^\perp$.

Proof. Let $v_1$ be the nearest element of $K$ to $v$; the standard proof of the existence of $v_1$ in a Hilbert space applies here as well, because it uses only the completeness of $K$ (e.g., Theorem I.2.5 in A Course in Functional Analysis by Conway). Let $v_2=v-v_1$. If there is $u\in K$ such that $\langle u,v_2\rangle>0$, then
$$\|v-(v_1+tu)\|^2=\|v_2- tu\|^2 = \|v_2\|^2 -2t\langle u,v_2\rangle +O(t^2)<\|v_2\|^2$$ for small $t>0$, contradicting the fact that $v_1$ is the nearest element of $K$ to $v$. $\Box$

Apply the lemma with $K=D$ and  $\widetilde H$ being the space generated by $D$ and another vector; the inner product is given by $B$. (Note that $D$ is complete with respect to $B$-inner product, because of coercitivity). The lemma provides a nonzero vector $v\in \widetilde H$ such that $v$ is $B$-orthogonal to $D$. Every element of $\widetilde H$ can be written as $d+tv$ for some $d\in D$ and $t\in \mathbb R$. We have
$$ B(d+tv,d+tv) = B(d,d)+t^2B(v,v) \tag1$$ and $$\|d+tv\|^2 \le 2 \|d\|^2 +2t^2\|v\|^2\tag2$$
Let $\gamma$ be the smallest of the numbers $\alpha/2$ and $\frac12 \|v\|^2/ B(v,v)$. Then
$$ B(d+tv,d+tv) \ge 2\gamma\|d\|^2 + 2 \gamma t^2\|v\|^2\ge \gamma \|d+tv\|^2 \tag3$$
as required. $\Box$

Title: Coercive bilinear form on Hilbert space


问:
bolt manufacturing normal distribution problem
The diameter of a bolt in a manufacturing process distributes $N(μ,σ^2)$, a bolt is considered fit if it's diameter's deviation from the mean is less than 2.
What's the probability that a single bolt is fit?

I still don't have a very good grasp on normal distribution and am trying to learn through this excercise, would love to get some help.

thank you.

答:
Hint: Denote the diameter with $D$. Then a bolt is fit iff $|D-\mu|\le2$. So$$P(|D-\mu|\le2)=P(|Z|\le~?)$$where $Z$ is the standard normal distribution $N(0,1)$. Can you complete?

As mentioned in the comments, since you seem to be expecting a numerical answer, I suspect by "deviation from the mean", your question could probably be meaning the quantity $\left|\frac{D-\mu}\sigma\right|$ instead of $|D-\mu|$.

In that case, we have $P\left(\left|\frac{D-\mu}\sigma\right|\le2\right)=P(|Z|\le2)=2P(0<Z<2)$. Now plug in the value of $P(0<Z<2)$ from the table. It is $0.4772$ so the answer is twice of that, i.e. $2\cdot0.4772=0.9544$.


问:
Why doesn't every random variable have null density function?
Can someone help me find the error in the following reasoning?

Let $Z$ be an absolutely continuous random variable.  I've been told that absolutely continuous random variable has $0$ probability in a single point $\mathbb{P}(Z=z)=0 ,\forall z \in \mathbb{R}$

but by definition of density $f_Z({z})=\mathbb{P}(Z=z)$

Hence $ f_Z({z})=0$, that is any absolutely continuous random variable has null density function? I know the result is absurd, I just can't find the error in the reasoning

答:
For a discrete random variable $Z$, the probability mass function $p_Z(z) = P(Z=z)$ is a complete description of the distribution. For a continuous random variable $Z$, the probabilities of single points no longer define the distribution. Instead one has to consider the probabilities of intervals. Usually one defines the cumulative probability function
$$
F_Z(z) = P(Z\le z).
$$
Then, the probability density function is defined as the derivative of $F_Z$, that is $f_Z(z) = \frac{\mathrm d F_Z}{\mathrm dz}(z)$.

This then allows to compute probabilites of intervals as
$$
P(a < Z \le b) = P(Z\le b) - P(Z\le a) = F_Z(b) - F_Z(a) = \int_a^b f_Z(z) \mathrm\,dz.
$$
Note that this is an integral over density values, not a sum of probabilities!

While probabilities are always $\le 1$, we can very well have densities $>1$. However, the area under the graph of $f_Z$, which is the integral $\int_{-\infty}^\infty f_Z(z)\,\mathrm dz$ is always equal to $1$.


问:
$M$ is compact and oriented/orientable if and only if top bundle trivial?
Is it true that for a smooth $m$-manifold $M$ and for $\bigwedge^m M := \bigwedge^m(T^{dual}M)$ (which I understand to be its 'top bundle')

$M$ is compact and oriented/orientable if and only if $\bigwedge^m M$ is trivial?

Well, for any smooth $m$-manifold $M$, I think $(\bigwedge^m M)_p \cong \bigwedge^m(T^{dual}_pM \cong \mathbb R^m) \cong \mathbb R^{\binom{m}{m}} = \mathbb R$, so I guess $\bigwedge^m M \cong M \times \mathbb R$, so I think specifically we have $\bigwedge^m M$ not only a trivial bundle (defined: isomorphic to $M \times \mathbb R^n$, for some $n$) but like, a trivial line bundle (defined: isomorphic to $M \times \mathbb R$). I don't quite see what compact or oriented/orientable has to do with this.

Also: I think $M$ is indeed manifold without boundary, so not sure if Stokes' theorem is relevant. I think what might be relevant is that

A top smooth form on a compact smooth $m$-manifold (smooth $m$-form on compact smooth $m$-manifold) which is never zero is never exact.

Smooth $m$-manifold $M$ is orientable if and only if $M$ has a smooth top form that is never zero.

Top deRham cohomology group of a compact orientable manifold is 1-dimensional

The relationship between de rham cohomology and exterior algebra, which I've forgotten. But I think it's to do with that forms are sections of wedge bundles

Maybe something to do with Poincaré duality and that compact de rham cohomology = regular de rham cohomology for compact manifolds or something

答:
If $M$ is a manifold of dimension $n$, then $\Lambda^n(T^*M)$ is a rank $\binom{n}{n}=1$ vector bundle over $M$, that is, a line bundle. But one does not have, in general, $\Lambda^n(T^*M) \simeq M \times \mathbb{R}$. This would imply that it is a trivial line bundle. On a given manifold, there could exist many non-trivial line bundles. For example, the Mobius band can be viewed as a non-trivial line bundle over the cirlce $\mathbb{S}^1$.

The answer to your question will depend on the definition of orientability you are using. One can say that $M$ is orientable if it has an orientation atlas, that is an atlas where change of charts have positive jacobian.

What is true, is that, with this definition of orientability $M$ is orientable if and only if $\Lambda^n(T^*M)$ is a trivial line bundle. The compactness has nothing to do with this.

The fact that $\Lambda^n (T^*M)$ is a trivial line bundle is equivalent to the fact that it has a non-vanishing section. Thus, $M$ is orientable if and only if there exists a nowhere-vanishing differential $n$-form, that is a volume form.

Moreover, Stokes theorem is a theorem about integration on manifolds, which is defined related to an orientation... So it seems irrelevant to apply Stokes theorem to a manifold to see if it is orientable (maybe I misunderstood what you were talking about while considering Stokes theorem).


答: As usual, the canonical example is $$A=\begin{bmatrix}0&1\\0&0\end{bmatrix}.$$ You have $\rho(A)=0$, so $\rho(A)<\|A\|$ for any norm.

Title: Example of a matrix satisfying $\rho(A) < \|A\|$

Body:
Give an example of a matrix $A \in \mathbf{C}^{n \times n}$ such that for all operator norms $\| \cdot \|$, one has $\rho(A) < \|A\|$.

$$\rho(A) := \max \left\{\left|\lambda_{1}\right|, \ldots,\left|\lambda_{n}\right|\right\}$$

where $\lambda_1, \dots , \lambda_n$ are the eigenvalues of $A$. I thought about it for a long time, but unfortunately I could not find an example.


问:
Is it a thereom or an axiom that two planes intersect only in lines?
I know that two planes always intersect in a line or are parallel to each other, and there's an question about two planes in a point: Can two planes intersect in a point?. But how can you prove that two planes can't intersect in any other shape, like a parabola, a circle, a rectangle or basically any function other than an line? Or is it an axiom that we can't prove?</p>

答:
As said in the comments there is a proof by analytic geometry:
Any plane is an solution to a linear equation, as such:
$$ax+by+cz=d$$
$$ex+fy+gz=h$$

Any line which lies in both planes thus must also be linear, so that's why it's a line.

However, I do not think it is possible to prove that in only Euclidean geometry,


问:
How do I find the subspace of $P_{2}$ from the following subsets?
One of the following is a subspace of $P_{2}$.

A. $W= \{ax^2+bx+c\mid a+b+c=1\}$

B. $W= \{ax^2+bx+c\mid a+b+c=0\}$

C. $W= \{ax^2+bx+c\mid a \geq 0\}$

D. $W= \{ax^2+bx+c \mid b,c \geq 0\}$

So it's either $B, C$, or $D$ because the zero vector condition won't work on $A$. What I don't understand is the difference between $C$ and $D$. How does having 'a' greater than or equal zero differ from having '$b$' and '$c$' greater than or equal zero? Can someone please explain how it works? This concept is a bit new to me and I couldn't find any examples of this particular case online.

答:
How does having 'a' greater than or equal zero differ from having 'b' and 'c' greater than or equal zero? Can someone please explain how it works?

So, for example, consider the polynomials
$$
p_1(x) = 2x^2 + 3x + 4,\\
p_2(x) = -3x^2 + 4x + 5,\\
p_3(x) = 0x^2 - 5x + 6,\\
p_4(x) = -5x^2 + 6x - 7.
$$
The polynomials $p_1$ and $p_3$ are elements of the set described in C because the $x^2$ coefficient (the "$a$" coefficient) is non-negative. The polynomials $p_1$ and $p_2$ are in the set described in D because the other coefficients (the "$b$" and "$c$" coefficients) are non-negative. $p_4$ is in neither set because it is not true that $a \geq 0$, and it is also not true that we have both $b \geq 0$ and $c \geq 0$.


问:
What is the relation between $\operatorname{tr}(\sigma \log(\sigma))$ and its projection?
Let $\sigma \in \mathbb R^{n\times n}$ be a positive semi-definite matrix ($n$ is an even number), and let the projections
\begin{equation*}
V_1 =
\begin{pmatrix}
I_{\frac{n}{2},\frac{n}{2}} & 0_{\frac{n}{2},\frac{n}{2}} \\
\end{pmatrix},
\end{equation*}
\begin{equation*}
V_2 =
\begin{pmatrix}
0_{\frac{n}{2},\frac{n}{2}} &I_{\frac{n}{2},\frac{n}{2}}
\end{pmatrix}
\end{equation*}

where $I_{n,n}$ and $0_{n,n}$ are the identity matrix and the zero matrix of size $n\times n$, respectively.

Now what is the relation between the two quantities:

$\operatorname{tr}(\sigma \log(\sigma))$, and

$\operatorname{tr}(V_1 \sigma V_1^T \log(V_1 \sigma V_1^T)) + \operatorname{tr}(V_2 \sigma V_2^T \log(V_2 \sigma V_2^T)) $

Is there any generalization for the relation bound when the projection involves more than $V_1$ and $V_2$?

The $\log$ here is the matrix logarithm.

答:
The matrix functional you are interested in can be expressed in terms of the Umegaki divergence
$$
D(\rho \| \sigma ) = \frac{\mathrm{Tr}[\rho (\log(\rho) - \log(\sigma))]}{\mathrm{Tr}[\rho]}.
$$
See Equation 4.61 for the full definition. We have $\mathrm{Tr}[\sigma \log \sigma] = \mathrm{Tr}[\sigma]D(\sigma \| I)$.

The Umegaki divergence satisfies something known as the data processing inequality, for each trace-preserving completely positive map $\mathcal{E}$, we have
$$
D(\rho\|\sigma) \geq D(\mathcal{E}(\rho) \| \mathcal{E}(\sigma)).
$$
Again I refer to the linked book for further details.

Now the thing to notice is that for any collection of linear operators $\{V_1, \dots, V_m\}$ such that $\sum_i V_i^* V_i = I$ the map $\rho \mapsto \sum_{i} V_i \rho V_i^*$ is completely positive and trace preserving. (I am using $^*$ to denote the adjoint / conjugate transpose or transpose when things are real) Indeed the two projections $V_1$ and $V_2$ form such a map. Let $\mathcal{F}$ be the map defined by the action $\sigma \mapsto V_1 \sigma V_1^T + V_2 \sigma V_2^T$ then we have

$$
\begin{aligned}
\mathrm{Tr}[\sigma \log \sigma] &= \mathrm{Tr}[\sigma] D(\sigma \|I) \\
&\geq \mathrm{Tr}[\mathcal{F}(\sigma)] D(\mathcal{F}(\sigma) \| \mathcal{F}( I)) \\
&= \mathrm{Tr}[\mathcal{F}(\sigma)] D(\mathcal{F}(\sigma) \| I ) \\
&= \mathrm{Tr}[\mathcal{F}(\sigma) \log \mathcal{F}(\sigma)] \\
&= \mathrm{Tr}[V_1 \sigma V_1^T \log V_1 \sigma V_1^T] + \mathrm{Tr}[V_2 \sigma V_2^T \log V_2 \sigma V_2^T].
\end{aligned}
$$
Given the scope of the data processing inequality there is definitely room for adding additional orthogonal projectors in if you increase the dimension of the space. For reference maps whose action is the sum of orthogonal projector sandwiches are sometimes referred to as pinching maps.


问:
Why WLLN implies this convergence?
I'm trying to understand this theorem which shows the convergence of the sample standard deviation to the standard deviation parameter. I didn't understand why $\frac{1}{n}\sum X_i^2\to \sigma^2+\mu^2$ because of WLLN. It should be some small detail I'm not seeing.

答: If $Y_1, \ldots, Y_n$ are i.i.d., the WLLN implies $\frac{1}{n} \sum_{i=1}^n Y_i$ converges in probability to $E[Y_1]$. Here, with $Y_i=X_i^2$ you have $E[Y_1]=E[X_1^2] = \text{Var}(X_1) + E[X_1]^2 = \sigma^2 + \mu^2$.

Title: Why WLLN implies this convergence?


问:
Is a bounded linear operator, which is bijective between dense subspaces, an isomorphism?
Suppose that we have Banach spaces $X$ and $Y$ and a bounded linear operator $T\colon X\to Y$ that maps a dense subspace $D_{X}\subset X$ bijectively into a dense subspace $D_{Y}\subset Y$. Can we conclude that $T$ is an isomorphism?

Here is what I tried for proving surjectivity: Pick any $y\in Y$. Then there is a sequence $D_{Y}\ni y_{n}\to y$. By assumption there is a (unique) $x_{n}\in D_{X}$ such that $y_{n}=T(x_{n})$. Now I tried to prove that $x_{n}$ is Cauchy, so that it has a limit $x\in X$, and then I was hoping that $y=T(x)$. But I cannot find an estimation of the form $\|x_{i}-x_{j}\|\leq C\|T(x_{i})-T(x_{j})\|$.

Any help would be greatly appreciated!

答:
Not even true when $X=Y$ is a Hilbert space. Any injective compact operator on an infinite-dimensional Hilbert space is an example.

For instance, take $X=Y=\ell^2(\mathbb N)$ and $$Tx=\big(x_1,\tfrac{x_2}2,\tfrac{x_3}3,\ldots\big),$$ on the dense subspace $$c_{00}=\{x\in X:\ \text{ finite support }\}.$$
The necessary and sufficient condition for $T$ to extend to an invertible operator on the full space is that $T^{-1}$ is bounded.


问:
Find the value of $\frac{\tan\theta}{1-\cot\theta}+\frac{\cot\theta}{1-\tan\theta}$
I want to know an objective approach to solve these type of expression in a quick time
Which of the expression equals to
$$\dfrac{\tan\theta}{1-\cot\theta}+\dfrac{\cot\theta}{1-\tan\theta}$$

a)$1-\tan\theta-\cot\theta$

b)$1+\tan\theta-\cot\theta$

c)$1-\tan\theta+\cot\theta$

d)$1+\tan\theta+\cot\theta$

I've tried it several ways like taking LCM,change whole into $\sin\theta$ and $\cos\theta$.but  I've stuck.

答:
Just to add the "cheating" method of solving these (say on an exam when you're pressed for time):

Use the symmetry to show that only options (a) and (d) are viable. Then plug in some value (say $x=\pi/8$), to find which one is of them is "right"


问:
Identifying this approximation of pi
I've come across this approximation of pi, which I'm struggling to put a name on. I want to do further research on it but I can't find any evidence of it on the internet. It is an infinite sum:
$\pi = 1 + 1/2 + 1/3 + 1/4 - 1/5 + 1/6 + 1/7 + 1/8 + 1/9 - 1/10 + 1/11 + 1/12 - 1/13 + \dots $

This is the description I've been given of it:

In this formula, each fraction 1/n has a sign (±) determined by: the first two terms have
positive signs; after that, if the denominator is a prime of the form 4m − 1 (for example,
n = 3, 7, 11, . . .), the sign is positive; if the denominator is a prime of the form 4m + 1 (for
example, n = 5, 13, 17, . . .), the sign is negative; if the denominator is a composite number,
then the sign is equal to the product of the signs corresponding to its factors (for example,
n = 9 = 3 × 3, so its sign is positive (a positive times a positive), while n = 10 = 5 × 2,
so its sign is negative (a negative times a positive)).

Can anyone help me identify which approximation this is? Is it a well-known approximation?

答:
Not an answer, but maybe this will help.  You can write the sum as an infinite product.  For primes $p$ that are $2$ or of the form $4n-1$ you have a factor
$$1+\frac 1p+\frac 1{p^2}+\frac 1{p^3}+\ldots=\frac 1{p-1}$$
For primes $q$ of the form $4n+1$ you have a factor
$$1-\frac 1q+\frac 1{q^2}-\frac 1{q^3}+\ldots=\frac 1{q+1}$$
So the whole sum becomes
$$\prod \frac 1{p_i-1} \cdot \prod \frac 1{q_i+1}$$
I don't know how to relate this to $\pi$

Title: Identifying this approximation of pi


问:
Maple calls absolute convergent sum divergent, but why?

Determine whether the sum

is divergent, absolute convergent or conditionally convergent.

First, I tried to solve this by hand. I tried to find a convergent majorant series for the series in question.

$$\bigg|\cos(n^2+1) \frac{2+n^2}{1+n!} \bigg| \leq\frac{n^2}{n!}$$

It turns out that $\sum_{n=1}^{\infty}\frac{n^2}{n!}=2e$, according to Maple that is. This result shows that the sum in the question is absolute convergent.

I also tried to use Maple to double check my answer, and it gave me this.

When Maple gives this sort of output, it usually means that the sum is not convergent. But didn't I just show that it was, or have I made a mistake?

Can a "Maple man" tell me why Maple gives this result, and perhaps how I can avoid it?

答: The sum converges absolutly. Your proof is correct (exept that inequlity $\frac{n^2+2}{1+n!} \leq \frac{n^2}{n!}$ is not true, but the last fraction may be replaced by $\frac{2+n^2}{n!}$, which give us convergent series).


问:
Why is Brouwer’s Fixed Point Theorem considered a result of algebraic topology?
Brouwer’s Fixed Point Theorem and some generalisations can be proved using properties of Barycentric subdivisions and easy special cases of Sperner’s Lemma relating to them plus basic properties of metric spaces and the fact that every compact convex subset of $\mathbb{R}^n$ is homeomorphic to an $m$-simplex for $m\leq n$ which can be proved by projecting both onto the unit sphere and rotating. For example, see the Wikipedia entry for its generalisation The Kakutani Fixed Point Theorem which gives a sketch. Not so much as a homotopy, let alone any algebra. Why then is Brouwer’s Fixed Point Theorem considered part of algebraic topology when it is not about the objects of algebraic topology and it does not require the methods of algebraic topology?

答: Sure, they can be proved combinatorially, but Brouwer invented degree to solve this problem. Brouwer certainly considered this to be a theorem of what we would now call algebraic topology. Is the fundamental theorem of algebra really a theorem of Lie theory because you can prove it by classifying compact Abelian Lie groups? Obviously not.</p>


问:
Property of natural numbers involving the sum of digits
How can you prove that every natural number $M$ or $M+1$ can be written as $k + \operatorname{Sum}(k)$, where $\operatorname{Sum}(k)$ represents the sum of the digits of some number k.

Example:
$$
248 = 241 + \operatorname{Sum}(241) = 241 + 2 + 4 + 1$$

答:
Hint: Each term in the sequence $k + \mbox{Sum}(k)$ either increases by 2, or decreases by some amount.

Why does that tell you that the image of $k + \mbox{Sum}(k)$ must include either $M$ or $M+1$?

Proof of Hint: If the last digit is not 9, then $k$ and $\mbox{Sum}(k)$ will both increase by 1, hence their sum increases by 2. If the last digit is 9, then $\mbox{Sum}(k)$ will decrease by at least 8, hence their sum decreases by sum amount.


答:
An alternative proof is to use Chebychev's inequality

$$\mathbb{P}\Big[|S_n^2-\sigma^2|\geq \epsilon\Big]\leq\frac{\mathbb{E}(S_n^2-\sigma^2)^2}{\epsilon^2}=\frac{\mathbb{V}(S_n^2)}{\epsilon^2}\xrightarrow{n\to \infty}0$$

this, with the sufficient condition that $\mathbb{V}(S_n^2)\xrightarrow{n\to \infty}0$

This proof is taken from Casella Berger, Chapter 5: Example 5.5.3

Title: Why WLLN implies this convergence?

Body: I'm trying to understand this theorem which shows the convergence of the sample standard deviation to the standard deviation parameter. I didn't understand why $\frac{1}{n}\sum X_i^2\to \sigma^2+\mu^2$ because of WLLN. It should be some small detail I'm not seeing.


问:
Show that $C$ is a convex subset of $\mathbb R^4$
So I'm given the following information:
Let $f:\mathbb R^4\rightarrow \mathbb R$ be given by

$$f(x_1,x_2,x_3,x_4)=(x_1-x_3)^2+(x_2-x_4)^2$$

and $C\subseteq \mathbb R^4$ by

$$C=\{(x_1,x_2,x_3,x_4)\in \mathbb R^4 \mid x_1^2+(x_2-2)^2\leq 1,\ x_3-x_4\geq 0\}$$

I'm asked to show that $C$ is a convex subset of $\mathbb R^4$.

I have the following definition:

A convex subset $C\subseteq \mathbb R^d$ is a subset that contains the line segment between any two of its points $x, y \in C$, i.e.,
$$(1-t)x+ty\in C$$
for every number $t$ with $0\leq t\leq 1$.

I'm just not sure how to put these things together, anything to help me get started is greatly appreciated.

答:
Let $A = \{(x_1,x_2) \in \mathbb R^2 \mid x_1^2+(x_2-2)^2\leq 1\}$ and $B  = \{(x_3,x_4)\in \mathbb R^2 \mid  x_3-x_4\geq 0\}$. Then $C = A \times B$. We show that $A, B$ are convex; this implies that their Cartesian product in convex.

$A$ is a closed disk with center $(0,2)$ and radius $1$; this is convex.

$B$ is the closed half-plane lying above the diagonal line $x_4 = x_3$; this is also convex.

You can prove the convexity of $A,B$ also formally. See Piquito's answer.


问:
A problem about $SS=G?$
Suppose $G$ is a Group, $S$ is a subset of $G$. Given that $\exists a,aS\cap S\neq \emptyset, aS\cup S=G$, prove or give a counterexample: $\forall g\in G,\exists x,y\in S,s.t.xy=g\ (i.e.\; SS=G)$

I can prove the proposition is true when G is a abelian group, the prove is as follow:

wts. $\forall g'\in G, g'S^{-1}\cap S\neq \emptyset $
, here $S^{-1}=\{s^{-1}: s\in S\}$

Anti-evidence method:
if $g'S^{-1}\cap S=\emptyset $
,then $ag'S^{-1}\cap aS=\emptyset$

given that $S\cap aS\ne \emptyset$ and $S\cup aS=G$, so $ag'S^{-1}\subsetneqq S$

proposition 2: if $pS^{-1}\subset S$, then $pS^{-1}=S$

prove: Anti-evidence method: if there is $s$ in S but not in $pS^{-1}$

then $ps^{-1} \in pS^{-1}\subset S$, $p(ps^{-1})^{-1} \in pS^{-1}$

$G$ is a abelian group, so $p(ps^{-1})^{-1}=s\in pS^{-1}$, contradictory.

So the proposition is proved.

so $ag'S^{-1}\subsetneqq S$ leads to a contradiction.

then SS=G is proved.

I wonder whether the proposition holds when G is not a abelian group. Can anyone help me?(thanks)

答:
As I wrote in the comments I don't understand how you get the main result from Proposition 2. However, here is a different way of getting the main result from proposition 2 (proposition 2 of the answer, not the question).

I use the notation of your answer not your question, so the 'special' element is called $g$, not $a$.

Let $s_1 \in S \cap Sg$, it exists by Proposition 2.
Let $s_2 \in S$ be such that $s_1 = s_2g$.

Let $B = \{e, g\}$. The premise that $S \cup gS = G$ is equivalent to $BS = G$.

Let $B' = s_2B$. Then on one hand we have:

$B'S = s_2BS = s_2G = G$

On the other hand we have $B' = \{s_2, s_2g\} = \{s_2, s_1\} \subset S$

So $B' \subset S$ and hence $G = B'S \subset SS$ as we wanted to show.

I already had thought up this argument a few days ago, but had no use for it because I couldn't prove Proposition 2. So I was really enthusiastic to read your elegant proof of Proposition 2 in the answer. This question was becoming a bit of an obsession, so I am happy it is settled now.

Title: A problem about $SS=G?$


问:
Question regarding analyticity, differentiability and the Cauchy-Riemann Equations ( Complex Analysis )

For non-analyticity, we can show that the Cauchy-Riemann equations don't hold anywhere. But here we can see that the Cauchy-Riemann equations hold when $y=2x$. I understand what the book did here but I just want to clarify something: If Cauchy-Riemann equations hold for $y=2x$, it doesn't necessarily mean that the function is differentiable. Do they just assume here that the function is differentiable along the line $y=2x$ and then talk about how the function won't be differentiable in any neighborhood of any complex number on the line $y=2x$ and so it won't be analytic anywhere? ( Since Cauchy-Riemann equations won't hold in the neighborhoods and hence not analytic). I just need to understand if they assumed that the function is differentiable along the line $y=2x$ or not. What's happening here? Thank you.

I cannot use "Sufficient Condition for Differentiability", since it is introduced later on in the book.

The problem below was solvable since I could use the "Sufficient Condition for Differentiability":

Show that the function $f(z) = x^2 - x + y + i(y^2 - 5y - x)$ is nowhere analytic but differentiable along $y = x + 2$. I tried doing this and so far the Cauchy-Riemann equations are only satisfied when we have $y = x + 2$. So when we have $y = x + 2$, all the demands for "Sufficient Condition for Differentiability" are met and so the function is differentiable along $y = x + 2$. However, the function won't be differentiable at every point in the neighborhood of any complex number along $ y = x + 2$, so it won't be analytic.

The question I have is, do we need differentiability along a curve to talk about differentiability at neighborhoods of points along the curve?

So far I've learnt that:

Analyticity at a point implies differentiability at a point but the converse is not true,
Analyticity at every point in a domain D implies differentiability at every point in a domain D and the converse is true,
Analytic at a point implies differentiable at a point implies Cauchy-Riemann equations hold at that point but the converses are not true.

Are all of these correct?

I know this is a lengthy question, but please help me understand this and clear my confusions. Thank you!

答: No, they did not assume that $f$ is differentiable at the pints of the form $x+2ix$. What happens is that the set $L=\{x+2ix\mid x\in\Bbb R\}$ is a line and therefore, for any $z\in L$, no disk centered at $z$ is contained in $L$. And so, for any disk centered at any point of $L$ there is a point of that disk that doesn't belong to $L$ and so, in particular, there is a point in that disk at which $f$ is not differentiable.