问:
Symmetrical Matrices with entities $0$ and $1$
Let $A$ be the set of all $3 × 3$ symmetric matrices all of whose entries are either $0$ or $1$. Five of these entries are $1$ and four of them are $0$.

Q.1 The number of matrices in A is

(A) 12

(B) 6

(C) 9

(D) 3

My approach is as follow

$\left[ {\begin{array}{*{20}{c}}
a&h&g\\
h&b&f\\
g&f&c
\end{array}} \right]$

CASE 1: $a,b,c\ \in{1,1,1}$
Then $f,g,h \in {0,0,1}$
Number of cases $\frac{3!}{2!}=3$

CASE 2: $a,b,c\ \in{1,0,0}$
Then $f,g,h \in {0,1,1}$
Number of cases $\frac{3!}{2!}*\frac{3!}{2!}=9$

Hence the number of matrix in set A =3+9=12 which is correct as per the official answer key.

Q.2 The number of matrices A in A for which the system of linear equations $A\left[ {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
0\\
0
\end{array}} \right]$  has a unique solution, is

(A) less than 4

(B) at least 4 but less than 7

(C) at least 7 but less than 10

(D) at least 10

Q.3 The number of matrices A in A for which the system of linear equations $A\left[ {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1\\
0\\
0
\end{array}} \right]$  is inconsistent , is

(A) 0

(B) more than 2

(C) 2

(D) 1

From continuation from Question1 how we will proceed Question2 and Question3.

As the set A has 12 matrices

答:
The equation has a unique solution if and only if $A$ is nonsingular.  Suppose then, that $A$ is nonsingular.

Can $A$ have three $1$s in the same row?  Then the corresponding column also has three $1$s, and the remaining four entries are $0$.  We see that the other two rows are identical, and $A$ is singular.  This eliminates three possibilities for $A$.

Therefore, $A$ must have two rows with two $1$s and one row with a single $1$.  Can the single $1$ lie on the diagonal?  Then the corresponding column also has only one $1$, so the other four entries are $1$ and again we have two identical rows. This eliminates another three possibilities for $A$.

Suppose that the first row of $A$ is $$\begin{bmatrix}0&1&0\end{bmatrix}$$  Each of the other rows has two $1$ and a $0$, and only one of them can have the $1$ on the diagonal.  One possibility is $$A=\begin{bmatrix}0&1&0\\1&0&1\\0&1&1\end{bmatrix}$$ which is nonsingular.

Permuting rows and columns gives six nonsingular matrices, which accounts for all twelve possibilities for $A$.

For part $3$, I can't think of anything better than to examine the cases we looked at in part $2$.  We know the equation is consistent in the six cases where $A$ is nonsingular.

The equation can be written as $$x\begin{bmatrix}{a\\h\\g}\end{bmatrix}+y\begin{bmatrix}{h\\b\\f}\end{bmatrix}+z\begin{bmatrix}{g\\f\\c}\end{bmatrix}=\begin{bmatrix}{1\\0\\0}\end{bmatrix}$$  For the case that one row has all $1$s, there are two cases to consider.  Either the first row has all $1$s, or one of the the other two rows has all $1$s.  This is because of the asymmetry on the right-hand side.  The second and third rows are identical, but the first is different.  In both cases, we easily find that the equation can be solved.

We have left the three cases where there are three $1$s on the diagonal.  I leave them to you.


问:
Prove $(x_1+x_2-x_3-x_4)(x_1-x_2+x_3-x_4)(x_1-x_2-x_3+x_4)$ is a symmetric polynomial.
The problem ask to prove what's in the title. From the book I know that, given the following polynomials,

$$h_1 = x_1x_2+x_3x_4$$
$$h_2 = x_1x_3+x_3x_4$$
$$h_3 = x_1x_4+x_2x_3$$

Any symmetric polynomial in $h_1, h_2$ and $h_3$ will be also symmetric in $x_1, x_2, x_3, x_4$. My idea is writing the polynomial $(x_1+x_2-x_3-x_4)(x_1-x_2+x_3-x_4)(x_1-x_2-x_3+x_4)$ in terms of $h_1, h_2$ and $h_3$, but I haven't found an easy way to show that, and I've spend a lot of hours on it.

I feel this is the way since $h_1-h_2=(x_2-x_3)(x_1-x_4), \ h_1-h_3=(x_1-x_3)(x_2-x_4), \ h_2-h_3=(x_1-x_2)(x_3-x_4)$ but when organizing the terms I can't get a way to get these products, also brute force and factorizing is really large.

答:
Hint:

I think it would simpler to check that the product is invariant by any transposition in $S_4$, since  transpositions  generate $S_4$. There are only $6$ of them.

Example: With the transposition $\tau=(1\,2)$
$$(x_1+x_2-x_3-x_4)(x_1-x_2+x_3-x_4)(x_1-x_2-x_3+x_4)\to (x_2+x_1-x_3-x_4)(x_2-x_1+x_3-x_4)(x_2-x_1-x_3+x_4)= (x_1+x_2-x_3-x_4)\bigl[-(x_1-x_2-x_3+x_4)\bigr]\bigl[-(x_1-x_2+x_3-x_4) \bigr],$$
which simply swaps the last two factors.

Edit: As pointed by @Edonbrayn in a comment, as $S_n$ is generated by the transpositions $(1\,k),\enspace k=2,\dots,n$, there remains only to check it is invariant by $(1\,3)$ and $(1\,4)$.

ContentLicense: CC BY-SA 4.0


问:
Partial fraction expansion two variables
How to expand

$$\frac{y}{(x-y)(y-1)}$$

by partial fraction expansion.

答:
It depends on whether you view the expression as a function of $x$ or as a function of $y$. If you view it as a function of $x$, the decomposition is already done.

If you view it as a function of $y$, make the usual ansatz:

$$\frac{y}{(x-y)(y-1)} = \frac{A}{x-y} + \frac{B}{y-1}.$$

$A$ and $B$ will depend on  $x$ but not on $y$.

ContentLicense: CC BY-SA 3.0


问:
Necessary and sufficient condition for weak convergence of gamma distribution
Let $(X_n)_n$ be a sequence of random variable such that $f_{X_n}(x)=\frac{1}{\Gamma(\alpha_n)}\lambda_n^{\alpha_n}x^{\alpha_n-1}e^{-\lambda_nx}1_{]0,+\infty[}(x),$ where $\alpha_n>0,\lambda_n>0.$

Suppose that $\alpha_n=1,\forall n \in \mathbb{N}.$ Find a necessary and sufficient condition on $(\lambda_n)_n$ such that $(X_n)_n$ converges in distribution.

More generally, find a necessary and sufficient condition on $\alpha_n,\lambda_n$ so that $(X_n)_n$ converges in distribution.

The first part is easy, it converges in distribution if and only if $0<\liminf_n\lambda_n=\limsup_n\lambda_n$.

Concerning part 2), is it true that a condition of weak convergence is the convergence of $(\alpha_n)$ and $(\lambda_n)$?

答:
"Is it true that a condition of weak convergence is the convergence of $\alpha_n$ and $\lambda_n$?" The answer is negative.

Counterexample:

Suppose $\lambda_n \to +\infty$, $\alpha_n = \bar{o}(\lambda_n)$, $n \to \infty$, and let $X_n$ have distribution $\Gamma(\lambda_n, \alpha_n)$. We have

$$\mathbb{E} e^{i t X_n}  = \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{-\alpha_n} = e^{-\frac{\alpha_n}{\lambda_n} \cdot \lambda_n \ln\bigl( 1 - \frac{it}{\lambda_n} \bigr)}.$$

Hence,

$$\lim_{n \to \infty} \mathbb{E} e^{i t X_n} =
\lim_{n \to \infty} e^{-\bar{o}(1) (-it)(1+\bar{o}(1))} = 1 = \mathbb{E} e^{i t \cdot 0},
$$
that is $X_n \to 0$ in distribution.

So we may take, for example, $\lambda_n = n$ and $\alpha_n = 2 + (-1)^n$ for our counterexample.

If we suppose that $\alpha_n$ and $\lambda_n$ are separated from zero and $\infty$, that is $0 < c \le \alpha_n, \lambda_n \le C < \infty$, than convergence of $X_n$ is equaivalent to pointwise convergence of functions  $\Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{-\alpha_n} $ for each $t \in \mathbb{R}$. In this case we may show that "there is convergence in distribution" iff "$\alpha_n$ and $\lambda_n$ are convergent" (and in this case $lim_n X_n \sim \Gamma(\lim_n \lambda_n, \lim_n \alpha_n)$).  The "if-part" follows immediately from the convergence of characteristic functions.

To prove the second part, let us take a convergent subsequence $\alpha_{n_k}$. We know that $X_n$ has a limit, hence $\exists \lim_n \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{\alpha_n}$. Put $t=1$. Sequences $\Bigl(1 - \frac{i}{\lambda_{n_k}} \Bigr)^{\alpha_{n_k}}$ and $\alpha_{n_k}$ are convergent. Hence, for some finite $\lambda, \alpha>0$ we have $\lambda_{n_k} \to \lambda$, $\alpha_{n_k} \to \alpha$  and hence $$\lim_n \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{\alpha_n} \to \Bigl(1 - \frac{it}{\lambda} \Bigr)^{\alpha}.$$

If there is another convergent subsequence $\alpha_{n_k^*} \to \alpha^*$, we have (as above): $\lambda_{n_k^*} \to \lambda^*$ and $$\lim_n \Bigl(1 - \frac{it}{\lambda_n} \Bigr)^{\alpha_n} \to \Bigl(1 - \frac{it}{\lambda^*} \Bigr)^{\alpha^*}.$$ Moreover, $\alpha^*, \lambda^* > 0$. We need to prove that $\alpha = \alpha^*$ and  $\lambda = \lambda^*$.

It's easy to see that

$$\Bigl(1 - \frac{it}{\lambda} \Bigr)^{\alpha} = \Bigl(1 - \frac{it}{\lambda^*} \Bigr)^{\alpha^*}$$
for all $t \in \mathbb{R}$. Put $C = \frac{\alpha}{\alpha^*} > 0$. We have $\Bigl(1 - \frac{it}{\lambda} \Bigr)^{C} = 1 - \frac{it}{\lambda^*} $. Put $u(t) = 1 - \frac{it}{\lambda} $. We have: $u^C = $ linear function of $u$, $u \in \mathbb{R}$. Hence, $C=1$ and $\alpha = \alpha^*$.

Moreover, $\Bigl(1 - \frac{it}{\lambda} \Bigr)^{C} = 1 - \frac{it}{\lambda^*} $, so $\lambda = \lambda^*$. The proof is finished.

For case when $\alpha_n$ and $\lambda_n$ are not separated from $0$ both we have the following theorem.

Theorem 1.

If $\alpha_n \ge c > 0$ and $\lambda_n \to 0$ then $X_n$ diverges.

If $\alpha_n \to 0$ and $\lambda_n \ge c > 0$ then $X_n \overset{w}{\to} 0$.

If $\alpha_n \to 0$ and $\lambda_n \to 0$ then

$$X_n \overset{w}{\to} \Longleftrightarrow X_n \overset{w}{\to} 0 \Longleftrightarrow \lim_{n \to \infty} \alpha_n \ln \lambda_n = 0.$$

Let us prove the theorem. If $\xi \sim \Gamma( \lambda, \alpha)$ then
\begin{gather}
\Bigl( 1 - \frac{it}{\lambda} \Bigr)^{-\alpha} = e^{-\alpha \ln(1 - \frac{it}{\lambda} )} = e^{-\alpha  \Bigl( \ln \sqrt{1^2 + \frac{t^2}{\lambda^2}} + i \cdot arg(-\frac{t}{\lambda}) \Bigr) } = \nonumber \\
= e^{ - \frac{\alpha}2 \ln \bigl(1 + \frac{t^2}{\lambda^2} \bigr)} \cdot e^{i \alpha arctg \frac{t}{\lambda} } = {\Bigl( 1 + \frac{t^2}{\lambda^2}\Bigr)}^{-\frac{\alpha}2} \cdot e^{i \alpha arctg \frac{t}{\lambda} }.\nonumber
\end{gather}
Hence $$Ee^{itX_n} = {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2} \cdot e^{i \alpha_n arctg \frac{t}{\lambda_n} }.$$

From Levy’s continuity theorem we have
$$ \Bigl( X_n \overset{w}{\to} \Bigr)  \Longleftrightarrow \Bigl(  \forall t \in \mathbb{R} \ \ \exists \lim_n Ee^{itX_n} = \phi(t) \text{and  } \phi(t) \text{  is continious at } 0 \Bigr).$$

If $\alpha_n \ge c > 0$ and $\lambda_n \to 0$ then $\lim_n |Ee^{itX_n}| = {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2} \to I_{t=0}$, where $I$ is an indicator function. If $X_n \overset{w}{\to} X$ then $\lim_n |Ee^{itX_n}| = |Ee^{itX}|$ is continuous (as it is the module of the characteristic function). But $I_{t=0}$ is not continuous.  Hence $X_n$ diverges. The first statement of the theorem is proved.

Here and everywhere below we will assume that $\alpha_n \to 0$. Thus $\alpha_n arctg \frac{t}{\lambda_n}  = \text{ bounded } \cdot \alpha_n = \bar{o}(1)$, $n \to \infty$. Therefore $e^{i \alpha_n arctg \frac{t}{\lambda_n} } = 1 + \bar{o}(1).$

Put $\phi_n(t) ={\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)}^{-\frac{\alpha_n}2}$. We proved that
\begin{gather} \Bigl( X_n \overset{w}{\to} \Bigr)  \Longleftrightarrow \Bigl(  \forall t \in \mathbb{R} \exists \lim_n \phi_n(t) = \phi(t) \text{ and } \phi(t) \text{ is continious at  } 0 \Bigr).          (1)
\end{gather}

If $\alpha_n \to 0$ and $\lambda_n \ge c > 0$ then $\phi_n(t) \to  1$ and moreover $Ee^{itX_n} \to 1$, hence $X_n \overset{w}{\to} 0$.  The second statement of the theorem is proved.

Here and everywhere below we will assume that $\lambda_n \to 0$. For $t \ne 0$ we have

\begin{gather}
\phi_n(t) =  e^{-\frac{\alpha_n}2 \ln {\Bigl( 1 + \frac{t^2}{\lambda_n^2}\Bigr)} } =  e^{-\frac{\alpha_n}2 (1 + \bar{o}(1))\ln {\Bigl(\frac{t^2}{\lambda_n^2}\Bigr)} } = \nonumber \\
= e^{-\alpha_n (1 + \bar{o}(1)) ( \ln t - \ln \lambda_n  )} = e^{\alpha_n \ln \lambda_n  (1 + \bar{o}(1))}
\end{gather}

Hence for $t \ne 0$  we have
$$ \exists lim_n \phi_n(t) \Longleftrightarrow  \exists lim_n \alpha_n \ln \lambda_n = A \in [-\infty, 0]$$
and if this condition holds true then $\phi(t) =  lim_n \phi_n(t) = e^{A}$, $t \ne 0$. Obviously $lim_n \phi_n(0) = lim_n 1 = 1$.

So
\begin{gather}
\Bigl(  \forall t \in \mathbb{R} \ \ \exists \lim_n \phi_n(t) = \phi(t) \text{ and } \phi(t) \text{ is continious at } 0 \Bigr) \Longleftrightarrow \nonumber \\
\exists lim_n (\alpha_n \ln \lambda_n) = A \text{ and } A=0.        (2)
\end{gather}
From (1) and (2) we get that $X_n$ converges iff $ \lim_{n \to \infty} \lambda_n \ln \alpha_n = 0$. Moreover, if $ \lim_{n \to \infty} \lambda_n \ln \alpha_n = 0$ then $Ee^{itX_n} =  \phi_n(t) \cdot e^{i \alpha_n arctg \frac{t}{\lambda_n} } \to 1 = Ee^{it\cdot 0}$, hence $X_n \overset{w}{\to} 0$.

Theorem is proved.


问:
If the derivative of a function is the zero function in $\mathbb R^n$, then the function is constant.
If the derivative of a function is the zero function in $\mathbb R^n$, then the function is constant. Basically I am considering $f: \Bbb R^m \to \Bbb R^n$ and $Df(x)=0$ for all $x\in \Bbb R^m$. The general fact is that If the derivative of a function is the zero function in $\mathbb R^n$, then the function is constant when the domain is path-connected but I am not getting any proof over math stack.
In $1$ dimension we can use LMVT but from $\Bbb R^m \to \Bbb R^n$ this is of no use. I think we have to deal with path and then locally constant. I don't know the proof of locally constant as well. If someone can help me from the scratch I can also fill up the details if I can follow it. Someone can also mention some books if those have the answer.

答:
If $x,y\in \mathbb R^m$ then the map $g:t\mapsto ty+(1-t)x$ is well-defined. Then, with $h:=f\circ g$, we can find $t\in (0,1)$ such that $|h(1)-h(0)|\le |h'(t)|.^{1}$

But $h(1)-h(0)=|f(y)-f(x)|$ and $h'(t)=Df(g(t))\circ g'(t)=0.$ It follows that $f(x)=f(y).$

$1.$ This is the mean-value inequality for vector-valued functions. Rudin proves it this way: define $z:=h(1)-h(0)$ and $\varphi(t)=\langle z,h(t)\rangle$ and apply the single-variable mean-value theorem to find a $c\in (0,1)$ such that $\varphi(1)-\varphi(0)=\varphi'(c)=\langle z,h'(c)\rangle.$ But we also have $\varphi(1)-\varphi(0)=|z|^2$ so the result follows by the Schwarz inequality.


问:
Prove that $\mathbb{Q}(\sqrt{a}, \sqrt{b})= \mathbb{Q}( \sqrt{a}+\sqrt{b})$
Let $a, b \in \mathbb{Q}$ be two elements such that $a,b$ and $ab$ are not
squares in $\mathbb{Q}$

Prove that $\mathbb{Q}(\sqrt{a}, \sqrt{b})= \mathbb{Q}( \sqrt{a}+\sqrt{b})$

答:
Usually, to prove equality of sets $A=B$ you show that $A \subseteq B$ and also $B \subseteq A$.

In your case, here are the two parts.

Part I.

Let $x \in \mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right)$. Then, ....
and you conclude $x \in \mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right)$. Hence, $\mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right) \subseteq \mathbb{Q}\left(\sqrt{a}+ \sqrt{b}\right)$.

Part II.

Let $x \in \mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right)$. Then, ....
and you conclude $x \in \mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right)$. Hence, $\mathbb{Q}\left(\sqrt{a} + \sqrt{b}\right) \subseteq \mathbb{Q}\left(\sqrt{a}, \sqrt{b}\right)$.

Please update your question or post comments about your progress.


问:
Is this a permissible way of proving/showing this statement?
The task is to show that $$\sin\phi=\frac{1}{2i}(e^{i\phi}-e^{-i\phi}) $$

with an emphasis on show. The task does not specifically state, prove this statement.

So, what I did, is to transform the right side until I got what is stated on the left side. $$ \sin\phi=\frac{1}{2i}(e^{i\phi}-e^{-i\phi}) = ... = ...=\sin\phi  $$

Was this a valid way to show the given statement, or should I have transformed the left side until I get the statement on the right side, like this: $$ \sin\phi=...=...=\frac{1}{2i}(e^{i\phi}-e^{-i\phi}) $$

Or, is there perhaps even no difference, and I can choose either way?

I am sorry for this seemingly trivial question, as I am still a Calc 1 newbie and would love to know the difference between showing and actually proving a statement. Because it seems to me like the first way is merely showing what is given, thus not a full proof, while the nature of the second one, namely taking the left side as the springboard and transforming it until we have what was to be shown, looks more reasonable. Maybe I'm also just overthinking. Any help would be appreciated.

答:
$$e^{i\phi}=\cos \phi +i \sin \phi,~~ e^{-i\phi}=\cos \phi -i\sin \phi$$
add these two to get $$\sin \phi= \frac{e^{i\phi}-e^{-i\phi}}{2i}$$


问:
Complex integral over two differents closed path
The first integral that I want to calculate is: $$\int_{|z-2|=5}\frac{z^3 +3}{z^2+1}\,dz$$

And the second is $$\int_{|z|=\frac{1}{2}}\frac{z^3 +3}{z^2+1}\,dz$$

The denominator can be factored into $(z+i)(z−i)$. As $z_0=i$ and  $z_1=-i$ are inside the path $|z−2|=5$ I can't use Cauchy and I don't know how to solve those integrals with other method.

In the second case $z_0$ and $z_1$ aren't inside $|z|=\frac{1}{2}$ so I can to use Cauchy

答:
You can use Cauchy's Residue theorem: if $f$ is analytic in a simply connected domain $\Omega$ except at a number of poles then the integral of $f$ along a closed contour in $\Omega$ that circulates the poles just once is $2\pi i$ times the sum of the residues at the poles lying inside the contour.  The residue at a pole $z_0$ is the the coefficient to the term $1/(z-z_0)$ in the Laurent expansion of $f$ about $z_0$.

In your first integral, the contour includes two poles at $z = \pm i$.

Writing $$\frac{z^3+3}{z^2+1} = z + \frac{1}{2}\Bigg(\frac{3i-1}{z+i}-\frac{3i+1}{z-i} \Bigg)$$
you can see the residues at the two poles.  At $z=-i$ it is therefore $\frac{1}{2}(3i-1)$ and that at $z=i$ is $-\frac{1}{2}(3i+1)$.  Now apply the theorem, noting that the sum of the two residues is $-1$, to obtain,
$$\int_{|z-2|=5} \frac{z^3+3}{z^2+1} dz = -2\pi i. $$

For the second integral, neither pole lies inside the contour and therefore the same theorem implies the integral will be zero.


问:
Center of mass of $4z^2=x^2+y^2$ within $x=0, y=0, z=0, z=3$
I have a solid limited by the cone: $4z^2=x^2+y^2$ and the following planes: $x=0, y=0, z=0, z=3$

This solid has a density = $1$

I already found the volume using the following triple integral ($r$ referring to radius):

$$\int_{0}^{\frac{\pi}{2}}\int_{0}^{6}\int_{\sqrt{r^2/4}}^{3} r dz dr d\theta =9\pi$$

From that, what would be the best way to find the moment of inertia and finally the center of mass / barycenter?

Thanks in advance

答:
The cone is symmetric for x and y .Z is the symmetry axis of the cone. It is comparable with revolving an isosceles triangle about z. The height of cone is from 0 t0 3 i.e, $h=3$ so the center of mass will have a distance
$\frac23 h=2$ from the vertex of the cone . The coordinates of this point is $C (0, 0, 2).$


问:
Proving $\displaystyle\prod_{k=1}^n(1+a_{k})\leq e$
If $a_{k}>-1$, $\forall 1\leq k\leq n$ such that :
$$
\sum_{k=1}^{n}\frac{a_{k}}{1+a_{k}x}\leq1\;\; ,\forall x\in[0,1]
$$
Then how can we show that :
$$
\prod_{k=1}^{n}(1+a_{k})\leq e
$$

答:
Integrating the identity
$$ \sum_{k=1}^n\frac{a_k}{1+a_kx}\leqslant 1 $$
between $0$ and $1$ gives
$$ \sum_{k=1}^n\log(1+a_k)\leqslant 1$$
Thus
$$ \prod_{k=1}^n(1+a_k)=\exp\left(\sum_{k=1}^n\log(1+a_k)\right)\leqslant e$$


问:
Does a strongly convex function in multiple variables have a "contraction" property?
Contraction is probably a bad term but I mean the following property: if $f(x)$ is $\lambda$-strongly convex and $x^*$ is its minimizer, then for all $x$ we have $\|x-x^*\|^2\leq \frac{2}{\lambda}(f(x)-f(x^*))$. What you get is that if the function value is close to the minimum, then the variables are also close.

I'm wondering what if $f(x,y)$ is strongly convex in $x$ and $y$, that is, for any $x$, $f(x,\cdot)$ is strongly convex and same hold for any fixed $y$ (but not necessarily jointly convex)? Say I find $x,y$ s.t $f(x,y)\leq f(x^*,y^*)+\epsilon$, where $(x^*,y^*)$ minimizes $f$. Can we say anything about $x$ being close to $x^*$ or $y$ being close to $y^*$? Thanks.

答:
No. The contraction property is a direct consequence of strong convexity. Specifically, from the first order character of strongly convex functions:
$$f(\mathbf{y})\geq f(\mathbf{x})+\partial f(\mathbf{x})^T(\mathbf{y}-\mathbf{x})+\frac{\lambda}{2}\|\mathbf{y}-\mathbf{x}\|^2.
$$
Without strong convexity, there is no contraction. Furthermore, convexity in individual coordinates might sound like a "nice" property, however it's almost meaningless from the perspective of convex analysis. Think of the function
$$f(x,y) = x^22^{y^2+1}$$
which is strongly convex for $x,y$ separately, but is a completely messy function otherwise.


问:
Does an irreducible $\mathbb CG$-module have a basis of the form $u,ug_1,\dots,ug_n$?
Suppose that $U$ is an irreducible $\mathbb CG$-module and $u\in U$. Let $\operatorname{span}(u_1,\dots,u_k)$ denotes the linear span of vectors $u_1,\dots,u_k\in U$.

I was thinking along these lines: there must exists some $g\in G$ such that $ug\notin\operatorname{span}(u)$, otherwise $\operatorname{span}(u)$ would be a nontrivial $\mathbb CG$-submodule. So I have two linearly independent vectors $u,ug\in U$.

Now $\operatorname{span}(u,ug)$ again is not closed under multiplication by group elements, otherwise it would be a nontrivial submodule. So I have some $(au+bug_1)g\notin \operatorname{span}(u,ug)$, which implies that either $ug$ or $ug_1g$ does not belong to $\operatorname{span}(u,ug)$.

I could continue in this way until I reach dimension of the $\mathbb CG$-module $U$.

However I have not seen this result in the introductory chapters of the book on representations I am studying. Which makes me think that either this result is not very useful, or that I have made a mistake somewhere. So is what I conjectured above true?

If $U$ is  irreducible $\mathbb CG$-module and $u\in U$, does there exists a basis for $U$ consisting of vectors of the form $u,ug_1,\dots,ug_n$, where $g_1,\dots,g_n\in G$?

答:
Sure.

If $u$ is any non-zero vector of an irreducible module $V$, then the subspace spanned by $X=\{gu:g\in G\}$ is a non-zero submodule. Irreducibility implies then that $X$ spans $V$ and therefore linear algebra tells you that a subset of $X$ is a basis $U$.


Title: Proving $\displaystyle\prod_{k=1}^n(1+a_{k})\leq e$

Body:
If $a_{k}>-1$, $\forall 1\leq k\leq n$ such that :
$$
\sum_{k=1}^{n}\frac{a_{k}}{1+a_{k}x}\leq1\;\; ,\forall x\in[0,1]
$$
Then how can we show that :
$$
\prod_{k=1}^{n}(1+a_{k})\leq e
$$

Body:
Note that :
$$
\int_{0}^{1}\sum_{k=1}^{n}\frac{a_{k}}{1+a_{k}x}\;\text{d}x=\sum_{k=1}^{n}\ln(1+a_{k})=\ln\left(\prod_{k=1}^{n}(1+a_{k})\right)\leq1
$$
Therefore :
$$
\prod_{k=1}^{n}(1+a_{k})\leq e
$$


问:
Proving that $X_n= 2^n 1_{(0,\frac{1}{n})}, n \geq 1 $ converges to zero in probability
Let $\Omega =[0,1],$ the sigma algebra: $F=B(\mathbb{[0,1]})$ and $\mathbb{P}=m$ (Lebesgue measure)
$X_n$ is a random variable: $X_n= 2^n 1_{(0,\frac{1}{n})}, n \geq 1 $

Prove directly, that is,  using the definition, that $X_n \overset{p}\to 0$

The definition of convergence in probability states that If $X, X_n$ are random variables   $\forall \varepsilon \geq 0, \mathbb{P}(|X_n| > \varepsilon) \overset{n \to \infty}\to 0$

My lecture notes give this short proof, that I am struggling to understand:

$\forall \varepsilon \geq 0, \mathbb{P}(|X_n| > \varepsilon)=\mathbb{P}(X_n=2^n)=\mathbb{P}((0,\frac{1}{n}))=\frac{1}{n} \overset{n \to \infty}\to 0$

I don't understand how they go from $\mathbb{P}(|X_n| > \varepsilon)$ to $\mathbb{P}(X_n=2^n)$ to $ \mathbb{P}((0,\frac{1}{n}))$. I trying to separate it into cases according to the value of $\varepsilon$, but I have to consider that the indicator function eventually becomes zero for sufficiently large n and at a fixed $\omega \in \Omega$. Can someone explain it in detail?

答:
Notice that for a fixed $n \in \mathbb{N}$,  $X_n(\omega) > 0$ if and only if $\omega \in (0, \frac{1}{n})$. Thus for any $\epsilon > 0$, $P(|X_n| > \epsilon)= P((0,\frac{1}{n})) = \frac{1}{n}$.

Edit: changed $x$ to $\omega$ to make notation a bit clearer


问:
How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?
Just a little bit unsure about this question:

How many 7 letter 'words' can be made from the English alphabet which contains at least 3 vowels?

And by 'words' I mean just a string of alphabet - does not have to be a real world in dictionary. I do know that you need to use Inclusion/Exclusion principle which says:

$|A \cup B| = |A| + |B| - |A\cap B|$

Let U be the set of all words containing 7 letters (26 letters to the power 7)
The only thing I'm not sure is what would be the sets A and B? Or do I need a third term C?

答:
It would be easier to count the number of words that have $2$ or fewer vowels, and then subtract this number from the total number of words (which you have already computed).

I assume that by "$3$" or more vowels" you mean $3$ or more occurrences of vowels, so in particular a word with 2 e's, 2 i's, and the rest consonants qualifies.

How many words are there with no vowels? Clearly
$$21^7$$
 if, as per usual convention, we agree that there are $5$ vowels.

How many words with $1$ vowel?  Where the vowel occurs can be chosen in $\binom{7}{1}$ ways. For each of these ways, the vowel can be chosen in $5$ ways.  And once you have done that, the consonants can be filled in in $21^6$ ways, for a total of
$$\binom{7}{1}(5)(21^6)$$

Finally, how many with $2$ vowels? The location of the vowels can be chosen in $\binom{7}{2}$ ways.  Once this has been done, the actual vowels can be put into these places in $5^2$ ways.  And then you can fill in the consonants in $21^5$ ways, for a total of
$$\binom{7}{2}(5^2)(21^5)$$

Add up the $3$ numbers we have obtained, subtract from $26^7$.

Our argument was a little indirect. We could instead find, using the same sort of reasoning, the number of words with $3$ vowels, with $4$ vowels, with $5$, with $6$, with $7$, and add up.  This is only a little more work than the indirect approach.  But any saving of work is helpful!  Also, the indirect approach that was described lets us concentrate on pretty simple situations, the most complicated of which is the $2$ vowel case.

Remark: The calculation we have done is closely connected to the Binomial Distribution, and if you have already covered this, it may be the point of the exercise.  So if you know about the Binomial Distribution, imagine the letters to be chosen at random.  Then the number of patterns with $3$ or more vowels is the probability of $3$ or more vowels, multiplied by $26^7$.


问:
How is the SVD of an item document matrix - $A$'s component equal to the eigenvector matrix of $AA^T$?
I'm trying to get an intuitive grip on SVDs to explore LSI as part of NLP. I understand that it involves the usage of SVDs. As I understand in SVD

$A = SΣU^T$

I understand that $S$ is a matrix of eigenvectors of $A^TA$ and $\Sigma$ is the square root of its eigenvalues as a diagonal matrix.

I don't understand how $U$ is the matrix of eigenvectors of $AA^T$

In LSI, we decompose this SVD matrix to get a shorter vector for each term and document to find out the relative distance between those vectors and the centroid

I don't understand how doing an SVD converts words to their own vectors, by which we can estimate the distance between documents

I'm very new to NLP, I'm finding it hard to get an intuitive sense of the math, I'd sincerely appreciate any help. Is there any books that I can refer to, thanks!

答:
I don't understand how U is the matrix of eigenvectors of A* A(trans)

Correction: $U$ is the matrix of eigenvectors of $AA^T$. If you want to understand this in general read on. Or if your interest in incidental of LSA, then skip to next section.

In SVD, we decompose a matrix $X$ three matrice each with useful properties.

$X = USV^T$. Here $U,V$ are unitary matrices: $U^T = U^{-1}$ and same for $V$. $S$ is a diagonal matrix.

The only property we need is that $U,V$ are unitary matrices.

I'm assuming real matrices every where.

Let's start with $XX^T$:

(please note the dimensions at each step as indicated)

$XX^T = USV^T (USV^T)^T$

$XX^T = USV^T_{n \times n} V_{n \times n}S^TU^T$

Here we use $V^TV = I$:

$XX^T = USI_{n \times n} S^T_{n \times m}U^T$

$I_{n \times n} S^T_{n \times m} = S^T$. Notice that $S$ is a diagonal matrix, so $SS^T$ is also diagonal:

$XX^T = U SS^T U^T$

$XX^T = U SS^T U^{-1}$

This is the same expression as the eigen-decomposition.

You can try a similar chain of proof for $X^TX$ to see that $V$ is a set of eigenvectors of $X^TX$.

Distance between documents

In LSI, we decompose this SVD matrix to get a shorter vector for each term and document to find out the relative distance between those vectors and the centroid

I think you mean "approximate" instead of "decompose."

I don't understand how doing an SVD converts words to their own vectors, by which we can estimate the distance between documents

SVD does not convert the words into vectors. The second part of your sentence is correct though: we can estimate distance between documents by using SVD and approximating it.

This is my understanding: Say you have a set of $n$ documents. To represent them in a mathematically, you strip out the common words such as the, a, here, there. In other words, these are words that you believe can appear in all these documents and have no significance.

After this, you count the remaining words in the document and create a column vector, where each word has a score (the number of times it occurs in this document). So far you have represented $1$ document as a column vector.

Now you do this for all the $n$ documents. Now stack them side by side and you have a matrix, while making sure that each word gets a row. Call this matrix $C$.

At this point what you have is a massive matrix $C$ which is humanly (and computationally) difficult to handle. This is where SVD is useful. It can decompose this matrix into three matrices each of which has an useful property.

$C = USV^T$

Here $S$ is a diagonal matrix. The diagonal elements will have different values, but all positive in this case.

Now you can think of approximating. Say you notice that most diagonal elements of $S$ close to zero. You set them to zero. Now with this thresholding done, you can reconstruct a matrix $C'$ by multiplying the three matrices again.

Mathematically, this $C'$ is an approximation of $C$.

More importantly, after we set threshold the diagonal elements of $S$, we'll have several columns of $S$ that are entirely zero. This means the corresponding rows in $V^T$ are actually useless because they do not contribute to the approximated $C'$.

Let's drop those irrelevant rows from $V^T$ and call this new matrix $V_{truncated}^T$.

$V_{truncated}^T$ contains the same number of columns as $C$ but very few rows. This means we started with $n$ documents and maybe a million rows but now we have $n$ documents and only a handful of rows in $V^T_{truncated}$.

In fact we could threshold the diagonal elements harshly leading to only $2$ rows in our $V^T_{truncated}$.

The product $SV^T_{truncated}$ is a low dimensional representation of our original documents. At this point we could easily draw the low-dimensional documents as 2D points on a piece of paper. Close together points represent similar documents or as you put it "distance between documents."

Some more material

Some intuitive answers here. After that, this and this explain how approximating the SVD achieves in this application i.e. LSI.


问:
Find limit of formula with binomial coefficcients

The Problem

Given:

$$f(n) = \frac{\sum_{i=1}^{n/2} \left[ \binom{n/2}{i} \sum_{j=0}^{i} \binom{n/2}{j} \right]}{2^n}$$

I want to prove that $\lim_{n \rightarrow \infty}f(n) = \frac{1}{2}$.

This relates to some research work I am doing and I am pretty sure it converges to $1/2$ from 'above' (or at least the data suggests that) but I would really like to have some solid proof about this limit.

What I've tried

I've tried to sandwich the limit by substituting the term $\sum_{j=0}^{i} \binom{n/2}{j}$ with a minimum value, i.e. $\binom{n/2}{0}+\binom{n/2}{1}=1+\frac{n}{2}$ (common to all subterms), which leads to $\lim_{n \rightarrow \infty}f(n) = 0$ and with a maximum value, i.e. $\sum_{j=0}^{n/2} \binom{n/2}{j}=2^{n/2}$ (adding extra subterms), which leads to $\lim_{n \rightarrow \infty}f(n) = 1$, but well, I need something better :) I've tried using the some other related formulas but couldn't find something better than that. Maybe it can be solved more straightforward though?

答:
Let $m=n/2$ for ease of notation and let a symbol $\mathbf x$ in bold denote $\binom mx$. Let's write out the numerator for $m=4$:
$$\begin{align}&\mathbf{1(0+1)}+\\
&\mathbf{2(0+1+2)}+\\
&\mathbf{3(0+1+2+3)}+\\
&\mathbf{4(0+1+2+3+4)}
\end{align}$$
By the symmetry of binomial coefficients, the above expression is equal to
$$\begin{align}\mathbf{(0+1+2+3+4)0}&\\
+\mathbf{(1+2+3+4)1}&\\
+\mathbf{(2+3+4)2}&\\
+\mathbf{(3+4)3}&
\end{align}$$
If we add the two together we get
$$\begin{align}
&\mathbf{0(0+1+2+3+4)}+\\
&\mathbf{1(0+1+2+3+4+1)}+\\
&\mathbf{2(0+1+2+3+4+2)}+\\
&\mathbf{3(0+1+2+3+4+3)}+\\
&\mathbf{4(0+1+2+3+4)}
\end{align}$$
$$=(\mathbf{0+1+2+3+4})^2+\mathbf 1^2+\mathbf 2^2+\mathbf 3^2=2^{2×4}+\mathbf 1^2+\mathbf 2^2+\mathbf 3^2$$
It is easy enough to derive the generalisation:
$$f(m)=\frac{(1/2)(2^{2m}+\sum_{k=1}^{m-1}\mathbf k^2)}{2^{2m}}$$
As $n\to\infty$ (and hence $m\to\infty$ too), $\sum_{k=1}^{m-1}\mathbf k^2$ does not grow as fast as $2^{2m}$ – it is smaller by a factor of $\sqrt{\pi m}$ – so it becomes negligible:
$$\lim_{m\to\infty}f(m)=\lim_{m\to\infty}\frac{(1/2)2^{2m}}{2^{2m}}=\frac12$$


Body:
How to put $(-\sqrt{3}-i)^{\frac{5}{7}}$ into polar form and find all roots.

What I tried:

$$w = -\sqrt{3}-i$$
$$\arg(w)=\arctan(\frac{1}{\sqrt 3})-\pi = \frac{\pi}{6} - \pi = \frac{-5\pi}{6}$$
$$w = 2(\cos(\frac{-5\pi}{6})+ i\sin(\frac{-5\pi}{6})) $$
$$w^5 = 2^5(\cos(-\frac{5\pi}{6}*5)+i\sin(-\frac{5\pi}{6}*5)) $$
$$z^7 = w^5$$
$$z = w^{\frac{5}{7}} $$
$$z = 32^{\frac{1}{7}}(\cos(\frac{-5\pi*5}{6*7}+\frac{2k\pi}{7})+i\sin(\frac{-5\pi*5}{6*7}+\frac{2k\pi}{7})) $$
Here I get stuck and I don't know how to continue...

Body:
you have :

$$(-\sqrt{3}-i)^{\frac{5}{7}}=\frac{1}{2^{5/7}}\left(\frac{-\sqrt{3}}{2}-\frac{1}{2}i\right)^{5/7}=\frac{1}{2^{5/7}} \left(\exp{\frac{i7\pi}{6}}\right)^{5/7}=\frac{1}{2^{5/7}}\exp{\frac{i5\pi}{6}} $$


问:
Show that a sequence of functions that converge uniformly are Riemann integrable. What if they only converge point wise?
Let $f_n$ be a sequence of Riemann integrable functions on $[a,b]$ that converge uniformly to a function f. Show that f is also Riemann integrable. What happens if $f_n$ only converges pointwise?

Considering this scenerio, show that

$$\lim_{n \to \infty} \int_{a}^{b} f_n(x) dx = \int_{a}^{b} f(x) dx$$

Not really sure how to start this problem. If they already converge to a function, uniformly, that function has to be continuous right? So then it is trivially Riemann integrable. The pointwise case I'm not sure about. And then the second part with the limits I'm not sure how to go about. Any help is appreciated!

答:
We can use the Riemann criterion to prove that the uniform limit $f$ of a sequence of Riemann integrable functions $(f_n)_n$ is also Riemann integrable.

By uniform convergence, for all $\epsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $n \geqslant N$ we have

$$-\frac{\epsilon}{3(b-a)} < f(x) - f_n(x) < \frac{\epsilon}{3(b-a)}$$

Let $P: a = x_0 < x_1 < \ldots < x_n = b$ be a partition. Since $f(x) = f(x) - f_n(x) + f_n(x),$ it follows that on any partition subinterval $I$,

$$\sup_I f(x) \leqslant \sup_I(f(x) - f_n(x)) + \sup_I f_n(x) < \frac{\epsilon}{3(b-a)}+ \sup_I f_n(x), \\ \inf_I f(x) \geqslant \inf_I(f(x) - f_n(x)) + \inf_I f_n(x) > -\frac{\epsilon}{3(b-a)}+ \inf_I f_n(x).$$

Thus, $ \inf_I f_n(x)- \frac{\epsilon}{3(b-a)} <\inf_I f(x) \leqslant \sup_I f(x) <  \sup_I f_n(x)+ \frac{\epsilon}{3(b-a)}.  $

Summing over all partition subintervals we get for upper and lower Darboux sums,

$$U(f,P) < \frac{\epsilon}{3} + U(f_n,P), \quad -L(f,P) < \frac{\epsilon}{3} - L(f_n,P),$$

and, hence,

$$U(f,P) - L(f,P) < \frac{2\epsilon}{3} + U(f_n,P) - L(f_n,P).$$

Since $f_n$ is Riemann integrable, there is a partition $P$ such that $U(f_n,P) - L(f_n,P) < \epsilon/3$ and it follows that  $U(f,P) - L(f,P) < \epsilon$ proving that $f$ is Riemann integrable.

Now you should be able to prove on your own that the limit of the sequence of integrals is the integral of the limit function by considering that  $|f_n(x) - f(x)| \to 0$ uniformly for all $x \in [a,b]$.


问:
$L^{2}(0,T; L^{2}(\Omega))=L^{2}([0,T]\times\Omega)$?
Suppose I have a function $u\in L^{2}([0,T]\times\Omega)$ for some bounded domain $\Omega$ in $\mathbb{R}^{2}$ or $\mathbb{R}^{3}$. I managed to prove that this implies $u\in L^{2}(0,T; L^{2}(\Omega))$. Now I want to show that opposite is not true. This is where I struggle.
According to Fubini's theorem, it is sufficient to give example of a function $u\in L^{2}(0,T; L^{2}(\Omega))$ which violates condition $u\in L^{2}(\Omega; L^{2}(0,T))$.
So could you please provide such function?
Thanks to all!

Let me show you how I see space $L^{2}(0,T; L^{2}(\Omega))$. The problem with it is that I couldn't find a single definition in any book whose rigorousness satisfied me.
So, as far as I understand, $L^{2}(0,T; L^{2}(\Omega))$ is a set of functions $\{f : [0,T]\times\Omega\to\overline{\mathbb{R}}|\text{for almost all } t\in[0,T] f(x,t)\in L^{2}(\Omega)|\int_{0}^{T}\int_{\Omega}f^{2}dxdt < +\infty\}$ and such that for any pair of functions $f, g$ from this set the function $\int_{\Omega}fgdx$ is measurable in time. The last condition is very crucial and is omitted everywhere like it's obvious or something. Note that condition that the integral is finite is well-defined, since $f^{2}$ is non-negative and to define Lebesgue integral for it you do not even have to make $f^{2}$ measurable. Also, integral $\int_{\Omega}fgdx$ is also well-defined for alsmot all $t$, since the integral of the absolute value is bounded via Holder's inequality and $f, g$ are assumed measurable since they are in $L^{2}$, so $\int_{\Omega}fgdx$ exists and is finite.

It troubles me that they keep using both $L^{2}([0,T]\times\Omega)$ and $L^{2}(0,T; L^{2}(\Omega))$ in literature. If these spaces are indentical, why different notations? So I assumed that $L^{2}(0,T; L^{2}(\Omega))$ is wider than $L^{2}([0,T]\times\Omega)$, which is why I ended up here at stackexchange.

My main concern is with PDEs. So, suppose I have some time-dependent second-order PDE in a form $Lu = f$. To make it simple, assume $u = 0$ on the boundary of $\Omega$ and $u(0) = 0$, $u_{t}(0) = 0$. You can think of $L$ as a wave operator. Now, if we multiply the equation by test function and integrate over entire $[0,T]\times\Omega$, we obtain some weak formulation. But we can instead integrate over $\Omega$ only and assume the integral relation holds for almost all $t$. Naturally, the test functions in both cases will be from different spaces, so are solutions we are looking for. One is, say, $L^{2}([0,T]\times\Omega)$, the other is $L^{2}(0,T; L^{2}(\Omega))$. So, the question is: if in both formulations we avoid all sorts of integrations by parts ( or do them similarly ), will we obtain the same existence-uniqueness results for $u$, as well as the same regularity results? I searched tons of literature, and looks like the only formulation people are interested in is the one with entire integration over $[0,T]\times\Omega$. But they keep using the second one when obtaining some a priori estimates in the papers devoted to Finite Element approximations. Which really pisses me off, since they don't bother proving existence and uniqueness for the second formulation.

答:
Without any literature I am not sure that everything I write down, is easily to understand. However this can bring some light into your thoughts...:-)
I directly follow the proof of Etienne Emmrich in his book
"Gewoehnliche und Operator-Differentialgleichungen".

Theorem: Let $1\leq p<\infty$ be given and $a,b\in\mathbb{R}$. Then we have
$$ L^{p}(0,T;L^{p}(a,b))\cong L^{p}((a,b)\times (0,T)).$$

For a proof of this result we need the following Lemma:

Lemma: Let $\tilde{u}=\tilde{u}(t):[0,T]\rightarrow L^{p}(a,b)$ be Bochner measurable $(1\leq p\leq\infty)$. Then the real valued function $u=u(x,t):=[\tilde{u}(t)](x)$ is Lebesgue measurable over $(a,b)\times (0,T)$.

Proof: Since $\tilde{u}$ is Bochner measurable there exists a sequence of simple functions $\{\tilde{u}_{n}\}$ where we have
$$\int_{a}^{b}|u_{n}(x,t)-u(x,t)|^{p}\rightarrow 0\ \hbox{ for } n\rightarrow\infty$$
and for the case $p=\infty$ the analogous result with the $L^{\infty}$-norm
for $t\in[0,T]$ a.e. Thereby we used $u_{n}(x,t):=[\tilde{u}_{n}(t)](x)$. Then there exists a subsequence, such that for $x\in[a,b]$ a.e. and $t\in[0,T]$ a.e.
we otain
$$|u(x,t)-u_{n'}(x,t)|\rightarrow 0\ \hbox{ for } n'\rightarrow\infty.$$
The functions $u_{n'}$ are measurable over $(a,b)\times(0,T)$ by construction, since the have only finitely many values with respect to $t$ and for every $t$ they are Lebesgue measurable with respect to $x$.
So we constructed a sequence of Lebesgue measurable functions converging a.e. to the function $u$. hence $u$ is Lebesgue measurable. $\square$

Let us come back to our "main" Theorem. We give the following proof:
Firstly we show $$L^{p}(0,T;L^{p}(a,b))\supseteq L^{p}((a,b)\times(0,T)).$$
For that let $u\in L^{p}((a,b)\times(0,T))$ be given. We define the "abstract" functio  $\tilde{u}$ by $[\tilde{u}(t)](x):=u(x,t)$. Since $u(\cdot,t)\in L^{p}(a,b)$ we also get $\tilde{u}$ maps into $L^{p}(a,b)$ as well.
Now pick $f\in (L^{p}(a,b))^{*}\cong L^{q}(a,b)$ ($\tfrac{1}{p}+\tfrac{1}{q}=1 \hbox{ and } q=\infty\hbox{ for } p=1$). Then
$$(x,t)\mapsto f(x)u(x,t)\in L^{1}((a,b)\times (0,T))$$ and by Fubini's Theorem the map
$$t\mapsto \int_{a}^{b}f(x)u(x,t)\ dx =\langle f,\tilde{u}(t)\rangle$$
on $[0,T]$ Lebesgue integrable such that also $t\rightarrow \langle f,\tilde{u}(t)\rangle$ Lebesgue measurable.
This is the weak measurability of $\tilde{u}$. Since $L^{p}$ is seperable for $1\leq p<\infty$ we have the Bochner measurability of $\tilde{u}$. $\tilde{u}$ is also Bochner integrable since by Fuibini's theorem also the map
$$t\rightarrow \int_{a}^{b}|u(x,t)|^{p}\ dx=\|\tilde{u}(t)\|^{p}_{L^{p}}$$
is Lebesgue integrable on $[0,T]$.

It remains to show $$L^{p}(0,T;L^{p}(a,b))\subseteq L^{p}((a,b)\times(0,T)).$$
Let $\tilde{u}\in L^{p}(0,T;L^{p}(a,b))$. We again define the real valued function $u(x,t):=[\tilde{u}(t)](x)$. The iterated integral
$$
\int_{0}^{T}\left(\int_{a}^{b}|u(x,t)|^{p}\ d x\right) \ dt<\infty
$$
exists. Since the function to integrate over is non negative, we can conclude that also
$$\int_{(a,b)\times(0,T)}|u(x,t)|^{p}\ d (x,t)$$ exists and is finite, if and only if $u$ is Lebesgue measurable on $(a,b)\times(0,T)$. This is the result of the aforementioned lemma. Finally
$$
\int_{0}^{T}\left(\int_{a}^{b}|u(x,t)|^{p}\ d x\right) \ dt=\int_{(a,b)\times(0,T)}|u(x,t)|^{p}\ d (x,t)
$$ and thus the norms are identical. $\square$


问:
Solve $x^TAx=k$ with $k=$cte?
How can we solve the following equation:
$$
x^TAx=k
$$
if $k$ = a constant and A is a positive-definite and symmetric matrix, $x$ is a vector. I can clearly see that in 3d we have an infinite set of solutions but I cannot formulate it in a mathematical way.
Thanks

答:
After diagonalization of the matrix (which is a rotation of the basis), the equation becomes

$$\lambda u^2+\mu v^2+\nu w^2=1,$$

which is the equation of an ellipsoid, as $\lambda,\mu,\nu>0$. (If you make $w$ constant, i.e. cut with a plane, you should recognize the equation of an ellipse.)

Title: Solve $x^TAx=k$ with $k=$cte?


问:
Radius of curvature at given point
The value of radius of curvature for the curve $$s = ce^{x/c}$$ at $(s,\psi)$

I know that radius of curvature is $$ \frac{(1+y_1)^2}{y_2}$$ how to apply these formula here

答:
Given $y=c e^{x/c}$ the curvature in cartesian coordinates is
$$\kappa(x)=\frac {|y''|}{\left(1+{y'}^{2}\right)^{\frac {3}{2}}}$$
So in this case
$$\kappa(x)=\frac{e^{x/c}}{\left| c\right|  \left(e^{\frac{2 x}{c}}+1\right)^{3/2}}$$
radius of curvature is
$r(x)=\frac{1}{\kappa(x)}$.

In this case
$$r(x)=\left| c\right|  e^{-\frac{x}{c}} \left(e^{\frac{2 x}{c}}+1\right)^{3/2}$$

Title: Radius of curvature at given point


问:
Indeterminate terms in a series
If a series has some kind of indeterminacy in any of its terms, is that enough to conclude that the series diverges? For example

$$\sum_{k=1}^{\infty} \left(-\frac{1}{ln(k)}\right)^k$$

at $k = 1$, we have an indeterminacy.

I know that if this series starts at $k=2$, it converges. So, my doubt is about the correctness of declaring that the series starting with $k = 1$ diverges by the presence of indeterminacy, or converges because from $k = 2$, the series converges.

答:
In such cases, you cannot say that the series converges or diverges. Both words have a precise meaning.

This series is just not defined.

Title: Indeterminate terms in a series


问:
How to prove $4 \cos(x) + x \sin(x) \geq 4 - x^2$ for all real $x$, with equality if and only if $x=0$?

Show $4 \cos(x) + x \sin(x) \geq 4 - x^2$ for all real $x$, with equality if and only if $x=0$.

LHS-RHS has a power series $x^6/(2×6!)−x^8/(4×8!)+x^{10}/(6×10!)−\ldots$, and I could try to prove that this is positive. But it seems messy. I was hoping for some cleverer trick, but I'm out of ideas right now.

答:
Let $h(x)=4\cos(x)+x\sin(x)+x^2-4$; note that $h(x)$ is even. For reference, we have
\begin{align}
h^\prime(x) &=2x+x\cos(x)-3\sin(x)\\
h^{\prime\prime}(x)& = 2-2\cos(x)-x\sin(x)\\
h^{(3)}(x)& =\sin(x)-x\cos(x)\\
h^{(4)}(x) &= x \sin(x)\\
h^{(5)}(x)& = \sin (x)+x \cos (x)\\
h^{(6)}(x)& =2 \cos (x)-x \sin (x)
\end{align}

Claim: It suffices to show $h(x)\ge 0$ on $[0,\pi]$.

We have $h'(x) = x(1+\cos(x)) + (x-3\sin(x))$. For $x>0$ the first term is non-negative and for $x>\pi$ the second one is as well (the unique positive real solution of $x=3\sin(x)$ is $x\approx 2.28$ but we take $x=\pi$ for convenience). In other words, for $x>\pi$ we have $h'(x)>0$ and $h(\pi)>0$.

Claim: $h'(0)=0$ (obvious) and $h$ has its global minimum at zero.

This is basically a corollary to the Second Derivative Test. Note that $h''''(x)\ge 0$ on $[-\pi,\pi]$ as the signs of the terms agree, which implies $h''$ has a minimum at $x=0$, i.e. that $h''(x)\ge 0$ on $[-\pi,\pi]$. Alternatively, since the sixth derivative is the first to evaluate to a nonzero number at zero and this value is positive, $h$ has a minimum at $x=0$.


问:
How to prove the correspondence between closed points of the spectrum of a ring and its maximal ideals?
Let $R$ be an entire ring.
How does one prove $\mathfrak{p}$ is a closed point of $\text{Spec} R$ if and only if $\mathfrak{p}$ is a maximal ideal of $R$?

答: By definition $\mathfrak{p}$ is a closed point iff there is no other prime ideals containing $\mathfrak{p}$ other than itself. But every prime ideal (in fact, every ideal) must be contained in some maximal ideal, and maximal ideals are prime.


问:
If $S \times \Bbb{R}^k$ is homeomorphic to $T \times \Bbb{R}^k$ and $S$ is compact, can we conclude that $T$ is compact?
If $S \times \Bbb{R}^k$ is homeomorphic to $T \times \Bbb{R}^k$ for some $k \geq 1$ and $S$ is compact, can we conclude that $T$ is compact?

The case where $k=1$ and $S$ and hence also $T$ are locally connected is dealt with in my answer to this question. I am now curious about the general case.

答:
Yes, if $S$ is compact, then $T$ is also compact.

In fact, the property that $S$ is compact can be expressed in terms of the topological space $X=S\times\mathbb{R}^ k$. The idea is natural enough, if $S$ is compact then it is possible to remove a compact set from $X$ which stops certain maps from the $(k-1)$-sphere $\mathbb{S}^{k-1}$ from being null-homotopic. The precise statement is a little tricky though.

The following are equivalent.

$S$ is compact.

There exists a compact $A\subseteq X$ such that, for every compact $B\supseteq A$ there exists a compact $C\supseteq B$ such that, for every $x\in X\setminus C$, there exists a continuous $f\colon \mathbb{S}^{k-1}\to X\setminus B$ containing $x$ in its image and which is null-homotopic as a map to $X$ but not null-homotopic as a map to $X\setminus B$.

First, suppose that 1 is true. We need to construct the sets $A$, $C$ and the function $f$ for any choices of $B$ and $x$.
Set $A=S\times\lbrace0\rbrace$. Given any compact $B\supseteq A$, let $\pi_{\mathbb{R}^k}(B)$ be its projection onto $\mathbb{R}^k$. As this is compact, it is contained in the closed ball $\bar B_R(0)$ for some $R > 0$. Set $C=S\times\bar B_R(0)$. Then, given any $x\in X\setminus C$, we can write $x=(s_0,y_0)$ where $\lVert y_0\rVert = r > R$. Define $f\colon \mathbb{S}^{k-1}\to X$ by $f(a)=(s_0,ra)$. This is null-homotopic as a map to $X$ via the homotopy $F(a,u)=(s_0,(1-u)ra)$. On the other hand, it is not null-homotopic as a map to $X\setminus B$. To see this, define $g\colon X\setminus B\to \mathbb{S}^{k-1}$ by $g(s,y)=y/\lVert y\rVert$. Then, $g\circ f$ is the identity on $\mathbb{S}^{k-1}$. However, it is standard that the (k-1)-sphere is not contractible or, equivalently, the identity on $\mathbb{S}^{k-1}$ is not null-homotopic, implying that $f$ is not null-homotopic.

Conversely, suppose that $S$ is not compact. We need to prove that 2 is false. To be explicit, the negation of 2 can be written as

2'. For any compact $A\subseteq X$ there exists a compact $B\supseteq A$ such that, for any compact $C\supseteq B$, there exists an $x\in X\setminus C$ such that every continuous $f\colon\mathbb{S}^{k-1}\to X\setminus B$ with $x$ in its image and which is null-homotopic as a map to $X$ is also null-homotopic as a map to $X\setminus B$.

So, we need to construct the set $B$ and point $x\in X\setminus C$ for any choice of $A$ and $C$, and show that any such function $f$ is null-homotopic as a map to $X\setminus B$.

Given a compact $A\subseteq X$, its projection onto $\mathbb{R}^k$ is compact, so contained in the closed ball $\bar B_R(0)$ for some $R > 0$. Also, its projection $S_1=\pi_S(A)$ is compact. Take $B=S_1\times\bar B_R(0)$. If $C$ is a compact set containing $B$ then the projection $\pi_S(C)$ is compact, so is not the whole of $S$. Therefore, we can choose $s_0\in S\setminus\pi_S(C)$. Set $x=(s_0,0)\in X\setminus C$.

Now, suppose that $f\colon\mathbb{S}^{k-1}\to X\setminus B$ has $x$ in its image and is null-homotopic as a map to $X$. We need to construct a null-homotopy for $f$ as a map to $X\setminus B$. Write $f(a)=(f_1(a),f_2(a))$ where $f_1,f_2$ are maps from $\mathbb{S}^{k-1}$ to $S$ and $\mathbb{R}^k$ respectively. First, there will exists continuous maps $f^\prime_2\colon\mathbb{S}^{k-1}\to\mathbb{R}^k\setminus\lbrace0\rbrace$ which are arbitrarily close to $f_2$. One way to see this is to first approximate $f_2$ by a smooth function. Then, as the image of a smooth map from $\mathbb{S}^{k-1}$ to $\mathbb{R}^k$ cannot contain any open ball about the origin (its image must have zero measure), there will be arbitrarily small $v\in\mathbb{R}^k$ such that $f^\prime_2=f_2+v$ does not pass through the origin. Then, $f$ will be homotopic to $(f_1,f_2^\prime)$ for $f_2^\prime$ close enough to $f_2$. Next, by scaling, $f$ is homotopic to $(f_1,rf^\prime_2)$ for any $r\ge1$. So, by scaling $f_2^\prime$, we can assume that $\lVert f^\prime_2\rVert > R$.

We have shown that $f$ is homotopic to $(f_1,f^\prime_2)$ where $f^\prime_2\colon\mathbb{S}^{k-1}\to\mathbb{R}^{k-1}\setminus\bar B_R(0)$. Next, by projection onto $S$, the condition that $f$ is null-homotopic as a map to $X$ implies that $f_1$ is null-homotopic. Also, as $f$ has $(s_0,0)$ in its image, $f_1$ has $s_0$ in its image so is homotopic to the constant map to $s_0$. This gives a homotopy of $f$ to the map $a\mapsto(s_0,f^\prime_2(a))$. Finally, by scaling, $f^\prime_2$ is homotopic to the constant map to $0$. So, we have constructed a null-homotopy for $f$ (as a map to $X\setminus B$) to the constant map $a\mapsto(s_0,0)$.


答:
Let $\mathfrak{p}$ be a prime ideal of $R$. The required preliminary result is to know the closure $\overline{\lbrace \mathfrak{p} \rbrace}$. Can you guess? Do you see a closed subset wrt the Zariski topology that obviously contains $\lbrace \mathfrak{p} \rbrace$?

Actually, the set of all prime ideals of $R$ containing $\mathfrak{p}$, usually denoted $V(\mathfrak{p})$, is the closure of $\lbrace \mathfrak{p} \rbrace$.

Let $\mathfrak{p}$ be a closed point, hence $\lbrace \mathfrak{p} \rbrace = \overline{\lbrace \mathfrak{p} \rbrace}$. Recall there exists a maximal ideal $\mathfrak{m}$ such that $\mathfrak{p} \subseteq \mathfrak{m}$ and from our preliminary result one has
$$
V(\mathfrak{p}) = \overline{\lbrace \mathfrak{p} \rbrace} = \lbrace \mathfrak{p} \rbrace.
$$
Since a maximal ideal is prime, one concludes $\mathfrak{m} = \mathfrak{p}$.

Conversely, let $\mathfrak{p}$ be a maximal ideal. Then, one has $V(\mathfrak{p}) = \lbrace \mathfrak{p} \rbrace$. From our preliminary result one concludes $\lbrace \mathfrak{p} \rbrace = \overline{\lbrace \mathfrak{p} \rbrace}$, i.e. $\mathfrak{p}$ is a closed point.

Title: How to prove the correspondence between closed points of the spectrum of a ring and its maximal ideals?

Body:
Let $R$ be an entire ring.
How does one prove $\mathfrak{p}$ is a closed point of $\text{Spec} R$ if and only if $\mathfrak{p}$ is a maximal ideal of $R$?


答:
Hint:

By Taylor's theorem with the mean-value forms of the remainder,
$$f(x)=x^2+4\cos x + x \sin x -4 = \frac{x^6}{360}+(-3\sin(\zeta)-\zeta \cos(\zeta))\frac{x^7}{5040}\tag1$$
where $\zeta$ is between $0$ and $x$, and $f^{(7)}(x)=-3 \sin (x) - x \cos x$.

So if $|x|$ is small enough (e.g., |x| < 11) $(1)$ is non-negative. If $|x|>11$ ($11$ can be made smaller), we can easily prove $x^2+4\cos x + x \sin x -4>0$. Can you take it from here?

Title: How to prove $4 \cos(x) + x \sin(x) \geq 4 - x^2$ for all real $x$, with equality if and only if $x=0$?

Body:
Show $4 \cos(x) + x \sin(x) \geq 4 - x^2$ for all real $x$, with equality if and only if $x=0$.

LHS-RHS has a power series $x^6/(2×6!)−x^8/(4×8!)+x^{10}/(6×10!)−\ldots$, and I could try to prove that this is positive. But it seems messy. I was hoping for some cleverer trick, but I'm out of ideas right now.


Title: How to put $(-\sqrt{3}-i)^{\frac{5}{7}}$ into polar form

Body:
It's easier to express in exponential, rather than cis, form.  Considering only the normalized complex number for brevity:

$$\begin{align*}
  w &= \text{e}^{(-\frac{5}{6} \pi +2 \pi k) i} \\
  w^5 &= \text{e}^{(-\frac{25}{6} \pi + 10 \pi k) i} = \text{e}^{(-\frac{25}{6} \pi + 2 \pi k) i} \\
  z^7 &= w^5 \\
  w^{\frac{5}{7}} &= \text{e}^{(-\frac{25}{42} \pi + \frac{2}{7} \pi k) i} \\
\end{align*}
$$

Now we enumerate through the 7 unique roots of unity, these are:

$$
\begin{align*}
  w^{\frac{5}{7}}_0 &= \text{e}^{-\frac{25}{42} \pi i} \approx -0.294755 - 0.955573 i \\
  w^{\frac{5}{7}}_1 &= \text{e}^{(\frac{2}{7} -\frac{25}{42}) \pi i}  = \text{e}^{-\frac{13}{42} \pi i} \approx 0.56332 - 0.826239 i \\
  w^{\frac{5}{7}}_2 &= \text{e}^{(\frac{4}{7} -\frac{25}{42}) \pi i}  = \text{e}^{-\frac{1}{42} \pi i} \approx 0.997204 - 0.0747301 i \\
  w^{\frac{5}{7}}_3 &= \text{e}^{(\frac{6}{7} -\frac{25}{42}) \pi i}  = \text{e}^{\frac{11}{42} \pi i} \approx 0.680173 + 0.733052 i \\
  w^{\frac{5}{7}}_4 &= \text{e}^{(\frac{8}{7} -\frac{25}{42}) \pi i} = \text{e}^{\frac{23}{42} \pi i} \approx -0.149042 + 0.988831 i \\
  w^{\frac{5}{7}}_5 &= \text{e}^{(\frac{10}{7} -\frac{25}{42}) \pi i} = \text{e}^{\frac{5}{6} \pi i} \approx -0.866025 + 0.5 i \\
  w^{\frac{5}{7}}_6 &= \text{e}^{(\frac{12}{7} -\frac{25}{42}) \pi i} = \text{e}^{-\frac{37}{42} \pi i} \approx -0.930874 - 0.365341 i \\
\end{align*}
$$

Beyond this, we wind up with $\frac{14}{7} \pi$ which is just $2 \pi$ and so the roots repeat.

And here they are plotted:

Finally, just scale by the magnitude.


问:
Find the limit of the sequence defined by $\forall n\in \mathbb{N}^*,u_n = \frac{2n(2n-1)}{4n^2+1} u_{n-1}$ and $u_0 =1$
Let $(u_n)$ be the sequence defined by $\forall n\in \mathbb{N}^*,u_n = \frac{2n(2n-1)}{4n^2+1} u_{n-1}$ and $u_0 =1$. I would like to find another way to find the limit of $(u_n)$.

What I have done:

We can see that $(u_n)$ is a strictly positive decreasing sequence so it converges, and by induction we can show that :

$$\forall n \in \mathbb{N}, u_n = \displaystyle \prod_{k=1}^{n} \dfrac{2k(2k-1)}{4k^2+1}$$

And I end up with:

$$\forall n \in \mathbb{N}, u_n = \displaystyle \prod_{k=1}^{n} \dfrac{2k(2k-1)}{4k^2+1} \leq \prod_{k=1}^{n} \dfrac{2k(2k-1)}{4k^2} = \prod_{k=1}^{n} \dfrac{2k-1}{2k} $$

I managed to show that:

$$ \prod_{k=1}^{n} \dfrac{2k-1}{2k}  = \dfrac{(2n)!}{(2^nn!)^2} \stackrel{\text{stirling}}{\sim} \dfrac{1}{\sqrt{n \pi}} $$

Therefore $\displaystyle \prod_{k=1}^{n} \dfrac{2k-1}{2k} \underset{n\rightarrow +\infty}{\longrightarrow} 0$ and by the squeeze theorem we deduce that $\displaystyle \lim_{n\rightarrow +\infty }u_n=0 $.

Do you see another approach?

答:
I would proceed as follows: you already know that $u_n$ converges to a non negative limit.
Also you can write $$u_n = \prod_{k=1}^{n}\frac{2k(2k-1)}{4k^2+1} = \prod_{k=1}^{n}\left(1-\frac{2k+1}{4k^2+1}\right)$$
By a well-known theorem this product is divergent, since $\sum_{k=1}^{\infty}\frac{2k+1}{4k^2+1}$ is divergent. Moreover, as $u_n$ converges being monotone decreasing and bounded below, it follows that this product must diverge to $0$, which is your limit.

This might be useful:
Equivalence of convergence of a series and convergence of an infinite product
Even though this link refers to positive $a_n$'s, it's still valid when all the $a_n$'s are negative.


问:
Directional derivative identity
Let $f:U\subset\mathbb{R}^n\rightarrow\mathbb{R}^m$ where $U$ is an open set, $p,v \in \mathbb{R}^n$ with $v \neq \vec{0}$. Let's recall the definition of the directional derivative of $f$ at $p$ in the direction of $v$:
$$
\frac{\partial f}{\partial v}(p) = \lim_{t\rightarrow0}\frac{f(p+tv)-f(p)}{t}
$$
Taking in consideration a norm (euclidean for example) in our vector space over the reals ($\mathbb{R}^n,\mathbb{R},|.|_e$), how can I prove the following identity?
$$
\frac{\partial f}{\partial v}(p) = |v|_e\frac{\partial f}{\partial\frac{v}{|v|_e}}(p)
$$
I tried to use the formal ($\epsilon, \delta$) definition of limit but couldn't get much further.

Any kind of advice is going to be highly appreciated. Thanks!

答:
After some struggle I can see why this is true... It's just a simple substitution in the limit variable...
$$
\frac{\partial f}{\partial v}(p) = \lim_{t\rightarrow0}\frac{f(p+t v)-f(p)}{t} =  \lim_{t\rightarrow0}\frac{f(p+t \frac{|v|}{|v|}v)-f(p)}{t}=
$$
now let $t' = t|v|$, therefore
$$
=\lim_{t'\rightarrow0}\frac{f(p+t'\frac{v}{|v|})-f(p)}{\frac{t'}{|v|}} = |v|\lim_{t'\rightarrow0}\frac{f(p+t'\frac{v}{|v|})-f(p)}{t'} = |v|\frac{\partial f}{\partial \frac{v}{|v|}}(p)
$$

Title: Directional derivative identity


问:
Expectation of number of balls drawn when it depends on the amount of green balls drawn on the last step
We have 60 green balls and 40 white balls in an urn.

Let $X_n$ be the number of balls drawn on step n defined as $ X_1 = 10 $ and $X_n =$ the number of green balls drawn in step $n-1$. The balls are then returned to the urn.

I need to calculate $\mathbb{E}(X_n)$ and show that $\lim_{n \to +\infty} \mathbb{E}(X_n) = 0 $. Suggestion: Find a suitable r.v. to condition $X_n$.

Intuitively it makes sense to me that the limit would be 0 since with high probability we're drawing each balls each time but I can't seem to find anything not convoluted for $\mathbb{E}(X_n)$.

Here's my work:

Let $Y_n$ be the amount of green balls drawn in step n. So we have that $X_n = Y_{n-1}$. $X_2 = Y_1 \sim Bi(10,0.6)$ so $\mathbb{E}(X_2) = 6$.

In general $\mathbb{E}(X_n) = \sum_{k=0} ^{10} k \frac{6}{10} P(Y_{n-2} = k)$

But I'm not sure how to go forward since each step depends on all the steps before and it gets very convoluted.

Any help would be greatly appreciated.

答: The idea is the same as in this other answer. Instead of drawing only as many green balls as were drawn previously, perform the following operation ten times: draw until you get a white ball. Given this scheme, the probability that you get to step $n$, where you then draw a green ball, is $(3/5)^n$, so the expected number of green balls drawn in step $n$ of the original formulation is $10(3/5)^n$, i.e. $E(X_n)=10(3/5)^{n-1}$. It is clear that this goes to zero as $n\to\infty$.


问:
Verification of ring isomorphism
I've to verify that $E(\mathbb{Z})$ and $\mathbb{Z}$ are isomorphic. $E$ is defined as
$$E(\mathbb{Z}) = \{f: \mathbb{Z} \to \mathbb{Z} | f(x+y) = f(x) +f(y)\}$$

This means there exists a function $\phi$ such that
$\phi(f+g) = \phi(f) + \phi(g)$
and
$\phi(fg) = \phi(f)\phi(g)$
with $\phi(x) \in \mathbb{Z}$

This comes as end chapter verification exercise of rings and rings' isomorphism. Groups and homorphisms are not introduced yet. So I Couldn't simply devise an ad-hoc isomorphism for the task. With your suggestion that $f: ℤ → ℤ$ is a group homorphism and to pick $f(n) = nf(1)$, for all $n ∈ ℤ$, it is straightforward. I developed like this:

Let $f+g$ defined as $f(n)+g(n) = n(f(1) + g(1)) so φ(n(f(1) + g(1))) = f(1) + g(1);$

Let $f∘g$ defiend as $f∘g = f((g(n))) = n⋅g(1)⋅f(1), φ(f((g(n)))) = g(1)⋅f(1)$, thus $\phi$ is indeed a ring isomorphism.

I'm not sure if the notation $φ(n(f(1) + g(1)))$ is correct.

答:
Hint: The question boils down to figuring out what exactly the $\Bbb Z\to\Bbb Z$ group homomorphisms are.

 They are all determined by their value at $1$.

Title: Verification of ring isomorphism


问:
Series expansion with remaining $\ln n$
I'm studying the asymptotic behavior $(n \rightarrow \infty)$ of the following formula, where $k$ is a given constant.
$$ \frac{1}{n^{k(k+1)/(2n)}(2kn−k(1+k) \ln n)^2}$$

I'm trying to do a series expansion on this equation to give the denominator a simpler form so that it is easier to make an asymptotic analysis.

I used mathematica/wolframalpha to expand the formula (the documents say Taylor expansion is used).
http://www.wolframalpha.com/input/?i=1%2F%28n%5E%28k+%28k+%2B+1%29%2F%282+n%29%29+%282+k+n+-+k+%281+%2B+k%29+Log%5Bn%5D%29%5E2%29

However in series expansion at $n \rightarrow \infty$, the result still has $\ln n$. This is actually a form I prefer, compared to the form
$$a_0 + a_1x + a_2 x^2+...$$
Does anyone see how the result may be produced? Any help is much appreciated. Thanks.

答:
Please be careful, because a Taylor expansion at $\infty$ is not a normal series expansion. For $k>0$ it will surely converge to zero, as $\ln(n)$ grows slower than $n$, and hence the denominator is not bounded, as the logarithm is much slower we can forget him for the asympotics

When we look at $$n^\frac{k(k+1)}{2n}$$ this converges to $1$ because the $n$-th root of $n$ converges to 1.
So the asymptotic behaviour should be like
\[  \frac{1}{(2k n)^2}=\frac{1}{4k^2n^2}\] which is also the coefficient in the taylor expansion at infinity.

Title: Series expansion with remaining $\ln n$


答: Notice that if $f:\mathbb{Z}\rightarrow\mathbb{Z}$ is a groupe homorphism then for all $n\in\mathbb{Z},f(n)=nf(1)$ thus $f$ is entirely determined  by the value of $f(1)$. Let $\phi:f\rightarrow f(1)$ then because of what said above $\phi$ is injective, it is also surjective because for all $a\in\mathbb{Z}$, $\phi(n\rightarrow na)=a$. I let you verify that $\phi$ is a ring homorphism.

Title: Verification of ring isomorphism

Body:
I've to verify that $E(\mathbb{Z})$ and $\mathbb{Z}$ are isomorphic. $E$ is defined as
$$E(\mathbb{Z}) = \{f: \mathbb{Z} \to \mathbb{Z} | f(x+y) = f(x) +f(y)\}$$

This means there exists a function $\phi$ such that
$\phi(f+g) = \phi(f) + \phi(g)$
and
$\phi(fg) = \phi(f)\phi(g)$
with $\phi(x) \in \mathbb{Z}$

This comes as end chapter verification exercise of rings and rings' isomorphism. Groups and homorphisms are not introduced yet. So I Couldn't simply devise an ad-hoc isomorphism for the task. With your suggestion that $f: ℤ → ℤ$ is a group homorphism and to pick $f(n) = nf(1)$, for all $n ∈ ℤ$, it is straightforward. I developed like this:

Let $f+g$ defined as $f(n)+g(n) = n(f(1) + g(1)) so φ(n(f(1) + g(1))) = f(1) + g(1);$

Let $f∘g$ defiend as $f∘g = f((g(n))) = n⋅g(1)⋅f(1), φ(f((g(n)))) = g(1)⋅f(1)$, thus $\phi$ is indeed a ring isomorphism.

I'm not sure if the notation $φ(n(f(1) + g(1)))$ is correct.


问:
Does this PDE has solutions
Consider the cauchy problem, $5u_x+2u_y=7$ with $u=1$ in line $2x-5y=0$.

Does it has no solution, unique solution or infinitely many solution?

Characteristic is parallel to the initial line. So I try to take different initial line and solve it but it seems something is not compatible and I couldn't find any solution. please give me a hint

答:
$$5u_x+2u_y=7$$
The general solution (from method of characteristics or other method) is :
$$u(x,y)=\frac72 y+F(2x-5y)$$
$F$ is an arbitrary function (to be determined according to some boundary condition).

You can check it directly in putting $u=\frac72 y+F(2x-5y)$ into $5u_x+2u_y=7$.

CONDITION : $u=1$ on $2x-5y=0$
$$1=\frac72 y+F(0)$$
This is impossible because $\frac72 y+F(0)$ which is not constant cannot be equal to a constant $=1$ , what ever the function $F$ is.

Thus there is no solution which satisfies both the PDE and the specified condition.

WITH a DIFFERENT CONDITION : $u(x,0)=h(x)$

$$h(x)=0+F(2x-0)$$
With $X=2x\quad \implies\quad h(X/2)=F(X).\quad$The function $F$ is determined. We put it into the above general solution where $X=2x-5y$.
$$u(x,y)=\frac72 y+h(\frac{2x-5y}{2})$$
This is valid what ever $h(0)$ is.

Moreover, if $h(0)=1$ then $u(0,0)=h(0)=1.\quad$ And $u(x,y)$ is not constant on the line $2x-5y=0$  but is : $u=\frac72 y+1=\frac75 x+1$.

Title: Does this PDE has solutions


问:
Symbol for a function from a group to itself
Is there a symbol for the function that, given a normal subgroup H of G, associates to $g\in G$ the element $g'=gh^{-1}$ with $h\in H$?

EDIT

I reformulate the question: given a subgroup $H$ of a group $G$, under what hypotheses is it possible to define a $\mathrm{mod}$ like function of $G$ in itself?

For $\mathrm{mod}$ I mean the modulo operation, i.e. $\mathrm{mod}\colon \mathbb{Z}\mapsto\mathbb{Z}$ such that, given a modulus, a integer is associated with the remainder of the division of the integer by the modulus.

If such a function can exist, how is it indicated? Will I write $g\, \mathrm{mod}\,h$ or is there another symbol?

答: As a function you could call it $t_{h^{-1}}$, meaning $t$ranslation by $h^{-1}$. Buuuut the fact that $h$ is in a normal subgroup is irrelevant, and this isn't a group homomorphism.


问:
why is this formula not logically valid?
Let a unary predicate $R$ be the only nonlogical symbol of $L$. Then the formula $R(x) \rightarrow(\forall x) R(x)$ is not logically valid. How can I show that?

答:
A formula is logically valid iff all structures and variable assignment functions satisfy the formula, and invalid iff there is at least one counter model + assignment under which it is false.

So to ensure that $R(x)$ is true under a certain assignment and $\forall x R(x)$ false, you just need to define the model such that $R$ is true of one object and false of another one, and specify the relevant variable assignment.


答:
Hint:

Consider a domain $\{a,b\}$ where $R(a)$ and $\lnot R(b)$ are true.

What is the truth value of $R(a) \to \forall x . R(x)$?

I hope this helps ^_^

Title: why is this formula not logically valid?

Body: Let a unary predicate $R$ be the only nonlogical symbol of $L$. Then the formula $R(x) \rightarrow(\forall x) R(x)$ is not logically valid. How can I show that?


问:
Confusion about Satisfiability by a denumerable sequence.
In "Introduction to Mathematical logic" by Elliot Mendelson , page $59$ . I couldn't understand the basic preliminary notations for the rigorous definition of Satisfiability.

For example , at a line they said,

"Moreover, instead of talking about the
$n$-tuples of objects that satisfy a wf that has $n$ free variables, it is much more
convenient from a technical standpoint to deal uniformly with denumerable
sequences. What we have in mind is that a denumerable sequence
$s = (s_1, s_2 , s_3 , ... )$ is to be thought of as satisfying a wf $\beta$ that has
${x_j}_1 , {x_j}_2 ,  ... , {x_j}_n$ as free variables (where $j_1 < j_2 < ... < j_n$ ) if the $n$-tuple
$\langle {s_j}_1 , {s_j}_2 ,  ... , {s_j}_n \rangle$ satisfies $\mathscr B$ in the usual sense."

I didn't quite understand why they used ${s_j}_1$ and ${x_j}_1$ instead of $s_1$ and $x_1$ . Maybe there is a critical detail I am missing out?

Also , Another passage for the rigorous definition:

"Let $M$ be an interpretation of a language $\mathscr L$ and let $D$ be the domain of
$M$. Let $\sum$ be the set of all denumerable sequences of elements of $D$. For a wf
$\mathscr B$ of $\mathscr L$ , we shall define what it means for a sequences $s= (s_1, s_2, ... )$ in $\mathscr L$ to
satisfy $\mathscr B$ in $M$. As a preliminary step, for a given s in $\sum$ we shall define a
functions $s^*$ that assigns to each term $t$ of $\mathscr L$ an element $s^{*}(t)$ in $D$."

They did say that for each $s = (s_1, s_2, ... )$ , there is a defined function $s^*$ , but even after further reading the text afterwards , I still did not understand how exactly $s^*$ is defined from $s$ . Can someone tell me where the gap in my understanding here is?

答:
A denumerable sequence is a function $s: \text {Var} \to D$, where $\text {Var}$ is the ordered list of variables of the language: $x_1,x_2,\ldots$ and $D$ is the domain of the interpretation.

What is e.g. $s(x_1)$? It is an element $d \in D$ such that: $s(x_1)=d$, and so on.

This function "induces" a function $s^* : \text {Term} \to D$ that assigns a reference to each term $t$ of the language, according to the following rule:

(i) if $t$ is a variable $x_i$, then $s^*(t)=s(x_i)$; as we know already, it is an element of $D$.

(ii) if $t$ is a constant symbol $c$, then its reference is fixed by the interpretation, i.e. an element $c^D \in D$. Thus: $s^*(c)=c^D$.

What is left?

The case (iii) with a compound term $f_n(x_1,\ldots,x_n)$.
In this case, we use the $n$-ary function $f_n^D$ on the domain $D$ assigned to the symbol $f_n$ by the interpretation, and we use $s$ to assign reference to the variable. Thus: $s^*[f_n(x_1,\ldots,x_n)]=f_n^D(s(x_1),\ldots,s(x_n))$.

Consider a simple example with the language of arithmetic, where the symbol $+$ is a binary function symbol, interpreted in the domain $D$ of natural numbers ($\mathbb N$) with the usual sum.

Let $s(x_1)=1$ and $s(x_2)=2$, and let $t$ the term: $+(x_1,x_2)$.

We have:

$s^*[+(x_1,x_2)]=+^D(1,2)=1+2$.

Regarding the first question, there is no "deep meaning".

The author is suggesting that the definition of satisfaction in terms of denumerable sequences (due originally to Alfred Tarski (1936)) is equivalent to the definition in terms of $n$-tuples currently used in model theory.

Consider a simple example with as $\mathscr B$ the formula: $\forall x_1 (x_1 < x_2)$.

In this case, the free variable in the formula is $x_2$ and thus what is relevant for the satisfaction of the formula by a sequence $s$ is the object assigned by $s$ to $x_2$.

Thus, the first quote above means: we have a denumerable sequence $s= (s_1,s_2,s_3,\ldots)$, that means a function: $s(x_1)=s_1, s(x_2)=s_2 \in D$, and so on.

We have that, regarding satisfiability, it is the same to consider the "full" sequence or the $n$-uple extracted from the sequence considering only the free variables occurring into $\mathscr B$: in this case: $⟨s_2⟩$.


问:
Decidability and undecidability of a set or language
I want to find out whether the following sets are decidable or not. Generally speaking, what exactly should be done about it? Doing some research, I think a language or set is decidable if a Turing machine exists that decides it, meaning the output for a given input is "yes" or "no". Please shed some light on the problem.

$\{x|x \text{is the code of a Turing machine that always halts in
    less than a finite number of steps}\}$

$\{x|x \text{is the code of a Turing machine that uses at most thousand cells on its tape}\}$

答:
I understood $A$ to be

$A = \{x : $ there exists a $n$ such that $\Phi_x(y)$ halts in less than $n$ steps for all $y\}$.

$A$ is not computable. $K$ is known to be incomputable. So we will make a reduction of $K$ to $A$. For each $e \in \omega$, we will make a Turing machine $T_e$ as follows: For any input $y$, $T_e$ will start running $\Phi_e(e)$. If $\Phi_e(e)$ does converge, then at this same stage $T_e(y)$ will converge. If $\Phi_e(e)$ does not coverge then $T_e(y)$ will not converge. There is a computable function $f$ that maps $e$ to the index of the Turing program $T_e$.

The claim is that this $f$ is the many to one reduction $K \leq_m A$. To see this, if $e \in K$, then $\Phi_e(e) \downarrow$. Then there is a least $n$ such that $\Phi_{e}(e)\downarrow$ in $n$ steps. Thus $\Phi_{f(e)} = T_e$ will converge on every input $y$ in $n$ steps. Hence $f(e) \in A$. If $e \notin K$, then $\Phi_e(e)\uparrow$. $\Phi_e$ does not converge on $e$. hence $T_e = \Phi_{f(e)}$ does not converge on any input. Thus $f(e) \notin A$.

Since $K \leq_m A$ and $K$ is not computable, $A$ is not computable.

Let $B$ denote the second set. Since $K$ is not computable, $\bar{K}$, its complement, is also not computable. We will strive to make a reduction $\bar{K} \leq_m B$.

For each $e \in \omega$, we will make a Turing machine $T_e$ as follows: On any input $y$, start running $\Phi_e(e)$. At every stage such that $\Phi_e(e)$ does not halt, the Turing program $T_e$ will stay put on the current box of the tape. For every step such that $\Phi_e(e)$ does halt, then $T_e$ will write a zero in the box and move to the right.

Let $f$ be the computable function taking $e$ to the index of the Turing pargram $T_e$. Observe that for any $e$ and $y$, $T_e(y) = \Phi_{f(e)}$ does not halt. However if $x \in \bar{K}$, then $\Phi_e(e) \uparrow$. Hence $T_e$ on any input will not halt but use only 1 box (the first original box). This shows that $f(e) \in B$. If $e \in \bar{K}$, then $\Phi_e(e)\downarrow$. $\Phi_e(e)$ will eventually halt. After this point $T_e$ on any input will move to the right forever. So $T_e$ will use infinitely many boxes on the tape. Thus $f(e) \notin B$.

This is the desired reduction $\bar{K} \leq B$. Since $\bar{K}$ is not computable, $B$ is not either.


问:
Is there a standard method for image rectification by vanish points?
My problem is to do a rectification of an image when I know two vanish points $v1$ and $v2$ in homogeneous coordinates. One of them is for vertical lines and another is for horizontal lines.
Is there a standard method to do this?

For example $v1 = (161283179, -310809790, 405443)$ - vertical vanish point and $v2 = (213625800,  24111240, 29076)$ - horizontal vanish point.

答:
I assume you want to perform metric rectification (rectify the image to the point where you can recover metric properties such as angles and ratios of lengths) as there's no way you can get a full Euclidean reconstruction with just images.

The operation that performs such a rectification is called a homography as a commenter has helpfully pointed out (although all homographies are not affine transforms, but an affine transform is a special type of homography). This homography is represented as a $3\times3$ matrix for the 2D case and has 8 degrees of freedom (as it is only defined up to scale). We only need to recover 4 of these to be able to rectify the image up to a similarity (which already imposes 4 constraints on the homography).

Section 2.7 of the book Multiple View Geometry in Computer Vision, provides two useful techniques to recover this matrix.

The first is a two step approach also called stratification. It first affinely rectifies the image and then performs metric rectification. Affine rectification is performed by mapping the vanishing line (cross product of any two distinct vanishing points) to its canonical position $(0, 0, 1)$ in the affine space. This recovers 2 degrees of freedom. If $(l_1, l_2, l_3)$ are the coordinates of the vanishing line then the homography that performs the affine rectification will be
$$H_1 = \begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ l_1 & l_2 & l_3 \end{pmatrix}$$
Metric rectification is a bit more involved. It is performed by using two distinct pairs of lines (4 lines in total), that are supposed to be orthogonal in the rectified image. We find the image of the dual conic (or equivalently the images of the circular points at infinity) and then retrieve the affine transformation that maps it to its canonical form. This further recovers 2 degrees of freedom
and rectifies the images so that metric properties can be recovered.

For each pair of lines $(l'_1, l'_2, l'_3)$ and $(m'_1, m'_2, m'_3)$ that are supposed to be orthogonal after rectification, we get the following constraint.
$$(l'_1m'_1, l'_1m'_2 + l'_2m'_1, l'_2m'_2)(s_{11}, s_{12}, s_{13})^T = 0$$
Where $s_{11}, s_{12}, s_{13}$ are elements of the matrix that represents the image of the dual conic
$$C^{*}_{\infty}{'} = \begin{pmatrix}s_{11} & s_{12} & 0 \\ s_{12} & s_{13} & 0 \\ 0 & 0 & 1\end{pmatrix}$$
Now, we can find the affine transformation that maps the image of the dual conic to its canonical position in metric space, $\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}$, using SVD as
$$C^{*}_{\infty}{'} = H_2\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}H_2^T$$
The final homography that will rectify the image is $H = H_2 * H_1$.

The second approach has a single step. We need at least 5 distinct pairs of lines (10 lines in total) that are supposed to be orthogonal in the rectified image. It is hard to find so many distinct pairs in day-to-day images and hence the stratified approach is usually favored. These pairs of lines are used to find the image of the absolute conic using constraints of the form
$$(l'_1m'_1, (l'_1m'_2 + l'_2m'_1)/2, l'_2m'_2, (l'_1m'_3 + l'_3m'_1)/2, (l'_2m'_3 + l'_3m'_2)/2, l'_3m'_3)(a, b, c, d, e, f)^T = 0$$
where $C^{*}_{\infty}{'} = \begin{pmatrix}a & b/2 & d/2 \\ b/2 & c & e/2 \\ d/2 & e/2 & f\end{pmatrix}$.

To find the mapping that transforms the image of the dual conic to its canonical position, we can rewrite $C^{*}_{\infty}{'}$ as
$$\begin{pmatrix} KK^T & KK^Tv \\ v^TKK^T  & v^TKK^Tv \end{pmatrix}$$
Next, using matrix decompositions, we can find the values of $K$ and $v$ that correspond to the values of the conic parameters $(a, b, c, d, e, f)$. This gives us the final homography
$$H = \begin{pmatrix}K & 0 \\ v^t & 1\end{pmatrix}$$

If you are not familiar with all of the concepts mentioned in the answer, I strongly recommend going through the second chapter of the Multiple View Geometry book linked above. I would also recommend reading up on the basics of projective geometry for a more solid understanding.


问:
Product of coset groups
Let $G$ be a group and $H, K$ two normal subgroups of $G$ such that $G = HK$ then for each $ (gH, g'K) \in (G / H) \times (G /K) $ there exists $a \in G$ such that $ ( gH, g'K) = (aH, aK) $

Hi, there I have no idea how to find this element $a$ I've tried this: $( gH, g'K) = (aH, aK) \Leftrightarrow gH = aH$ and $ g'K = aK \Leftrightarrow ga^{-1} \in H $ and $g'a^{-1} \in K$ but I don't know how to continue I would appreciate any advice.

答:
If we want $gH = aH$ and $g'K = aK$, then by the definition of cosets we must have
$$a = gh,\ a = g'k$$
for some $(h,k)\in H \times K$. It is not immediately obvious that this is possible, but we can try to work backward to the solution.

To have $a = gh$ and $a = g'k$ we must have $gh = g'k$. This is equivalent to
$$hk^{-1} = g^{-1}g'$$
But because $HK = G$ we know it is always possible to find $h,k$ that satisfy this.

Now to construct the solution, we start with the result that $HK = G$ implies
$$\exists (h,k)\in H \times K \mid g^{-1}g' = hk$$
Thus we have $gh = g'k^{-1}$. If we let $a = gh = g'k^{-1}$ then we get our solution.

Title: Product of coset groups


问:
Computing surface integral for $F(x,y,z) = (xy,-x^2,x+z)$

Let $F: \mathbb{R}^3 \to \mathbb{R}^3, F(x,y,z) = (xy,-x^2,x+z)$ be a vector field. Compute the surface integral over the set $S$ which is bounded by the plane $2x+2y+z=6$ in the set $\{(x,y,z) \in \mathbb{R}^3 : x,y,z\geqslant 0 \}.$

So I would need to calculate $$\iint_{T}f(\varphi(x,y)) \|\frac{\partial\varphi}{\partial x} \times \frac{\partial\varphi}{\partial y} \|  \ dx \ dy$$

Parameterizing $F$ as $\varphi(x,y) = (x,y, 6-2x-2y)$ I have that $$\frac{\partial\varphi}{\partial x}=(1,0,-2) \text{ and } \frac{\partial\varphi}{\partial y}=(0,1,-2)$$

so $\|\frac{\partial\varphi}{\partial x} \times \frac{\partial\varphi}{\partial y} \| = \| (2,2,1)\| = 3.$

Computing the composition $f\circ\varphi$ I get that $f(x,y,6-2x-2y)=(x^2y,-y^2,x+6-2x-2y)$, thus the integral I would need to compute is $$\iint_{T} (x^2y,-y^2,x+6-2x-2y)\cdot 3 \ dx \ dy$$

but this seems wrong, this is just $f(\varphi(x,y)$ scaled by $3$. What am I missing here? Any hints would be appreciated.

答:
I think that you should evaluate the flux of the vector field $\vec{F}$ through the given surface $S$, a triangle where the normal vector $\vec{n}$ is constant:
$$\begin{align}\iint_S \vec{F} \cdot \vec{n} \ dS&=\iint_{T}\vec{F}(\varphi(x,y)) \cdot \left(\frac{\partial\varphi}{\partial x} \times \frac{\partial\varphi}{\partial y}\right) \ dx dy
\\&=\iint_T (xy,-x^2,x+(6-2x-2y))\cdot(2,2,1)\, dx\,dy
\end{align}$$
where $\varphi(x,y)=(x,y, 6-2x-2y)$ and $T=\{(x,y): x+y\leq 3,x\ge 0,y\ge 0\}$.

As a reference, see wikipage.


问:
Help understanding a $\frak {sl}(2)$ Lie algebra representation
${\frak sl}(2)= \left\{ \begin{bmatrix}a & b \\ c & d \end{bmatrix}: a + d=0\right\}= \text{span}\left\{H,E,F\right\}$

where $H = \begin{bmatrix}1 & 0 \\ 0 & -1 \end{bmatrix},$ $E = \begin{bmatrix}0 & 1 \\ 0 & 0 \end{bmatrix}$ and $F = \begin{bmatrix}0 & 0 \\ 1 & 0 \end{bmatrix}.$

Also, the brackets $[H,E]=2E,$ $[H,F]=-2F$ and $[E,F]=H$ provide an alternative (isomorphic) way of understanding the Lie algebra. With some handwringing after the last sentence it seem clear, but then the pictorial creativity is unleashed at the beginning of this nice lecture with this representation of the $5$-dimensional representation:

The dots represent "basis elements" and $E$ (in orange) sends each element to the next one from left to right, while $F$ (green) does the opposite, and $H$ in purple sends each element to itself. The numbers denote multiplicity.

The lecturer at that point picks the fourth white dot from the left calling it a vector, and he checks the basic arithmetic following the arrows and multiplicity notations to confirm that $[E,F]=H$ as $EF - FE=6 - 4= 2.$

Very elegant, but can you please show me what that vector ("basis elements") - the white dots - would look like so that I can reproduce what's being done with basic linear algebra matrix to vector multiplication?

As it turns out, a lot of meaningful information as it pertains to the actual issue behind the question is buried in comments, so here is a summary. Besides the matrices provided in the accepted answer, there is a different set that matches - up to some possible sign adjustment of doubtful conceptual relevance - the diagram (thank you @Torsten Schoeneberg) as such:

$$\begin{align}
H&=\Tiny{\begin{bmatrix}\color{blue}4&0&0&0&0\\0&\color{blue}2&0&0&0\\0&0&\color{blue}0&0&0\\0&0&0&\color{blue}{-2}&0\\0&0&0&0&\color{blue}{-4}\end{bmatrix}}\\
E&=\Tiny{\begin{bmatrix}0&\color{orange}4&0&0&0\\0&0&\color{orange}3&0&0\\0&0&0&\color{orange}2&0\\0&0&0&0&\color{orange}1\\0&0&0&0&0\end{bmatrix}}\\
F&=\Tiny{ \begin{bmatrix}0&0&0&0&0\\\color{green}{1}&0&0&0&0\\0&\color{green}{2}&0&0&0\\0&0&\color{green}{3}&0&0\\0&0&0&\color{green}{4}&0\end{bmatrix}}
\end{align}
$$

These matrices do fulfill the bracket relationships, and follow the diagram to a sign change, reflecting the multiplicity values written in colors on the board. For example $e_4=\Tiny\begin{bmatrix}0&0&0&1&0\end{bmatrix}^\top,$ corresponding to the fourth white dot, and performing the bracket $[E,F]:$ This is the composition $E\circ F(e_4)$. Starting with $EF,$ $F\cdot e_4=4 e_5$ and composing this result with $E,$ we obtain $4 e_4.$ Whereas $FE$ will entail starting with $E\cdot e_4 = 2 e_3$ which then becomes $2F\cdot e_3=6e_4.$ Therefore $4e_4-6e_4=-2e_4,$ as opposed to the positive $2$ in the lecture.

The leftward extreme example $\Tiny e_1=\begin{bmatrix}1&0&0&0&0\end{bmatrix}^\top,$ under the action of the same bracket should result in $4e_1-\mathbf 0=4e_1.$ Following the diagram, $EF$ would turn out $0,$ because there is no green arrow originating in the first dot, and $FE$ would be $4e_1,$ which would pick up a $-$ in the commutator.

答: Consider these three $5\times5$ matrices:$$\mathcal H=\begin{bmatrix}4&0&0&0&0\\0&2&0&0&0\\0&0&0&0&0\\0&0&0&-2&0\\0&0&0&0&-4\end{bmatrix},\ \mathcal E=\begin{bmatrix}0&4&0&0&0\\0&0&6&0&0\\0&0&0&6&0\\0&0&0&0&4\\0&0&0&0&0\end{bmatrix}\text{, and }\mathcal F=\begin{bmatrix}0&0&0&0&0\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{bmatrix}.$$If $aE+bF+cH\in\mathfrak{sl}_2$, consider the matrix $a\mathcal E+b\mathcal F+c\mathcal H$. It's a $5\times5$ matrix and this defines a $5$-dimensional representation of $\mathfrak{sl}_2$ since, as you can check, $[\mathcal H,\mathcal E]=2\mathcal E$, $[\mathcal H,\mathcal F]=-2\mathcal F$, and $[\mathcal E,\mathcal F]=\mathcal H$.


问:
Suppose $\|x_n\|\leq 1$ for all $n\geq 1$ and $x_n\to x$ weakly as $n\to\infty$. Is it true that $\|x\|\leq 1$?
$\{x_n\}_{n=1}^\infty$ is a sequence in $\mathcal H$ (an infinite dimensional Hilbert space with associated norm $\|\cdot\|$) and  $x\in \mathcal H.$

答:
Yes, the norm is (sequentially) weakly lower-semicontinuous: if $x_n$ converges weakly to $x$, then $||x|| \le \liminf_n ||x_n||$.

Weak lower semicontinuity of a functional on Hilbert space?

The proof is here.


问:
Proving two continuous functions coincide at 0 using a continuous piece function of the two
I'm reviewing some old undergrad notes of mine and came across this continuity question.

"Let $f:(-\infty,0]\to\mathbb{R}$ and $g:[0,\infty)\to\mathbb{R}$ be continuous functions on the entirety of their respective domains. Now let us define the function $h:\mathbb{R}\to\mathbb{R}$ by:

$$h(x)=
    \begin{cases}
      f(x) & x\le{}0\\
      g(x) & x>0\\
    \end{cases}       $$
Show that $h$ is continuous at $c=0$ iff $f(0)=g(0).$" The backward implication was no trouble at all but I am having trouble in the forward direction.

My attempt:

Suppose $h$ is continuous at $c=0$.

Fix $\epsilon>0$. Choose $\delta_f,\delta_g,\delta_h>0$ that satisfy the definition of continuity at $f,g$ and $h$, respectively, at $c$ w.r.t the fixed $\epsilon$. Define $\delta:=\min(\{\delta_f,\delta_g,\delta_h\})$.

Now fix $x\in(c-\delta,c+\delta)$.

Case 1 $(x>0)$:
$$|f(0)-g(0)|=|h(0)-g(0)|=|h(0)-g(x)+g(x)-g(0)|\le{}|h(0)-g(x)|+|g(x)-g(0)|$$
$$=|h(0)-h(x)|+|g(x)-g(0)|<2\epsilon$$

Which is true for any choice of $\epsilon>0$ which forces $f(0)=g(0).$ So we are done.

Case 2 $(x\le0)$:

$$|f(0)-g(0)|=|f(0)-h(x)+h(x)-g(0)|\le{}|f(0)-h(x)|+|h(x)-g(0)|=|f(0)-f(x)|+|h(x)-g(0)|$$

This is were I get stuck. Can this proof work in this way? Is there a nicer way to achieve the same result?

Hints would be more appreciated.

答:
Why don't you just say, if $h$ is continuous at $0$ then

$$f(0)=h(0)= \lim\limits_{x \to 0^+} h(x) = \lim\limits_{x \to 0^+} g(x) = g(0)?$$


问:
$x^2+x+1$ is the cube of a prime.
Please help me find all natural numbers $x$ so that $x^2+x+1$ is the cube of a prime number.
(Used in here)

答:
By Ljunggren (1942), the equation $x^2+x+1=y^3$ has only these integer solutions:
$(x,y)=(0,1)(-1,1)(18,7)(-19,7).$</p>

Title: $x^2+x+1$ is the cube of a prime.


问:
Suppose $f(x)-f(y)=h(g(x)-g(y))$, then $h$ is linear?
Background $f,g$ are real functions, $h$ is increasing, satisfying:

$$f(x)-f(y)=h(g(x)-g(y)),$$

What function is $h$?

Motivation: At a first glance, $h$ can be any function, since $f,g$ are arbitrary functions. However, things are not that simple.

Hypothesis: $h$ must be linear.

My try: Let $y=c$, where $g(c)=0$ (Suppose this zero exist)

$$f(x)-f(c)=h(g(x)-g(c))$$
So,
$$f(x)=h(g(x)-g(c))+f(c)$$

Then, $h(g(x)-g(y))=h(g(x)-g(c))+f(c)-(h(g(y)-g(c))+f(c))$

$$h(g(x)-g(y))=h(g(x)-g(c))-h(g(y)-g(c))$$

Since $g(c)=0$,

$$h(g(x)-g(y))=h(g(x))-h(g(y))$$

By definition, $h$ is linear functional. That is, $h(x)=ax+b$.

However, I have no idea how to prove this without assuming that $g(c)=0$ for some constant $c$.

答:
Fix $c$. If

$$f(x)-f(y)=h(g(x)-g(y))$$

then

$$f(x)-f(y)=h(g_{2}(x)-g_{2}(y))$$

where

$$g_{2}(x)=g(x)-g(c)$$.

I think that completes your argument.

Neat question.


问:
$S=\{x\in X: \|x\|=1\}$ cannot be covered by a finite family of closed balls in X
I am working with an exercise and have som questions.
The exercise is

Let X be an infinite dimensional normed vector space  over $\mathbb{K}$, where $\mathbb{K}=\mathbb{R}$ or $\mathbb{K}=\mathbb{C}$

Show that one cannot cover the unit sphere $S=\{x\in X: \|x\|=1\}$ with a finite family of closed balls in $X$ such that none of the balls contains 0.

I am thinking to show that S is not covered by a finite family of closed balls in X such that none of the balls contains 0. To show this I am thinking that I have to show $S\not\subset \cup \{x\in X| d(x,p)\leq r\}$ and to show this I am thinking to take a $x\in S$ and show that $x \notin \cup \{x\in X| d(x,p)\leq r\}$.

I note that for $x\in S$ I have that $\|x\|=1$. And to show that $x\notin \cup \{x\in X| d(x,p)\leq r\}$ I am thinking that I have to show that $d(x,p)>r$ . And to show this I am thinking that I somehow have to use Riesz lemma, but to be able to use this lemma I have to have that $\cup \{x\in X| d(x,p)\leq r\}$ is a closed proper subspace of X. It is obviously closed but I don't know how to show that it is a proper subspace of X.
But after I have argued that it is a proper subspace of X then I am thinking that I have from Riesz lemma that there exist a $x\in X$ with $\|x\|=1$ such that $\|x-y\|\geq a$ for all $y\in
\cup \{x\in X| d(x,p)\leq r\}$, where $0<a<1$. Since we have that $p\in \cup \{x\in X| d(x,p)\leq r\}$ then we have that $\|x-p\| \geq a$, and what we want to show is that $d(x,p)>r$ i.e. $\|x-p\|>r$. So dont now how I get from $\|x-p\| \geq a$ to $\|x-p\|>r$

答: I found another, easier proof using the Hahn-Banach theorem that works in the general case. Let $B_i$ for $i=1,\ldots,n$ be closed balls not containing $0$. These balls are closed convex sets. Hence, we can find continous functionals $\lambda_i$, s.t. $\mathrm{Re}\,\lambda_i(x) \geq 1$ for $x \in B_i$. The vector space $V = \cap_{i=1}^n \mathrm{ker}(\lambda_i)$ does not intersect any of the $B_i$. But $V \neq 0$, because $X$ is infinite-dimensional. So we find an $x\in V \cap S$. Therefore, no finite number of closed balls not containing $0$ can cover $S$.


问:
Let $ z\in \mathbb{C} $ and $\sin(z)=0$. Prove that $z\in \mathbb{R}$
Assume that $$ z\in \mathbb{C},\ \sin(z)=0 $$ and follow that $$ z\in \mathbb{R}. $$

What I have tried so far: I went for comparing with the Euler's form. Since we can say that for all z $$ \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}, \ \cos(z)=\frac{e^{iz}+e^{-iz}}{2} $$ we can follow that $$ e^{iz}=\cos(z)+i\sin(z). $$ so, if we let sin(z) be zero, conforming to the task, we can transform $$ e^{iz}=\cos(z)+i*0 = \cos(z). $$ and since $$ Re(e^{iz}) = \cos(z), \ \cos(z)\in \mathbb{R}$$ we can say that also $$ z\in \mathbb{R} $$ is this a valid way of showing the argument, or would I have to use the following as a vantage point: $$ z= r e^{i\phi}, \ for \ z=x+iy $$ and somehow transform that term to get the solution. I don't really see how I would get the sin(z)=0 to aid me when it's not in the Euler form of the $r e^{i\phi}$ form. Any help would be super appreciated.

Edit: Thanks a lot for the replies, everyone. I'm sorry for my onesided thinking since I'm still a calc 1 newbie. I'll take a look and re-evaluate.

Body:
We know that $sin(z) = \frac{e^{iz} - e^{-iz}}{2i} = 0$. Put $t = e^{iz}$, then $t = 1/t$, $t^2 = 1$, $t = \pm 1$. Hence $e^{iz} = \pm 1$.

As $z = a+bi$ where $a,b \in \mathbb{R}$, we have $e^{iz} = e^{-b+ia} = \pm1$.

Thus $|e^{-b+ia}| = |\pm 1| = 1$, but $|e^{-b+ia}| = |e^{-b}| \cdot |e^{ia}| = e^{-b} \cdot 1$. Hence $e^{-b} = |e^{-b+ia}|  = 1$ and $b=0$. That means that $z = a+bi$ and $b=0$, q.e.d.


问:
A good reference to learn the concept of partition function
I am looking for a good reference and easy to learn the concept of partition function in mathematics. Can anyone help me?</p>

答: One of the standard texts in this field is The theory of Partitions by George Andrews. http://books.google.com/books/about/The_Theory_of_Partitions.html?id=Sp7z9sK7RNkC</p>


答:
First, we have
$$\sin(a + bi) = \sin(a)\cos(bi) + \cos(a)\sin(bi).$$

Then
$$\sin(bi) = \frac{e^{-b}-e^{b}}{2i} = \sinh(b)i, \text{ and } \cos(bi) = \frac{e^{-b}+e^b}{2}=\cosh(b).$$

So we get

$$\sin(a + bi) = \sin(a)\cosh(b)+\cos(a)\sinh(b)i.$$

If this is a real number then either $\cos(a) = 0$ or $\sinh(b) = 0 \iff b = 0$. Maybe you can work out the rest from here?

Title: Let $ z\in \mathbb{C} $ and $\sin(z)=0$. Prove that $z\in \mathbb{R}$

Body:
Assume that $$ z\in \mathbb{C},\ \sin(z)=0 $$ and follow that $$ z\in \mathbb{R}. $$

What I have tried so far: I went for comparing with the Euler's form. Since we can say that for all z $$ \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}, \ \cos(z)=\frac{e^{iz}+e^{-iz}}{2} $$ we can follow that $$ e^{iz}=\cos(z)+i\sin(z). $$ so, if we let sin(z) be zero, conforming to the task, we can transform $$ e^{iz}=\cos(z)+i*0 = \cos(z). $$ and since $$ Re(e^{iz}) = \cos(z), \ \cos(z)\in \mathbb{R}$$ we can say that also $$ z\in \mathbb{R} $$ is this a valid way of showing the argument, or would I have to use the following as a vantage point: $$ z= r e^{i\phi}, \ for \ z=x+iy $$ and somehow transform that term to get the solution. I don't really see how I would get the sin(z)=0 to aid me when it's not in the Euler form of the $r e^{i\phi}$ form. Any help would be super appreciated.

Edit: Thanks a lot for the replies, everyone. I'm sorry for my onesided thinking since I'm still a calc 1 newbie. I'll take a look and re-evaluate.


问:
To show that the integral $\int_{-\infty}^{\infty}\frac{(p'(x))^2}{(p(x))^2+(p'(x))^2}dx$ converges and is less or equal than $n^{3/2}\pi$
Consider a polynomial $p \in \mathbb{R}[x]$ of degree $n$ and with no real roots. Prove that$$\int_{-\infty}^{\infty}\frac{(p'(x))^2}{(p(x))^2+(p'(x))^2}dx$$converges, and is less or equal than $n^{3/2}\pi$

My approach

Now let $x_1, x_2, \dots, x_n$ be the roots of $p$. By Cauchy-Schwarz

$$(\sum_{k=1}^{n}{\frac{1}{x-x_k}})^2\leq n\sum_{k=1}^{n}{\frac{1}{|x-x_k|^2}}$$

I don't know what to do next.
If I am wrong kindly provide a detailed answer in the answer section.
I have shown what I have thought of or what I have done .

Can anyone confirm if my thought process is right?

Just a reminder... This question has been lying unanswered for a long time

Thank you.

答:
First of all, we can define:
$$p_n(x)=\sum_{k=0}^na_kx^k\tag{1}$$
$$p_n'(x)=\sum_{k=0}^nka_kx^{k-1}$$

Now by the multinomial theorem:
$$\left(\sum_{i=1}^mx_i\right)^n=\sum_{\sum j_i=n}{n\choose{j_1,j_2...j_m}}\prod_{t=1}^mx_t^{j_t}$$
From this you should be able to come up with an expression for: $p_n^2$ and $p_n'^2$.

Now also note that from what we know (due to there being no real roots):
$$p_n(x_0)=\sum_{k=0}^na_kx_0^k\ne0\,\,\,\,x_0\in\mathbb{R}$$
we know that:
$$p_n(x)^2=O(x^{2n})$$
$$p_n'(x)^2=O(x^{2(n-1)})$$
and so it is clear that:
$$\frac{p_n'(x)^2}{p_n(x)^2+p_n'(x)^2}=O\left(\frac1{x^2}\right)$$


Body:
$$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}=0$$implies$$e^{iz}=e^{-iz}$$
Use $$z=x+iy$$ with $x,y\in\mathbb R$
Then $$e^{i(x+iy)}=e^{-i(x+iy)}\\e^{ix}e^{-y}=e^{-ix}e^{y}$$
Now take the absolute value, and notice that $|e^{ix}|=1$:
$$|e^{ix}e^{-y}|=|e^{-ix}e^{y}|\\|e^{ix}||e^{-y}|=|e^{-ix}||e^{y}|\\|e^{-y}|=|e^y|$$
Since $y\in \mathbb R$ it means that $$e^y=\frac1{e^y}=1$$or $y=0$.


问:
Removing an edge from a strongly connected graph
Let $G$ be a strongly connected graph with at least $2n-1$ edges. Prove that it's possible to remove an edge such that G will still be strongly connected.

I found this question in discrete math book under "advanced graph" section and tried solving it using Euler's theorem and this corollary:

Let $G$ be a planar simple connected graph with n nodes and e edges. For $e \geq 3$
$$e \leq 3v-6$$

But I can't quite figure out how to use this theorem for strongly connected graphs. I'm not sure if this is the correct approach tho. Any other alternative solutions is very much appreciated.

答:
The comparison to planar graphs won't fly. Planar graphs are undirected graphs, so you can't apply the theorem. Even if you could, there is nothing saying that $G$ is planar, either.

There's an argument by induction using the following idea.

Let $C$ be a (directed) $k$-cycle in $G$. Then one of two things must hold:

There is an additional edge between two vertices of $C$. If we delete that edge, $G$ remains strongly connected.

If we contract all of $C$ to a single vertex, the remainder is a strongly connected graph $G'$ with $n-k+1$ vertices and at least $(2n-1)-k$ edges. Finding an edge to delete in $G'$ will give you an edge to delete in $G$.

(The contraction replaces the $k$ vertices of $C$ by a single vertex $v_C$; every edge $(v,w)$ for $v \in V(C)$ will become an edge $(v_C,w)$, and the same for edges going the other way. We don't keep the edges of $C$. This can create multiple copies of an edge.)

As a bonus, by carefully analyzing the proof you get, we can figure out that the result also holds for $2n-2$ edges, except when all the cycles are $2$-cycles. That case corresponds to the directed graphs we get by starting with an undirected tree, and replacing each edge $vw$ by two directed edges $(v,w)$ and $(w,v)$.


答:
$0 = \sin(z) = (e^{iz} - e^{-iz})/(2i) \iff e^{iz} = e^{-iz} \iff e^{2iz} = 1$

Write $z$ as $a + bi$

$1e^{0} = 1 = e^{2iz} = e^{2i(a + bi)} = e^{-2b+2ai} = e^{-2b} e^{2ai}$

Equating lengths and angles we have:

$1 = e^{-2b} \implies b = 0$

$0 = 2a + 2 \pi n \implies a = \pi k$

So $z = \pi k$ - all of these are real