问:
Neukirch's Algebraic Number Theory Field Theory (pgs. 10 and 11)
I'm working through Neukirch's Algebriac Number Theory for a reading course, but I'm getting hung up on some of the field theory stuff on pages 10 and 11.

On page 10, during the proof that $Tr_{L/K} \circ Tr_{M/L} = Tr_{M/K}$ for a tower of field extensions $K \subseteq L \subseteq M$, Neukirch partitions Hom$_K(L,\overline{K})$ by $\sigma \sim \tau$ if $\sigma|_L = \tau|_L$ into $m$ classes, each with representative $\sigma_i$. Now the part that's throwing me off is when he uses the fact that $$\sum_{\sigma \sim \sigma_i} \sigma x = Tr_{\sigma_i M/ \sigma_i L}(\sigma_i x).$$ I've got no idea where this could come from. It seems reasonable, but I've not been able to prove it using the definitions and propositions given.

On a different note, towards the top of page 11, Neukirch says that for the special case of a basis of the form $\{1, \theta, \dots, \theta^{n-1}\}$, we can determine that $d(1,\theta,\dots,\theta^{n-1}) = \prod_{i<j}(\theta_i - \theta_j)^2$ by "successively multiplying each of the first $n-1$ columns of the Vandermonde matrix by $\theta_1$ and subtracting it from the following", where $\theta_i = \sigma_i\theta$.

I have several questions here. Most fundamentally, what does he mean by $\prod_{i<j}(\theta_i - \theta_j)^2$? The discriminant is certainly a value in $L$, but this product leaves the variable $j$ open. My assumption was that this intended to mean $\prod_{j=1}^n\prod_{i < j}(\theta_i - \theta_j)^2$, but I'm not sure about this at all.

On the same note, I don't see what he means regarding multiplying the first $n-1$ columns by $\theta_i$ and then subtracting from the following. Does he mean that we transform the $i$th column $c_i$ by $c_i \mapsto (\theta_ic_i - c_{i+1})$? I tried to do the calculation using that, but I still had trouble with it.

答:
On page 10, during the proof that $Tr_{L/K} \circ Tr_{M/L} = Tr_{M/K}$ for a tower of field extensions $K \subseteq L \subseteq M$, Neukirch partitions Hom$_K(L,\overline{K})$ by $\sigma \sim \tau$ if $\sigma|_L = \tau|_L$ into $m$ classes, each with representative $\sigma_i$.

Neukirch partitions $\operatorname{Hom}_K(\color{red}M, \overline{K})$ into $m$ classes using that relation.

Now the part that's throwing me off is when he uses the fact that $$\sum_{\sigma \sim \sigma_i} \sigma x = Tr_{\sigma_i M/ \sigma_i L}(\sigma_i x).$$

To see this, you need to know that $\operatorname{Hom}_{\sigma_i L}(\sigma_i M, \overline{K}) = \{ \sigma \circ \sigma_i^{-1} \mid \sigma \sim \sigma_i \}$ (where $\sigma_i^{-1} : \sigma_i M \to M$ is an abuse of notation), which we now proof:

Take $\tau \in \operatorname{Hom}_{\sigma_i L}(\sigma_i M, \overline{K})$. We now define $\sigma: M \to \overline{K}$ to be $\tau \circ \sigma_i$. Then clearly it is a field homomorphism that fixes $K$, so $\sigma \in \operatorname{Hom}_K(M, \overline{K})$. Now, for any $y \in L$, we have $\sigma(y) = \tau(\sigma_i(y)) = \sigma_i(y)$ because $\tau$ fixes $\sigma_iL$. Therefore, $\sigma \sim \sigma_i$, so $\tau = \sigma \circ \sigma_i^{-1}$. This proves the forward inclusion.

The backward inclusion is left as an exercise.

Once you have this, you can calculate:
$$Tr_{\sigma_iM/\sigma_iL}(\sigma_i x) = \sum_{\tau \in \operatorname{Hom}_{\sigma_i L}(\sigma_i M, \overline{K})} \tau(\sigma_ix) = \sum_{\sigma \sim \sigma_i} (\sigma \circ \sigma_i^{-1})(\sigma_ix) = \sum_{\sigma \sim \sigma_i} \sigma x$$

I have several questions here. Most fundamentally, what does he mean by $\prod_{i<j}(\theta_i - \theta_j)^2$? The discriminant is certainly a value in $L$, but this product leaves the variable $j$ open. My assumption was that this intended to mean $\prod_{j=1}^n\prod_{i < j}(\theta_i - \theta_j)^2$, but I'm not sure about this at all.

Indeed, $\prod_{i<j}(\theta_i - \theta_j)^2$ does mean $\prod_{j=1}^n\prod_{i < j}(\theta_i - \theta_j)^2$. Rather, my interpretation is that $\prod_{i<j}$ means $\prod_{\{(i,j) \mid i<j\}}$. You can rationalize this notation in any way you want.

On the same note, I don't see what he means regarding multiplying the first $n-1$ columns by $\theta_{\color{red}1}$ [correction mine] and then subtracting from the following. Does he mean that we transform the $i$th column $c_i$ by $c_i \mapsto (\theta_ic_i - c_{i+1})$?

In English, when we subtract $3$ from $5$, we get $2$, and this is seen as performing an operation on $5$.

So, what he means (Man sieht dies, indem man in der Vandermondeschen Matrix jede der $(n-1)$ ersten Spalten mit $\theta_1$ multipliziert und von der folgenden subtrahiert und so fortfährt.), is to transform $c_{i+1} \mapsto c_{i+1} - \theta_1 c_i$. Let's take $n=3$ as an example. Our Vandermonde matrix is:

$$\begin{pmatrix} 1 & \theta_1 & \theta_1^2 \\ 1 & \theta_2 & \theta_2^2 \\ 1 & \theta_3 & \theta_3^2 \end{pmatrix}$$

If we perform the operation $c_{i+1} \mapsto c_{i+1} - \theta_1 c_i$, we get:

$$\begin{pmatrix} 1 & \theta_1 - \theta_1 \times 1 & \theta_1^2 - \theta_1 \times \theta_1 \\ 1 & \theta_2 - \theta_1 \times 1 & \theta_2^2 - \theta_1 \times \theta_2 \\ 1 & \theta_3 - \theta_1 \times 1 & \theta_3^2 - \theta_1 \times \theta_3 \end{pmatrix}$$

which we simplify to get:

$$\begin{pmatrix} 1 & 0 & 0 \\ 1 & \theta_2 - \theta_1 & \theta_2(\theta_2 - \theta_1) \\ 1 & \theta_3 - \theta_1 & \theta_3(\theta_3 - \theta_1) \end{pmatrix}$$

We expand the determinant on the first row to get the same determinant as the matrix:

$$\begin{pmatrix} \theta_2 - \theta_1 & \theta_2(\theta_2 - \theta_1) \\ \theta_3 - \theta_1 & \theta_3(\theta_3 - \theta_1) \end{pmatrix}$$

We can then factorize $(\theta_2 - \theta_1)$ from the first row and $(\theta_3 - \theta_1)$ from the second row to get:

$$\begin{pmatrix} 1 & \theta_2 \\ 1 & \theta_3 \end{pmatrix}$$

which is the Vandermonde matrix with $n$ being $1$ smaller, so we are done by induction.


问:
What is the answer to this limit
what is the limit value of the power series:

$$ \lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$$

where $m>1$.

答:
Here is a start. Using the integral representation derived in a previous problem and the identity

$$ (xD)^m=\sum_{r=0}^{m}\begin{Bmatrix} m\\r \end{Bmatrix}x^r D^r, $$

where $D=\frac{d}{dx}$ and $ \begin{Bmatrix} m\\r \end{Bmatrix} $ are Stirling numbers of the second kind.

$$\sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k}}  = - \int_{0}^{1} x \cdot u^{x u} \, du .$$

$$ \implies (xD)^m\sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k}}=\sum_{k=1}^\infty (-1)^k k^m\frac{x^k}{k^{k}}=-(xD)^m \int_{0}^{1} x \cdot u^{x u} \, du.$$

$$ \implies \sum_{k=1}^\infty (-1)^k k^m\frac{x^k}{k^{k}}=\sum_{r=0}^{m}\begin{Bmatrix} m\\r \end{Bmatrix} \int_{0}^{1} f(x,u) du , $$

where $f(x,u)$ is a function in $x$ and $u$, see the note. Now, it is your job to try analyzing these integrals and see what the limit is.

Notes:

1) We used the identity

$$ D^r xe^{ax}= {a}^{r-1}{{\rm e}^{ax}} \left( ax+r \right)  .$$

For finding the $n$th derivative, see section 6.

2)

$$ f(x, u) = { {m{x}^{2\,m} \left( \ln  \left( u \right)\right) ^{m-1}{u}^{ux+m
-1}}{ }}+{x}^{2\,m+1} \left( u\ln  \left( u
 \right)  \right) ^{m}{u}^{ux}.$$

ContentLicense: CC BY-SA 3.0

Title: What is the answer to this limit


问:
Help-me, please! Solve $\oint_{c} xy ds$ where $C$ is the intersection of the surfaces $x^2+y^2=4$ and $ y+z=8 $
Solve $\oint_{c} xy ds$ where $C$ is the intersection of the surfaces $x^2+y^2=4$ and $ y+z=8 $

Hi. I tried to resolve the issue below using the definition of line integral, but I couldn’t solve it. Could someone appreciate my resolution and help me finish it? Below follows what I did:

$$\oint_{c} xy ds = {\int }_{C}f\left(x,y,z\right)ds= \int_{a}^{b}f\left(\text{r}\left(t\right)\right)\sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}dt.$$

\begin{align*}
  x&=2\cos t \\
  y&=2\sin t \\
  z&=8-2\sin t
 \end{align*}

\begin{align*}
f\left(\text{r}\left(t\right)\right)&=\langle 2\cos t, 2\sin t, 8-2\sin t \rangle \\
ds = \left\Vert\left(\text{r}\left(t\right)\right)^{\prime} \right\Vert&= \sqrt{{\left({x}^{\prime }\left(t\right)\right)}^{2}+{\left({y}^{\prime }\left(t\right)\right)}^{2}+{\left({z}^{\prime }\left(t\right)\right)}^{2}}dt
\end{align*}

Thus,

\begin{align*}
\oint_{c} xy ds ={\int }_{C}f\left(x,y,z\right)ds&={\int_{0}^{2\pi}}(4\cos t\sin t)\sqrt{{\left({-2\sin t}\right)}^{2}+{\left({2\cos t}\right)}^{2}+{\left({-2\cos t}\right)}^{2}} dt \\
  &={\int_0^{2\pi}} 8\cos \left(t\right)\sin \left(t\right)\sqrt{-\sin ^2\left(t\right)+2} dt \\
  &= -\left[\frac{8}{3}\left(1+\cos ^2\left(t \right)\right)^{\frac{3}{2}}\right]_{_{0}}^{^{2\pi}} \\
  &=0.
\end{align*}

Thanks!

答:
The integrand is an odd function of $x$ and the curve is invariant under the reflection $x\leftrightarrow -x$ thus

$$I = \oint_Cxy\:ds = \oint_C (-x)y\:ds = -I \implies I=0$$

by odd symmetry.

ContentLicense: CC BY-SA 4.0


问:
Geometrical construction for Snell's law?
Snell's law from geometrical optics states that the ratio of the angles of incidence $\theta_1$ and of the angle of refraction $\theta_2$ as shown in figure1, is the same as the opposite ratio of the indices of refraction $n_1$ and $n_2$.

$$
\frac{\sin\theta_1}{\sin \theta_2} = \frac{n_2}{n_1}
$$

(figure originally from wikimedia)

Now let $P$ be a point in one medium (with refraction index $n_1$) and $Q$ a point in the other one as in the figure. My question is, is there is a nice geometrical construction (at best using only ruler and compass) to find the point $O$ in the figure such that Snell's law is satisfied. (Suppose you know the interface and $n_2/n_1$)?

Edit
A long time ago user17762 announced to post a construction. However until now no simple construction was given by anybody. So, does anybody know how to do this?

答:
Yes, right. When entering a denser medium light slows down..  Adding another answer keeping only to the construction method. Drawn on Geogebra, removed the axes and grid to trace a ray inside a medium of higher refractive index $\mu$.

Choose an arbitrary point X such that

$$ \mu= \dfrac{XP}{XQ} \left( =\dfrac{n_2}{n_1}=\dfrac{v_1}{v_2} \right) $$

which is  possible using Ruler & Compass using suitable segment lengths.

From $X$ draw bisector to angle $PXQ$ cutting interface at $O,$ which is the required point of incidence. Due to refraction, the point of incidence $O$ always shifts to the right, when stat point $P$ is at top left, compared to the straight unrefracted ray $PQ$ point of impingement on interface.

The ratio of sides and times spent in each medium is maintained due to the constant bisector property of Apollonian Circle ( no need to draw it).

The diagram represents a lightray refraction into  a medium of refractive index 1.5


问:
How can I show that $\sum \limits_{n=2}^\infty\frac{1}{n\ln n}$ is divergent without using the integral test?
How can I show that $\sum \limits_{n=2}^\infty\frac{1}{n\ln n}$ is divergent without using the integral test?

I tried using the comparison test but I could not come up with an inequality that helps me show the divergence of a series. I also tried using the limit comparison test but I was not successful.

Please do not give me solutions; just a hint so that I can figure it out myself.

答:
Due to Chebyshev's estimates there exists constant $A$ and $n_0$ such that the $n$'th prime $p_n$ is bounded below by $An\log n$ for all $n>n_0$ Consequently, we have

$$
\sum_{n>n_0}^\infty{1\over n\log n}\ge\sum_{n>n_0}^\infty{A\over p_n}
$$

and due to Euler the series on the RHS diverges, so the series $\sum{1\over n\log n}$ diverges.


问:
Let $x_n \in[0,1)$ s.t $x_n - \sqrt{n}\in \mathbb{Z}$. Prove that $X=\{x_n | n\in \mathbb{N}\}$ is dense in $[0,1]$.

Let $x_n \in[0,1)$ s.t $x_n - \sqrt{n}\in \mathbb{Z}$. Prove that $X=\{x_n | n\in \mathbb{N}\}$ is dense in $[0,1]$.

It it suffices to prove that $\overline{X}=[0,1].$ I was trying to use the fact that $p+q\sqrt{2}$ $(p,q \in \mathbb{N})$ is dense in $\mathbb{R}$. So I can choose $n=2q^2$ and $x_n=p+q\sqrt{2}$ close enought of $1$. Hence, $x_n=p+q\sqrt{2} - q\sqrt{2} = p \in \mathbb{Z}$.

I am pretty sure that this argument is invalid but I feel that the idea can be useful somehow.

答: $p+q\sqrt{2}$ with $p,q \in \mathbb{N}$ is not dense in $\mathbb{R}$, you probably mean $p,q \in \mathbb{Z}$. That's true, and an easy consequence of the irrationality of $\sqrt{2}$, but certainly not easier than a direct proof of your claim that the fractional part of $\sqrt{n}$ is dense in $[0,1]$: when $n$ grows from $m^2$ to $(m+1)^2-1$, the fractional part of the square root grows from $0$ to almost $1$, with differences $$\sqrt{n+1}-\sqrt{n}=\frac1{\sqrt{n+1}+\sqrt{n}}<\frac1{2m}.$$


问:
Solution to $p^n \equiv -1 \pmod d$
Here is an observation I found when fiddling around with numbers:

Let $G$ be the multiplicative group of by $p^n \bmod d$, and if there exists some $k$ that satisfies $3^k \equiv -1\pmod d$, then it seems that $ k = \frac{|G|}{2}$.

For example, the multiplicative group $\{3^n, \times_{\bmod 25}\}$ has order 20, and $3^{10} \equiv 24\pmod{25}$.

Is this true in general? Or does it only works with a set of values? Why/why not?

Thanks! If there is something not clear, pls tell me and I will reiterate what I mean.

答:
In general this is not true, consider e.g. $p=3, d=8$.

If $p$ and $d$ are coprime and $\pm1\pmod d$ are the only solutions of $x^2\equiv 1\pmod d$ (which is for example the case when $d$ is an odd prime power) then this is true. Note that $G$ is a cyclic group generated by $p$. So in particular the order of $p$ is $|G|=:k$. Since $p^{k}\equiv 1\pmod d$, we get $$(p^{k/2})^2\equiv 1\pmod d$$
and hence $p^{k/2}\equiv\pm1\pmod d$ by assumption. Since the order of $p$ is $k$ we then necessarily get $p^{k/2}\equiv -1\pmod d$.

Edit: (for the changed question) If $p^k\equiv-1\mod d$, then $p^{2k}\equiv 1\mod d$, so $2k$ is a multiple of $|G|$. If $k$ is minimal, then $k<|G|$, so the only possibility is $k=\frac{|G|}{2}$. If $k$ is not minimal then we might also have e.g. $k=\frac{|G|}{2}+|G|$.

Title: Solution to $p^n \equiv -1 \pmod d$


问:
Algebraic solution to $\frac{a}{1+bc}+ \frac{b}{1+ac}+ \frac{c}{1+ab} \le 2$
With $0 \le a \le b \le c \le 1$.

I know that this question has been answered here : If $0\le a,b,c\le 1$, then $\frac a{1+bc}+\frac b{1+ac}+\frac c{1+ab}\le 2$.

But I’m looking for a solution using some simple algebra .

My attempt:

We know that :
$$a^2+1 \geq 2a, a^2\le bc \implies \frac{bc+1}{a}\geq 2$$
$$\iff \frac{a}{bc+1}\le \frac{1}{2}$$

But this trick doest works on:
$$ \frac{b}{1+ac} $$
Because $b^2\geq ac$

答:
$WLOG$ $a\ge b\ge c$.

Then we have $\frac{a}{bc+1}+\frac{b}{ca+1}+\frac{c}{ab+1}\le \frac{a+b+c}{bc+1}$

so we only need to prove $\frac{a+b+c}{bc+1}\le 2$ , which is equivalent to $a+b+c\le 2bc+2$ , and this is equivalent to $(b-1)(c-1)+bc+(1-a)\ge 0$ , which is obviously true.


问:
Bisector of the acute angle formed between the lines $4x-3y+7=0$ and $3x-4y+14=0$

The bisector of the acute angle formed between the line $4x-3y+7=0$ and $3x-4y+14=0$ has the equation...

By calculating the intersection point, we get is as $(2,5)$.
But I couldn't proceed because I don't know how to find the equation on the acute side. Please help.

I would prefer a more geometrical approach

答: since  their bisector pass the point $(2,5)$ too, so after shifting their bisectors to meet whose intersection points at axis $Y$, we can deduce, the signature in bisector equation $\frac{\vert{k-4/3}\vert}{1+(4/3)k}=\frac{\vert{3/4-k}\vert}{1+(3/4)k}$, is between $3/4$ and $4/3$. thus one is positive and another is negative, then computation lead $k=1$. lastly, use the fact that their bisector pass through $(2,5)$ too, to deduce $2a-5b=-c$, thus $c=3$ between $7/3$ and $14/4$.


问:
Given $x,y\in\mathbb{C}^n$ s.t. $f(x,y)=\sup_{\theta,\phi}\{\|e^{i\theta}x-e^{i\phi}y\|^2,\theta,\phi\in\mathbb{R}\}$
Given$$x,y\in\mathbb{C}^n,\quad f(x,y)=\sup_{\theta,\phi}\{\|e^{i\theta}x-e^{i\phi}y\|^2,\theta,\phi\in\mathbb{R}\}$$

Then which is/are the following are true?

$1.\ f(x,y)\le \|x\|^2+\|y\|^2-2Re|\langle x,y\rangle|$

$2.\ f(x,y)\le \|x\|^2+\|y\|^2+2Re|\langle x,y\rangle|$

$3.\ f(x,y)= \|x\|^2+\|y\|^2+2Re\langle x,y\rangle$

$4.\ f(x,y)\ge \|x\|^2+\|y\|^2-2Re\langle x,y\rangle$

I have no idea how to start with or how to solve, could any one help me?

答:
First, an important equality!

$$\begin{align*}\|a-b\|^2 &= \langle a - b, a - b\rangle \\
&= \langle a,a \rangle - \langle a,b \rangle - \langle b,a \rangle + \langle b ,b \rangle \\
&= \|a\|^2 - (\langle a,b\rangle + \langle b,a\rangle) + \|b\|^2 \\
&= \|a\|^2 - (\langle a,b\rangle + \overline{\langle a,b\rangle}) + \|b\|^2 \\
&= \|a\|^2 - 2\mbox{Re}(\langle a,b\rangle) + \|b\|^2\end{align*}$$

Now, let's attack the norm inside the sup.

$$\begin{align*}\|e^{i\theta}x - e^{i\phi}y\|^2 &= \|x\|^2 - 2\mbox{Re}(\langle e^{i\theta}x, e^{i\phi}y\rangle) + \|y\|^2 \\
&= \|x\|^2 - 2\mbox{Re}(e^{i(\theta-\phi)}\langle x, y\rangle) + \|y\|^2 \end{align*}$$

Finally, use this nice fact about complex numbers: given any complex number $\lambda$, there is some $e^{ip}$ so that $\mbox{Re}(e^{ip}\lambda) = |\lambda|$. Geometrically, we are rotating the vector $\lambda$ so that it lies on the real axis. (We could also rotate to be $-|\lambda|$...)

Put these facts together, noting that $\mbox{Re}|\langle x,y\rangle| = |\langle x,y\rangle|$, since the norm is already real!


问:
Example of a product $\sigma$-algebra
I've been researching the definition of the product $\sigma$-algebra.

Could someone provide me with a concrete example. For example, if $S=\{1,2\}, T=\{2,3\}$ and we take the $\sigma$ algebras of both to be their respective power sets, then what would be the product $\sigma$-algebra of $S \times T$?

答:
That example is a bit too trivial IMO. Any set of the form $\{(s,t)\}$ where $(s,t) \in S \times T$ is a measurable rectangle, $\{s\} \times \{t\}$, so in the product $\sigma$-algebra. It follows, as everything is a finite union of singletons, that the product $\sigma$-algebra on $S \times T$ is also just the power set.

Maybe the finite-cofinite algebra on $\Bbb R$, and its square, are more interesting...

Title: Example of a product $\sigma$-algebra


问:
A linear system with inequalities constraint
For each $m,n \in \mathbb{N}$ think $\mathbb{R}^m$ as the space of real column vectors of size $m$ and $\mathbb{R}^{m \times n}$ as the space of matrices of size $m\times n$.

Let $d \in \mathbb{N}$.

Let $a:\{1,\dots,2^d\} \to \{1\}\times\{0,1\}^d$ be an enumeration (injective and surjective map).

Let $A \in \mathbb{R}^{(d+1)\times 2^d}$ be the matrix whose columns are $a(1),\dots, a(2^d)$.

Is it true that for each $b \in \{1\} \times [0,1]^d$ there exists $x \in [0,+\infty)^{2^d}$ such that
$$
Ax=b?
$$

It is clear that without the constraint on the sign of the components of $x$ the problem has a solution since $\operatorname{rank}(A) = d+1$. Also, for each specific instance of $b$ we could run the simplex algorithm to determine if it as a solution or not. However, I'm not quite sure about how to tackle this problem theoretically to give a proof that has always a solution (running various time the simplex algorithm on various instances always returned a solution, but, of course, it could be just luck). Any ideas?

答:
Yes, there is an $x$ with $x_i\in[0,1]$.

Essentially, you want to solve
$$
\sum_{n=1}^{2^d} x_n=1,\quad \sum_{\lfloor\frac{n}{2^k}\rfloor\equiv 1\pmod{2}}x_n=b_k
$$
and you can see
$$
x_n=\prod_{k=1}^d f_k(n),\quad f_k(n):=\begin{cases}b_k&\lfloor\frac{n}{2^k}\rfloor\equiv 1\pmod{2}\\1-b_k&\lfloor\frac{n}{2^k}\rfloor\equiv 0\pmod{2}\end{cases}
$$
is a solution.

How to come up with this  Induct on $d$ and you see each $x_i$ splits into two --- one that contributes to the new equation $\dots=b_{d+1}$ and one doesn't.  So naturally you scale those that contribute by $b_{d+1}$ and the other gets the $1-b_{d+1}$ leftover, and the base case is obvious.


问:
Probability of CDC data

All of this is my work, for (c) I believe its $1 - pbinom(144, 138, 2.9)$ which yields $.0192 $

For (d) I know that when D is negative, that would mean the woman is taller than the man, so I think it is just $ 69.2/63.8$ ?

Please correct me if I am wrong, Thank you in advance!

答:
(a) $P(X_1 > 72) = 0.1671.$  In R: I did not standardize because the CDF pnorm in R accepts
the actual means and standard deviations. [Irrelevant pbinom must be a typo--one I have sometimes made myself.]

```<code>1 - pnorm(72, 69.2, 2.9)
[1] 0.1671429

hdr = "Dist'n of Heights of Men"
curve(dnorm(x, 69.2, 2.9), 60, 80, lwd=2, ylab="PDF", main=hdr)
 abline(h=0, col="green2"); abline(v=72, col="red")
</code>```

(b) OK

(c) $P(S > 144) = 0.0037.$

```<code>1 - pnorm(144, 69.2+63.8, sqrt(2*2.9^2))
[1] 0.00365776

hdr = "Dist'n of Sums of Heights of Couples"
curve(dnorm(x, 69.2+63.8, sqrt(2*2.9^2)), 120, 150,
      lwd=2, ylab="PDF", main=hdr)
  abline(h=0, col="green2"); abline(v=144, col="red")
</code>```

(d) OK

(e) $P(D < 0) = 0.0940.$

```<code>pnorm(0, 69.2-63.8, sqrt(2*2.9^2))
[1] 0.09397269

hdr = "Dist'n of Man's Ht - Woman's Ht"
curve(dnorm(x, 69.2-63.8, sqrt(2*2.9^2)), -10, 20,
            lwd=2, ylab="PDF", main=hdr)
 abline(h=0, col="green2"); abline(v=0, col="red")
</code>```

Also, by simulation of a million couples, which should give
a probability correct to about two places.

```<code>set.seed(2021)
m = rnorm(10^6, 69.2, 2.9)
w = rnorm(10^6, 63.8, 2.9)
mean(m < w)
[1] 0.094211
</code>```

Title: Probability of CDC data

Body:
(a) $P(X_1 > 72) = 0.1671.$  In R: I did not standardize because the CDF pnorm in R accepts
the actual means and standard deviations. [Irrelevant pbinom must be a typo--one I have sometimes made myself.]

1 - pnorm(72, 69.2, 2.9)
[1] 0.1671429

hdr = "Dist'n of Heights of Men"
curve(dnorm(x, 69.2, 2.9), 60, 80, lwd=2, ylab="PDF", main=hdr)
 abline(h=0, col="green2"); abline(v=72, col="red")

(b) OK

(c) $P(S > 144) = 0.0037.$

1 - pnorm(144, 69.2+63.8, sqrt(2*2.9^2))
[1] 0.00365776

hdr = "Dist'n of Sums of Heights of Couples"
curve(dnorm(x, 69.2+63.8, sqrt(2*2.9^2)), 120, 150,
      lwd=2, ylab="PDF", main=hdr)
  abline(h=0, col="green2"); abline(v=144, col="red")

(d) OK

(e) $P(D < 0) = 0.0940.$

pnorm(0, 69.2-63.8, sqrt(2*2.9^2))
[1] 0.09397269

hdr = "Dist'n of Man's Ht - Woman's Ht"
curve(dnorm(x, 69.2-63.8, sqrt(2*2.9^2)), -10, 20,
            lwd=2, ylab="PDF", main=hdr)
 abline(h=0, col="green2"); abline(v=0, col="red")

Also, by simulation of a million couples, which should give
a probability correct to about two places.

set.seed(2021)
m = rnorm(10^6, 69.2, 2.9)
w = rnorm(10^6, 63.8, 2.9)
mean(m < w)
[1] 0.094211


问:
Prove $ \sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$
Let $ABC$ be acute triangle. Prove that
$$\sum \sqrt{\tan A} \geq \sum \sqrt{\cot\frac{A}{2}}$$
My attempt:
$$\sqrt{\tan A} + \sqrt{\tan B}\geq 2\sqrt[4]{\tan A\cdot\tan B}$$
At here I think I need to prove $$2\sqrt[4]{\tan A\cdot\tan B}\ge2\sqrt{\cot\frac{A}{2}}$$
$$\Leftrightarrow \tan A\cdot\tan B\ge \cot^2\frac{A}{2}$$
And I was stuck here.
Help me... Thanks

答:
Remark: $\tan A,\ \tan B,\ \tan C,\ \cot\frac{A}{2},\ \cot\frac{B}{2},$ and $\cot\frac{C}{2}$ are all positive because $0<A,\ B,\ C<\frac{\pi}{2}$. This is used in this proof, for example in cancellings or taking square roots of both sides of inequalities.

The function $f\left(x\right)=\tan x$ is convex on the interval $\left(0,\ \frac{\pi}{2}\right)$, so we can use Jensen's Inequality to show that

$\begin{align}
\sum_\text{cyc}\frac{\tan A+\tan B}{2}\ge\sum_\text{cyc}\tan\frac{A+B}{2}=\sum_\text{cyc}\cot\frac{C}{2}\Rightarrow\sum_\text{cyc}\tan A\ge\sum_\text{cyc}\cot\frac{A}{2}\ \left(\#\right)
\end{align}$

It remains to prove that $\sum_\limits\text{cyc}\sqrt{\tan A\tan B}\ge\sum_\limits\text{cyc}\sqrt{\cot\frac{A}{2}\cot\frac{B}{2}}\ \left(\#\#\right)\Leftrightarrow\\$
$\begin{align}
&\Leftrightarrow\sum_\text{cyc}\tan A\tan B\ge\sum_\text{cyc}\cot\frac{A}{2}\cot\frac{B}{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\frac{2\cot\frac{A}{2}}{\cot^{2}\ \frac{A}{2}-1}\cdot\frac{2\cot\frac{B}{2}}{\cot^{2}\ \frac{B}{2}-1}\ge\sum_\text{cyc}\cot\frac{A}{2}\cot\frac{B}{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\frac{4}{\left(\cot^{2}\frac{A}{2}-1\right)\left(\cot^{2}\frac{B}{2}-1\right)}\ge0\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\left(\cot^{2}\frac{A}{2}-1\right)\left(\cot^{2}\frac{B}{2}-1\right)\ge0
\end{align}$

which is true because $0<A,\ B<\frac{\pi}{2}\Leftrightarrow$

$\Leftrightarrow1<\cot\frac{A}{2},\ \cot\frac{A}{2}<+\infty$.

From $\left(\#\right)$ and $\left(\#\#\right)$ we have

$\begin{align}
&\sum_\text{cyc}\tan A+2\sum_\text{cyc}\sqrt{\tan A\tan B}\ge\sum_\text{cyc}\cot\frac{A}{2}+2\sum_\text{cyc}\sqrt{\cot\frac{A}{2}\cot\frac{B}{2}}\Leftrightarrow\\
&\Leftrightarrow\left(\sum_\text{cyc}\sqrt{\tan A}\right)^{2}\ge\left(\sum_\text{cyc}\sqrt{\cot\frac{A}{2}}\right)^{2}\Leftrightarrow\\
&\Leftrightarrow\sum_\text{cyc}\sqrt{\tan A}\ge\sum_\text{cyc}\sqrt{\cot\frac{A}{2}}\ \cdot
\end{align}$

Done!


问:
Integrate $\oint_{C} \sec z \mathrm{d}z$
The literature I read says that

$$\oint_C \sec z \mathrm{d}z = 0$$

This is what I know;
From the Cauchy integral theorem $\oint_C f(z)\mathrm{d}z = 0$ if $f(z)$ is analytic in a simple connected domain $D$ for every simple closed path $C$ in $D$.

But $f(z) = \sec z$ is not analytic at $\frac{(2n + 1) \pi}{2}$, $n \in \mathbb{Z}$

答:
Actually, if you have to evaluate the following integral $~\oint_C \sec(z)~dz~$ in counter clockwise direction around the unit circle $C:|z|=1$, then you get the value of the integral is zero. The reason behind this is as follows:

Clearly, $~\sec(z)~$ is analytic at all pint except at $~\cos(z)=0~$ i.e., at $~z=\frac{(2n + 1) \pi}{2}~, n \in \mathbb{Z}.$

So poles are at $~z_0=\cdots,~-\frac{3\pi}{2},~-\frac{\pi}{2},~\frac{\pi}{2},~\frac{3\pi}{2},~ \cdots~$

But none of these poles lie inside the unit circle $C:|z|=1$.

Hence the sum of the residues at the poles is zero.

Therefore by residue theorem,
$$I=2\pi i\left[\cdots+\text{Res}(f, -3\pi/2)+\text{Res}(f, -\pi/2)+\text{Res}(f, 3\pi/2)+\text{Res}(f, \pi/2)+\cdots\right]=0$$


问:
Closed form coefficients of Elementary Symmetric Polynomial specialized to the natural numbers
I am writing a program that computes a Lagrange polynomial using parallel processing for data points situated at every natural number. But I can have $n$ many data points, which will potentially create a polynomial of degree $n$. Doing some math and research I found out that what I need is a general form for the $ith$ coefficient of the Elementary Symmetric Polynomial specialized to the natural numbers. I just asked a question about about a double sum inside one of the solving for one of the coefficients partial sum of $\sum^h_{i=1}\sum^i_{j=1}ij$. But I thought I would ask the whole thing here.

To give some examples the coefficients of the cubic case would be $x^3-(1+2+3)x^2+(1*2+2*3+1*3)x-1*2*3$ and the quartic case would be $x^4-(1+2+3+4)x^3+(1*2+2*3+1*3+1*4+2*4+3*4)x^2-(1*2*3+1*3*4+1*2*4+2*3*4)x-1*2*3*4$

Because the polynomials would be like $(x-3)(x-2)(x-1)$ as an example.

I spent all morning coming up with only the first three partial sums of for the coefficients.

Which are $1,\frac{n(n+1)}{2},\frac{(n-1)n(n+1)(3n+2)}{24},\frac{n^2(n+1)^2(n-1)(n-2)}{48}$. Before I embark on trying to get the next coefficient then trying to generalize from there, is there a solution (I am not an expert in what seems to be combinatoric algebra) that someone can point me too for the closed form for each of the coefficients or a proof that someone can show.

I read this but this is a little over my head meaning that I don't know if my answer is in there or not. I am computer guy, so any simplified explanation is greatly appreciated.

Any help is much appreciated! Thanks in advance!

Edit: I did more research and I am came across Vieta's formula, maybe I can use this for a closed form.

Edit 2: I found this post talking about Stirling numbers Coefficients of polynomial $(x+1)(x+2)...(x+n)$ I am wondering if I can manipulate it to my answer

答:
I found the answer after a while. The coefficients are calculated using the signed Stirling numbers of the first kind.

These are the signed stirling numbers of the first kind

In order to get the coefficients you select the row by the degree of the polynomial and read it from right to left.

To calculate it, the facts I used were

$$s(n,k)=-(n-1)\cdot s(n-1,k)+s(n-1,k-1)$$
$$s(n,n) = 1$$
$$s(n,0) = 0, n >0$$

And what is interesting is the polynomials that I found
$1,\frac{n(n+1)}{2},\frac{(n-1)n(n+1)(3n+2)}{24},\frac{n^2(n+1)^2(n-1)(n-2)}{48}$
describe the diagonals.

So for instance $\frac{n(n+1)}{2}$ is the diagonal 0,1,3,6,10,15,... (at positions (2,1),(3,2),(4,3),...)

And another example would be for $\frac{(n-1)n(n+1)(3n+2)}{24}$ where the value on the diagonal are 2,11,35,...

So a future follow up question would be is there a closed form for the diagonals of the stirling numbers of the first kind?


问:
conjugate function of log trace of matrix exponential
Given an n-by-n real symmetric matrix X, define
$$e^X = \sum_{k=0}^{n}\frac{X^k}{k!}$$
Derive the Fenchel conjugate of
$$f(X) = log(Tr(e^X))$$
as a function on n-by-n real symmetric matrices. Here Tr denotes the trace function.

答:
By definition, the Fenchel conjugate of $f(X)$ is given by:
\begin{equation}
\label{eq:f}\tag{1}
f^{*}(Y) = \sup_{X} X\circ Y - f(X).
\end{equation}

It can be shown that $g(X) \overset{\text{def}}{=} X\circ Y - f(X)$ is a concave function in terms of $X$.
(Actually, you can use the second derivative of $f(X)$ with respect to $X$ to show that $f(X)$ is a convex function.)
To compute the supremum of $g(X)$, we take derivative of $g(X)$ with respect to $X$, we have
\begin{equation*}
\begin{aligned}
Y= \frac{(\exp X) }{\text{Tr} (\exp X) }.
\end{aligned}
\end{equation*}
Thus, $I\circ Y=  1$ and $y_{ii}>0$ for $1\le i\le n$.
Using the fact $\log (AB) =\log (A)  + \log (B) $ if $AB=BA$, it follows that
\begin{equation}
\label{eq:y}\tag{2}
Y = \frac{(\exp X) }{\exp f(X)} \implies \log Y + f(X) I  =  X.
\end{equation}

Substitute \eqref{eq:y} into \eqref{eq:f}, we have

\begin{align}
\label{eq:f:final}
f^{*}(Y) & = (\log Y + f(X)  I  ) \circ Y - f(X)\notag \\
& = \log Y \circ Y + f(X)  I \circ Y - f(X) \notag\\
& = \log Y \circ Y. \tag*{using $I\circ Y=  1$.}
\end{align}


问:
Finding a solution for a polynomial
Let $p$ belongs to IR, find the number of real solutions of the equation of $x^5 - 5 x = p$.
how can we solve the problem ?

答: Hint : Study the function $x \mapsto x^5-5x$ and apply IVT on the intervals where the function is monotonous.

Title: Finding a solution for a polynomial


答:
Hint:

It is an easy matter to show the the polynomial has a single maximum at $(-1,4-p)$ and a single minimum $(1,-4-p)$. We have the table of variations

$$\begin{array}{}-\infty&&-1&&1&&\infty\\\hline-\infty&\nearrow&4-p&\searrow&-4-p&\nearrow&\infty\end{array}$$

From this you determine where there can be roots, depending on the values of $p$.

Title: Finding a solution for a polynomial

Body:
Let $p$ belongs to IR, find the number of real solutions of the equation of $x^5 - 5 x = p$.
how can we solve the problem ?


问:
Let $F_g: F(X) \to F(X)$ be a linear map such that $f \mapsto gf$. Compute the determinant of $F_g.$

Let $X =\{a,b,c\}$ and let $g : X \to \mathbb{R}$ be a function for which $g(a)=2,g(b)=4,g(c)=-1.$ Let $F_g: F(X) \to F(X)$ be a linear map such that $f \mapsto gf$ where $F(X)$ is the set of all functions $f:X \to \mathbb{R}$. Compute the determinant of $F_g.$

I have a slight problem when trying to express $F_g$ as a matrix. If I understood the problem correctly I have that $$F_g(f(a))= g(a)f(a) =2f(a) \\ F_g(f(b)) =g(b)f(b) = 4f(b) \\F_g(f(c)) =g(c)f(c) = -f(c)$$ but I would need to express this as a matrix in order to find the determinant. The fact that every linear transformation can be expressed as a matrix should apply here as well since $F_g$ is a linear map. How can I proceed with this?

答: Assuming that $F(x)$ is the set of all maps from $X$ to $\Bbb R$, then a basis of $F(x)$  will be $B=\{e_a,e_b,e_c\}$, where $e_x(y)$ (with $x,y,\in\{a,b,c\}$) is $1$ if $x=y$ and $0$ otherwise. Now, $F_g(e_a)=g\times e_a=2e_a$. You also have $F_g(e_b)=4e_b$ and $F_g(e_c)=-e_c$. So, the matrix of $F_g$ with respect to $B$ is$$\begin{bmatrix}2&0&0\\0&4&0\\0&0&-1\end{bmatrix}.$$


问:
Show that $f(z) = z^n$ is an entire function
Show that $f(z) = z^n$ is an entire function where $n = 0,1,2...$

I can show that $z$, $z^2$ and $z^3$ are entire functions but what about $z^n$ how do you crack that one.

答:
Hint: the product of two entire function is entire (product rule).

Now use induction for your problem.


问:
Example for monotone convergence
I'm searching for an example of sequence of function in which if doesn't exist a Lebesgue integrable function g that increases the sequence, then the equality in the convergence theorem is not anymore maintained.

Thank you.

答:
Take
$$f_n(t) = \left\{
    \begin{array}{ll}
       1 & \mbox{if } t \geq n \\
        0 & \mbox{otherwise.}
    \end{array}
\right.$$

Then $$\int \lim_{n \rightarrow +\infty} f_n = 0 \neq +\infty = \lim_{n \rightarrow +\infty} \int f_n$$

You see here that the theorem is wrong if the sequence is decreasing.

Title: Example for monotone convergence


问:
Infinite convex combination is in the closure of the convex hull.
Let $V$ be a topological vector space and $\{x_i\}_{i \in I}$ be a net in $V$. Further, let $\{\lambda_i\}_{i\in I}$ be a net in $[0,1]$ such that
$$\sum_{i \in I}\lambda_i =1.$$ Assume $x= \sum_{i \in I} \lambda_i x_i$ exists in $V$. That is, $$x= \lim_F \sum_{i \in F} \lambda_i x_i$$
where $F$ ranges over all finite subsets of $I$.

Is it true that $x$ is in the closure of the convex hull of $\{x_i:i \in I\}$?

Intuitively, this seems clear, but I would need to show that $x$ is the limit of (finite) convex combinations of $\{x_i: i \in I\}$, but I can't see why this is true.

If necessary, I can assume that $V$ is a normed space although I think the result holds more generally.

答:
Yes, it holds in any topological vector space: $x=\lim (\sum_{i \in F} \lambda_i) \frac {\sum_{i \in F} \lambda_ix_i} {\sum_{i \in F} \lambda_i}$ and $\frac {\sum_{i \in F} \lambda_i x_i} {\sum_{i \in F} \lambda_i}$ is in the convex hull of $\{x_i: i \in I\}$.

As pointed out by QuantumSpace we must take limit along  subnet to avoid division by $0$.


问:
Representation of Linear Map in Matrix Form
In the Matrices section of the book Linear Algebra Done Right, there's a section I'm finding hard to understand. The passage is:

The $k^{th}$ column of M(T) consists of the scalars needed to write
$Tv_k$ as a linear combination of ($w_1$,.....,$w_m$):

$Tv_k$ = $\sum^{m}_{j=1} A_{j,k}w_j$

I'm having hard time understanding the summation part. Let for example we have linear map from $\mathbb{R}^2\to \mathbb{R}^2$ of this form.

$T\begin{pmatrix}1\\0\end{pmatrix}$ = $\begin{pmatrix}3\\4\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}$ = $\begin{pmatrix}6\\9\end{pmatrix}$.

In this example the matrix $A$ is \begin{pmatrix}
3 & 6\\
4 & 9\\
\end{pmatrix}

Then how we are reaching to $Tv_k$ = $\sum^{m}_{j=1} A_{j,k}w_j$ in this example?

Let $T\begin{pmatrix}1\\0\end{pmatrix}$. Now, how will I represent it in terms of

$\sum^{m}_{j=1}A_{j,1}$$w_j$ where $A_{j,1} = \begin{pmatrix}
3\\
4\\
\end{pmatrix}$ and $w_j = \begin{pmatrix}3\\4\end{pmatrix}$

Basically, how the summation is getting into this portion?
Thanks!

答:
Linear mapping $T:\Bbb R^n\rightarrow\Bbb R^m$ can be represented by the matrix

$A = \left( \begin{array}{cccc} w_1 & w_2 &\ldots & w_n \end{array} \right)$

where $w_i = Tv_i$ is the image (column vector) of basis element $v_i$, where $\Bbb R^n$ has basis $\{v_1,\ldots, v_n\}$.

Then matrix multiplication

$Ax = \left( \begin{array}{cccc} w_1 & w_2 &\ldots & w_n \end{array} \right)\left(\begin{array}{c} x_1 \\\vdots\\ x_n \end{array}\right)$ gives the column vector
$x_1w_1+\ldots,x_nw_n$, which is a linear combination of the columns of $A$.


问:
In a ring $R$, if $xRy\subseteq I$ where $I\unlhd R$ is an ideal, then $(x)(y)\subseteq I$
I came to the part of the proof that i can't quite crack. If $xRy\subseteq I$ where $I\unlhd R$ is an  ideal, then $(x)(y)\subseteq I$.

Attempt: Let $z \in (x)(y)$. Then $z$ is of the form:
$$z=\sum_{i=1}^nx_iy_i,\quad x_i\in (x), y_i\in(y)$$
Every $x_iy_i$ can be written in the form:
$$\left(\sum_{k=1}^{n_x}r_kxs_k + rx+xs + m_xx\right)\left(\sum_{k=1}^{n_y}l_kyt_y + ly + yt + m_yy\right)$$
where all $r_k,s_k,l_k,t_k,r,l,t,s\in R$ and $m_x,m_y \in \mathbb Z$. I realised that if you multiply any two terms in the two parentheses you get something in $I$ (it comes from the fact that $xRy \subseteq I$ and $I$ being an ideal), except the last two terms $m_xx \cdot m_yy= (m_x\cdot m_y)xy$ which is dont have to be in $xRy$ since $R$ doesn't need to have a $1$.

答:
The result is just not true in general if $R$ does not have a unit.

As an example, $R=2\mathbb{Z}$, $I=8\mathbb{Z}$, and $x=y=2$. Then $xRy=I$, but $xy=4\not\in I$.


问:
Use integration by parts twice $\int_{0}^{\frac{\pi}{8}}x^2\cos(4x)dx$
I am confuse about this integration by two parts I don't get the steps and any help would be much appreciated  i know the answer is 0.00730 but don't know how to get to it. This is the work

Body:
I'll do the first iteration for you. Set
\begin{align}
u &= x^2 & \frac{dv}{dx}&=\cos(4x) \\
\frac{du}{dx} &= 2x & v&=\frac{1}{4}\sin(4x)
\end{align}
Hence, the integral equals
\begin{align}
\int_{0}^{\pi/8} u\frac{dv}{dx} \, dx &= \left[uv\right]_{0}^{\pi/8} - \int_{0}^{\pi/8} v\frac{du}{dx} \, dx \\
\int_{0}^{\pi/8} x^2\cos(4x) \, dx &= \left[\frac{1}{4}x^2\sin(4x)\right]_{0}^{\pi/8} - \int_{0}^{\pi/8}\frac{1}{2}x\sin(4x) \, dx
\end{align}
Now set $u=\frac{1}{2}x$ and $\frac{dv}{dx}=\sin(4x)$. Let me know how you get on.

Let $$I=\int_{0}^{\pi/8}\frac{1}{2}x \sin(4x) \, dx \, .$$
If we set $u=\frac{1}{2}x$ and $\frac{dv}{dx}=\sin(4x)$, then we get that $\frac{du}{dx}=\frac{1}{2}$ and $v=-\frac{1}{4}\cos(4x)$. Hence,
\begin{align}
I &= \left[-\frac{1}{8}x\cos(4x)\right]_{0}^{\pi/8} + \int_{0}^{\pi/8}\frac{1}{8} \cos(4x) \, dx \\
&= 0 + \int_{0}^{\pi/8}\frac{1}{8} \cos(4x) \\
&= \left[\frac{1}{32}\sin(4x)\right]_{0}^{\pi/8} \\
&= \frac{1}{32}\sin(\pi/2) \\
&= \frac{1}{32} \, .
\end{align}
Hence,
\begin{align}
\int_{0}^{\pi/8} x^2\cos(4x) &= \left[\frac{1}{4}x^2\sin(4x)\right]_{0}^{\pi/8} - \frac{1}{32} \\
&= \frac{\pi^2}{256}-\frac{1}{32} \\
&= \frac{\pi^2 - 8}{256} \, .
\end{align}


问:
Probability of picking $z-n$ uniquely colored balls successively and $n$ repeats without replacement
A follow up on this post.

In a bucket, there are $x$ different colors of balls, $y$ of each color
($x \cdot y$ total balls). If you pick $z$ balls at random without replacement, what is the probability that you pick a different colored ball each time?

Generalized answer by Bram28:

$x$: number of colors

$y$: number of balls per color

$z$: number of picks

$P$: chance of all $z$ picks having a different color

Now, if $z > x$, then obviously $P = 0$.

Otherwise (i.e. if $z \leq x$):
$$
P = \frac{\binom{X}{Z} \cdot Y^Z}{\binom{X \cdot Y}{Z}}
$$

Question:

Lets say $n$ of the picks $z$ are repeated colors, how would I get the probability to pick $z-n$ unique colors and $n$ repeated colors? I'm having trouble including the number of balls per color $y$ to calculate the probability, since I want $n$ repeats but there are only $y$ balls per color.

(This problem is "without replacement", and the repeats I'm trying to calculate come from the colors having multiple balls.)

Finally, I want to add all probabilities with  $0 \leq n \leq y$.

.

.

Also, I plotted $P$ as $x$ increases and I get values greater than $1$, I don't understand how that can happen for this problem.

答:
The solution to the modified problem is:
$$
P = \frac{\binom{x}{1}\binom yn\cdot \binom{x-1}{z-n}\binom y1^{z-n}}{\binom{x \cdot y}{z}},\tag1
$$
where $1<n\le y$ is assumed.

The general case

The most general statement of the problem would be the following: what is the probability of the event that the drawn balls are of $k$ (unspecified) colors with the number of balls for each of these colors being represented by the multiset $Z=\{z_1,z_2,\dots,z_k\}$ with $z_i>0$ and $\sum z_i=z$.

The probability in question will be (equal number $y$ of balls of each color is assumed, generalization is obvious):
$$
P_Z=\frac{\binom xk\prod\limits_{i=1}^k\binom{y}{z_i}}{\binom{x \cdot y}{z}}M(Z),\tag2
$$
with $M (Z) $ being the number of ways to permute the elements of $Z$:
$$
M(Z)=\frac{k!}{\prod\limits_{i\in Z}\mu_Z(i)!},\tag3
$$
where $\mu_Z(i)$ is the multiplicity of the number $i$ in the multiset $Z$.

All previously considered cases are particular instances of the formula (2). The formula can be slightly simplified if one adds $x-k$ $0$'s to the multiset $Z$.


答:
The integral is $\int_0^{\frac{\pi}{8}}x^2\cos(4x)dx$.
Let $u(x)=x^2$ and $v'(x)=\cos(4x)$. In those case, always differentiate the polynomial and integrate the trigonometric function.

We have $u'(x)=2x$, and a possible candidate for $v$ is $v(x)=\frac14\sin(4x)$.
Then
\begin{align}
\int_0^{\frac{\pi}{8}}x^2\cos(4x)dx
&=\int_0^{\frac{\pi}{8}}u(x)v'(x)dx\\
&=\left[u(x)v(x)\right]_0^{\frac{\pi}{8}}-\int_0^{\frac{\pi}{8}}u'(x)v(x)dx\\
&=\left[x^2 \times\frac14\sin(4x)\right]_0^{\frac{\pi}{8}}-\frac12\int_0^{\frac{\pi}8}x\sin(4x)dx
\end{align}

Evaluate the bracket, then use again an integration by parts for the second integral.

Title: Use integration by parts twice $\int_{0}^{\frac{\pi}{8}}x^2\cos(4x)dx$

Body: I am confuse about this integration by two parts I don't get the steps and any help would be much appreciated  i know the answer is 0.00730 but don't know how to get to it. This is the work


问:
Munkres Topology -- Solution too straight forward? (S13.5)
Throughout my time studying math, I've generally realized that when an exercise feels "too straightforward" it likely means I've made a jump in my logic and did not justify myself properly. I was asked to prove the following fact:

Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$.

There is also a secondary part to prove this for a subbasis, but I am ignoring it temporarily. My proof is as follows:

The topology generated by $\mathcal{A}$ (denoted $\mathcal{T}_{\mathcal{A}})$ is a topology containing $\mathcal{A}$, and thus $\mathcal{T}_{A} \in \lbrace\mathcal{T}_{i} \mid \mathcal{A} \subset \mathcal{T}_{i} \rbrace$. $\mathcal{T}_{\mathcal{A}} \subset \mathcal{T}_{i} ~ \forall i$ as $\mathcal{A} \subset \mathcal{T}_{i}$ and topologies are closed under unions and intersections of their elements. It follows then that $\bigcap_{i \in I} \mathcal{T}_{i} = \mathcal{T}_{\mathcal{A}}$

Am I beating myself up for no reason? Self studying this book after I graduated with my degree has me feeling nervous (prepping for grad school and have never taken topology). I also apologize if this question has been asked, I find more people having trouble with the subbasis part.

答:
Your proof is fine and complete IMO.

We could fancify (if that is a word) it a little bit, and add more explanations:

Let $\mathscr{T}$ be the set of all topologies on $X$ that contain $\mathcal{A}$.

It’s clear that $\mathcal{T}_{\mathcal{A}} \in \mathscr{T}$ as members of $\mathcal{T}_{\mathcal{A}}$ are unions of subfamilies of $\mathcal{A}$ (also unions of singleton subfamilies) and so in particular $\mathcal{A} \subseteq \mathcal{T}_{\mathcal{A}}$. So trivially, as the intersection of a non-empty family is a subset of each of its members:

$$\bigcap \mathscr{T} \subseteq \mathcal{T}_{\mathcal{A}}$$

On the other hand, if $\mathcal{T} \in \mathscr{T}$ we know that as topologies are closed under unions, that $\mathcal{T}_{\mathcal{A}} \subseteq \mathcal{T}$. As the latter was arbitrary in $\mathscr{T}$ we also have:

$$\mathcal{T}_{\mathcal{A}} \subseteq \bigcap \mathscr{T}$$

And these inclusions show the required equality.


问:
Ball properties with different metrics
I am stuck with the following intuitive question:

Let $M \subset \mathbb{R}^n$ be an open bounded set (with standard Euclidean topology). Let $g_1$ be the induced Euclidean metric from $\mathbb{R}^n$ to $M$ and let $g_2$ be another (arbitrary) metric on $M$ such that $g_2$ induces the same topology on $M$.

Question:

For any $\delta >0$, is it always true that there is $\epsilon >0$ such that
$$B_{g_1}(x,\epsilon) \subset B_{g_2}(x,\delta),\forall x \in M?$$

Here $B_{g_1}(x,\epsilon):=\{y\in M: d_{g_1}(x,y) <\epsilon\}$ and $B_{g_2}(x,\epsilon):=\{y\in M: d_{g_2}(x,y) <\epsilon\}$ are open balls of radius $\epsilon$ w.r.t. each metrics.

I think this is not true but I couldn't find a counterexample.

答: Not true: Take $n=1, M=(0,1)$ and $g_2(x)=|\frac  1 x-\frac 1 y|$. Then $g_2$ has the same open sets as $g_1$ and the inclusion you have written says that $\frac  1 x$ is uniformly continuous on $(0,1)$ which is false.

Title: Ball properties with different metrics


问:
How do we solve $y''=e^{2y}$?
I tried to make the substitution $v=y'(x)$, which would ( hopefully ) make this into a separable ODE as per the following. $$y''=e^{2y}$$ by the chain rule, we have $v'=v'(y(x)) \cdot y'(x)=\frac{dv}{dy}\frac{dy}{dx}$. If we make the substitution, then $$\frac{dv}{dy}\frac{dy}{dx}=e^{2y}$$

So we could cancel the differential terms $dy$? I wonder if we can really do so, and also a better substitution might be just $v=y(x)$. Am I on the right track here?

答:
Multiply both sides by $y'$

$$y'y''=y'e^{2y} \implies (y')^2 = e^{2y}\pm C_1^2$$

Then square root and separate variables

$$\frac{e^{-y}dy}{e^{-y}\sqrt{e^{2y}\pm C_1^2}}= \pm dx$$

$$ \begin{cases} \frac{\sinh^{-1}(C_1e^{-y})}{C_1} = \pm x + C_2 \\ \frac{\sin^{-1}(C_1 e^{-y})}{C_1} = \pm x + C_2 \\ e^{-y} = \pm x + C \end{cases}$$ depending on the sign of the $C_1^2$ term. Each of these is invertible to get

$$\begin{cases} y = -\log(\pm \sinh(C_1 x + C_2) ) + \log C_1 \\ y = -\log(\pm \sin(C_1 x + C_2)) + \log C_1 \\ y = -\log(\pm x + C) \end{cases}$$

Note the first and second sets of $C_2$s are not equivalent.

Title: How do we solve $y''=e^{2y}$?


问:
Is the function $f(x,y)$ describing the given graph is symmetric for any $x\neq y$? Or, only for $x=y$?
We have an infinite periodic graph with the two lengths $x$ and $y$

and we have a function $f(x,y)$ that describes a specific property of this periodic structure.

Then, according to the symmetry of the graph, should the function $f(x,y)$ be symmetric under the permutation $\;x\leftrightarrow y\;$ for any $x\neq y$? Or, only for $x=y$?

答: Assuming the "periodic structure" has no notion of a "starting point" (or at least, if $f$ is invariant under horizontal translations of the structure), then $f$ should be invariant under swapping the lengths, since this just amounts to translating the structure by length $x$.


答:
The tabular method (or the DI method) is great for integrals that require integration by parts multiple times:

\begin{array}{ccc}
& \text D & \text I \\
+ & \color{blue}{x^2} & \cos4x \\
- & \color{red}{2x} & \color{blue}{\frac14\sin4x} \\
+ & \color{green}{2} & \color{red}{-\frac1{16}\cos4x} \\
- & 0 & \color{green}{-\frac1{64}\sin4x}
\end{array}

From this we know that $$\int x^2\cos4x\,dx=\frac14x^2\sin4x+\frac18x\cos4x-\frac1{32}\sin4x+C.$$

Since this is a definite integral, simply evaluate the antiderivative from the bounds of integration (I will leave the work to you) and we have the result $$\frac1{256}(\pi^2-8).$$


问:
Stream functions and divergence?

We see that the existence of a stream function guarantees that the vector field has zero
  divergence or, equivalently, is source free. The converse is also true on simply connected
  regions of $\mathbb{R}^2$

Why does the region have to be simply connected on $\mathbb{R}^2$ for the converse to be true?

If the vector field has zero divergence,
$$\frac{\partial{f}}{\partial{x}} + \frac{\partial{g}}{\partial{y}} = 0$$
$$\frac{\partial{f}}{\partial{x}} = -\frac{\partial{g}}{\partial{y}}$$

Integrating either side with respect to the appropriate variable would give you the stream line function. Doesn't this show that if a vector field has zero divergence it has a stream function no matter what?

答:
This being said, there is a result in non simply connected domain. If
$$ \textrm{div}D=0,$$ and on the boundary $\Gamma_i$ of each connected component of $\Omega$ there holds $$\int_{\Gamma_i} D\cdot n \textrm{d}\sigma =0,$$ then there exists a function $H$ such that $$D=(-\partial_{x_2} H,\partial_{x_1} H).$$

See Lemma I.1 in Brezis, Bethuel & Hélein's book. The proof is that you can solve the Neumann problem on each hole, and then supplement $D$ with the gradient of the solutions, obtaining a divergence free problem on the corresponding simply connected set.

Title: Stream functions and divergence?


问:
What are the finite subgroups of $SU_2(C)$?
Is there any reference which classifies the finite subgroups of $SU_2(C)$ up to conjugacy ?

What I know is that lifting the finite subgroups of $SO_3(R)$ by the map $SU_2(C) \rightarrow PSU_2(C)$ gives rise to the groups : cyclic groups of even order, dicyclic groups, binary tetrahedral/octahedral/icosahedral groups. But I dont know if they are the only one.

答:
Yes, the ADE classification enumerates all possible finite subgroups of $SU(2)$. Just a trivial correction, when it comes to the Abelian cyclic groups, A, the order may be both even and odd because $U(1)$ inside $SU(2)$ has all these $Z_N$ subgroups. The odd ones aren't linked to subgroups of $SO(3)$ in the same way.

The dicyclic groups, D, and the three exceptions, E, are as you wrote.


问:
Simple proof to show $\mathbb{Q}_2((-2)^{\frac{1}{2}})$ is contained in a cyclotomic extension
Let $(-2)^{\frac{1}{2}}$ be any element in $\overline{\mathbb Q_2}$ satisfying $x^2+2=0$, then is there a simple way to show that $\mathbb{Q}_2((-2)^{\frac{1}{2}})$ is contained in a cyclotomic extension of $\mathbb{Q}_2$?

What I know is that the polynomial $x^2+2$ is Eisenstein over $\mathbb{Q}_2$, hence irreducible, but I don't see how to continue.

This is a step in the proof of  local Kronecker-Weber theorem, so we can't use this theorem.

Thanks.

答:
$\newcommand{\Qp}{\Bbb Q_p}$Let's first examine cyclotomic extensions of $\Qp$. For any $k$, let $\mu_k$ denote the set of $k$th roots of unity.

If $k$ is coprime to $p$, then $\Qp\left(\mu_k\right)$ is an unramified extension, and so it is equal to $\Qp\left(\mu_{p^f-1}\right)$ where $f$ is the degree of the residue field extension. In fact, $f$ is the multiplicative order of $p$ mod $k$. In any case, it suffices to consider $p^f-1$ in the case of being coprime to $p$.

On the other end of the spectrum, $k$ is a power of $p$, say $k = p^d$. The minimal polynomial of $\mu_k$ over $\Bbb Q$ is $\Phi_{p^d} = \dfrac{X^{p^d}-1}{X^{p^{d-1}}-1} = \displaystyle \sum_{i=0}^{p-1} X^{ip^{d-1}}$. We don't know if it is irreducible over $\Qp$, so let $g(X) := \Phi_{p^d}(X+1)$, and we shall show that $g$ is Eisenstein at $p$:

The leading coefficient of $g$ is $1$.

The constant coefficient of $g$ is $p$.

To show that the other coefficients are divisible by $p$, we reduce modulo $p$ and get $\Phi_{p^d} = \dfrac{(X-1)^{p^d}}{(X-1)^{p^{d-1}}} = (X-1)^{p^d - p^{d-1}}$, so $g = X^{p^d-p^{d-1}}$, so indeed the other coefficients are divisible by $p$.

Therefore, $\Phi_{p^d}$ is irreducible in $\Qp$ by Eisenstein, so $\Qp\left(\mu_{p^d}\right)$ is totally ramified over $\Qp$ with degree $p^d - p^{d-1}$.

In summary, here are our ingredients:

$\Qp\left(\mu_{p^f-1}\right)$ is unramified with $f = f$.

$\Qp\left(\mu_{p^d}\right)$ is totally ramified with $e = p^d - p^{d-1}$.

Since the extension we seek, namely $\Bbb Q_2\left(\sqrt{-2}\right)$, has $e=2$ and $f=1$, we see that we must try $\Bbb Q_2\left(\mu_{2^d}\right)$ first.

If $d=1$ then $e = 2 - 1 = 1$, which is not enough.

So let's examine $d=2$ where $e = 4 - 2 = 2$, i.e. we examine $\Bbb Q_2\left(\mu_4\right)$.

Let $\alpha$ be a primitive $4$th root of unity, so $\alpha^2+1 = 0$. From the discussion above, we see that $\alpha-1$ is a uniformizer, so its square with differ from $-2$ by a unit. Indeed, $(\alpha-1)^2 = \alpha^2+1-2\alpha = -2\alpha$.

Now we see that if $-2$ is to be a square, then so must $\alpha$; but if $\alpha$ is a square, then its square root would be a primitive $8$th root of unity, which cannot be found in $\Bbb Q_2\left(\mu_4\right)$ (since $\Bbb Q_2\left(\mu_8\right)$ has $e=4$).

So we know that $\Bbb Q_2\left(\mu_8\right)$ is the right place to find our square root of $-2$. Indeed, letting $\zeta_8$ be a primitive 8th root of unity, we see that $\alpha := \zeta_8^2$ is a primitive 4th root of unity, so $(\alpha-1)^2 = -2\alpha = -2\zeta_8^2$, so $-2 = \left[\zeta_8^{-1}(\alpha-1)\right]^2$.

Therefore, $\sqrt{-2} = \zeta_8^{-1}(\alpha-1) = \zeta_8 - \zeta_8^{-1}$


Title: How do we solve $y''=e^{2y}$?

Body:
I tried to make the substitution $v=y'(x)$, which would ( hopefully ) make this into a separable ODE as per the following. $$y''=e^{2y}$$ by the chain rule, we have $v'=v'(y(x)) \cdot y'(x)=\frac{dv}{dy}\frac{dy}{dx}$. If we make the substitution, then $$\frac{dv}{dy}\frac{dy}{dx}=e^{2y}$$

So we could cancel the differential terms $dy$? I wonder if we can really do so, and also a better substitution might be just $v=y(x)$. Am I on the right track here?

Body:
You have

$$y^{\prime \prime}(x) y^{\prime}(x) = y^{\prime}(x) e^{2y}$$
hence

$$(y^{\prime}(x))^2 -(y_0^\prime)^2= e^{2y(x)} - e^{2y_0}$$

From there
$$\frac{y^\prime(x)}{\sqrt{e^{2y(x)} - e^{2y_0}+(y_0^\prime)^2}} = \pm 1$$ and

$$\int_{y_0}^{y(x)}\frac{dt}{\sqrt{e^{2t} - e^{2y_0}+(y_0^\prime)^2}}= \pm (x - x_0)$$


答:
Observe that with $v\equiv\frac{dy}{dx}$ you have:
$$
\frac{d^2y}{dx^2}=\frac{dv}{dx}=\frac{dv}{dy}\frac{dy}{dx}=v\frac{dv}{dy}\tag1
$$

With this your equation transforms to:
$$
v\frac{dv}{dy}=e^{2y}\implies v^2=e^{2y}+C\implies\frac{dy}{dx}=\pm\sqrt{e^{2y}+C}.
$$

Can you take it from here?

Title: How do we solve $y''=e^{2y}$?


答:
Apart from IBP, a much nicer way of doing it is Feynman's trick (differentiation under the integral sign).

Let $I(k) = -\int_0^{\frac{\pi}8} \cos kx dx = - \frac 1k\sin \frac{k\pi} 2$

Then the required integral is $(\frac{d^2}{dk^2}I(k)) \vert_{k=4}$

which would after differentiating twice and substituting $k=4$ give the required answer of $\frac 1{256} (\pi^2 - 8)$


问:
prove $\inf S = -\sup (-S)$
Let $-S$ be the set $\left\{ -s:s \in S \right\}$ where $-S$ is the set that contains negatives of the members of $S$. We want to prove that $\inf(S) = -\sup(-S)$

Here is how I proved it
Let $s_0= \sup(-S)$. That is for all $-s_1\in -S$ then $-s_1 \leq s_0$. Multiplying both sides by $-1$ we get $-s_0 \leq s_1$ for all $s_1 \in S$. So $\inf(S)=-s_0=-\sup(-S)$ It looks short and sweet. Not sure if its right though.

答: It’s fine as far as it goes, but it’s incomplete. You’ve shown that $-\sup(-S)$ is a lower bound for $S$, but not that it’s the greatest lower bound. If you assume that it’s not the greatest lower bound, you can use the same basic idea to get a contradiction; I’ll leave it to you to try that on your own, but I’ll be happy to say more if necessary.

Title: prove $\inf S = -\sup (-S)$


问:
Confusions about definition of the sum of two or several random variables

Description of the Questions:

I have some trouble understanding the definition of the sum of two random variables and hope someone could help me to get it straight. And my main confusions could be divided into the following questions:

First of all, the usual definition of the sum of two random variables is given by:

Definition: Sum of two random variables

Let $X$ and $Y$ be random variables defined on a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, then define a new random variable $Z(\omega)$ on the same probability space as: $Z(\omega) = X(\omega)+Y(\omega). \quad \forall \omega \in \Omega $

Notice: It is necessary to verify that this definition is well defined. i.e. we need to verify that $Z(\omega)$ is indeed a random variable. But we could simply leave it aside since this proof is not required in my question.

Now, if we consider the following simple example:

Suppose we have a given probability space $(\Omega, \mathcal{F}, \mathbb{P})$, where $\Omega = \{\omega_1, \omega_2, \omega_3, \omega_4\}$ & $\mathcal{F} = \mathcal{P}(\Omega)$ and a proper probability measure $\mathbb{P}$. Also, suppose we have two same discrete random variable $X_1(\omega)$ and $X_2(\omega)$ defined as follow:

$$X_1(\omega_i) = X_2(\omega_i) = i \quad \forall \omega_i \in \Omega $$
Finally, we could define the sum of these two variables as:

$$Z(\omega) = X_1(\omega) + X_2(\omega) \quad \forall \omega \in \Omega$$
But, According to this definition, when $\omega_i$ goes through all the possible value in $\Omega \quad$ i.e.  $i \in \{ 1,2,3,4\}$, $Z(\omega) \in \{ 2,4,6,8\}$

I'm fully aware that this result is due to the understanding that restricting $\omega_i$ to be the same event in both $X_1$ and $X_2$

Question 1:  From the angle of strictly following the definition, Am I somehow wrong or not in understanding the definition?

But when calculating of probability measure $\mathbb{P}(Z(\omega)=z) = \displaystyle{\sum_{x_{1}}p_{X_1,X_2}(x_1,z-x_2)}$, we could see that $\mathbb{P}(Z(\omega)=3)$ exists and equals to $ \displaystyle\sum_{x_{1}}p_{X_1,X_2}(x_1,3-x_2) = p_{X_1,X_2}(1,2) + p_{X_1,X_2}(2,1)$

I also know that if we allow that $\omega_i$ to be different in $X_1$ and $X_2$, then we should probably consider a 2-dimensional space $\Omega \times \Omega$, i.e. $Z(\delta = (\omega_1,\omega_2)) = X_1(\omega_1) + X_2(\omega_2) \quad \forall \omega_1, \omega_2 \in \Omega, \delta \in \Omega \times \Omega$

Question 2: this definition might work, and also conform to probability calculation, but here $\delta \notin \Omega$, which means $Z(\delta)$ is defined in a different probability space, and thus we need to re-evaluate the well-definedness of such definition, is this idea correct?

答:
By definition of a random variable it maps to $\mathbb{R}$ so that is where $Z(\omega)$ lands. And yes, when you look at the sum of two random variables you look at a new variable whose domain is a product set of the original $\Omega$. Measurability is defined in a broad way relating image-subsets to domain-subsets. There is no need at all to re-evaluate the well-definedness but rather simply to prove measurability of $Z$. Which you can do like it is done in this post:

Proving that sum of two measurable functions is measurable.

Edit:

In general measurability is defined like this:
https://en.wikipedia.org/wiki/Measurable_function

In the case of random variables the definition is that X needs to be measurable w.r.t. $(\mathbb{R},\mathfrak{B})$ where $\mathfrak{B}$ is the Borel-$\sigma$-Algebra. This means that for any $A \in \mathfrak{B}$ we have that

$X^{-1}(A) \in \mathfrak{F}$

where $\mathfrak{F}$ is the $\sigma$-Algebra of your probability space $(\Omega, \mathfrak{F}, P)$. A prime reason why we demand this property for random variables is that it allows us to use a Pushforward-Measure on $\mathbb{R}$:
https://en.wikipedia.org/wiki/Pushforward_measure

That means we can assign probabilities to random events in $\mathbb{R}$  (e.g. the sum of two throws of a dice) using the probability measure we already have on $\Omega$.


问:
Expected squared deviation of the James Stein estimator
Let $z\sim\mathcal{N}_N(\mu,I)$, the multivariate $N$-dimensional normal with mean $\mu=(\mu_1,\dots,\mu_N)^T$ and the identity matrix $I$ as the covariance matrix. In the textbook that I'm reading, there is a proof for why the James-Stein estimator $\hat{\mu}^{(\text{JS})}=(1-\frac{N-2}{S})z$ outperforms the MLE $\hat{\mu}^{(\text{MLE})}=z$ in terms of total squared error, that is $$\textbf{E}_\mu[\|\hat{\mu}^{(\text{JS})}-\mu\|^2]<\textbf{E}_\mu[\|\hat{\mu}^{(\text{MLE})}-\mu\|^2]$$ for every choice of $\mu$. The textbook is pretty famous and I highly doubt that there is an error in the proof. However, I do not under stand the following step:

It obviously holds that $$(\hat{\mu}_i-\mu_i)^2=(\hat{\mu}_i-z_i)^2-(\mu_i-z_i)^2+2(z_i-\mu_i)(\hat{\mu}_i-\mu_i).$$ Now the text book says that summing over $N$ and taking expectation on both sides gives $$\textbf{E}_\mu[\|\hat{\mu}-\mu\|^2]=\textbf{E}_\mu[\|\hat{\mu}-z\|^2]-N+2\sum_{i=1}^N\textbf{Cov}_\mu[z_i,\hat{\mu}_i]$$

$\hat{\mu}$ is not defined in the textbook. Given the context, I assume that $\hat{\mu}$ is supposed to be an unbiased estimator for $z$ ($\textbf{E}_\mu[\hat{\mu}_i]=z_i$). Otherwise, I do not understand where the covariance term comes from.

Later in the proof, $\hat{\mu}$ ist swapped by $\hat{\mu}^{\text{(JS)}}$. However, $$\textbf{E}_{\mu}[\hat{\mu}^{\text{(JS)}}]=\textbf{E}_{\mu}[(1-\frac{N-2}{S})z_i]=\mu_i-(N-2)\textbf{E}_{\mu}[\frac{z_i}{S}]$$ and $$\textbf{E}_{\mu}[\frac{z_i}{S}]=\textbf{E}_{\mu}[\frac{-(z_i-\mu_i)+\mu_i}{S}]=-\textbf{E}_{\mu}[\frac{z_i}{S}]+2\mu_i\textbf{E}_\mu[\frac{1}{S}]$$ by symmetry. Since the last term is positive, $\hat{\mu}^{\text{(JS)}}$  is only an unbiased estimator if $\mu=0$, but not for every choice of $\mu$.

I tried to figure out where I went wrong, but I am lost. I would appreciate any help.

答:
For any (square-integrable) estimator $\hat{\mu}$ of $\mu$,
\begin{align}
\mathsf{E}[\hat{\mu}_i-\mu_i]^2&=\mathsf{E}[\hat{\mu}_i-z_i]^2+\mathsf{E}[z_i-\mu_i]^2+2\mathsf{E}[(z_i-\mu_i)(\hat{\mu}_i-z_i)] \\
&=\mathsf{E}[\hat{\mu}_i-z_i]^2-1+2\operatorname{Cov}(z_i,\hat{\mu}_i)
\end{align}
because
\begin{align}
\mathsf{E}[(z_i-\mu_i)(\hat{\mu}_i-z_i)]&=\mathsf{E}[(z_i-\mu_i)(\hat{\mu}_i-z_i)]-\underbrace{\mathsf{E}[z_i-\mu_i]}_{=0}\mathsf{E}[\hat{\mu}_i-z_i] \\
&=\mathsf{E}[(z_i-\mu_i)(\hat{\mu}_i-z_i-\mathsf{E}[\hat{\mu}_i-z_i])] = \\
&=\operatorname{Cov}(z_i,\hat{\mu}_i-z_i)=\operatorname{Cov}(z_i,\hat{\mu}_i)-\operatorname{Var}(z_i).
\end{align}
Therefore,
$$
\mathsf{E}\|\hat{\mu}-\mu\|^2=\mathsf{E}\|\hat{\mu}-z\|^2-N+2\sum_{i=1}^N\operatorname{Cov}(z_i,\hat{\mu}_i).
$$


答:
Hint:

When trying $$\int u'v\,dx=uv-\int uv'\,dx$$ you need to choose what functions $u$ and $v$ are. Here, you have two choices:

$u=\dfrac{x^3}3,v=\cos(4x)$ or

$u=\dfrac{\sin(4x)}4,v=x^2$.

In the first case, the new integral will even be harder because of the factor $x^3$. In the second case, $x^2$ will be differentiated to $2x$, and the degree lowers. (The other factor always remains an easy trigonometric function.) You can hope that a second application of integration by parts will reduce $x$ to $1$. Try that.

Using this trick, all integrals of the forms

$$\int P(x)\,e^{ax}\,dx,\int P(x)\cos(ax)\,dx,\int P(x)\sin(ax)\,dx$$

where $P$ is a polynomial can be solved.


答: This argument is incomplete.  If $s_0=\sup(-S)$, the certainly for all $s_1\in S$ we have $-s_1\le s_0$.  But that says only that $s_0$ is an upper bound of $-S$, not that is is the least upper bound.  You have to account for "leastness".

Title: prove $\inf S = -\sup (-S)$

Body:
Let $-S$ be the set $\left\{ -s:s \in S \right\}$ where $-S$ is the set that contains negatives of the members of $S$. We want to prove that $\inf(S) = -\sup(-S)$

Here is how I proved it
Let $s_0= \sup(-S)$. That is for all $-s_1\in -S$ then $-s_1 \leq s_0$. Multiplying both sides by $-1$ we get $-s_0 \leq s_1$ for all $s_1 \in S$. So $\inf(S)=-s_0=-\sup(-S)$ It looks short and sweet. Not sure if its right though.


问:
Number of reduced residue classes with given order?
In general, is there a way to compute the number of reduced residue classes $\pmod n$ of a certain order?

For example, say $n=385=5\times7\times11$. Then
$$
(\mathbb Z/385\mathbb Z)^\times\cong(\mathbb Z/5\mathbb Z)^\times\times (\mathbb Z/7\mathbb Z)^\times\times (\mathbb Z/11\mathbb Z)^\times.
$$
In $(\mathbb Z/5\mathbb Z)^\times$ there are $1,1,2$ elements of orders $1,2,4$.

In $(\mathbb Z/7\mathbb Z)^\times$ there are $1,1,2,2$ elements of orders $1,2,3,6$.

In $(\mathbb Z/11\mathbb Z)^\times$ there are $1,1,4,4$ elements of orders $1,2,5,10$.

Then, if I wanted to see how many elements of order $4$ (say) there were in $(\mathbb Z/385\mathbb Z)^\times$, I can 'brute' force and check all the triples $(a\pmod5,b\pmod7,c\pmod{11})$ for which $\text{lcm}(\text{ord}_5a,\text{ord}_7b,\text{ord}_{11}c)=4$ and conclude that there are $8$ elements of order $4$.

Is there a faster method/theorem to compute this? And if there isn't, what about for $\lambda(385)=60$, the maximum order? Is there a way to compute how many elements there are of order $60$?

答:
I am not entirely sure about the first part. The "nice" thing is that you can just focus on the orders and then calculate how many elements of a given order there are. Since the group of units mod $p$ is cyclic, the number of elements of order $n$ is equal to the number of solutions of the equation

$$ kn  \equiv 0 \pmod{p-1}$$

such that $k m \neq 0 \pmod{p-1}$ for any $m < n$.

For the last part, note that you want to calculate the number of generators for each prime. The number of generators $\mod p$ is equal to $\phi (\phi(p)) = \phi(p-1)$. Hence, if $n$ is a square-free product of primes then

$$\lambda (n) = \prod _{p \mid n} \phi(p-1)$$

Note the square-free condition ensures that the $\mod p$ groups are cyclic. Otherwise, it gets a bit messy and I haven't given it much thought.


答:
You seem to be using for the bases $v_1,v_2$ and $w_1,w_2$ the standard basis $\pmatrix{0\\1}$ and $\pmatrix{1\\0}$ of $\mathbb{R}^2$.

You calculate the matrix $A$ correctly.

But when you write $w_j=\pmatrix{3\\4}$ I think you are writing nonsense.

What we get (and expect to get) is

$T v_1=A_{11}w_1 +A_{21}w_2$ which is $T\pmatrix{1\\0}=3\pmatrix{1\\0}+4\pmatrix{0\\1}$.

Title: Representation of Linear Map in Matrix Form

Body:
In the Matrices section of the book Linear Algebra Done Right, there's a section I'm finding hard to understand. The passage is:

The $k^{th}$ column of M(T) consists of the scalars needed to write
$Tv_k$ as a linear combination of ($w_1$,.....,$w_m$):

$Tv_k$ = $\sum^{m}_{j=1} A_{j,k}w_j$

I'm having hard time understanding the summation part. Let for example we have linear map from $\mathbb{R}^2\to \mathbb{R}^2$ of this form.

$T\begin{pmatrix}1\\0\end{pmatrix}$ = $\begin{pmatrix}3\\4\end{pmatrix}$ and $T\begin{pmatrix}0\\1\end{pmatrix}$ = $\begin{pmatrix}6\\9\end{pmatrix}$.

In this example the matrix $A$ is \begin{pmatrix}
3 & 6\\
4 & 9\\
\end{pmatrix}

Then how we are reaching to $Tv_k$ = $\sum^{m}_{j=1} A_{j,k}w_j$ in this example?

Let $T\begin{pmatrix}1\\0\end{pmatrix}$. Now, how will I represent it in terms of

$\sum^{m}_{j=1}A_{j,1}$$w_j$ where $A_{j,1} = \begin{pmatrix}
3\\
4\\
\end{pmatrix}$ and $w_j = \begin{pmatrix}3\\4\end{pmatrix}$

Basically, how the summation is getting into this portion?
Thanks!


问:
Total work compressing a conical spring
A conical spring's stiffness varies linearly with displacement from rest. Its stiffness when uncompressed is 45 N/m, and 150 N/m when fully compressed. The uncompressed spring stretches 30 cm further than the compressed strength. Find the total work compressing the spring.

The first thing I did was outline my total work formula. The total work from $t=0$ to $t=x$ (in meters), $W(0, x)$, is $$\int_0^xr_w(t)dt$$ where $r_w(t)$ is the rate of change of work. I know that work is an amount of force (which varies) applied over some distance, $w=Fd$, so then rate of change of work is $dw=dF'=tF'$. The rate of change of force is given by $$\biggr(\frac{150-45}{0.3}\biggr)t+45=350t+45$$

Then $$r_w(t)=t(350t+45)$$
Which follows that \begin{align*}W(0,x)&=\int_0^x(350t^2+45t)dt \\ W(0, 0.3)&= 5.175\end{align*}
I believe this is in N/m. Is this correct? It sounds way too low for me given the rates provided at $x=0$ and $x=0.3$.

答:
$dw = tF'$ is wrong. It should be $dW = F(s)ds$, so
$$
W = \int_{x_0}^{x_1}F(s) ds
$$
Note that $F(s)ds$ is force times distance, so it generalizes $W=Fd$.

Also the unit isn't $N/m$, it's $Nm=J$, once again because we multiply $F$ and $ds$ (in a manner of speaking).

BONUS: To find the correct integral yourself, you can approximate by assuming constant $F$ over small distances. Divide the interval $[x_0,x_1]$ into $n$ small intervals $[s_i,s_{i+1}]$ for $i=0\ldots n$. Then:
$$
W \approx \sum_{i=0}^{n-1} F(s_i) \Delta s_i
\underset{n\to\inf}{\to} \int_{x_0}^{x_1}F(s) ds
$$

Title: Total work compressing a conical spring


问:
How to simplify the summation of a recurrence relation
After solving the recurrence relation

$$T(n) = 3T(\frac{n}{3}) + n\log(n)$$

I get following equation

$$T(n)=3kT(\frac{n}{3k})+ n\log(n) + n\log(\frac{n}{3}) + n\log(\frac{n}{3^2})+\dots+n\log(\frac{n}{3^k})$$

I don't know how to simplify the summation and how to know the asymptotic function?

答:
Assuming $\log n = \ln n$ we have

$$
T\left(3^{\log_3 n}\right)=3T\left(3^{\log_3 \left(\frac n3\right)}\right)+n\frac{\log_3 n}{\log_3 e}
$$

now calling $\mathcal{T}(\cdot)=T\left(3^{(\cdot)}\right)$ and $z = \log_3 n$ we follow with the recurrence

$$
\mathcal{T}(z)= 3 \mathcal{T}(z-1)+z3^z c_0
$$

with solution

$$
\mathcal{T}(z)= 3^{z-1}c_1+z(z+1)3^z\frac{c_0}{2}
$$

and now going backwards with $z = \log_3 n$ we arrive at

$$
T(n) = \frac n3 c_1 +n(\log_3 n) (1+\log_3 n)\frac {c_0}{2}
$$

NOTE

The recurrence $\mathcal{T}(z)= 3 \mathcal{T}(z-1)+z3^z c_0 $ is linear so the solution can be composed as $\mathcal{T}(z)=\mathcal{T}_h(z)+\mathcal{T}_p(z)$. The homogeneous solution is direct

$$
\mathcal{T}_h(z) = 3^{z-1}c_1
$$

now taking as particular $\mathcal{T}_p(z) = 3^{z-1}c_1(z)$ and substituting into the complete recurrence we get

$$
c_1(z) = c_1(z-1)+3z c_0
$$

hence

$$
c_1(z) = \frac 32z(z+1)c_0
$$

etc.


问:
Expectation of the product of two matrices
I recently viewed this question and I decided it to give it a try. I got stuck in the computation of a certain expectation and variance. Summarizing the problem, this is the issue:

Let $X\in R^{n\times d}$ be a data matrix, $R\in R^{n\times d}$ a
matrix such that $ R_{ij}\sim\text{Bern}(p)$ is sampled i.i.d. from
the Bernoulli distribution, $M=X\odot R$ the element-wise product
of the previous two matrices, and $\Gamma$ be a diagonal matrix with
$\Gamma_{ii}=(X^\top X)_{ii}^{1/2}$.

Show that $E[M]=pX$, and $var(M)=p(1-p)\Gamma²$

I first went considering both the columns of $X$ and $R$ as independent vectors. Using a specific example where both $X$ and $R$ have three columns, i.e. $X=[A\ B\ C]^T$, and $R=[\alpha\ \beta\ \delta]^T$, we have that $M=[\alpha A\ \beta B\ \delta C]^T$. Taking the expectation of this:
$$E[M]=
\begin{bmatrix}
E[\alpha A] \\
E[\beta B]\\
E[\delta C] \\
\end{bmatrix} \stackrel{\star}{=}
\begin{bmatrix}
E[\alpha]E[A] \\
E[\beta]E[B]\\
E[\delta]E[C] \\
\end{bmatrix} =
\begin{bmatrix}
pE[A] \\
pE[B]\\
pE[C] \\
\end{bmatrix}=pE[X]
$$

where I used that $\star$ the variables are independent.

It doesn't look like the answer is correct and I am not sure what I am misunderstanding. For the variance, I don't even know where to start.

答:
I just realised what is the answer (and also thanks to the reflection questions of @Gateau-Gallois).

Mean: $X$ is taken as a matrix of constants whereas $R\sim Bern(p)$. That's why $E[M]=pX$.

Variance: Following a similar argument, one can see then that the covariance matrix is:
$$
\begin{align}
var(M_{ij})&=var(X_{ij}R_{ij})\stackrel{\star}{=}X_{ij}^2var(R_{ij})
\end{align}$$
where $\star$ is the same assumption as in the question and the fact that $var(aX)=a^2var(X)$. The expression results in 0 if $i\neq j$ since $R_{ij}$ are i.i.d. and otherwise is $X^2_{ii}var(R_{ii})=X^2_{ii}p(1-p)$. Taking, then, the matrix form:
$$
\begin{align}
var(M)=p(1-p)(X^TX)_{ii}\stackrel{\star^2}{=}p(1-p)\Gamma^2
\end{align}
$$
where $\star^2$ is the definition $\Gamma=(X^TX)_{ii}^{1/2}$stated in the question.


问:
Show that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$ cannot be a square when $x\neq3$ and $x\neq2$ respectively.
As the title says, I need help showing that $x^4+x^3+x^2+x+1$ and $x^4+2x^3+2x^2+2x+5$  cannot be a square when $x\neq3$  and $x\neq2$  respectively, where $x$ is a natural number.

In order to do this, I have to use the fact that if $x$ is any natural number >1, any natural number $n\geq x$ can be represented uniquely in the form

$n=c_0+c_1x+c_2x^2+...+c_mx^m$,

where $0\leq c_i \leq x-1$ for $i=0,1,2,...,m-1$  and
$0< c_m \leq x-1$.

答:
If $x > 3$ and $n = x^4+x^3+x^2+x+1$ is a square then $4n$ is also a square.  However

$$ \begin{eqnarray}4(x^4+x^3+x^2+x+1) &=& (2x^2 + x)^2 + 3x^2 + 4x+4 \\ &=& (2x^2+x+1)^2 - (x+1)(x-3) \end{eqnarray}$$

which shows that

$$(2x^2 + x)^2 < 4n < (2x^2+x+1)^2.$$

Since $4n$ is between two consecutive squares it cannot be a square itself.  In the same way

$$ \begin{eqnarray}x^4+2x^3+2x^2+2x+5 &=& (x^2 + x)^2 + x^2 + 2x+5 \\ &=& (x^2+x+1)^2 - (x+2)(x-2) \end{eqnarray}$$

which shows that $x^4+2x^3+2x^2+2x+5$ is not a square when $x>2$.


问:
Projecting $\mathbb R^3$ onto $\mathbb RP^2$
I am reading Introduction to Algebraic Geometry by Justin R. Smith, and Exercise 2 of section 1.6 (p.33) states the following:

In computer graphics, after a scene in $\mathbb R P^3$ has been constructed, it is
viewed — i.e., there is a “camera” that photographs the scene with proper perspective.
Suppose this camera lies at the origin and is pointed in the positive $z$
direction. Describe the mapping that shows how the scene looks.
How do we handle the situation where the camera is not at the origin and
pointed in the $z$-direction?

The answer for the first part (camera at origin) says it is given by (from $\mathbb R^3\subset\def\R{\mathbb R}\R P^3$ to $\R P^2$)

$$\begin{bmatrix}x\\y\\z\\1\end{bmatrix}\mapsto\begin{bmatrix}x\\y\\z\end{bmatrix}$$

I'm very confused by where this comes from, especially because this answer seems totally independent of the fact that the camera is facing $z$-axis.

答: OK, I realize that the point is we are making the plane $z=0$ become points at infinity, and this is achieved by making $z$ (implicitly) the last coordinate in $\mathbb R P^2$.


问:
$a(a + 2)$ is $a^2 + 2a$, but why? Should not we start with Parentheses?
The result of $a(a + 2)$ is $a^2  + 2a$, but why? Shouldn't we start with parentheses first?

答: If you are computing, say, $7\times(7+2)$, yes, you compute first $7+2(=9)$, and then you compute $7\times 9(=63)$. However, by the distributive property, $7\times(7+2)$ is also equal to $7\times7+7\times2=7^2+7\times2$. And, by the same reason, $a(a+2)=a^2+a\times2=a^2+2a$.


问:
Polynomial with coefficients 1.
How can I solve (only for real solutions) an equation of the form $c = x + x^2 + ... + x^n$, where $c \in \mathbb{N}$ and $n \in \mathbb{N}$ are constants.

My first instinct was to rewrite this as:

$c = x(1 + x(1 + ... (x)))$

But I'm not sure if that's of any use.

Thanks :)

答:
@Wuestenfux gave the most convenient form for the problem.

Assuming $c>0$, consider that you look for the zero of function
$$f(x)=x^{n+1} - (c+1)x + c $$ for which
$$f'(x)=(n+1)x^n- (c+1)\qquad \text{and} \qquad f''(x)=n(n+1)x $$ The first derivative cancels at
$$x_*= \Bigg[\frac {c+1}{n+1}\Bigg]^{\frac 1n}$$ To get an estimate of  the solution, expand $f(x)$ as Taylor series around $x=x_*$. This will give
$$f(x)=f(x_*)+\frac 12 f''(x_*) (x-x_*)^2+O\left((x-x_*)^3\right)$$ This will give
$$x_0=x_*+\sqrt{-2 \frac{f(x_*)}{f''(x_*)}}$$ from which you can start Newton iterations
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

Since you wrote in comments that, usually $c<10^3$, let me be generous using $c=10^4$. I report in the table the value of the $x_0$, $x_1$ and the "exact" solution.

$$\left(
\begin{array}{cccc}
n & x_0 & x_1 & \text{solution} \\
 5 & 6.58133 & 6.18432 &  6.08727 \\
 10 & 2.53756 & 2.42381 &  2.37859 \\
 15 & 1.83624 & 1.77471 &  1.74604 \\
 20 & 1.56196 & 1.52078 &  1.50023 \\
 25 & 1.41831 & 1.38771 &  1.37193 \\
 30 & 1.33069 & 1.30652 &  1.29384 \\
 35 & 1.27200 & 1.25210 &  1.24160 \\
 40 & 1.23008 & 1.21323 &  1.20433 \\
 45 & 1.19874 & 1.18417 &  1.17649 \\
 50 & 1.17448 & 1.16166 &  1.15494
\end{array}
\right)$$ You can conceive that, with such an $x_0$, Newton method or, better, Halley or househloder methods will converge within very few iterations.

Title: Polynomial with coefficients 1.


问:
What is the answer to this limit
what is the limit value of the power series:

$$ \lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$$

where $m>1$.

答:
For $\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \frac{x^k}{k^{k-m}}$,
the ratio of consecutive terms is

$\begin{align}
\frac{x^{k+1}}{(k+1)^{k+1-m}}\big/\frac{x^k}{k^{k-m}}
&=\frac{x k^{k-m}}{(k+1)^{k+1-m}}\\
&=\frac{x }{k(1+1/k)^{k+1-m}}\\
&=\frac{x }{k(1+1/k)^{k+1}(1+1/k)^{-m}}\\
&\approx\frac{x }{ke(1+1/k)^{-m}}\\
&=\frac{x (1+1/k)^{m}}{ke}\\
\end{align}
$

For large $k$ and fixed $m$,
$(1+1/k)^{m} \approx 1+m/k$,
so the ratio is about
$x/(ke)$,
so the sum is like
 $\lim_{x\rightarrow +\infty} \sum_{k=1}^\infty (-1)^k \big(\frac{x}{ek}\big)^k$.

Note: I feel somewhat uncomfortable
about this conclusion, but I will continue anyway.

Interesting how the $m$ goes away (unless I made a mistake, which is not unknown).

In the answer pointed to by Mhenni Benghorbal,
it is shown that this limit is $-1$,
so it seems that this is the limit of this sum also.

Title: What is the answer to this limit


问:
Simple Integration of Exponential Function
I would like to integrate $3.5 \times 10 ^{-8} \times 2.737^{x}$

Any help is appreciated.

答:
Simply remember that, for any $a>0  $, $a^x=\mathrm  e^{x\ln a}$ , so
$$\int a^x\,\mathrm dx=\frac{a^x}{\ln a}\qquad(a\ne 1)$$


答:
This is basically $\int ab^x dx=a \frac{b^x}{\ln b},$
where $a=3.5(10)^{-8}, b=2.737$

Title: Simple Integration of Exponential Function

Body:
I would like to integrate $3.5 \times 10 ^{-8} \times 2.737^{x}$

Any help is appreciated.


Body:
$$3.5  \times 10 ^{-8}\int   2.737^{x}\,dx=3.5  \times 10 ^{-8} \frac{2.737^x}{\log 2.737}+C=$$
$$=3.476\times 10^{-8}\times 2.737^x+C$$


问:
Integrability of a function (Darboux)
Question:

Let $f: [a,b] \to \mathbb{R}$ and assume $0 \leq f(x) \leq B$ for  $x\in [a,b]$.   Show that
  $$U(f^2,P) -L(f^2,P) \leq 2B\ ( U(f,P) - L(f,P) )$$
  for all partitions $P$ of $[a,b]$.

Show that if $f$ is integrable on $[a,b]$, then so is $f^2$ [no positivity assumption here].

I found this hint buried in the latex file my prof gave me.

(Hint: $f(x)^2 -f(y)^2 = (f(x) + f(y)) \cdot (f(x) -f(y))$.)

It really wasn't easy to put it to good use...

$$\begin{align}
U(f^2,P) -L(f^2,P) &= (U(f,p) + L(f,p)) \cdot (U(f,p) - L(f,p)) \\
&\leq (|U(f,p)| + |L(f,p)|) \cdot (U(f,p) - L(f,p))
\end{align}$$

We know $|U(f,p)|\geq |L(f,p)|$, and since $f(x) \leq B$, $|f(x)| \leq B$ for all $x \in [a,b]$ where $B \in \mathbb {R}$.

Knowing that  $|L(f,p)| \leq |U(f,p)|\leq |f(x)| \leq B$, we get:

$$(B + B) \cdot (U(f,p) - L(f,p)) = 2B\ (U(f,P) - L(f,P))$$

Hence $U(f^2,P)-L(f^2,P) \leq 2B\ (U(f,P) - L(f,P))$.

Anyone agree with me?

Part 2)

Theorem: Let $f:[a,b] \to [c,d]$ be Daraboux-integrable and $g:[c,d] \to \mathbb {R}$ be continuous. Then the composition $g \circ f$ is Daraboux-integrable.

Let $g=x^2$, so that $g \circ f=f^2$. Since $g$ is continous on $[c,d]$ for all $c,d \in \mathbb {R}$, $f^{2}$ is integrable as long as $f$ is integrable on $[a,b]$ by the theorem above.

答:
The first question (without positivity assumpation) is answerred here.

As for the second part it is easy modulo results of the first question. Let $\small P$ be some partition of $\small P$. Since $\small f$ is integrable then
$$
\small
\lim\limits_{\mu(P)\to 0}(U(f,P)-L(f,P))=0
$$
From inequality of the first question we get
$$
\small
0\leq \lim\limits_{\mu(P)\to 0}(U(f^2,P)-L(f^2),P)\leq \lim\limits_{\mu(P)\to 0}2B(U(f,P)-L(f),P))=2B\lim\limits_{\mu(P)\to 0}(U(f,P)-L(f),P))=0
$$
Hence $\small \lim_{\mu(P)\to 0)}(U(f^2,P)-L(f^2),P)=0$. Since partition $\small P$ was arbitrary this means that $\small f^2$ is integrable.

Title: Integrability of a function (Darboux)


问:
If $3a+4b=x^2$ and $4a+3b=y^2$ Show that $7\mid a,b$
$a,b\in \mathbb N$ And such that $3a+4b$ and $3b+4a$ are both perfect squares.

Show that $7\mid a,b$.

I don’t know how to solve this I’ve just made a system of equations:
$$\cases{3a+4b=x^2 \\ 3b+4a=y^2} \implies (3a+4b)(3b+4a)=(xy)^2$$
$$\iff 12a^2+25ab+12b^2=x^2y^2$$

Then I’ve tried to solve the quadratic equation but it didn’t help me.

答:
Add the two equations, we have:
$$x^2+y^2 = 7(a+b)$$
Thus, $7 \mid (x^2+y^2)$. Use the fact that the quadratic residues modulo $7$ are only $0, 1,2,4$, and if two of these add to $0 \bmod{7}$, then both $x$ and $y$ are $0 \bmod{7}$. Thus, $7 \mid x,y$.

Next, use the fact that $7 \mid x,y$. Since $7 \mid (3a+4b)$, we get $a \equiv b \pmod{7}$. Moreover, from earlier, we can see that $49 \mid (x^2+y^2)$, so $7 \mid (a+b)$. Thus, $7 \mid a,b$.


问:
How do we solve $y''=e^{2y}$?
I tried to make the substitution $v=y'(x)$, which would ( hopefully ) make this into a separable ODE as per the following. $$y''=e^{2y}$$ by the chain rule, we have $v'=v'(y(x)) \cdot y'(x)=\frac{dv}{dy}\frac{dy}{dx}$. If we make the substitution, then $$\frac{dv}{dy}\frac{dy}{dx}=e^{2y}$$

So we could cancel the differential terms $dy$? I wonder if we can really do so, and also a better substitution might be just $v=y(x)$. Am I on the right track here?

答:
Another way to do it.

Switch vcariables
$$y''=e^{2y} \implies -\frac{x''}{[x']^3}=e^{2y}$$ Reduction of order $p=x'$ leads to
$$p'=-p^3 \,e^{2y} \implies p=\pm\frac{1}{\sqrt{e^{2 y}+ c_1}}$$ One more integration
$$x+c_2=\mp \frac{\tanh ^{-1}\left(\frac{\sqrt{e^{2 y}+c_1}}{\sqrt{c_1}}\right)}{\sqrt{c_1}}$$ which is easy to inverse.

Title: How do we solve $y''=e^{2y}$?


问:
Prove if $n^2$ is even, then $n$ is even.
I am just learning maths, and would like someone to verify my proof.

Suppose $n$ is an integer, and that $n^2$ is even.  If we add $n$ to $n^2$, we have $n^2 + n = n(n+1)$, and it follows that $n(n+1)$ is even.  Since $n^2$ is even, $n$ is even.

Is this valid?

答:
For natural numbers:
If $n$ is even: $n=2k$ for $k =...,-2,-1,0, 1, 2,...$. $n^2=n\cdot n = 2k2k = 4k^2 = 2\cdot 2k^2$. This number is an even number.

If $n$ is odd: $n=2k + 1$ for $k =...,-2,-1,0, 1, 2,...$. $n^2=n\cdot n = (2k+1)(2k+1) = 4k^2+4k+1 = 2\cdot \left(2k\left(k+1\right)\right)+1$. This number is an odd number.

Square preserves the even or odd character of the number.

By simply using the above rules, you can easily make sure that this is true for any number raised to any power.
For even $n$ the $n^a$ is even and for odd $n$ the $n^a$ is odd for $a = 2,3,4,...$.


问:
Direct Sum of Finite ring $\mathbb{Z}/k\mathbb{Z}$
Consider $R = \bigoplus_{i=1}^{n} \mathbb{Z}/k\mathbb{Z}$ for some $n\in\mathbb{N}$ and some composite $k \in\mathbb{N}$. Clearly it forms a module over $\mathbb{Z}/k\mathbb{Z}$.

I want to count maximal cardinality of a subset $B \subset R$ such that for every $b \in B$ there exists some $i \in\{1, 2, \ldots , k-1\}$ such that $i.r \not\in B$. I worked through some examples to possibly arrive at a pattern, but have not succeeded yet. I would appreciate any help with this.

答:
If $k=p$ is prime then any largest subset $B_p$ will not contain $0$ and for each line $\in (\Bbb{F}_p{}^n -0)/\Bbb{F}_p^\times$ exactly one element will be removed, which means that $$|B_p| = p^n-1-\frac{p^n-1}{p-1}$$

If $k=p^e$ is a prime power then an optimal subset will be $$B_{p^e}=(\Bbb{Z}/p^e \Bbb{Z})^n - (p^{e-1} \Bbb{F}_p{}^n - p^{e-1} B_p), \qquad |B_{p^e}|=p^{en}-(1+\frac{p^n-1}{p-1})$$

If $k=\prod_{j=1}^J p_j^{e_j}$ then I'm not sure. Maybe
$$B_k=((\Bbb{Z}/p_j^{e_j} \Bbb{Z})^n-0)\oplus (\Bbb{Z}/\frac{k}{p_j^{e_j}} \Bbb{Z})^n$$ where $p_j^{e_j}$ is the largest prime power dividing $k$.


问:
About the axioms for vector spaces. Can we derive the axiom which guarantees the existence of the zero vector from these $7$ axioms?
I am reading an applied mathematics book (in Japanese) by Kenichi Kanatani.

The author wrote as follows:

If non-empty set $\mathcal{L}$ along with an addition on $\mathcal{L}$ and a scalar multiplication on $\mathcal{L}$ has the following properties, then $\mathcal{L}$ is called a vector space.

For any $u,v\in\mathcal{L}$, $u+v=v+u$.

For any $u,v,w\in\mathcal{L}$, $u+(v+w)=(u+v)+w$.

For any $u,v\in\mathcal{L}$, there exists $x\in\mathcal{L}$ such that $u+x=v$. (We write $x$ as $v-u$.)

For any $u,v\in\mathcal{L}$ and any $\lambda\in\mathbb{R}$, $\lambda(u+v)=\lambda v+\lambda u$.

For any $u\in\mathcal{L}$ and any $\lambda,\mu\in\mathbb{R}$, $(\lambda+\mu)u=\lambda u + \mu u$.

For any $u\in\mathcal{L}$ and any $\lambda,\mu\in\mathbb{R}$, $(\lambda\mu)u=\lambda(\mu u)$.

For any $u\in\mathcal{L}$, $1u=u$.

And the author wrote as follows:

For any $u\in\mathcal{L}$, $u-u\in\mathcal{L}$. We write this as $0$. By definition, for any $u\in\mathcal{L}$, $u+0=u$.

By the way, the author is not a mathematician.

The author wrote "For any $u,v\in\mathcal{L}$, there exists $x\in\mathcal{L}$ such that $u+x=v$. (We write $x$ as $v-u$.)". But the author didn't prove that $x$ is unique.

The author didn't prove that for any $u,v\in\mathcal{L}$, $u-u=v-v$ held.

Can we really prove that $u-u=v-v$ for any $u,v\in\mathcal{L}$?

Suppose that we can prove that for any $u,v\in\mathcal{L}$, there exists $x\in\mathcal{L}$ such that $u+x=v$ uniquely.

Then, for any $u,v\in\mathcal{L}$, $$u+v+0u=(u+0u)+v=(1u+0u)+v=(1+0)u+v=1u+v=u+v,$$
and $$u+v+0v=u+(v+0v)=u+(1v+0v)=u+(1+0)v=u+1v=u+v.$$

So, by uniqueness, $$0u=0v.$$

Let $z$ be an arbitrary element of $\mathcal{L}$.

Let $u$ be an arbitrary element of $\mathcal{L}$.

Then, $$u+0z=u+0u=1u+0u=(1+0)u=1u=u.$$

So, $0z$ is zero vector.

答:
Let $v$ be an arbitrary element of $\mathcal{L}$.

Let $u$ be an arbitrary element of $\mathcal{L}$.

There exists $x\in\mathcal{L}$ such that $u+x=v$.

$v+0u=(u+x)+0u=(u+0u)+x=(1u+0u)+x=(1+0)u+x=1u+x=u+x=v$.

So, $0u$ is zero vector.


问:
Confusion over conditional probabilities

The Question

A company is considering a marketing campaign for a product which
currently is purchased by $4\%$ of people. If the company decides to
run a newspaper campaign then based on previous similar campaigns, the
Market Research Company states that the probability that an arbitrary
person (person selected at random) sees the advert is $0.55$.

Also, the Market Research Company assumes that a person who sees the
advert will be twice as likely to purchase the product (i.e. $8\%$ of
people). Given that a person had bought the product, what is the
conditional probability that they had seen the adverts?

My Understanding

I'm just practicing for a basic maths exam and would just appreciate some clarity on how to calculate conditional probabilities. This is my attempt:

$$\frac{0.55 \cdot 0.08}{ 0.08} + \frac{0.45 \cdot 0.04}{0.04} = 0.2475$$

Am I heading in the right direction or am I completely wrong? Thanks!

答:
Given the data

$\mathbb{P}[\text{Purchase}|\overline{\text{Adv}}]=0.04$

$\mathbb{P}[\text{Purchase}|\text{Adv}]=0.08$

You are requesting to calculate

$$\mathbb{P}[\text{Adv}|\text{Purchase}]=\frac{\mathbb{P}[\text{Adv}\cap\text{Purchase}]}{\mathbb{P}[\text{Purchase}]}=\frac{0.55\cdot0.08}{0.55\cdot0.08+0.45\cdot0.04}\approx 70.97\%$$


答:
Now since you have changed the question

As noted by Haran $7|x,y$ so $$7|(4a+3b-3a-4b) \to 7|(a-b)$$ Now as $$7|x,y \to 49|x^2+y^2 \to 7|a+b$$ Thus $$7|(a+b+a-b)=2a\to 7|2a\to 7|a$$ similarly for $b$....

Title: If $3a+4b=x^2$ and $4a+3b=y^2$ Show that $7\mid a,b$

Body:
$a,b\in \mathbb N$ And such that $3a+4b$ and $3b+4a$ are both perfect squares.

Show that $7\mid a,b$.

I don’t know how to solve this I’ve just made a system of equations:
$$\cases{3a+4b=x^2 \\ 3b+4a=y^2} \implies (3a+4b)(3b+4a)=(xy)^2$$
$$\iff 12a^2+25ab+12b^2=x^2y^2$$

Then I’ve tried to solve the quadratic equation but it didn’t help me.


问:
Let $\mathbb{R}$ be a commutative ring with $1$ and $M$ = $(a)$ be a maximal ideal of $R$. I need to show that $a$ is irreducible.
Let $R$ be a commutative ring with $1$ and $M$ = $(a)$ be a maximal ideal of $R$. I need to show that $a$ is irreducible.

I am thinking along the lines that an ideal $M$ is maximal in $R$ iff $R/M$ is a field but I did not get anywhere. Any help would be appreciated.

Also, is the converse true?

答: Hint: A reducible element $a$ is the product of two non-units $a=bc$ and so $(a)\subset (b)$.


问:
Primality of the numbers in the form of $2n^2-1$
I have a question about primality of integers in the form of $2n^2-1$.

I can prove that for the certain type of n such integers are always composite. For example, if $n=7k+2$ or $n=7k+5$, the whole expression would be always divisible by $7$.

The same is applicable to the whole (probably infinite) set of numbers in the form of $n=ak+b$, where $a$ is $7$, $17$, $23$ etc. and $b$ usually has two values (like $2$ and $5$ for $a=7$).
($a$ is prime here and $b \le a-1$)

I also suspect that the only composite values of $2n^2-1$ are those whose factors are from that set of a (like $7$, $17$, $23$ etc.).

I am trying to see if there anything else can be said about primality or compositness of those numbers $n$. Is there any other forms of n that can guarantee compositness (or primality)?

I would appreciate any ideas.

Thanks!

答:
Take any $a$ and consider $n_a=a\pmod{2a^2-1}.$ Then $2n_a^2-1=2a^2-1\pmod{2a^2-1}$ and you get that $2n_a^2-1$ is divisible by $2a^2-1.$ Latter leads to the conclusion that all numbers of the form $2n_a^2-1$ with $n_a>a$ are composite.
As to the whether there exists infinitely many primes of the form $2n^2-1,$ I believe it is an open problem.</p>


答:
The proof by contrapositive is rather straightforward, if I am not mistaken

Suppose that $a$ is not irreducible. Then it means that $a = xy$ for for some non-unit elements $x, y \in R$.

But then, $(a) = (xy) \subsetneq (x)$ (you might want to formally prove the fact that the last inclusion is strict), which is a contradiction.

More details here

Title: Let $\mathbb{R}$ be a commutative ring with $1$ and $M$ = $(a)$ be a maximal ideal of $R$. I need to show that $a$ is irreducible.

Body:
Let $R$ be a commutative ring with $1$ and $M$ = $(a)$ be a maximal ideal of $R$. I need to show that $a$ is irreducible.

I am thinking along the lines that an ideal $M$ is maximal in $R$ iff $R/M$ is a field but I did not get anywhere. Any help would be appreciated.

Also, is the converse true?


问:
Consider the linear system $ \begin{Bmatrix} x - z = -2a \\ 2x- y+ 2z= 0\\ 3x+ 4y -19z= -a \\ \end{Bmatrix}$

$$
   \begin{Bmatrix}
    x - z = -2a \\
    2x- y+ 2z= 0\\
   3x+ 4y -19z= -a \\
   \end{Bmatrix}
$$

For which values of a does the linear system have

a) a unique solution

b) infinitely many solutions

c) no solution

答:
In $AX=B$ form $|A|=\begin{vmatrix} 1 & 0 &-1 \\ 2 & -1 & 2 \\ 3 & 4 & -19 \end{vmatrix}=0,$ which means this system of Eqs. cannot have unique solution. It will have either many solution of no solution. Let $z=t$,
$\implies y=4a+4t$, putting these in third equation $t$ disappears and one gets $22a=-a \implies a=0$ for consistency. So when $a=0$ yoy have many solutions ans when $a=0$, there will be no solution.


问:
r-sided polygons are formed by joining the vertices of an n-sided polygon.
Find the number of polygons that can be formed, none of whose sides coincide with those of the n-sided polygon.

Since we have to form an r sided polygon, it is obvious that we would have to select r vertices from the n vertices of the n-sided polygon the restriction being, that no consecutive vertices should be chosen.
So to ensure that these vertices are separated we can select $n-r \choose r$ vertices that are between the selected vertices as separators.

This is where I got stuck. I tried searching the web but couldn't find any satisfactory explanations. Could someone please help me out with the logical approach to solve this?

答:
Assuming you start with a convex polygon and you want to select convex polygons, you have to select $r$ vertices from $n$ where $ r \ge 3$ and you do not select adjacent vertices.

Suppose one of the vertices is $A$.  Then you want $r-1$ non-adjacent vertices from the $n-3$ vertices which are not $A$ or its neighbours.

You could use induction or a form of stars and bars to show this is ${n - r-1 \choose r-1}$

But you need to multiply this by $n$ as the number of choices for $A$, and divide by $r$ as each polygon is counted multiple times, suggesting the answer may be $$\frac nr{n - r-1 \choose r-1} $$ at least when $r \ge 3$.  As  a check, this give the obvious answer of $2$ when $n=2r$.


问:
Adjoint operator, bijective
Let $A\in\mathcal{L}(X,Y)$, where $X,Y$ are normed vector spaces. Define the adjoint operator
$$\begin{array}{ll}
A^{\prime}\ : & Y^{\prime}\rightarrow X^{\prime},\\
 & G \mapsto A^{\prime}(G)\ =\ G\circ A.
\end{array}$$
Its easy to show that $A^{\prime}\in\mathcal{L}(Y^{\prime}, X^{\prime})$ and $\|A\| = \|A^{\prime}\|$.

Now, suppose that $A$ is bijective, then show that $A^{\prime}$ is also bijective.

First at all, I have proven that $[\mbox{Im}(A)]^{\circ} = \mbox{Ker}(A^{\prime})$, and then $\mbox{Ker}(A^{\prime}) = \{O\}$. Therefore $A^{\prime}$ is injective. My problem is that I don't know how I can show that $A^{\prime}$ is onto.

Please I need help.

Thanks in advance.

答:
I don't think the statement is true when the spaces are not complete. Let $X$ be the space of differentiable functions on $[-1,1]$ with the $C^1$ norm, and let $Y$ be the same set of functions, but equipped with the $C^0$ norm. Let $A$ be the identity map. It is certainly continuous, since $\|f\|_0 \leq \|f\|_1$.

Now, let $T(f) = f'(0)$. This is in $X'$, but is not the image of any operator under $A'$. Suppose $S: Y \rightarrow \mathbb{R}$ were such an operator. Then $S(f) = T(f)$. But this means that differentiation at $0$ is a continuous map on $Y$. But there are functions bounded by 1 that have arbitrarily large derivatives at 0.

Title: Adjoint operator, bijective


问:
Word problem in the braid group
The geodesic length of elements can be defined as the length of a minimal path from 1 to w in the Cayley graph of G. This length is dependent on the particular generating set X.

The Word Problem is decidable in G if and only if the geodesic length function is computable.

Sufficiency is obvious, but why is the necessity needed? Further, "computable" I interpret as the calculation terminates in a finite number of steps, with the correct answer, is this correct, or is it more nuanced?

The braid group has a quadratic Dehn function, i.e., there exist constants $C_n$, $C'_n$ such that, if w is an n-strand braid word of length $l$ that represents the unit braid, then the number of braid relations needed to transform $w$ into the empty word is at most $C_n l^{2}$ and, on the other hand, there exists for each $l$ at least one length $l$ word $w$ such that the minimal number of such braid relations is at least $C'_n l^2$. [This paragraph is copied from Combinatorial Distance between Braid Words, P. Dehornoy, which mentions it is a well-known result]

Because the first result is if and only if, this, together with the braid group having a quadratic Dehn function, (presumably) means that word problem in the braid group is decidable. Does this mean that the Dehn function can be used to calculate the geodesic length in the braid group?

答:
Yes, computable means that there is an algorithm to solve the problem that is guaranteed  to succeed on all inputs.

If the word problem in  $G$ is solvable, then you can compute the geodesic length of $g \in G$ as follows.

For $n := 0, 1,2,3 \ldots$, write down all words of length $n$ and check whether any of them are equal in $G$ to $g$. When you find such an element, stop: the geodesic length of $g$ is $n$.

Title: Word problem in the braid group