问:
Trigonometric equation $1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$
The equation is $$1-\frac{1}{\sin(x)}-\frac {1}{\cos(x)}=0$$ I have tried multiplying both sides by $\sin(x)\cos(x)$ and I got $$\sin(x)\cos(x)-\cos(x)-\sin(x)=0$$ but honestly I think the only way out here is by a graph?

Body:
$$\sin(x)\cos(x)=\sin(x)+\cos(x)$$
$$\frac12 (\sin(2x))=\sin(x)+\cos(x)$$ $$\frac{\sin(2x)}{2}=\sin(x)+\cos(x)$$ $$\frac12 (\sin(2x))=\frac{2}{\sqrt2}(\sin(x+\frac{π}{4}))$$ $$\sin(2x)=\frac{4}{\sqrt2}\left(\sin(x+\frac{\pi}{4}) \right)$$ $$\sin(2x)=2\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)$$ $$2\sin^2\left(x+\frac{π}{4}\right)-1=2\sqrt{2}\sin\left(x+\frac{π}{4}\right)$$ $$2\sin^2\left(x+\frac{π}{4}\right)-1-2\sqrt{2}\sin\left(x+\frac{π}{4}\right)=0$$ $$2{\color{green}{\sin^2\left(x+\frac{π}{4}\right)}}-2\sqrt2 {\color{green}{\sin\left(x+\frac{π}{4}\right)}}-1=0$$


问:
5-set scaled Venn diagram
Let's assume that we have $5$ sets, each having up to $100$ elements. Some elements belong to multiple sets. Even more, there is at least one element for each of the $2^5=32$ configurations (by configuration I mean a vector of $0$s and $1$s indicating whether a specific element belongs to the corresponding set or not). Now I would like to represent these sets on a scaled Venn diagram. In other words, I want each of the $32$ regions of the Venn diagram have an area proportional to the number of elements that belong to it.

Is it always possible?

Now, more practical question: is there a (possibly, approximative) algorithm that would help to construct such a diagram?

Thank you.

答: Sure... and just stretch each region, as needed.

ContentLicense: CC BY-SA 4.0

Title: 5-set scaled Venn diagram


问:
How do I calculate the flag that a coset of $SL(n) \backslash B$ stabilzes?
Let $SL(n)$ be the special linear group of rank $N$ and $B$ a borel subgroup. It is known that the quotient group $SL(n) \backslash B$ is a flag variety of complete flags: https://mathoverflow.net/questions/144287/a-question-about-flag-variety-of-sln-mathbbc. I understand how the "standard" borel subgroup of upper triangular matrices will stabilize the "standard flag".

My question is, given a coset in $SL(n) \backslash B$, say $xB$, how do I compute the flag that it represents?

Thank you very much.

答:
Here is an alternative to Claudius' description in the comments. A coset $xB$ has a stabilizer $xBx^{-1}$ which is a conjugate of the Borel. Everything I'm about to say about this conjugate can be checked by checking it for the standard Borel $B$ itself.

This conjugate of the Borel fixes a unique $1$-dimensional subspace $V_1$ of $\mathbb{C}^n$; this is the first space in the flag. This induces an action on $\mathbb{C}^n/V_1$, which again fixes a unique $1$-dimensional subspace, corresponding to a $2$-dimensional subspace $V_2 \supseteq V_1$; this is the second space in the flag. Etc.


问:
Differentials on the Fermat curve $x^3 + y^3 = z^3$
Consider the Fermat curve $x^n + y^n = z^n$ in $\mathbb{P}^2$, where we'll assume we work over a field of characteristic $0$. I am able to infer from its Hilbert polynomial that this curve has genus $g = (n-1)(n-2) / 2$. In particular, if $n = 3$, then $g = 1$, and through Serre duality, one infers for this case that the sheaf $\Omega^1_{C}$ of $1$-forms is isomorphic to $\mathscr{O}_C$. I would like to confirm this by hand, but I do not know how to approach this.

Take the three usual affine opens $U_i = \operatorname{Spec} k[x_{0/i},x_{1/i},x_{2/i}] / (x_{i/i} - 1)$ on $\mathbb{P}^2$. We have
$$\begin{split}
\Omega^1_{C}\big|_{U_0} &= \big\langle dx_{1/0},dx_{2/0} : 3x_{1/0}^2 \,dx_{1/0} = 3x_{2/0}^2\,dx_{2/0}\big\rangle, \\
\Omega^1_{C}\big|_{U_1} &= \big\langle dx_{0/1},dx_{2/1} : 3x_{0/1}^2 \,dx_{0/1} = 3x_{2/1}^2\,dx_{2/1}\big\rangle, \\
\Omega^1_{C}\big|_{U_2} &= \big\langle dx_{0/2},dx_{1/2} : 3x_{0/2}^2 \,dx_{0/2} = -3x_{1/2}^2\,dx_{1/2}\big\rangle.
\end{split}$$
This should somehow define a line bundle, and I am aware that a section such as $dx_{1/0}$ on $U_0$ should transform to $-(1/x_{0/1}^2)\,dx_{0/1}$ on $U_1$. Is it possible to see from these equations that $\Omega^1_C \cong \mathscr{O}_C$?

答:
Here's a way to cheat to get the answer. Since the Fermat curve $C: U^3 + V^3 = W^3$ has genus $1$ and a rational point ($[1:0:1]$, for instance), then it is an elliptic curve, hence has a short Weierstrass model $E: y^2 = f(x)$. It is well-known that $\omega := \frac{dx}{y}$ is a nonvanshing regular differential on $E$, so we can pullback $\omega$ along the isomorphism $\varphi: C \overset{\sim}{\to} E$ to obtain a nonvanishing regular differential $\eta := \varphi^* \omega$ on $C$.

So let's find an isomorphism $\varphi$. Let $F = U^3 + V^3 - W^3$. We make the following changes of variable.
\begin{align*}
  \begin{pmatrix}
    U'\\ V'\\ W'
  \end{pmatrix}
  &=
  \begin{pmatrix}
    1/3 & 0 & 0\\
    0 & 1/18 & -2Z\\
    0 & 1/18 & 2Z
  \end{pmatrix}
  \begin{pmatrix}
    U\\ V\\ W
  \end{pmatrix}
\end{align*}

Then $27 F(U', V', W') = U^3 - V^2 W - 432 W^3$. Letting $E: Y^2 Z = X^3 - 432 Z^3$, then we have an isomorphism
\begin{align*}
  \varphi: C &\to E\\
  [U:V:W] &\mapsto [X:Y:Z] = \left[3U:9(V+W):-\frac{1}{4}(V-W)\right]
\end{align*}
with inverse $[X:Y:Z] \mapsto [U:V:W] = \left[\frac{1}{3}X : \frac{1}{18}Y - 2Z :  \frac{1}{18}Y + 2Z\right]$. To pull back $\omega$, let's write things in the affine coordinates $u = U/W, v = V/W$ and $x = X/Z, y = Y/Z$ on the affine opens where $W \neq 0$ and $Z \neq 0$. On these patches we have
\begin{align*}
  \varphi: (u,v) \mapsto (x,y) = \left(\frac{-12u}{v-1}, \frac{-36(v+1)}{v-1}\right)
\end{align*}
Letting $\eta = \varphi^* \omega$, then
\begin{align*}
  \eta &= \varphi^* \omega = \varphi^*\left(\frac{dx}{y}\right) = \frac{d\left(-12\frac{u}{v-1}\right)}{\frac{-36(v+1)}{v-1}}
\end{align*}
which after some simplification becomes
\begin{align*}
  \eta = \frac{1}{3(v+1)} du - \frac{u}{3(v^2-1)} dv \, .
\end{align*}

We know that $\eta$ has no zeroes or poles since $\omega$ doesn't, but we can also check this explicitly. For instance, on the distinguished affine open where $u \neq 0$ we have
\begin{align*}
  du = -\frac{v^2}{u^2} dv
\end{align*}
and after some calculation (using the relation $u^3 + v^3 = 1$) one can show that $\eta = \frac{dv}{3u^2}$. Since $\frac{1}{3u^2}$ has no zeroes or poles on this patch, that means $\eta$ doesn't either. Similarly when $v \neq 0$ then $\eta = -\frac{1}{3v^2} du$ so there are no zeroes or poles on this patch, either. That shows there are no zeroes or poles on $U_2$ and one should be able to check on $U_0$ and $U_1$, as well.

As discussed in the comments, this shows that $\Omega_C$ has a nonvanishing global section, which implies that $\Omega_C \cong \mathscr{O}_C$.


问:
Finding a square root by division method
A gardener has 1000 plants. He wants to plant in such a way that the number of rows and columns remain same. Find the minimum number of plants he needed more for this.</p>

答:
The method of calculating square root that you are searching for is probably the one described in this video - http://youtu.be/UnshaiwljGY

Anyhow, coming to your question,

sqrt(1000) = 31.6227

  So, the next perfect square is 322 which is 1024

  => The farmer needs 24 plants more.

  This way the there will be 32 equal rows and columns of plants.

ContentLicense: CC BY-SA 3.0


问:
does the series $ \sum _ { n = 0 } ^ \infty \frac 1 { \sqrt n + \ln n } $ converge or diverge? having trouble

Does the series
$$ \sum _ { n = 0 } ^ \infty \frac 1 { \sqrt n + \ln n } $$
converge or diverge?

I believe this is divergent but having a hard time proving it is.

答:
\begin{align}
\require{cancel}
& \sum_{\xcancel{n\, =\, 0}}^\infty \frac 1 {\sqrt n + \ln n} \qquad \ln0 \text{ is undefined} \\[8pt]
& \sum_{n\, =\, 1}^\infty \frac 1 {\sqrt n + \ln n} \ge \sum_{n\,=\,1}^\infty \frac 1 {\sqrt n + \sqrt n} = +\infty.
\end{align}


问:
Proving that $S_n/n$ converges in probability
I have been struggling with the practice problem below for the past week. Could you help me how to solve this?

$X_1,X_2, \cdots$ are sequence of independent random variables. $X_1=0$, and for $n\geq 2, P(X_n=0)=1-\frac{1}{n\textrm{ln}(n)}, P(X_n=n)=P(X_n=-n)=\frac{1}{2n\textrm{ln}(n)}$. We also define $S_n=X_1+\cdots+X_n$.

Then,

Does $S_n/n$ converge in probability to a limit?

Does $S_n/n$ converge almost surely to a limit?

I have been trying to use the law of large number to solve this problem, but I can't really apply the law of large number because $X_i$'s are not iid. On the other hand, I found this theorem below which I think can be used to solve this problem, but I am having trouble applying the theorem. The theorem is:

Let $\{X_{n,k}\}_{1\leq k\leq n}$ be a triangular array of random variables such that $X_{n,1},...X_{n,n}$ are independent for each $n$. Let $b_n > 0$ such that $b_n \rightarrow \infty$, and let $\bar{X}_{n,k} = X_{n,k}1_{|X_{n,k}| \leq b_n}$. (Here, $1_{|X_{n,k}| \leq b_n}$ is an indicator variable). Suppose as $n \rightarrow \infty$,
(1) $\sum_{k=1}^{\infty}P(|X_{n,k}| >b_n) \rightarrow 0$
(2) $b_n^{-2}\sum_{k=1}^{n}E({\bar{X}_{n,k}}^2) \rightarrow 0$
Define $S_n=X_{n,1}+ \cdots + X_{n,n}$ and $a_n = \sum_{k=1}^{n}E(\bar{X}_{n,k})$,
then $\frac{S_n-a_n}{b_n}$ converges in probability to $0$.

How can I solve this?

Thank you,

答:
The answer to question 1 is that $S_n/n \to 0$ in probability.  This can be proved with a simple application of Chebyshev's inequality.  Since the $X_i$ are independent and satisfy $\mathbb{E}(X_j) = 0$, we have
$$
\mathbb{E}(S_n^2) = \sum_{j=1}^{n} \mathbb{E}(X_j^2) = \sum_{j=2}^{n} j^2 \cdot \frac{1}{j\log j} = \sum_{j=1}^{n} \frac{j}{\log j} \leq n \cdot \frac{n}{\log n} = \frac{n^2}{\log n}.
$$
Therefore Chebyshev's inequality says
$$
\mathbb{P}(|S_n| > n\epsilon) \leq \frac{\mathbb{E}(S_n^2)}{(n\epsilon)^2} \leq \frac{1}{\epsilon^2 \log n} \to 0
$$
for any fixed $\epsilon > 0$.

I don't know the answer to question 2...it's very interesting and I'll keep thinking about it


问:
Gaussian integral with a complex exponent
In physics, one often encounters the integral:
$$I = \int_{-\infty}^\infty e^{-\frac{1}{2}zx^2}dx $$
for $z\in\mathbb{C}$. Inputting this into Mathematica, I get that $I=\frac{\sqrt{2\pi}}{\sqrt{z}}$ so long as $\text{Re}(z)>0$. I'm a little confused by this (isn't $\sqrt{z}$ at risk of being multivalued since there are two square roots?), so I would like to derive it.

I followed a derivation which claims to do this in polar coordinates. Squaring the above and switching to polar coordinates using $r^2=x_1^2+x_2^2$, one finds that:
$$ I^2= \int_{-\infty}^\infty e^{-\frac{1}{2}zx_1^2}dx_1 \int_{-\infty}^\infty e^{-\frac{1}{2}zx_2^2}dx_2 =\int_0^{2\pi}\int_0^\infty re^{-zr^2/2} dr d\phi$$
which, under a change of variables $u=zr^2/2$ yields:
$$I^2=\frac{2\pi}{z}$$
So far so good. It is the next step that confuses me, where the square root is taken on both sides to yield:
$$I=\frac{\sqrt{2\pi}}{\sqrt{z}}$$
Suppose $z\in \mathbb{R}$ - then I'm fine with this, since $I$ must be real and positive, so this can only hold for $z>0$ and it must be the positive root, so there's no ambiguity. But I don't understand (a) how to derive the above result for $z\in\mathbb{C}$, and (b) why the expression $\sqrt{z}$ for $z\in\mathbb{C}$ is well-defined at all. How do I resolve these concerns?

答:
The integral is an analytic function of $z$ as long as it converges, so as soon as you know

the formula works for real positive $z$

the integral converges when $z$ is in the open right half plane

$1/\sqrt{z}$ has a branch that is analytic in the open right half plane

you can conclude that the formula (using that branch of $1/\sqrt{z}$) is valid in the open right half plane.


问:
Internal and orthogonal direct sum isomorphic in Hilbert spaces

Let $H$ be a Hilbert space and let $E$ and $F$ be two closed subspaces of $H$ such that $H = E \oplus F$ as an internal direct sum. Can we conclude that $H$ is isomorphic to $E \times F$, where $E \times F$ is considered as a Hilbert space with the inner product $$\langle (x_1, y_1), (x_2, y_2) \rangle = \langle x_1, y_1 \rangle_H + \langle x_2, y_2 \rangle_H $$ via the map $i:x \mapsto (e_x, f_x)$, where $x = e_x + f_x$ is the decomposition in the internal direct sum?

If $F = E^\perp$, then this question is trivial.

However, we know that $F$ need not be equal to $E^\perp$ (Does orthogonal decomposition characterize direct sums in Hilbert space?). In this case, are we able to conclude they are isomorphic as Hilbert spaces via the canonical map? I do not know how to approach this.

I cannot conclude that $i$ is bounded directly, since $$||i(x)|| = \sqrt{||e_x||^2 + ||f_x||^2}, $$ which is not equal to $||e_x + f_x|| = x$, and it is always larger than this (by the triangle inequality).

Is it important that $H$ is a Hilbert space? We can ask the same question if $H$ is a Banach space. Will the answer change?

答:
It is a consequence of the closed graph Theorem, hence a non-trivial fact, that, whenever a Banach space $X$ is the (algebraic) direct sum of two closed subspaces $E$ and $F$, then the natural projections $\pi_E$ and $\pi_F$ are continuous.  Therefore, for every $x$ in $E$, the new norm
$$
   |||x||| = \|\pi_E(x)\| + \|\pi_F(x)\|
   $$
is equivalent to the original norm on $X$, and hence $X$ is isomorphic and homeomorphic to $E\times F$ via the map described by the OP.


问:
A hard 'if and only if' trigonometric identity proof
Prove
$$
\frac{-2+2\tan A+2\cos B\cdot\sin B+\cot^2 A\cdot({\sec^4A-\operatorname{cosec}^2A-2)}}{2+\tan^2A-2\sin^2A} =(\sin A+\cos A)^2
$$

if and only if B is the double angle of A, or $B=2A+2k\pi$, $k=0,1,2,3...$

Advice is welcome as to improve notation and format.

i understand the RHS can be simplified to 1+sin(2A), but that doesn't seen to go anywhere .

Also, I have been able to simplify the left hand side of the expression down a little, but it's not taking me anywhere at all. I have a feeling that the difficulty of this proof lies in using the 2's in the LHS.

again, many thanks,
Yun Fei

答:
Abbreviating your equation as

$$\frac{P + 2\sin B \cos B}{Q} = R$$

we have

$$\sin 2B = Q R - P$$

where

$$\begin{align}
R &= \left(\cos A + \sin A \right)^2 = \cos^2 A + \sin^2 A + 2 \sin A \cos A \\
&= 1 + \sin 2A \\[4pt]
Q &= 2 - 2 \sin^2 A + \tan^2 A = 2 \cos^2 A + \frac{\sin^2 A}{\cos^2 A} \\
&= \frac{1}{\cos^2 A}\left(2 \cos^4 A + \sin^2 A\right)
\end{align}$$

For $P$, we'll conveniently add and subtract $\csc^4 A$:

$$\begin{align}
P &= 2 \tan A - 2 + \cot^2 A \left(- 2 + \sec^4 A - \csc^2 A \right) + \left( \csc^4 A - \csc^4 A \right)\\
&= 2 \tan A - 2 \left(1 +\cot^2 A \right) + \frac{\cos^2 A}{\sin^2 A}\left(\frac{1}{\cos^4 A} - \frac{1}{\sin^2 A} \right) + \frac{1}{\sin^4 A} &- \csc^4 A\\
&= 2 \tan A - 2 \csc^2 A + \frac{1}{\sin^2 A \cos^2 A} + \left(-\frac{\cos^2 A}{\sin^4 A} + \frac{1}{\sin^4 A}\right) &-\csc^4 A\\[4pt]
&=2\tan A + \frac{1}{\sin^2 A \cos^2 A} - \frac{2}{\sin^2 A} + \frac{1}{\sin^2 A} &-\csc^4 A \\[4pt]
&=2\tan A + \frac{1}{\sin^2 A \cos^2 A} - \frac{1}{\sin^2 A} &-\csc^4 A \\[4pt]
&=\frac{2\sin A}{\cos A}+ \frac{1-\cos^2 A}{\sin^2 A\cos^2 A} &-\csc^4 A \\[4pt]
&= \frac{2\sin A \cos A}{\cos^2 A} + \frac{1}{\cos^2 A} &-\csc^4 A \\[4pt]
&= \frac{1+\sin 2A}{\cos^2 A} &-\csc^4 A
\end{align}$$

Therefore,

$$\begin{align}
QR-P &= \frac{1}{\cos^2A}\left( 2 \cos^4 A + \sin^2 A \right)\left( 1 + \sin 2A \right) - \frac{1}{\cos^2 A}\left( 1 + \sin 2 A \right) + \csc^4 A \\[4pt]
&= \frac{1}{\cos^2A}\left( 2 \cos^4 A + \sin^2 A - 1\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
&= \frac{1}{\cos^2A}\left( 2 \cos^4 A - \cos^2 A\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
&=\left( 2 \cos^2 A - 1\right)\left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
&=\cos 2A \left( 1 + \sin 2A \right) + \csc^4 A \\[4pt]
\end{align}$$

So, your equation reduces to

$$\sin 2 B = \cos 2A ( 1 + \sin 2 A ) + \csc^4 A$$

which is decidedly not equivalent to $B = 2 A + 2k\pi$. Perhaps your original equation has a typo or something.

Note that, even had the equation become

$$\sin 2B = \sin 4A$$

(which seems closest to what you might be anticipating), we'd have $2B = 4A + 2k\pi$ OR $2B = \pi - 4A + 2k\pi$, whence $B = 2A+k\pi$ OR $B = \frac{\pi}{2}-2A + k\pi$.


问:
Integrals exercise.
How can I calculate the following integral $$\int_{0}^{\infty}\lfloor x\rfloor e^{-3x}\ \text{d}x$$
One way to do it is to express the integral as a geometric sum, but I don't know how to do that part, some help.

Title: Integrals exercise.

Body:
$\sum_0^\infty n\int_n^{n+1}e^{-3x}dx=\frac{1}{3}\sum_0^\infty n(e^{-3n}-e^{-3(n+1)})=$$\frac{1-e^{-3}}{3}\sum_0^\infty ne^{-3n}$

$ne^{-nx}=-\frac{d}{dx}e^{-nx}$.  Sum to get $\frac{d}{dx}\frac{-1}{1-e^{-x}}=\frac{e^{-x}}{(1-e^{-x})^2}$.

Let $x=3$ to get final result $\frac{e^{-3}}{3(1-e^{-3})}=\frac{1}{3(e^3-1)}$.


问:
Given matrices $A$ and $B$ solve for $P^{-1}AP=B$
Given
$A =
\begin{bmatrix}
0 & 1 & 0\\
0 & 1 & 4 \\
0 & 0 & 2
\end{bmatrix}
$ and
$B=\begin{bmatrix}
0 & -2 & 3\\
0 & 1 & -1 \\
0 & 0 & 2
\end{bmatrix}$. Find $P$ such that $P^{-1}AP=B$.

Initially I wanted to set this up in the same way you would solve this as a eigenvectors/diagonalizable problem. Since $B$ is not a diagonal matrix, setting the columns of $P$ as the eigenvectors won't help us.

One thing that I noticed is that $A$ and $B$ have the same eigenvalues.  (I am not sure if this is important or not.)

答:
Similarity is an equivalence relation on the space of square matrices.

If you can diagonalize $A$ and $B$ to the same matrix $D=Q^{-1}AQ=R^{-1}BR$,

then $RQ^{-1}AQR^{-1}=B$, so take $P=QR^{-1}$.

You can change coordinates from one basis to another

if you can change both of them to eigenvector coordinates.


问:
Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$?
How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$</p>

答:
\begin{align}
\int_0^{\pi/2} \dfrac{dx}{1+\sin^2(x)} & = \int_0^{\pi/2}\sum_{k=0}^{\infty}(-1)^k \sin^{2k}(x) dx = \sum_{k=0}^{\infty}(-1)^k \int_0^{\pi/2}\sin^{2k}(x) dx\\
& = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{4^k} \dbinom{2k}k\dfrac{\pi}2 = \dfrac1{\sqrt{1-4\times \left(\dfrac{-1}4\right)}} \dfrac{\pi}2 = \dfrac{\pi}{2\sqrt2}
\end{align}
where we used

$\dfrac1{1+r} = \displaystyle \sum_{k=0}^{\infty}(-r)^k$; $\displaystyle \int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$; $ \dfrac1{\sqrt{1-4x}} = \displaystyle\sum_{k=0}^{\infty} \dbinom{2k}k x^k \,\, \forall x \in \left[-\dfrac14,\dfrac14 \right)$


问:
Example of a point that is not a Lebesgue point.
I learned about Lebesgue points which are points $x$ such that

$$\lim\limits_{r\to 0} \frac{1}{\lvert B(x,r)\rvert}\int_{B(x,r)}\lvert f(y)-f(x) \rvert dy=0$$

On wikipedia, they say that Lebesgue points are points that don't oscillate too much. So I am trying to find an example that is not a Lebesgue point. Let $f(x)=\sin {1 \over x}$ if $x\neq 0$. Is this Lebesgue integrable on $(x-\epsilon,x+\epsilon)$? Then how to show

$$\lim\limits_{r\to 0} \frac{1}{2\epsilon}\int_{-\epsilon}^{\epsilon}\lvert \sin {1 \over y} \rvert dy=0$$?

答:
First, notice that for $t\in [\pi/4 + n\pi, 3 \pi/4 + n \pi]$ we have $|\sin t|\ge \frac{1}{\sqrt{2}}$. Therefore, on the interval $[\frac{1}{3\pi/4 + n \pi}, \frac{1}{\pi/4 + n \pi}]$  we have $|\sin \frac{1}{x}|\ge \frac{1}{\sqrt{2}}$. We conclude
$$\int_0^{\frac{1}{\pi/4 + n \pi}} |\sin \frac{1}{x}| \ge \frac{1}{\sqrt{2}}\sum_{k\ge n} ( \frac{1}{\pi/4 + k \pi}-\frac{1}{3\pi/4 + k\pi})$$

It is not very hard to see that the sum is larger than $\frac{1}{2} \frac{1}{\pi/4 + n \pi}$. This shows that $0$ is not a Lebesgue point.


问:
Prove a function is continuous on the union of closed sets if it is continuous on each set.
I have a function $f: (A,d) \to (B,s)$ be a function between two metric spaces. If $$A=\bigcup_{i=1}^n C_i$$ where $C_i$ are closed sets in $A$ and $f$ is continuous on each $C_i$.

Prove $f$ is continuous on $A$ and give an example where $f$ can fail to be continuous on all $A$ if instead I had a countably infinite union of closed sets $$A=\bigcup_{i=1}^{\infty }  C_i$$

So I take the open sets $O_i$ in $B$ then $f^{-1} (O_i)$ is continuous in $C_i$ so
$$f^{-1} \left( \bigcup_{i=1}^n O_i \right)$$ gives me $$\bigcup_{i=1}^n f^{-1} (O_i)$$ which is open in $A$ as union of open sets.

Is this correct? I think I am missing something.

Also I wasn't able to find an example where $f$ can fail to be continuous on all $A$ if I had a countably infinite union of closed sets $$A=\bigcup_{i=1}^{\infty } $$
Can you give an example?

答:
In fact, you can only conclude that $f^{-1}(O_i)$ is open relative to $C_i$. It is an open set in a subspace instead of a "globally" open set.

The correct proof involves a variation of the definition: A function $f:X\to Y$ is continuous if $f^{-1}(F)$ is closed in $X$ for every closed set $F$ in $Y$. (You can try proving it by yourself. Almost one-line.)

Then for each closed set $F$ in $Y$, we have $$f^{-1}(F)=f^{-1}(F)\cap X=f^{-1}(F)\cap\left(\bigcup_{i=1}^{n}{C_i}\right)=\bigcup_{i=1}^{n}{\left(f^{-1}(F)\cap C_i\right)}.$$ Since $f$ is continuous on each $C_i$, it follows that $f^{-1}(F)\cap C_i$ is closed in $C_i$ and even in $X$ (because $C_i$ is closed!!). Note that here we have a finite union of closed sets, so $f^{-1}(F)$ is closed. By the theorem we mentioned above, $f$ is continuous.

As for the counterexample, I think you have realized the point: countable union of closed sets is not necessarily closed. Try to find one by yourself!


问:
Why is this solution for the birthday problem wrong?

How many people do you need in the same room for the probability $p$ to be at least $0.5$ for two people to have the same birthday?

My idea is that for every pair of people the probability that their birthday is the same day is $\frac{1}{365}$, thus we need to count the pairs of people we can match and divide by $365$. If there are $n$ people in the same room there are $\binom{n}{2}$ pairs we can match. Therefore if we set $\frac{\binom{n}{2}}{365}=0.5$ and solve for $n$ we should find how many people are needed.
$$\frac{\binom{n}{2}}{365}=\frac{\frac{n\cdot (n-1)(n-2)!}{2(n-2)!}}{365}=\frac{n(n-1)}{2\cdot 365}=0.5$$
$$n(n-1)=365$$
$$n^2-n-365=0$$
Using abc formula we get: $n \approx 19.5$  (since n must be positive).

However the correct solution is $n\approx 23$. So I am "close", but wrong. Where is my mistake?

答:
You are finding an $n$ where the expected number of shared birthday pairs is 0.50. Sometimes, there will be more than one pair with a shared day. This means that the average number of shared pairs GIVEN at least one shared pair is greater than $1.$ So, for that $n,$ the probability of at least one shared pair is less than $.50.$ You have found a lower bound for the birthday problem.

Put it another way, when $n=28,$ you get $1.03.$ Does this mean there is a $103\%$ chance of a shared birthday? No. It means there are $1.03$ pairs who share a birthday on average. Some groups of $28$ will have no shared birthday pairs. Some will have many.


问:
Pursuing stacks
What does it mean in $n$-categories the saying pursuing stacks ?

答:
It's a manuscript by Grothendieck detailing a program for developing a theory of "higher stacks" and related objects. It started off as a letter to Quillen, and an English translation can be found online here.

In the text quoted in the OP - Simpson's Homotopy theory of higher categories - the manuscript is introduced at the bottom of page $7$ ('A major turning point in the history of higher categories was Alexander Grothendieck's famous manuscript "Pursuing Stacks"...').

Title: Pursuing stacks


问:
malnormal subgroup of amalgamated free product
Consider the amalgamated free product $\Gamma = K\ast_{H\simeq H'} L$. Let $A$ be a malnormal subgroup of K i.e, for all $k\in K\setminus A$, $k^{-1}Ak \cap A ={1}$.
Is A malnormal in $\Gamma$?

I was trying to do this by writing elements in normal form but completely stuck.
For example let $g\in \Gamma\setminus A$ and $m\in A$, then i was trying to prove that $g^{-1}mg$ does not belong to A.
if we write $g$ in normal form in $\Gamma$ using right cosets representative of $H,H'$ in $K,L$ respectively  then $g^{-1}$ will be in normal form using left cosets of $H,H'$ in $K,L$, then $g^{-1}mg$ will no longer be in a normal form using right cosets or left cosets.
I don't know how to handle this.
Thanks for any help.

答:
Assuming that you meant what I suggested in my comment regarding malnormality of $A$ in $K$, this isn't true. The idea for a counterexample is that $A$ could have a nontrivial intersection with $H$, and $A \cap H$ could be identified with some subgroup of $H'$ that is a proper, normal subgroup $L$.

To turn that into an actual counterexample:

Let $K$ be the rank $3$ free group $K = \langle a \rangle * \langle b\rangle * \langle c \rangle$,

Let $A < K$ be the rank $2$ free factor $A = \langle b \rangle * \langle c \rangle$ which is malnormal in $K$,

Let $H < A$ be the rank $1$ free factor $H = \langle c \rangle$,

Let $L$ be the rank $2$ free abelian group $\langle c' \rangle \oplus \langle d \rangle$,

Let $H < L$ be the rank $1$ direct factor $\langle c' \rangle$.

And now amalgamate $K$ and $L$ using an isomorphism $\langle c \rangle \approx \langle c' \rangle$.


问:
Prove orthogonality of x and y
if ||x + y|| = ||x - y|| then x and y must be orthogonal</p>

Title: Prove orthogonality of x and y

Body: From $$\|x+y\|^2=\|x-y\|^2$$ we get $$\|x\|^2+2\langle x,y \rangle +\|y\|^2=\|x\|^2-2\langle x,y \rangle +\|y\|^2.$$ Therefore, $$\langle x,y\rangle=0.$$


答:
Just square the relation and explicitly write in terms of the scalar product:
$$||x+y||^2=||x-y||^2\\(x+y)\cdot(x+y)=(x-y)\cdot(x-y)\\||x||^2+||y||^2+2x\cdot y=||x||^2+||y||^2-2x\cdot y$$
This yields $x\cdot y=0$

Title: Prove orthogonality of x and y

Body: if ||x + y|| = ||x - y|| then x and y must be orthogonal</p>


问:
$\mu$ measurable functions and separable metric spaces
I was reading a math textbook and the author gives the following without proof. I have no clue on how to proceed.

Let $(X, \mathcal{F}, \mu)$ be a measure space and $(Y,d)$ be a separable metric space ($d$ is the metric). If $f:(X,\mathcal{F}) \rightarrow (Y, d)$ is a $\mu$-measurable function prove that there exists an $\mathcal{F}$ measurable function which coincides with $f$ everywhere except on a $\mu$-negligible set.

Any help is greatly appreciated.

EDIT: The textbook is "Functions of Bounded Variation and Free Discontinuity Problems" by Luigi Ambrosio et. al.

答:
Let $(V_n) \in Y$ be a countable base for the separable metric space $Y$. Then define
$V_n^\prime = V_n \cap V_{n-1}^c \cap V_{n-2}^c... \cap V_1^c$ and $V_1^\prime = V_1$. Clearly, $V_n^\prime$ are all
measurable. Now $f^{-1}(V_n^\prime) = E_n \cup N_n = E_n + (E_n^c \cap N_n) = E_n + N_n^\prime$, where $E_n$ is
$\mathcal{F}$-measurable and $N_n, N_n^\prime$ are
$\mu$-negligible. As all $V_n^\prime$ are all pairwise disjoint, $f^{-1}(V_n^\prime) = E_n + N_n^\prime$ are all disjoint.
Define a new function $g: X \rightarrow Y$ such that $g(x) = f(x)$ on all $x \in E_n$ and on $x \in (\cup_{i=0}^\infty E_i)^c$ to be any
value. Then $g$ is $\mathcal{F}$-measurable and is a.e. equal to $f$.

I hope this proof is correct.


问:
Generalization of Boolean OR?
I have been looking at the Boolean OR function and Im trying to find its integral analogue.

What I mean is:

Boolean AND (x, y) where x and y are Boolean Values with 0 = False, 1 = True is equivalent to multiplication:

xANDy = x*y

xORy = ?

It appears to be equivalent x + y - xANDy = x + y - x*y

Is this a correct intuition? My defense of AND and multiplication comes from their relationship in probability where p of A and B = pA * pB if A and B are independent. Likewise probability p of A or B = pA + pB - pA*pB

答:
If you have algebra where $true \gt false$, then the binary operation $AND$ corresponds to the minimum of the operands. And $OR$ corresponds to the maximum of the operands.

I think, the probability you not need to define $AND$ and $OR$ operations.

Title: Generalization of Boolean OR?


问:
How to find the Laplace transformation
Given $\mathscr{L}\left\{\cos(8t)/\sqrt{\pi t}\right\}=e^{-8/s}/\sqrt{s}$

How to find the Laplace transformation

$\mathscr{L}\{\cos(8t)\sqrt{t/\pi}\}$?

答:
$$\cal L\left\{\frac{ \cos 4\sqrt{2 t}} {\sqrt{\pi t}} u(t) \right\} = \frac{e^{-8/s}}{\sqrt{s}}. $$

Since $$\frac{ \sqrt{t} \cos 4\sqrt{2 t}} {\sqrt{\pi }} =\frac{ \cos 4\sqrt{2 t}} {\sqrt{\pi t}} \cdot t,$$

Integrating-by-parts $\Longrightarrow$ Laplace transform of $t \cdot f(t) $ is $- \cal L \left\{f(t)\right\}$.

We have:

$$\cal L\left\{\frac{ \sqrt{t} \cos 4\sqrt{2 t}} {\sqrt{\pi }} u(t) \right\} =-\frac{d}{ds} \cal L\left\{\frac{ \cos 4\sqrt{2 t}} {\sqrt{\pi t}}u(t) \right\} =-\frac{d}{ds} \frac{e^{-8/s}}{\sqrt{s}} =\frac{e^{-8/s}(s-16)}{2s^{5/2}}.$$

Title: How to find the Laplace transformation


问:
hyperbolic Sine and Cosine Mclaurin series
I know the McLaurin series expansion for the hyperbolic Sine and Cosine functions; they are defined as

$$\sinh x =\left(x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots \right)=\sum_{n=0}^{\infty} \frac{x^{2n +1}}{\left( 2n +1\right)!,} \tag{1}$$
$$ \cosh x =\left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots \right)=\sum_{n=0}^{\infty} \frac{x^{2n}}{\left( 2n \right)!,}, \tag{2}$$

My question: is there some definition for these series?:
$$?=\left(x + \frac{x^3 c}{3!} + \frac{x^5 c^2}{5!} + \frac{x^7 c^3}{7!} + \cdots \right) =\sum_{n=0}^{\infty} \frac{x^{2n +1}c^n}{\left( 2n +1\right)!}, \tag{3}$$
$$?=\left(1 + \frac{x^2 c}{2!} + \frac{x^4 c^2}{4!} + \frac{x^6 c^3}{6!} + \cdots \right)=\sum_{n=0}^{\infty} \frac{x^{2n} c^n}{\left( 2n \right)!}. \tag{4}$$

答:
If $c>0$, we have
$$
\sum_{n=0}^{\infty} \frac{x^{2n +1}c^n}{\left( 2n +1\right)!}
=\frac1{\sqrt{c}}\sum_{n=0}^{\infty}\frac{(x\sqrt{c})^{2n +1}}{\left( 2n +1\right)!}
=\frac{\sinh(x\sqrt{c})}{\sqrt{c}},
$$
and similarly
$$
\sum_{n=0}^{\infty} \frac{x^{2n} c^n}{\left( 2n \right)!}
=
\cosh(x\sqrt{c}).
$$
On the other hand, if $c<0$, then we have
$$
\sum_{n=0}^{\infty} \frac{x^{2n +1}c^n}{\left( 2n +1\right)!}
=\frac1{\sqrt{-c}}\sum_{n=0}^{\infty}(-1)^n\frac{(x\sqrt{-c})^{2n +1}}{\left( 2n +1\right)!}
=\frac{\sin(x\sqrt{-c})}{\sqrt{-c}},
$$
and similarly
$$
\sum_{n=0}^{\infty} \frac{x^{2n} c^n}{\left( 2n \right)!}
=
\cos(x\sqrt{-c}).
$$


Title: hyperbolic Sine and Cosine Mclaurin series

Body:
I know the McLaurin series expansion for the hyperbolic Sine and Cosine functions; they are defined as

$$\sinh x =\left(x + \frac{x^3}{3!} + \frac{x^5}{5!} + \frac{x^7}{7!} + \cdots \right)=\sum_{n=0}^{\infty} \frac{x^{2n +1}}{\left( 2n +1\right)!,} \tag{1}$$
$$ \cosh x =\left(1 + \frac{x^2}{2!} + \frac{x^4}{4!} + \frac{x^6}{6!} + \cdots \right)=\sum_{n=0}^{\infty} \frac{x^{2n}}{\left( 2n \right)!,}, \tag{2}$$

My question: is there some definition for these series?:
$$?=\left(x + \frac{x^3 c}{3!} + \frac{x^5 c^2}{5!} + \frac{x^7 c^3}{7!} + \cdots \right) =\sum_{n=0}^{\infty} \frac{x^{2n +1}c^n}{\left( 2n +1\right)!}, \tag{3}$$
$$?=\left(1 + \frac{x^2 c}{2!} + \frac{x^4 c^2}{4!} + \frac{x^6 c^3}{6!} + \cdots \right)=\sum_{n=0}^{\infty} \frac{x^{2n} c^n}{\left( 2n \right)!}. \tag{4}$$

Body:
hint

If $ c>0 $ then

$$x^{2n}c^n=(x\sqrt{c})^{2n}$$

and

$$x^{2n+1}c^n=\frac{1}{\sqrt{c}}(x\sqrt{c})^{2n+1}$$


问:
Proving a set contains the product algebra
The stuff in this question came from a measure theory course notes pdf (link below). The proof of the theorem I am interested is also included in this pdf but it is quite short and skips over details. I filled in the details and present my proof and it would be great if some experience proof writers can give it a look.

https://www.math.ucdavis.edu/~hunter/measure_theory/measure_notes_ch5.pdf

Defn 1: Measurable Rectangles

Suppose that $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces, then a measurable rectangle is a subset $A\times B$  of $X\times Y$ where $A\in \mathcal{A}$ and $B\in \mathcal{B}$ are measurable subsets of $X$ and $Y$ respectively.

Defn 2: Product $\sigma$-algebra

Suppose that $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces. The product $\sigma$-algebra $\mathcal{A}\otimes \mathcal{B}$ is the $\sigma$-algebra on $X\times Y$ generated by the collection of all measurable rectangles,
$$
A\otimes \mathcal{B}=\sigma(\{A\times B:A\in \mathcal{A},B\in \mathcal{B}\}).
$$
The product of $(X,\mathcal{A})$ and $(Y,\mathcal{B})$ is the measurable space $(X\times Y,\mathcal{A}\otimes \mathcal{B})$

Defn 3: Sections

Suppose that $E\subset X\times Y$. For any $x\in X$ and $y\in Y$, we define the $x$-section $E_x\subset Y$ and then $y$-section $E^y\subset X$ of $E$ by
$$
E_x=\{y\in Y: (x,y)\in E\},\ \ \ E_y=\{x\in X: (x,y)\in E\}
$$

Thm : Generator of Product $\sigma$-algebra

If $(X, \mathcal{A})$ and $(Y,\mathcal{B})$ are measurable spaces and $E\in \mathcal{A}\otimes \mathcal{B}$, then $E_x\in \mathcal{B}$ is for every $x\in X$ and $E_y\in \mathcal{A}$ for every $y\in Y$. In mathematical symbols:
$$
{A}\otimes \mathcal{B}\subset \mathcal{M}= \{E\subset X\times Y: E_x\in \mathcal{B},\ \forall x\in X\text{ and }E_y\in \mathcal{A},\ \forall y \in Y\}
$$

Proof

Fix $E=A\times B\subset X\times Y$. Let $E\in \{A\times B:A\in \mathcal{A},B\in \mathcal{B}\}$.

If $(x,y)\in E$, then $E_x=B$ and $E_y=A$. This proves that $\mathcal{M}$ contains the generator of $\mathcal{A}\otimes \mathcal{B}$ (the collection of measurable rectangles).

Next, we prove that $\mathcal{M}$ is a $\sigma$-algebra, this completes the proof that
$$
A\otimes B\subset \mathcal{M}
$$

Let $A\times B\in \mathcal{M}$.

Closure under complements. Recall
$$
(A\times B)^c=(A^c\times B)\cup (A\times B^c)\cup (A^c\times B^c)
$$
It is clear that either $(E^c)_x=E_x$ or $(E^c)_x=E_x^c$. Since $\mathcal{B}$ is a $\sigma$-algebra, $E_x\in \mathcal{B}\implies E_x^c\in \mathcal{B}$. The arugument is similar for $E_y$.

Closure under countable union. We have
$$
\Big[\bigcup_i^n (A_i\times B_i)\Big]_x=\bigcup_i^n E_{x,i}\in \mathcal{B}
$$
because $\mathcal{B}$ is $\sigma$-algebra. The arugument is similar for $E_y$. $\blacksquare$

答:
Since $\mathcal A\otimes\mathcal B$ is by definition the smallest $\sigma$-algebra containing the measurable rectangles, proving that $\mathcal M$ is a $\sigma$-algebra will indeed imply $\mathcal A\otimes\mathcal B\subseteq\mathcal M$, as it also contains every measurable rectangles.

However in your proof you start out with specific elements of $\mathcal M$ (the rectangles) instead of generic elements.

What you should rather do is

Assume $E\in\mathcal M$ is arbitrary, and prove that $E^c\in\mathcal M$.

Assume $E_n\in\mathcal M$ are arbitrary, and prove that $\bigcup_nE_n\in\mathcal M$.


问:
Can the map sending a presentation to its group be considered as a functor?
It is well-known that the functor $Grp \to Set$ sending a group $G$ to its underlying set $UG$ has a left adjoint, the functor $Set\to Grp$ sending a set $X$ to the free group $FX$.

I wonder whether one can consider the following modification of the functor $F$: to each presentation $\langle S\mid R\rangle$ we assign the group $FS/N$, where $N$ is the normal subgroup induced by $R$.

Is this a functor?

For this to be true, there need to be a category of presentations. Is there such a category?

Does the functor $\langle S\mid R\rangle\mapsto FS/N$ have a right adjoint (maybe similar to $U$)?

Of course, my questions are not precise, because I haven't specified what the morphisms between presentations are. The question is if one can do all this in a sensible way.

答:
Giving a presentation $\langle S \mid R \rangle$ of a group amounts to describing it as the cokernel of a map $F(R) \to F(S)$ between free groups. There is a category $C$ whose objects are such maps and whose morphisms $(F(R_1) \to F(S_1)) \to (F(R_2) \to F(S_2))$ are maps $f : S_1 \to S_2$ such that the induced map $F(f) : F(S_1) \to F(S_2)$ sends every relation in $R_1$ to an element in the image of $F(R_2)$. Taking the cokernel to get the group is a functor on this category, which has a right adjoint sending a group $G$ to the map $1 \to F(G)$.

Explicitly this adjunction means the following: we have

$$\text{Hom}_{\text{Grp}}(\langle S \mid R \rangle, G) \cong \text{Hom}_C(F(R) \to F(S), 1 \to F(G))$$

which follows since both sides describe the set of functions $f : S \to G$ such that every element of $F(R)$ is sent to the identity.


问:
basis of orthogonal complement
Why is the basis of the orthogonal complement $U^\perp$ found by computing the null space of the row space of the subspace $U$?

答: Because you are looking for all the vectors that are orthogonal to each of the vectors of the basis of your subspace $U$; hence when you multiply any vector in $U^\perp$ by any other vector in the basis of $U$, you should get $0$. This is equivalent to findind all $X$ such that $AX=0$, where $A$ is a matrix with the vectors of the basis of $U$ as rows. Hence why we solve fore the $\ker{A}$.

Title: basis of orthogonal complement


问:
limits , need help
How would I go about finding the following limit:

$$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}}$$

Title: limits , need help

Body: $$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}} = \lim_{n \to \infty} (2^n(1-n^2 2^{-n}))^{\frac{1}{n}} = 2 \lim_{n \to \infty} (1-n^2 2^{-n}))^{\frac{1}{n}} = 2$$


问:
Let $G$ be a finite group and $A:=\{a\in G\mid a\neq a^{-1}\}$. Prove that $|A|$ is even.

Let $G$ be a finite group and $A:=\{a \in G\mid a \neq a^{-1} \}$ a set that contains all the elements of $G$ that are not equal to their respective inverses. Prove that $A$ contains an even number of elements.

I have seen some posts here here about this proof, but none of which were similar to my attempt.

Here's my attempt:

Since $G$ is finite, then $A$ is also finite.

In addition, every element of $A$ has an inverse because $G$ is a group.

Now, divide $A$ in two sets called $X$ and $Y$, such that $X\subseteq A$ and $Y\subseteq A$, so that every element of $X$ has its inverse in $Y$.

Let $k_{1},k_{2} \in \mathbb{N}$, such that $\left | X \right | = k_{1}$ and $\left | Y \right | = k_{2}$.

Since there is no element equal to its inverse in $A$, then $ \left | A \right | = \left | X \right | + \left | Y \right |$.

Moreover, $\left | X \right | = \left | Y \right |$ because $A$ only contains elements which are different from their respective inverses.

So,
\begin{aligned}
\left | A \right | &= \left | X \right | + \left | Y \right | \\
&= k_{1} + k_{2} && \text{[$\left | X \right | = k_{1}$ and $\left | Y \right | = k_{2}$]} \\
&= k_{1} + k_{1} && \text{[$\left | X \right | = \left | Y \right |$]} \\
&= 2\cdot k_{1}
\end{aligned}

$2k_{1}$ is an even number, by the definition of even number.

Therefore, the set $A$ contains an even number of elements.

Does my proof look fine? Every help is appreciated!

答:
There is a simpler division of $A$ that does the trick. Instead of dividing $A$ into two disjoint subsets of equal cardinality, divide it into a pairwise disjoint collection of two element subsets:
$$A = \bigcup_{x \in A} \{x,x^{-1}\}
$$


问:
Why is a full turn of the circle 360°? Why not any other number?
I was just wondering why we have 90° degrees for a perpendicular angle. Why not 100° or any other number?

What is the significance of 90° for the perpendicular or 360° for a circle?

I didn't ever think about this during my school time.

Can someone please explain it mathematically? Is it due to some historical reason?

答:
In military (well, I don't know if this is true for all countries), a full turn of a circle is 6400' (I don't know the translation of "milésimos", thousandth?).

This has the advantages already mentioned (divisible by 2, 3, 4, ...) and they can easily use it to measure things. 1 milésimo is about 1 meter at 1 km. So, if they know that a target is at about 1 km and it has 3 milésimos, the target is about 3 meters wide. They use it the other way round too. For example, a tank is about 7-8 meters by 2-3 meters. If they see a tank with their binocular and measure about 7 milésimos, they know that the tank is about 1 km away.

They have several ways to measure milésimos, from graduated binocular to "rules of thumb".


问:
Locus problem on Hyperbola: Tangents are drawn from a point P to the Hyperbola...

Tangents are drawn to ---$3x^2-2y^2=6$ from a point P. If these tangents intersect the coordinate axes at concyclic points then what is the locus of P?

Here is my approach:

I took the general slope format of the tangent as: $y=mx ± \sqrt{(2m^2-3)}$
and taking P as: (h,k), adjusting and squaring the equation:

$(h^2-2)m^2 - 2hkm + k^2 +3$ -(i) .....[as (h,k) lies on the tangent]

This is where I got stuck. When I viewed my book,

It took the slope of these tangents as $m_1$ and $m_2$ and then took: $m_1.m_2=1$

After this it was simple to take in values from the equation (i) but what I did not understand is why they took $m_1.m_2=1$

Can someone please explain this to me?

答:
Tangents intersect $y$ axis at $y=q_1$, $y=q_2$, and intersect $x$ axis at $x=-q_1/m_1$, $x=-q_2/m_2$. Intersecting chords theorem implies then
$$
q_1q_2={q_1q_2\over m_1m_2},
\quad\text{that is:}\quad
m_1m_2=1.
$$


问:
Supremum and infimum of a sequence- lim sup
Do we agree that the sequence $x_n$ as defined by $\begin{cases}7^\frac{1}{n} &, \impliedby n \equiv 0 \mod 2\\n^{-7} &, \impliedby n\equiv 1 \mod 2\end{cases}$ has infimum $0$ (as the second sequence goes to zero and they are both strictly decreasing), and supremum $\sqrt{7}$ (because they are both strictly decreasing and the first starts at $\sqrt{7}$ while the second starts at $1$ ? The exercice I'm doing seems to suggest the supremum is $1$. I should note that it is asking for the $\limsup$, and my problem may stem from the fact that I don't understand the difference between those if there is one.

EDIT : Actually I asked the teacher and we didn't see it in class this year, might be why I don't get that exercise. Eh, I learned something new anyway.

答: Ok si as far as I understand (intuitively), the limit inferior and limit superior are the same as supremum and infimum, only it is when $n\to \infty$. Hence why we don't care about the first terms and see easily that $\lim \sup =1$.


问:
Let E be the standard basis and S another basis for $\mathbb{R}^2$. Assume the matrix M is understood to be in terms of E. Find $[M]_S.$
The question: Let $E = \left\{\begin{pmatrix} 1\\ 0\end{pmatrix}, \begin{pmatrix} 0\\ 1\end{pmatrix} \right\}$ be the standard basis for $\mathbb{R}^2$. The set $S = \left\{\begin{pmatrix} 1\\ 2\end{pmatrix}, \begin{pmatrix} -3\\ 2\end{pmatrix} \right\}$ is another basis for $\mathbb{R}^2$, Assume the matrix $M =\begin{bmatrix} 3 & 5\\ -3 & 2 \end{bmatrix}$ is understood to be in terms of $E$. Find $[M]_S.$

I am really confused on what this is asking of me and how I should approach this. I think that it is saying that I have $[M]_E$, so then to find $M$. I have tried
$
\left[
\begin{array}{cc|cc}
3 & 5 & 1 & 0 \\
-3 & 2 & 0 & 1 \\
\end{array}
\right]
\rightarrow
\left[
\begin{array}{cc|cc}
1 & 0 & \frac{2}{21} & \frac{-5}{21}\\
0 & 1 & \frac{1}{7}& \frac{1}{7}\\
\end{array}
\right]
$, so that I am left with $M = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}$. I then used this new $M$ to solve for $[M]_S$ by doing $MS = \frac{1}{21}\begin{bmatrix}
2 & -5\\
3& 3\\
\end{bmatrix}
\begin{bmatrix}
1 & -3\\
2& 2\\
\end{bmatrix}=
\frac{1}{21}
\begin{bmatrix}
-8 & -16\\
9& -3\\
\end{bmatrix}
$.

But this seems extremely wrong. My professor didn't do a great job going over bases, so any advice would be appreciated.

答:
It is known that the components of a vector $v$ in a new base $S$ are given by $S^{-1}v$.

So for the transformation $M$ having $v\mapsto Mv$, we do
$S^{-1}Mv$ to get the new components for $Mv$ and observe that
$$S^{-1}v\mapsto S^{-1}Mv=(S^{-1}MS)S^{-1}v,$$
then we see how the matrix $S^{-1}MS$ relates the new versions $S^{-1}v$ and $S^{-1}Mv$.


问:
How can I make sense of the equation $\mathrm{d}=\mathrm{d}_{\mathbb R^3}+\mathrm{d}t\wedge\partial_t$? Is it valid for 1-forms only?
I need to understand the equation $\mathrm{d}=\mathrm{d}_{\mathbb R^3}+\mathrm{d}t\wedge\partial_t$ (especially the term $\mathrm{d}t\wedge\partial_t$). The author claims that it is valid in Minkowski Space. But I don't understand how to make sense of that formula, if anything only for 1-forms.

答:
I interpret the identity as
$\mathrm{d}f
= \mathrm{d}\mathbf{x}\cdot\nabla f + \mathrm{d}t\,\partial_t f$
for a function $f(\mathbf{x}, t),$
i.e. $\mathrm{d}_{\mathbf{R}^3} = \mathrm{d}\mathbf{x}\cdot\nabla$ and $\mathrm{d}t \wedge \partial_t = \mathrm{d}t \, \partial_t$.


问:
About the eigenvalues
I know maybe it is so simple, but I stuck at it. Suppose that the matrix $A$ satisfies $A^2=Identity$. How can we derive something about the eigenvalues of $A$?

答: If $\lambda$ is an eigenvalue of $A$, then $\lambda^2$ is an eigenvalue of $A^2$ (prove this!)  If $A^2$ is the identity, the only eigenvalue of $A^2$ is $1$.  What does that tell you about the eigenvalues of $A$?

Title: About the eigenvalues


问:
Probability and dependent events
An unfair coin is flipped twice. $A$ is the event that heads comes up first followed by tails. $B$ is the event that tails comes up first followed by heads. The probability heads will come up is $p$. The likelihood that tails will come up is $1 - p$. What is $$P(A| A \cup B)$$

Thanks for the help!

答:
Hint

If $~p(A) = x~$ and $~p(B) = y$

then $p(A|A\cup B) = \frac{x}{x + y}.$

Title: Probability and dependent events


答:
Yuo have a nice idea, but let's highlight some issues:

Now, divide $A$ in two sets called $X$ and $Y$, such that $X\subseteq A$ and $Y\subseteq A$, so that every element of $X$ has its inverse in $Y$.

Ok, here are a couple of instances: (a) $X=\emptyset$ and $Y=A$; (b) $X=A$ and $Y=A$. In order for the idea to work, you have to add: (i) that, for all $x\in A$, $x\in X$ or $x\in Y$; (ii) that $X\cap Y=\emptyset$

If you add the conditions above, $\lvert X\rvert=\lvert Y\rvert$ is mainly due to the fact that $x\mapsto x^{-1}$ is an injection from $X$ to $Y$ (which is true regardless of (i) or (ii)), plus surjectivity thanks to (i). Condition (ii) or the fact that no element is equal to its inverse are irrelevant for this end.

On the other hand, (ii) is needed to prove that $\lvert A\rvert=\lvert X\rvert+\lvert Y\rvert$.

The attentive reader could also question the existence of a pair of sets such as in (1).

Title: Let $G$ be a finite group and $A:=\{a\in G\mid a\neq a^{-1}\}$. Prove that $|A|$ is even.

Body:
Let $G$ be a finite group and $A:=\{a \in G\mid a \neq a^{-1} \}$ a set that contains all the elements of $G$ that are not equal to their respective inverses. Prove that $A$ contains an even number of elements.

I have seen some posts here here about this proof, but none of which were similar to my attempt.

Here's my attempt:

Since $G$ is finite, then $A$ is also finite.

In addition, every element of $A$ has an inverse because $G$ is a group.

Now, divide $A$ in two sets called $X$ and $Y$, such that $X\subseteq A$ and $Y\subseteq A$, so that every element of $X$ has its inverse in $Y$.

Let $k_{1},k_{2} \in \mathbb{N}$, such that $\left | X \right | = k_{1}$ and $\left | Y \right | = k_{2}$.

Since there is no element equal to its inverse in $A$, then $ \left | A \right | = \left | X \right | + \left | Y \right |$.

Moreover, $\left | X \right | = \left | Y \right |$ because $A$ only contains elements which are different from their respective inverses.

So,
\begin{aligned}
\left | A \right | &= \left | X \right | + \left | Y \right | \\
&= k_{1} + k_{2} && \text{[$\left | X \right | = k_{1}$ and $\left | Y \right | = k_{2}$]} \\
&= k_{1} + k_{1} && \text{[$\left | X \right | = \left | Y \right |$]} \\
&= 2\cdot k_{1}
\end{aligned}

$2k_{1}$ is an even number, by the definition of even number.

Therefore, the set $A$ contains an even number of elements.

Does my proof look fine? Every help is appreciated!


问:
show that a set called ordered pair is unique
For any sets $x$, $y$ we write $\langle x,y \rangle$ for the set $\{\{x\},\{x,y\}\}$ Show that this set is unique.

My proof:

By axiom ZF3 of pairs there exists a set (I call it $z:=\langle x,y \rangle$) whose elements are $x$ and $y$. If now $z^{\prime}$ is another such set, then any element of  $z^{\prime}$ $x$ or $y$ is also an element of $z$ and vice versa and by the ZF1 (extensionality) axiom we must have that $z=z^{\prime}$.

Am I correct?

答:
Formally one can write $$\forall z(z\in<x,y>\leftrightarrow (z=\{x\}\lor z=\{x,y\}))$$ (and also the statements $z=\{x\}$ and $z=\{x,y\}$ can be properly expressed as sentences in the language of ZF). Now, if $w$ is another set such that $$\forall z(z\in w\leftrightarrow (z=\{x\}\lor z=\{x,y\})),$$ then obviously the two sets ($w$ and $<x, y>$)  are equal by the axiom of extensionality. (I assumed so far you already know (how to prove) that there exists a set whose elements are exactly $\{x\}$ and $\{x,y\}$, given sets $x$ and $y$. If not, it is easy to show that this existence is ensured by the axiom of pair and the separation schema).

I hope this answers your question (I'm not pretty sure I actually understood what you were asking for).


答:
After taking logarithm with base $e$, it suffices to calculate:
$$\lim_{x\rightarrow \infty}\frac{\mathrm{ln}(2^x-x^2)}{x}=\lim_{x\rightarrow\infty}\frac{\mathrm{ln}2 \cdot 2^{x} - 2x}{2^x - x^2}= \lim_{x\rightarrow\infty}\frac{\mathrm{ln}2 - \frac{2x}{2^{x}}}{1 - \frac{x^2}{2^x}}=\mathrm{ln}2.$$

Hence the original limit is 2.

Note. We used L'Hôpital's rule in the first equality.

Title: limits , need help

Body:
How would I go about finding the following limit:

$$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}}$$


答:
hint

Let $ \lambda $ be an eigenvalue.

this means there exists a vector $ X\ne 0 $ such that

$$A.X=\lambda.X$$
thus

$$A.A.X=A^2.X=\lambda.A.X=(\lambda)^2.X$$

Title: About the eigenvalues

Body: I know maybe it is so simple, but I stuck at it. Suppose that the matrix $A$ satisfies $A^2=Identity$. How can we derive something about the eigenvalues of $A$?


答:
Try to think about what it means for a matrix to be its own inverse. ANswer :

Let $Ax=\lambda x$. Then we have $A^2x=\lambda Ax \implies x=\lambda A x \implies x= \lambda^2x \implies \lambda = \pm 1.$

Title: About the eigenvalues


问:
Why is the correct value set of functions so arbitrary?
When my professor describes a function, for example $f(x) = \sin(x)$, he is allowed to describe it as a function $f: A \rightarrow B$ with $A = B = \mathbb{R}$, even tough sin(x) only has values in $[-1, 1]$

However, when we are getting asked which is the value set of $f(x) = e^x$, the correct answer is not $\mathbb{R}$, but $\mathbb{R}_+$, because only positive values occur.

What do I miss here? I don't understand why sometimes you can be unprecise with your sets, and sometimes not.

答:
There is a difference between the range, or the image, of a function, and the 'codomain'.

When we write $f: A \rightarrow B$, this means that for every $a \in A$ there is a unique value $b \in B$ with $f(a) = b$.  However not every $b \in B$ needs to have an $a \in A$ with $f(a) = b$.  If the latter property is satisfied, $f$ is called surjective, or onto.  $B$ is called the codomain of $f$.  Sometimes the definition function includes a codomain, so that the codomain must be specified.  But mathematicians tend to be loose with this, so that we could write $\sin : \mathbb{R} \rightarrow \mathbb{R}$ just as well as $\sin : \mathbb{R} \rightarrow \mathbb{C}$.

On the other hand, the image, or the range, of $f$ is the set $\{b \in B: \exists a \in A \, f(a) = b\}$ is the set of all elements of $B$ that are actually mapped onto by $f$.  So the image of $\sin$ is $[-1,1]$, regardless of whether the codomain is considered to be $\mathbb{R}$ or $[-1,1]$.

I'm not sure about the definition of value set, so I can't say which definition of meant, but judging from the answer I assume your professor is asking about the range.


问:
Raabe-Duhamel test for $\sum_{n=1}^{+\infty}\frac{a^nn!}{n^n}$
I need to prove the convergence/divergence of this summation using Raabe-Duhamel test.
I reached to this point after writing up $\frac{a_n}{a_{n+1}}$:

$$\frac{(n+1)^n(n+1)}{an^n}$$

I don't know how to continue it.
$a\ge0$.

答:
The ratio $a_n/a_{n+1}$ is the OP is not quite correct.  Instead, note that we have for $a\ne0$

$$\begin{align}
\frac{a_n}{a_{n+1}}&=\frac{a^nn!/n^n}{a^{n+1}(n+1)!/(n+1)^{n+1}}\\\\
&=\frac1a\,\,\frac{n!}{(n+1)!}\,\,\frac{(n+1)^{n+1}}{n^n}\\\\
&=\frac1a\,\,\frac1{n+1}\,\,\frac{(n+1)(n+1)^n}{n^n}\\\\
&=\frac1a\,\,\frac{(n+1)^n}{n^n}\\\\
&=\frac1a \left(1+\frac1n\right)^n
\end{align}$$

And inasmuch as $\lim\limits_{n\to \infty}\left(1+\frac1n\right)^n=e$, we find that

$$\lim_{n\to \infty }\frac{a_n}{a_{n+1}}=\frac ea$$

and the ratio test reveals that the series converges for $a<e$ and diverges for $a>e$.

But the ratio test is inconclusive for $a=e$.  For $a=e$ the Raabe-Duhamel test reveals

$$\lim_{n\to\infty}\left(n\left(\frac1e \left(1+\frac1n\right)^n-1\right) \right)=-\frac12$$

and we can conclude that the series diverges for $a\ge e$ and converges for $a<e$.


问:
Would consecutive prime integers be 2 number's apart like 1,2,3 and 5?
I know it's kinda stupid but I would like to learn about integers, asking here cause I couldn't find it online. Just want an explanation, please.</p>

答:
Interpreting your question as : why are some of the first primes two numbers apart.

Because $2$ is prime so after $2$ no multiple of $2$ can be prime, so a few of the next primes after $2$ are going to be consecutive odd numbers.


答:
It is NOT wrong to say  $\sin(x): \mathbb{R} \rightarrow \mathbb{R}$ since [-1,1] is in $\mathbb{R}$ but perhaps the you could say "it is not honest" after we prove that $\sin(x)$ only takes on values in $[-1,1]$

Math generally only cares about correctness, and once you're correct you generally look for secondary goals of elegance and convenience so hence your professor doesn't bother restricting that set every time he/she/ze writes it. (Another example being $y = x^2: \mathbb{R} \rightarrow \mathbb{R}^+$ but usually they drop the $+$.)

Now while thats frustrating we could see it as motivation to create some new math. Given a TRUE mathematical sentence $S(s_1 ... s_n)$ which accepts sets $s_1 ... s_n$ as an input: we can say it is $\text{Minimal}$ if there doesn't exist an indexed set $s_i$ and another  $u$ such that $u \subset s_i$ but $S(s_1 ... u ... s_n)$ is also true.

To make this concrete, we let $s_1 = \mathbb{R}, s_2 = \mathbb{R}$. Then $S(s_1, s_2)$ is the statement $\sin(x): s_1 \rightarrow s_2$

The triple $\left( s_1 = \mathbb{R}, s_2 = \mathbb{R}, S(s_1, s_2) \right) $ is NOT minimal since as you noted $[-1,1] \subset \mathbb{R}$ that still lets $S$ be true so we can reduce the triple to:  $\left( s_1 = \mathbb{R}, s_2 = [-1,1] , S(s_1, s_2) \right) $.

Now this is minimal.

The general problem then of "minimizing" sentences can probably lead to some interesting in complex math. For example, what would it take to build an automated minimization program? Even for elementary sets this starts to be an interesting exercise in theorem proving and a hard software engineering challenge.

Title: Why is the correct value set of functions so arbitrary?

Body:
When my professor describes a function, for example $f(x) = \sin(x)$, he is allowed to describe it as a function $f: A \rightarrow B$ with $A = B = \mathbb{R}$, even tough sin(x) only has values in $[-1, 1]$

However, when we are getting asked which is the value set of $f(x) = e^x$, the correct answer is not $\mathbb{R}$, but $\mathbb{R}_+$, because only positive values occur.

What do I miss here? I don't understand why sometimes you can be unprecise with your sets, and sometimes not.


问:
$n$-th derivative change of variable
I have a change of variables from $\alpha$ to $\beta$ such such that $$\frac{d}{d\beta} = \frac{1}{f(\alpha)}\frac{d}{d\alpha}$$
therfore, the second derivative will read
$$\frac{d^2}{d\beta^2} =\frac{1}{f(\alpha)}\frac{d}{d\alpha}\times \frac{1}{f(\alpha)}\frac{d}{d\alpha}= \frac{1}{f^2(\alpha)}\frac{d^2}{d\alpha^2}-\frac{f'(\alpha)}{f^3(\alpha)}\frac{d}{d\alpha}$$
and the third order reads
$$\frac{d^3}{d\beta^3} =-\frac{f''(\alpha)}{f^4(\alpha)}\frac{d}{d\alpha}+\frac{3 f'^2(\alpha)}{f^5(\alpha)}\frac{d}{d\alpha}-\frac{3 f'(\alpha)}{f^4(\alpha)}\frac{d^2}{d\alpha^2}+\frac{1}{f^3(\alpha)} \frac{d^3}{d\alpha^3}$$
I need to find the $n$-th derivative in $\beta$:
$$\frac{d^n}{d\beta^n} =\, ?$$
is there a general formula for this case?

答:
Yes, known as Leibnitz formula. It gives the $n^{\text{th}} $ derivative of a product

$$(F.G)^{(n)}=\sum_{k=0}^n\binom{n}{k}F^{(k)}G^{(n-k)}$$

Title: $n$-th derivative change of variable


问:
Fourier transform of $\log|x|$ in $\mathbb{R}^2$ and the solution to $\Delta u = \delta_0$
In my analysis class, we are covering tempered distributions now. I was given this two-part question in Fourier transforms of distributions.

a. We are asked to compute the Fourier transform of $\log |x|$ as a tempered distribution in $\mathbb{R}^2$. Here, $|x|$ is the 2d Euclidean norm of the vector $x$.

b. We are asked to find the fundamental solution to the Laplace equation in $\mathbb{R}^2$, $\Delta u = \delta_0$, where $u$ is a distribution and understood in the sense of weak solutions.

Here, we take the Laplacian $\Delta u = -\frac{\partial^2 u}{\partial x^2}-\frac{\partial^2 u}{\partial y^2}$. We take the Schwartz space of functions $S(\mathbb{R}^2)$ and its continuous dual space, $S'(\mathbb{R}^2)$, the space of tempered distributions. For $T \in S'(\mathbb{R}^2)$ and $\phi \in S(\mathbb{R}^2)$, we use the notation
$$\langle T,\phi \rangle = T(\phi)$$
and when $T(x)$ is a function, we define $$\langle T,\phi \rangle = \int_{\mathbb{R}^2} T(x)\phi(x)dx$$ For the Foruier transform on tempered distributions, we define $$ \langle \mathcal{F}T,\phi \rangle = \langle T,\mathcal{F}\phi \rangle. $$
To be honest, I have no idea how to compute the Fourier transform of $\log|x|$ in $\mathbb{R}^2$ and how to use it to find the fundamental solution in part b, I do think I need to move to the Fourier domain in b but other than that I am lost. I thank anyone who can help with parts A and B.

答:
The most coherent/memorable approach I know is not the most direct: first, compute that $\Delta |x|^s=s(s+n-1)\cdot |x|^{s-2}$ on $\mathbb R^n$. One can also check that the residue of the meromorphic family of tempered distributions $s\to |x|^s$ at $s=-n$ is an explicit constant multiple of $\delta$. Thus, for $n\not=2$, the limit as $s\to -n$ shows that $\Delta |x|^{2-n}$ is a constant multiple of $\delta$.

The squared factor $s^2$ disrupts this, moving the residue to the degree-one term in the (distribution-valued) Laurent expansion, but differentiating in $s$ gives reduces the degree. Thus,
$$
\Delta(\log|x|\cdot |x|^s) \;=\; \Delta {\partial\over \partial s}|x|^s \;=\; {\partial\over \partial s}\Delta |x|^s
\;=\; {\partial\over \partial s} s^2|x|^{s-2}
$$
Taking the limit as $s\to 0$ gives $\Delta \log|x|$ in the first term, and captures the residue of $|x|^s$ in the last. Namely, $\Delta \log|x|=2\pi\cdot \delta$.

Taking Fourier transform of this shows that $|x|^2\cdot \widehat{\log|x|}$ is a constant multiple of $1$. Since $1/|x|^2$ is not locally integrable in $\mathbb R^2$, we cannot say that the Fourier transform of $\log|x|$ is (a constant multiple of) $1/|x|^2$...


问:
Finding critical points of a function that is bounded by a condition
I got stuck trying to solve a critical points calculation problem of the following function :

f(x,y)=3x−y

Thing is, I calculated the partial derivatives and came up with no results for any critical points.
The function is bounded by the following term:

x^2+2y^2=38

which I immediately assumed was the equation of an ellipse, and tried to bound the 'x' and the 'y' in the original function by the ellipse's 'a' and 'b' values.

I think I got utterly lost in the process of solving this problem, help would be much appreciated

答:
$f(x,y) = 3x-y$

Given constraint is $x = \sqrt{38 - 2y^2}$

$f(x,y) = 3\sqrt{38 - 2y^2} - y$

Now at critical points, $ \, \, \frac{-6y}{\sqrt{38-2y^2}} - 1 = 0$ (taking derivative and equating to zero)

So $y = \pm1$ and now you can find values of $x$ from the constraint and complete the solution.


问:
Parametric equations for given line
How would you find the parametric equations for:

1) a line through $(3,1)$ and $(-5,4)$.

2) a segment joining $(1,1)$ and $(2,3)$.

Can anyone show me the steps of doing it cause the way my textbook shows it, it has gotten me in a loop.

答:
The equation of the first line $$\frac{y-1}{x-3}=\frac{1-4}{3-(-5)}\iff \frac{y-1}{-3}=\frac{x-3}8=a\text{(say)}$$ where $a$ is any real number

$$\implies y=1-3a,x=3+8a$$

Title: Parametric equations for given line


问:
Prove that for all $\phi, \psi \in M$, $S(\phi \circ \psi) = S(\phi) S(\psi)$
Problem:
Let $[n] = \{1,2, \dots, n\}$. We define $M$ to be the set of bijections $\phi : [n] \rightarrow [n]$, and $S: M \rightarrow \mathbb{R}$.

The map $S(\phi) = \prod_{i>j=1}^n \dfrac{\phi(i) - \phi(j)}{i-j}$.

I must prove that $S(\phi\psi) = S(\phi)S(\psi)$

I've tried defining $S(\phi)$ and $S(\psi)$ and multiplying them together to see if I can get something of the form $S(\phi\psi)$, but I'm having big issues with the $\Pi$ symbol. I would like some hints, or if this problem looks familiar in terms of $S$, some general direction.

Edit: Perhaps, is the range of this function $\{1, -1\}$ and how would I go about showing that, if so? I suppose there's a proof by contradiction approach?

答:
We have
$$S(\phi\psi) = \prod_{i > j} \frac{\phi(\psi(i)) - \phi(\psi(j))}{i-j} = \prod_{i > j} \frac{\phi(\psi(i)) - \phi(\psi(j))}{\psi(i)-\psi(j)} \cdot \prod_{i > j} \frac{\psi(i)-\psi(j)}{i-j} = S(\phi)S(\psi)$$

Indeed, we have
$$\prod_{i > j} \frac{\phi(\psi(i)) - \phi(\psi(j))}{\psi(i)-\psi(j)} = \prod_{i>j} \frac{\phi(i)-\phi(j)}{i-j}$$
because $\psi$ is a bijection. If $\psi(i)>\psi(j)$ then leave the term as it is, and if $\psi(i)<\psi(j)$ we use the fact that
$$\frac{\phi(\psi(i)) - \phi(\psi(j))}{\psi(i)-\psi(j)} = \frac{\phi(\psi(j)) - \phi(\psi(i))}{\psi(j)-\psi(i)}.$$


答:
First: $1$ is generally not considered to be prime. To be fair to you, it hasn't always been this way- many mathematicians considered $1$ prime until about 1900.

Second: No, not all consecutive prime numbers are two numbers apart. The first example is $2,3$, which are only one number apart. Another example is $7,11$, in which the number right in the middle of them, $9$, is not prime- $9=3^2$.

Third: However, as Ahmad mentioned in the comments, whether there are infinitely many primes that are two numbers apart is actually not known! It's called the twin prime conjecture, and it's one of the most famous unsolved problems in math today.

Title: Would consecutive prime integers be 2 number's apart like 1,2,3 and 5?

Body: I know it's kinda stupid but I would like to learn about integers, asking here cause I couldn't find it online. Just want an explanation, please.</p>


问:
What is the solution to $\arctan -i$?
I know that $\operatorname{Arctan} z = \frac{1}{2i} \operatorname{Ln} \frac{1+iz}{1-iz}$ but in this case we have $$\frac{1}{2i} \operatorname{Ln}\frac{1+1}{1-1}$$
Is it possible to deal with this formula or what other methods should I try?

答: If $\tan z=-i$ then $\sin^2z=-\cos^2z$, contradicting $\sin^2z+\cos^2z=1$.

Title: What is the solution to $\arctan -i$?


问:
Is there notation for the logical and/or of lots of items?
You can add lots of things together using summation notation $$\sum_{n=1}^5 n = 15$$

Is there a similar operator for logical "and" and "or"?

I considered $\exists$/$\forall$, but that doesn't allow me to easily specify how to determine the answer in some cases.

答:
Use \bigvee and \bigwedge ,
$$  \bigvee_{i=1}^{100} X_i  $$
and
$$  \bigwedge_{i \in I} X_i  \text{,}  $$
respectively.

See also What is the meaning of  $\bigvee$ (bigvee) operator

Be aware that these are also used for meets and joins (lattice theory).  Depending on your context, it could be a good idea to explicitly introduce this notation.


问:
How to solve this sum of normal distribution question?
Assume the weight of a person follows a normal distribution N(71,7). What is the probability of 4 people weighing more than 300kg?

I tried solving this by multiplying the values by 4, so it'd be N(284,28). I converted that into $x=284+28z$ which lead to  $z=\frac{x-284}{24}$.
To solve for $P(x>300)$, I converted it around to $ 1-P(x<300)$ $$ 1-P\left(\frac{300-284}{24}\right) $$
$$ 1-P\left(z<\frac{2}{3}\right) $$
Then i used the normal distribution table and got the following result: $1-0.2546 = 0.745$, however I have a suspicion I started from the wrong track and this result isn't correct.

This was an exam question at a college statistics class

答:
I assume heights are independently normally distributed with mean $\mu = 71$ and standard deviation $\sigma = 7.$ [Sometimes the second parameter is the variance and sometime the variance, you should always say which convention you are using.]

Then $S = X_1+X_2+X_3+X_4$ has $E(S) = E(X_1)+\cdots+E(X_4) = 4(71) = 284.$
and (because of independence) $V(S) = V(X_1)+\cdots+V(X_4) = 4(7^2) = 196.$

So $P(S > 300) = 1 - P(S_4 \le 300 = 0.1265.$ Computation in R (where the second parameter is the SD). You can get approximately the same answer by standardizing and using printed tables of the standard normal CDF (which
involves some rounding.).

```<code>1 - pnorm(300, 284, sqrt(196))
[1] 0.126549
</code>```

So an elevator with safe weight load 300, serving a population with
weights distributed $\mathsf{Norm}(\mu = 71, \sigma=7)$ should have
a limit of three (not four) passengers.

Simulation: With a million sums of four we get $0.1266\pm 0.0007.$

```<code>set.seed(1120)
x1 = rnorm(10^6, 71, 7)
x2 = rnorm(10^6, 71, 7)
x3 = rnorm(10^6, 71, 7)
x4 = rnorm(10^6, 71, 7)
s = x1 + x2 + x3 + x4

summary(s);  sd(s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  217.9   274.6   284.0   284.0   293.4   351.7
[1] 13.99282
mean(s > 300)
[1] 0.126589         # aprx P(S > 300) = 0.126549
2*sd(s > 300)/1000
[1] 0.0006650243     # aprs 95% margin of simulation err
</code>```

```<code>hist(s, prob=T, br=30, col="skyblue2", main="Sim Dist'n Sum of Four")
 curve(dnorm(x, 284, 14), add=T, col="red", lwd=2)
 abline(v=300, lty="dotted", lwd=2)
</code>```

You can get a good approximation by standardizing and using printed tables of
the standard normal CDF (a proxess which requires some rounding).

Body:
I assume heights are independently normally distributed with mean $\mu = 71$ and standard deviation $\sigma = 7.$ [Sometimes the second parameter is the variance and sometime the variance, you should always say which convention you are using.]

Then $S = X_1+X_2+X_3+X_4$ has $E(S) = E(X_1)+\cdots+E(X_4) = 4(71) = 284.$
and (because of independence) $V(S) = V(X_1)+\cdots+V(X_4) = 4(7^2) = 196.$

So $P(S > 300) = 1 - P(S_4 \le 300 = 0.1265.$ Computation in R (where the second parameter is the SD). You can get approximately the same answer by standardizing and using printed tables of the standard normal CDF (which
involves some rounding.).

1 - pnorm(300, 284, sqrt(196))
[1] 0.126549

So an elevator with safe weight load 300, serving a population with
weights distributed $\mathsf{Norm}(\mu = 71, \sigma=7)$ should have
a limit of three (not four) passengers.

Simulation: With a million sums of four we get $0.1266\pm 0.0007.$

set.seed(1120)
x1 = rnorm(10^6, 71, 7)
x2 = rnorm(10^6, 71, 7)
x3 = rnorm(10^6, 71, 7)
x4 = rnorm(10^6, 71, 7)
s = x1 + x2 + x3 + x4

summary(s);  sd(s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
  217.9   274.6   284.0   284.0   293.4   351.7
[1] 13.99282
mean(s > 300)
[1] 0.126589         # aprx P(S > 300) = 0.126549
2*sd(s > 300)/1000
[1] 0.0006650243     # aprs 95% margin of simulation err

hist(s, prob=T, br=30, col="skyblue2", main="Sim Dist'n Sum of Four")
 curve(dnorm(x, 284, 14), add=T, col="red", lwd=2)
 abline(v=300, lty="dotted", lwd=2)

You can get a good approximation by standardizing and using printed tables of
the standard normal CDF (a proxess which requires some rounding).


问:
A basic doubt on the definition of a Poisson random variable
What is the significance of "large city" in the definition of the following Poisson variable :

"Number of phone calls placed during a ten second interval in a large city"

I guess either $n \to \infty$ or $p$ very small is not implied by "large city".

答:
A Poisson r.v. $N$ follows a Poisson distribution, which has only one parameter $\lambda$; and takes values in $\mathbb{N}$ — what do you refer to with $n$ and $p$? (are you thinking instead of a binomial distribution $\operatorname{Bin}(n,p)$?)

I assume the "large city" assumption is just so that you don't have to worry about the number of calls being too big wrt the number of people in the city (technically, if you model it with a Poisson-distributed r.v. $N$, it could take any non-negative integer value with non-zero probability, including very large ones).


答: Interestingly, you can quite easily use a statement like $$(\forall i\in I)(\exists a_i\in A_i)$$ to get repeated "there exist" statements and similarly you could use $$(\forall i\in I)(\forall a_i \in A_i)$$ to get repeated "for all" statements. You would just need to define the set $I$ and the sets $A_i$. Note that each $a_i$ is not necessarily unique, and in the second statement, $a_i$ is unique only when $|A_i|=1$.

Title: Is there notation for the logical and/or of lots of items?

Body:
You can add lots of things together using summation notation $$\sum_{n=1}^5 n = 15$$

Is there a similar operator for logical "and" and "or"?

I considered $\exists$/$\forall$, but that doesn't allow me to easily specify how to determine the answer in some cases.


答: Your mistake is that variances add, not standard deviations. If $7$ is the one-person variance, work with $300=284+\sqrt{28}z$; if $7$ ($49$) is the one-person srandard deviation (variance), work with $300=284+\sqrt{4\times49}z=284+14z$.

Title: How to solve this sum of normal distribution question?

Body:
Assume the weight of a person follows a normal distribution N(71,7). What is the probability of 4 people weighing more than 300kg?

I tried solving this by multiplying the values by 4, so it'd be N(284,28). I converted that into $x=284+28z$ which lead to  $z=\frac{x-284}{24}$.
To solve for $P(x>300)$, I converted it around to $ 1-P(x<300)$ $$ 1-P\left(\frac{300-284}{24}\right) $$
$$ 1-P\left(z<\frac{2}{3}\right) $$
Then i used the normal distribution table and got the following result: $1-0.2546 = 0.745$, however I have a suspicion I started from the wrong track and this result isn't correct.

This was an exam question at a college statistics class


问:
Pdf about conic optimization
Because of COVID I can't go in library to do some research, so I ask you any reference to books dealing with conic optimization, conic duality and Fenchel's duality. I state that I don't know anything about these arguments, so I have to find a pdf not too complicated.

Could you please provide me any links? Thanks in advance.

答:
The topics that you've mentioned are covered in the textbook by Stephen Boyd and Lieven Vandenberghe:

https://web.stanford.edu/~boyd/cvxbook/

Lecture notes by Aharon Ben-Tal and Arkadi Nemirovsky are also available online:

https://www2.isye.gatech.edu/~nemirovs/

These are both posted with the permission of the publisher.

Title: Pdf about conic optimization


问:
If the result of my integral is of the form $ \ln(\infty)-ln(\infty) $ (they are the same functions), can i say that the integral evaluates to $0$?
$$ \int f(t)dt=ln(\infty)-ln(\infty)= 0 $$

Can this be true?

If this is true, then these terms fall out and i can evaluate my integral better.

答:
I'm not quite sure what the question is asking but you may be interested in Cauchy Principal Value which evaluates integrals of this form, take for example:
$$\lim_{h\to\infty}\int_{-h}^h\frac1xdx$$


问:
limits , need help
How would I go about finding the following limit:

$$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}}$$

Title: limits , need help

Body:
$\lim_{n \to \infty} (2^n-n^2)^{\frac{1}{n}}=\lim_{n \to \infty} (2^n)^{\frac{1}{n}}(1-\frac{1}{\frac{2^n}{n^2}})^{(\frac{1}{n})(\frac{-2^n}{n^2})(\frac{-n^2}{2^n})}=\lim_{n \to \infty}2(e)^{\frac{-n}{2^n}}=2$.

The idea is to regain the special limit=$ \lim_{n \to \infty} (1+\frac{1}{n})^n=e$


问:
Best approach for Undecidability proof
Context:

Hi, my professor sent me this challenge and I got stuck. I thought using Rice's Theorem for this question, since $M$ is non-trivial, but he told me to use a reduction.

Is he right? Should I try to follow his advice or try my own?

Question

Let $M$ be a Turing machine and for every non-trivial property $X$ in a semidecidable language we define the following decision problem:

Is it true that the language $\textrm{ACCEPT}(M)  \cup \textrm{REJECT}(M)$ satisfies the property $X$?

Prove that the decision problem is undecidable.

So far:

I was told to build a reduction $r$ using Property Testing for $X$ that for every instance $M$ builds another instance $M'$ that satisfies the following:

$$\text{ACCEPT}(M) \cup \text{REJECT}(M)$$

How should I proceed?

答:
I assume that $\text{ACCEPT}(M)$ denotes the set of strings accepted by Turing machine $M$, and similarly that $\text{REJECT}(M)$ denotes the set of strings rejected by Turing machine $M$. The decision problem must have a parameter and be

Given a Turing machine $\langle M \rangle$, is it true that the language $\textrm{ACCEPT}(M)  \cup \textrm{REJECT}(M)$ satisfies the property $X$?

Let us call this new decision problem $\mathcal{N}_X$.
Moreover, when we speak of properties, the usual definition is that a property $X$ is a family of semidecidable languages.

Rice's theorem tells us that for any non-trivial property $X$, the decision problem $\mathcal{R}_X$

Given a Turing machine $\langle M \rangle$, is it true that $\textrm{ACCEPT}(M) \in X$?

We can easily reduce $\mathcal{R}_X$ to $\mathcal{N}_X$ by the following reduction:

Given an $\mathcal{R}_X$ instance $\langle M \rangle$, build the machine description of a machine $M'$ that behaves as follows:

On input $x$

Simulate $M$ with input $x$.

If $M$ accepted, then accept.

Else loop forever.

Clearly, $\textrm{ACCEPT}(M') = \textrm{ACCEPT}(M)$ and $\textrm{REJECT}(M') = \emptyset$ and therefore $\textrm{ACCEPT}(M') \cup \textrm{REJECT}(M') = \textrm{ACCEPT}(M)$.

Title: Best approach for Undecidability proof


问:
Are all ordinals higher than $ω$ epsilon numbers?
By definition,  $α$ is an epsilon number iff $α=ω^α$. Indeed, the $ε$, $ζ$, and $η$ numbers are all epsilon numbers. But is $ω_1$ also an epsilon number? It seems that $ω^{ω_n}=ω_n$ for $n > 0$, which would mean that even ordinals $ω_1$ to $α↦ω_α$ are all epsilon numbers.

Edit: What I meant was ordinals greater than all the $ω_0$ numbers.

Edit 2: Yes, every initial ordinal greater than $ω$.

Thank you.

答:
Well, as written the answer is obviously no: $\omega+1$ is an ordinal greater than $\omega$ which is not an $\epsilon$-number.

However, I suspect you meant to ask whether every cardinal greater than $\omega$ (or if you prefer, every initial ordinal greater than $\omega$) is an $\epsilon$-number. The answer to this, as you suspect, is yes. The key point is that as long as $\alpha$ is infinite we have $\vert\alpha\vert=\vert\omega^\alpha\vert$ (where ordinal exponentiation is being used here, of course). This means that any uncountable cardinal is a limit of $\epsilon$-numbers, and any limit of $\epsilon$-numbers is again an $\epsilon$-number.


问:
Is this a correct way to show convergence for $\sum_{k=1}^{\infty} x^k (arctan(\frac{1}{k}))^2$?
Is this a correct way to show convergence for $\sum_{k=1}^{\infty} x^k (arctan(\frac{1}{k}))^2$?

Ratio test with $lim_{k->\infty} |\frac{a_{k+1}}{a_k}|: \sum_{n=0}^\infty (\frac {\frac{(\frac{1}{k+1})^{2n+1}}{2n+1}}{\frac{(\frac{1}{k})^{2n+1}}{2n+1}})^2=...=\sum_{n=0}^{\infty}(1-\frac{1}{k+1})^{4n+2}=\sum_{n=0}^{\infty}((1-\frac{1}{k+1})^n)^4(1-\frac{1}{k+1})^2$

According to the binomial theorem we have: $\lim_{n->\infty} \lim_{k->\infty}(1-\frac{1}{k+1})^n=\lim_{n->\infty} \lim_{k->\infty} 1-\frac{1}{k+1}+(-\frac{1}{k+1})^2)+...+(-\frac{1}{k+1})^n=\\lim_{n->\infty} 1+0+0^2+...+0^n=1$

So $R=1$ which means that it converges for $|x|<1$.
For $x=1$ we have exactly $\sum_{k=1}^{\infty} (arctan(\frac{1}{k}))^2$, which converges $\implies$ convergence for $x=-1$ also. So convergence $\forall x: |x|\leq 1$.

答:
Set up your limit, then apply L'Hopital's rule.
\begin{align*}
\lim_{k \rightarrow \infty} \left| \frac{a_{k+1}}{a_k} \right|
    &= \lim_{k \rightarrow \infty} \left| \frac{x^{k+1} \arctan^2 \left( \frac{1}{k+1} \right)}{x^{k} \arctan^2 \left( \frac{1}{k} \right)} \right|  \\
    &= \lim_{k \rightarrow \infty} \left|x  \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)} \right|  \\
    &= \lim_{k \rightarrow \infty} |x|  \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)}  &  &  \text{squares are automatically nonnegative}  \\
    &= |x| \lim_{k \rightarrow \infty} \frac{ \arctan^2 \left( \frac{1}{k+1} \right)}{ \arctan^2 \left( \frac{1}{k} \right)}  \\
    &\overset{L'H}= |x| \lim_{k \rightarrow \infty} \frac{ \frac{1}{1 + \left( \frac{1}{k+1} \right)^2 }}{ \frac{1}{1 + \left( \frac{1}{k} \right)^2 }}  \\
    &= |x| \cdot 1  \\
    &= |x|  \text{.}
\end{align*}
Now the ratio test gives you a condition on the value of this limit to ensure convergence: $|x| < 1$.

Now to the endpoints.  Suppose $x = 1$.  Then we consider $\sum_{k=1}^\infty \arctan^2 \frac{1}{k}$.  One way to proceed is to note $\arctan x < x$ for $x > 0$.  So
$$\sum_{k=1}^\infty \arctan^2 \frac{1}{k} < \sum_{k=1}^\infty \frac{1}{k^2}  \text{,}  $$
which is a convergent series.

Now suppose $x = -1$.  From the previous, we know the sum we obtain is absolutely convergent, hence convergent.  Therefore, the sum converges for $x \in [-1,1]$.


问:
Calculate $\sum_\limits{n=-\infty}^{\infty} \left(\frac{\sin(w_0n)}{\pi n}\right)^4$ where $w_0<\frac{\pi}{2}$
By Parseval's theorem, this sum will be equal to the power in the Fourier transform of $$\left(\frac{\sin(w_0n)}{\pi n}\right)^2$$

$\left(\dfrac{\sin(w_0n)}{\pi n}\right)^2$ can be written as $x[n]\cdot x[n]$, where $x[n]= \dfrac{\sin(w_0n)}{\pi n}$   So we can compute the Fourier transform as the convolution:

$$\frac{1}{2\pi} \int_{-\pi}^{\pi} X(t)X(w-t)\mathrm dt$$

I'm stuck on finding this convolution. I know that $X(t)$ is $1$ when $-w_0\leq t\leq w_0$. I don't know when the product $X(t)X(w-t)$ will be $1$ and when $0$. Can someone help me?

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
Lets study the behavior of
$\ds{\,\,\,
\bracks{\sin\pars{w_{0}z} \over \pi z}^{4}_{\,z\ \in\ \mathbb{C}}}$ as $\ds{\,\,y \equiv \Im\pars{z} \to \pm\infty}$.

Namely,
\begin{align}
&\expo{-2\pi\verts{y}}\,
\bracks{\sin\pars{w_{0}z} \over \pi z}_{\,x\ =\ \Re\pars{z}}^{4}
\\[5mm]
\stackrel{\mrm{as}\ y\ \to\ \pm\infty}{\sim}\,\,\,&
\expo{-2\pi\verts{y}}\,\,
{\expo{\pm 4\ic x}\expo{4\verts{w0}\verts{y}} \over 16\pi^{4}y^{4}}
\\[5mm] = &\
{\expo{\pm 4\ic x} \over 16\pi^{4}}\,\,
y^{-4}\,\exp\pars{\vphantom{\huge A}-4\verts{y}
\bracks{\color{red}{\vphantom{\LARGE A}{\pi \over 2} - \verts{w_{0}}}}}
\\[5mm]
\stackrel{\mrm{as}\ y\ \to\ \pm\infty}{\to} &
\,\,\,\,\,{\large 0}\quad
\mbox{whenever}\quad \color{red}{\verts{w_{0}} <
{\pi \over 2}}.
\end{align}
It enforces the validity of the Abel-Plana Formula I'm using in the following evaluation. Additional details are given in the above cited link.

Then,
\begin{align}
&\bbox[5px,#ffd]{\sum_{n = -\infty}^{\infty}\ \bracks{{\sin\pars{w_{0}\,n} \over \pi n}}^{4}} =
{w_{0}^{4} \over \pi^{4}}\sum_{n = -\infty}^{\infty}\ \on{sinc}^{4}\pars{w_{0}\,n}
\\[5mm] = &\
-\,{w_{0}^{4} \over \pi^{4}} + {2w_{0}^{4} \over \pi^{4}}\sum_{n = 0}^{\infty}\ \on{sinc}^{4}\pars{w_{0}\,n}
\\[5mm] = &\
-\,{w_{0}^{4} \over \pi^{4}} +
{2w_{0}^{4} \over \pi^{4}}\bracks{\int_{0}^{\infty}\ \on{sinc}^{4}\pars{w_{0}\,n}\,\dd n +
{1 \over 2}\,\on{sinc}\pars{0}}
\\[5mm] = &\
{2w_{0}^{3} \over \pi^{4}}\
\underbrace{\int_{0}^{\infty}\ {\sin^{4}\pars{n} \over n^{4}}\,\dd n}_{\ds{\pi \over 3}} =
\bbx{{2 \over 3\pi^{3}}\,w_{0}^{3}}
\end{align}


问:
Where $p$ is a prime number and $n$ is an integer, let $f(n)=\sum_{i=1}^{p-1}[in^{i-1}]$. Prove that if $f(m)=f(n)$(mod $p$) then $m=n$(mod $p$)
Any and all help is greatly appreciated, I received a hint to "set $f(n) = s'(n)$ and compute $s(x)$ by computing $s(x) - x*s(x)$, now plug in $x=n$ and you'll have a formula for $f(n)$" but still not sure where to go from here.

答:
Let $ s(n) = \ldots $

 $\sum_{i=1}^{p-1} n^i = \frac{ n^p - n } { n - 1 } $.

So $f(n) = s'(n) = \ldots $

 $$\frac{ (p-1)n^{p+1} - p n^p + n } { (n-1)^2 n}$$

So if $ f(n) \equiv f(m) \pmod{p} $, then ...

 Do some work. If you get stuck, show your work and explain why you're stuck.

Hence, $n -1  \equiv m -1  \pmod{p}$

Note: Because of the denominator, you might want to deal with the $ n \equiv 0, 1 \pmod{p} $ cases separately.


问:
How many distinct multiplications?
I have a multiset $\left\{ {2,3,3,3,4,4,5,5,6,7} \right\}$. How many distinct multiplications of three numbers chosen from the multiset?

e.g. $2×3×3$; $2×3×4$; $4×4×3$;... not $2×3×3$; $3×3×2$;... and $4×3×3$; $6×2×3$; ... (equal multiplication) are considered as the same.

答:
You can solve this using generating functions. For each number in your multiset, you can write a polynomial where the power of $x$ represents the number of times you choose a certain element from the set.

So you would have the polynomial $(1+x)$ representing the possibility of choosing or not choosing the number 2. Similarly $1+x+x^2+x^3$ represents the possibilities of choosing from the three available 3's in your list. Similarly, you have $1+x+x^2$ for the two 4's.

The answer you want is the coefficient of $x^3$ in the polynomial $(1+x)^3(1+x+x^2)^2(1+x+x^+x^3)$ which is 36.

If you are not familiar with generating functions, check out Herbert Wilf's generatingfunctionology for a detailed exposition.

EDIT: After Piseth changed his question to specify that the product has to be unique, I have to agree with Gerry's answer that there is no simple way to solve this other than to subtract repetitions from my answer.

Title: How many distinct multiplications?


答: This is not quite a definition, just an illustration of a quantity, well modeled by a Poisson random variable. I am not sure what $n,p$ in your question are, but I think that you need a large city because if you have a small population sample, they don't call each other according to the same frequency distribution...</p>

Title: A basic doubt on the definition of a Poisson random variable

Body:
What is the significance of "large city" in the definition of the following Poisson variable :

"Number of phone calls placed during a ten second interval in a large city"

I guess either $n \to \infty$ or $p$ very small is not implied by "large city".


问:
Verify the inf of a subset S of $\mathbb R$
Let S=$\{a_n=\frac{3n+1}{n}:n\geq 0\}$ , verify that $infS=3$.

MY IDEA is:  $\lim_{n \to \infty }a_n=3$, so by the definition of the limit we can say: $\forall \epsilon>0 \exists \bar{n}: |\frac{3n+1}{n}-3|<\epsilon$.

Now we recall the following property of the infimum of a generic $A\subseteq \mathbb R$:

Let $A\subseteq\mathbb R$ is bounded from below, $l=infA\iff\left.\Bigg\lbrace\begin{aligned}
 & (i) l\leq a\enspace \forall a\in A \\
 &  (ii) \forall \epsilon>0\enspace \exists\enspace a \in A: l+\epsilon>a          \\
 \end{aligned}
 \right.$

The definition of limit tell us that: $-\epsilon<\frac{3n+1}{n}-3<\epsilon\Longrightarrow 3+\epsilon>\frac{3n+1}{n}$. Therefore from the property above we conclude that $infS=3$.

It's all right?

答: Use the definition of the infimum. Let $\inf S =x$ then by definition for all $\epsilon>0$, one needs to find a $y\in S$ such that $y < x+\epsilon.$ In other words you need to find $n\in \mathbb{N}$ such that $a_n=\frac{3n+1}{n}<x+\epsilon.$ Our guess for the infimum is $x=3$ and so we need to find an $\epsilon>0$ that works. So we try, $\frac{3n+1}{n}<3+\epsilon\implies \frac{1}{n}<\epsilon$ and so if we choose any $n>\frac{1}{\epsilon}$ then we have that $a_n<3+\epsilon$ and thus $\inf S = 3.$


问:
Compute the $n$-fold iteration of an indefinite integral operator using integration by parts and induction.
Consider the following integral operator $T:L^2([0,2])\longrightarrow L^{2}([0,2])$ defined by $$(Tf)(x)=\int_0^x f(y)\,dy.$$ Page 230 of Peter Lax's functional analysis claimed that the $n$-fold iterate of $T$ is given by the formula: $$(T^n f)(x)=\dfrac{1}{(n-1)!}\int_0^x (x-r)^{n-1}f(r) \, dr, \text{ for } x\in [0,1].$$ And he said that this can be proved using induction and integration by parts.

So I followed his instruction:  Suppose this holds for a fixed $n$, and we prove it is true for $n+1$. By definition, for $x\in [0,1]$,
\begin{align*}
& (T^{n+1}f)(x)=T((T^{n}f)(x))=\int_0^x (T^{n}f)(y) \, dy \\[8pt]
= {} &\int_0^x \Big(\frac{1}{(n-1)!}\int_0^y(y-r)^{n-1}f(r) \, dr\Big) \, dy\\[8pt]
= {} & \frac{1}{(n-1)!}\int_0^x \int_0^y (y-r)^{n-1}f(r) \,dr\,dy.
\end{align*}

For fixed $y\in [0,1]$, we deal with the inner integral $\int_0^y (y-r)^{n-1} f(r) \, dr$.  Consider $u:=f(r)$ so that $du=f'(r) \, dr$, and $dv:=(y-r)^{n-1} \, dr$ so that $v=-\frac{1}{n}(y-r)^{n}.$ Then, using integration by parts, we have $$\int_{0}^{y}(y-r)^{n-1}f(r) \, dr=-\frac{1}{n}(y-r)^n f(r)+\int_0^1\frac{1}{n}(y-r)^n f'(r) \, dr.$$

Then, I focused on $\frac{1}{n}\int_0^1 (y-r)^{n}f'(r) \, dr$. Consider $dv:=f'(r)\,dr$ so that $f(r)=v$, and $u:=(y-r)^{n}$ so that $du=-n(y-r)^{n-1} \, dr.$ Using integration by parts, we have $$\dfrac{1}{n} \int_0^1 (y-r)^{n}f'(r) \, dr = \frac{1}{n}(y-r)^n f(r)+\frac{1}{n}\cdot n\int_0^1 (y-r)^{n-1} \, dr.$$ So if we plug this back into above, it seems I just proved $0=0$.

What should I do? Did I even have a wrong start? Thank you!

答:
Note that in principle you don't know that $f$ is differentiable. This is probably not a big deal, as differentiable functions are dense. But it complicates the proof.

In your use of integration by parts, if you do integration by parts twice while exchanging the roles of the two functions, so indeed you get nothing.

Here the argument becomes much easier if instead you do the following (where the exchange of integrals is valid due to Fubini); no parts, tough.
\begin{align}
T^{n+1}f(x)
&=T^n(Tf)x
=\frac1{(n-1)!}\,\int_0^x(x-r)^{n-1}\,Tf(r)\,dr\\[0.3cm]
&=\frac1{(n-1)!}\,\int_0^x(x-r)^{n-1}\,\int_0^rf(s)\,ds\,dr\\[0.3cm]
&=\frac1{(n-1)!}\,\int_0^rf(s)\,\int_s^x(x-r)^{n-1}\,dr\,ds\\[0.3cm]
&=\frac1{(n-1)!}\,\int_0^rf(s)\,\frac{(x-s)^n}n\,ds\\[0.3cm]
&=\frac1{n!}\,\int_0^rf(s)\,(x-s)^n\,ds\\[0.3cm]
\end{align}


问:
Amazing link between $\int_0^{\infty } \frac{x^2}{\sinh ^2 x} \, dx$ and $\sum _{n=1}^{\infty } \frac{1}{n^2}$?
Given $f(x)=\frac{x^2}{\sinh ^2 x}$

I computed the Laplace transform $\mathscr{L}\left(f(x)\right)(s)=\frac{1}{4} \left(4 \psi ^{(1)}\left(\frac{s}{2}\right)+s \psi ^{(2)}\left(\frac{s}{2}\right)\right)$

Then $$\int_0^{\infty } \frac{x^2}{\sinh ^2 x} \, dx=\underset{s\to 0}{\text{lim}}\mathscr{L}\left(f(x)\right)(s)=\frac{\pi ^2}{6}=\zeta(2)$$
which is $$\sum _{n=1}^{\infty } \frac{1}{n^2}=\zeta(2)$$

Do you think there is a deeper link between these two results or is it just a coincidence?

Body:
In fact,

$$
\begin{align}
\int_0^{\infty } \frac{x^2}{\sinh ^2 x} \, dx&=\int_0^\infty\frac{4x^2}{(e^x-e^{-x})^2}\,dx\\[1ex]
&=\int_0^\infty\frac{4e^{-2x}x^2}{(1-e^{-2x})^2}\,dx\\[1ex]
&=\int_0^\infty\sum_{n=1}^\infty 4ne^{-2nx}x^2\,dx\\[1ex]
&=\sum_{n=1}^\infty4\int_0^\infty ne^{-2nx}x^2\,dx\\[1ex]
&=\sum_{n=1}^\infty\frac{1}{n^2}.
\end{align}$$


问:
Cannot decipher symbols and question.
Rasta calls a number like a Tavas if and only if $1 ≤ a ≤ n$ and the sum of all primes (like $p$) that $p | a$ is exactly equal to $k$.

He asks you to find the number of Tavases.

Given example

$n=20 k=7$

The only Tavases are $7$, $10$ and $20$.

what does $p | a$ mean? and how did it work out $7$, $10$ and $20$

https://www.hackerearth.com/problem/algorithm/rasta-and-tavas/

答:
To explain the question:

For any natural number $m$ we define $F(m)$ to be the sum of the distinct prime divisors of $m$.  Thus $F(p)=p$ if $p$ is prime.  Indeed, $F(p^r)=p$ if $p$ is prime.  $F(20)=7$ since the prime divisors of $20$ are $2,5$ and $2+5=7$ and so on.

Given $n, k$ the questions asks us to find all natural numbers $m$ such that $$m≤n\quad \&\quad F(m)=k$$

For reasonably sized $n,k$ this should be a fairly easy computation.

Title: Cannot decipher symbols and question.


问:
Find Tangents to a Rotated Translated Ellipse at 2 points through which a straight line passes.
This image (view in this link) is a problem I can't seem to complete (for a personal project).

I have a 2D array of points (or pixels) that represent an ellipse and I'm trying to to find tangents to the ellipse at 2 points on the ellipse (through which a straight line passes not necessarily parallel to the horizontal). See picture for an understanding of the problem.

I can't seem to figure this out because of the rotation. I imagine it would involve a change in coordinate systems perhaps (I may be wrong).

答:
The general equation of an ellipse, like other conics, is given by a 2nd degree equation that can be given the form:

$$ax^2+2bxy+cy^2+2dx+2ey+1=0 \tag{1}$$

depending on $5$ unknown parameters.

One needs $5$ equations for determining them. These equations will be obtained by finding $5$ points of the ellipse, the first two of them $(x_1,y_1)$ and $(x_2,y_2)$ being the points of intersection with the line.

Once you have solved the linear system, i.e., obtained $a,b,c,d,e$, it is easy to obtain the equations of the tangent by the a kind of bilinear form associated to the "kind of" quadratic expression (1):

$$ax_kx+b(xy_k+x_ky)+cy_ky+d(x+x_k)+e(y+y_k)+1=0 \ \ \ \text{for} \ \ k=1,2\tag{2}$$

Remark 1: how to take points $P_3,P_4,P_5$ ? The most apart one from the other... It is clear that taking different points will give (slightly) different equations for the ellipse.

Remark 2: The last term "1" in equation (1) is arbitrary ; it could be any non-zero coefficient; for example, if we replace "1" by "5", with the same points we would obtain values for $(a,b,c,d,e)$ that are five times the values found with "1". We cannot replace "1" by "0" because it would force the ellipse to pass through the origin, which is not the case here (I think you take the origin on the lower left of the picture).

(Not essential at all for your study)

Remark 3: Explanation of the "kind of" : the "true" quadratic form is the so-called homogeneized version of (1)

$$ax^2+2bxy+cy^2+2dxt+2eyt+t^2=0 \tag{3}$$

and its "true" associated bilinar form (generalized dot product) is:

$$ax_kx+b(xy_k+x_ky)+cy_ky+d(xt_k+x_kt)+e(yt+y_kt)+tt_k=0\tag{4}$$

being understood that we set $t=t_k=1$ for retrieving (1) from (3) and (2) from (4).


问:
Finite Direct product of solvable groups is solvable

I found this proof. But among my colleagues it is rumored that there is a mistake. Could you tell me if I did, what would the mistake be? I really don't have much idea how this proof works.

答:
where $H_{11}$ is the smallest nontrivial subgroup of $G_1$ in a composition series...

That makes sense only when each $G_i$ is finite. Infinite solvable groups don't even have composition series: A solvable group with composition series is finite and don't have to have minimal subgroups.

The general solution would be due to:

Lemma. Let $G$ be a group and $H\subseteq G$ a normal subgroup. Then $G$ is solvable if and only if both $H$ and $G/H$ are.

For a proof see this: For $G$ a group and $H\unlhd G$, then $G$ is solvable iff $H$ and $G/H$ are solvable?

With that we automatically obtain that $G\times H$ is solvable if and only if both $G$ and $H$ are. And then we apply a simple induction to conclude that $G_1\times\cdots\times G_n$ is solvable if and only if each $G_i$ is.


问:
Suppose $F(x+y) \geq \frac{x}{x+y} F(x) + \frac{y}{x+y}F(y)$ for all $x ,y \in \mathbb{R}_+$. Is $F$ convex?
Let $F : \mathbb{R}_+ \to \mathbb{R}$ be a function which is controlled in the following way: $$F(x+y) \geq \frac{x}{x+y} F(x) + \frac{y}{x+y} F(y).$$ Is this function convex?

This estimate is satisfied by convex functions, i.e., $F(x) = e^x$ and $F(x)=x^2$: \begin{eqnarray*}
(x+y) e^{x+y} &=& xe^{x+y} + y e^{x+y} \ \geq \ x e^x + ye^y, \\
(x+y)(x+y)^2 & =& x \cdot x^2 + y \cdot y^2 + (\text{positive cross terms})
\end{eqnarray*}

It might be worth noting that if $t = \frac{x}{x+y} \in (0,1]$, then the defining estimate reads: $$F\left( \frac{x}{t} \right) \geq t F(x) + (1-t) F \left( \frac{x}{t} -1 \right).$$

答:
$f$ does not have to be convex.

Try for example $f(x)=-x^{-½}$, for $x>0$, which is concave and satisfies
$$
f(x+y)>\frac{x}{x+y}f(y)+\frac{x}{x+y}f(y)
$$


问:
Can't understand part of proof of Catalan conjecture
So I'm trying to understand this proof or at least part of it and I'm stumped at a point that seems pretty basic.

In the ring $Z[i]$ we have $x^p = y^2 + 1 = (y - i)(y + i)$. It is known that $Z[i]$ is a unique factorisation domain. Now there is no prime $\pi \in Z[i]$ such that $\pi|y + i$ and $\pi|y - i$. For suppose there is such a prime., then it has to divide $y + i - (y - i) = 2i$, so $\pi$ divides 2, since $i$ is a unit. It follows that 2 divides $x$, which we have proven to be impossible.

that last step is what confuses me ("it follows that 2 divides x"). I can see how that fact would follow if 2 divided $\pi$, as then, since $\pi$ divides, say $y - i$, it divides $x^p$ and hence x itself. But how the contradiction follows from the fact that $\pi$ divides 2 eludes me. I can also see how that would follow if $\pi$ dividing 2 meant that $\pi$ equals 2. And that would be the case in the normal ring of integers but since these are Gaussian integers, I'm not sure that's the case...

答: Norms show you that the only possible primes dividing $2$ are $\pm (1 \pm i)$, and since $x$ is real, if one of these primes divides $x$, so too must its conjugate.  That's because complex conjugation is a ring automorphism.  If both $1+i$ and $1-i$ divide $x$, then so too must their product, so $2=(1+i)(1-i) = (-1+i)(-1-i) \vert x$.


答:
Since $\pi \mid y+i$ then $N(\pi) \mid N(y+i) = y^2+1 = x^p$

But $\pi \mid 2$ implies $\pi \sim 1+i$, so $N(\pi) = 2$

Title: Can't understand part of proof of Catalan conjecture

Body:
So I'm trying to understand this proof or at least part of it and I'm stumped at a point that seems pretty basic.

In the ring $Z[i]$ we have $x^p = y^2 + 1 = (y - i)(y + i)$. It is known that $Z[i]$ is a unique factorisation domain. Now there is no prime $\pi \in Z[i]$ such that $\pi|y + i$ and $\pi|y - i$. For suppose there is such a prime., then it has to divide $y + i - (y - i) = 2i$, so $\pi$ divides 2, since $i$ is a unit. It follows that 2 divides $x$, which we have proven to be impossible.

that last step is what confuses me ("it follows that 2 divides x"). I can see how that fact would follow if 2 divided $\pi$, as then, since $\pi$ divides, say $y - i$, it divides $x^p$ and hence x itself. But how the contradiction follows from the fact that $\pi$ divides 2 eludes me. I can also see how that would follow if $\pi$ dividing 2 meant that $\pi$ equals 2. And that would be the case in the normal ring of integers but since these are Gaussian integers, I'm not sure that's the case...


问:
Proving $B=(A\cap B^c)\cup(A^c\cap B)\iff A=\emptyset$
I would like to know how to prove the second part of the biconditional I want to prove that $B=(A\cap B^c)\cup(A^c\cap B)\iff A=\emptyset$

I already demonstrated this part:

If $A=\emptyset$ then $A\cap B^c=\emptyset$ and since $A^c=\emptyset^c=U$ and since $B\subseteq U$ then $B\cap U=B$ and $B\cap A^c=B$ and therefore $(A\cap B^c)\cup(A^c\cap B)=\emptyset\cup B=B$ which is a part of the mathematical proof

But how can I prove that if $B=(A\cap B^c)\cup(A^c\cap B)$ then $A=\emptyset$ ?

I appreciate your time and help

答:
If $B=(A\cap B^c\cap B)\cup(A^c\cap B\cap B)$, then notice that:

$B=B\cap B=(A\cap B^c\cap B)\cup(A^c\cap B\cap B)=A^c\cap B$; therefore you have $A^c\supseteq B$, i.e. $A\subseteq B^c$.

$\emptyset=B\cap B^c=(A\cap B^c\cap B^c)\cup(A^c\cap B\cap B^c)=A\cap B^c$. By the previous point, $A\cap B^c=A$.


问:
Difference between Norm and Distance
I'm now studying metric space. Here, I don't understand why definitions of distance and norm in euclidean space are repectively given in my book.

I understand the difference between two concepts when i'm working on non-euclidean space, but is there any even slight difference between these two concepts when it is $\mathbb{R}^k$?

答:
The metric space/distance metric is a mathematical construction built up on topological space (definition of convergence, continuity, completeness, separability, connectedness, compactness etc. etc.). The distance operates with point sets and doesn't know the vectors. The Hilbert space is built up on metric space and defines new operation as inner product, vectors and using norms. The metric space itself doesn't know norms and vectors, since as is introduced on Hilbert space. Below is a derivation how to utilize metric space in Hilbert space.

Reference: Muscat, J. Functional Analysis An Introduction to Metric Spaces, Hilbert Spaces, and Banach Algebras, Springer, 2014.

Title: Difference between Norm and Distance