答:
In order to have a clean complete solution, the "steps in between" have to be mentioned, and moreover it should be clear if we work with $\Rightarrow$ (only), or if we work with equivalences all the time.

For instance, if we take the initial equation, and move some terms "on the other side", then take squares, then we are going into one direction (only). From the given equation we obtain the new equation, (i.e. the given equation implies the new equation), but not necessarily in the converse direction also. With such a strategy, one must finally verify the obtained solutions. (Or do something instead.)

Best, we apply squaring on quantities that are both positive. Then squaring leads to a new equation, but we have an equivalence, a particular value $z_0$ is solution of the first equation if and only if it is a solution of the second equation.

In this sense, we can deliver the solution as follows (so that no final check is needed).

Let $z$ be a solution of the equation $$\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0\ ,$$ where all square roots are taken from values $\ge 0$. (And the results are $\ge 0$.)

(In other words, $5z+5\ge 0$, and $3-3z\ge 0$, and $z\ge 0$. Equivalently, $0\le z\le 1$. And it is the duty of the person writing down the problem to insert this condition in the problem. But well, teachers often mention "so that the equation makes sense". And more often they tacitly (i.e. do not) mention it. Well, for me, the stuff makes sense also for $z=1+i$, and i will take my own branch of the square root to have more solutions. Long story short: Didactically, the problem must come with $0\le z\le 1$. So...)

We restrict to values of  $z$ (and thus also to solutions $z$)
with $0\le z\le 1$. (The only values, that "make sense in the school" while writing the equation. (Well, only in some very special schools respecting the tradition of old centuries, i.e. those schools in those countries where students, parents, teachers, and the piece with educational responsibility in the government insist to live in the older centuries. Such schools deny the knowledge of $i$, and insist to put all the responsibility for the claims and proofs on the student=solver.))

Now we rewrite equivalently at each step, i.e. from line to next line:
$$
\begin{aligned}
\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} &= 0\ ,\\
\sqrt{5z+5} &= \sqrt{3 - 3z} + 2\sqrt{z} \ ,\\
&\qquad\text{(squaring quantities $\ge 0$)}\\
5z+5 &= 3-3z + 4z +4\sqrt{3z(1-z)}\ ,\\
4z+2 &= 4\sqrt{3z(1-z)}\ ,\\
2z+1 &= 2\sqrt{3z(1-z)}\ ,\\
&\qquad\text{(squaring quantities $\ge 0$)}\\
4z^2+4z+1 &= 12z(1-z)\ ,\\
4z^2+4z+1 &= 12z-12z^2\ ,\\
16z^2-8z+1 &= 0\ ,\\
(4z-1)^2 &= 0\ ,\\
(4z-1) &= 0\ ,\\
4z&=1\ ,\\
z&=\frac 14\ .
\end{aligned}
$$
In this moment, we know that $1/4\in[0,1]$, makes sense for the given equation, is a solution, and the only solution. (There is no need to check it.)

Note: In this special case it is easier to work in only one direction, using implications $\Rightarrow$, and finally verify this solution:
$$
\underbrace{\sqrt{5\cdot \frac 14+5}}_{=\frac 52}
-
\underbrace{\sqrt{3-3\cdot \frac 14}}_{=\frac 32}
-
2\sqrt{\frac 14}
=0\ .
$$
(Yes, $\frac 52-\frac 32-1=0$.)

Body:
Find all $z$ such that $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$.

After much trial and error, I was able to rearrange the above equation
as $80z^2 - 112z-4 = 0$.

$z = \frac{7\pm3\sqrt{6}}{10}$.

I substituted $\frac{7+3\sqrt{6}}{10}$ as $z$ in the $\sqrt{5z+5} - \sqrt{3 - 3z} - 2\sqrt{z} = 0$. Things worked out.

Am I still not done? Must I check for extraneous solutions in this situation?
I wonder if it might not require it because if ever we know one of the roots of a quadratic with rational coefficients is irrational, we know the other.

I suppose I'm also trying to avoid going through another round of substitution and test with these irrational conjugates. I'm not very trusting of my algebra yet. So I end up rewriting the same equations time and time again.


问:
How do I solve for $dy/dx$ if $y=\ln (\sin x+\ln x)$?
Solve for $\frac{dy}{dx}$ if $y=\ln(\sin x+\ln x)$.

I know how to solve for integrals involving $du$ and $u$, but how do I do this type of problem (I think it's the opposite of the integral problem)?

答: You use the chain rule, which says $[f(g(x))]'=f'(g(x))g'(x)$</p>

ContentLicense: CC BY-SA 3.0


问:
Material derivative of a vector field
I have this velocity field, of which I have to get the material derivative:
$$\mathbf u(x,y)=\sin(x-y)\mathbf i-\cos(x-y)\mathbf j$$
Where i & j are the vector field components.
Seems pretty straightforward:
$$\mathbf a=\mathbf u \cdot \nabla \mathbf u$$

$$\mathbf a=(\cos(x-y)-\sin(x-y))(\sin(x-y)\mathbf i-\cos(x-y)\mathbf j$$
But apparently the correct answer is:
$$\mathbf a=(\sin(x-y)+\cos(x-y))(\cos(x-y)\mathbf i+\sin(x-y)\mathbf j$$
Did I not simplify right? Or am I missing something?

答:
$$ \boldsymbol u \cdot \nabla \boldsymbol u = (u_1 \partial_x + u_2\partial_y)\boldsymbol{u} $$

I think you have the order wrong.

ContentLicense: CC BY-SA 4.0

Title: Material derivative of a vector field


问:
How many ways are there in the case that there is ONLY ONE adjacent couple?
"Three couple will be arranged in a row.How many ways are there in the case that there is ONLY ONE couple which are adjacent."

For example , let $(a_1,a_2),(b_1,b_2),(c_1,c_2)$ be the couples, so they cannot be arranged in the form of $a_1,c_1,c_2,b_2,b_1,a_2$.  There can be only one couple adjacent such as $b_1,a_1,a_2,c_1,b_2,c_2$ is acceptable.

I found this question in my textbook.I tried that $C(3,1).5!.2!-[C(3,2).6.2!.2!.2!+C(3,3).2!.2!.2!.3!]$
but my answer is wrong and I am confused. Can you help me?

答:
Suppose the couples are $A,B,C$. Since exactly one couple is to be together, the number will be $3$ times the number only couple $A$ is together.

To apply inclusion-exclusion, we shall have to exclude the $2$ ways in which two couples including $A$ can be formed  and be together and add one formation in which all three couples are together.

For the computation, it is convenient to consider couples who are together as "individuals" for this purpose.

Thus we get  $\;3[2^1\cdot5! - 2\cdot2^2\cdot4! +1.2^3\cdot 3!] = 288$

Added explanation for first term

$2^1$ because a couple can sit in $2$ ways, and $5!$ gives permutations of $4+1$ "individual"


问:
What's the value of the inverse square and inverse of an unknown exponential?
The problem is as follows:

Given:

$$x^{-x}=(-8)^{-3^{-1}} \cdot (-8^{0})^{3^{4^{5}}}$$

Find the result of

$$(x^{-2}-x^{-1})$$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\frac{3}{4}\\
2.&\frac{1}{4}\\
3.&-\frac{1}{4}\\
4.&\frac{1}{8}\\
5.&\frac{3}{8}\\
\end{array}$

What I attempted to do was as follows:

From what I can understand this whole expression equals to $-1$ as the sign is taken out of the exponential and the $8$ powered to the rest becomes just $1$.

$(-8^{0})^{3^{4^{5}}}=-1$

Then if I'm not mistaken the thing is with the rest:

$(-8)^{-3^{-1}}=(-8)^{-\frac{1}{3}}=-\left(\frac{1}{2^3}\right)^{\frac{1}{3}}$

$-\left(\frac{1}{2^3}\right)^{\frac{1}{3}}=-\frac{1}{2}$

Then this means:

$\left(-\frac{1}{2}\right)\times (-1)=\frac{1}{2}$

Then this means:

$x^{-x}=2^{-1}$

But this doesn't seem right. I'm stuck there. Can someone help me here?. Maybe is my interpretation of the problem not accurate?. The official answer is the first option but I don't know how to get there.

Body:
If
$$x^{-x}=8 \cdot (-8)^{-3^{-1}} \cdot (-8^{0})^{3^{4^{5}}}$$
or

$$x^{-x}=-(8)^{\left(\frac{3}{2}\right)^{-1}} \cdot (-8^{0})^{3^{4^{5}}},$$

then we have
$$(x^{-2}-x^{-1})=\frac{3}{4}.$$


问:
Set of all finite subsets of $\mathbb{N}$ not equal to the to set of subsets of $\mathbb{N}$
I can kind of grasp why this is the case as if we take the union of all finite subsets of cardinality $i$ as $i$ runs through every natural number, we are listing finitely many elements each time.

Whereas with the set of all subsets we can just keep adding new elements via Cantor's argument.

However I do not feel very secure on this type of understanding, I was just wondering if someone could deliver a good explanation as to why they aren't equivalent.

答:
The first set is a strict subset of the second: if for example you consider the set of natural numbers $\mathbb{N}$, then the set of even numbers or the set of all natural numbers above a certain value, are all infinite sets and thus are only contained in the set of all subsets, but not in the set of finite subsets.

Actually the cardinality of the first is equal to $\mathbb{N}$, whereas the cardinality of the second (which is the power set or $\mathbb{N}$) is equal to the one of the real numbers.


问:
prove inequality $\frac{\sin \pi x}{\pi x} \geq \frac{1-x^2}{1+x^2}$ for all R
Let $F(x) = \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} $, is even, so just prove $x > 0$.

When $ x \geq 1 $, $\frac{\sin\pi x}{\pi x} \leq 1$, so
$$ \frac{\sin \pi x}{\pi x} - \frac{1-x^2}{1+x^2} \\= -\frac{\sin\pi (x-1)}{\pi(x-1)} \frac{x-1}{x} - \frac{1-x^2}{1+x^2} \\ \geq \frac{1-x}{x} - \frac{1-x^2}{1+x^2} \\ = \frac{(x-1)^2}{x(1+x^2)} \\ \geq 0 $$

So how to prove it when $0 < x <1 $?

答:
Disclaimer: Here is a solution which is not mine. I just slightly  adapted it from the very same question, asked as Problem 5642, by R. Redheffer, American Mathematical Monthly 76, (1969), p.422, entitled "A delightful inequality"...

Let us fix $x$ with $0 < x < 1$:

Starting from the classical infinite product for the cardinal sine:

$$\frac{\sin \pi x}{\pi x}=\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2}\right)=(1-x^2)\underbrace{\prod_{k=2}^{\infty}\left(1-\frac{x^2}{k^2}\right)}_{\lim_{n \to \infty} P_n(x)}$$

where $P_n(x)$ is defined by:

$$P_n(x):=\prod_{k=2}^{n}\left(1-\frac{x^2}{k^2}\right),$$

it is enough to prove that $$\text{for any} \ \ n \ge 2, \ \ P_n(x) \ge \dfrac{1}{1+x^2} \ \ \iff \ \ (1+x^2)P_n(x) \ge 1 \ \ \tag{1}$$

Let us remark the evident relation:

$$P_{n+1}(x)=\left(1-\frac{x^2}{(n+1)^2}\right)P_n(x) \tag{2}$$

Actually we get, for all $n \ge 2$:

$$(1+x^2)P_n(x) \ge 1+\frac{x^2}{n}$$

by a simple induction argument based on (2). Therefore, (1) is established.

An interesting article where I have found the reference to this question can be found here.

Edit: I found the proof by River Li very interesting.

I noticed a possible alternative treatment beginning at the level where a third degree polynomial appears between square brackets.

Setting $a:=\pi^2$ and $Y:=y^2$,

$$\begin{align}
&\Big[2\pi^2(2y^2 + 6\pi^2) - (\pi^2 - y^2)(\pi^2 + y^2)^2\Big]\\
\ge & \ a(2Y+6a)-(a-Y)(a+Y)^2\\
= & \ Y^3 + aY^2 - a^2Y + 4aY  + 12a^2- a^3\\
= & \ \underbrace{(Y+a/3)^3}_{>0}+4a\underbrace{\Big[Y(1-a/3)+a(3-7a/27)\Big]}_{p(Y)}
\end{align}$$

with first degree polynomial $p(Y)>0$ for $Y \in [0,1]$ because

$$p(0)=a(3-7a/27)>0 \ \ \text{and} \ \ p(1)=a(8/3-7a/27)+1>0$$

as a consequence of the fact that $a=\pi^2<10$.


答:
Use the "good old" chain rule: remember that?

$$y = \ln(f(x)) \iff y' = \dfrac{f'(x)}{f(x)}$$

$$y = \ln(\sin x + \ln x) \implies y' = \dfrac{\left(\cos x + \frac 1x\right)}{\sin x + \ln x}$$

Title: How do I solve for $dy/dx$ if $y=\ln (\sin x+\ln x)$?

Body:
Solve for $\frac{dy}{dx}$ if $y=\ln(\sin x+\ln x)$.

I know how to solve for integrals involving $du$ and $u$, but how do I do this type of problem (I think it's the opposite of the integral problem)?


问:
Let $A=B\oplus C$. If $X\subset A$, $V\not\subset B$, $\dim(A)\neq\dim(C)$ is then $\dim(V)>\dim(A)-\dim(B)$?
Let $A=B\oplus C$ with $\dim(A)=\dim(B)+\dim(C)$ and $C$ being the complementary vector space to $B$. If $V\subset A$, $\dim(A)\neq\dim(C)$ and $V\not\subset B$, does that necessarily imply $\dim(V)>\dim(A)-\dim(B)$?

答:
Not necessarily, for example take $A=\mathbb{R^3}$ and $B=\text{span}(e_1),\, C=\text{span}(e_2,e_3)$ where $\{e_1,e_2,e_3\}$ the standard basis of $\mathbb{R^3}$. Take as $V=\text{span}(e_1+e_2)$. Then $V\subsetneq C$ and $V\subsetneq B,\, \text{dim}(V)=1$ and
$$\text{dim}(V)=1<3-1=2=\text{dim}(A)-\text{dim}(B)$$

And
$$\text{dim}(V)=1\leq 3-2=\text{dim}(A)-\text{dim }(C)$$


问:
Complex integral help involving $\sin^{2n}(x)$
Show that $$\int_0^\pi \sin^{2n} \theta d\theta=\dfrac{\pi(2n)!}{(2^n n!)^2} $$

So far I have came up with:

$$\sin^{2n} \theta = \left(\dfrac {z-z^{-1}}{2i} \right)^{2n}$$ and I know I should be using:

$$(a+b)^n=\sum_{k=0}^n \dfrac{n!}{(n-k)!k!}a^kb^{n-k}$$

but I'm not sure how to get the conclusion. Any help will be greatly appreciated.

答:
This is my favorite example of integration by parts.

Let $I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^n(x) dx$.

$I_n = \displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-1}(x) d(-\cos(x)) = -\sin^{n-1}(x) \cos(x) |_{0}^{\frac{\pi}{2}} + \int_{0}^{\frac{\pi}{2}} (n-1) \sin^{n-2}(x) \cos^2(x) dx$

The first expression on the right hand side is zero since $\sin(0) = 0$ and $\cos(\frac{\pi}{2}) = 0$.

Now rewrite $\cos^2(x) = 1 - \sin^2(x)$ to get

$$I_n = (n-1) \left(\displaystyle \int_{0}^{\frac{\pi}{2}} \sin^{n-2}(x) dx - \int_{0}^{\frac{\pi}{2}} \sin^{n}(x) dx \right) = (n-1) I_{n-2} - (n-1) I_n$$

Rearranging we get  $n I_n = (n-1) I_{n-2}$, $I_n = \frac{n-1}{n}I_{n-2}$.

Using this recurrence we get
$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} I_1$$

$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} I_0$$

$I_1$ and $I_0$ can be directly evaluated to be $1$ and $\frac{\pi}{2}$ respectively and hence,

$$I_{2k+1} = \frac{2k}{2k+1}\frac{2k-2}{2k-1} \cdots \frac{2}{3} = \dfrac{4^k (k!)^2}{(2k+1)!}$$

$$I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k+1}}$$

Your integral is twice the above integral, since the sine function is symmetric about $x=\pi/2$. Hence, the value of your integral is

$$J_{2k+1} = \dfrac{2^{2k+1} (k!)^2}{(2k+1)!}$$

$$J_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{1}{2} \frac{\pi}{2} = \dfrac{(2k)!}{(k!)^2}\dfrac{\pi}{2^{2k}}$$


问:
$\text{Var}(Y|Z)$ where $Y,Z$ has normal distribution
Let $X_1,...X_{25}$ be a sequence of i.i.d with $N(0,1)$ distribution and let $Y=X_1+...X_{10}$ and $Z=X_1+...+X_{25}$. I have to find $\text{Var}(Y|Z)$. What is the easiest way to do it?

答:
It is easy to see, that $(Y,Z)$ follows the multivariate normal distribution:
$$N(\begin{pmatrix}0\\ 0\end{pmatrix} , \begin{pmatrix} 10 & 10 \\ 10 & 25 \end{pmatrix})$$
Using the formula for conditioning in a multivariate normal distribution we get the conditional variance to be $Var(Y|Z) = (1-\rho^2)Var(Y)$, where $\rho$ is the correlation coefficient between $Y$ and $Z$. And since
$$Var(Y) = \sum_{i=1}^{10} Var(X_i) = 10$$
and
$$\rho = \frac{Cov(Y,Z)}{\sqrt{Var(Y)}\sqrt{Var(Z)}} = \frac{10}{\sqrt{10}\sqrt{25}}=\sqrt{\frac{10}{25}},$$
we get, that
$$Var(Y|Z) = (1-\frac{10}{25})10 = 6.$$

(see https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Bivariate_case_2 for reference about the conditional distribution in a bivariate normal distribution)


问:
How to interpret the square root of inner product over an arbitrary field?
In an inner product space, the norm $\|x\| = \sqrt{\langle x,x\rangle}$ is induced. I realise I have almost always dealt with either real or complex numbers, so took the square root for granted.

Reading the Inner Product entry on
Wolfram (https://mathworld.wolfram.com/InnerProduct.html), it says "A vector space together with an inner product on it is called an inner product space. This definition also applies to an abstract vector space over any field." This was said after introducing the axioms in the context of a space over $\mathbb{R}$.

I am confused because I do not know how to interpret the square root in an arbitrary field. I assume the obvious way would be to define it to be the element $a \in \mathbb{F}$ such that $a^2 = \langle x,x\rangle$. But the issue I am having is how we even know whether such an element exists in the field? Is this a standard result from ring theory?

My understanding has always been that inner product spaces (and normed spaces) are only defined over either the real or complex numbers. How do you construct them (or something equivalent) over some arbitrary field?

答:
The claim doesn't make sense. On vector spaces $V$ over arbitrary field $k$ we have bilinear forms $b(x,y)$. When $k=\Bbb{C}$ we also look at sesquilinear forms, which means that the second argument is linear after applying an automorphism $\sigma$ of the field (the complex conjugaison). But then we can consider $V$ as a $k^\sigma$ vector space to make it linear, so assume that $b$ is truly linear.

$q(x) = b(x,x)$ is a quadratic form.

A first desirable property is that $b(x,y)=b(y,x)$ (when $char(k)\ne 2$ there is a one-to-one correspondence between quadratic forms and symmetric bilinear forms).

A second one is that $q(x)=0$ iff $x=0$. In that case $q$ is said anisotropic.

When $k$ is an ordered field there is a 3rd one: that $\forall x,q(x)\ge 0$. With the previous ones this is the definition of "$b$ is an inner product". When it is the case then $\|x\|=\sqrt{q(x)}$ is some kind of norm (when $k$ is not a subfield of $\Bbb{R}$ then $\|x\|$ is not real valued so this is a bit different). Do you think we always have $\|x+y\| \le \|x\|+\|y\|$ ?

$\sqrt{q(x)}$ is an element of the algebraic extension of $k$ obtained by adding all the square roots of elements $\ge 0$, it is ordered too, through $\sqrt{a}\ge \sqrt{b}$ iff $a\ge b$, then applying the law of orders.

Note that real valued norms exist other any fields, for example $\|x\| = 0$ if $x_1=x_2=0$ and $=1$ otherwise is a real-valued norm over $k^2$ for any field, a norm for the trivial absolute value  $|a|_{tr}= 0$ if $a=0$ and $=1$ otherwise, such that $\|ax\|=|a|_{tr} \|x\|$.


问:
Milne Thomson method for determining an analytic function from its real part
What is the logic behind taking  $z = {\bar {z}}$ while finding analytic function in Milne-Thomson Method ?

I mean we write
$ { f(z)=u(x,y)+iv(x,y)} $

as
$
{\displaystyle f(z)=u\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar {z}}}{2i}}\right)+iv\left({\frac {z+{\bar {z}}}{2}}\ ,{\frac {z-{\bar {z}}}{2i}}\right)} {\displaystyle } $

Its fine uptil now but then we say $f(z)$ can be regarded as an identity in two independent variables $z$ and $ {\bar {z}}$

so we take $z = {\bar {z}}$, this gives us

$ { f(z)=u(z,0)+iv(z,0)} $

Source

I am not getting the the part $z = {\bar {z}}$ . How do we regard it an identity of 2 independent variable? Please guide me with an intuitive explanation. ( Please use example if you can ) .

I went through this Link  but could not understand how $w$ is independent of $z$. I mean whatever value of $x$ and $y$ we take $z$ and $w$ are obtained from the same x and y, so how can we say both are independent!

Thanks in advance !

答:
I know I'm a little late to the party but I think this will feel more intuitive.

\begin{align}
f(z) &= u(x,y) + \imath v(x,y) \\
f(x + \imath y) &= u(x,y) + \imath v(x,y)
\end{align}

Put $ y=0 $,
\[ f(x) = u(x, 0) + \imath v(x,0) \]
Now, $x$ becomes just a dummy variable for the function $f$. Replacing $x$ by $z$ we get,
\[ f(z) = u(z,0) + \imath v(z,0) \]

This exact thing is reflected when we take $ z=\bar{z} $, that is, we are putting $y=0$. We are first expressing the function using only a single real variable and then saying that the variable can also take on complex values.


问:
SO(3) invariant metric on two sphere
I read that the $SO(3)$ invariant metric on the two sphere $S_2$ is given by

$ds^2=d\theta^2+\sin^2\theta d\phi^2$

How can I derive this? and how does $g_{ij}$ look like?

答:
The 2-sphere can be parameterized by
$$\begin{cases}
x = \sin\theta \cos\phi \\
y = \sin\theta \sin\phi \\
z = \cos\theta
\end{cases}$$
where $0 \leq \theta \leq \pi$ and $0 \leq \phi \leq 2\pi.$

From this we get
$$\begin{cases}
dx = \cos\theta\,d\theta \cos\phi - \sin\theta \sin\phi\,d\phi \\
dy = \cos\theta\,d\theta \sin\phi + \sin\theta \cos\phi\,d\phi \\
dz = -\sin\theta\,d\theta
\end{cases}$$
so the ordinary $SO(3)$-invariant metric is given by
$$
ds^2
= dx^2 + dy^2 + dz^2
\\= (\cos\theta\,d\theta \cos\phi - \sin\theta \sin\phi\,d\phi)^2
+ (\cos\theta\,d\theta \sin\phi + \sin\theta \cos\phi\,d\phi)^2
+ (-\sin\theta\,d\theta)^2
\\= d\theta^2 + \sin^2\phi\,d\phi^2
.
$$

In the coordinate system $(\theta,\phi)$ we therefore have
$$\begin{array}{ll}
g_{\theta\theta} = 1 & g_{\theta\phi} = 0 \\
g_{\phi\theta} = 0 & g_{\phi\phi} = \sin^2\phi
\end{array}$$

Title: SO(3) invariant metric on two sphere


问:
Prove there is a tree with $n$ vertices having degrees $d_1, d_2....d_n$
For $n ≥ 2$ suppose $d_1, d_2,....d_n$ are positive integers with sum $2n - 2$. Prove there is a tree with n vertices having degrees $d_1, d_2....d_n$. I'm at a loss on this one. I'm sure it's pretty simple but I just can't get it.</p>

答:
you can use induction : it's obvious for n = 2; prove that there is at lease an $a_i = 1$ in this sequence and also there is at least one $a_i\geq 2$ and use this to fact.

P.S: since it's homework I would rather not to solve the problem but just give hints.


问:
Limit of $\frac{1}{\sqrt[n] n}$
I would like to show that the limit of the sequence $(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).

Definition: A sequence $(a_n)_{n=1}^{\infty}$ converges to a real number A iff for each $\epsilon > 0$ there is a positive integer N such that for all $n \geq N$ we have $|a_n - A| < \epsilon$.

Edit: Here is my work.

Let $\epsilon > 0$. There exists $N$ such that for $n \geq N$,

$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} - 1| = |\frac{n^{\frac{n-1}{n}} - n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$

Here is where I am stuck.

答:
Not sure if that would help

your equation $= \dfrac{1}{n^\frac{1}{n}} = n^\frac{-1}{n} = \exp\left(\dfrac{-1}{n}*\ln(n)\right)$

$\dfrac{-1}{n}*\ln(n)$ goes to $0$ when $N$ goes to $\infty$

Therefore $\exp(0) = 1$

Title: Limit of $\frac{1}{\sqrt[n] n}$


答:
Proceed along the following lines.

First show that there exists one $d_i$ such that $d_i = 1$. Why is this true? What will happen if all the $d_j$'s are greater than $1$?

Remove this vertex. What happens to the sum $\sum_{j \neq i} d_j$?

Now induction to the rescue.

Title: Prove there is a tree with $n$ vertices having degrees $d_1, d_2....d_n$

Body: For $n ≥ 2$ suppose $d_1, d_2,....d_n$ are positive integers with sum $2n - 2$. Prove there is a tree with n vertices having degrees $d_1, d_2....d_n$. I'm at a loss on this one. I'm sure it's pretty simple but I just can't get it.</p>


问:
How does Cohen's theorem of transfinite induction relate to the the reagular (or more modern) one?
My understanding of transfinite induction comes from Weaver where he says:

Theorem: Suppose that for every ordinal $\alpha$

$(\forall\beta<\alpha)\phi(\beta)\rightarrow\phi(\alpha)$

holds. Then $(\forall\alpha)\phi(\alpha)$ holds.

I understand this statement - it is a pretty clear extension of "regular" induction - essentially that if knowing $\phi$ is true for all smaller ordinals tell you that $\phi$ is true for the "next" ordinal then $\phi$ is true for all ordinals.

However, Cohen formulation is very different and I'm having trouble following it. He says (note: Cohen also uses $\phi$ in his formulation, but I think it means something different, so I replaced it with $\psi$ to try to avoid confusion):

Definition: let $A_n(x,y,t_1,...t_k)$ run through all formulas in ZF with at least two free variables

Theorem: Let $t_i$ be given and assume $\forall x \exists!yA_n(x,y,t_1,...,t_k)$ so that $A$ defines a function $y=\psi(x)$. For every ordinal $\alpha$ and set $z$ there is a unique function $f$ defined on $\{\beta|\beta\leq\alpha\}$ (this is actually the set $\alpha+1$), such that $f(0)=z$ and for all $\beta\leq\alpha$, $f(\beta)=\psi(h)$ where $h$ is $f$ restricted to $\beta$.

I'm not able to relate these two theorems. I think Weaver's $\phi$ is Cohen's $A_n$ but I'm not sure. Here are some thoughts I've tried:

Thought 1: I've tried to start at $\alpha=0$. Since $0=\emptyset$, then $h$ being $f$ restricted to $0$ is empty i.e. $h=\emptyset$. That means that:

$f(0)=z=\psi(\emptyset)$

But here, I'm already stuck. This is supposed to hold for every $z$ but it can't be true because $\psi(\emptyset)$ is unique.

Thought 2: But, okay, let's say that $f(0)=z$ and the statement $f(\beta)=\psi(h)$ only holds for ordinals greater than $0$. So, then, $f(1)=\phi(h)$ where $h$ is the function mapping $0$ to $z$ (I suppose $h=\{\{0, \{0,h\}\}\}$). So now we know what $f(1)$ is. And we can keep going to $f(2), f(3), ...$ and then $f(\omega)$, etc.... So we've got some function $f$ defined and being "built up" from how $\psi$ "tells us" what to do next (given that we started from $z$). But I still don't see how to relate this back to Weaver's formulation and a way of proving some statement is true for all ordinals.

Any help relating the two, or just understanding what Cohen's formulation is saying, is appreciated.

P.S. before anyone asks, I'm just reading Cohen for fun.

答:
I checked with some friends and got an answer to this myself.

Thought 1: since $f(0) = z$, the statement of the theorem really should be "for all $\beta$ such that $0<\beta\leq\alpha$, $f(\beta)=\psi(h)$".

Thought 2: Cohen's version is really more "Transfinite Recursion" than it is "Transfinite Induction." In other words, given a statement $A_n$ and a set $z$ you can prove that there is a function $f$ on the ordinals where $f(\beta)$ only depends on how $f$ works on ordinals less than $\beta$.

The two versions (what I'm calling Weaver's and Cohen's) are equally powerful since you can get each from the other.


问:
Comparing two Diagonalisable matrices with identical specific features

Hello, I was wondering if anyone could help me with this Linear Algebra question that I'm stuck on.

So far this is what I've worked out:

I know that if a Matrix is Diagonalisable then it's diagonal entries are the eigenvalues of the matrix.

However having only been told that the matrices have the same set of eigenvalues, I believe this means that I cannot assume that they are ordered the same in each matrix and hence the determinants may differ. So b) cannot be correct.

However for the trace, I know that the trace for a Diagonalisable matrix is the addition of all the diagonal entries. Therefore the order of the eigen values along the diagonal doesn't matter and hence c) must be correct.

For a) I have read up in my notes that $X_A(\lambda)=det(A-(\lambda)I_n)$ and if I am saying that the determinants may not equal each other then I believe that $X_A(\lambda)$ cannot equal $X_b(\lambda)$.

And therefore I think my answer should be that c) and d) are correct.

Can anyone advise me if I'm correct or not/where I'm going wrong.

Thanks in advance.

答:
First, a quick response to what you've "worked out".

I know that if a Matrix is Diagonalisable then it's diagonal entries are the eigenvalues of the matrix.

This is incorrect

However having only been told that the matrices have the same set of eigenvalues, I believe this means that I cannot assume that they are ordered the same in each matrix and hence the determinants may differ. So b) cannot be correct.

Also incorrect

However for the trace, I know that the trace for a Diagonalisable matrix is the addition of all the diagonal entries. Therefore the order of the eigen values along the diagonal doesn't matter and hence c) must be correct.

There is nothing special about diagonalizable (US spelling, hope you don't mind) matrices in this regard. It is true that c is correct.

For a) I have read up in my notes that $X_A(\lambda)=det(A-(\lambda)I_n)$ and if I am saying that the determinants may not equal each other then I believe that $X_A(\lambda)$ cannot equal $X_b(\lambda)$.

Also incorrect (but consistent with your earlier statement)

And therefore I think my answer should be that c) and d) are correct.

You haven't said anything about d).

Now, to answer the questions. a,b,c are all correct, d is not correct.

If $A,B$ are diagonalizable with the same multiplicities, then there is a single diagonal matrix $D$ for which we have $A = PDP^{-1}$ and $B = QDQ^{-1}$ for some invertible matrices $P,Q$. In other words, $A$ and $B$ are similar to the same diagonal matrix (and are therefore similar to each other).

$A$ and $B$ have the same eigenvalues (the diagonal elements of $D$). The determinant is the product of these eigenvlaues. So, $A,B$ have the same determinant.

Likewise, the trace is the sum of these eigenvalues. So, $A,B$ have the same trace.

The algebraic multiplicity of the eigenvalue $\lambda$ of $A$ is the exponent of $(x - \lambda)$ in the characteristic polynomial of $A$. Since the eigenvalues of $A,B$ have the same algebraic multiplicities, $A,B$ must have the same characteristic polynomial.

The eigenspace $E_{\lambda}$ depends on the eigenvectors, not eigenvalues. $A$ and $B$ do not necessarily have the same eigenvectors. For example, if we consider the similar matrices
$$
A = \pmatrix{2&1\\1&2}, \quad B = \pmatrix{1&2\\0&3},
$$
then we see that the eigenvalues of $A,B$ are $1,3$. $E_1(A)$ is spanned by $(1,-1)$, and this is not equal to $E_1(B)$, which is spanned by $(1,0)$.


问:
Find eigenvalues of $I - uv^T$
I want to show that the eigenvalues of $I - uv^T$, where $u,v \in \mathbb{R}^n$ are given by 1 with multiplicity $n-1$ and $1-v^Tu$ with multiplicity 1. I have tried setting up the eigenvalue equation $(I-uv^T)x = \lambda x$ which gives

$(1-\lambda)x - uv^Tx = 0$,

but I don't know how to proceed from here. Any help is appreciated!

答: Hint: Start by finding the eigenvalues of $uv^T$, a matrix of rank $1$.

Title: Find eigenvalues of $I - uv^T$


答:
First of all, the statement that you are trying to show is correct in the case that $v^Tu \neq 0$, but the case that $v^Tu = 0$ should be handled separately.

One approach to your problem in the case that $v^Tu \neq 0$ is to simply "guess" what the correct eigenvectors look like. In particular, I claim that if $x$ is orthogonal to $v$, then $x$ is an eigenvector. What is the associated eigenvalue? I also claim that if $x$ is parallel to $u$, then $x$ is an eigenvector. What is the associated eigenvalue?

Because $\operatorname{span}(u)$ and $\{v\}^\perp$ are complementary subspaces, we can then conclude that the operator $A = I - uv^T$ is diagonalizable and that all eigenvalues have been accounted for.

For the case that $v^Tu = 0$, it suffices to note that $[A - I]^2 = 0$. It follows that $A$ has $1$ as its only eigenvalue.

Title: Find eigenvalues of $I - uv^T$

Body:
I want to show that the eigenvalues of $I - uv^T$, where $u,v \in \mathbb{R}^n$ are given by 1 with multiplicity $n-1$ and $1-v^Tu$ with multiplicity 1. I have tried setting up the eigenvalue equation $(I-uv^T)x = \lambda x$ which gives

$(1-\lambda)x - uv^Tx = 0$,

but I don't know how to proceed from here. Any help is appreciated!


问:
Combining two or more PDFs into one
I have a right triangle, with two sides with lengths $x$ and $y$ respectively. Their lengths are randomly taken from two different pdfs. Simple diagram of triangle

The sides of the triangle have the following pdfs (can be any other pdf):

$f_{X}(x) = -\frac{6}{a^{3}}x^{2} + \frac{6}{a^{2}}x $

$f_{Y}(y) = \frac{1}{a} $

Where $a$ is a constant that defines the maxima of the range (give it any number that you want).
How can I get the pdf of the hypotenuse $h$.

$h = \sqrt{x^2+y^2}$

$f_{H}(h) = \ ? $

答:
You might be interested to read this section on Wikipedia. You can retrieve some of the formulas mentioned there without too much computation.

If $g$ is a borelian function,

$$\mathop{\mathsf{E}} g(h)=\mathop{\mathsf{E}} g\left(\sqrt{x^2+y^2}\right)=\int_0^a\int_0^a g\left(\sqrt{x^2+y^2}\right)f_X(x) f_Y(y) \mathop{dx} \mathop{dy}.$$

Now changing to polar coordinates:

$$\mathop{\mathsf{E}} g(h)=\int_0^a g(r) \left(\int_0^{2\pi}f_X(r\cos\varphi) f_Y(r\sin\varphi)\mathop{d\varphi} r\right)\mathop{dr}.$$

So by definition of the density function $f_H(h)=h\int_0^{2\pi}f_X(h\cos\varphi) f_Y(h\sin\varphi)\mathop{d\varphi}$. Computing this integral in your case is just a calculus exercise.

Title: Combining two or more PDFs into one


问:
Understanding some square root notation in LMFDB occuring in the definition of a p-adic field
I looked at the LMFDB page for the following $p$-adic field:

There I noted the notation $\mathbb{Q}_3(\sqrt{*})$ under "unramified subfield". Could you explain me what $*$ is supposed to be?

I think the square root makes sense as the extension is of degree $2$, so it is cyclic. Maybe the square root should just signify that the extension is cyclic but I am not sure.

答:
That notation only seems to be used in the LMFDB for quadratic extensions. I don't see anywhere in the LMFDB where the meaning of this notation is spelled out; I'll ask around and see if we can add a line somewhere explaining it. Anyway, to work out what it means, let's look at the structure of quadratic extensions of $\newcommand{\QQ}{\mathbb{Q}}\newcommand{\ZZ}{\mathbb{Z}}\QQ_p$.

As with any field not of characteristic $2$, quadratic extensions of $\mathbb{Q}_p$ correspond to non-identity elements of $\QQ_p^\times/(\QQ_p^\times)^2$. So, let's look at the structure of this group. For $p \neq 2$, we have
$$\QQ_p^\times \cong \underbrace{p^\ZZ \times \mu_{p-1} \times (1 + p\ZZ_p)}_{\text{written multiplicatively}} \cong \underbrace{\ZZ \times (\ZZ/(p-1)\ZZ) \times \ZZ_p}_{\text{written additively}},$$
where the isomorphism is given by writing each $x \in \QQ_p$ in the form $p^n \zeta (1 + py)$, where $\zeta^{p-1} = 1$. For $p = 2$, we have the slight variant
$$\QQ_2^\times \cong \underbrace{2^\ZZ \times \{\pm 1\} \times (1 + 4 \ZZ_2)}_{\text{written multiplicatively}} \cong \underbrace{\ZZ \times (\ZZ/2\ZZ) \times \ZZ_2}_{\text{written additively}}.$$
So we have
$$\QQ_p^\times/(\QQ_p^\times)^2 \cong \begin{cases}
(\ZZ/2\ZZ)^2 & \text{if } p \neq 2, \\
(\ZZ/2\ZZ)^3 & \text{if } p = 2.
\end{cases}$$

In particular, for $p \neq 2$, there are three quadratic extensions of $\QQ_p$:
$$\QQ_p(\sqrt{*}), \quad \QQ_p(\sqrt{p}), \quad \QQ_p(\sqrt{p*}),$$
where $*$ stands in for any non-square in $\ZZ_p^\times$ (they all produce the same quadratic extensions). The extension $\QQ_p(\sqrt{*})$ is unramified and the other two are ramified.

The case for $p = 2$ is similar except that we have seven extensions:
$$\QQ_2(\sqrt{*}), \QQ_2(\sqrt{2}), \QQ_2(\sqrt{-1}), \QQ_2(\sqrt{-2}), \QQ_2(\sqrt{2*}), \QQ_2(\sqrt{-*}), \QQ_2(\sqrt{-2*}),$$
where $*$ stands in for any non-square in $1 + 4\ZZ_2$. The first of these is unramified and the rest are ramified.


问:
There are eight coins in a row, all showing heads.
There are eight coins in a row, all showing heads. In each move, we can flip over two adjacent coins provided that they are both showing heads or both showing tails. How many different patterns of heads and tails can we obtain after a number of moves?

My solution: We can replace each head with the number '1', and each tail with the number '0', we will then end up with a binary sequence with a starting decimal value of 255. Each move will add or subtract $3*2^n$, $(6 \geq n \geq 0)$, thus there will be $255 / 3 + 1 = 86 $ solutions.

There is already a solution on StackExchange here: How many different patterns of heads and tails can we obtain after a number of moves?

Although I understand the accepted solution for the above link, I still don't understand why my solution is wrong, can anyone help continue my solution or give me a hint?

答:
I wrote a python script to count them by brute force.  It turns out that there are $70$ strings, and they are exactly the nonnegative multiples of $3$ less than $256$ with an even number of $1$ bits in their binary representation.

Before trying to prove this, I tested the hypothesis, that if their are $n$ coins, and $n$ is even, then the reachable positions are the nonnegative multiples of $3$ less than $2^n$ with an even number of $1$ bits, and if $n$ is odd, the then the reachable positions are the nonnegative numbers less than $2^n$ and $\equiv1\text{mod } 3$ with an odd number of $1$ bits.

This turned out to be true for $n=2,3,...,10$ but false for $n=11,12,13,14,15,16$. It appears likely then, that some ad hoc argument is required for the $n=8$ case, which will essentially come down to brute force.


问:
How to solve system equations that are composed of fractions
I'm trying to figure how to solve this particular system of equations.

$$
A = \frac{1}{x}, B = \frac{1}{y}
$$

$$
\frac{12}{x}-\frac{12}{y} = 7\\
   \frac{3}{x}+\frac{4}{y} = 0
$$

Body:
$12A - 12B = 7$         (1)

$12A + 16B = 0$         (2)

Use (1) - (2) : $B = -1/4$

then pass $B$ to (2) (or (1)),  you can get $12A - 16 \cdot 1/4 = 0$, so $A = 1/3$).


问:
Converse of little Desargues theorem
My book tells me the following:

Little Desargues Theorem: If two triangles are in perspective from a point $P$, and if two pairs of corresponding sides meet on a line $\ell$ through $P$, then the third pair of corresponding sides also meet on $\ell$.

As an exercise in the book, I'm asked to state and explain why the converse is true (supposedly, the converse should follow from the original statement itself).

My formulation of the converse is as follows: If two triangles are in perspective from a point $P$ and if one pair of corresponding sides meet on $\ell$, then the other two pairs of corresponding sides must meet on the same line $\ell$ through $P$.

Is my formulation correct? If so, I'm pretty confused as to why this follows from the original statement. Can someone please explain this exercise to me?

答:
Your formulation of the converse of the Desargues’ Little Theorem is wrong, because what you have done is nothing but expressing the same theorem in a different way. A converse of a given theorem is a proposition whose premise and conclusion are the conclusion and premise of the given theorem.

Here is the correct formulation.

If two triangles are perspective from a line $\ell$, which is usually called the perspective axis, and if two straight lines joining the two of the three pairs of corresponding vertices of these two triangle meet on $\ell$ at a point $P$, then the line joining the remaining pair of vertices also intersects $\ell$ at $P$.

By the way, point $P$ is called the perspective center of the two triangles in question.

All the propositions in projective geometry occur in dual pairs which have the property that, starting from either proposition of a pair, the other can be immediately inferred by interchanging the parts played by the words $\mathbf{line}$ and $\mathbf{point}$. In your case, the given theorem, which is a very special case of the Desargues' theorem, states that some triangles are perspective from a line because they are perspective from a point. Then its converse must state that some triangles are perspective from a point because they are perspective from a line.

Title: Converse of little Desargues theorem


问:
there are k edge-disjoint paths from a to b in this graph
I draw examples to convince myself that this actually seems to be true.
I have no idea how to prove it, maybe some kind of proof by contradiction is possible here, I mean assuming there are no k disjoint paths...

However, the given graph is directed. $a,b \in V$. In-degree and out-degree equal for any other vertices: $d^+(v)=d^-(v)$ if $v \in V $ \ $ ${a,b}.
Also we have $d^+(a)-d^-(a)=k>0$.

答:
I can think of two ways you can solve this, I will hint at both.

Add $k$ edges between $a$ and $b$ such that the new graph $G'$ is Eulerian. Then find your edge-disjoint paths in $G'$ that correspond to edge-disjoint paths in $G$ (What do you know about Euler cycles?)

Find a path $P$ from $a$ to $b$ in $G$, and remove the edges of $P$ from $G$ to obtain $G'$. What can you say about $G'$?


答:
Let
$$ 1-\frac{1}{\sqrt[n]{n}}=a_n. \tag{1} $$
Clearly $a_n\in(0,1)$. From (1), one has
$$ \frac{1}{\sqrt[n]{n}}=1-a_n. $$
So
$$ n=\frac1{(1-a_n)^n}=(1+a_n+a_n^2+\cdots)^n\ge(1+a_n)^n\ge\binom{n}{2}a_n^2 $$
from which one has
$$ 0<a_n\le\sqrt{\frac{2}{n-1}}. $$
Thus, for $\forall \epsilon>0$, letting $\sqrt{\frac{2}{n-1}}<\epsilon$ gives $n>\frac{2}{\epsilon^2}+1$. Define $N=\lfloor\frac{2}{\epsilon^2}+1\rfloor+1$ and then when $n\ge N$, one has $0<a_n<\epsilon$ or
$$ \left|\frac1{\sqrt[n]n}-1\right|<\epsilon,$$
namely
$$ \lim_{n\to\infty}\frac1{\sqrt[n]n}=1. $$

Title: Limit of $\frac{1}{\sqrt[n] n}$

Body:
I would like to show that the limit of the sequence $(\frac{1}{\sqrt[n] n})_{n=1}^{\infty}$ is equal to 1 using the definition of the limit of a sequence. (Another way to say this is that it converges to 1).

Definition: A sequence $(a_n)_{n=1}^{\infty}$ converges to a real number A iff for each $\epsilon > 0$ there is a positive integer N such that for all $n \geq N$ we have $|a_n - A| < \epsilon$.

Edit: Here is my work.

Let $\epsilon > 0$. There exists $N$ such that for $n \geq N$,

$$|\frac{1}{\sqrt[n] n}-1| = |\frac{n^{\frac{n-1}{n}}}{n} - 1| = |\frac{n^{\frac{n-1}{n}} - n}{n}| = \frac{n-n^{\frac{n-1}{n}}}{n}$$

Here is where I am stuck.


答:
The unramified subfield of $K/\Bbb{Q}_p$ is $\Bbb{Q}_p(\zeta_{p^f-1})$ where $p^f$ is the cardinality of $O_K/m_K$. It is also $\Bbb{Q}_p(a)$ for any $a$ with a monic $\Bbb{Z}_p[x]$ minimal polynomial $h$ such that $\Bbb{F}_{p^f} =\Bbb{F}_p[x]/(h)$.

At the bottom of your link they gave an example of $h \in \Bbb{Z}[x]$, because the minimal polynomial of $\zeta_{p^f-1}$ doesn't have  coefficients in $\Bbb{Q}$.

Title: Understanding some square root notation in LMFDB occuring in the definition of a p-adic field

Body:
I looked at the LMFDB page for the following $p$-adic field:

There I noted the notation $\mathbb{Q}_3(\sqrt{*})$ under "unramified subfield". Could you explain me what $*$ is supposed to be?

I think the square root makes sense as the extension is of degree $2$, so it is cyclic. Maybe the square root should just signify that the extension is cyclic but I am not sure.


问:
np hard and np complete simple question
I need a very short and simple description for my doubts about some concept in following:

all NP-Complete problems can be reducible to all problems in NP-Hard.

any np-hard problem can be reducible to one  NP-complete problem.

first is true and second is false.

I have three question:

what is the role of "all" words here?

why the first is true?

if we say "some np-hard problem can be reducible to one  NP-complete problem. for the second fact, now is it correct?

答:
I think better wording will be "any NP-complete problem can be reduced to any NP-hard problem" and "any np-hard problem can be reduced to an NP-complete problem".

The first is true, because all NP-complete problems are in NP, and NP-hard problems are exactly such that any problem from NP can be reduced to them.

The second is false, because NP-hard problem can be harder than NP. For example, Halting problem is NP-hard (any computable problem can be reduced to Halting problem), but it's not in NP, and thus can't be reduced to NP-complete problem.

Title: np hard and np complete simple question


问:
How to calculate the area of the given circle segmentation highlighted in green?
How to calculate the area of any irregular circle segmentation (highlighted in green)? Central angle is 90 degrees and the radiuses are 98/99 and 100/100 respectively.

答:
Area of half a circular segment of radius $r=\sqrt{c^2+a^2}=\sqrt{99^2+1}$, angle $\theta$ with $\tan\theta=c/a$, offset $a=1$, and half-chord length $c=99$ is the sector area $\frac{1}{2}r^2\theta$ minus the triangle's area $\frac{1}{2}ca$, that is, $$A=\frac{1}{2}(r^2\theta-ca)$$

Hence green shaded area is $\pi R^2/4-\frac{1}{2}(r^2\theta-ca)$ where $R=100$.

Answer: Approximately 255 sq.units.


问:
Find a point equidistant from two points and a line
Given a line: $x-\frac{2}{2} = y-\frac{1}{6} = z+\frac{2}{2}$ and two points $P_1(1,1,0)$ and $P_2(0,1,-1)$, identify the point $V$ which resides on the line and is equidistant from $P_1$, and $P_2$.

答:
The locus of all the point $P(x,y)$ such that $PP_1=PP_2$ is the plane
$$(x-1)^2+(y-1)^2+z^2=x^2+(y-1)^2+(z+1)^2\to x+z=0$$
Find the intersection between the line and the plane solving the system
$$\begin{cases}
\frac{x-2}{2} =\frac{y-1}{6}\\
\frac{x-2}{2} = \frac{z+2}{2}\\
x+z=0\\
\end{cases}
$$
Solution should be $V(0,-5,0)$


问:
An integration question which involves the unit one
When integrating a function, and the integration function is multiplied by 1, do I multiply by one or integrate the one?

I say this because, if the function is 1, then this integrates?

for example:

$$ \int^{2}_{-2} cos\frac{n\pi t}{2}1dt $$

as opposed to

2. $$ \int^{2}_{-2} 1dt $$

Why is it, that 1) I can only multiply it by 1 as opposed to integrating the cos function by t and 1 over t

答:
Typically, you can just multiply the one; there is never any harm in  that. However, you can make use of the one in integration by parts. Anything multiplied by one is itself, so you can always makes use of it like so:

With integration by parts, we can integrate the x and differentiate the ln(x), which is helpful because we can't integrate what we are already trying to find the integral of.
I know this wasn't an exact answer, but I hope this puts some insight into the one.


问:
Why is $x^4-3x^2+18$ irreducible over the $3$-adics?
Let $f = x^4 - 3x^2 + 18 \in \mathbb{Q}_3[x]$. Since there is an LMFDB page of an extension defined by this polynomial, I assume that $f$ is irreducible. Could you verify if this is true and also why that is (not) the case?

It is not an Eisenstein polynomial (as $3^2$ divides $18$, the constant coefficient of $f$). I also tried the substitution $y = x^2$ and asked myself if $g = y^2 - 3y + 18$ is irreducible. Its reduction mod $3$ is $y^2$ which is not irreducible, so I cannot say whether $g$ is irreducible or not.

答:
If it's reducible, it has a factor of degree $1$ or $2$.  I'll leave it up to you to
show there is no factor of degree $1$ (i.e. no root in $\mathbb Q_3$).  So suppose
$x^2 + \alpha x + \beta$ divides $x^4 - 3 x^2 + 18$.  The remainder on this division
is $$(-\alpha^3 + 2 \alpha \beta + 3 \alpha) x - \alpha^2 \beta + \beta^2 + 3 \beta + 18$$
If $\alpha = 0$ we'd have $\beta^2 + 3 \beta + 18 = 0$, which has no root in $\mathbb Q_3$,
so $\alpha \ne 0$ and $\beta = (\alpha^2-3)/2$.  Substituting in to the remainder,
this gives us $$-\frac{\alpha^4}{4} + \frac{3}{2} \alpha^2 + \frac{63}{4} = 0$$
which has no solution in $\mathbb Q_3$.

EDIT: Here's one way to see $\beta^2 + 3 \beta + 18=0$ has no solution in $\mathbb Q_3$. If $\beta$ has $3$-adic order $m$, then $\beta^2$ has order $2m$ and $3\beta$ has order $m+1$, while of course $18$ has order $2$.
So the only possibility is $m=1$.
Thus we have the $3$-adic expansion $\beta = \beta_1 3^{1} + \beta_2 3^{2} + \ldots$, with $\beta_i \in \{0,1,2\}$.  But these three possibilities make $\beta^2 + 3\beta + 18 =
2 \cdot 3^2 + \ldots, 1 \cdot 3^2 + \ldots, 2 \cdot 3^2 + \ldots$ respectively, and not $0$.


问:
How does $f^{(3)} (0) \geq 3\ $?

Let $f(x)$ be a real polynomial of degree $4.$ Suppose $f(-1) = 0,f(0) = 0, f(1) = 1$ and $f^{(1)} (0) = 0,$ where $f^{(k)} (a)$ is the value of $k^{\text {th}}$ derivative of $f(x)$ at $x = a.$ Which of the following statements are true?

$(1)$ There exists $a \in (-1,1)$ such that $f^{(3)} (a) \geq 3$

$(2)$ $f^{(3)} (a) \geq 3,$ for all $a \in (-1,1)$

$(3)$ $0 \lt f^{(3)} (0) \leq 2$

$(4)$ $f^{(3)} (0) \geq 3$

My attempt $:$ Since $f$ is a real polynomial of degree $4$ having $-1$ and $0$ as roots it follows that there exist $p,q,r \in \Bbb R$ with $p \neq 0$ such that $$f(x) = x(x+1)(px^2 + qx + r).$$ By the given hypothesis $f(1) = 1$ which yields $p + q+ r = \frac {1} {2}.$ Now by expanding $f(x)$ we get $$f(x) = p x^4 + q x^3 + (p + r) x^2 + qx  + r.$$ So $$f^{(1)} (x) = 4 p x^3 + 3 q x^2 + 2 (p + r) x + q.$$ Since $f^{(1)} (0) = 0$ it follows that $q = 0.$ Hence we have $p + r = \frac {1} {2}.$ Therefore $f^{(1)} (x) = 4px^3 + x.$ Thus $f^{(3)} (x) = 24 p x.$ So $f^{(3)} (0) = 0,$ which violates $(3)$ and $(4).$ Also if we take $p = \frac {1} {24}$ then for all $a \in (-1,1)$ we have $f^{(3)} (a) \lt 1,$ which violates $(1)$ and $(2)$ as well. So neither of the given options happens to be true. But the answer key to this question suggests that $(1)$ and $(4)$ are the only correct options. Am I doing something wrong here? Can anybody please check my solution?

Thanks for your time.

Source $:$ CSIR NTA NET DECEMBER $2019.$

答: The conditions $f(0)=0$ and $f'(0)=0$ tell us that $f(x)=mx^4+ax^3+bx^2$ for some $a,b\in\Bbb R$. Furthermore, $f(-1)=0$ implies $b-a=-m$ and $f(1)=1$ implies $a+b=1-m$. Thus the polynomial is $$f(x)=mx^4+\frac12x^3+\left(\frac12-m\right)x^2.$$

Title: How does $f^{(3)} (0) \geq 3\ $?


问:
Calculation of permutation in tensor of rank 4
I am trying to compute the symmetric part of a 4th order tensor $A_{ijkl}$

From a previous post (Symmetric Part of Product of 2 tank 2 tensors), I saw that I need to compute the permutations of $A_{ijkl}$ (4!=24 permutations).

I would like to know if anyone can give some hints on how to compute these permutations.

If $$A_{ij}=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 &1 \end{bmatrix}\quad\text{and}\quad A_{kl}=\begin{bmatrix} 1 &0 &0 \\ 0 &1 &0 \\ 0 & 0 &1 \end{bmatrix}$$

Then

$$A_{ijkl}=\begin{bmatrix}
1 &0 &0 \\
0 &1 &0 \\
0 & 0 &1
\end{bmatrix}
 \begin{bmatrix}
1 &0 &0 \\
0 &1 &0 \\
0 & 0 &1 \end{bmatrix} \\
= \begin{bmatrix}
1.& 0.& 0.& 0.& 1.& 0.& 0.& 0.& 1. \\
 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
1.& 0.& 0.& 0.& 1.& 0.& 0.& 0.& 1.\\
0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.& 0.\\
 1.& 0.& 0.& 0.& 1.& 0.& 0.& 0.& 1.\\
\end{bmatrix}
$$

How can I compute the remainder, e.g., $A_{ikjl},A_{jikl}, A_{lkij}, \ldots$

Kind Regards!

答:
Your $2$ dimensional array is constructed by taking $M_{(i,j),(k,l)} = A_{ijkl}$, where the tuples $(i,j)$ and $(k,l)$ are ordered lexicographically. At least, that is what I suspect.

Likewise, the corresponding array for $A_{ikjl}$ would be found by taking $M_{(i,k),(j,l)} = A_{ijkl}$. In particular, we have
$$
A_{ikjl} = \left[\begin{array}{ccc|ccc|ccc}
1 & 0 & 0 & 0&0&0&0&0&0\\
0&1&0&0&0&0&0&0&0\\
0&0&1&0&0&0&0&0&0\\
\hline
0&0&0&1&0&0&0&0&0\\
0&0&0&0&1&0&0&0&0\\
0&0&0&0&0&1&0&0&0\\
\hline
0&0&0&0&0&0&1&0&0\\
0&0&0&0&0&0&0&1&0\\
0&0&0&0&0&0&0&0&1
\end{array}\right].
$$
In the above, the "blocks" of each column correspond to $k=1,2,3$, and the blocks of each row correspond to $l = 1,2,3$.

For example: to get the $5,7$ entry in the above array, note that the $5$th tuple in $\{1,2,3\}^2$ is $(2,2)$, and the $7$th tuple is $(3,1)$. So, the number in the $5,7$ spot corresponds to $i=2,k=2,j=3,l=1$. That is, we have $A_{23} \cdot A_{21} = 0 \cdot 0 = 0$.

Note that by the symmetry of the $2 \times 2$ arrays $A_{ij}$ and $A_{kl}$, switching $i$ and $j$ and switching $k$ and $l$ do not change the resulting permuted tensor. For instance, we will have $A_{ikjl} = A_{jkil}$, and $A_{ikjl} = A_{iljk}$. Similarly, making the simultaneous switch $i \leftrightarrow k,j \leftrightarrow l$ does not change the resulting permuted tensor, so that $A_{ikjl} = A_{kilj}$.

I hope that helps.

An alternative approach: note that $A_{ij} = \sum_{i,j = 1}^3 \delta_{ij} e_i \otimes e_j = \sum_i e_i \otimes e_i$. Similarly, $A_{kl} = \sum_{k=1}^3 e_k \otimes e_k$. With that, we have
$$
A_{ijkl} = \sum_{i,k = 1}^3 (e_i \otimes e_i) \otimes (e_k \otimes e_k)
= \sum_{i,k = 1}^3 e_i \otimes e_i \otimes e_k \otimes e_k.
$$
Note that each permutation corresponds to a reordering of this product. For instance,
$$
A_{ikjl} = \sum_{i,k=1}^3 e_i \otimes e_k \otimes e_i \otimes e_k.
$$

Note that each permutation will yield either the original matrix from your post, the matrix $A_{ikjl}$ in this answer above, or the matrix
$$
A_{kjil} =
\left[\begin{array}{ccc|ccc|ccc}
1 & 0 & 0 & 0&0&0&0&0&0\\
0&0&0&1&0&0&0&0&0\\
0&0&0&0&0&0&1&0&0\\
\hline
0&1&0&0&0&0&0&0&0\\
0&0&0&0&1&0&0&0&0\\
0&0&0&0&0&0&0&1&0\\
\hline
0&0&1&0&0&0&0&0&0\\
0&0&0&0&0&1&0&0&0\\
0&0&0&0&0&0&0&0&1
\end{array}\right]
$$


问:
drawing open balls for the radar screen metric
What do the unit balls of this metric look like for radii 1/2, 1 and 3/2 please ?

Radar screen metric is $d(x,y):= \min(1, \|x - y\|_2)$, where the subscript $2$ after the last $\|$ indicates it's over $\mathbb{R}\times\mathbb{R}$.

Thanks

答:
HINT:

The metric $d$ is identical to the usual metric when $\|x-y\|\le 1$; what does that tell you about the $d$-balls of radii $\frac12$ and $1$?

Now let $x\in\Bbb R^2$ be arbitrary. The open $d$-ball of radius $\frac32$ centred at $x$ contains those points $y\in\Bbb R^2$ such that $d(x,y)<\frac32$. Can you find any point $y\in\Bbb R^2$ such that $d(x,y)$ is not less than $\frac32$?


问:
Limit $\lim_{n\to\infty} \frac{n!}{n+2^n}$
In an exercise I'm asked to evaluate the following limit:

$$\lim_{n\to\infty} \frac{n!}{n + 2^n}$$

I arrived at the conclusion that:

$$\frac{n!}{n + 2^n} \geq \frac{n!}{2^n}$$

So my plan is to prove that $\frac{n!}{2^n}$ diverges and as a consequence show that $\frac{n!}{n + 2^n}$ also diverges, But I'm not being able to do so.

How can I prove that $\lim n! / 2^n$ is infinity?

Body:
Hint:

$$\forall n>3, \frac{n!}{n+2^n} \geqslant \frac{n(n-1)2^{n-3}}{n+2^n} = \frac{n-1}{\frac{1}{2^{n-3}}+\frac 8n} > \frac{n-1}{1+8}.$$


问:
Choose the correct options from the following options given...

I know that $f$ is strictly monotone. Please give some hint from this.

答:
No, No, No, Yes: Consider
$$ f(x)=\max\{x,x^3\}+1.$$


问:
Prove that $\sup_{(x,y) \in D} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}} < \sup_{(x,y) \in D} (|x-y|^{1-\alpha}(|\log|y-x|| + C) )$
I am working on the following qual prep question.

Suppose $f:[0,1] \to \mathbb{R}$ is absolutely continuous and there exists $C>0$ such that
$$\int_0^{1}e^{|f'(x)|}dx \leq C$$
Let $D=\{(x,y) \in [0,1] \times [0,1] : x \neq y\}$. Prove that
$$\sup_{(x,y) \in D} \frac{|f(x)-f(y)|}{|x-y|^{\alpha}} < \sup_{(x,y) \in D} (|x-y|^{1-\alpha}(|\log|y-x|| + C) )$$

I have a hint that we can apply Jensen's inequality to
$$\frac{1}{x-y} \int_x^y|f'(t)|dt, \phi(t)=e^t$$
But I could not figure out how.

答:
Jensen's inequality tells you that $$\phi \left( \frac 1{y-x} \int_x^y g(t) \,dt \right) \le \frac 1{y-x} \int_x^y \phi(g(t)) \, dt$$ whenever $g$ is integrable on $[0,1]$ because $\phi$ is convex.  Since $f$ is absolutely continuous on $[0,1]$ its derivative is integrable and for any $x,y \in [0,1]$  the fundamental theorem of calculus gives you
$$f(x) - f(y) = \int_x^y f'(t) \, dt.$$
Now let $x,y \in [0,1]$ and assume without loss of generality that $x < y$.  Then
$$\frac{|f(x) - f(y)|}{|x-y|} \le \frac 1{y-x}\int_x^y |f'(t)| \, dt$$ so that
$$\exp\left(\frac{|f(x) - f(y)|}{|x-y|}\right) \le \exp\left(\frac 1{y-x}\int_x^y |f'(t)| \, dt\right) \le \frac 1{y-x} \int_x^y e^{|f'(t)|} \, dt \le \frac{C}{y-x} = \frac{C}{|x-y|}.$$
Now apply the logarithm to both sides to get
$$\frac{|f(x) - f(y)|}{|x-y|} \le \log|x-y| + \log C$$
and
$$\frac{|f(x) - f(y)|}{|x-y|^\alpha} \le |x-y|^{1-\alpha} \left( \log |x-y| + \log C \right).$$


问:
Find the vector equation of the line of points that satisfy the equation
```<code>x + 2y - 3z  = 0
3x + 2y + z  = 0
5x + 6y - 5z = 0
</code>```

I was thinking about solving this using Gaussian Elimination, but the result is going to be x = y = z = 0. So could anyone give me a hint how to go about this issue, please?

答: The determinant of your matrix vanishes. Thus its rank is less than $3$. By inspection, it is also greater than $1$, so it must be $2$. Thus, when you apply Gaussian elimination, you should end up with one row of zeros at the end, which allows you to choose $z=\lambda$ arbitrarily and then express $y$ and $x$ in terms of $\lambda$ using the non-zero rows. That gives you the equation of a line through the origin containing the solutions to the system of equations.</p>


问:
Doubt about why I can't treat this as a Bernoulli process
I know the title is not descriptive enough, but I don't know how else to say it. I don't know why I can't use the Binomial distribution to get the result I'm looking for. The teacher solved it long ago with a more straightforward method, but I tried today to solve the same problem using Binomial distribution and the result is not the same.

So, the problem is:

A total of $17$ balls are distributed in $20$ baskets, where every basket can hold as many balls as I want to, and every basket has the same probability of being filled with each ball. Each ball is indistinguishsable from another. What is the probability of every ball falling into the first basket?

The teacher solved it through combinatorics, using:

$$\dfrac1{\dbinom{20+17-1}{17}}$$

I wanted to solve it using Binomial distribution:

$$\dbinom{n}k p^k (1-p)^{n-k} = \dbinom{17}{17} \left(\dfrac1{20}\right)^{17} \left(\dfrac{19}{20} \right)^0$$

Knowing that there are $17$ throws, and I want $17$ successes. And as there are $20$ baskets, and only one is the successful one, the probability of each ball of going into that basket is $1/20$.

The problem is, it doesn't give the same result. And I don't know why, because I think I'm not using the binomial distribution wrong.

Thank you for your time!

答:
Your answer $\left(\frac{1}{20}\right)^{17}$ makes sense to me, given your description of the problem.

I recognize the pattern of the teacher's answer, and that is an answer to a different question. There are $\binom{20+17-1}{17}$ ways to distribute $17$ indistinguishable balls into 20 urns. If all such distributions were equally likely, then that would explain the teacher's answer. However as you describe the problem, not all such outcomes are equally likely. As you describe it, $$16,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$$ is $17$ times more likely to occur than $$17,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0$$ but your teacher's answer counts these outcomes as equally likely. Perhaps that was an error or perhaps that was somehow stated in the original wording.


问:
Help in writing a piecewise step function?
I have the following step function I am working with:

List of piecewise functions

Upon plotting these, I realize I have a step function, specifically a floor function. Basically, for every power of 2, I use the function $$n^3 - n^2$$

For instance, values between 2 and 4 (exclusive) should equal 4, as $ 2^3 - 2^2 = 4$.

Similarly, values from 4 to 8 (exclusive) should equal 18, as $ 3^3 - 3^2 = 18$.

And finally, values from 8 to 16 (exclusive) should equal 48, as $ 4^3 - 4^2 = 18$.

For a range of values, namely up to but excluding what $ 2^n $ actually is, should equal $n^3 - n^2$ as the output. I've included a Desmos graph as well: Desmos graph. I'm envisioning this to be a floor function but am just unsure of how to write it. Thanks in advance.

答:
I figured out the solution after playing around with logarithms in floor functions this morning! I believe it's

$$ f(x) = (\lfloor{lg (2x)}\rfloor)^3 -  (\lfloor{lg (2x)}\rfloor)^2$$ in which $ lg$ is the binary logarithm.


问:
What is the difference between linear and affine function
I am a bit confused. What is the difference between a linear and affine function? Any suggestions will be appreciated</p>

答:
Answer for any confused French reader.

In France/French, the distinction between linear and affine appears to be different from other countries.

An affine function in the French sense is a any function that may be written

$f(x) = a + bx$

where $a$ and $b$ are independent of $x$ (not necessarily Real). For example, $f(n) = a + bn$, where $n \in N$ is an affine function over the Natural number set $N$.

A linear function in the French sense is an affine function that passes through the origin, that is $a=0$ and $f(x)=bx$ for some number $b$ independent of $x$.

Reference: wiki/Fonction_affine


答: A factor $1$ can always be implied. $f(x)=1\cdot f(x)=1\cdot1\cdot f(x)=\cdots$. So if you tried to integrate these factors, you would get a nonsense.

Title: An integration question which involves the unit one

Body:
When integrating a function, and the integration function is multiplied by 1, do I multiply by one or integrate the one?

I say this because, if the function is 1, then this integrates?

for example:

$$ \int^{2}_{-2} cos\frac{n\pi t}{2}1dt $$

as opposed to

2. $$ \int^{2}_{-2} 1dt $$

Why is it, that 1) I can only multiply it by 1 as opposed to integrating the cos function by t and 1 over t


问:
Surface Integral with vector field question?
Let $V$ be a volume in $\mathbb{R}^3$ bounded by a simple closed piecewise-smooth surface $S$ with outward pointing normal vector $\mathbf{n}$. For which one of the following vector fields is the surface integral $\iint_S f\cdot\mathbf{n} \space dS$ equal to the volume of $V$?

A: $\mathbf{f}(\mathbf{r})=(1,1,1)$

B: $\mathbf{f}(\mathbf{r})=(2x,-y^2,2yz-z)$

C: $\mathbf{f}(\mathbf{r})=(z^2,y^2,1-2yz)$

D: $\mathbf{f}(\mathbf{r})=\frac{1}{2}(x,y,z)$

答:
Hint: For the surface integral for outward pointing normal vector to be equal to volume, $ div F$ (divergence of the vector field) should be positive constant $1$ independent of $x, y, z$.

$\frac{\partial (2x)}{\partial x} + \frac{\partial (-y^2)}{\partial y} + \frac{\partial (2yz-z)}{\partial z}$ is the one that gives $1$.

$\\\int_V div F dv$ becomes $\int_V 1 dv = V$


问:
If $\langle A,B \rangle=0$ for two symmetric matrices, what about $\langle A,B^{-1}\rangle $?
You are given two square real symmetric matrices $A$ and $B$. They are mutually orthogonal $A \perp B$, or in other words their frobenius inner product is zero, or in other words $\langle A ,\, B\rangle_F=0$.

My intuition says $A\perp B^{-1}$, i.e. $\langle A, B^{-1}\rangle_F=0$, but I don't really know how to approach proving such an expression. Can you prove or disprove such a claim?

答: Take $A = \begin{pmatrix} 2 & 0 \\ 0 & -1 \end{pmatrix}$ and $B = \begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}$. Then $\langle A, B\rangle = 0$ but $\langle A, B^{-1}\rangle = \frac{3}{2}$.


问:
Components of signed measure are greater than or equal to positive and negative variation of the signed measure.
Let $\mu = \mu_1 - \mu_2$, where $\mu_1$ and $\mu_2$ are measures. Show that $\mu_1 \geq \mu^+$ and $\mu_2 \geq \mu^-$.

答:
Consider $N,P$ a Hahn decomposition. Considering $\mu^+,\mu^-$ the positive and negative variations of $\mu$ we have by definition that for every set $E \in \mathbf{X}$
$$
\mu^+(E)=\mu(E \cap P)=\mu_1(E \cap P)-\mu_2(E \cap P) \leq \mu_1(E \cap P) \leq \mu_1(E)
$$
and
$$
\mu^-(E)=-\mu(E \cap N)=\mu_2(E \cap N)-\mu_1(E \cap N) \leq \mu_2(E \cap N) \leq \mu_2(E).
$$


问:
When is this norm defined ( continuous functions and entire functions)
Note : This question was asked 7 hours ago but closed so I thought about it again and asking it with my attempts . Hope , it on -site now.

The following question was asked  in a masters exam for which I am preparing and I was unable to solve it. So, I am asking for help here.

Both continuity and entire function implies that sup will be attained as set is compact. entire implies further that sup will be attained at the boundary. Also, $||f||_K$ =0 implies that f(z)=0.But By this I marked all options are correct but answer is

 D only

So, Please tell what mistake I am making .
I have done a course on functional analysis and followed Krieszig.

Can you please give outline of solution?

答: $\Vert f\Vert_K=0$ does not imply that $f=0$. First, it is possible that $K=\{z_0\}$, a single point. For example, if $K=\{0\}$ and $f(z)=z^2$, then $\sup_{z\in K}|f(z)|=0$, so $\Vert f\Vert_K=0$, but $f\neq 0$. This invalidates options A and B. If $K$ has non-empty interior, then it is still possible that a continuous function $f$ is zero on all of $K$, but nonzero off of $K$. For example, $K=\{z\in\mathbb C:|z|\leq 1\}$ and $$f(z)=\begin{cases} 0 & |z|\leq 1\\ |z|-1 & |z|>1\end{cases}.$$ This invalidates C. But if an entire function is zero on a nonempty open set (such as the interior of $K$), then it is zero on all of $\mathbb C$. Therefore, $D$ is correct.


问:
Why is this expectation calculation wrong?
Let

$E(X|A)=2+E(X)$

$E(X|B)=3+E(X)$

$E(X)=0.5E(X|A)+0.3E(X|B)$

Now to find $E(X^2)$ I did

$E(X^2|A)=4+E(X^2)$

$E(X^2|B)=9+E(X^2)$

$E(X^2)=0.5E(X^2|A)+0.3E(X^2|B)$

What did I do wrong here?

Edit.

For a recursive conditional expectation like this,

$E(X^2|A)=E[(2+X)^2]$

$E(X^2|B)=E[(3+X)^2]$

$E(X^2)=0.5[4+4E(X)+E(X^2)]+0.3[9+6E(X)+E(X^2)]$

where $E(X)=9.5$

答:
This is not an answer, as the question is not stated. I'm merely trying to show what information can be gleamed from the question. I'd delete this answer later.

Judging from your comments, the probability space is $\{ A, B, C \}$. Hence,

$\begin{align}
E[X] & = P(X=X(A) ) \times E[X|A] + P(X=X(B)) \times E[X|B] + P(X=X(C)) \times E[X|C] \\
& = P(X=X(A)) \times X(A) + P(X=X(B)) \times X(B) + P(X=X(C)) \times X(C) \\
\end{align}$

However, we do not know what $P(X=X(A))$ is.

From the first 3 equations, we get that

$$E[X] = 0.5 (2+E[X]) + 0.3 (3 + E[X]) \Rightarrow E[X] = 9.5, E[X|A] = 11.5, E[X|B]=12.5 $$


问:
Combining two or more PDFs into one
I have a right triangle, with two sides with lengths $x$ and $y$ respectively. Their lengths are randomly taken from two different pdfs. Simple diagram of triangle

The sides of the triangle have the following pdfs (can be any other pdf):

$f_{X}(x) = -\frac{6}{a^{3}}x^{2} + \frac{6}{a^{2}}x $

$f_{Y}(y) = \frac{1}{a} $

Where $a$ is a constant that defines the maxima of the range (give it any number that you want).
How can I get the pdf of the hypotenuse $h$.

$h = \sqrt{x^2+y^2}$

$f_{H}(h) = \ ? $

答: Continuing @md5's calculation, the support$$h\in[0,\,a\sqrt{2}],\,h\cos\varphi\in[0,\,a],\,h\sin\varphi\in[0,\,a]$$splits into $h\in[0,\,a],\,\varphi\in[0,\,\pi/2]$ and $h\in[a,\,a\sqrt{2}],\,\varphi\in[\arccos\tfrac{a}{h},\,\arcsin\tfrac{a}{h}]$. These respectively give $h$ the PDF$$\int_0^{\pi/2}\tfrac{6h^2}{a^3}\cos\varphi(1-\tfrac{h}{a}\cos\varphi)d\varphi=\tfrac{6h}{a^2}[\sin\varphi-\tfrac{h}{2a}(\varphi+\tfrac12\sin2\varphi)]_0^{\pi/2}=\tfrac{6h^2}{a^3}(1-\tfrac{h\pi}{4a})$$and$$\tfrac{6h^2}{a^3}[\sin\varphi-\tfrac{h}{2a}(\varphi+\tfrac12\sin2\varphi)]_{\pi/2-\arcsin\tfrac{a}{h}}^{\arcsin\tfrac{a}{h}}=\tfrac{6h^2}{a^3}\left(\tfrac{a}{h}-\sqrt{1-\tfrac{a^2}{h^2}}-\tfrac{h}{2a}(2\arcsin\tfrac{a}{h}-\tfrac{\pi}{2})\right).$$The case $a=1$ provides an instructive sanity check:$$\begin{align}\int_0^1(6x^2-\tfrac32x^3\pi)dx&=2-\tfrac38\pi,\\\int_1^\sqrt{2}(6x-6x^2\sqrt{1-1/x^2}-6x^3\arcsin\tfrac1x+\tfrac32x^3\pi)dx\\=\tfrac32[3x^2-\tfrac12\sqrt{1-1/x^2}x(5x^2-2)-\tfrac32x^4\arcsin\tfrac1x+\tfrac38\pi x^4]_1^\sqrt{2}&=\tfrac38\pi-1.\end{align}$$(I'll admit I cheated with that last part.)

Title: Combining two or more PDFs into one


问:
Definite integral $\int_{0}^{1}( \frac{1}{\sqrt{x\ln x}}+\sqrt{\frac{\ln x}{x}}) dx$
$$\int_{0}^{1}\left( \frac{1}{\sqrt{x\ln x}}+\sqrt{\frac{\ln x}{x}}\right) dx$$

Here is what I tried and failed

$$\int_{0}^{1}\frac{1}{\sqrt{x} }\left(\frac{1}{\sqrt{\ln x}}+\sqrt{\ln x}\right)dx$$

$$\int_{0}^{1} \frac{1}{\sqrt{x}}\left( \frac{1 + \ln x}{\sqrt{\ln x}}\right)dx$$

After that, I got stuck.

答:
The first thing to do here is to realize that we are taking square roots of negative numbers, so we need to proceed carefully, so as not to be led astray by ostensible identities like $1/\sqrt z=\sqrt{1/z}$, which is correct for positive reals but problematic for negative reals. The best thing to do (in my experience) is to reexpress things so as to get rid of the square root symbol altogether.

One way to do this is to let $x=e^{-u^2}$ with $0\le u\lt\infty$. We see that

$${1\over\sqrt{x\ln x}}+\sqrt{\ln x\over x}={1\over\sqrt{-u^2e^{-u^2}}}+\sqrt{-u^2\over e^{-u^2}}={u\sqrt{-1}\over e^{-u^2/2}}+{1\over ue^{-u^2}\sqrt{-1}}=ie^{u^2/2}\left(u-{1\over u} \right)$$

(Note the crucial minus sign in the final step!)

With $x=e^{-u^2}\implies dx=-2ue^{-u^2}du$, we find

$$\int_0^1\left({1\over\sqrt{x\ln x}}+\sqrt{\ln x\over x}\right)dx=-2i\int_\infty^0e^{u^2/2}\left(u-{1\over u} \right)ue^{-u^2}du=2i\int_0^\infty(u^2-1)e^{-u^2/2}du$$

Finally, integration by parts (with $u=u$ and $dv=ue^{-u^2/2}du$, so that $v=-e^{-u^2/2}$) tells us

$$\int_0^\infty u^2e^{-u^2/2}du=-ue^{-u^2/2}\big|_0^\infty+\int_0^\infty e^{-u^2/2}du=(0-0)+\int_0^\infty e^{-u^2/2}du=\int_0^\infty e^{-u^2/2}du$$

so that

$$\int_0^\infty(u^2-1)e^{-u^2/2}du=\int_0^\infty u^2e^{-u^2/2}du-\int_0^\infty e^{-u^2/2}du=\int_0^\infty e^{-u^2/2}du-\int_0^\infty e^{-u^2/2}du=0$$

Remark: I've spelled things out in more steps than may be necessary. The key steps are the appearance of the negative sign in the first display and the separation of $u^2$ from $-1$ for integration by parts to get to the cancelation of integrals. A close reading of Deepesh Meena's answer should reveal that the same two steps are in effect taken there as well, without the change of variable, but I have to admit I haven't read that answer closely enough to see how it avoids making a $1/\sqrt{z}=\sqrt{1/z}$ type error when it manipulates square roots of the things that are negative.


问:
Differentiate $\log_{10}x$
My attempt:

$\eqalign{
  & \log_{10}x = {{\ln x} \over {\ln 10}}  \cr
  & u = \ln x  \cr
  & v = \ln 10  \cr
  & {{du} \over {dx}} = {1 \over x}  \cr
  & {{dv} \over {dx}} = 0  \cr
  & {v^2} = {(\ln10)^2}  \cr
  & {{dy} \over {dx}} = {{\left( {{{\ln 10} \over x}} \right)} \over {2\ln 10}} = {{\ln10} \over x} \times {1 \over {2\ln 10}} = {1 \over {2x}} \cr} $

The right answer is: ${{dy} \over {dx}} = {1 \over {x\ln 10}}$ , where did I go wrong?

Thanks!

答:
if someone doesn't know how to write $\log_{10}x$ in simpler form then there is a direct differential formula:
$$\frac{d}{dx}\log_ax=\frac 1x\log_{e}{a}$$
so here $$\frac{d}{dx}\log_{10}x=\frac 1x\log_e10$$
using identity :
$$\log_nm=\frac{1}{\log_mn}$$

we can write:
$$\frac 1x\log_e10\implies\frac 1{x\log_{10}e}$$

Title: Differentiate $\log_{10}x$


问:
How many ways are there in the case that there is ONLY ONE adjacent couple?
"Three couple will be arranged in a row.How many ways are there in the case that there is ONLY ONE couple which are adjacent."

For example , let $(a_1,a_2),(b_1,b_2),(c_1,c_2)$ be the couples, so they cannot be arranged in the form of $a_1,c_1,c_2,b_2,b_1,a_2$.  There can be only one couple adjacent such as $b_1,a_1,a_2,c_1,b_2,c_2$ is acceptable.

I found this question in my textbook.I tried that $C(3,1).5!.2!-[C(3,2).6.2!.2!.2!+C(3,3).2!.2!.2!.3!]$
but my answer is wrong and I am confused. Can you help me?

答:
There are three ways to select the adjacent couple.  We then have five objects to permute, the couple and the other four people.  The five objects can be arranged in $5!$ ways.  The members of the couple can be arranged internally in $2!$ ways.  Hence, there are
$$\binom{3}{1}5!2!$$
arrangements in which a couple is together.

However, we have counted each case in which two couples in which the members of a couple are adjacent twice, once for each way of designating one of the couples as the couple that is together.  We do not want to count such cases at all, so we must subtract twice the number of cases in which there are two couples in which the members of a couple are adjacent.

There are $\binom{3}{2}$ ways to select two couples in which the members of a  couple are adjacent.  That gives us four objects to arrange, the two couples and the two individuals from the third couple.  The objects can be arranged in $4!$ ways.  Each adjacent couple can be arranged internally in $2!$ ways.  Hence, there are
$$\binom{3}{2}4!2!2!$$
such cases.

That gives us a running total of
$$\binom{3}{1}5!2! - 2\binom{3}{2}4!2!2!$$
but we have not yet accounted for those arrangements in which there are three couples in which the members of the couple are adjacent to each other.

We counted the cases in which there are three couples are together three times when we counted arrangements in which the members of a couple are together, once for each way of designating one of the three couples as the couple with adjacent members.  Observe that we counted cases in which there are three couples with adjacent members three times when we counted cases in which there are two couples with adjacent members, once for each of the $\binom{3}{2}$ ways of designating two of the three couples as the couples which are together.  Since we subtracted twice the number of arrangements in which there are two couples with adjacent members, we subtracted cases in which there are three couples with adjacent members six times.  We don't want to count such arrangements at all, so we need to add three times the number of arrangements in which there are three couples with adjacent members.

If there are three couples with adjacent members, there are $3!$ ways to arrange the couples and $2!$ ways to arrange each couple internally.  Hence, there are
$$3!2!2!2!$$
such arrangements.

Consequently, by the generalized Inclusion-Exclusion Principle, there are
$$\binom{3}{1}5!2! - 2\binom{3}{2}4!2!2! + 3\binom{3}{3}3!2!2!2!$$
arrangements of three couples in which there is only one adjacent couple.


问:
Prove that quotient space is compact space
I am studying general topology. And this is problem I faced.

Problem : Let ~ on $S^1 \times [-1, 1]$ : $(a, b)$~$(c,d)$ iff either $(a, b)=(c,d)$ or $b=d=1$ or $b=d=-1$. Show that $S^1 \times [-1, 1]$ is homeomorphic to $S^2$.

My attempt : Let $g$ : $S^1 \times [-1, 1] \rightarrow S^2$ : $g(cos(\theta), sin(\theta), y)=(\sqrt{1-y^2}cos(\theta),\sqrt{1-y^2}sin(\theta), y) $. Then $g$ is continuous and surjective. which induces a continuous bijection  $f$ : $S^1 \times [-1, 1] /$~ $\rightarrow S^2$.

$S^2$ is obviously Hausdorff. So, if $S^1 \times [-1, 1] /$~ is a compact space, $f$ is a homoemorphism.

But I have trouble proving compactness. How can I show it?

答: $S^1$ is compact and $[-1,1]$ is compact, therefore $S^1 \times [-1,1]$ is also compact. Now $S^2$ is Hausdorff being a subspace of a Hausdorff space, hence the map $f: S^1\times [-1,1]\setminus\sim\to S^2$ is continuous bijection and it is a closed map. It is thus a quotient map and since it is an injective quotient map, it will induces a homeomorphism $\tilde{f}: S^1\times [-1,1] \setminus \sim \to S^2$.


问:
$ \bigcap\limits_{k \in \mathbb{N}} \bigcup\limits_{n \in \mathbb{N}} A_{n, k}$ prove the properties of injection.
For $n, k \in \mathbb{N}$ we describe $A_{n, k}$ as a set of all functions $f \in \mathbb{N}^\mathbb{N}$, such that for for every function $f(n) = k$. For each of sets determined below (A and B) decide which descriptions (from 1 to 6) can be applied and prove that.

$$ (A) \bigcup\limits_{n \in \mathbb{N}} \bigcap\limits_{k \in \mathbb{N}} A_{n, k}$$

$$ (B) \bigcap\limits_{k \in \mathbb{N}} \bigcup\limits_{n \in \mathbb{N}} A_{n, k}$$

Descriptions:

$\emptyset$

$\mathbb{N}^\mathbb{N}$

all functions $f \in \mathbb{N}^\mathbb{N}$ that are injective

all functions $f \in \mathbb{N}^\mathbb{N}$ that are surjective

all functions $f \in \mathbb{N}^\mathbb{N}$ that are constant

none of those

FIrst of all, as I understand, we can interpret the sets as:

$$ (A) \bigcup\limits_{n \in \mathbb{N}} \bigcap\limits_{k \in \mathbb{N}} A_{n, k} = \exists n \forall k$$

$$ (B) \bigcap\limits_{k \in \mathbb{N}} \bigcup\limits_{n \in \mathbb{N}} A_{n, k} = \forall k \exists n$$

Descriptions with my thinking and answers:

$\emptyset$ $\implies $ I think that it is $A$ because there are no $n$ (arguments of function) that would return all $y$ (possible values of function)

$\mathbb{N}^\mathbb{N}$ $\implies $ none

all functions $f \in \mathbb{N}^\mathbb{N}$ that are injective $\implies $ none

all functions $f \in \mathbb{N}^\mathbb{N}$ that are surjective $\implies $ I am sure that it is $B$ because if for all $y$ (possible values of function) exists some $n$ (argument of function) that returns that value, then that is a surjection

all functions $f \in \mathbb{N}^\mathbb{N}$ that are constant $\implies $ none

none of those  $\implies$ X

The problem is that I need to prove that my answers are correct (if they are). I don't know how to do that but it seems pretty repetitive (I suppose it has to be done "both ways") I would be very grateful if  somebody could prove one of them so that I can try to do similarly all the others by myself.

答:
For A suppose there is some $f \in \bigcup_{n \in \mathbb{N}} \bigcap_{k \in \mathbb{N}} A_{n,k}$. Then, by definition, there is some $n_0\in \mathbb{N}$ such that $f \in A_{n_0,k}$ for all $k \in \mathbb{N}$. Hence by definition of $A_{n_0,k}$,
$$
f(n_0)=k, \ \forall k \in \mathbb{N}.
$$
Which is a contradiction. Hence $\bigcup_{n \in \mathbb{N}} \bigcap_{k \in \mathbb{N}} A_{n,k}$ is empty.

For B suppose there is some $f \in \bigcap_{k \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} A_{n,k}$. Then by definition for all $k \in \mathbb{N}$ there is some $n_k \in \mathbb{N}$ such that $f\in A_{n_k,k}$...(Try to show that f is surjective and then you have $\bigcap_{k \in \mathbb{N}} \bigcup_{n \in \mathbb{N}} A_{n,k} \subseteq \{f \in \mathbb{N}^{\mathbb{N}} \mid f \ is \ surjective\}$). For the other inclusion suppose $f$ is surjective. Hence by definition for all $k \in \mathbb{N}$ there is some $n_k$ such that $f(n_k)=k$...


问:
To prove that $2^{3n}+2^n +1$ is not a perfect square.
Question: Prove that $2^{3n} + 2^n + 1$ cannot be a perfect square for any natural $n$.

I attempted this question and failed in two different ways.

1) I considered a polynomial $p(x) = x^3+ x + 1 - m^2$ (for some natural $m$) and factorized the polynomial assuming $2^n$ is a root. I, then, tried substituting some numbers to get a contradiction in divisibility (since $x - 2^n$ is a factor). Alas, all of that was in vain.

2) I wrote $2^{3n} + 2^n + 1 = m^2$ and thus $(m-1)(m+1) = 2^n(2^{2n} + 1)$. So one can conclude that if $xy = 2^{2n} + 1$ such that $(x,y) = 1$, then one of the $m-1$ or $m+1$ must be $2^{n-1}x$ and the other must be $2y$ (this follows from the fact that $m$ is odd and $(m-1,m+1) = 2$). I am stuck at this point.

I would appreciate a hint in either direction. I am hoping that 1) will work out. It will teach me a new proofing (err... proving) technique.

Thanks!

答: Hint: Consider 2 cases. If $n$ is odd, take $\pmod{3}$. If $n$ is even, bound between 2 squares.


问:
if $f' = f - g$ and $g' = g - f$ and $f(0) = g(0) = 1$ find$ f$ and $g$
If $f' = f - g$ and $g' = g - f$ and $f(0) = g(0) = 1$ find$ f$ and $g$.

To me, this seems to be a differential equation question.

$$f''=f'-g'=f'-(g-f) \rightarrow f''-f'-f=-g$$

Is this the right idea?

答:
Here is a suggested approach.  I left the details for you to work out.

$f'+g'=0$, so $f=-g+constant$.

Solve for $constant$ from $f(0)=g(0)=1$ and then solve $f'=f-g=2f-constant$ for $f$,

and then $g$ follows.


问:
In complex analysis $f(\partial K) = \partial f(K)$ always for for compact $K$ and holomorphic $f$?
For a compact set $K \subset \Bbb{C}$ under the standard metric, does it hold that if $f$ is a non-constant holomorphic function on $K$, that $f(\partial K) = \partial f(K)$? If not, does $f(\partial K) \supset \partial f(K)$ hold? (here $\partial S:=\overline{S}\setminus \textrm{int}(S)$ denotes the boundary of $S$)

答:
Hints:

To prove $\partial f(K)\subseteq f(\partial K)$ use that $f(K)$ is closed and apply the invariance of domain theorem.

To see that the converse inclusion fails, consider $f(z)=z^2$ and the intersection of the closed unit disc with the closed upper half plane.


问:
If $\alpha = \beta$, why can't the entropy-regularized Wasserstein distance equal $0$?
In optimal transportation theory, the optimal re-allocation of probability distribution $\alpha$'s mass to another distribution $\beta$ is solved by minimizing the Wasserstein distance with respect to the transport plan.

$$W (\alpha, \beta) = \min_{\pi\in \Pi(\alpha\beta)} \int c(x,y) \mathrm{d}\pi(x,y) $$

Alternatively, the relative entropy-regularized Wasserstein distance, also called Sinkhorn distance, can be used:

$$W_\epsilon (\alpha, \beta) = \min_{\pi\in \Pi(\alpha\beta)} \int c(x,y) \mathrm{d}\pi(x,y) + \epsilon H(\pi \| \alpha \otimes \beta)$$
where $\epsilon$ is the regularization parameter, and relative entropy is$$H(\pi \| \alpha \otimes \beta) = \int \ln \left(\frac{\mathrm{d}\pi (x,y)}{\mathrm{d}\alpha(x)  \mathrm{d}\beta(y)  } \right) \mathrm{d}\pi (x,y) $$
Aude Genevay said that if you try the extreme case where both the source and target distributions are identical, $\alpha = \beta$, then we would expect the entropy-regularized Wasserstein distance (Sinkhorn distance) to equal $0$ since there is nothing to move, however it is incapable of doing so. Because of this she proposes the Sinkhorn divergence instead, a normalization which does equal $0$ if $\alpha = \beta$:

$$\bar{W}_\epsilon (\alpha, \beta) = W_\epsilon (\alpha, \beta) - \frac{1}{2} [W_\epsilon (\alpha, \alpha) + W_\epsilon (\beta, \beta) ]$$
In other words, $\bar{W}_\epsilon (\alpha, \alpha) = 0$.

Questions

Why (or for what levels of regularization) can't the Sinkhorn
distance, shown earlier, achieve $0$?

Does standard optimal transport, which uses the unregularized
Wasserstein distance, also suffer from this incapability (even
though I know that the Wasserstein distance by itself, without OT,
will achieve $0$)?

and why, mathematically, does the Sinkhorn divergence?

答:
To show why this is the case I think the way to go about things is to consider the first term in

$$W_\epsilon (\alpha, \beta) = \min_{\pi\in \Pi(\alpha,\beta)} \int c(x,y) \mathrm{d}\pi(x,y) + \epsilon H(\pi \| \alpha \otimes \beta)$$

and note it is only zero when $\pi(A,B)=\pi(A\cap B,A\cap B)$ (i.e when $\pi$ puts all the mass onto the diagonal $x=y$), otherwise it is positive. Then you are just left to check what happens  to $\epsilon H(\pi \| \alpha \otimes \beta)$ when $\pi$ is of such a form.

When you say "even though I know that the Wasserstein distance by itself, without OT, will achieve 0 " this doesn't really make sense the Wasserstein distance is an OT problem itself. To see why $W_2(\alpha,\alpha)=0$ take as a $\gamma=f_\#\alpha$ where $f:\mathbb{R}^d\to \text{diag}(\mathbb{R}^d)$ is defined as $f(x)=(x,x)$, and define as a candidate $\pi\in\Pi(\alpha,\beta)$ (not necessarily optimal) $\pi(A)=\gamma(A \cap \text{diag}(\mathbb{R}^d))$. Note $\text{diag}(\mathbb{R}^d)=\{(x,x)\in\mathbb{R}^{2d}~:~x\in\mathbb{R}^d\}$

Are you asking how $\tilde{W}_\epsilon(\alpha,\alpha)=W_\epsilon(\alpha,\alpha)-\frac{1}{2}(W_\epsilon(\alpha,\alpha)+W_\epsilon(\alpha,\alpha))=0$?

Edit : for 1) and $c(x,y)=\|x-y\|^2$ proceed like this :

Note that since $H(\pi|\gamma)=\int h(\frac{d\pi}{d\gamma})d\gamma$ for $h(x)=x\log x$ which is strictly convex. Hence using Jensen $H(\pi|\gamma)\geq 0$ with equality IFF $\frac{d\pi}{d\gamma}$ is constant, i.e. $\gamma=\pi$.

So we are forced to consider (where we set $M_i$ equal to the i-th moment of $\alpha$)
$$ \frac{1}{2}\int \|x-y\|^2 d\alpha(x)d\alpha (y) = M_2-\int x\cdot y d\alpha(x)d\alpha(y) \geq M_2 -M_1^2 >0,$$

where in the last line we used Jensens inequality for the strictly convex function $x\mapsto x^2$.


答:
We have
$$
\begin{cases}
f' = f- g\\
g' = g - f \\
\end{cases}
$$

If we sum, we obtain:
$$
f' + g' = (f+g)' = 0
$$
this means that $$f(x) = k - g(x), \qquad \text{for some $k \in \mathbb R$}$$
Now,
$$
1 = f(0) = k - g(0) = k - 1 \implies k = 2
$$

So the relationship is $f(x) = 2 - g(x)$.

Can you continue from here?

Title: if $f' = f - g$ and $g' = g - f$ and $f(0) = g(0) = 1$ find$ f$ and $g$

Body:
If $f' = f - g$ and $g' = g - f$ and $f(0) = g(0) = 1$ find$ f$ and $g$.

To me, this seems to be a differential equation question.

$$f''=f'-g'=f'-(g-f) \rightarrow f''-f'-f=-g$$

Is this the right idea?


问:
Basic question regarding order of dx, dy and dz
So, the question I am asking is so fundamental (and might be very obvious) that I could not find any previous  question similar to this particular question, which is this:
given a function $f(x,y,z)$ , we need to calculate its triple integral, so is it always necessary  that we start off by $$\int\int\int f(x,y,z)dxdydz  \tag{1}$$ and then change order of integral to calculate
(suppose:$\int\int\int f(x,y,z)dzdxdy [2]$) or we can directly start from $[2]$ given we know its appropriate upper limits and lower limits.

P.S.: I am not asking how to change order of $dx$, $dy$, $dz$ just asking if we can start by any arbitrary permutation of them.

答:
As long as you are careful to change the limits of integration appropriately, yes.

The easiest case is where the limits of integration are constants.  $\int_a^b\int_c^d\int_e^f F(x,y,z)dxdydz$ can be written in all 3!= 6 permutations and give the same result.  Its even simpler if F(x,y,z)= f(x)g(y)h(z): $\int_a^b\int_c^d\int_e^f F(x,y,z)dxdydz= \left(\int_a^b f(x)dx\right)\left(\int_c^d g(y)dy\right)\left(\int_e^f h(z)dz\right)$ and of course those three quantities can be permuted at will since multiplication of numbers is commutative.

If the limits of integration contain variables, it a little harder.  Suppose the integral is $\int_0^1\int_0^1\int_0^{\sqrt{1-y}} F(x,y,z)dxdydz= \int_0^1\int_0^{1-x^2}\int_0^1 F(x,y,z)dzdydx$.


问:
Is the power function strictly convex?
Let $p\in(1,+\infty)$ and

\begin{equation*}
    \begin{aligned}
    \phi_p:  \ & \mathbb{K} \longrightarrow [0,+\infty)\\
    & s \ \longmapsto \
    \phi_p(s):=|s|^p,
    \end{aligned}
\end{equation*}
$(\mathbb{K}=\mathbb{R}$ o $\mathbb{K}=\mathbb{C})$.
For all $u, v\in[0,+\infty)$, $u\neq v$, and $\lambda\in(0,1)$,
\begin{equation*}
    |\lambda u+(1-\lambda)v|^p<\lambda |u|^p+(1-\lambda)|v|^p.
\end{equation*}
I have proved for $\mathbb{K}=\mathbb{R}$ using Hölder's inequality but I cannot prove for
$\mathbb{K}=\mathbb{C}$.

答:
Hölder's inequality also holds for complex numbers so your proof probably works for that case as well.

There's also a simple reduction from $\mathbb{C}$ to $\mathbb{R}$ as follows.

Suppose that $u = a+ib$ and $v = c+id$. The inequality becomes
\begin{equation}
\left|\lambda a + (1-\lambda)c + i[\lambda b + (1-\lambda)d]\right|^p \le \lambda\left|a + ib\right|^p + (1-\lambda)\left|c + id\right|^p,
\end{equation}
or equivalently
\begin{equation}
\left((\lambda a + (1-\lambda)c)^2 + (\lambda b + (1-\lambda)d)^2\right)^{p/2} \le \lambda\left(a^2 + b^2\right)^{p/2} + (1-\lambda)\left(c^2 + d^2\right)^{p/2}.
\end{equation}
Notice that $(\lambda a + (1-\lambda)c)^2 \le \lambda a^2 + (1-\lambda)c^2$ and $(\lambda b + (1-\lambda)d)^2 \le \lambda b^2 + (1-\lambda)d^2$, it suffices to prove that
\begin{equation}
(\lambda x + (1-\lambda)y)^{q} \le \lambda x^q + (1-\lambda)y^q,
\end{equation}
where $x=a^2 + b^2, y=c^2+d^2$ and $q=p/2$.

Title: Is the power function strictly convex?


问:
Definition of Index Set?
The most widely used Index Sets are $\mathbb{N}$, $\mathbb{R}$. Both of which are totally-ordered-set.

Is it important for Index-Sets to be totally ordered set?

答:
Do not confuse index set with an indexed set. The sets N and R are simple sets, you can define a total-order upon their elements, but even if you do, they are not yet index sets. Indexed sets are build when you define a function between a totally-ordered set (as the domain) and an arbitrary set (in the codomain), then, the totally-ordered set you used becomes the index set, because you used it to index the arbitrary set. Does that makes sense?

To your question: Is it important for Index-Sets to be totally ordered set?

You can define a function between a set with another, but there would be no ordered define, so, consider the set {1, 2}, it is no different than the set {2,1}, the only way you can tell the element 1 goes "first", and then the element 2 is because we have defined a total-order in the set {1,2}.

So, I would say, no, is not important that index-sets are totally ordered sets, it simply makes the indexation meaningful.

Title: Definition of Index Set?


问:
Find a polynomial function given local extrema
I'm struggling to find a polynomial function with given local extrema.

For instance, consider the problem:

Find a function with a local maximum at $(0,0)$ and a local minimum at $(1,−5)$.

I start with $f(0)=0$, $f(1)=−5$, $f′(0)=0$ (or is it undefined?), $f′(1)=0$ and then try to solve $f(x)=ax^3+bx^2+cx+d$ with the above values.

So far I have $f(1)=-5$ gives $a+b+c+d=-5$ and $f'(1)=0$ gives $3a+2b+c=0$, but I'm unsure how to go about working with $f(0)$ and $f'(0)$.

Is it $f(0)=0$ so $a+b+c=d=0$ so $d=0$ and $f'(0)=a+b=c=0$ so $c=0$? This seems incorrect, but I'm not sure else I can do here.

答:
How to proceed

as you mentioned $f(0)=0$, $f'(0)=0$, $f(1)=5$, $f'(1)=0$, because those are local minimas and maximas and therefore, there derivates are equal to $0$.

Second step now is to use $f(x)=a.x^3 + b.x^2 + c.x +d$ and use the values we previously talked about in this function

Replace for each equation the x by its value. You will end up having $d=c=0$

The rest will keep it as a challenge for you.


答:
$$
\begin {cases}
f'=f-g \\
g'=g-f
\end{cases}
$$
Sum both differential equations:
$$(f+g)'=0$$
$$\implies f+g=k$$
Then:
$$f'=f-g =f-(k-f)=2f-k$$
Note that $f'(0)=0 \implies k=2$
$$f'=2(f-1)$$
$$\int \dfrac {d(f-1)}{f-1}=2 \int dx$$
That's easy to solve.


问:
Evaluate the price of asset in log Black and Scholes model
Context:

I'm given Black and Scholes model.

Problem:

For $T > 1$ the asset returns $$Y_T = \ln(S_T)\cdot 1_{S_T>1} $$
where $S_t$ is solution for Black and Scholes model:
$$S_t = S_0 \cdot exp[(\mu - \frac{\sigma^2}{2})t +\sigma W_t]$$
$W_t$ is Wiener process and $1_{S_T>1} = 1$ if $S_T>1$ and $1_{S_T>1} = 0$ otherwise.

It is given that today $t$ is equal to $1$.

Evaluate the current price $Y_1$.

Can you give some ideas?

答:
If you choose the money-market instrument as the numeraire, you get

$$Y_t = e^{-r(T-t)}E\left[Y_T \vert \mathcal{F}_t\right]$$
where the expectation is taken under the risk-neutral measure. Under the risk neutral measure $S_t$ follows the SDE
$$dS_t = rS_tdt + \sigma S_t dW_t$$
where $W$ is Brownian Motion. This is predicated of course on the Black-Scholes model.

Note that you can write $S_T$ as follows.
$$S_T = S_t e^{\left(r-\frac{1}{2}\sigma^2\right)(T-t) + \sigma(W_T - W_t)}$$

$$E\left[Y_T \vert \mathcal{F}_t\right] = \left(\log{S_t} + \left(r-\frac{1}{2}\sigma^2\right)(T-t)\right)E[\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t] + E[\sigma(W_T - W_t)\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t]$$

Let's calculate the first conditional expectation. Define $d(t, S_t) := \log{S_t} + \left(r-\frac{1}{2}\sigma^2\right)(T-t)$. Since $W_T - W_t$ is independent of $\mathcal{F}_t$,

$$E[\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t] = P\{d(t, S_t) + \sigma(W_T-W_t)\vert\mathcal{F}_t\} = \Phi\left(\frac{d(t, S_t)}{\sigma\sqrt{T-t}}\right)$$
where $\Phi$ is the distribution function of standard normal r.v.

The second conditional expectation is calculated similarly.Replace $W_T - W_t$ by $\sqrt{T-t}X$ where X is standard normal independent of $\mathcal{F_t}$. This is purely for notational convenience. You can then write
$$E[\sigma(W_T - W_t)\mathbb{1}_{\log{S_T} > 0} \vert \mathcal{F}_t] = E[\sigma\sqrt{T-t}X\mathbb{1}_{d(t, S_t) + \sigma\sqrt{T-t}X > 0} \vert \mathcal{F}_t] = \sigma\sqrt{T-t}E[X\mathbb{1}_{X > -\frac{d(t, S_t)}{\sigma\sqrt{T-t}}} \vert \mathcal{F}_t]$$

We will now focus on
$$E[X\mathbb{1}_{X > -\frac{d(t, S_t)}{\sigma\sqrt{T-t}}} \vert \mathcal{F}_t]$$

We are taking the conditional expectation of something that is a mixture of r.v.s that are either measurable wrt to the conditioning sigma algebra or independent of it. We can then treat the measurable ones as if they were constants and compute the conditional expectation of the independent r.v.s as if it is an unconditional expectation.

For a standard normal r.v. $X$ define the function $g$ as
$$g(a) := E[X\mathbb{1}_{X>a}] = \frac{1}{\sqrt{2\pi}}e^{-\frac{a^2}{2}}$$

We can then write
$$E[X\mathbb{1}_{X > -\frac{d(t, S_t)}{\sigma\sqrt{T-t}}} \vert \mathcal{F}_t] = g\left(-\frac{d(t,S_t)}{\sigma\sqrt{T-t}}\right)$$

Putting everything together

$$Y_1 = e^{-r(T-1)}d(S_1)\Phi\left(\frac{d(1, S_1)}{\sigma\sqrt{T-1}}\right) + \sigma\sqrt{T-1}g\left(-\frac{d(1,S_1)}{\sigma\sqrt{T-1}}\right)$$


问:
Finding inverse modulo
I'm trying to find "the smallest positive multiple of 100" that leaves remainder 9 when divided by 19.

Here is what I have done before I got stuck:

$x ≡ 9 \mod 19$

$\gcd(9,19) = \gcd(19,9)\\
          19 = 9*2 + 1\\
          9 = 1*9 + 0$

Bezout's Theorm

$\gcd (19,9) = m*19 + n*9\\
      1 = 19 - 9*2\\
        = 1*19 -2*9$

THUS, x = -2 mod 19 = 17 mod 19 = 17

BUT! 17 is not a multiple of 100, so from here, what do I do to find "the smallest positive multiple of 100" that leaves remainder 9 when divided by 19?

Thanks!

答:
We have $100 N \equiv 9 \pmod{19}$, and want to find the smallest value of $N$.

This is equivalent to $5 N \equiv 9 \pmod{19}$.

This is equivalent to $ N \equiv 36 \equiv 17 \pmod{19}$, where we multiplied by $4 = 5^{-1} \pmod{19}$.

Hence, the answer is 1700.

Title: Finding inverse modulo


问:
Homogenization, what it is in inequalities and how to utilize it to its fullest.
I have just learnt about homogenization, however I don't really understand it and I can't really utilize it (https://artofproblemsolving.com/wiki/index.php/Homogenization this is where I learnt about it). Could you please explain to me how to use it optimally? Moreover I found a question on IMO 2020, which I believe can be solved using homogenization. It goes as follows:

Let the real numbers $a, b, c, d$ be such that $a\ge b\ge c \ge d >0$ and $a+b+c+d=1$. Prove that:

$(a+2b+3c+4d)a^ab^bc^cd^d<1$

I attempted to solve it by saying that RHS is equal to $(a+b+c+d)^a$, but then I realized that this doesn't help and it doesn't homogenize the inequality either. Then I tried using AM-GM on LHS and getting something which can be homogenized, however I did not succeed there either. Could you please explain to me how to use homogenization optimally and how this question could be solved using it?

答:
clean solution without any expansion

As others have noted, it is sufficient to show $$(a+b+c+d)^3 \ge (a^2+b^2+c^2+d^2)(a+2b+3c+4d)$$
write it as $$\dfrac{(a+b+c+d)^2}{a^2+b^2+c^2+d^2} \ge \dfrac{a+2b+3c+4d}{a+b+c+d}$$
subtracting 1,
$$\dfrac{2\sum_{sym} ab}{\sum a^2}\ge \dfrac{b+2c+3d}{a+b+c+d}$$
equivalent to
$$ \dfrac{2\sum_{sym} ab}{b+2c+3d}\ge \dfrac{a^2+b^2+c^2+d^2}{a+b+c+d}$$
which is true bcoz
$$ \dfrac{2\sum_{sym} ab}{b+2c+3d}\ge a \ge \dfrac{a^2+b^2+c^2+d^2}{a+b+c+d}$$


问:
Why is this map surjective??
I have the following Lie group homomorphism:

$\phi: Sp(1) \rightarrow SO(3)$ given by $\phi(q)x=qxq^{-1}$, where $x \in \mathbb{R}^3$ and $Sp(1)$ is the set of quaternions with module $1$.

I want to prove it is surjective, i.e., every matrix in $SO(3)$ comes by $\phi$ from an element $q \in Sp(1)$.

答: The way to do this is to see that your map $\phi(q)x=qxq^{-1}$ is actually a linear isometry and orientation preserving. The quaternion $qxq^{-1}$ has real part $0$ for any quaternion $q$ with modulus $1$. Now the point is the map $\phi$ sends $q$ and $-q$ to the same rotation of $\mathbb{R^3}$. The following document has the formula for a rotation matrix associated to a unit quaternion, which I will not attempt to derive here....Rotation Matrix (incidentally that document says the expression for this matrix in $SO(3)$ is due to Euler...)

Title: Why is this map surjective??


问:
Sum of multiples of 4 fibonacci numbers
I'm trying to prove this identity
$$\sum_{i=1}^n F_{4i} = F^2_{2n+1} -1$$
So, I try to use PIM to prove.

Base case: $n=2$

$$\sum_{i=1}^2 F_{4i} =  3 +21=24=F_5^2 -1$$

Inductive step: suppose this sum is true for $k \in \mathbb{N}$
\begin{equation*}
            \sum_{i=1}^k F_{4i} = F^2_{2k+1} -1
        \end{equation*}
We want to show
\begin{equation*}
        \sum_{i=1}^{k+1} F_{4i} = F^2_{2k+3} -1
    \end{equation*}
Then
\begin{equation*}
        \sum_{i=1}^{k+1} F_{4i} =  \sum_{i=1}^k F_{4i} + F_{4k + 4} = F^2_{2k+1} -1 + F_{4k + 4}
    \end{equation*}
I'm stuck here

答:
Almost there.

To finish,
you need to show that

$F^2_{2k+1} -1 + F_{4k + 4}
=F^2_{2k+3} -1
$
or
$F_{4k + 4}
=F^2_{2k+3}-F^2_{2k+1}
$.

Now use
d'Ocagne's identity
$F_{2n}
=F_{n+1}^2-F_{n-1}^2
$
with
$2n+2$ for $n$
and you are done.

Title: Sum of multiples of 4 fibonacci numbers


问:
Prove that $\frac AB \leq \frac{A+C}{B+D} \leq \frac CD$
Assume $\frac{A}{B} \le \frac{C}{D}$     and $A,B,C,D   \in  \mathbb{R}^+$

Prove that $\frac{A}{B} \le \frac{A+C}{B+D} \le\frac{C}{D}$

(this is a homework exercise I have to do, so please help me :(

答:
In general, when you are asked to prove $m<n$ usually it is easier to start with working out $m-n$. Assuming that $A$,$B$,$C$,$D$ are positive:

$$\frac{A}{B}-\frac{A+C}{B+D}=\frac{AB+AD-AB-BC}{B(B+D)}=\frac{AD-BC}{B(B+D)} = \frac{-BD(\frac{C}{D}-\frac{A}{B})}{B(B+D)}\le0$$

$$\frac{C}{D}-\frac{A+C}{B+D}=\frac{BC+CD-AD-CD}{D(B+D)}=\frac{BC-AD}{D(B+D)}= \frac{BD(\frac{C}{D}-\frac{A}{B})}{B(B+D)}\ge 0$$


问:
understanding Skorokhod's theorem and convergences
I have started with master course in probability and confused with Skorokhod's theorem.

Let us start with some sequence $X_{n}$ of, say, positive random variables, taking values from the same probability space. Assume, that $X_{n}$ converges in distribution to random variable $X$. Next, let us consider some random positive sequence $Y_{n}$, defined on the same probability space as $X_{n}$ and assume that there exists random $n_{0}$ such that for all $n > n_{0}$ we have almost surely
$$
Y_{n} \leq X_{n}.
$$

Next, since we have a weak convergence for $X_{n}$ and the real line is separable space, then we  can construct a sequence $\tilde{X}_{n}$, which has the same distributions as $X_{n}$ and converges a.s. to some $\tilde{X}$ with the same distribution as $X$.

Questions I am faced with:

Does this make sense to write
$$
\limsup_{n} Y_{n} \leq X
$$
if all the sequence is defined on the same probability space?

Also, is the following

$$
\limsup_{n} Y_{n} \leq \tilde{X}
$$

and
$$
E[\limsup_{n} Y_{n}] \leq E[\tilde{X}]
$$

correct?

答:
Your conjecture is not true.The counterexmaple is in following:

By taking $X_n$ to be a sequence of iid random variable whose law is $\mathcal{U}([0,1])$. $Y_n$ is chosen to be equal to $X_n$, that is $Y_n:= X_n$.

So on one hand you see that $$X_n \xrightarrow{ \text{weakly}} \mathcal{U}[0,1]$$
While $$\limsup Y_n = 1 \quad a.e$$