问:
probability of guessing a k-digit number sequentially from n trials
I recently encountered the following problem.
Assume that we need to guess a $k$-digit number. We guess it digit by digit and there are $n$ guesses in total. What is the probability of success?

Say, $k=5$, and $n=25$. We can spend, e.g., 9 trials guessing the first digit, then guess the second digit from the 3rd trial etc. But we cannot make more than 25 guesses in total.

I believe we should proceed as follows:

guessed from 5 trials: $p=1/10^5$; -- we cannot make less than 5 trials when guessing step by step

guessed from 6 trials: $p=1/10^4\cdot(1-1/10)\cdot 1/9\cdot 5$

guessed from 7 trials: $p=1/10^3\cdot(1-1/10)^2\cdot 1/9^2\cdot \binom{5}{2}+1/10^4\cdot(1-1/10)\cdot(1-1/9)\cdot 1/8\cdot 5$, etc.

but I'm failing to generalize this scheme. And there is an additional problem that we have to set an upper bound to the number of trials at some point...

UPDATE (09.02): As @JMoravitz suggested, one can write an iterative scheme for computing the probability of guessing exactly $\kappa$ digits from $\eta$ trials, $f(\kappa,\eta)$. I believe that this scheme can be written as
$$f(\kappa,\eta)=\frac{1}{10}\sum_{i=1}^{10}f(\kappa−1,\eta−i),$$
where we use the fact that the probability of guessing the digit from $1\le \kappa\le 10$ guesses is always the same:
$$p(i=1)=\frac1{10},\quad p(i=2)=\frac{9}{10}\frac{1}{9}=\frac{1}{10},...$$
We also need to add boundary conditions $$\begin{cases}f(\kappa,\eta)=0,&\eta<\kappa\mbox{ (we cannot guess $\kappa$ digits from less than $\eta$ trials)},\\
f(1,\eta)=\frac{1}{10},&0<\eta\le 10\mbox{ (probabilities of guessing the first digit from $\eta$ trials)},\\
f(1,\eta)=0,&\eta=0\mbox{ or }\eta>10.\end{cases}$$
The final probability is to be found as the sum $$p^*=\sum_{i=k}^n f(k,i)=\sum_{i=5}^{25} f(5,i).$$
This scheme looks pretty neat and can be easily computed numerically.

However, I wonder if there is any chance to solve this problem analytically? Perhaps there exists a different approach?...

答:
Let $\ G_i\ $ be the number of guesses taken to determine the $\ i^\text{th}\ $ digit. Then $\ G_1,G_2,\dots\ $ are independent discrete random variables, all uniformly distributed over the integers $\ 1,2,\dots,10\ $, and the total number of guesses taken to determine every one of $\ k\ $ digits is $\ \displaystyle T_k=\sum_{i=1}^kG_i\ $.  The probability mass function $\ p_{T_k}\ $of $\ T_k\ $ is therefore the $\ k-$fold convolution of the uniform distribution over the integers $\ 1,2,\dots,10\ $:
\begin{align}
p_{T_j}(h)&=\sum_{i=1}^{10}\frac{p_{T_{j-1}}(h-i)}{10}\\
&=\sum_{i=\max(1,h-10(j-1))}^{\min(10,h+1-j)}\frac{p_{T_{j-1}}(h-i)}{10}\ ,
\end{align}
because $\ p_{T_{j-1}}(s)=0\ $ for $\ s<j-1\ $ or $\ s> 10(j-1)\ $.

If you're allowed $\ n\ $ guesses to to get all $\ k\ $ digits correct, then your probability of success is
$$
\sum_{i=1}^nT_k(i)=\sum_{i=k}^nT_k(i)\ .
$$
One convenient way of obtaining $\ p_{T_k}\ $ is to use generating functions.  The generating function $\ g_0(x)\ $ of the uniform distribution over the integers $\ 1,2,\dots,10\ $ is given by
$$
g_0(x)=\frac{1}{10}\sum_{i=1}^{10}x^i\ ,
$$
and the generating function of the convolution of two probability mass functions is the product of their generating functions.  The generating function $\ g_k(x)\ $ of $\ p_{T_k}\ $ is therefore given by
$$
g_k(x)=g_0(x)^k=\frac{1}{10^k}\left(\sum_{i=1}^{10}x^i\right)^k\ .
$$
The generating function of $\ p_{T_5}\ $ is
\begin{align}
\frac{1}{10^5}\big(x^{50}&+5x^{49}+15x^{48}+35x^{47}+70x^{46}+126x^{45}\\
&+210x^{44}+330x^{43}+495x^{42}+715x^{41}+996x^{40}\\
&+1340x^{39}+1745x^{38}+2205x^{37}+2710x^{36}+3246x^{35}\\
&+3795x^{34}+4335x^{33}+4840x^{32}+5280x^{31}+5631x^{30}\\
&+5875x^{29}+6000x^{28}+6000x^{27}+5875x^{26}+5631x^{25}\\
&+5280x^{24}+4840x^{23}+4335x^{22}+3795x^{21}+3246x^{20}\\
&+2710x^{19}+2205x^{18}+1745x^{17}+1340x^{16}+996x^{15}\\
&+715x^{14}+495x^{13}+330x^{12}+210x^{11}+126x^{10}\\
&+70x^9+35x^8+15x^7+5x^6+x^5\big)\ .
\end{align}
Therefore, the probability of determining $5$ digits with $25$ or fewer guesses is
\begin{align}
\frac{1}{10^5}\big(1&+5+15+35+70+126+210+330+495+715\\
&+996+1340+1745+2205+2710+3246\\
&+3795+4335+4840+5280+5631\big)\\
&=\frac{38125}{10^5}\ .
\end{align}


问:
properties of real differentiable function
problem

I have tried to investigate f according to given criteria but dont seem to go nowhere,

supposedly i have could show that according to given criteria f must be either strictly positive and growing or strictly negative and decreasing

but i do not know how to go further in proving the given statement

thanks in advance for your help

答:
Note that $$
\left(f(x)e^{-cx}\right)'=f'(x)e^{-cx}-cf(x)e^{-cx}\geq 0.
$$
Which implies that
$$
f(x)\geq e^{cx} \frac{f(x_0)}{e^{cx_0}}=f(x_0) e^{c(x-x_0)}
$$
for all $x> x_0$. Applying continuity at $a$ gives you the desired. You can prove the other inequality similarly.

ContentLicense: CC BY-SA 4.0


问:
Question about the proof of $e$ irrationality
I found an article with the following elementary proof of $e$ irrationality
https://people.ohio.edu/diao/papers/Irrationality%20of%20e.pdf.
But I am having trouble understanding one step.

It states that, from:

\begin{equation}
 \frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}
\end{equation}

follows:

\begin{equation}
 q_n \geq q_{n+1}
\end{equation}

I can not understand why this is so. Can anyone explain in more detail?

答:
The proof states that $p_n$ and $q_n$ are relatively prime.
So, if you rewrite $$\frac{p_{n+1}}{q_{n+1}} = \frac{n p_{n} - q_{n}}{q_n}$$
as $$p_{n+1}q_{n} = (np_{n}-q_{n})q_{n+1}$$ you can see that $p_{n+1}$ has to divide $np_{n}-q_{n}$ in words $np_{n}-q_{n} \ge p_{n+1}$

Now just plug in this estimation to the first equation to the what you need.
By the way, nice proof, thanks for bringing it up.


问:
The tangent bundle of the 2-sphere $TS^2$ is an embedded submanifold of $\mathbb{R^6}$
Define the tangent bundle of $S^2$ as:
$$TS^2 = \{(u,v)\ |\ u,v \in \mathbb{R}^3, |u| = 1, \left = 0\}$$
I want to show that this is an embedded sub-manifold of $\mathbb{R}^6$.

My intuition leads me to say this will be a 4-manifold, but the proof that it is an embedded sub-manifold is leaving me stumped.

Edit: I noticed here: Tangent bundle $TM$ is a submanifold of $\mathbb{R}^{2n}$

How do I extend this to also give embedding?

Body: $$f((u_1,u_2,u_3),(v_1,v_2,v_3)) = (u_1,u_2,u_3,v_1,v_2,v_3)$$


问:
If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$

If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$

If we take $t=\tan \theta$ then $x=2\theta$ and for $y$  ,I am getting two values as follows ,

if we apply identity $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$ then we get $y=2\theta$

and if we take $\tan \theta $ common from $\dfrac{1-\tan^2 \theta}{2\tan\theta}$  and simplify then we get $y= \dfrac{\pi}{4} + \theta$ ?

how we are getting two values of $y$ ?

thankyou

答: I'm not sure where you're getting $y=\theta+\pi/4$ but it is definitely incorrect as $$\cot(\theta+\pi/4)=\frac{1-\tan\theta\tan\pi/4}{\tan\theta+\tan\pi/4}=\frac{1-\tan\theta}{1+\tan\theta}$$ is not the same as $$\frac{1-t^2}{2t}=\frac{1-\tan^2\theta}{2\tan\theta}.$$


答:
The Complete solution to this problem is to be found here

 Substitute t by tan x, then use double angle formulae. Full solution HERE or HERE

As per your doubt of getting two answers is concerned, it definitely has some error, and it would be better if you elaborate. But I suppose you have done some error in evaluating the inverse functions with their stringent rules......

Body:
If $\sin x=\frac{2 t}{1+t^{2}}$ and $\cot y=\frac{1-t^{2}}{2 t}$, then the value of $\frac{d^{2} x}{d y^{2}}$

If we take $t=\tan \theta$ then $x=2\theta$ and for $y$  ,I am getting two values as follows ,

if we apply identity $\tan 2 A=\frac{2 \tan A}{1-\tan ^{2} A}$ then we get $y=2\theta$

and if we take $\tan \theta $ common from $\dfrac{1-\tan^2 \theta}{2\tan\theta}$  and simplify then we get $y= \dfrac{\pi}{4} + \theta$ ?

how we are getting two values of $y$ ?

thankyou


问:
How can one prove that a function $f(x,t)$ cannot be expressed by the form $X(x)\cdot T(t)$?
I was reading about ODE variable separation solving and the book says that assuming that a function can be expressed as the product of two single-variable functions loses generality, which I understand. I cannot, however, prove that it does. For example: the function $\sqrt{x+t}$ cannot be expressed as $X(x)\cdot T(t)$. I tried proving it by contradiction and got to the result $X_1(x)\cdot T_1(t) = (x+it)\cdot(x-it)$ but I'm not sure if this is sufficient. That is, what guarantees that $(x+it)\cdot(x-it)$ cannot be rearranged in some way so that it only has the variable $x$ in one of the factors and only $t$ in the other?

答:
If $f(x,t) = \sqrt{x+t} = X(x) \cdot Y(t)$ for all $x, t \ge 0$ then

$$X(x)\cdot Y(0) = \sqrt{x} \text{ and } X(0) \cdot Y(t) = \sqrt t$$

But $X(0) \cdot Y(0) = \sqrt{0+0} = 0$ and therefore the contradiction $X(0) = 0$ or $Y(0) = 0$.


问:
$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$
I tried to solve this inequality:
$$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}+\frac{d}{1-d}+\frac{e}{1-e}\ge\frac{5}{4}$$
with
$$a+b+c+d+e=1$$
I am stuck at this. I don't want the full solution, a hint would be enough.</p>

ContentLicense: CC BY-SA 3.0

Body:
Hint:

You have $1-a=b+c+d+e$ and
$a=1-(b+c+d+e)$

$\dfrac{1-(b+c+d+e)}{b+c+d+e}=\dfrac{1}{b+c+d+e}-1$ and $b+c+d+e=1-a$

Using AM $\ge$ HM

$(\dfrac{1}{1-a}+\dfrac{1}{1-b}+\dfrac{1}{1-c}+\dfrac{1}{1-d}+\dfrac{1}{1-e}) \ge \dfrac{5^2}{1+1+1+1+1-(a+b+c+d+e)} $


问:
Every plane in $\mathbb{R}^3$ is a smooth surface
This is from Pressley's Differential Geometry. Let $a$ be a point on the plane and let $p$ and $q$ be two unit vectors parallel to the plane and orthogonal to each other. If $\mathbf{r}$ is any point on the plane then $\mathbf{r-a}$ is parallel to the plane, and so $$\mathbf{r}-\mathbf{a}=u\mathbf{p}+v\mathbf{q}$$ for scalars $u,v$. So now we have the surface patch $\sigma(u,v) = \mathbf{a} + u\mathbf{p}+v\mathbf{q}$. Now the book states that its inverse is $\sigma^{-1}(\mathbf r) = ((\mathbf r-\mathbf a)\cdot\mathbf p,(\mathbf r- \mathbf a)\mathbf q)$.

Now my problem is that I have no idea how the inverse was found. Could someone explain it to me.

答:
We know that
$$ \newcommand{\v}\mathbf \v r-\v a = u \v p + v \v q$$
and we want to solve for $u$ and $v$. Of course we can rearrange and get
$$ \newcommand{\v}\mathbf  u \v p =\v r-\v a - v \v q \tag{1}$$
How can we  get rid of the $\v p$? Well recall the definition of $\v p, \v q$ being orthogonal and unit length,
$$ \v p \cdot \v p = 1 = \v q \cdot \v q, \\ \v p \cdot \v q = 0.$$
So if we take the dot product of $(1)$ with $\v p$, we are left with
$$ u = \v r\cdot\v p-\v a \cdot\v p  - v \v q \cdot\v p = (\v r - \v a) \cdot \v p$$
which is precisely the first component of the inverse given to you. Similarly with the second.


问:
Given some of the eigenvalues and corresponding eigenvectors reconstruct the entire matrix.

Given a symmetric matrix $A$ that has eigenvalues $4, 3, 2,$ and $2$ and the eigenvectors belonging to the eigenvalues $4$ and $3$. Provide a procedure
to reconstruct the entire matrix.

Since the matrix is symmetric then it can be diagonalized. Let $Av = 4v, Aw = 3w$. Since the eigenvalue $2$ is repeated we want to find two linearly independent eigenvectors corresponding to the eigenvalue $2$.

Since the matrix is symmetric then the eigenvectors are orthogonal. If $z, h$ are eigenvectors corresponding to $2$ then $z^Tv = z^Tw = h^Tv = h^Tw = 0$. Therefore, both $z,h$ should be orthogonal to a two dimensional vector space $S$ constructed from the basis vectors $v,w$.

How can $z,h$ be found?

答:
I will assume that the $v,w$ given are unit-vectors. Because $A$ is symmetric, they must be orthogonal.

If you want to find $z$ and $h$, then it suffices to find an orthonormal basis for the nullspace of the $2 \times 4$ matrix whose rows are $v^T, w^T$. However, there's a way to reconstruct $A$ that does not require the computation of $z$ and $h$.

Verify that the matrix $B = 2vv^T + ww^T$ is symmetric with eigenvalues $2,1,0,0$. $v$ is an eigenvector of $B$ associated with the eigenvalue $2$ and $w$ is an eigenvector of $B$ associated with the eigenvalue $1$.

Using this, conclude that $A = B + 2I$ (where $I$ denotes the identity matrix) has all the properties required of $A$.


问:
Examples of principal ideal local Artinian rings
I want to find some examples of principal ideal local Artinian rings.
That is, a commutative ring $R$ with identity with a unique maximal ideal, and this ideal is generated by a nilpotent element. (There is not any idempotent element in a local ring except $0$ and $1$, so under the Artinian condition, the maximal ideal cannot be generated by an idempotent one, and hence must be generated by a nilpotent one).

I have found three examples:

(1) Quotient ring $\mathbf{Z}/ p^d \mathbf{Z}$ with $d \ge 2$ and prime $p$;

(2) Quotient ring $K [x]/ \langle x^2 \rangle$ with a field $K$;

(3) Localization $(\mathbf{Z}/ p^d q \mathbf{Z})_q$ of the quotient ring $\mathbf{Z}/ p^d q \mathbf{Z}$ on $q$ with $d \ge 2$ and distinct primes $p$ and $q$.

But I realized that (3) is isomorphic to (1).

For some application, I need more examples (finite ring is more useful for me). But I cannot figure out any one else.

Many many thanks!

答:
Well, most generally, if you take any principal maximal ideal $M$ of a commutative ring $R$, and select a nonzero integer $n > 1$, then $R/M^n$ is going to be an example.

Considering that all local principal ideal rings look like that (a ring whose ideals are precisely the powers of its maximal ideal) it's unlikely you'll get a much broader class of examples.

The trick, I suppose, is to be able to choose so that $M$ does interesting things. For example, if $M^2=M$ you only get a field.  But if $R$ is a domain, then you're guaranteed that $M^{i+1}\subsetneq M^i$.

As it happens the DaRT query only yields two things, both of which you've covered already.  But you could take any of the results of the PIDs that aren't fields query and do the construction I'm describing.


问:
Condition of concavity for a function
I have a continuous function $f:\mathbb{R}^n\to\mathbb{R}$ such that $\forall x,y\in\mathbb{R}^n$ I have that $f(x)\ge\cfrac{1}{2}f(x-y)+\cfrac{1}{2}f(x+y)$.

How can I prove that $\forall x,y\in\mathbb{R}^n$ and $\forall t\in[0,1]$ I have that $f(tx+(1-t)y)\ge tf(x)+(1-t)f(y)$ ?

答:
Firstly, $f(x)\ge\frac{1}{2}f(x-y)+\frac{1}{2}f(x+y)$ is no different from $f(\frac{x+y}{2})\ge\frac{1}{2}f(x)+\frac{1}{2}f(y)$.

Secondly, Midpoint-Convex and Continuous Implies Convex

Thirdly, A Lebesgue measurable function on an interval C is concave if and only if it is midpoint concave: https://en.wikipedia.org/wiki/Concave_function

Title: Condition of concavity for a function


问:
$\widetilde{F}=\bigoplus\limits_{x\in X}F_{\{x\}}$ is not isomorphic to $F$
I want to find a sheaf $F$ and a topological space $X$ such that $\widetilde{F}=\bigoplus\limits_{x\in X}F_{\{x\}}$ is not isomophic to $F$.

Where $F_{\{x\}}$ is the skyscraper sheaf, this is
\begin{align}
F_{\{x\}}(U)=\begin{cases}F_x \quad\text{if}\quad\ U\ni x,\\0 \quad\text{else}. \end{cases}
\end{align}

Note that $\widetilde{F}_y=F_y$ for all $y\in X$.

Any hit would be appreciated.

答:
As user10354138 suggested if we consider $F$ the constant sheaf $\mathbb{Z}/2\mathbb{Z}$ on the indiscrete two-point space $X=\{x,y\}$ we have the following:

$\widetilde{F}(X)=F_{\{x\}}(X)\oplus F_{\{y\}}(X)=F_x\oplus F_y=\mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}.$

While, $F(X)=\{s:X\rightarrow \mathbb{Z}/2\mathbb{Z}:\text{s is locally constant}\}=\mathbb{Z}/2\mathbb{Z}.$

Is this correct?


问:
Integral of $\sin(x)^{\frac{4}{3}}$
How does one solve the integral $\int_{0}^{2\pi}(\sin(x))^{4/3}\ \mathrm{d}x$, or more generally $\int_{0}^{2\pi}(\sin(x))^{(m+n)/m}\ \mathrm{d}x$ ? Standard CAD math as Maple did not provide answers.

答:
For $0 \leq t \leq \pi$ and $a>0$
$$\int_0^t \big[\sin(x)\big]^a\,dx=\frac{\sqrt{\pi }\,\, \Gamma \left(\frac{a+1}{2}\right)}{a\,\, \Gamma
   \left(\frac{a}{2}\right)}- \,
   _2F_1\left(\frac{1}{2},\frac{1-a}{2};\frac{3}{2};\cos ^2(t)\right)\cos (t)$$ where appears the gaussian hypergeometric function.

For other conditions, use the  symmetry properties .

Edit

For $t=\pi$,  we have
$$\int_0^\pi \big[\sin(x)\big]^a\,dx=\frac{ \pi\,\,  \Gamma (a+1)}{2^a\,\,\Gamma \left(\frac{a}{2}+1\right)^2}$$ which most of the times will not make any problem.

For large $a$, a good approximation is

$$\sqrt {\frac  {2\pi}a}\,\exp\Bigg[-\frac{1}{4 a}+\frac{1}{24 a^3}-\frac{1}{20 a^5}+O\left(\frac{1}{a^7}\right) \Bigg]$$ which is in a relative error of less than $0.1$% for $a>2$ and less than $0.001$% for $a>4$.

Title: Integral of $\sin(x)^{\frac{4}{3}}$


答:
Note that $f(x,y)=X(x)T(t)$ implies things like
$$\tag1f(x_1,t_1)f(x_2,t_2)=f(x_1,t_2)f(x_2,t_1)$$
for all $x_1,x_2,t_1,t_2$.

In the concrete example $f(x,t)=\sqrt{x+t}$, nearly all choices show that $(1)$ does not hold, e.g.
$$f(1,1)f(2,2)=\sqrt2\sqrt 4=2\sqrt 2\ne 3=\sqrt3\sqrt 3 =f(1,2)f(2,1) $$
or even simpler
$$f(0,0)f(1,1)=\sqrt 0\sqrt 2=0\ne 1=\sqrt 1\sqrt 1 =f(0,1)f(1,0).$$

Title: How can one prove that a function $f(x,t)$ cannot be expressed by the form $X(x)\cdot T(t)$?

Body: I was reading about ODE variable separation solving and the book says that assuming that a function can be expressed as the product of two single-variable functions loses generality, which I understand. I cannot, however, prove that it does. For example: the function $\sqrt{x+t}$ cannot be expressed as $X(x)\cdot T(t)$. I tried proving it by contradiction and got to the result $X_1(x)\cdot T_1(t) = (x+it)\cdot(x-it)$ but I'm not sure if this is sufficient. That is, what guarantees that $(x+it)\cdot(x-it)$ cannot be rearranged in some way so that it only has the variable $x$ in one of the factors and only $t$ in the other?


问:
What's the difference between 'for any' and 'for all'?
There are lots of expressions like, for all x, for any x, for some x, etc.

It may be because of the fact that I am not a native English speaker, but I can't tell the difference between 'for all' and 'for any'.

(I think 'for some x' means 'there exists at least one point', right?)

Could you explain it? thank you!

答:
I just want to point out the difference between "for all" and "for any":

1) "for all" usually used in the end of the sentence, meaning the condition is always satisfied. For example, "$x=x$ for all $x\in\mathbb{R}$".

2) "for any" usually is placed in the beginning of the sentence, stressing that you are choosing an arbitrary element. For example, "for any $x\in\mathbb{R}$, we have $x=x$".


答:
To prove that $\sqrt{x+t}=X(x)T(t)$ is not valid in general it is suffisant to show a conter-example.

$$\text{Suppose that}\quad  \sqrt{x+t}=X(x)T(t)\quad
\text{is true any }(x,t) \tag 1$$
$$\text{then}\quad \sqrt{2x+t}=X(2x)T(t)\quad
\text{which implies}$$
$$\frac{\sqrt{2x+t}}{\sqrt{x+t}}=\frac{X(2x)}{X(x]}$$
The term on the right is a function of $x$ only while the term on the left is a function involving $t$. This is impossible. Thus the supposition $(1)$ is false.


答:
Note that $\sigma(u,v)=\mathbf{r}$ and $\sigma^{-1}(\mathbf{r})=(u,v)$. On the other hand, the formula $\mathbf{r}-\mathbf{a}=u\mathbf{p}+v\mathbf{q}$ implies that
$$
u=(\mathbf{r}-\mathbf{a})\cdot \mathbf{p},\quad v=(\mathbf{r}-\mathbf{a})\cdot \mathbf{q}
$$

by orthogonality of the vectors $\mathbf{p}$ and $\mathbf{q}$ (and that they are of length $1$).

Title: Every plane in $\mathbb{R}^3$ is a smooth surface

Body:
This is from Pressley's Differential Geometry. Let $a$ be a point on the plane and let $p$ and $q$ be two unit vectors parallel to the plane and orthogonal to each other. If $\mathbf{r}$ is any point on the plane then $\mathbf{r-a}$ is parallel to the plane, and so $$\mathbf{r}-\mathbf{a}=u\mathbf{p}+v\mathbf{q}$$ for scalars $u,v$. So now we have the surface patch $\sigma(u,v) = \mathbf{a} + u\mathbf{p}+v\mathbf{q}$. Now the book states that its inverse is $\sigma^{-1}(\mathbf r) = ((\mathbf r-\mathbf a)\cdot\mathbf p,(\mathbf r- \mathbf a)\mathbf q)$.

Now my problem is that I have no idea how the inverse was found. Could someone explain it to me.


问:
How to solve an integral inside a summation (with a divergent term)
I have a question that might be silly, but I really don't understand what is going on. I have to solve the following integral:

$$ \sum_{n \in \mathbb{Z}} \int_{m}^{m+1}e^{inx}dx $$

However if we try to solve it in normal fashion, its solution is divergent when $n=0$. The solution in the paper that I am reading is:

$$ \sum_{n \in \mathbb{Z}} \int_{m}^{m+1}e^{inx}dx =  \sum_{n \in \mathbb{Z}}\delta_{n,0} + \frac{1-\delta_{n,0}}{in} \left(e^{in(m+1)}-e^{in(m)}\right) $$

However I think that the second term on the RHS is still divergent in $n=0$. Moreover, I do understand that the $\delta^{'}$s are introduced as an identity to cure the divergence, but I really don't understand how that is equal to the LHS.

Hopefully you can give me some insight :)

答:
Ok, I guess I'll spell it out completely.

$$S_m=\sum_{n\in\Bbb{Z}}\int_{m}^{m+1}e^{inx}\mathrm{d}x$$
If $n=0$, the integrand is just $1$. In other words
$$\int_{m}^{m+1}e^{i\cdot0\cdot x}\mathrm{d}x=\int_{m}^{m+1}1\mathrm{d}x=m+1-m=1.$$
Otherwise,
$$\int_m^{m+1}e^{inx}\mathrm{d}x=\left(\frac{e^{inx}}{in}\right)\bigg|^{m+1}_m=\frac{e^{in(m+1)}-e^{inm}}{in}$$
So our sum can be written as
$$S_m=1+\sum_{n\in\Bbb{Z}\backslash \{0\}}\frac{e^{in(m+1)}-e^{inm}}{in}$$
The sum given in your question is
$$\sum_{n \in \mathbb{Z}}\delta_{n,0} + (1-\delta_{n,0})\frac{e^{in(m+1)}-e^{inm}}{in}  $$
The zero term is
$$\delta_{0,0}+(1-\delta_{0,0})\cdot(...)=1-(1-1)\cdot(...)=1$$
And the other terms are
$$\delta_{n,0}+(1-\delta_{n,0})\frac{e^{in(m+1)}-e^{in(m)}}{in}=\frac{e^{in(m+1)}-e^{inm}}{in}$$
So the sum is
$$S_m=1+\sum_{n\in\Bbb{Z}\backslash \{0\}}\frac{e^{in(m+1)}-e^{inm}}{in}.$$
Anything unclear?


问:
how to solve this multivariate quadratic equation?
Any hope to acquire an analytic solution to such equations:

Solve:
$$\sum_{j=1}^n a_{ij} x_i x_j = b_i$$

for $i=1,\ldots,n$, where $a_{ij}$'s and $b_i$'s are known constants and $x_i$'s are unknowns to be solved.

Thanks a lot!

P.S. Thanks for alex.jordan's interesting comment! Let's consider this problem in a positive setting, i.e. let's require all coefficients($a_{ij}$'s and $b_i$'s) to be positive so that the solution seems to exist. Also $a$ is symmetric, i.e. $a_{ij}=a_{ji}$. If there could be any fast numerical solution it's also useful.

P.S. Just for the information of all readers: the original form of this problem is formulated as:
$$ x_i\cdot\left(\sum_{j=1}^n a_{ij}x_j\right) = \sum_{j=1}^n a_{ij}q_{ij} $$ where $q_{ij}\in(0, 1)$. I simplified the RHS into $b_{ij}$ but now it seems better not to do so(sorry for that!) to at least give the existence of real solutions a better chance.

答:
Let $A=\left( a_{ij} \right)_{n \times n} $, $B=\mathrm{diag}\{b_1, b_2, \cdots, b_n\}=\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}$, $C=A^{-1}B$.

There is a possible numerical solution when $A$ is invertible and $b_i\neq 0$ ($i$ = 1, 2, $\cdots$, $n$ ).

From the conditions, it will result

$$
\begin{pmatrix}
x_1 a_{11} & \cdots & x_1 a_{1n} \\
\vdots & \ddots & \vdots \\
x_n a_{n1} & \cdots & x_n a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix}  =
\begin{pmatrix}
b_1 \\
\vdots \\
b_n
\end{pmatrix}.
 $$
or
$$ \begin{pmatrix}
x_1 \\
& \ddots \\
& & x_n
\end{pmatrix}
\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix}  =
\begin{pmatrix}
b_1 \\
\vdots \\
b_n
\end{pmatrix}.
 $$
As $\sum\limits _{j=1} ^n a_{ij} x_i x_j=b_i$, i.e., $x_i \sum\limits _{j=1} ^n a_{ij}  x_j=b_i$, since $b_i \neq 0$, then $x_i \neq 0$.
so
$$
\begin{pmatrix}
a_{11} & \cdots & a_{1n} \\
\vdots & \ddots & \vdots \\
a_{n1} & \cdots & a_{nn}
\end{pmatrix}
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix}  =
\begin{pmatrix}
x_1^{-1} \\
& \ddots \\
& & x_n^{-1}
\end{pmatrix}
\begin{pmatrix}
b_1 \\
\vdots \\
b_n
\end{pmatrix} =
\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix}.
 $$

As $A=\left( a_{ij} \right)_{n \times n} $, $B=\mathrm{diag}\{b_1, b_2, \cdots, b_n\}=\begin{pmatrix}
b_1 \\
& \ddots \\
& & b_n
\end{pmatrix}$, $C=A^{-1}B$, then
$$ A
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix}   =
B
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix},
 $$
so

$$
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix}   =
A^{-1} B
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix},
 $$
that is to say,

$$
\begin{pmatrix}
x_1 \\
\vdots \\
x_n
\end{pmatrix}   =
C
\begin{pmatrix}
x_1 ^{-1} \\
\vdots \\
x_n^{-1}
\end{pmatrix}.
 $$

Let $
\begin{pmatrix}
x_1^{(0)} \\
\vdots \\
x_n^{(0)}
\end{pmatrix}   =
\begin{pmatrix}
1 \\
\vdots \\
1
\end{pmatrix}$,

 by the recurrence
$$
\begin{pmatrix}
x_1^{(k+1)} \\
\vdots \\
x_n^{(k+1)}
\end{pmatrix}   =
C
\begin{pmatrix}
{\left(x_1^{(k)}\right)} ^{-1} \\
\vdots \\
{\left(x_n^{(k)}\right)}^{-1},
\end{pmatrix},
 $$
we can get a numerical solution.

If the result does not converge, we can use another one
$$
\begin{pmatrix}
{\left(x_1^{(k+1)}\right)} ^{-1} \\
\vdots \\
{\left(x_n^{(k+1)}\right)}^{-1},
\end{pmatrix}   =
C^{-1}
\begin{pmatrix}
x_1^{(k)} \\
\vdots \\
x_n^{(k)}
\end{pmatrix}.
 $$

If you want the analytic solution, you may have a try by using the Groebner-Shirshov Bases. (The software Maple or Mathematica can do it.) But I am not sure the analytic solution can be find easily.


问:
The equation of a sphere tangent to a plane at a point whose center belongs to a plane
Determine the equation of a sphere that is tangent to the plane $x+y+z-9=0$ at the point $M(2,3,4)$ and whose center belongs to the plane:
$$ (P) : 7x-4y+5z-14=0 $$
My solution: it is clear that the point $A(2,5,4)$ belongs to the plane $(P)$. Then, let $d$ be the line
passing through $A$, $d$ orthogonal to $(P)$ (so that the director vector is equal with the normal vector of the plane, $v(7,-4,5)$ ). From the parametric equations of this line I found that the center of the sphere is of the form $C(2+7t, 5-4t, 4+5t)$. At this point, if I calculate the distance from the center of the sphere to the first plane I get ${|8t+11|\over\sqrt{3}}=r$, where $r$ is the radius of the sphere. And I'm stuck at this point because I don't really see how I can use the point $M$ to solve this problem. Any ideas? Thanks!

答:
It is possible to work backwards. Consider the equation of a sphere:

$$F(x,y,z) = (x-a)^2 + (y-b)^2 + (z-c)^2$$
$$\nabla F =  \left<2(x-a), 2(y-b), 2(z-c)\right>\Rightarrow \nabla(2,3,4) = \left<2(2-a), 2(3-b), 2(4-c)\right>$$

so the tangent plane is thus $2(2-a)(x-2) + 2(3-b)(y-3) + 2(4-c)(z-4) = 0$. Comparing coefficients with $x+y+z - 9 = 0$, the centre is at $a=1, b=2, c=3$, which does belong to the given plane.

Thus the squared distance from the centre $(1,2,3)$ to $(2,3,4)$ is $3$, which yields $(x-1)^2+(y-2)^2+(z-3)^2=3$.


答:
I think $\log$ and partial derivative can do it.

$\frac{\partial}{\partial x}\log(\sqrt{x+t})=\frac{1}{2}\frac{\partial}{\partial x}\log(x+t)=\frac{1}{2(x+t)}$ is still related to $t$.

If $f(x,t)=h(x)g(t)$, $\frac{\partial}{\partial x}\log(f(x,t))=\frac{\partial}{\partial x}\log(h(x)g(t))=\frac{\partial}{\partial x}\log(h(x))+\frac{\partial}{\partial x}\log(g(t))=\frac{h'(x)}{h(x)}$ is unrelated to $t$.


问:
How can one find an orthogonal frame $X_1, X_2$?
If we consider $\phi : \mathbb{R^2} → \mathbb{R^3}$ defined by $\phi(x, y) = (x, y,sin(x + y))$

How can one find an orhogonal frame $X_1, X_2 : \mathbb{R^2} → \mathbb{R^2}$ with respect to the Riemannian metric g and in addition how can I write down the volume 2-form on corresponding to your frame $X_1, X_2$?

With $g_{i,j} (p) = g(p)(e_i, e_j )$ where $g = \phi∗g_E$ is the pull-back of the Euclidean metric on $\mathbb{R^3}$

First, I proved that the jacobian $J_\phi(p)$ has no kernel for all $p \in \mathbb{R^2}$ and then I was able to show that this defines a Riemann metric on $\mathbb{R^2}$ with the following computations.

$\phi*g_{11} = 1 + cos^2(x+y)$, $\phi*g_{12}=\phi*g_{21} = cos^2(x+y)$ and $\phi*g_{22} = 1 + cos^2(x+y)$

From here I am stuck on how to proceed. Any help would be grateful. Thanks in advance.

答:
Assume that your vector fields take the form $X=X^1 \frac{\partial}{\partial x}+X^2\frac{\partial}{\partial y}$, $Y=Y^1\frac{\partial}{\partial x}+Y^2\frac{\partial}{\partial y}$ for some functions $X^i:\mathbb{R^2}\to \mathbb{R}$ and $Y^i:\mathbb{R^2}\to \mathbb{R}$. Then, in coordinates you have $$g(X,Y)=g_{11}X^1Y^1+g_{12}X^1Y^2+g_{22}X^2Y^2+g_{21}X^2Y^1=0$$ You can solve this functional equation for $X^i$ and $Y^i$.

In order to get a volume form, you can compute the musical isomorphism on these vector fields, namely, let $\alpha=g(X,\cdot)$ and $\beta=g(Y,\cdot)$. Since $X,Y$ are linearly independent, $dA=\alpha\wedge\beta$ will give you a non-vanishing two-form on $\mathbb{R^2}$.

Edit: Here's a little more info:

If we plug in our coefficients, we get $$(1+\cos^2(x+y))X^1Y^1+\cos^2(x+y)(X^2Y^1+Y^2X^1)+(1+\cos^2(x+y))X^2Y^2=0$$
We first can make an arbitrary choice for $X$, we'll just take the constant coordinate vector field $X=\frac{\partial}{\partial x}$, hence $X^1=1,X^2=0$. This simplifies our equation to $$(1+\cos^2(x+y))Y^1+\cos^2(x+y)Y^2=0$$ From here, you can get a relation between $Y^1$ and $Y^2$, and then by assuming some form of $Y^1$ you can get $Y^2$ using said relation.


问:
Solve the following Linear Congruences
Solve the following Linear Congruences:
 $$x\equiv 4\pmod {12}$$
 $$x\equiv 7\pmod {21}$$
 $$x\equiv 10\pmod {15}$$</p>

答:
HINT:

$x\equiv4\pmod {12}\implies x\equiv0\pmod 4$ and $x\equiv1\pmod 3$

Similarly,
$x\equiv7\pmod {21}\implies x\equiv0\pmod 7$ and $x\equiv1\pmod 3$

$x\equiv10\pmod {15}\implies x\equiv0\pmod 5$ and $x\equiv1\pmod 3$

So, $x\equiv1\pmod 3$ and $x\equiv0\pmod {4\cdot5\cdot 7}$

Use CRT as $(3,140)=1$

Title: Solve the following Linear Congruences


问:
How to prove that a hyperbolic geodesic triangle induces a tesselation
Let $m_1,m_2,m_3$ be integers greather than $2$ such that $\frac{1}{m_1}+\frac{1}{m_2}+\frac{1}{m_3}<1$. Consider a hyperbolic triangle $\tau$ with internal angles given by $\frac{\pi}{m_1},\frac{\pi}{m_2}$ and $\frac{\pi}{m_3}$ on each respectively vertex. I would like to prove that if $G$ is the subgroup of $Aut(\mathbb{H})$ generated by the three reflexions along the sides of the triangle, say $r_1,r_2$ and $r_3$, then $\{g(\tau)\}_{g\in G}$ is a tesselation of the Hyperbolic Plane $\mathbb{H}$.

My approach was to define the quotient $$X:=\frac{\bigcup_{g\in G}\overline{\tau}\times\{g\}}{\sim},$$
where $(x,h)\sim (y,hr_i)$ if $i=1,2,3$ and $y=r_{i}^{-1}x$, in order to prove that the function $f:X\rightarrow\mathbb{H}$ given by $f([x,g])=gx$ is an $G$-equivariant homeomorphism. Intuitively, the quotient makes the desired tesselation identifying the correct sides.

There's a way to give a metric to $X$ in a way that its completness is equivalent to $f$ being a surjective covering map, but i could not be able to fulfill the details. Any hint or an easier aproach is appreciated!.

答: This is a straightforward application of the Poincare polygon theorem, which is a substantial theorem. For references, take a look at the answer to this mathoverflow question, which includes generalizations to higher dimensions and other geometries.</p>


问:
Proving that $1+λ^2$ is an eigenvalue of $I+A^2$ where $λ$ is an eigenvalue of matrix $A$
So $Ax = λx$;

$A(Ax) = λ(Ax) \to (A^2)x = (λ^2)x$

I kind of dont know how to get the $1$ and $I$ here...

Any help is appreciated

答:
$Ax= \lambda x$ implies that

$$(I+A^2)x=x+A^2x=x+A(\lambda x)=x+\lambda Ax=x+\lambda^2x= (1+ \lambda^2)x$$


答:
$$Ax=\lambda x\to A^2x=\lambda^2x$$

You already did that. From here, sum $Ix=x$ on both sides and get

$$A^2x=\lambda^2x\to A^2x+Ix=\lambda^2x+x\to (A^2+I)x=(1+\lambda^2)x.$$

Title: Proving that $1+λ^2$ is an eigenvalue of $I+A^2$ where $λ$ is an eigenvalue of matrix $A$

Body:
So $Ax = λx$;

$A(Ax) = λ(Ax) \to (A^2)x = (λ^2)x$

I kind of dont know how to get the $1$ and $I$ here...

Any help is appreciated


问:
Seminorm defined by maximum of sublinear functional
Let $p$ be a sublinear functional defined in a vector space $X$. I'm trying to prove that the function $q$ defined by:
$$q(x) = \max_{|\alpha|=1}p(\alpha x)$$
is a seminorm. I'm having trouble at proving the homogeneity property though, that is, $q(\lambda x) = |\lambda|q(x)$. It is pretty obvious for $\lambda >0$, but for $\lambda<0$ I'm not sure how to proceed. Any tips?

答:
For $ \lambda <0$ observe that

$$p( \alpha \lambda x)=p((-\alpha)( - \lambda)x)$$

and

$$|-\alpha|=1 \iff |\alpha |=1.$$


问:
Lebesgue measure nullsets
I have a question concerning nullsets of the Lebesgue measure $\lambda^d$ on $\mathbb{R}^d$.
I think the set
 $$\{(x_1,\ldots,x_n)\in\mathbb{R}^n: \sum_{i=1}^n x_i=0\}$$
is a nullset of the n-dimensional Lebesgue measure. Any advice how prove this? I have my difficulties.

Regards,
Analyst77

答:
Another proof would use Fubini's theorem.  Since
$$
E:=\left\{(x_1,\ldots,x_n)\in\mathbb{R}^n: \sum_{i=1}^n x_i=0\right\}
$$
is a Lebesgue measurable set (it is a closed set), and for each $(x_1,\dots,x_{n-1}) \in \mathbb R^{n-1}$ the set
$$
\left\{t \in \mathbb R : (x_1,\dots,x_{n-1},t) \in E\right\}
$$
has $1$ dimensional measure zero (it is a single point), we conclude that $E$ itself has $n$ dimensional measure zero.</p>

Title: Lebesgue measure nullsets


问:
probability of guessing a k-digit number sequentially from n trials
I recently encountered the following problem.
Assume that we need to guess a $k$-digit number. We guess it digit by digit and there are $n$ guesses in total. What is the probability of success?

Say, $k=5$, and $n=25$. We can spend, e.g., 9 trials guessing the first digit, then guess the second digit from the 3rd trial etc. But we cannot make more than 25 guesses in total.

I believe we should proceed as follows:

guessed from 5 trials: $p=1/10^5$; -- we cannot make less than 5 trials when guessing step by step

guessed from 6 trials: $p=1/10^4\cdot(1-1/10)\cdot 1/9\cdot 5$

guessed from 7 trials: $p=1/10^3\cdot(1-1/10)^2\cdot 1/9^2\cdot \binom{5}{2}+1/10^4\cdot(1-1/10)\cdot(1-1/9)\cdot 1/8\cdot 5$, etc.

but I'm failing to generalize this scheme. And there is an additional problem that we have to set an upper bound to the number of trials at some point...

UPDATE (09.02): As @JMoravitz suggested, one can write an iterative scheme for computing the probability of guessing exactly $\kappa$ digits from $\eta$ trials, $f(\kappa,\eta)$. I believe that this scheme can be written as
$$f(\kappa,\eta)=\frac{1}{10}\sum_{i=1}^{10}f(\kappa−1,\eta−i),$$
where we use the fact that the probability of guessing the digit from $1\le \kappa\le 10$ guesses is always the same:
$$p(i=1)=\frac1{10},\quad p(i=2)=\frac{9}{10}\frac{1}{9}=\frac{1}{10},...$$
We also need to add boundary conditions $$\begin{cases}f(\kappa,\eta)=0,&\eta<\kappa\mbox{ (we cannot guess $\kappa$ digits from less than $\eta$ trials)},\\
f(1,\eta)=\frac{1}{10},&0<\eta\le 10\mbox{ (probabilities of guessing the first digit from $\eta$ trials)},\\
f(1,\eta)=0,&\eta=0\mbox{ or }\eta>10.\end{cases}$$
The final probability is to be found as the sum $$p^*=\sum_{i=k}^n f(k,i)=\sum_{i=5}^{25} f(5,i).$$
This scheme looks pretty neat and can be easily computed numerically.

However, I wonder if there is any chance to solve this problem analytically? Perhaps there exists a different approach?...

答:
May I try to offer an approach to have some estimation to this problem (and my apology in advance if what I am writing is trivial or wrong):

Let $X_i$ denote the number of guesses required to find the $i$ digit. Hence, lets denote $X = X_1+..+X_k$ as the total number of guesses. $X$ is sum of independent random variables.

It can be calculated that $\mathbb{E}X = 5.5k \; V[X] = 8.25k$

And if we denote $n = t + \mathbb{E}X$ ($t$ is not necessarily natural number, or you can take a floor of this equation) then we can have some estimation by Chevishev ineq.:
$$\mathbb{P}(|X-\mathbb{E}X|>t)\leq \frac {8.25k}{t^2}$$

Then we can get some conclusions about the problem, for instance, that for large enough numbers if the number of guesses allowed is in order of $O(k^{0.5+\epsilon})$ then we get that $$\mathbb{P}(|X-\mathbb{E}X|>t)\leq \frac {8.25k}{(k^{0.5+\epsilon})^2}$$
is small as we wish, and hence we get good lower bound to the probability of success:
$$1- \mathbb{P}(|X-\mathbb{E}X|>t) = \mathbb{P}(\mathbb{E}X - t\leq X\leq\mathbb{E}X+t = n) \leq \mathbb{P}(X\leq n)$$


问:
How to find the $\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$
im struggeling to find $$\lim _{x\to 0}\left(2-e^{\arcsin^{2}\left(\sqrt{x}\right)}\right)^{\frac{3}{x}}$$

Ive tried the following:
$$\lim_{x \to x_0} ax^{bx} = \lim_{x \to x_0} e^{ax^{bx}} =  \lim_{x \to x_0} e^{bx \ln(ax)} =  e^{\lim_{x \to x_0} bx \ln(ax)}$$
wich leads me to
$$ = e^{\lim_{x \to 0} \frac{3\ln(2-e^{\arcsin^2(\sqrt{x})})}{x}}$$
Is this the right way to go? If yes how to get rid of the division by x?
Thanks for any help!

答:
Hints:

$\lim_{x\to 0}(1+a(e^{-x}-1))^{-\frac{1}{x}}=e^a$

$\lim_{x\to 0}\frac{\arcsin x}{x}=1$

You can see the list of special limits in the form of $f(x)^{g(x)}$ here:
https://en.wikipedia.org/wiki/List_of_limits#Logarithmic_and_exponential_functions_Anchor01

We have the limit: $\lim_{x\to 0}(2-e^{\arcsin^2 \sqrt{x}})^{\frac{3}{x}}$. By doing algebraic manipulation, we get: $$\begin{align} & \lim_{x\to 0}(1+(-1)(e^{\arcsin^2 \sqrt{x}}-1))^{\frac{3}{x}} \\
& = \lim_{x\to 0}\Biggl((1+(-1)(e^{\arcsin^2 \sqrt{x}}-1))^{\frac{1}{\arcsin^2 \sqrt{x}}}\Biggl)^{\frac{3\arcsin^2 \sqrt{x}}{x}} \\ \end{align}$$ We know that: $$\lim_{x\to 0}(1+(-1)(e^{\arcsin^2 \sqrt{x}}-1))^{\frac{1}{\arcsin^2 \sqrt {x}}}=e^{-1} \space \space \text{and} \space \space \lim_{x\to 0}3\Biggl(\frac{\arcsin \sqrt{x}}{\sqrt{x}}\Biggl)^2=3 $$ This boils down to the fact that overall limit is: $$e^{-3}$$


问:
what is meant by a distributed method in the context of solving (convex) optimization problems?
Wanted to double-check my understanding of the following:

"one of the theoretical advantages of formulating a problem as a convex optimization problem is that the associated dual problem sometimes leads to an efficient or distributed method for solving it"

What is a distributed method is it the same as a distributed algorithm?
https://en.wikipedia.org/wiki/Distributed_algorithm

答:
Yes, distributed method and distributed algorithm mean the same thing here. Roughly speaking, in these algorithms, the tasks are 'distributed' across multiple devices, and their results are combined together through some mechanism.

Let me give a canonical example in optimization. Consider the following problem:
\begin{equation}
\min f(x) + g(z) \quad \mbox{s.t}\quad Ax + Bz = c.
\end{equation}
The Lagrangian is
\begin{equation}
L(x, z, y) = f(x) + g(z) + y^\top(Ax + Bz - c),
\end{equation}
and the dual function is given by
\begin{equation}
g(y) = \min_{x,z} L(x, z, y) = \min_{x,z}\left\{ f(x) + g(z) + y^\top(Ax + Bz - c)\right\}.
\end{equation}
Now assume that strong duality holds, then instead of solving the original problem, one can solve the dual problem: $\max_y g(y)$. Starting from an initial solution $y_0$, the subgradient method solves this problem by performing the following updates for $k = 0,1,2\ldots$:
\begin{align}
x_{k+1}, z_{k+1} &\gets \arg\min_{x,z} L(x,z,y_k),\tag{1} \\
y_{k+1} &\gets y_k + t_k(Ax_{k+1} + Bx_{k+1} - c). \tag{2}
\end{align}
In this algorithm, the most expensive step is solving the subproblem $(1)$. It is straightforward that $(1)$ is separable, so we can minimize over $x$ and $z$ independently:
\begin{align}
x_{k+1} &\gets \arg\min_{x} \left\{f(x) + y_k^\top Ax \right\} \tag{1a}\\
z_{k+1} &\gets \arg\min_{z} \left\{g(z) + y_k^\top Bz \right\}. \tag{1b}
\end{align}
Now, suppose that our computer only has one processor, then we have to use it to solve sequentially $(1a) \to (1b) \to (2)$ (or $(1b) \to (1a) \to (2)$). On the other hand, if our computer has 2 processors, then we can solve $(1a)$ and $(1b)$ in parallel (on each processor), which means the task $(1)$ has been done in a distributed manner.

The algorithm described in the above example is often referred to as dual decomposition, as we are solving the dual problem (instead of the primal one) and in addition, the steps of this task can be decomposed into smaller subproblems that are easier to solve.


问:
Transformation from [0, 1] to [0, 1] pushing values above x towards 1 and values below x towards 0.
I have a vector of 64 elements between [0, 1]. Those elements represent weights in a model. I would like to define the value t as the 16th maximum weight (75th percentile) and then push the weights above towards 1 and the weights below towards 0. The goal is to increase the separation, the distinction between 'high' weights and 'low' weights. Defining t is not an issue, but I do not know which transformation I could apply to achieve my goal.

I am also interested in a more general solution where the input vector elements are between [a, b] and where you can control the 'rate' at which elements are pushed towards one side or the other.

At first, I thought using logarithmic scaling would help, but it only pushes every value from [0, 1] towards 1 or towards 0 depending on the implementation. The formula for the log scaling was given to me by user1337:

$$x \mapsto \frac{\ln  \left(\frac{a \theta -b}{a-b}+\frac{(1-\theta ) x}{a-b}\right)}{ \ln (\theta) }. $$

答:
An imperfect idea, but you tagged python, so perhaps you can make it work for your purpose:

$$f(x) = a+(b-a)\frac{\left(\frac{x-a}{b-a}\right)^{cd}}{ \left(\frac{x-a}{b-a}\right)^{cd} + \left(1- \left(\frac{x-a}{b-a}\right)^c \right)^d }$$

Positives:

$f$ maps $[a,b]$ to $[a,b]$ with $f(a)=a$ and $f(b)=b$ (exactly).

$d$ allows to tune the "strength".

Negatives:

$c$ is fiendishly complex to calculate as a function of $d$ and what you called $t$. So this is only useful if you can approximate $c$ numerically by looking for the $c$ that minimizes $\vert f(t)-t \vert$.


问:
K-Theory Equivalence Classes
Let $M$ be a finite dimensional compact manifold and $(Vect(M),\oplus)$ be the abelian monoid of complex vector bundles on $M$. I just read that it is possible to construct an equivalence relation on $Vect(M)^2$ by
$$(E_1,F_1)\sim (E_2,F_2) \iff \exists N\in \mathbb N \, s.t. \, E_1\oplus F_2\oplus \mathbb C^N\cong E_2\oplus F_1\oplus \mathbb C^N$$
In the book i'm reading (Salomon's notes on SW theory), it says that "the additional summand $\mathbb C^N$ is needed to obtain an equivalence relation". It seems to me that the only problem of not having it could be the transitivity property: is it not necessarily true that if $E_1\oplus F_2\cong E_2\oplus F_1$ and $E_2\oplus F_3\cong E_3\oplus F_2$, then $E_1\oplus F_3\cong E_3\oplus F_1$? I'm not able to prove the transitivity property also for the adjusted relation above. Can someone help me with that?

答:
The conceptual way to define the equvalence relation is that
$$
(E_1,F_1)\sim (E_2,F_2) \Leftrightarrow \exists G\in Vect(M) s.t. E_1\oplus F_2\oplus G\cong E_2\oplus F_1\oplus G.
$$
In this formulation the transitivity is easier to prove. This relation is the same as the one you used since it is well known that for $G\in Vect(M)$, there is $\tilde G\in Vect(M)$ such that $G\oplus\tilde G\cong \mathbb C^N$ for some $N\in\mathbb N$.

Edit (in view of the comment by @MichaelAlbanese): Without "stabilization" as expressed by the definition of the equivalence relation, transitivty is lost. Indeed start with arbitrary bundles $E$, $\tilde E$ and $F$ and then put $E_1:=E$, $F_1=0$, $E_2=E\oplus F$, $F_2=F$, $E_3=\tilde E$ and $F_3=0$. Then $E_1\oplus F_2\cong E_2\oplus F_1$ is satisfied automatically, while $E_2\oplus F_3\cong E_3\oplus F_2$ exactly means that $E\oplus F\cong \tilde E\oplus F$. If transitivity would hold then this would imply that $E\cong\tilde E$. But it is well known that vector budnels can be stably isomorphic without being isomorphic.The standard example (for real vector bundles) is $TS^2$ which is non-trivial by the hairy ball theorem, but its sum with the normal bundle to $S^2$ (which is easily seen to be trivial) is $T\mathbb R^3|_{S^2}$, which is also trivial. Hence $TS^2\oplus\mathbb R\cong\mathbb R^3\cong\mathbb R^2\oplus\mathbb R$ but $TS^2$ is not isomorphic to $\mathbb R^2$.

Title: K-Theory Equivalence Classes


问:
Show that a solution to an equation set is also a solution to another equation
$y=\frac12x^2;\;
y^2+(x-3)^2=3^2$ (circle)

The function and circle intersect at two points, one is $(0,0)$ and lets say the other is $(x_0,y_0)$
Show that $x=x_0$ is also a solution to the equation $x^3+4x=24$.

I dont understand at all how to solve this. I tried solving
$(x^3+4x-24)/(x-x_0)$ and try and get a new expression for $x_0$(?) since there can't be any remainder, but then I just end up with the exact same equation, in terms of $x_0$ instead.

答:
Given $y=\frac12x^2\iff x=\sqrt{2y} \tag{1}$
$y^2+(x-3)^2=3^2 \tag{2}$

$$\quad y^2+(x-3)^2=3^2\\
\implies (x^2-6x+9)+y^2-9=0\\
\implies x^2-6x+y^2=0\\
\implies \frac{6\pm\sqrt{36-4y^2}}{2} \\
x=3\pm\sqrt{3^3-y^2}\qquad y^2\le 3^3\\
\implies x\in\big\{0,3,6\big\}$$

We now try $x$-values into $(1)$ and we get we get
$$y=\frac{x^2}{2}\implies y=0\quad y=4.5 \quad y=18$$

Since $y\le 3$, the only integer solution is $\space (0,0)\space$ but we can use your results $\space  (x^3 + 4 x - 24) = 0\quad$
with WolframAlpha as shown here $\text{to find}\space x\approx 2.42682\quad$ Equation $(1)$ now shows $y\approx2.94474$

We could also feed the original equations to WolframAlpha for both values or use the cubic equation (tedious) to find the one real solution for $x.\qquad$ For $x^2+4x-24=0\space$ we let
$$a=1,\space b=0,\space c=4,\space d=-24$$
and
$$x=\sqrt[\Large{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)+\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}\\
+\sqrt[\Large{3}]{\biggl(\frac{-b^3}{27a^3 }+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)-\sqrt{\biggl(\frac{-b^3}{27a^3}+\frac{bc}{6a^2}-\frac{d}{2a}\biggr)^2+\biggl(\frac{c}{3a}-\frac{b^2}{9a^2}\biggr)^3}}-\frac{b}{3a}$$

I have used the cubic equation and it works but there are sometimes intermediate complex components that require robust software such as WolframAlpha or Python to get results.


问:
How do you solve $3^x=9\cdot27$?
I don't understand how to calculate the following expression $3^x=9\cdot27$.

Thank you a lot for the help.

答: you simply re-write $9\cdot 27=3^2\cdot 3^3 $. Hence, it gives $3^5$. Therefore $3^x=3^5$ implies $x=5$.

Title: How do you solve $3^x=9\cdot27$?


问:
Matrix exponential for non-commutative operator entries of matrix
I would like to find the matrix exponential $e^{iHt}$ of the Hermitian matrix $H$ where

$$
H=\begin{pmatrix}
\delta& \sqrt{2}a & 0\\
\sqrt{2}a^\dagger &0& \sqrt{2}a\\
0 &\sqrt{2}a^\dagger&-\delta
\end{pmatrix}
$$
$$[a,a^\dagger]=a a^\dagger-a^\dagger a=1$$
$$[a,\delta]=[a^\dagger,\delta]=0.$$

The dagger $\dagger$ denotes complex conjugate. My thought is to make use of the definition of matrix exponential $e^{X}=\sum_{k=0}^{\infty} \frac{1}{k !} X^{k}$ and calculate $H^2$ and $H^3$ to find some pattern to exploit. For example, in the special case where $\delta=0$, we can find a diagonal matrix $D$
$$
D=\begin{pmatrix}
4a^\dagger a+6 &0&0\\
0&4a^\dagger a+2&0\\
0&0&4a^\dagger a-2
\end{pmatrix}
$$
such that $H^{2n+1}=D^{n}H$ and $H^{2n+2}=D^{n}H^2$, with which we can obtain a neat result of $e^{i H t}$. However, it seems that no such diagonal matrix exists for $\delta \neq 0$ case. Are there any other ways to calculating matrix exponential like this with non-commuting entries? Any input is much appreciated!

答:
A tip of the hat to @Semiclassical 's insightful suggestion. Now that he has unravelled the group structure of H through the symmetry operator N, you may classify its eigenstates of eigenvalue n and specify them further to find eigenstates of the hamiltonian.

Recall $a^\dagger |n\rangle=\sqrt{n+1}|n+1\rangle$ and $a|n\rangle= \sqrt{n}|n-1\rangle$, so impose $H V(n)=\epsilon(n) ~ V(n)$ to fix the labile coefficients b and c for the unnormalized N eigenstate,
$$
V(n)\equiv \begin{pmatrix} \sqrt{2} b(n) |n-1\rangle \\ |n\rangle \\\sqrt{2} c(n) | n+1\rangle    \end{pmatrix}.
$$

The Hamiltonian eigenvalue condition then restricts
$$
\delta b +\sqrt{n}=\epsilon b \\
2\sqrt{n} b + 2\sqrt{n+1} c = \epsilon\\
\sqrt{n+1}-\delta c =\epsilon c, \leadsto \\
b=\frac{\sqrt n}{\epsilon -\delta}, ~~ c=\frac{\sqrt {n+1}}{\epsilon +\delta} ,\\
\epsilon (\epsilon^2 -\delta^2 -4n-2)+2\delta =0.
$$
The (readily depressed!) cubic equation may be solved for $\epsilon$, but I won't go into it... For vanishing δ, you have
$$
\epsilon =\pm \sqrt{4n+2},\\
b(n) = 1/\sqrt{4+2/n}, ~~~ c(n)= 1/\sqrt{4-2/(n+1)},
$$
so that
$V(n,t) = \exp \left( \pm it \sqrt{4n+2}\right )~~ V(n,0) $, etc. Even if $D-H^2\neq 0$, $V(n)$ is in its kernel.

I have not fully explored the non-vanishing δ solutions of the (depressed!) cubic.

If you wished to find the operator exponentiation (but why?) ,
you could write its spectral resolution in terms of these states $V(n)$, after you normalize them.


问:
what is running yield?
Suppose an investor who pays £120 on 15 July 2007 for £100 nominal of a six-year stock with half-yearly coupon payments of 13% pa and is not subject to tax. The stock is redeemable at par. Calculate the running yield?

I understood entire question but what actually is running yield?

答:
"Used to describe the income investors get from their portfolio as a percentage of market value of the securities."

http://www.investopedia.com/terms/r/runningyield.asp

Title: what is running yield?


问:
Evaluating the limit $\lim_{n\to \infty} \frac{n^n}{(2n)!}$
I'm trying to evaluate this limit. $$\lim_{n\to \infty} \frac{n^n}{(2n)!}$$ I know that $n^n$ grows faster than $n!$. So the result should be $+\infty$. But the real result is zero. Why?

答:
$n^n$ grows faster than $n!$ but slower than $(2n)!$, since:

$$(2n)! = (2n)(2n-1)\dots(n+1)n! > n^nn!$$

Hence $$\lim_{n\to\infty}\frac {n^n}{(2n)!} \le \lim_{n\to \infty}\frac {n^n}{n^nn!} = \lim_{n\to \infty}\frac {1}{n!} = 0$$

The confusion may be from the fact that $(2n)! \ne 2n!$.


答: Note that $$0 \le \dfrac{n^n}{(2n)!} = \dfrac{1}{n!}\cdot\dfrac{n \cdot n \cdots n}{(n + 1)\cdot(n+2)\cdots(2n)} \le \dfrac{1}{n!} \to 0.$$

Title: Evaluating the limit $\lim_{n\to \infty} \frac{n^n}{(2n)!}$

Body: I'm trying to evaluate this limit. $$\lim_{n\to \infty} \frac{n^n}{(2n)!}$$ I know that $n^n$ grows faster than $n!$. So the result should be $+\infty$. But the real result is zero. Why?


问:
Equivalent metrics in $\ell ^\infty$ space
$$\ell^{\infty}=\{z=(z_n):{\sup}_{n}|z_n|<\infty\} \ and\ x,y\in \ell^{\infty}$$
$$d_{\infty}(x,y)=sup_{n}|x_n-y_n| $$ and $$d(x,y)=\sum_{n=1}^{\infty}{\frac{1}{2^n}.{\frac{|x_n-y_n|}{1+|x_n-y_n|}}}$$ show that these metrics is not equivalent metrics. How can I do this without using topology

Body: $(x^{(k)}=(1,1,...1,0,0,...): \forall n\geq k; x_n^k=0)\subset \ell ^\infty$ $$\lim_{k \to \infty}{x^k}=(1,1,.....)\in \ell ^ \infty$$ $$x^k \nrightarrow^{d_\infty} (1,1,1,....)$$ We must show $d_\infty(x^k,(1,1,...))\geq \epsilon_0 \ \exists \epsilon_0 >0$ $$Let's\ see \ this.$$ $Take \ \epsilon_0=\frac{1}{2} \  \exists N \in \Bbb{N} \forall k>N ; d_{\infty}(x^k,(1,1,...))=sup_{n}{|x_n-1|}=1\geq\epsilon_0=\frac{1}{2}$ $$x^k \rightarrow^{d} (1,1,1,....)$$ $Given \ \epsilon>0 \ take \ \  N>log_{(\frac{1}{2})}^{2.\epsilon} \ and \ \forall \ k\geq N $  $$d(x^k,(1,1,....))=\sum_{n=1}^{\infty}{\frac{1}{2^n}}{\frac{|x_{n}^{k}-1|}{1+|x_{n}^{k}-1|}}=\sum_{n=1}^{k}{\frac{1}{2^n}}{\frac{|x_{n}^{k}-1|}{1+|x_{n}^{k}-1|}}+\sum_{n=k+1}^{\infty}{\frac{1}{2^n}}{\frac{|x_{n}^{k}-1|}{1+|x_{n}^{k}-1|}}=\sum_{n=k+1}^{\infty}{\frac{1}{2^n}}{\frac{1}{2}}=\sum_{n=1}^{\infty}{\frac{1}{2^{n+1}}}-\sum_{n=1}^{k+1}{\frac{1}{2^{n+1}}}<\frac{1}{2}.\frac{1}{1-\frac{1}{2}}-\sum_{n=1}^{k+1}{\frac{1}{2^{n+1}}}<1-\sum_{n=1}^{k+1}{\frac{1}{2^{n+1}}}=1-{\frac{1}{2}}.\frac{1-(\frac{1}{2})^{k+1}}{1-(\frac{1}{2})}=(\frac{1}{2})^{k+1}<(2^{-1-N})<\epsilon$$ Because  $$\sum_{n=1}^{k}{\frac{1}{2^n}}{\frac{|(1=x_{n}^{k})-1|}{1+|x_{n}^{k}-1|}}=\sum_{n=1}^{k}{\frac{1}{2^n}}{\frac{|1-1|}{1+|1-1|}}=\sum^{n=1}_{k}{\frac{1}{2^k}}.0=0$$ and this means $x^{k} \rightarrow ^ {d} (1,1,...)$ then $d_{\infty}\neq d$


问:
Antichains and weighted sums
Given an antichain $S_1,S_2,\dots,S_m$ of subsets of $\{1,2,\dots,n\},$ does there always exist positive integers $x, y_1,\dots,y_n$ so that for $i=1,\dots,m$ we have:

$$\sum_{k\in S_i} y_k =x\tag 1$$

(Antichain means that for for $i\neq j,$ we have $S_i\not\subseteq S_j.$)

Just a question that occurred to me while working on another problem.

Certainly, given $x,y_1,\dots,y_n,$ the set
$$\mathcal A =\left\{S\subseteq\{1,\dots,n\}\mid \sum_{k\in S} y_k =x\right\}$$ is an antichain, so this question asks if all antichains are a subset of an antichain of the form $\mathcal A.$

答:
This is not true.

Counterexample

Let $n=5$ and $$\mathcal A=\{45,123,124,125,134,135,234,245\}
$$

By the symmetry of $\mathcal A,$ we can see that if there is a solution $x,y_1,\dots,y_5,$ there is a solution with $y_1=y_2=y_3$ and $y_4=y_5.$

But then $x=3y_1=2y_1+y_4=2y_4.$ The second equality means $y_1=y_4$ while the third equality means that $2y_1=y_4.$ This means $y_1=y_4=0,$ violating the requirement that the $y_i$ are positive.

Title: Antichains and weighted sums


Body: $\dfrac{\frac{(n+1)^{n+1}}{(2n+2)!}}{\frac{n^n}{(2n)!}}=\dfrac{(2n)!\cdot (n+1)^{n+1}}{n^n\cdot (2n+2)!}=\dfrac{n+1}{(2n+1)(2n+2)}\bigg(1+\dfrac{1}{n}\bigg)^n\to0<1$


答:
$(2n)!\approx\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}=2\sqrt{\pi}\left(\frac2e\right)^{2n}n^{2n+1/2}$
so:
$$\frac{n^n}{(2n)!}=\frac{n^n}{2\sqrt{\pi}\left(\frac2e\right)^{2n}n^{2n+1/2}}=\frac{1}{2\sqrt{\pi}}\left(\frac e2\right)^{2n}n^{\frac12-n}$$
now as we take the limit notice that the $e/2$ term grows slower than the $n$ term when both raised to a power of $n$ and so it goes towards $0$


问:
Question about Spivak's proof of how to use u-substitution when the derivative of the inner function does not appear in the integral
Spivak (3rd edition) proposes solving the integral $$\int \frac{1+e^x}{1-e^x} dx$$ by letting $u=e^x$, $x=\ln(u)$, and $dx=\frac{1}{u}du$. This results in the integral $$\int \frac{1+u}{1-u}\frac{1}{u}du\\=\int \frac{2}{1-u}+\frac{1}{u}du=-2\ln(1-u)+\ln(u)=-2\ln(1-e^x)+x$$

From this example, Spivak argues that a similar method will work on any integral of the form $\int f(g(x))dx$ whenever $g(x)$ is invertible in the appropriate interval. Because this method is not a simple application of the substitution theorem, Spivak provides the following justification for his claim.

Consider continuous $f$ and $g$ where $g$ is invertible on the appropriate interval. Applying the above  method to the arbitrary case, we let $u=g(x)$, $x=g^{−1}(u)$, and $dx=(g^{−1})′(u)du$. Thus, we need to show that $$∫f(g(x))dx=∫f(u)(g^{−1})′(u)du$$ To prove this equality Spivak uses a more typical substitution $u=g(x)$, $du=g′(x)dx$ and applies it by noting that $$∫f(g(x))dx=∫f(g(x))g′(x)\frac{1}{g′(x)}dx$$ Presumably using the substitution theorem, which roughly states that $∫f(g(x))g'(x)dx=∫f(u)du$, Spivak asserts that  $$∫f(g(x))g′(x)\frac{1}{g′(x)}dx=∫f(u)\frac{1}{g′(g^{−1}(u))}du$$ Then, because $(g^{-1})'(u)=\frac{1}{g'(g^{-1}(u))}$ Spivak concludes $$∫f(u)\frac{1}{g′(g^{−1}(u))}du=∫f(u)(g^{−1})′(u)du$$

I lose track of the argument when Spivak argues that $$∫f(g(x))g′(x)\frac{1}{g′(x)}dx=∫f(u)\frac{1}{g′(g^{−1}(u))}du$$ In the original example, it was clear to me how we could apply the substitution theorem to make this equality true because $\frac{1}{g'(x)}$ was in fact a function of $g(x)$ as $g'(x)=g(x)$. But this is not necessarily true in all cases, or so it seems. How do we know that $f(g(x))\frac{1}{g'x}$ can be written in the form $h(g(x))$ for some continuous function $h$? To sum up, my main questions is, how do we use the substitution theorem to justify the equality $$∫f(g(x))g′(x)\frac{1}{g′(x)}dx=∫f(u)\frac{1}{g′(g^{−1}(u))}du$$

答:
Believe it or not,
$$
\int f(g(x)) \, dx
$$
is still of the correct form to admit substitution, provided that $g$ is invertible. To begin with, consider what we are aiming for: we know that any integral of the form
$$
\int \phi'(\mu(x))\mu'(x) \, dx
$$
is equal to $\phi(\mu(x))+C$, and making the substitution $u=\mu(x)$ works out just the same (hence why integration by substitution is valid in the first place). This means that if $f(g(x))$ can be written as $\phi'(\mu(x))\mu'(x)$ for some functions $\phi'$ and $\mu$, then substitution is applicable. First, rewrite the integral as
$$
\int \frac{f(g(x))}{g'(x)}g'(x) \, dx
$$
Now, the derivative of the inner function $g$ appears elsewhere, which is what we wanted. From here, it is simply a matter of rearrangement:
\begin{align}
\int \frac{f(g(x))}{g'(x)}g'(x) \, dx &= \int \frac{f(g(x))}{g'(g^{-1}(g(x)))}g'(x) \, dx \\
&= \int \phi'(g(x))g'(x) \, dx
\end{align}
where $\phi'$ is the function defined by
$$
\phi'(x) = \frac{f(x)}{g'(g^{-1}(x))} \, .
$$
(It should now be clear why it is necessary that $g$ is invertible.) We find that the integral is equal to
$$
\int f(g(x)) \, dx = \int \phi'(g(x))g'(x) \, dx = \phi(g(x)) + C
$$
Compare this method with how we usually make substitutions. When we set $u=g(x)$ so that $x = g^{-1}(u)$ and $dx = (g^{-1})'(u) \, du$, it may seem that we are doing something completely different to the method shown above. However, if we realise that
$$
(g^{-1})'(u) = \frac{1}{g'(g^{-1}(u))}
$$
then it becomes clear that the two methods are virtually the same:
\begin{align}
\int f(g(x)) \, dx &= \int f(u) (g^{-1})'(u) \, du \\
&= \int \frac{f(u)}{(g'(g^{-1}(u))} \, du \\
&= \int \phi'(u) \, du \\
&= \phi(u) + C \\
&= \phi(g(x)) + C \, .
\end{align}


问:
1D Heat equation $x\in(0,L)$ with mixed boundary conditions $u_x(t, 0)=-1$, $u(t, L)=0$
I am attempting to find an efficient (less computationally demanding than solving with numeric integration) solution for the 1D heat equation:

$u_t - \alpha u_{xx} = 0$

IC: $ u(x, 0) = 0$

BC 1: $-u_x(0, t) = 1$

BC 2: $u(L, t) = 0$

I have attempted to solve this using separation of variables and have arrived at

$u(x, t) = L - x -\sum_{n=1}^{\infty}A_n e^{-\lambda_n^2\alpha t} \cos(\lambda_n x)$

with $ \lambda_n = \frac{(2n-1)\pi}{2L}, n = 1, 2,  \dots $ and $  A_n = \frac{{(2n-1)(L-1)} \pi \left(-1\right)^{n} - 2}{L \lambda_{n}^{2}}
$

Unfortunately, this series does not seem to converge. I can show my working to arrive at the above if it is required, but I feel certain that this problem has been solved before, so any pointers would be appreciated.

I include the solution steps I took below:

Let's say
\begin{align}
u(x, t) = w(x, t) + v(x)
\end{align}
where $w$ is the transient solution and $v$ is the steady state solution.
Now, clearly $v(x) = c(L - x)$.
This satisfies BC2 directly.
BC1 leads to $-u_x(0, t) = -v_x = c = 1$, so
\begin{align}
  v(x) = L-x
\end{align}

For the transient solution, we go back to original problem, which transforms simply to $w_t = \alpha w_{xx}$.

IC: $u(x, 0) = 0 = w(x, 0) + v(x) = w(x, 0) + L - x$, so $w(x, 0) = x-L$

BC1: $-u_x(0, t) = -(w_x(0, t) + v_x(x)) = -w_x(0, t) + 1 = 1$, therefore $w_x(0, t)=0$.

BC2: $u(L, t) = w(L, t) + v(L) = w(L, t) = 0$.

We assume that $w(x, t) = X(x)T(t)$, so that $w_t=XT'$ and $w_{xx} = X''T$.
The original problem therefore becomes $XT'=\alpha X''T$, which leads to
\begin{align}
  \frac{X''}{X} = \frac{1}{\alpha}\frac{T'}{T} = -\lambda^2
\end{align}
since these can only be equal if they are constant. Other signs or zero value for the constant leads only to trivial solutions.

We now have two separate ODEs:
\begin{align}
  X'' + \lambda^2X = 0 \quad \text{and} \quad T  + \lambda^2 \alpha  T' = 0
\end{align}
The general solutions to these are
\begin{align}
X = c_1 \cos (\lambda x) + c_2 \sin(\lambda x) \quad \text{and} \quad Y=c_3e^{-\lambda^2\alpha t}
\end{align}
Therefore
\begin{align}
w &= XT  \\
  &= c_3e^{-\lambda^2\alpha t}[ c_1 \cos(\lambda x) + c_2 \sin(\lambda x) ] \\
  &= e^{-\lambda^2\alpha t}[A \cos(\lambda x) + B \sin(\lambda x)]
\end{align}
Applying the boundary conditions:
\begin{align}
  BC1: w_x(0, t) = 0 \rightarrow -e^{-\lambda^2\alpha  t}(A \lambda {\sin(\lambda 0)} + B \lambda \cos(\lambda 0)) = 0 \rightarrow B=0
\end{align}
So we know that $w = Ae^{-\lambda^2\alpha t}\cos(\lambda x)$
\begin{align}
  BC2: w(L, t) = 0 \rightarrow w = Ae^{-\lambda^2\alpha t}\cos(\lambda L) = 0
\end{align}
Obviously there are infinitely many $\lambda$s that satisfy this condition, which we indicate by writing
\begin{align}
  \lambda_n = \frac{(2n-1)\pi}{2L}, n = 1, 2,  \dots
\end{align}
Since all these values are solutions to the original problem, but do not in general satisfy the boundary conditions individually, we assume that some linear combination of the solutions is the actual solution, and write
\begin{align}
  w = \sum_{n=1}^\infty A_ne^{-\lambda_n^2\alpha t}\cos(\lambda_nx)
\end{align}

We solve for $A_n$ from the initial condition:
\begin{align}
  w(x, 0) = x - L = \sum_{n=1}^\infty A_n\cos(\lambda_nx)
\end{align}

If we multiply by $\cos(\lambda_m x)$ (note the different subscript) and integrate along the length of the rod (0 to $L$), we find
\begin{align}
  \int_0^L \cos(\lambda_m x)x - L\cos(\lambda_mx) \mathrm{d}x = A_n \sum_{n=1}^\infty \int_0^L \cos(\lambda_nx) \cos(\lambda_m x) \mathrm{d}x
\end{align}
The integral on the right vanishes except when $n=m$, so we have
\begin{align}
  \int_0^L \cos(\lambda_n x)x\mathrm{d}x - L\int_0^L\cos(\lambda_nx) \mathrm{d}x = A_n \sum_{n=1}^\infty \int_0^L \cos^2(\lambda_nx) \mathrm{d}x
\end{align}
Which evaluates to
\begin{align}
\frac{L \lambda_{n} \sin(L \lambda_{n}) + \cos(L \lambda_{n}) - 1}{\lambda_{n}^{2}}%
-\frac{L^{2} \sin(L \lambda_{n})}{\lambda_{n}}%
=%
A_n\frac{2L\lambda_n + \sin(2L\lambda_n)}{4\lambda_n}
\end{align}
We can solve for $A_n$ as
\begin{align}
  A_n = -\frac{4 \, {\left({\left(L^{2} \lambda_{n} - L \lambda_{n}\right)}
\sin\left(L \lambda_{n}\right) - \cos\left(L \lambda_{n}\right) +
1\right)}}{2 \, L \lambda_{n}^{2} + \lambda_{n} \sin\left(2 \, L
\lambda_{n}\right)}
\end{align}
Applying trigonometric identites yields
\begin{align}
 A_n = -\frac{2 \, {\left({\left(L^{2} \lambda_{n} - L \lambda_{n}\right)}
\sin\left(L \lambda_{n}\right) - \cos\left(L \lambda_{n}\right) +
1\right)}}{L \lambda_{n}^{2} + \lambda_{n} \sin\left(L
\lambda_{n}\right) \cos\left(L \lambda_{n}\right)}
\end{align}
But we found $\lambda_n$ by requiring $\cos(L\lambda_n)=0$, so
\begin{align}
A_n = -\frac{2 \, {\left({\left(L^{2} \lambda_{n} - L \lambda_{n}\right)}
\sin\left(L \lambda_{n}\right) +
1\right)}}{L \lambda_{n}^{2}}
\end{align}

Now we substitute our solution for $\lambda_n$ and simplify to find
\begin{align}
  A_n = \frac{{(2n-1)(L-1)} \pi \left(-1\right)^{n} - 2}{L \lambda_{n}^{2}}
\end{align}

答:
I'm sorry, but I cannot tell where you went wrong; your answer makes no sense from a dimensional standpoint.  As for the convergence of the series, I will answer below, but first let's actually solve the equation correctly.

EXACT SOLUTION OF THE EQUATION

In my humble opinion, I think these particular mixed BCs do not lend themselves easily to a simple solution by separation of variables (although the final solution is indeed of that form).  I chose, rather, to solve this equation using a Laplace transform over $t$; that is

$$\hat{u}(x,s) = \int_0^{\infty} dt \, u(x,t) e^{-s t}$$

Then the PDE becomes on ODE on $x$:

$$\alpha \frac{d^2}{d x^2} \hat{u}(x,s) - s \hat{u}(x,s)=0$$

$$\left[\frac{d}{dx} \hat{u}(x,s)\right]_{x=0} = -\frac{1}{s}$$

$$\hat{u}(L,s) = 0$$

Note that the ODE already incorporates the initial condition.

The solution to this equation (the details of which solution I leave to the reader) is

$$\hat{u}(x,s) = \sqrt{\alpha} \frac{\sinh{\left[\frac{L-x}{\sqrt{\alpha}} \sqrt{s}\right]}}{s^{3/2} \cosh{\left[\frac{L}{\sqrt{\alpha}} \sqrt{s}\right]}}$$

Getting the solution to the original equation is then a matter of taking an inverse Laplace transform:

$$u(x,t) = \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, e^{s t} \hat{u}(x,s)$$

Although the integration contour appears to be complicated by the branch points in the LT expression, the ILT turns out to be simply equal to the sum of the residues of the poles of $\hat{u}(x,s)$.  The reason this is so is because the functions of $\sqrt{s}$ are all even, so that there is no effect of crossing any branch cuts.  The poles of $\hat{u}$ are at $s = -(n+1/2)^2 \pi^2 \alpha/L^2$ for integers $n$ and at $s=0$.  Using the fact that, for a function of the form $f(s) = p(s)/q(s)$, the residue of a simple pole $s_0$ of $f$ is $p(s_0)/q'(s_0)$, I get for the ILT, after some algebra:

$$u(x,t) =L-x- 2 \frac{L}{\pi^2} \sum_{n=0}^{\infty} \frac{1}{(n+1/2)^2} \cos{\left[ \left ( n+\frac12 \right) \pi \frac{x}{L}\right]} e^{-\alpha (n+1/2)^2 \pi^2 t/L^2}$$

The series obviously converges, and you can perform some rudimentary checks to see that the initial and boundary conditions are satisfied, e.g., note that

$$\sum_{n=0}^{\infty} \frac{1}{(n+1/2)^2} = \frac{\pi^2}{2}$$

Also, you should check that the solution jibes with intuition at the steady state (i.e., as $t \to \infty$): here, $u(x,t) \sim L-x$, which makes physical sense (the temperature gradient at $x=0$ implies heat flow out of the left end of the bar).

RATE OF CONVERGENCE

The series above, while clearly convergent, does so painfully slow, especially for small $t$.  Even for $n=50$, there is significant oscillation in the sum for $t=0$ which should ultimately converge to zero everywhere in $(0,L)$.  Nevertheless, as $t$ increases, I do see increasingly fast convergence. The following is the behavior of the solution for $L=\alpha=1$, $n=10$ terms for times between $t=0.01$ and $t=5$:

$n=20$ is identical.


问:
System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$
This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right.
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?

答:
Observe that your system is equivalent to $$\begin{cases}f(x)=y\\ f(y)=z\\f(z)=x\end{cases}$$ Where $$f(x):=3\cdot \left[\left(\frac{3}{6}\right)^x+\left(\frac{4}{6}\right)^x+\left(\frac{5}{6}\right)^x\right]$$ Let $g(x)=f(x)-x$. Adding up the three equations yields $\sum g(x)=0$, so at least one of $g(x), g(y), g(z)$ has to be $\leqslant 0$. It's easy to see that $g(x)$ is strictly decreasing and has its only root at $x=3$.

Case 1. $g(x)\leqslant 0\iff x\geqslant 3$. Since $f(x)$ is also strictly decreasing, this implies that $y=f(x)\leqslant f(3)=3$. Thus $y\leqslant 3\implies z=f(y)\geqslant f(3)=3$. Hence, $z\geqslant 3\implies x=f(z)\leqslant f(3)=3$. So $x\leqslant 3$. But this is absurd unless $x=y=z=3$.

Cases 2 and 3 are analogous.


问:
induction proof on bezout's identity $d=ax+by$
I am reading the proof of the following theorem from Apostol's Introduction to Analytic Number Theory.

Theorem 1.2 Given any two integers a and b, there is a common divisor $d$ of a and b of the form $d=ax+by$, where $x$ and $y$ are integers. Moreover, every common divisor of $a$ and $b$ divides this $d$.

I'm stuck on the first step. The author used induction on $n$, where $n=a+b$ and assumed that the theorem has been proved for $0,1,2, \cdots, n-1$.

I think the induction on $n$ should be an independent process where $n$ should be independent on other variables such as $a$ and $b$ here. For example, when we prove that binomial theorem:

$$(x+y)^n=\sum_{k=0}^n\binom{n}{k}x^{n-k}y^k=\sum_{k=0}^n\binom{n}{k}x^ky^{n-k}$$

We would use the induction on $n$, and assume this equality holds for $0,1,2, \cdots, n-1$. However, $a,b$ are all constants, not relevant to $n$.

What's wrong with my thought?

For reference, the following is the whole proof from the book:

Here is a possible relevant: link.

答:
We are being asked to prove a statement of the form

"$P(a, b)$ is true for all $(a, b)$ in $\{(0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), \dots\}$"

To prove this statement, we first chunk this set into smaller groupings based on the height $a + b$:
$$\{(0, 0), (1, 0), (0, 1), (2, 0), (1, 1), (0, 2), \dots\} = \{(0, 0)\} \cup \{(1, 0), (0, 1) \} \cup \{(2, 0), (1, 1), (0, 2) \} \cup \cdots.$$

Then, we argue that:

If $(a, b)$ is in the 0th chunk $\{(0, 0)\}$, then the proposition is true.

If $P(x)$ is true for all $(a, b)$ in the first $n$ chunks, then $P(x)$ is true for all $(a, b)$ in the $(n + 1)$th chunk.

By inducting on $n$, the number of chunks being considered, we can demonstrate that the proposition is true for $(a, b)$ coming from any chunk.

Does this address your questions?


问:
Why I got extra wrong solutions during solving the system of equations?
I was trying to solve this system of equations:

$${ \begin{cases}{x^4+y^4=1} \\ {x^3+y^3=1}\end{cases} }$$
I subtracted the second equation from the first one:$$x^4-x^3+y^4-y^3=0$$
$$x^3(x-1)+y^3(y-1)=0$$
We can easily check that $(x,y)=(1,1)$ and $(x,y)=(0,0)$ are roots of this equation. but if we plug in these values in the original system of equations we can see they are not the answer. So why this is happening?

答:
Your new equation is indeed valid, but it alone doesn't determine solutions to the original equations. You must still have a second companion equation.

For example, look at the first original equation. There are lots of solutions to that equation, but they aren't necessarily solutions to the second original equation as well.

Any solution of the original system must also satisfy your new equation. But only those solutions which satisfy both your new equation and a companion equation (whatever you use for the new second equation) will also be solutions of the original equations, provided the process of obtaining the two new equations is reversible (such as making linear combinations of the two old equations, which you did).


问:
Proving that $1+λ^2$ is an eigenvalue of $I+A^2$ where $λ$ is an eigenvalue of matrix $A$
So $Ax = λx$;

$A(Ax) = λ(Ax) \to (A^2)x = (λ^2)x$

I kind of dont know how to get the $1$ and $I$ here...

Any help is appreciated

答:
This is just a special case of a general statement that I wish people learned early and by heart (it is not that hard): if $v$ is an eigenvector of a square matrix $A$ with eigenvalue $\lambda$, then for every polynomial $P\in K[X]$, the polynomial $P[A]$ in $A$ acts on $v$ as multiplication by the scalar $P[\lambda]$ (or equivalently, $v$ is also an eigenvector of $P[A]$ with eigenvalue $P[\lambda]$). It applies here with $P=1+X^2$.

The proof of this property uses linearity in $P$ of $P[A](v)$, after the simple proof that $A^k(v)=\lambda^kv$ for all$~k\in\Bbb N$.


问:
Prove cardinality of a finite set is different than that of its power set
I was hoping someone could check my work and confirm whether it's a valid proof by induction.

For this problem we're allowed to use that $\left|P(A)\right|$ = $2^\left|A\right|$ for part of our proof

We want to prove $\left|P(A)\right|$ > $\left|A\right|$ for any finite set

That is that: $2^\left|A\right|$ > $\left|A\right|$

Base case $\left|A\right|$ = 1

$2^1$ > $1$, therefore this relation holds for $\left|A\right|$ = 1

Let us assume that the relation is true for the value $n = k$

That is that $2^k$ > $k$

For $2^{k+1} > k + 1$ we can rearrange this algebraically to get

2 > $({k + 1})^{1 / (k + 1)}$

lim k → ∞ $({k + 1})^{1 / (k + 1)} = 1$

As such, it is true that $2^{k+1} > k + 1$

Therefore $\left|P(A)\right|$ > $\left|A\right|$ and as such $\left|P(A)\right|$ != $\left|A\right|$

答: No, your proof's not valid. Yes, it's true that $\lim_{k\to\infty}(k+1)^{1/(k+1)}=1$, but this doesn't imply $(k+1)^{1/(k+1)}<2$ for all $k$. It's much easier to note $2^k>k\implies2^{k+1}=2\cdot2^k>2k\ge k+1$ provided $k\ge1$, giving an inductive step that works for $k\ge1$. We then just need $k=0,\,k=1$ to both be include in the base case.


问:
What is rigorous notation for functions?
I have seen many ways to denote a function: $f(x)=x^2, y=x^2, f: x\mapsto x^2$ and so on. What is exact notation for functions? Please include lethal doses of rigor, set theory, and of course notational exactness.

Note: I am very familiar with functions in general. I just know that a lot of mathematical literature abuses notation when it comes to functions.

答:
You should include the following:

1: The name of your function e.g. $f$

2: The domain and codomain (not the image!) of your function. e.g. $A \rightarrow B$

3: The rule of your function, e.g. $x \mapsto x^2$.

so this function would be:

$f:A \rightarrow B,\ x \mapsto x^2 $


答:
We have:

$$\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{y}{3} \tag{1}$$
$$\left( \frac{3}{6} \right)^y+\left( \frac{4}{6} \right)^y+\left( \frac{5}{6} \right)^y =\frac{z}{3}\tag{2}$$
$$\left( \frac{3}{6} \right)^z+\left( \frac{4}{6} \right)^z+\left( \frac{5}{6} \right)^z =\frac{x}{3}\tag{3}$$
Let's denote $f(t) =\left( \frac{3}{6} \right)^t+\left( \frac{4}{6} \right)^t+\left( \frac{5}{6} \right)^t  $, this function is decreasing in $t$ (you can prove it easily).
WLOG, suppose $x=\min{\{x,y,z\}}$. This implies that $x \le y$.

From (1) and (3), we have: $x \le y \implies f(x) \ge f(z) \implies x \le z$.

From (2) and (3), we have: $ x \le z \implies  f(z) \le f(y) \implies z \ge y$.

From (1) and (2), we have: $ y \le z \implies  f(x) \le f(y) \implies x \ge y$.

But we already suppose $x=\min{\{x,y,z\}}$ , hence, $x=y$. And from (1) and (2), we have $f(x) = f(y) \implies y = z$. So, $x=y=z$.

It suffices now to solve the equation
$$f(x)=\left( \frac{3}{6} \right)^x+\left( \frac{4}{6} \right)^x+\left( \frac{5}{6} \right)^x =\frac{x}{3}$$
We notice that $f(x)$ is decreasing function while $=\frac{x}{3}$ is an increasing function. Thus, there is at most 1 solution. And because $x=3$ is the solution, then $x=y=z=3$ is the unique solution of the system equation.

Title: System of equations: $3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y$

Body:
This was taken from a local mathematical olympiad in Romania. It is from the year 2003.
Solve for $x, y, z \in \mathbb{R}$:
$$
\left\{
\begin{array}{c}
3^x + 4^x + 5^x = 2^x \cdot 3^{x -1} \cdot y \\
3^y + 4^y + 5^y = 2^y \cdot 3^{y -1} \cdot z \\
3^z + 4^z + 5^z = 2^z \cdot 3^{z -1} \cdot x
\end{array}
\right.
$$
So, the first thing I notice was $4^x = 2^{2x} = a$ so we get a system of quadratic equations. This proved to be quite hard to deal with, so I gave it up. I also tried dividing by $2^x \cdot 3^{x - 1} \cdot y$ and the others, but it was weird to deal with. Do you have any ideas?


问:
Help me get my head around this simple linear regression problem
Given a very simple linear projection model $Y=X\beta+e$ with $E(e|X)=0$ and $X$ a scalar. Notice that this is a simple linear model with no intercept.

Then $\beta = E(XY)/E(X^2)$ from the least squares formula.

On the other hand, if I take expectation on both sides of the original model, I get $E(Y)=\beta E(X)+E(e)=\beta E(X)$. Therefore, $\beta=E(Y)/E(X)$. (Of course, suppose $E(X)\neq 0$.)

I just don't see how the two quantities could be identical. Did I do something wrong?

Body:
$E[f(X)Y]=E[f(X)X]\beta+E[f(X)e]=E[f(X)X]\beta$, so $\beta=\frac{E[f(X)Y]}{E[f(X)X]}=\frac{E[Y]}{E[X]}=E[\frac{Y}{X}]=E[\frac{f(X)}{E[f(X)]}\frac{Y}{X}]$ ...

Strange ...


问:
Show that matrix $A$ is NOT diagonalizable.
Let $A$ be a square matrix $A^2=0$ and $A\neq0$ and show that it is not diagonalizable.

I decided to use the sample matrix of $$A = \begin{bmatrix}0 & 1\\0 & 0 \end{bmatrix}$$ which satisfies the conditions above.

So my question is: how would I prove this is not diagonalizable. The matrix leads to a eigenvalue of $\lambda=0$ with an algebraic multiplicity of $2$.

I know that if the algebraic multiplicity and geometric multiplicity are equal, then it is diagonalizable.

But I am kind of stuck from here since when I use $\det(A-0I)=0$, it just leads to get $x_2=0$ but then also that $x_2=t$ so I don't really know what to do. Any help would be appreciated.

答: If $A^k=0$ (in your case with $k=2$), then any eigenvalue of $A$ satisfies $\lambda^k=0$ (check what happens to an eigenvector) and therefore $\lambda=0$. Then if $A$ is diagonalisable, the only possibility is that $A=0$.


问:
What are interesting properties of totally bounded uniform spaces?
I work on reloids, a generalization of uniform spaces.

As such I am interested about properties of totally bounded uniform spaces.

My question: What are interesting properties of totally bounded uniform spaces? (The more theorems/propositions you give, the better.)

答:
As far as I remember, these are equivalent for a uniform space $(X,\mathcal D)$:

$(X,\mathcal D)$ is totally bounded, that is, for each entourage $D\in \mathcal D$, there are $x_1,...,x_n\in X$ with $D[x_1]\cup...\cup D[x_n]=X$.

$(X,\mathcal D)$ is precompact, that is, it has a compact completion.

Every completion of $(X,\mathcal D)$ is compact.

Every net in $X$ has a Cauchy subnet.

Also every Tychonoff space has a compatible totally bounded uniformity.


问:
Expected average gain from implementing a policy via majority vote
Suppose a country with $n$ citizens decides whether to implement a new policy. The value of implementing that policy for each individual citizen is known only to them and is uniformly distributed on $[-1; 1]$ (it is positive, when they would gain from this policy being implemented and negative if they will lose). If the policy is not implemented nobody will either gain anything or loose anything. Whether to implement the policy is decided via a majority vote: the policy will be implemented if and only if at least half of the citizens votes for it.

A weakly dominant strategy for each citizen is to vote for the policy if their gain from it is positive and against it, if their gain from it is negative. What is the average expected gain of citizens if all of them follow this strategy?

The policy will be selected only if and only if more citizens gain from it than lose from it. Suppose $A$ is the set of all citizens with positive gain. Than conditional expectation for fixed $A$ will be $\frac{2|A|-n}{n 2^{n+1}}$. That means that total expected gain will be:

$\sum_{k=\lceil \frac{n}{2} \rceil }^n \frac{C_n^k (2k-n)}{n 2^{n+1}}$

It is quite obvious, that this value is bounded by 1 from above. It is also monotonously non-decreasing with growing $n$, because:

$\frac{C_{n+1}^k (2k-n-1)}{(n+1) 2^{n+2}} - \frac{C_{n}^k (2k-n)}{n 2^{n+1}} = \frac{C_{n+1}^k (2k-n-1)n - C_{n}^k (2k-n)(n+1)}{n (n+1) 2^{n+1}} = \frac{C_n^k}{{n 2^{n+1}}} (\frac{n(2k-1-n)}{n+1-k} - 2k - n) \geq 0$ when $k \geq \frac{n}{2}$

That means, it converges to some number on $[0;1]$. I would like to find that number, but I don't know how.

答:
Let's consider this for very large odd $n$.

The number of positive votes is binomially distributed with mean $\frac n2$ and variance $\frac{n}{4}$, and for large $n$ will be approximately normally distributed.

So conditioned on being a majority, i.e. exceeding $\frac n 2$, this will be approximately half-normally distributed, and the conditional mean will be about  $\frac{n}2+{\frac{\sqrt n}{\sqrt{2\pi}}}$.

The expected conditional gain for an individual majority voter is distributed on $[0,1]$ and so the conditional expected sum of their gains is $\frac{n}4+{\frac{\sqrt n}{2\sqrt{2\pi}}}$.

A similar analysis for the minority voters is an expected $\frac{n}2-{\frac{\sqrt n}{\sqrt{2\pi}}}$ number and a conditional expected net (negative) gain of about $-\frac{n}4+{\frac{\sqrt n}{2\sqrt{2\pi}}}$.

So overall, the conditional expected net gain is about ${\frac{\sqrt n}{\sqrt{2\pi}}}$

Half the proposals will have a positive decision

So the overall expected net gain is about ${\dfrac{\sqrt n}{2\sqrt{2\pi}}}$


答:
If you can use that $|\mathcal{P}(A)|=2^{|A|}$ things can become easier than in the proof you show. We must show that $2^{|A|}>|A|$, or what is equivalent, that for any $k\in\mathbb{N}\setminus\{0\}$, $2^{k}>k$ (where $|A|=k$; I do not know why you haven't considered the case $|A|=0$, but this one is quite simple, so I will ignore it).

Base case: $k=1$

In this case, $2^{k}=2^{1}=2$, which is greater than $k=1$.

Induction hypothesis: assume the result holds for some $j\in\mathbb{N}$ such that $j\geq 1$

Then, $2^{j+1}=2.2^{j}$, and since $2^{j}>j$ (by induction hypothesis), we obtain
$$2^{j+1}=2.2^{j}>2.j\geq j+1$$
(where $2.j\geq j+1$ since $j\geq 1$), meaning $2^{j+1}>j+1$. This ends the proof by induction that $|\mathcal{P}(A)|>|A|$ whenever $A$ is a finite, non-empty set.

Notice, however, that this is not the standard proof of such a result: one shows that $|\mathcal{P}(A)|\geq |A|$ by considering the injective function $f:A\rightarrow\mathcal{P}(A)$ taking an element $a\in A$ to the singleton $\{a\}$; one then shows that we cannot have $|A|\geq|\mathcal{P}(A)|$ (and therefore $|\mathcal{P}(A)|>|A|$) by a diagonal argument à la Cantor.

Title: Prove cardinality of a finite set is different than that of its power set

Body:
I was hoping someone could check my work and confirm whether it's a valid proof by induction.

For this problem we're allowed to use that $\left|P(A)\right|$ = $2^\left|A\right|$ for part of our proof

We want to prove $\left|P(A)\right|$ > $\left|A\right|$ for any finite set

That is that: $2^\left|A\right|$ > $\left|A\right|$

Base case $\left|A\right|$ = 1

$2^1$ > $1$, therefore this relation holds for $\left|A\right|$ = 1

Let us assume that the relation is true for the value $n = k$

That is that $2^k$ > $k$

For $2^{k+1} > k + 1$ we can rearrange this algebraically to get

2 > $({k + 1})^{1 / (k + 1)}$

lim k → ∞ $({k + 1})^{1 / (k + 1)} = 1$

As such, it is true that $2^{k+1} > k + 1$

Therefore $\left|P(A)\right|$ > $\left|A\right|$ and as such $\left|P(A)\right|$ != $\left|A\right|$


问:
Convergence of $\sum_{k = 1}^{\infty} \cos (\log k)$ and $\sum_{k = 1}^{\infty} \sin (\log k)$
I was trying to determine if

$$\sum_{k = 1}^{\infty} \cos (\log k)$$

(with $k = 1, 2, \ldots$) converges. $\log$ is the natural logarithm.

I tried the ratio test,

$$\lim_{k \to \infty} \frac{\cos \left[ \log ( k + 1 ) \right]}{\cos \left[ \log ( k ) \right]} = 1$$

but it is not useful to determine the convergence.

According to Wolfram Alpha, it diverges. How to prove it?

And what about $\displaystyle \sum_{k = 1}^{\infty} \sin (\log k)$?

What I can empirically observe is that, regardless of how big $\log k$ can become, its cosine will always be a number $x$ such that $-1 \leq x \leq 1$. But I can't immediately see that this leads to an infinite sum.

Trivially, the term test is:

$$\lim_{k \to \infty} \cos (\log k)$$

and this limit is not defined, due to the cosine. Maybe this is sufficient to state that the series diverges.

答:
Suppose the series converges
$$
\sum_{k=1}^\infty \cos(\log k) = s \in \Bbb R
$$
and let $s_n = \sum_{k=1}^n \cos(\log k)$.

This would imply
$$
\cos(\log(n)) = s_{n} - s_{n-1} \to s-s = 0
$$
by arithmetic of limits, but $\lim_{n\to\infty} \cos(\log n)$ doesn't exist.

The argument that $\sum \sin(\log k)$ diverges should be similar.


问:
Prove $t^{1\over t-1}$ is rational only when $t=2$ in the context of solving $x^y=y^x$ where $x\neq y$
Prove $t^{1\over t-1}$ is rational only when $t=2$ where $t$ is any real number.

This problem arises from me watching the blackpenredpen video solving $x^y=y^x$ where $x\neq y$. https://www.youtube.com/watch?v=PI1NeGtJo7s, in which he gives the solution $x=t^{1\over t-1}$.

First I considered the integer solutions where $x\neq y$ other than $2^4=4^2$ but realized it is trivial because $1<t^{1\over t-1}<2$ for any $t\geq 2$ and $-1>t^{1\over t-1}>-2$ for any $t\leq-2$.

The next question is then, are there any rational solutions other than the $2,4$ pair. It does not seem trivial to me as inequalities do not help.

My guess is that $t^{1\over t-1}$ cannot be rational other than when $t=2$ but I wan't able to prove.

答:
The range of $f(t)=t^{1/(t-1)}$ is $(1,\infty)$ and the function is continuous and monotonic, hence invertible. So there is an infinite (but countable) number of solutions of

$$t^{1/(t-1)}=q$$ where $q$ is rational.

$$t=-\frac1{\log q}W\left(-\frac{\log q}q\right).$$


问:
What does $B \to O$ mean in Peter Freyd's book Abelian Categories
I am trying to read “Abelian Categories” of Peter J. Freyd. I was reading proposition 2.12 on page 37. The author states that if $A \to B$ is an epimorphism, then $B \to O$ is its cokernel. My problem is, that I cannot find what $B \to O$ means. On page 26, he defined the zero object $O$ (“an object with precisely one map to and from each object”). Furthermore, he defines the zero map $B \overset {0} \to A$ to be the unique map $B \to O \to A$, but he doesnot mention $B \to O$ (and dually $O \to A$).  So my question is: how is $B \to O$ defined and why is it the cokernel of the epimorphism $A \to B$ ?
Can anyone help me with this ?

答:
$B \to O$ is simply the unique map from $B$ to $O$ as given in the definition of a zero object which you give ("an object with precisely one map to and from each object"). So in particular there is precisely one map from $B$ to $O$ and that is what Freyd means by $B \to O$. Same thing for $O \to A$, that's the unique map from $O$ to $A$ guaranteed by the definition of $O$ as a zero object.
By the dual of the definition of kernel on p26-7, a map $B \overset {x} \to C$ is the cokernel of a given map $A \overset {f} \to B$ if and only if $xf=0$, and for any map $B \overset {y} \to D$ for which $yf=0$ there is a unique map $C \overset {z} \to D$ for which $zx=y$. So to show that the zero map $B \overset {0} \to O$ is the cokernel of a given epimorphism $A \overset {f} \to B$, I have to show that $0f=0$, and that for any map $B \overset {y} \to D$ for which $yf=0$ there is a unique map $O \overset {z} \to D$  for which  $z0=y$.  As $f$ is an epimorphism and $yf=0f=0$, it follows that $y$ must be a zero map, so we can take $z$ to be the zero map from $O$ to $D$.  Uniqueness of this map follows since it is the only map from $O$ to $D$.


问:
If $m(A^\circ)=m(\overline A)$ then $m(\overline A\setminus A^\circ)=0$?
Let $A$ be a subset of $\mathbb R^n$ and let $A^\circ$ and $\overline A$ denote the interior and the closure of $A$ respectively. Let $m$ be the Lebesgue measure on $\mathbb R$.

Is it true that if $m(A^\circ)=m(\overline A)$ then $m(\overline A\setminus A^\circ)=0$?

I can show this easily if $m(\overline A)$ is finite: $m(\overline A)=m\Big((\overline A\setminus A^\circ)\cup A^\circ\Big)=m(\overline A\setminus A^\circ)+m(A^\circ)$. Now since $m(\overline A)$ is finite, I can subtract on both sides and I have the required result.

But what about when $m(\overline A)=\infty$?

答: Then it is false. Take, say, $A=[0,\infty)\cup\bigl((-\infty,0)\cap\Bbb Q\bigr)$. Then $\overline A=\Bbb R$ and $\mathring A=(0,\infty)$. So,$$m\left(\overline A\setminus\mathring A\right)=m\bigl((-\infty,0]\bigr)=\infty.$$


问:
Random time vs stopping time & expectation
Let $B_t$ be a standard Brownian motion. Let us introduce a stopping time $\tau$ as
$$\tau \ = \ \ \inf\Big\{t:\ |B_t|=1\Big\}.$$
Now, what I am interested in are random variables $\tau_x$ defined as
$$\tau_x(\omega) \ = \ x\cdot\tau(\omega),$$
for $x> 0$. Obviously, $\tau_x$ is not a stopping time anymore. Nevertheless, we can still
consider a random variable $$B_{\tau_x}(\omega) \ = \ B_{\tau_x(\omega)}(\omega),$$
and it is (if I understand correctly) still well defined, i.e. measurable - as claimed in

The strong Markov property with an uncountable index set.

My question is, are basic quantities such as
$$\mathbb{E}\Big(B_{\tau_\frac{1}{2}}\Big|B_{\tau}=1\Big)$$
computable? How can we calculate it? My crude intuition is, that
$$\mathbb{E}\Big(B_{\tau_x}\Big|B_{\tau}=1\Big) \ = \ x,$$
as $\Delta:=B_{\tau}-B_{0}=1$ and on average we should make $x$ of it in the first $x$ part of random $\tau$ time.

I would be glad for any help and insight.

答:
The following R-code serves to compute the mean of $B_{\tau/2}$ conditionally on $B_\tau = 1$. Paths of Brownian motions on $[0,1]$ are simulated via independent Gaussian increments. For the subsample where $\tau \leq 1$ and $B_{\tau} = 1$, the values $B_{\tau/2}$ are computed, if necessary via linear interpolation. The mean is then approximated by taking the average of these values.

(Intuitively, conditioning on $\tau \leq 1$, shouldn't be too problematic; but a rigorous argument escapes me.)

I find that $\mathbb{E}[B_{\tau/2} \, | \, B_\tau = 1] \approx 0.26$. Replacing $\tau$ by $\bar{\tau} := \inf\{t > 0 : B_t = 1\}$, one also finds $\approx 0.26$. I find these numerical results to be somewhat consistent with an earlier comment of mine.

```<code>set.seed(2)

N <- 5000
n <- 5000

h <- 1/n

X <- apply(rbind(rep(0,N),matrix(rnorm(n*N, mean = 0, sd = sqrt(h)), nrow = n, ncol = N)), 2, cumsum)

tau <- apply(X, 2, function(y){Position(function(x){x>=1 | x<=-1}, y)})

B_tau <- vector()
B_tau_half <- vector()
for (i in 1:N){
  if(!is.na(tau[i]) & X[tau[i],i] >= 1){
    B_tau <- c(B_tau,X[tau[i],i])
    if(tau[i] %% 2 == 0){
      B_tau_half <- c(B_tau_half,X[tau[i]/2,i])
    }
    else{
      B_tau_half <- c(B_tau_half,X[(tau[i]-1)/2,i]/2 + X[(tau[i]+1)/2,i]/2)
    }
  }
}

mean(B_tau_half)
</code>```

Body:
The following R-code serves to compute the mean of $B_{\tau/2}$ conditionally on $B_\tau = 1$. Paths of Brownian motions on $[0,1]$ are simulated via independent Gaussian increments. For the subsample where $\tau \leq 1$ and $B_{\tau} = 1$, the values $B_{\tau/2}$ are computed, if necessary via linear interpolation. The mean is then approximated by taking the average of these values.

(Intuitively, conditioning on $\tau \leq 1$, shouldn't be too problematic; but a rigorous argument escapes me.)

I find that $\mathbb{E}[B_{\tau/2} \, | \, B_\tau = 1] \approx 0.26$. Replacing $\tau$ by $\bar{\tau} := \inf\{t > 0 : B_t = 1\}$, one also finds $\approx 0.26$. I find these numerical results to be somewhat consistent with an earlier comment of mine.

set.seed(2)

N <- 5000
n <- 5000

h <- 1/n

X <- apply(rbind(rep(0,N),matrix(rnorm(n*N, mean = 0, sd = sqrt(h)), nrow = n, ncol = N)), 2, cumsum)

tau <- apply(X, 2, function(y){Position(function(x){x>=1 | x<=-1}, y)})

B_tau <- vector()
B_tau_half <- vector()
for (i in 1:N){
  if(!is.na(tau[i]) & X[tau[i],i] >= 1){
    B_tau <- c(B_tau,X[tau[i],i])
    if(tau[i] %% 2 == 0){
      B_tau_half <- c(B_tau_half,X[tau[i]/2,i])
    }
    else{
      B_tau_half <- c(B_tau_half,X[(tau[i]-1)/2,i]/2 + X[(tau[i]+1)/2,i]/2)
    }
  }
}

mean(B_tau_half)


问:
Classic geometry problem with algebra

Given right $\triangle ABC$ such that $\angle ACB=90^\circ$ with height $CH$. If the length of all three side of the triangle are integer and $BH=29^3$ then find $\cos \angle ABC$

I have an incomplete solution that using Wolfram alpha to help. First let $AB=c$, $BC=a$, $AC=b$ so we have $a^2=29^3c$. That mean I could pick $a=29^2n$, $c=29n^2$ for some integer $n$ but I can't solve for anything useful at all. But as for $b^2=(c-29^3)c$, I could get $a=29^2×421$, $b=60×29×421$ $c=29×421^2$ ( by Wolfram alpha ). So then I could solve for $\cos \angle ABC=\frac{29}{421}$. I wonder if there is any other approach to this without involving calculator or heavy algebra at all. And if you know the good way to solve $b^2=(c-29^3)c$ please let me know too ( sorry for asking two questions ). Thanks!

答:
You can get something useful from letting $a=29^2 n$ and $c= 29n^2$. We have:

$$a^2 + b^2 = c^2 \leadsto 29^4 n^2 + b^2 = 29^2n^4$$

Since $b$ is an integer, it must be of the form $29nk$. Then we have:

$$29^4 n^2 + 29^2 n^2 k^2 = 29^2n^4 \leadsto 29^2 + k^2 = n^2$$

and so $(29,k,n)$ forms a Pythagorean triple. We know that Pythagorean triples are of the form:

$$(p^2-q^2, 2pq, p^2+q^2)$$

so we have $29 = p^2-q^2 = (p+q)(p-q)$. Since $29$ is prime, $p+q=29$ and $p-q=1$. Thus $$p=15, q=14, k = 2pq = 420, n = p^2 + q^2 = 421$$

Therefore $$\cos\angle ABC = \frac ac = \frac {29^2 n }{29n^2} = \frac{29}n = \frac {29}{421}$$

Title: Classic geometry problem with algebra


答: Hint: For $n$ even, the sum is exactly $\binom{n}{n/2}2^{-n-2} \approx (8 \pi n)^{-1/2} $

Title: Expected average gain from implementing a policy via majority vote

Body:
Suppose a country with $n$ citizens decides whether to implement a new policy. The value of implementing that policy for each individual citizen is known only to them and is uniformly distributed on $[-1; 1]$ (it is positive, when they would gain from this policy being implemented and negative if they will lose). If the policy is not implemented nobody will either gain anything or loose anything. Whether to implement the policy is decided via a majority vote: the policy will be implemented if and only if at least half of the citizens votes for it.

A weakly dominant strategy for each citizen is to vote for the policy if their gain from it is positive and against it, if their gain from it is negative. What is the average expected gain of citizens if all of them follow this strategy?

The policy will be selected only if and only if more citizens gain from it than lose from it. Suppose $A$ is the set of all citizens with positive gain. Than conditional expectation for fixed $A$ will be $\frac{2|A|-n}{n 2^{n+1}}$. That means that total expected gain will be:

$\sum_{k=\lceil \frac{n}{2} \rceil }^n \frac{C_n^k (2k-n)}{n 2^{n+1}}$

It is quite obvious, that this value is bounded by 1 from above. It is also monotonously non-decreasing with growing $n$, because:

$\frac{C_{n+1}^k (2k-n-1)}{(n+1) 2^{n+2}} - \frac{C_{n}^k (2k-n)}{n 2^{n+1}} = \frac{C_{n+1}^k (2k-n-1)n - C_{n}^k (2k-n)(n+1)}{n (n+1) 2^{n+1}} = \frac{C_n^k}{{n 2^{n+1}}} (\frac{n(2k-1-n)}{n+1-k} - 2k - n) \geq 0$ when $k \geq \frac{n}{2}$

That means, it converges to some number on $[0;1]$. I would like to find that number, but I don't know how.


问:
Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

Show that $S\cap(T\,\cup \,T') = (S\,\cap\,T)\cup(S\,\cap\,T')$

I did this by creating 3 sets:

$S\,\,=\{a,b,c,d\}$

$T\,\,=\{c,d,e,f\}$

$T'=\{a,b,c,d\}$

And then I performed the operations and checked they are the same. Is there something else I should do or only this will suffice?

答: HINT: is it possible to write the left hand side otherwise? Maybe draw it first, so you have a good idea.</p>


问:
Finding all cluster points of a sequence and proving it rigorously
This is a problem from Kenneth Brown's Intro to Real Analysis:

Find all the cluster points of the sequence:
$$x_n = \cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{2}\right) + 2^{-n} + (-1)^n$$

I know that as n gets large, $2^{-n}$, goes to zero, but from here, how do you reconcile the different intervals for the trig functions and the oscillating negative one?

I need to find all cluster points and prove them. Thanks!

答:
First, you can show that if $(v_n)_{n\in\mathbb{N}}$ is a sequence such that $$\lim\limits_{n\to+\infty}v_n=0,$$ then the cluster points of $(u_n+v_n)_{n\in\mathbb{N}}$ and $(u_n)_{n\in\mathbb{N}}$ are the same. I will give the proof of one inclusion: if $a\in \mathbb{R}$ is a cluster point for $(u_n)_{n\in\mathbb{N}}$, then by definition for all $\varepsilon'>0$ and for all $N'\in\mathbb{N}$, it exists $n\in\mathbb{N}$ such that $n\geq N'$ and $|u_{n}-a|<\varepsilon'$. Now, for $\varepsilon>0$ and $N\in\mathbb{N}$, take $N''\geq N$ such that $n\geq N''\implies|v_n|<\frac{\varepsilon}{2}$ (which holds by hypothesis on $(v_n)_{n\in\mathbb{N}}$), then take $n$ given by the previous property with $(\varepsilon',N')=(\frac{\varepsilon}{2},N'')$. Thus, you have a $n$ greater than $N$ such that
$$|u_n+v_n-a|\leq |u_n-a|+|v_n|<\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon,$$
which shows that $a$ is a cluster point of $(u_n+v_n)_{n\in\mathbb{N}}$.

So we only have to find the cluster points of the sequence defined by $y_n=\cos\left(\frac{n\pi}{4}\right) + \sin\left(\frac{n\pi}{2}\right) + (-1)^n$. But the value of $y_n$ only depends of the remainder of the division of $n$ by $8$, so this sequence is $8$-periodic, and the cluster points are $\{y_0,y_1,\dots,y_7\}$. You finally juste have to compute these eight values to get the cluster points.


问:
Is this ODE a well-known and studied one?
I came across this ordinary differential equation
$$x \,(y')^2+a=0\,,$$
where $a$ is a real number, $x$ is a real variable and $y$ is a real algebraic function of $x$. I was wondering whether this is a well-known and studied ODE or not.

答: If you make the change of variable $x = t^4$, $y(x) = u(t)$, you find $u$ satisfies Chrystal's equation.


答:
$$\log \lfloor e^{2\pi k} \rfloor = \log( e^{2\pi k}+O(1))=
\log e^{2\pi k} +\log(1+ O(e^{-2\pi k}))=2\pi k+O(e^{-2\pi k})$$

Thus $$\lim_{k\to \infty}\cos(\log \lfloor e^{2\pi k} \rfloor )=1$$

Similarly $$\lim_{k\to \infty}\sin(\log \lfloor e^{2\pi (k+1/4)} \rfloor )=1$$

Title: Convergence of $\sum_{k = 1}^{\infty} \cos (\log k)$ and $\sum_{k = 1}^{\infty} \sin (\log k)$

Body:
I was trying to determine if

$$\sum_{k = 1}^{\infty} \cos (\log k)$$

(with $k = 1, 2, \ldots$) converges. $\log$ is the natural logarithm.

I tried the ratio test,

$$\lim_{k \to \infty} \frac{\cos \left[ \log ( k + 1 ) \right]}{\cos \left[ \log ( k ) \right]} = 1$$

but it is not useful to determine the convergence.

According to Wolfram Alpha, it diverges. How to prove it?

And what about $\displaystyle \sum_{k = 1}^{\infty} \sin (\log k)$?

What I can empirically observe is that, regardless of how big $\log k$ can become, its cosine will always be a number $x$ such that $-1 \leq x \leq 1$. But I can't immediately see that this leads to an infinite sum.

Trivially, the term test is:

$$\lim_{k \to \infty} \cos (\log k)$$

and this limit is not defined, due to the cosine. Maybe this is sufficient to state that the series diverges.