Title: Find the sum of the series $1+(\frac 12 +\frac 13)\frac 14 +(\frac 14 +\frac 15)\frac{1}{4^2}.....$ upto $\infty$

Body:
$$S = \frac 12 .\frac 14 + \frac 14.\frac{1}{4^2} + \frac 16 . \frac{1}{4^3} ....... + \frac 13. \frac 14 + \frac 15 . \frac{1}{4^2}.....$$

The first part of the series is a standard logarithmic series shoes sum is $\frac {-1}{2} (\log \frac 32 +\log \frac 12)$, but I don’t not know how to deal with the second part, which I also believe is a logarithmic expansion

Body:
Hint. What is the series for $\frac1{2x}(\log(1+x)-\log(1-x))$?

$$S'=1+\frac13\cdot\frac1{2^2}+\frac15\cdot\frac1{2^4}...$$and$$\frac1{2x}(\log(1+x)-\log(1-x))=1+x^2/3+x^4/5...$$Now keep $x=1/2$.


问:
Fixed points and iterates of an invertible function
Suppose that $g : [0,1] \rightarrow [0,1]$ is a continuous and strictly increasing function such that $g(0)=0$ and $g(1)=1$. Under these hypotheses $g(x)$ has an inverse function $g^{-1} :[0,1] \to [0,1]$ such that $g^{-1}(g(x)) =x$ and $g(g^{-1}(x)) =x$ for all $x \in[0,1]$. How do I show that the fixed point of $g(x)$ is also a fixed point of $g^{-1}(x)$. Also would $g^{-1}(x)$ have the same iterates of $g(x)$?</p>

答: Suppose $a$ is a fixed point of $g(x)$.  Then $g(a)=a$.  Applying $g^{-1}$ to both sides we get $g^{-1}(g(a))=g^{-1}(a)$, but the LHS is $a$, so $a$ is a fixed point of $g^{-1}(x)$.</p>

ContentLicense: CC BY-SA 3.0


问:
Research question - prove inequality for $n \log n$.
The following inequality has come up in some research and appears true when plotted, but I'm not sure how to to prove it. If $g(z) = z \log z$, then for $x>3$ ($x$ real),
$$
g(x) < g(x-1)\cdot \left(1 + \frac{g(x-1) - g(x-2)}{g(x-1)-1} \right).
$$

答:
I believe the following outline works:

It suffices to show that the function $\dfrac{g(x-1)-g(x-2)}{g(x)-g(x-1)} g(x-1)-g(x-1)+1$ is positive for $x>3$ (since then one can multiply through by $\dfrac{g(x)-g(x-1)}{g(x-1)-1}$ which is positive).

The generalized mean value theorem tells us that if $f$ and $g$ are differentiable, then
$$
\frac{f(x)-f(x-1)}{g(x)-g(x-1)} = \frac{f'(\xi)}{g'(\xi)}
$$
for some $\xi\in(x-1,x)$. Applying this with $f(x) = g(x-1)$ gives
$$
\frac{g(x-1)-g(x-2)}{g(x)-g(x-1)} = \frac{g'(\xi-1)}{g'(\xi)} = \frac{1+\log(\xi-1)}{1+\log\xi}
$$
for some $\xi\in(x-1,x)$. And since $\dfrac{1+\log(\xi-1)}{1+\log\xi}$ is increasing, we can replace it by its lower bound $\dfrac{1+\log(x-2)}{1+\log(x-1)}$.

We have thus shown that it suffices to show that $\dfrac{1+\log(x-2)}{1+\log(x-1)} g(x-1)-g(x-1)+1$ is positive for $x>3$, or equivalently that $1 + \log(x-1) - g(x-1) \big(\!\log(x-1) - \log(x-2)\big)$ is positive there (since we can then divide by $1+\log(x-1)$ which is positive).

By the ordinary mean value theorem, $\log(x-1) - \log(x-2) = \dfrac1{\eta}$ for some $\eta\in(x-2,x-1)$, and in particular $\log(x-1) - \log(x-2) < \dfrac1{x-2}$.

It therefore suffices to show that $1 + \log(x-1) - \dfrac{g(x-1)}{x-2} > 0$ is positive for $x>3$, or equivalently that $x - 2 - \log(x-1)$ is positive there (since we can then divide by $x-2$ which is positive).

But finally this is a simple calculus calculation, since $x-2-\log(x-1)=0$ when $x=2$ and $\dfrac d{dx}(x-2-\log(x-1)) = 1-\dfrac1{x-1}>0$ for $x>2$.

ContentLicense: CC BY-SA 4.0


问:
How to compare two variables whose differential entropy are both negative?
I know that a higher positive value for entropy indicates greater uncertainty, but not sure how this works when comparing two negative values

If two continuous random variables $X$ and $Y$ have differential
entropies of $-1.3$ and $-0.6$ respectively, which one has more
uncertainty/disorder?

Also, given that the differential entropy of the uniform
distribution $\sim U(0,\frac{1}{2})$  is well-known to be $$-\log(2)
= -0.693147$$ and for the Gaussian $\sim N(0,1)$ it's $$\frac{1}{2}\log(2\pi e \sigma^2) = 1.42$$, what can be said about the previous two values compared to the uniform entropy and Gaussian entropy?

(Differential entropy here should not be confused with discrete Shannon entropy, which is always non-negative.) The differential entropy article does not explain how to compare two negative values.

答:
Chapter 8 of Cover and Thomas' book treats differential entropy. One of the main points is the following:

While the differential entropy $h(X)$ can be negative, the quantity $2^{h(X)}$ is the volume of the support set, which is always nonnegative.

In parallel with the discrete theory, there is an AEP property, so by the weak law of large numbers the sample entropy converges to $h(X)$ and we have, in particular for $X=(X_1,\ldots,X_n)$ an i.i.d. vector
$$
\mathrm{Vol}(A_{\varepsilon}^{(n)})\leq 2^{n(h(X)+\varepsilon)}, \quad \forall n\geq 1,
$$
where the typical set $A_{\varepsilon}^{(n)}$ is the subset of $\mathbb{R}^n$ where the sample entropy per symbol and the source entropy differ by at most $\varepsilon.$

So to answer your question, since $h(X)=-1.3,$ and $h(Y)=-0.6$, $X$ is supported on a set of volume $2^{-1.3}$ which is smaller than the volume supporting $Y$, namely $2^{-0.6}.$ Thus as in the discrete case
$h(X)<h(Y)$ implies that $Y$ has higher uncertainty than $X.$


Title: Fixed points and iterates of an invertible function

Body: Suppose that $g : [0,1] \rightarrow [0,1]$ is a continuous and strictly increasing function such that $g(0)=0$ and $g(1)=1$. Under these hypotheses $g(x)$ has an inverse function $g^{-1} :[0,1] \to [0,1]$ such that $g^{-1}(g(x)) =x$ and $g(g^{-1}(x)) =x$ for all $x \in[0,1]$. How do I show that the fixed point of $g(x)$ is also a fixed point of $g^{-1}(x)$. Also would $g^{-1}(x)$ have the same iterates of $g(x)$?</p>

Body: If $g(x)=x$, then $g^{-1}(x)=g^{-1}(g(x))=x$. If $g^{-1}(x)=x$ then $g(x)=g(g^{-1}(x))=x$.


问:
Not understanding how to solve this question?

If $z=x+iy$ lies in the third quadrant, then prove that $\frac{\bar{z}}{z}$ also lies in the third quadrant when $y<x<0$.

My attempt:  I got equation $$\frac{x^2 - y^2 - 2ixy}{x^2+y^2}$$

Now , I got $x^2 - y^2<0 $ and $-2xy<0$ , $x^2 < y ^2$. Therefore $y$ is greater than $x$. I got $y<x$ this I did not get.

答:
Using Euler's identity, we have that for $z = re^{i \theta}$:

$$\frac{\bar{z}}{z} = \frac{re^{-i \theta}}{re^{i \theta}} = e^{-2 i \theta} = \cos(-2 \theta)+i\sin(-2\theta).$$

$z$ lies in the third quadrant, hence $\pi < \theta < 3 \pi/2$. $y < x < 0$ restricts this to $5 \pi / 4 < \theta < 3 \pi/2$ (sketch this).

This means $-3 \pi < -2\theta < -5 \pi/2$ (do each inequality separately), or adding $4 \pi$ to both sides, $\pi < -2 \theta < \frac{3 \pi}{2}$. Hence $\bar{z}/z$ also lies in the third quadrant.


问:
Expectation of product of combination of random variables
Let's say $X$, $Y$ and $Z$ are three independent and identical random variables.

Let $V = 2X-Y$ and $W= 2X-2Z+5$.

I have to find covariance of $V$ and $W$.

I can find $E(V)$ and $E(W)$, but I am confused on how to find the expectation of $V*W$, which will be $E[(2X-Y)(2X-2Z+5)]$.

Can we apply basic algebraic rules here and multiply all variables by opening the brackets or is there any other way?

答:
Rather than calculating $\operatorname{E}[VW]$, you can directly compute the covariance using the identity
$$\operatorname{Cov}[aX + bY, cV + dW] = ac \operatorname{Cov}[X,V] + ad \operatorname{Cov}[X,W] + bc \operatorname{Cov}[Y,V] + bd \operatorname{Cov}[Y,W],$$  where $a, b, c, d$ are scalar constants and $X, Y, V, W$ are random variables.  In other words, the covariance operator behaves like multiplication.
The other property of covariance is that it is unchanged under location translations of random variables; i.e., $$\operatorname{Cov}[X + a, Y + b] = \operatorname{Cov}[X,Y],$$ for scalar constants $a, b$.  This should be familiar if you already know that $$\operatorname{Var}[X + a] = \operatorname{Var}[X].$$

Then with these properties in mind, we have $$\begin{align}
\operatorname{Cov}[2X - Y, 2X - 2Z + 5] &= \operatorname{Cov}[2X - Y, 2X - 2Z] \\
&= 4 \operatorname{Cov}[X, X] - 4 \operatorname{Cov}[X, Z] - 2 \operatorname{Cov}[Y, X] + 2 \operatorname{Cov}[Y, Z]. \end{align}$$
Finally, use the independence of $X, Y, Z$ to simplify, and recall $\operatorname{Cov}[X,X] = \operatorname{Var}[X]$.


问:
Why is this not a equivalence relation in the $\mathbb{Z}$ set?

The relation $xRy$ on the set $\mathbb{Z}$ if $x + y$ is divisible by $3$.

$xRx = x + x = 2x$. In order to $2x$ be divisible by $3$, $x$ itself need to be divisible by $3$, it is not a problem is it?

$xRy = yRx$

$xRy = yRz = xRz$

I think the last two are immediate, so probably my error is in the first.

答: $xRx$ is not true for $x=1$. Therefore, the relation is not reflexive.


问:
Show specific properties of function
Let $f : (0, 1) \to \mathbb{R}$ be defined by $f(x) =  0$ if $x$ is irrational and $f(x) = 1/n$ if $x = \frac{m}{n}$ $m,n \in \mathbb{Z}_{+}$ with $g.c.d(m,n) = 1$.

Then show that
a) $f$ is discontinuous at all rational points

b) for any $\epsilon > 0$ the set $E = \{x \in (0,1) : f(x) \geq \epsilon\}$ is finite.

c) Let $E$ be as above. Show that if $a$ is irrational then $inf\{|a -x| : x \in E\} > 0$

d) $f$ is continuous at all irrational points.

I have done parts a and b but couldn’t figure out how to do c and d can anyone provide a proof of these parts please

答:
a) $\forall x \in Q \cap (0,1), \exists \epsilon>0, \forall \delta_\epsilon>0$ such that $|x-x_0|<\delta_\epsilon$ then $|f(x)-f(x_0)|\ge\epsilon$

Let $x=m/n$, then $f(x)=1/n$, $\epsilon = 1/(n-1)$, $x_0 \in (x-\delta,x+\delta) \cap (0,1) \cap R/Q$, $f(x_0)=0$

Then $|f(x)-f(x_0)|=|1/n-0|\ge1/(n-1)=\epsilon$

b) $\forall \epsilon > 0$ the set $E = \{x \in (0,1) : f(x) \geq \epsilon\}$ is finite.

There is a property of the natural number (Peano) such that $\forall \epsilon > 0$, $\exists n \in N$ such that $n \ge 1/\epsilon > n-1$

you just have to notice that for such n we have at most:

0/1, 1/1

0/2, 1/2, 2/2

0/3, 1/3, 2/3, 3/3

...

0/n, 1/n, ..., n/n

(n^2+n-2)/2 numbers, which is finite (some are repeated 1/2 = 2/4)

c) You just have to prove that there is a Cauchy succession such that $\lim_{n\to\infty} x_n-x_{n+m} = 0$ and $\lim_{n\to\infty} x_n = a$

d) For a sufficiently small $\delta$ you can avoid all the rationals whose demoninator is smaller than some n (since the number of rationals with the denominator smaller than same $n \in N$ is finite as proven in b)

Title: Show specific properties of function


问:
Detailed proof that no essential singularity at infinity implies polynomial

Suppose $f(z)$ is holomorphic in the whole plane, and that $f(z)$ does not have an essential singularity at $\infty$. Prove that $f(z)$ is a polynomial.

I've tried following the hint given in this question. Since $f(z)$ has a nonessential singularity at $\infty$, so $g(z)=f(1/z)$ has a nonessential singularity at $0$. There are two cases:

1) $g(z)$ has a removable singularity at $0$. This means $\lim_{z\rightarrow 0}zg(z)=0$.

2) $g(z)$ has a pole at $0$. This means $g(z)=h(z)/z^k$ for some analytic function $h(z)$ such that $h(0)\neq 0$.

How can I finish each of these cases?

答:
Since we have a non-essential singularity at $\infty$ we have an $h$ such that $z^{-h}f(z)$ has a limit as $z$ tends to $\infty$ that is neither $0$ nor $\infty$. This means that $f$ cannot have more than $h$ zeroes at $0$ because otherwise $f(z) = z^{h+k}f_k(z)$ and hence limit of $z^{-h}f(z)$ would not be finite. This also means that $f^{(h+1)}(0) \neq 0$ (the $h+1$st derivative is not $0$).
Now
$$f(z) = f(0) + f'(0)z + ... + z^{h+1}g(z)$$
where $g(z)$ is analytic in all of complex plane (including $0$). Hence $g(0)$
is finite and $\neq 0$ (because $g(0) = f^{h+1}(0)$)
Also observe that we can write:
$$z^{-h}f(z) = z^{-h}(f(0) + f'(0)z + ... + z^{h+1}g(z))$$

Now since $\lim_{z \to \infty} z^{-h}f(z)$ exists we see that
$\lim_{z \to \infty} zg(z)$ also exists.

Also observe that $g(z)$ cannot be $0$ at infinity because otherwise $g(1/t) = t^{k}g_2(t) \implies g(z) = \frac{g_2(1/z)}{z^{k}}$ and would mean $g(0)$ is not defined.

All this implies that $g(z) = 0$


问:
Method to partial fractions.
For Example:

$$\frac{ax^2 + bx+c}{(dx+e)(fx^2+g)}\equiv\frac{A}{dx+e}+\frac{Bx+C}{fx^2+g}$$

and

$$\frac{ax^4 + bx^3+cx^2+dx+e}{(x+f)(x^2+g)}\equiv Ax+B+\frac{C}{x+f}+\frac{Dx+E}{x^2+g}$$

How do you know how to format the right hand side, in the equation below in partial fractions? (what is the 'pattern'?)

$$\frac{2ax}{(x-2a)(x^2+a^2)}$$
where a is a non zero constant

Could you look at it as the top of the fraction being a simple derivative of the bottom? If so how would you know to add the Ax+B in the latter example?

答:
Step 1: Look at the orders of the numerator and the denominator. For instance, in the second example, the order of the numerator is 4 ($x^4$ is the largest power) and the order of the denominator is 3 ($x\cdot x^2=x^3$ is the largest power). If $n\geq m$, where $n$ is the order of the numerator and $m$ is the order of the denominator, then you need to add linear terms up to order $n-m$. In this case, up to order $1$, so you have
$$
Ax+B+\frac??+\frac??
$$

Step 2: Separate out the factors in the denominator into separate fractions. So in your question case, you have
$$
\frac{2ax}{(x-2a)(x^2+a^2)} = \frac?{x-2a}+\frac?{x^2+a^2}
$$

Step 3: Give each numerator powers up to, but not including the order of the denominator. So the denominator of the first fraction is order 1, so you have up to order 0, meaning just a constant, $A$. The denominator of the second fraction is order 2, so you have up to order 1, so it's $Bx+C$, giving
$$
\frac{2ax}{(x-2a)(x^2+a^2)} = \frac{A}{x-2a}+\frac{Bx+C}{x^2+a^2}
$$
If a denominator had a cubic, order 3, then the numerator would be up to order 2, so $Ax^2+Bx+C$.

EDIT: Note that you may include a fourth step if you have a repeated factor. If you have $(x-a)^2$ as a factor of the denominator, then you could write
$$
\frac{Ax+B}{(x-a)^2}
$$
But you could also write it as
$$
\frac{A}{x-a}+\frac{B}{(x-a)^2}
$$
which is equivalent for different values of $A$ and $B$. It is often viewed as better to use the latter expression.

Title: Method to partial fractions.


问:
Is there a positive series of rational numbers that converge to $\pi$?
I'm looking for a series of rational numbers that approach $\pi$ that are all non-negative. In addition, I also want this series to have a general rule. So something like $\dfrac31+\dfrac1{10}+\dfrac4{100}+\ldots$ is not gonna cut it.

Does such a series exist?

Body:
This one is not bad
$$\pi=\sum_{n=0}^\infty \frac{ (2 n)\text{!!}}{2^{n-1}\,(2 n+1)\text{!!}}$$
The terms form the sequence
$$\left\{2,\frac{2}{3},\frac{4}{15},\frac{4}{35},\frac{16}{315},\frac{16}{693},
   \frac{32}{3003},\frac{32}{6435},\frac{256}{109395},\frac{256}{230945},\frac{512}{969969}\right\}$$ and the partial sums the sequence
$$\left\{2,\frac{8}{3},\frac{44}{15},\frac{64}{21},\frac{976}{315},\frac{10816}
   {3465},\frac{141088}{45045},\frac{47104}{15015},\frac{2404096}{765765},\frac{45693952}{14549535}\right\}$$ It does not converge very fast sine
$$\frac{a_{n+1}}{a_n}=\frac{n+1}{2 n+3}=\frac{1}{2}-\frac{1}{4 n}+O\left(\frac{1}{n^2}\right)$$

A better one is
$$\pi=\sum_{n=0}^\infty \frac{ (5 n+3)}{2^{n-1}\,(3 n+1) (3 n+2) \binom{3 n}{n}}$$
The terms form the sequence
$$\left\{3,\frac{2}{15},\frac{13}{1680},\frac{3}{6160},\frac{23}{720720},\frac{
   1}{466752},\frac{11}{75246080},\frac{1}{99095040}\right\}$$ and the partial sums the sequence
$$\left\{3,\frac{47}{15},\frac{1759}{560},\frac{2419}{770},\frac{205837}{65520}
   ,\frac{153966181}{49008960},\frac{23402860601}{7449361920},\frac{989459183   }{314954640}\right\}$$ It converges much faster that the previous one since

$$\frac{a_{n+1}}{a_n}=\frac{(n+1) (2 n+1) (5 n+8)}{3 (3 n+4) (3 n+5) (5 n+3)}= \frac{2}{27}-\frac{1}{27 n}+O\left(\frac{1}{n^2}\right)$$


问:
If $|1-z^n|\le\frac12$, then $\frac1{2|1+z+\cdots+z^{n-1}|}\le\frac1n$
This should actually be a simple question: For $z\in\mathbb C$ with $$|1-z^n|\le\frac12\;\;\;\text{for all }n\in\mathbb N\tag4$$ it follows from $$1-z^n=(1-z)(1+z+\cdots+z^{n-1})\tag5$$ that $$|1-z|\le\frac1{2|1+z+\cdots+z^{n-1}|}\tag6,$$ but why is the rhs less than $n^{-1}$?

Maybe we need to use the inequality $(a_1+\cdots+a_n)^2\le n(a_1^2+\cdots+a_n^2)$ for all $a\in\mathbb R^n$ and the fact that $|z|\ge\Re|z|$ for all $z\in\mathbb C$.

EDIT: The following inequality can be found in the proof of Proposition 2.4.2 in Linde's Probability in Banach Spaces: Stable and Infinitely Divisible Distributions, but I don't understand why this inequality holds:

答:
Let's prove the following slightly stronger statement: Let $z\in \mathbb{C}$ and $n\in \mathbb{N}$. If $|1-z^k|\leq \frac 1 2$ for $k\leq n-1$, then
$$
\mathrm{Re}(1+z+\dots+z^{n-1})\geq \frac n 2.
$$
In fact, $\mathrm{Re}(z^k)=1-\mathrm{Re}(1-z^k)\geq 1-|1-z^k|\geq \frac 1 2$. This readily implies the desired inequality.


问:
Probability of getting a piece longer than $1/2$ on cutting a rope of length $1$ at two randomly chosen points
Here's a bootstrap approach I used to find the answer (python 3).

```<code>def num_pieces(num,lenght):
    ot = list(range(1,lenght+1))[::-1]
    nig = []
    for i in range(lenght-1):
        n = random.randint(1, num-ot[i])
        nig.append(n)
        num -= n
    nig.append(num)
    r = []
    for i in nig:
         r.append(i/10)
    return r

res = []
n1 = 1000
n2 = int(n1/10)

for i in range(n1):
    a = []
    for i in range(n2):
        r = num_pieces(10,3)
        if r[0] > 0.5 or r[1] > 0.5 or r[2] > 0.5:
            a.append(1)
        else:
            a.append(0)
    res.append(sum(a)/n2)

pd.DataFrame(res).hist();
</code>```

And here's the result

If my code is correct (I believe it is), the probability is about 50%

Question - I do not know how to solve this task using plain math. In other words - what is the math base under the bootstrap result?

UPDATE

After receiving those awesome answers below, I checked the code and fixed the num_pieces generator. It worked incorrectly. All other code works just fine. So, here's the final solution

```<code>def get_random():
    a = [random.random(), random.random()]
    c = 1 - (max(a))
    b = max(a) - min(a)
    a = min(a)
    return [a,b,c]
res = []
n1 = 1000
n2 = int(n1/10)
for i in range(n1):
    a = []
    for i in range(n2):
        r = get_random()
        if r[0] > 0.5 or r[1] > 0.5 or r[2] > 0.5:
            a.append(1)
        else:
            a.append(0)
    res.append(sum(a)/n2)
pd.DataFrame(res).hist();
</code>```

The probability is 75% indeed. Thank you all for answering!

答:
The answer is 75%. It can be evaluated via
$$
2 \int_0^1 \int_0^y \begin{cases}
1, &\text{if } \max\{x,y-x,1-y\} > \frac12 \\
0, &\text{if } \max\{x,y-x,1-y\} \le \frac12
\end{cases} \bigg\} \,dx \,dy
$$
(where we use symmetry to declare $x$ the left cut point and $y$ the right cut point).

In the graph below, the black triangles are the points $(x,y)$ from the integral above that satisfy $\max\{x,y-x,1-y\} > \frac12$ (and the light gray triangles are their symmetric twins $(y,x)$), while the white triangles are the points that satisfy $\max\{x,y-x,1-y\} \le \frac12$.


问:
Centralizers of connected linear group and its Lie algebra
If we have that $G$ is a connected linear group and $H<G$, where $H$ is also connected, with $\mathfrak{h}$ the lie algebra of $H$ and we define the centralizers of the elements in the following way:

$Z(H):=\{a\in G| aha^{-1}=h\forall h\in H\}$ and $Z(\mathfrak{h})=\{a\in G|ad(a)Y=Y\forall Y\in \mathfrak{h}\}$

I have a question which is asking me to show that these two are the same but I am unsure how to go about doing this, is there some property of the adjoint that I am not seeing here that may be useful?

Thanks for any help

答:
For these sorts of questions, you want to transition between Lie algebras and Lie groups using the exponential map.  The useful facts for this problem are:

$\exp(Ad_a Y) = a (\exp Y) a^{-1}$. This follows from $Ad_a$ being the derivative of conjugation and the naturality of $\exp$.

If $X \in \mathfrak g$ then $X \in \mathfrak h \subset\mathfrak g$ if and only if $\exp X \in H$.

A connected Lie group is generated by the image of the exponential map.

So for this problem, suppose first $a \in Z(H)$ and let $Y \in \mathfrak h$.  Then by 1.,
$$
\exp(Ad_a Y) = a \exp(Y) a^{-1} \in H
$$
so that $Ad_a Y \in \mathfrak h$ by 2.  Thus $a \in Z(\mathfrak h)$.

Conversely, suppose $a \in Z(\mathfrak h)$ and let $h \in H$.  Then since $H$ is connected, by 3. we can write $h = \exp(Y_1) \cdots \exp(Y_n)$ for $Y_j \in \mathfrak h$, giving us
$$
a h a^{-1} = a\exp(Y_1) \cdots \exp(Y_n)a^{-1} = (a \exp(Y_1) a^{-1})(a \exp(Y_2) a^{-1}) \cdots (a \exp(Y_n) a^{-1})
$$
which lies in $H$ since each $a \exp(Y_j) a^{-1} \in H$.  Thus $a \in Z(H)$.


答:
Think about it this way.  Say $x$ is a number between $0$ and $1$, representing the position of the first cut.  For instance, suppose $x = 0.25$.  Now the second cut might occur on the short piece with probability $0.25$, and no matter where it is cut there, the long piece exceeds $0.5$.  Or the long piece is cut with probability $0.75$, but if the second cut occurs too close to one of the ends of the long piece, the remaining length will still be greater than $0.5$; otherwise, if cut near the middle, the two pieces will be less than $0.5$.  So we just have to figure out what we mean by being "too close" to one of the ends.

For the case $x = 0.25$, it is easy to see that if $y$ is the position of the second cut, if $0.25 < y < 0.5$, that is "too close" to an end--the remaining piece will be longer than $0.5$.  And if $0.75 < y < 1$, then again, the second cut is too close to the other end, and the middle piece will be longer than $0.5$.  So, given that $x = 0.25$, the probability that our second cut will give one section of rope that is longer than $0.5$ is the total length of all intervals satisfying $0 < y < 0.5$, or $0.75 < y < 1$; i.e., it is $0.5 + 0.25 = 0.75$.

Naturally, the next step is to generalize this to any choice of $x$.  Without loss of generality, let $0 < x < 0.5$ (as the situation is symmetric about $x = 0.5$).  Then as we saw above, every choice of $y$ between $0$ and $x$ will make the interval $[x, 1]$ longer than $0.5$, and every choice of $y$ between $x$ and $0.5$, or between $x + 0.5$ and $1$, will also make a segment longer than $0.5$.  So the probability of the second cut occurring in the interval $[0, 0.5] \cup [x + 0.5, 1]$ is $1 - (x + 0.5) + (0.5 - 0) = 1 - x.$  That's a nice result!

Finally, we use calculus to integrate over all $x$ between $0$ and $0.5$, then multiply by $2$ because we need to account for symmetry.  We have $$2 \int_{x=0}^{0.5} 1-x \, dx = \frac{3}{4}.$$  This is the desired probability.

Title: Probability of getting a piece longer than $1/2$ on cutting a rope of length $1$ at two randomly chosen points

Body:
Here's a bootstrap approach I used to find the answer (python 3).

```<code>def num_pieces(num,lenght):
    ot = list(range(1,lenght+1))[::-1]
    nig = []
    for i in range(lenght-1):
        n = random.randint(1, num-ot[i])
        nig.append(n)
        num -= n
    nig.append(num)
    r = []
    for i in nig:
         r.append(i/10)
    return r

res = []
n1 = 1000
n2 = int(n1/10)

for i in range(n1):
    a = []
    for i in range(n2):
        r = num_pieces(10,3)
        if r[0] > 0.5 or r[1] > 0.5 or r[2] > 0.5:
            a.append(1)
        else:
            a.append(0)
    res.append(sum(a)/n2)

pd.DataFrame(res).hist();
</code>```

And here's the result

If my code is correct (I believe it is), the probability is about 50%

Question - I do not know how to solve this task using plain math. In other words - what is the math base under the bootstrap result?

UPDATE

After receiving those awesome answers below, I checked the code and fixed the num_pieces generator. It worked incorrectly. All other code works just fine. So, here's the final solution

```<code>def get_random():
    a = [random.random(), random.random()]
    c = 1 - (max(a))
    b = max(a) - min(a)
    a = min(a)
    return [a,b,c]
res = []
n1 = 1000
n2 = int(n1/10)
for i in range(n1):
    a = []
    for i in range(n2):
        r = get_random()
        if r[0] > 0.5 or r[1] > 0.5 or r[2] > 0.5:
            a.append(1)
        else:
            a.append(0)
    res.append(sum(a)/n2)
pd.DataFrame(res).hist();
</code>```

The probability is 75% indeed. Thank you all for answering!


问:
Proving $\bigcap_n I_n \neq \emptyset$ for monotonically smaller closed intervals $(I_n)_n$
Theorem: Let $(I_n)_n$ be monotonically smaller closed intervals. i.e. $I_{n+1}\subseteq I_n$
Then $$\bigcap_n I_n \neq \emptyset$$
I easily grasped this geometrically, But I doubt of my proof. This is my proof:

Since $I_n$'s are real closed intervals, $I_n \neq \emptyset$. Then we can take $x \in I_n$. Since $I_n\subseteq I_{n-1}$, $x \in I_{n-1}$.
And generally from that,
$$x \in I_{n-2} \Rightarrow x \in I_{n-3} \Rightarrow x \in I_{n-4} \Rightarrow...\Rightarrow x \in I_2 \Rightarrow x \in I_1$$

Thus $x \in I_{n}$ is also in other closed intervals $I_i$s where $i < n-1$. Thus,
$$\bigcap_{i \leq n} I_i \neq \emptyset$$ Then, if we just take $\lim_{n\to \infty}$ equation will still hold and consequently,
$$\bigcap_{n} I_n \neq \emptyset$$

To me, it looks right, but I could'nt be sure. I would appreciate your short-time look. Thanks from now.

答:
Then, if we just take $\lim_{n\to \infty}$ equation will still hold
and consequently, $$\bigcap_{n} I_n \neq \emptyset$$

This is bad reasoning. Just because some relation holds for all values of $n$, that does not mean the relation holds for the limit.

For example, the relation $\frac1n > 0$ holds for all $n\in\mathbb N$, but the relation $$\lim_{n\to\infty} \frac 1n > 0$$

does not hold.

Another way to see your proof is not OK is this:

If $I_n$ are open intervals, then your "proof" can be used to prove that $$\bigcap_n I_n \neq \emptyset$$

even though it is possbile to find a sequence of intervals such that their intersection is empty (for example, $I_n = (0, 10^{-n})$).

To improve your proof: well, it's not really a case of improving it, I'm afraid your best bet is to go back to the drawing board. Try to find a value $x$ that is in all intervals $I_n$. To find it, think about $I_n=[a_n, b_n]$, and then think about the sequences $a_1, a_2,\dots$ and $[b_1, b_2, \dots]$. What are some of their interesting properties?


问:
Is the product of two elements in a monoid also an element of that monoid?
Let $(M, ·)$ be a monoid and $x$ and $y$ any two elements of said monoid. Is the following true?

$$∀ x, y ∈ M, \qquad  x · y ∈ M$$

答:
A monoid is a triple $(M,\circ,e)$ consisting of a non-empty set $M$, an operation $\circ:M\times M\rightarrow M:(x,y)\mapsto x\circ y (=xy)$ and a neutral element $e\in M$.
So the multiplication of two elements $x,y\in M$ yields another element $xy\in M$.


问:
Solving Euler-equation alike 2nd order DE with disturbing RHS
For a homework problem, I have to solve

$$ t^2 \ddot{x} - 3 t \dot{x} + 3x  = t^2 $$

which seems quite similar to the Euler Equation, which I would know how to solve, apart from the disturbing $ t^2 $ on the RHS instead of a constant $0$.

How does one approach such types of DE? Would it make a difference if there would be a constant $ a \neq 0$ on the RHS?

Thanks for any directions!

答:
In general, an equation of the form:

$$\mathcal{L}(x)=f(t)$$

Where $\mathcal{L}$ denotes some differential operator can be solved by finding the form of a general solution "$y_h(t)$"to the homogeneous equation (where the sum of any two solutions is again a solution):

$$\mathcal{L}(y_h)=0$$

And then finding any particular solution $y_p(t)$ that satisfies the inhomogeneous equation (where the sum of any two solutions is not necessarily another solution):

$$\mathcal{L}(y_p)=f(t)$$

Then the full solution to the original problem will be $x(t) = y_h(t)+y_p(t)$.

Whilst finding solutions to inhomogeneous problems can usually be approached in a formulaic way, finding particular solutions are usually more of a case by case basis. We frequently call the function $f$ the "forcing" of the equation, and it is frequently best to attempt to find a particular solution by looking for something that "looks like" the forcing, plug it into the left hand side and see if it works.

When the forcing is a polynomial, try plugging in a general polynomial of the same degree into the left hand side $(x=at^2+bt+c)$, and then compare coefficients with the right hand side to determine if there exists some real numbers $(a,b,c)$ such that the equation works.


问:
Weighted sum of two normal distributions
Let $X,Y,Z$ be three standard normal distribution : $X,Y,Z \overset{d}{=} N(0,1)$ but $X,Y,Z$ are not necessarily jointly distributed.

Let $w_X, w_Y,w_Z \in \mathbb{R}$.

We know that if $w_X =w_Y$, we can easily construct $Y=-X$ such as the sum $S=w_X X +w_Y Y=X+Y =0$.

If $w_X \neq w_Y$, can we construct $X$ and $Y$ such as the sum $S=w_X X +w_Y Y = 0$ (or equal to a constant)  as well?

What about the sum $S=w_X X +w_Y Y + w_Z Z$ ?

答:
Partial answer: $X=tY$ with $t \neq \pm 1$ is impossible. This is because $X=t^{n}Y$ and $|X|=|t|^{n}|Y|$ which tends to $\infty$ if $|t|>1$ and to $0$ is $|t|<1$.

The last one is possible in some cases and not possible in others. For example, we can find $X,Y,Z$ if $w_x=\frac 1 {\sqrt 2}=w_y$ and $w_z=-1$. Take $X,Y$ i.i.d  $N(0,1)$ and $Z=\frac {X+Y} {\sqrt 2}$. In general this is not possible and this is is clear from what we proved above. [Take $w_z=0$ and $w_y=2w_x$ for example].


问:
What's the exact/closest approximate to the inverse of getting 70%?
To clarify: I'm creating a computer program and I'm trying to find the actual mathematical way of approaching my problem. Given a number x, a is 0.7x. After doing some testing, it seems that to get back up to x from a, I can multiply by 1.42857143, but sometimes this isn't enough precision and gets a number that is very wrong.

Is there a blatantly obvious solution to this that I can't see? I want to find a better way to solve this without repeatedly adding 1 until it's right.

Thanks!

答: Ooooookay so it looks like I'm an idiot. 1.42857143 is 10÷7, and somehow I didn't notice that. I can simply just repeat the value if I want more precision.


答: We have a piece of length $>{1\over2}$ iff (a) both cuts are on the same half of the rope, or (b) are on different halfs, but the right cut is further right on its half than the left cut on its  half. The probability that one of these happens is ${1\over2}+{1\over2}\cdot{1\over2}={3\over4}$.


问:
real numbers a vector space over rational numbers?
Let $V$ be set of real numbers and $K$ the field of rational numbers.

Is $V$ a vector space over $K$, with ordinary addition of real numbers
and multiplication by rational numbers?

答: Anytime you have a field extension $V/K$ you can consider $V$ as a vector space over $K$.</p>


问:
Proving image of $f:D^n\rightarrow \mathbb{R}^n \;$ contains $\;D^n$
Consider a closed disc $D^n\subseteq \mathbb{R}^n$ and let $f:D^n\rightarrow \mathbb{R}^n$ be a continuous map with $f|_{S^{n-1}}$ is the identity on the boundary of $D^n$. Prove that $D^n \subseteq \operatorname{Im}(f)$. Is my following argument correct?

Suppose $x\in D^n$ and $x\not \in f(D^n)$. Then we have $\operatorname{id}_{S^{n-1}}$ factors through $f\circ i:S^{n-1} \hookrightarrow D^n \rightarrow \mathbb{R}^n-\{x\}$. Recalling the fact that $\mathbb{R}^n-\{x\}$ deform retracts to $S^{n-1}$, the contradiction here comes from applying the $H_{n-1}$ functor to obtain that $H_{n-1}(f\circ i):\mathbb{Z}\rightarrow 0 \rightarrow \mathbb{Z}$ is the zero homomorphism, certainly
not equal to the identity homomorphism $H_{n-1}(\operatorname{id}_{S^{n-1}}):\mathbb{Z}\rightarrow \mathbb{Z}$. Hence $f$ must contain its image.

Also is there a way to prove this directly? I think I can smell Brouwer's fixed point theorem and/or Borsuk-Ulam but I'm not sure how to employ it.

答:
Your proof is correct. You show that if $D^n$ is not contained in the image of $f$, then we would get a retraction $\phi : D^n \to S^{n-1}$ which would show that $id: S^{n-1} \to S^{n-1}$ is null-homotopic. There are various approaches to falsify this, you do it via homology.

Brouwers's fixed point theorem is usually proved via homology (using the non-existence of a retraction $\phi : D^n \to S^{n-1}$), but there are also other proofs. See for example Why is Brouwer’s Fixed Point Theorem considered a result of algebraic topology?

Anyway, if you take Brouwers's fixed point theorem as a given result, then you can easily show that $S^{n-1}$ is not a retract of $D^n$: Assume there exists a retraction $\phi : D^n \to S^{n-1}$. Then $\psi :  D^n \to S^{n-1}, \psi(x) = -\phi(x)$, is well-defined and does not have a fixed point. In fact, no $x$ in the open unit ball can be a fixed point, and no $x \in S^{n-1}$ can be a fixed point because $\psi(x) = - \phi(x) = -x \ne x$.


问:
expected number of flips until $k$ consecutive misses or number of flips finish
I followed the argument given in the answers for this question to find the expected number of misses in a sequence of coin-tosses with success probability $p$. The stopping condition for tossing is either until the $k^{th}$ consecutive miss or until the number of rounds reach $n$, whichever comes first. We can consider tail as a miss and head as hit.

The question in the link differs from my question. In my question stopping happens only when $k$ consecutive miss happens or rounds reach $n$. In the linked question stopping happens when $k$ miss happens.

The probability that $k^{th}$ consecutive miss happen at $i^{th}$ trial can not be simply represent in terms of $i$ and $k$ as the number of hits or misses before the start of the sub-sequence of $k$-consecutive miss is still unknown. There can be many misses and hits before the start of the sub-sequence which results in stopping. Similarly there can be unknown number of misses or hits before reaching the end of round ($n$).

Is it even possible to represent this number in a closed form ? Any help and suggestions ?

答:
Consider first an auxilary problem. In how many ways can we arrange $b$ black balls and $w$ white balls so that every sequence of black balls is shorter than $B$?

The problem is equivalent to that of placing $b$ balls into $w+1$ bins of the capacity $B-1$ balls each. The problem has very well-known solution:
$$
N(w,b,B)=\sum_{i\ge0}(-1)^i\binom{w+1}i\binom{b+w-Bi}w,\tag1
$$
with the convention $\binom nk=0$ for  $n<0$.

Using this one obtains the following expression for the expected number of rounds in your problem:
$$\sum_{i=0}^n n\,N(i,n-i,k)p^i(1-p)^{n-i}+\sum_{l=k}^n l\,\sum_{i=0}^{l-k} N(i-1,l-k-i,k)p^i(1-p)^{l-i},\tag2$$
where the first term accounts for the events where no $k$-row of misses happens until reaching the $n$-th round, whereas the second term accounts for the events where $l-k-1$ rounds ($k+1\le l\le n$) with no $k$-row of misses is followed by a single hit and the $k$-row of misses ($i$ in both sums counts the number of hits). To incorporate into the second sum the event with no hit (pure $k$-row of misses) we redefine $N$ for the case $w=-1,b=0$:
$$N(-1,0,B)=1.
$$


问:
Proving that a non-empty subset of a vector space is itself a subspace iff it satisfies the closure axioms
I have a problem understanding a certain part in the proof. The proof goes as follows:-

The proof :-

If S is a subspace then trivially by definition it satisfies the closure axioms.
Conversely, if S satisfies the closure axioms then the axioms 3,4,7,8,9,10 are trivially satisfied.

Axiom 5:- Let V be a linear space

$$S \subseteq V  \implies \exists O (\forall u \in S \implies u+O = u)$$

let $u$ be any element in $S$ , then by hypothesis  $ku \in S$ since it satisfies the closure axioms. Let $k$=0 then by a previous theorem which states as follows:-

In a given linear space , let $u$ and $v$ be any arbitrary vectors and k and d arbitrary scalars then we have the following properties :-

[1] $0u=O$
 [2] $cO = O$
......

and since  $ku \in S$ then it follows that $O \in S$  from [1].
 The proof follows from here proving also that axiom 6 is true in the same sense.

My problem :-

How can we apply the theorem that states $0u=O$ while that theorem only holds in a Linear Space while we haven't proved that $S$ is indeed a linear space.

答: The key is that $S\subseteq V$, and $V$ is a linear space. As the property you mention is universal, i.e., holds for every element of $V$, then it is true for all elements in $S$.


问:
Regarding Finite group theory, Core of subgroups, and $\Delta_H(G)$.
This was a challenge question posted by my professor in algebra course last semester, and I've been trying to solve it. I think there is probably a solution online but I couldn't find much information about this challenge.

Let $G$ be a non-trivial finite group, and let $H\le G$ be s.t. $|\text{Core}_G(H)|=1$, where $\text{Core}_G(H):=\underset{g\in G}{\cap}gHg^{-1}$.

Prove that $\Delta_H(G):=G-\underset{g\in G}{\cup}gHg^{-1}$ is nonempty;

Can $\Delta_H(G)$ contain an element of prime power order?

My thought:

I was discussing by case. Case 1: $Z(G)$ is non-trivial; Case 2: otherwise.

In case 1, say we have $x\in Z(G)$ with $x\ne e$. Then $x\notin H$, otherwise, we would have $x\in\text{Core}_G(H)$ which makes $|\text{Core}_G(H)|>1$. Thus $x\in\Delta_H(G)$, as $x\notin gHg^{-1}$ for any $g\in G$. Therefore $\Delta_H(G)$ is nonempty. Moreover, it turns out that $Z(G)\subset\Delta_H(G)$, so there must be an element of prime power order.

I have no clues on how to deal with case 2. I tried to think in terms of conjugacy classes (so $\text{Core}_G(H)$ is the intersection of all conjugacy classes of $h\in H$ and $\Delta_H(G)$ is the set of elements which are not in any conjugacy classes of $h\in H$), but it seems to not lead to a solution.

答: With respect to your question 1. the classical approach is a counting argument as pointed out by Arturo. But if you know some character theory of groups, notably induced characters and Frobenius Reciprocity, you can proceed as follows. Let $H \lt G$, with $|G:H|$ finite and assume $\bigcup_{g \in G}g^{-1}Hg=G$. Induce the trivial character $1_H$ of $H$: then $(1_H)^G(x)=\frac{1}{|H|}\sum_{g \in G}1_H^{\circ}(g^{-1}xg)=|G:H|$. (here $1_H^{\circ}(t)=1_H(t)=1$ if $t \in H$ and $1_H^{\circ}(t)=0$ if $t \notin H$). Hence $|G:H|=[(1_H)^G,1_G]_G=[1_H,(1_G)_H]_H=[1_H,1_H]_H=1$ by Frobenius Reciprocity. So, $G=H$, a contradiction.


问:
Self-Study Sufficient Statistics, Pdf with Indicator Function
The example is from the Book Hogg Introduction to Mathematical Statistics Page 384, Chapter 7.2 Sufficient Statistics. Please let me know if my argument for the solution is correct, since I used a different one then presented in the book.

Problem:

Let $X_{(1)} < X_{(2)} < \dots < X_{(n)}$ denote the oder statistics of a random sample of sice n from the distribution with pdf:

$$
f(x;\theta) = exp(-(x-\theta)) \mathbb{I}_{[\theta, \infty)}
$$

Solution:

By factorization theorem: $L(X;\theta) = g(T(X), \theta)h(x)$:

\begin{align*}
L(x;\theta) &= \prod_{i=1}^nexp(-(x_i-\theta)) \mathbb{I}\{\theta < x_i < \infty\} \\
&= exp(-(\sum_{i=1}^nx_i- n\theta)) \cdot \mathbb{I}\{\theta < {x_i}_{(i=1,\dots,n)} < \infty\} \\
&= \mathbb{I}\{max\{X_i\} < \infty\}exp(-\sum_{i=1}^nx_i)\cdot exp(n\theta) \cdot \mathbb{I}\{\theta < min\{X_i\}\}
\end{align*}

By factorization theorem:

$$
h(x) = \mathbb{I}\{max\{X_i\} < \infty\}exp(-\sum_{i=1}^nx_i)
$$

and:
$$
g(T(X), \theta) = exp(n\theta) \cdot \mathbb{I}\{\theta < min\{X_i\}\} \Rightarrow T(X) = min(X)
$$

答:
Yes correct, but in a very simple way

$$f_X(x|\theta)=e^{\theta}e^{-x}\cdot\mathbb{1}_{[\theta;\infty)}(x)$$

Now simple observing that (here $X_1,...,X_n$ are not ordered...)

$\theta \leq X_1<\infty$

$\theta \leq X_2< \infty$

...

$\theta \leq X_n< \infty$

thus it is self evident that

$$\theta\leq X_{(1)}$$

Thus the likelihood is

$$L(\theta)=\underbrace{e^{-\Sigma_iX_i}}_{h(\mathbf{x})}\cdot\underbrace{ e^{n\theta}\cdot\mathbb{1}_{(-\infty;x_{(1)}]}(\theta)}_{g[t(\mathbf{x}),\theta]}$$

thus $T=X_{(1)}$

I edited your question because the order satistic is always written as $X_{(i)}$


问:
Midpoint convex bounded function is continuous
I'm reading a proof of the following statement:

Let $f: (a,b) \to \mathbb{R}$ be a midpoint convex (i.e. $f(1/2(x+y)) \leq 1/2f(x) + 1/2f(y)$) function that is bounded. Then $f$ is continuous.

Here is a proof that I found here: Proving continuity of $f$

To prove that a bounded midpoint convex is continuous, argue by contradiction. Supose $f$ is discontinuous at $x_0\in(a,b)$. Without loss of generality we may assume $x_0=0$, $f(x_0)=0$.

First step.  There exists a sequence $\{x_n\}\subset(a,b)$, such that $\lim_{n\to\infty}x_n=0$ and $\lim_{n\to\infty}f(x_n)=m\ne0$. We may assume that $m>0$.

Second step. The sequence $\{2\,x_n\}$ also converges to $0$ and
$$
f(x_n)=f\Bigl(\frac{0+2\,x_n}2\Bigr)\le\frac{f(0)+f(2\,x_n)}2\implies f(2\,x_n)\ge2\,f(x_n)\implies\liminf f(2\,x_n)\ge2\,m.
$$
Iteration shows that
$$
\liminf f(2^k\,x_n)\ge2^k\,m,
$$
which is impossible since $f$ is bounded.

Question: Why can we assume $m > 0$?

答:
Suppose
$$\lim_{n\to\infty}f(x_n)=m<0.$$
Let $y_n=-x_n$, then
$$0=f(0)=f\left(\frac{x_n+y_n}{2}\right)\leq\frac{f(x_n)+f(y_n)}{2},$$
so $f(y_n)\geq-f(x_n)$, then
$$\limsup_{n\to\infty}f(y_n)\geq\limsup_{n\to\infty}(-f(x_n))
=\lim_{n\to\infty}(-f(x_n))=-m>0.$$
So there exists subsequences $\{f(y_{_{n_k}})\}$ such that
$$\lim_{k\to\infty}f(y_{_{n_k}})=\limsup_{n\to\infty}f(y_n)\geq-m>0.$$


问:
Steady state of Markov Process
I would like to have acknowledgment about this exercise.

Find, less than a constant, the steady state of the following Markov process
$$\begin{bmatrix}
w_{k+1}\\
s_{k+1}\\
d_{k+1}
\end{bmatrix} = \begin{bmatrix}
1 & 1/4 &0\\
0 & 3/4 &1/2\\
0 & 0 & 1/2
\end{bmatrix} \begin{bmatrix}
w_{k}\\
s_{k}\\
d_{k}
\end{bmatrix}.
$$
Here it is my solution: with direct calculations I found that the matrix has 3 distinct eigenvalues $\lambda_1 =1, \lambda_2=3/4$ and $\lambda_3 = 1/2$, whose corresponding eigenvectors are $(1, 0, 0), (1, -1, 0)$ and $(-2, 1, -2)$ respectively.
Setting $u_k:= \begin{bmatrix}
w_k\\
s_k\\
d_k
\end{bmatrix}$, the previous equality becomes $u_{k+1} = A u_k$ which gives, iterating, that $u_k = A^k u_0$. But the matrix $A$ is diagonalizable, so $A^k =S \Lambda^k S^{-1}$, i.e.
$$
\begin{bmatrix}
1 & 1/4 &0\\
0 & 3/4 &1/2\\
0 & 0 & 1/2
\end{bmatrix} = \begin{bmatrix}
1 & 1 &-2\\
0 & -1 & 1\\
0 & 0 & -2
\end{bmatrix}\begin{bmatrix}
1 & 0 &0\\
0 & (3/4)^k & 0\\
0 & 0 & (1/2)^k
\end{bmatrix}\begin{bmatrix}
1 & 1 &-1/2\\
0 & -1 &-1/2\\
0 & 0 & -1/2
\end{bmatrix}.
$$
Thus
$$u_k = S\Lambda^k S^{-1} u_0 \iff \begin{bmatrix}
1\\
0\\
0
\end{bmatrix} (w_0 +s_0 +d_0) +\begin{bmatrix}
1\\
-1\\
0
\end{bmatrix} (3/4)^k(-s_0 -1/2 d_0) +\begin{bmatrix}
-2\\
1\\
-2
\end{bmatrix}(1/2)^k (-1/2) d_0.
$$
The steady state is obtained evaluating the limit for $k\to +\infty$ of $u_k$, which is, in this case,
$$\lim_{k\to +\infty} u_k = w_0.$$

Could anyone please tell me if it is all true or I missed something?

Thank you in advance!

答:
The terminology is a bit unusual, since there is not really a Markov chain here. If you wish to get the limit of $(u_k)$, then what you did is almost correct, except that you get
$$
\lim_{k \to + \infty} u_k = (w_0 + s_0 + d_0) [1, 0, 0].
$$

However, if you are really in the framework of Markov chains, you do not need all this. Denote by $M$ a stochastic matrix, and $Y_k$ the vector describing the probabilities to be in each state after $k$ steps. If $M$ is irreducible and aperiodic, then $Y_k = Y M^k$ converges to the stationary probability, i.e. the unique left-eigenvector of $M$ corresponding to the eigenvalue 1.

In your context, everything is written for right-stochastic matrices, and thus the limit of $(u_k)$ is the unique right-eigenvector of the matrix corresponding to the eigenvalue 1, which is indeed $(1,0,0)$. This is consistent with what is written above since $w_0 + s_0 + d_0 = 1$ for probability vectors.

All calculations apart, this is quite clear: from state 3, you eventually go to state 2; and from state 2, you eventually go to state 1, which is absorbing. Therefore, regardless of where you start, you always end up at state 1 and stay there.

Title: Steady state of Markov Process


答:
A vector space over a field $K$ is any set $V$ equipped with functions $V\times V\to V$ and $K\times V\to V$ satisfying the axioms (I sometimes call them "computation rules").

So, since $\mathbb{R}$ with the two functions you mention satisfies the axioms, it is a vector space over the rationals. It is infinite dimensional, by reasons of cardinality: a finite dimensional vector space over an infinite field $K$ has the same cardinality as $K$.

In the same vein, $\mathbb{C}$ is a vector space over $\mathbb{Q}$ as well. This has important applications in the theory of algebraic numbers, because one can use the powerful machinery of vector spaces. Note that $\mathbb{C}$ and $\mathbb{R}$ are vector spaces over any of their subfields. One of the most useful results in this respect is the formula

$$
[M:L][L:K]=[M:K]
$$

where $M$, $L$ and $K$ are subfields of $\mathbb{C}$ (any field would do, actually) with $K\subseteq L\subseteq M$ and with $[L:K]$ we denote the dimension of $L$ as a vector space over $K$.

Title: real numbers a vector space over rational numbers?

Body:
Let $V$ be set of real numbers and $K$ the field of rational numbers.

Is $V$ a vector space over $K$, with ordinary addition of real numbers
and multiplication by rational numbers?


问:
What kind of maps preserve convexity?
Linear maps seem to preserve convexity of an object. Are there more maps with this property? What subset of projections preserve convexity?</p>

答:
I do not understand what you mean by object, if a set or a function. In the case of a set, note that there is no necessary condition as explained in this answer.

If for object you mean a function, it is easier:

$-f$ is concave if and only if $f$ is convex. Nonnegative weighted sums:

if $w_{1}, \ldots, w_{n} \geq 0$ and $f_{1}, \ldots, f_{n}$ are all convex, then so is $w_{1} f_{1}+\cdots+w_{n} f_{n}$. In particular, the sum of two convex functions is convex.

this property extends to infinite sums, integrals and expected values as well (provided that they exist). Elementwise maximum: let $\left\{f_{i}\right\}_{i \in I}$ be a collection of convex functions. Then $g(x)=\sup _{i \in I} f_{i}(x)$ is convex. The domain of $g(x)$ is the collection of points where the expression is finite. Important special cases:

If $f_{1}, \ldots, f_{n}$ are convex functions then so is $g(x)=\max \left\{f_{1}(x), \ldots, f_{n}(x)\right\}$

Danskin's theorem: If $f(x, y)$ is convex in $x$ then $g(x)=\sup _{y \in C} f(x, y)$ is convex in $x$ even if $C$ is not a convex set.
Composition:

If $f$ and $g$ are convex functions and $g$ is non-decreasing over a univariate domain, then $h(x)=g(f(x))$ is convex. As an example, if $f$ is convex, then so is $e^{f(x)} .$ because $e^{x}$ is convex and monotonically increasing.

If $f$ is concave and $g$ is convex and non-increasing over a univariate domain, then $h(x)=g(f(x))$ is convex.

Convexity is invariant under affine maps: that is, if $f$ is convex with domain $D_{f} \subseteq \mathbf{R}^{m}$, then so is $g(x)=f(A x+b),$ where $A \in \mathbf{R}^{m \times n}, b \in \mathbf{R}^{m}$ with domain $D_{g} \subseteq \mathbf{R}^{n}$.

Minimization: If $f(x, y)$ is convex in $(x, y)$ then $g(x)=\inf _{y \in C} f(x, y)$ is convex in $x,$ provided that $C$ is a convex set and that $g(x) \neq-\infty$

If $f$ is convex, then its perspective $g(x, t)=t f\left(\frac{x}{t}\right)$ with domain $\left\{(x, t) \mid \frac{x}{t} \in \operatorname{Dom}(f), t>0\right\}$ is convex.

Title: What kind of maps preserve convexity?


问:
Prove $g : (0, 1) \rightarrow \mathbb{R}$ , $g_{(x)}=\frac{f_{(x)}}{x}$ is an increasing function over $(0,1)$
$f:[0,1] \rightarrow \mathbb{R}$ is continuous over $[0,1]$ and $f_{(0)} = 0$, if $f^{\prime}$ exists over $(0,1)$ and is an increasing function then prove $g : (0, 1) \rightarrow \mathbb{R}$ , $g_{(x)}=\frac{f_{(x)}}{x}$ is an increasing function over $(0,1)$.

My attempt:
Suppose we approximate $f^{\prime}$ using polynomials, it's an increasing function so it has a positive leading coefficient and degree of at least 0, so: $f_{(x)}^{\prime} = a_0x^{n} + ...$ where $a_0 > 0$ and $n \geqslant 0$ then after integration: $f_{(x)} = b_0x^{n+1} + ...$ where $b_0 > 0$ and $n+1 \geqslant 1$ so dividing by $x$: $g_{(x)}=\frac{f_{(x)}}{x} = b_0x^{n} + ...$ where $b_0 > 0$ and $n \geqslant 0$ so $g_{(x)}$ is an increasing function.

Is this proof correct? If YES, what could go wrong? If NO, why? and what's the correct way to prove this?

and finally, is there a better way you can think of? and can you simply explain the reason for open and closed intervals and $f_{(0)} = 0$ conditions?

答: Since $f'$ is increasing it follows that $f$ is a convex function. Let $0 <x<y <1$. Then $x=\frac x y y +(1-\frac  x y) 0$. This gives $f(x) \leq \frac x y f(y) +(1-\frac  x y) f(0)$. Hence $\frac {f(x) } x\leq \frac {f(y)} y$.


问:
Show that $R_{w}$ is bounded/compact using no arbitrage or non-redundancy arguments
Consider a financial market with $d+1$ assets in a one-period model. The $0-$th asset is considered the risk-free asset, the others are risky. The vector $\overline{\pi} \in \mathbb R^{d+1}$ denotes the initial price vector at $t=0$, and the random vector $\overline{S} \in \mathbb R^{d+1}$ denotes prices at $t=1$. Further we have $\pi^{0}=1,\; S^{0}=1+r$, as the $0-$th asset is risk-free.

Here a strategy $\overline{\xi}\in \mathbb R^{d+1}$ is called an arbitrage opportunity if $\overline{\xi}⋅\overline{\pi}≤0$ but $\overline{\xi}⋅\overline{S}\geq 0$ and $\mathbb P(\overline{\xi}⋅\overline{S}> 0)>0$. Further the market model is called non-redundant if
$\overline{\xi}\cdot \overline{S}=0 \; \;\mathbb P\text{-a.s.}\implies \overline{\xi}=0$.

By the Law of One price we know that in an arbitrage-free model, for any $\overline{\rho},\; \overline{\xi}$ such that $\overline{\xi}\cdot\overline{S}=\overline{\rho}\cdot\overline{S}$ a.s. it must follow that $\overline{\xi}\cdot\overline{\pi}=\overline{\rho}\cdot\overline{\pi}$.

Now as in  https://quant.stackexchange.com/q/17839/42184 we need to show that for $w>0$,

$$\mathcal{R}_{w}:=\{\overline{\xi}\in \mathbb R^{d+1}:\overline{\xi}\cdot \overline{\pi}=w,\;\overline{\xi}\cdot\overline{S}\geq 0\; \text{a.s.}\}\;\; \text{ is compact} $$

As we are in finite-dimensions, I assume that we need to use $\text{compact}\iff \text{closed and bounded}$.

Closedness is no problem at all, using convergence arguments. However, I am struggling to show boundedness. As I have not used the no arbitrage arguments or the non-redundancy condition, I assume that I will need to use them in my argument. Any ideas?

答: Here is a contradiction idea (inspired from proof of Theorem 1.32 from Follmer's Stochastic Finance which is the source material of this question): Take a sequence in your set $R_w$ such that $\limsup_{n \rightarrow \infty} |\overline{\xi_n}| = \infty$, and consider looking at $\eta_n = \frac{\overline{\xi_n}}{|\overline{\xi_n}|}$. Note that $\eta_n$ is a bounded sequence in $\mathbb R^n$, so a subsequence of it (by Bolzano-Weierstrass) will converge to an $\eta$ with $|\eta|=1$ (norm is continuous). However, this is a problem, since $\eta$ now satisfies $\eta \cdot\overline{\pi} = 0$, and $\eta\cdot\overline{S}\geq 0 \, \,\mathbb P$almost surely . Now, note that by nonredundancy, it must be the case that $\mathbb P(\eta ⋅\overline{S}> 0)>0$, because otherwise $\eta = 0$, which contradicts $|\eta| = 1$. Thus, $\eta$ is an arbitrage, which should not happen in an arbitrage-free market.


答:
Use the mean value theorem: for $x \in (0,1)$
$$f(x) - f(0) = xf'(\xi)$$
where $0 < \xi < x$.  Then for $x \in(0,1)$
\begin{align}
g'(x) &= \frac{f'(x)}{x} - \frac{f(x)}{x^2} \\
&= \frac{1}{x^2}\Big(xf'(x) - f(x)\Big) \\
&= \frac{1}{x^2}\Big(xf'(x)-xf'(\xi)\Big)
\end{align}
and because $f'$ is increasing, the term on the right is non-negative, so that $g'(x) \geqslant 0$ and $g$ is increasing.

Title: Prove $g : (0, 1) \rightarrow \mathbb{R}$ , $g_{(x)}=\frac{f_{(x)}}{x}$ is an increasing function over $(0,1)$

Body:
$f:[0,1] \rightarrow \mathbb{R}$ is continuous over $[0,1]$ and $f_{(0)} = 0$, if $f^{\prime}$ exists over $(0,1)$ and is an increasing function then prove $g : (0, 1) \rightarrow \mathbb{R}$ , $g_{(x)}=\frac{f_{(x)}}{x}$ is an increasing function over $(0,1)$.

My attempt:
Suppose we approximate $f^{\prime}$ using polynomials, it's an increasing function so it has a positive leading coefficient and degree of at least 0, so: $f_{(x)}^{\prime} = a_0x^{n} + ...$ where $a_0 > 0$ and $n \geqslant 0$ then after integration: $f_{(x)} = b_0x^{n+1} + ...$ where $b_0 > 0$ and $n+1 \geqslant 1$ so dividing by $x$: $g_{(x)}=\frac{f_{(x)}}{x} = b_0x^{n} + ...$ where $b_0 > 0$ and $n \geqslant 0$ so $g_{(x)}$ is an increasing function.

Is this proof correct? If YES, what could go wrong? If NO, why? and what's the correct way to prove this?

and finally, is there a better way you can think of? and can you simply explain the reason for open and closed intervals and $f_{(0)} = 0$ conditions?


问:
How good is the approximation $e^{-x} = \left(1- \frac{x}{n}\right)^n$?
We know that for $n\rightarrow \infty$, the above converges. But how good of an approximation is it for non arbitrarily large $n$ (i.e. $n << \infty$)?

答:
In the same spirit as @Gary's answer, consider
$$y=\left(1 - \frac{x}{n} \right)^n\implies \log(y)=n \log\left(1 - \frac{x}{n} \right)$$ Using Taylor series when $n$ is large
$$\log(y)=n\left(-\frac{x}{n}-\frac{x^2}{2 n^2}-\frac{x^3}{3
   n^3}+O\left(\frac{1}{n^4}\right)\right)$$
$$\log(y)=-x-\frac{x^2}{2 n}-\frac{x^3}{3 n^2}+O\left(\frac{1}{n^3}\right)$$
$$y=e^{\log(y)}=e^{-x}\left(1-\frac{x^2}{2 n}+\frac{x^3 (3 x-8)}{24
   n^2}+O\left(\frac{1}{n^3}\right)\right)$$


问:
If $a^3=a$ in a ring, prove: the ring is commutative
Let $R$ be a ring, not necessarily with a unit element. $R$ is not necessarily integral.
If for any $a \in R$, $a^3=a$, prove: $R$ is commutative:
Any $a, b  \in R$, $ab=ba$.

My efforts on it:
I can use the relation $(a+b)^3=a+b$ and $(a-b)^3=a-b$. But the relation is cubic and I do not know how to reduce it to a relation about $ab$ and $ba$.

答:
This is a cute problem that one can find in Herstein's book on Algebra.

Start by proving that if $xy=0$, then $yx=0$.
Then,let us begin by noting that if $x^2=x$, then we have that $yx=xy$. This follows since
$x^2y-xy=x(xy-y)=x(y-y)=0$, and then $(xy-y)x=0$, so that $xyx = yx$. You can do it in a similar way for $(y-yx)x$ to show it is zero and deduce that $xy = xyx$ and as such, $xy=yx$.

Now, we have that for any $x \in R$, we have that $x^2$ satisfies $(x^2)^2=x^4=x^2$. So, $x^2$ commutes with any element of R. We have that $(x+x^2)$ commutes with every element of R too. Indeed, we have that $(x+x^2)^2 = 2(x+x^2)$. You can show that if $y \in R$ satisifes $y^2 = n y$, then y commutes with any element of R.

Now, finish by using the identity $(x+x^2)-x^2=x$ and that the sum of elements that commutes with any element of R commutes with any element of R.


问:
Let $a,b\in\mathbb{R}$ and $f(x)=\lim_{n\to\infty}\dfrac{2x^{2x+1}+ax+b}{x^{2n+2}+4x^{2n+1}+5}$ if $f$ is continuous on $\mathbb{R}$, find $a$, $b$
Let $a,b\in\mathbb{R}$ and $\displaystyle f(x)=\lim_{n\to\infty}\dfrac{2x^{2x+1}+ax+b}{x^{2n+2}+4x^{2n+1}+5}$ if $f$ is continuous on $\mathbb{R}$, find $a$, $b$

My attempt:
\begin{align}
f(x)&=\lim_{n\to\infty}\dfrac{2x(x^{2n})+ax+b}{x^2(x^{2n})+4x(x^{2n})+5}\\
f(x)&=\begin{cases}
\dfrac{2}{x+4}&;|x|>1\\
\dfrac{2+a+b}{10}&;x=1\\
\dfrac{ax+b}{5}&;1>|x|>0\\
\dfrac{b}{5}&;x=0
\end{cases}
\end{align}
Are there any theorems that will help solve this problem?

答: Actually, it's$$f(x)=\begin{cases}\frac2{x+4}&\text{ if }|x|>1\\\frac{ax+b}5&\text{ if }|x|<1\\\frac{2+a+b}{10}&\text{ if }x=1\\\frac{-2-a+b}2&\text{ if }x=-1.\end{cases}$$In order that $f$ is continuous at $1$, you must have$$\frac25=\frac{2+a+b}{10}=\frac{a+b}5.$$Therefore, $a+b=2$. And, in order that $f$ is continuous at $1$, you must have$$\frac23=\frac{-2-a+b}2=\frac{-a+b}5.$$Therefore, $-a+b=\frac{10}3$. So $a=b=\cdots$


问:
What are the isomorphisms in the category of vector spaces, $Vec_K$?
Here is an interesting question from "The Rising Sea" book of Algebraic Geometry: http://math.stanford.edu/~vakil/216blog/FOAGapr2915public.pdf

1.2.3 Example. Consider the category $Vec_K$ of vector spaces over a field $K$. The objects are $K$-vector spaces, and the morphisms are linear transformations. What are the isomorphisms?

I would say that the isomorphisms are the invertible linear transformations, is that correct?

答:
Question: "I would say that the isomorphisms are the invertible linear transformations, is that correct?"

Answer: If $C$ is any category and $U,V$ objects in $C$, we define $U$ to be isomorphic to $V$ (written $U \cong V$) iff there are morphisms (in $C$) $f:U \rightarrow V$ and $g: V\rightarrow U$ such that $f\circ g=Id_V$ and $g \circ f=Id_U$. In your case the maps $f,g$ are invertible linear maps/linear transformations between vector spaces.


问:
Solve Least Squares with Tikhonov Regularization Linear Equality and Non Negative Constrains
How to solve the following constrained minimization problem:
$$
\arg_S min_\; \frac{1}{2}\left \{ \left \| K_2SK_1^T-M \right \|_F^2 +\lambda \left \| S \right \|_F^2\right \} \\
s.t. \sum_{1}^{col}S=Sum1 \\
\sum_{1}^{row}S=Sum2 \\
s_{i,j} \in S \geq 0 \\
$$

where $K_1$,$K_2$,$M$ and $S$ are 2d Matrix, and only $S$ is unknown. In the constraints, $Sum1$ is the sum along the column of $S$, which is a row vector. $Sum2$ is the sum along the row of $S$, which is a column vector.  All elements in $S$ are non-negative.

Here is the data stored in mat format. How to solve this kind of problem? Are there any MATLAB routines I can use out-of-box?

```<code>    load('matlab.mat');
%　 min norm( K2*X*K1'-M,'fro')^2+lambda*norm(X,'fro')^2
%   s.t. sum(X,1) = Sum1 ;   sum(X,2) = Sum2; all(X>=0) = true
</code>```

What I've Tried:

CVX:

Failed to find the optimal solution with all supported solvers. Here is the cvx script I use:

```<code>cvx_begin
variable S(size(K2,2),size(K1,2))
minimize(0.5*(square_pos(norm(K2*S*K1'-M,'fro'))+lambda*square_pos(norm(S,'fro'))))
sum(S,1) == Sum1
sum(S,2) == Sum2
S>=0
cvx_end
</code>```

Edit:

A synthetic data with much smaller scale was attached here. DR was also normalized.

答:
The Objective Function is given by:

$$\begin{aligned}
\arg \min_{S} \quad & \frac{1}{2} {\left\| {K}_{2} S {K}_{1}^{T} - M \right\|}_{F}^{2} + \frac{\lambda}{2} {\left\| S \right\|}_{F}^{2} \\
\text{subject to} \quad & {S}^{T} \boldsymbol{1} = \boldsymbol{u} \\
& S \boldsymbol{1} = \boldsymbol{v} \\
& S \geq 0
\end{aligned}$$

While one could use many specialized solvers for the above problem I'd start with the Projected Gradient Descent.

In order to use the Projected Gradient Descent we need the projection onto the intersection of all 3 constraints sets.
Yet, since the set $ \left\{ S \mid S \geq 0 \right\} $ isn't a sub space we can't use the alternating projection method (See Projections onto Convex Sets and Projection of $ z $ onto the Affine Half Space $ \left\{ x \mid A x = b, \; x \geq 0 \right\} $).
It can be solved using the Dykstra's Projection Algorithm from Orthogonal Projection onto the Intersection of Convex Sets.

The Dykstra's Projection Algorithm requires the projection onto each set to generate an iterative algorithm to computer the projection onto the intersection of all sets.

Hence, the atoms of the projected gradient Descent are:

The Gradient of the objective function: $ {K}_{2}^{T} \left( {K}_{2} S {K}_{1}^{T} - M \right) {K}_{1} + \lambda S  $.

Projection onto the 1st constraint: $ \operatorname{Proj}_{ \left\{ X \mid {X}^{T} \boldsymbol{1} = \boldsymbol{u} \right\} } \left( Y \right) = Y - \boldsymbol{1} \frac{ \boldsymbol{1}^{T} Y - \boldsymbol{u}^{T} }{
\boldsymbol{1}^{T} \boldsymbol{1} } $.

Projection onto the 2nd constraint: $ \operatorname{Proj}_{ \left\{ X \mid X \boldsymbol{1} = \boldsymbol{v} \right\} } \left( Y \right) = Y - \frac{ Y \boldsymbol{1} - \boldsymbol{v} }{
\boldsymbol{1}^{T} \boldsymbol{1} } \boldsymbol{1}^{T} $.

Projection onto the 3rd constraint: $ \operatorname{Proj}_{ \left\{ X \mid X \geq 0 \right\} } \left( Y \right) = \max \left( Y, 0 \right) $.

Remark: Since $ \boldsymbol{1} $ is a vector of ones (With the appropriate dimension in each use above) the projections steps could be optimized. For example $ \boldsymbol{1}^{T} \boldsymbol{1} = m $ or $ \boldsymbol{1}^{T} \boldsymbol{1} = n $ (Number of rows / columns) and using sum() and repamt() to imitate the multiplication by the vector.

I implemented the solution in MATLAB and got the following results compared to CVX:

The iterative methods are slower than CVX for this task (I used MOSEK).

Yet, this is on small size of the data. Probably they will be able to be faster for large scale problem (Still you'll have to give a solver a lot of time to get the result).

I'd also recommend adding Line Search. As this problem is not well poised.


问:
How can I find Sobolev generalized derivative of function $f(t)$
Given following function
$$
f(t) :=
\begin{cases}
 0 & \text{ if } t \in (-\infty ,1]  \\
 2x -2& \text{ if } t \in (1, 2]  \\
 2& \text{ if } t \in (2, 3)  \\
 8 - 2x& \text{ if } t \in (3, 4]  \\
 0 & \text{ if } t \in (4, \infty)
\end{cases}
$$
How can I find Sobolev generalized derivative of function $f(t)$

答:
In general, if $f$ is continuous and piecewise continuously differentiable (say on the intervals $(-\infty,t_1],[t_1,t_2],\dots,[t_{n-1},t_n],[t_n,\infty)$), then
$$
g(t)=\begin{cases}f'(t)&\text{if }t\neq t_k\\0&\text{otherwise}\end{cases}
$$
is a weak derivative of $f$. This can be easily seen by splitting up the integral in the definition of weak derivatives and integrating by parts in each of the pieces. I leave it up to you to apply this result to the function at hand.


问:
Question about power of square root
I am currently prepping for next year's math course. Currently, I am practicing complex numbers and I have come across something I don't understand.

The problem is how to write the following complex number in rectangular form:
$(1+i)^{13}$

I know the argument(z) is $\dfrac{(13 \cdot \pi)}{4}$

but when trying to find modulus, the solution says it is $\sqrt{2}^{13}$. I follow this so far but I don't get what the rule is for the next step, which is $2^6 \cdot \sqrt(2)$.
What is happening here?

Any help is very much appreciated!

Title: Question about power of square root

Body: Hint: $x^{13}=x^{12} x = (x^2)^6 x$


Title: Question about power of square root

Body:
I am currently prepping for next year's math course. Currently, I am practicing complex numbers and I have come across something I don't understand.

The problem is how to write the following complex number in rectangular form:
$(1+i)^{13}$

I know the argument(z) is $\dfrac{(13 \cdot \pi)}{4}$

but when trying to find modulus, the solution says it is $\sqrt{2}^{13}$. I follow this so far but I don't get what the rule is for the next step, which is $2^6 \cdot \sqrt(2)$.
What is happening here?

Any help is very much appreciated!

Body: The absolute value is$$\left|(1+i)^{13}\right|=|1+i|^{13}=\sqrt2^{13}$$and\begin{align}\sqrt2^{13}&=\sqrt2^{12}\times\sqrt2\\&=\left(2^{1/2}\right)^{12}\times\sqrt2\\&=2^6\sqrt2.\end{align}


问:
Simplify $\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}$
The double square roots can be removed by $8+2\sqrt7 = (\sqrt7+1)^2$ and $$4-\sqrt7 = \frac{1}{4}(16-4\sqrt7) = \frac{1}{4}(14+2-2\sqrt2\cdot\sqrt14) = \frac{1}{4}(\sqrt14-\sqrt2)^2$$The expression simplifies to $\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2}$. This problem was given to me as an MCQ, the four options were:

A) $1$,

B) $2$,

C)$32\sqrt3$

D) $36\sqrt3$.

I am not getting any of these options, $$\begin{eqnarray*}\frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2} &=& \frac{1}{2}\cdot \frac{\sqrt14-\sqrt2}{\sqrt7+1-\sqrt2} \cdot \frac{\sqrt7+1+\sqrt2}{\sqrt7+1+\sqrt2} \\ &=& \frac{1}{2}\cdot \frac{6\sqrt2+2\sqrt7-2}{6+2\sqrt7} \\ &=& \frac{1}{4}(9\sqrt2-3\sqrt14+4\sqrt7-10) \end{eqnarray*}$$ Is my calculation incorrect ?

答:
Let $\sqrt{8+2\sqrt{7}}=\sqrt{x}+\sqrt{y} \implies x+y+2\sqrt{xy}, \implies x+y=8, xy=7 \implies x=1,y=7$ and let $\sqrt{4-\sqrt{7}}=\sqrt{p}-\sqrt{q} \implies p+q=4, pq=7/4 \implies p=7/2, q=1/2.$
Then
$$F=\frac{\sqrt{4-\sqrt{7}}}{\sqrt{8+2\sqrt{7}}-\sqrt{2}}= \frac{\sqrt{7/2}-\sqrt{1/2}}{1+\sqrt{7}-\sqrt{2}}$$

Something is amiss!


问:
an exponent problem
I need the value of $x$, but I cannot figure out how to deal with ${\sqrt x-1}$ in the following equation.

$x^{\sqrt x-1} =3/2$

Could anyone please help me to figure this out.

答:
Let's substitute, $x=y^2$,

$$
x^{\sqrt{x}-1}=(y^2)^{y-1} = y^{2y-2} = \frac{3}{2}
$$

$$
\implies y^{y-1} = \sqrt{\frac{3}{2}}
$$

This is still an implicit equation but we got rid of the $\sqrt{x}$.

Here is the Numerical Solution by Wolfram. This problem cannot be solved analytically using elementary operations. You need to follow some algorithm like [Bisection Method][2]. They work by taking some initial interval at the start say $[1, 100]$. Check the function value at both points. Then bisect the current interval i.e. check the value at $y=50$. We see that the root lies in $[1, 50]$. This way we go on bisecting the interval to get to the solution.

The solution using any of the algorithms is $x=0.3616$ or  $x=2.25$.

[2]: https://en.wikipedia.org/wiki/Root-finding_algorithms#:~:text=The%20simplest%20root%2Dfinding%20algorithm,point%20that%20bisects%20the%20interval).

Title: an exponent problem


问:
Integral equals zero and empty interior
While trying to prepare for my Real Analysis exam I encoutered the following exercise:

Problem: Let $f:[a, b] \rightarrow \mathbb{R}$ be Riemann integrable. If the set $$
X=\{x \in[a, b] ; f(x) \neq 0\}
$$ has empty interior (notation: $\operatorname{int}(x)=\varnothing$) then $$\int_{a}^{b}|f(x)| d x=0$$

What have i tried? I started by trying to write what empty interior means and it went like this:

Since $\operatorname{int}(x)=\varnothing$ there does not exist $\epsilon > 0$ such that $$(\bar{x}-\varepsilon, \bar{x}+\varepsilon) \subset X$$ which means that $\forall \epsilon > 0$:

$$(\bar{x}-\varepsilon, \bar{x}+\varepsilon) \cap X^{c} \neq \varnothing$$

Let's call it $z$ the element in the set $(\bar{x}-\varepsilon, \bar{x}+\varepsilon) \cap X^{c}$

Since $z$ is in the intersection of the aforementioned set, we have that $z \in X^{c}$ and therefore $$\left\{\begin{array}{l}z \in[a, b] \\ f(z)=0\end{array}\right.$$

Finally, one is able to infer that $|z-\bar{x}|<\varepsilon$ since $z \in (\bar{x}-\varepsilon, \bar{x}+\varepsilon)$

But now I am stuck. I did find the same exercise already answered here If the set on which a Riemann integrable function $f$ is nonzero has empty interior then the integral of $|f|$ is $0$. but i am looking forward to someone looking at what i have tried so far. Also, i am very curious to know whether one can finish the exercise using what i have already done.

Thanks in advance, Lucas

答:
This community wiki solution is intended to clear the question from the unanswered queue.

Brian M. Scott has answered your question in his comments. Let $[r,s]$, $r < s$, be any closed subinterval of $[a,b]$. Then $\inf_{x \in [r,s]} \lvert f(x) \rvert = 0$; otherwise $(r,s) \subset X$ which would show that $X$ has non-empty interior. Hence for each
partition $P = (x_0,\ldots, x_n)$ of $[a,b]$ the lower sum is
$$L(\lvert f \rvert;P) = \sum_{i=1}^n \inf_{x \in [x_{i-1},x_i]} \lvert f(x) \rvert (x_i - x_{i-1}) = 0 .$$
Since $f$ is Riemann integrable, also $\lvert f \rvert$ is and we get
$$\int_a^b \lvert f(x) \rvert dx = \sup_P L(\lvert f \rvert;P) = 0 .$$

Title: Integral equals zero and empty interior


问:
Factor of a product is product of factors
I'm trying to prove the uniqueness of prime factorization of any natural number $n\ge 2$ (I take $0\in\mathbb N$). I've shown the existence of a prime factorization for any such number (by infinite descent on the statement that 'for each $n\ge 2$, there exists a greatest prime that divides $n$').

To show uniqueness, I realize that I will be done if I can show that
$$
\forall a\;\forall b\;\forall c\;\left(c\text{ divides } ab\implies\exists m\;\exists n\; (m\text{ divides } a\;\wedge\; n\text{ divides } b\;\wedge\; c=mn)\right).
$$

But after a lot of futile attempts, I'm not able to "see" any path leading me to this statement. Any help?

Note: I haven’t proven Euclid’s division lemma, and am not working with gcd’s. Hence I’ll appreciate a way that avoids these.

答:
Proof of the uniqueness of prime factorization :

Let's suppose we have two factorizations : $p_1 ^{\alpha_1}...p_r ^{\alpha_r} = q_1 ^{\beta_1}...q_t ^{\beta_t}$.

Then for $1\leqslant i \leqslant r$, we have $p_i | q_1 ^{\beta_1}...q_t ^{\beta_t}$, and the $q_l$ are prime, so there exists one $1\leqslant j \leqslant t$ such that $p_i | q_j$ so $p_i = q_j$.

By doing the same with the $q_j$, we get that $\{p_1,...,p_r\} = \{q_1,...q_l\}$ and as all the $p_i$ and $q_j$ are different, we have that $r=l$.

We can then reorder the $(p_i)$ and $(q_j)$ to have $p_i = q_i$.

Let's prove we have $\alpha_i=\beta_i$ for $1\leqslant i\leqslant r$.

For $1\leqslant i\leqslant r$, $p_i^{\alpha_i} |q_1 ^{\beta_1}...q_r ^{\beta_r}$ and by Gauss Lemm, since $p_i$ is prime with all the $q_j$ for $j\geqslant 2$, we have $p_i^{\alpha_i} | q_i^{\beta_i} = p_i^{\beta_i}$, and we can do the same with $q_i$, so $p_i^{\alpha_i} = p_i^{\beta_i}$.

So $\alpha_i = \beta_i$, and this for all $i$.

Then the decomposition is unique.

For a proof of Euclid Lemma :

Let's suppose there exists $p$ prime and $a,b\in \mathbb{N}$ such that $p$ does not divide $a$ and $b$, but $p|ab$. Let's take, for $p$ and $a$ given, among all the possibilities of $b$, the smallest.

Then $0<b<p$ (because if not we could replace $b$ by a smaller one). We note $r$ the rest of Euclid's division of $p$ by $b$, which is not null since $p$ is prime.
Then $p=mb+r$ so $ar = ap -mab$ is a multiple of $p$.

But we know that $0<r<b<p$ so $ar$ is not a multiple of $p$ because of the minimality of $p$.

So we've got a contradiction, and that proves Euclid's Lemma.


问:
Proving the homology version of Cauchy's Theorem through the homotopy version
I wish to know if one can prove the homology version of Cauchy's Theorem through the homotopy version, and if moreover,  the following argument correct.

The theorem is as follows

Let $\gamma$ be a closed chain in an open set $U$, and assume $\gamma$ is homologous to $0$ in $U$. Let $f$ be holomorphic in $U$. Then $$\oint_\gamma f=0$$

My proof:

Since $\gamma$ is homologous to $0$, it follows that $\gamma$ has a winding number of 0 with respect to all points that are outside of the natural domain of $f$ (i.e the compliment of $U$). This implies that we either have $(1)$ $\gamma$ doesn't enclose any points outside the natural domain of $U$ or $(2)$ $\gamma$ encloses $n$ many points/regions an even amount of times, half clockwise and half anticlock wise.

For $(1)$ we see that the region enclosed by $\gamma$ is therefore simply connected and hence the result follows due to the homotopy form of Cauchy's Theorem.

For $(2)$ we partition the how many loops we have, say $2k$, into $k$ single clockwise loops and $k$ anti clockwise loops. We see that the integral is given by

$$\oint_\gamma f=\sum_\text{clockwise} \oint_{\gamma_i} f +\sum_\text{anticlockwise} \oint_{\gamma_i} f$$

For every single such closed path we can find a function such that each of these paths is homotopic to a small disc enclosing the corresponding region/point. We can next apply Cauchy's Integral formula to get values for each of the integrals. Since half of them will be equal to the other half but negative we get the desired result.

Note: In the case where a path encloses say $j$ many regions/points outside of $U$, we can construct a homotopy between the path and a path consisting of a small circle around each point and two way line connecting  each disc. We can use Cauchy's Integral Formula for the discs, and in regards to the inner closed paths that connect the discs, the integral along them is zero since the enclosure is simply connected. Thus, appling the same argument as in $(2)$ we get the desired outcome.

My question:
Is this proof correct and if it is not where have I gone wrong?

All help is appreciated.

答:
There are two problems with this approach:

The first one is that "the region enclosed by $\gamma$ is therefore simply connected" is true but highly non trivial (it follows from the Jordan curve theorem).

The second one is a real mistake: it is not necessarily true that you can partition your chain as you described. Consider, as a counterexample, the chain composed by the single curve pictured here:

In $\mathbb{C}-\{z_0,z_1\}$, the curve is homologous to $0$, but it is not homotopically trivial$^1$, and thus one cannot apply the decomposition you proposed. Also, it is not necessarily true that the chain encloses only a finite
number of points outside $U$, making the sum you write a series.

There is another straightforward proof of the homological version of Cauchy, given
the homotopical one (actually, some corollaries of it), by Dixon (see Lang's "Complex analysis" for a complete exposition of it). The basic idea is to prove Cauchy's formula directly (and from that we easily obtain Cauchy's theorem), by considering
$$g(z,y):=\begin{cases}\frac{f(z)-f(y)}{z-y}&\text{if }z\neq y \\
f'(z)&\text{if }z=y\end{cases}\\
h(z):=\begin{cases}\int_\Gamma g(z,y)dy &\text{if }z\in U\\
\int_\Gamma \frac{f(z)-f(y)}{z-y}dy &\text{if } W(z,\Gamma)=0
\end{cases}
$$
It is not hard to prove that $h$ is holomorphic (Hint: use Morera) on $U\cup \{z\in \mathbb{C}:W(z,\Gamma)=0\}=\mathbb{C}$, and that $\lim_{z\to \infty}h(z)=0$. By Liouville's theorem, $h(z)\equiv 0$ and thus
$$
\int_\Gamma \frac{f(y)}{z-y}dy=\int_\Gamma \frac{f(z)}{z-y}dy=2\pi if(z)W(z,\Gamma)$$

$^1$:If you know what $\pi_1$ and $H_1$ are (the fundamental and the first homology group) this has a nice interpretation: the curve pictured is in the homotopy class of the commutator of the two generators of $\pi(\mathbb{C}-\{z_0,z_1\})$. Since this group is not abelian, the curve is not homotopically trivial. However, $H_1$ is abelian, and thus the curve is homologous to $0$.


问:
Understanding probability of independent events
I am trying to learn probability, and having a hard time to be honest. What I try to do is to break a problem in terms of event space. For eg, event space for rolling a die is {1,2,3,4,5,6}. For rolling two dice it would be 36 elements with pairs from 1,1 to 6,6. Then I try to calculate probability of a given event using simple basic formula of occurrences/total events.

On topic of probability of independent events, suppose I roll 2 dice and want to calculate probability of 2 on first die and 5 on second die. Book I am reading uses "And" rule for independent events:

```<code>P(2 and 5) = P(2) * P(5)
</code>```

I am not sure of this approach. When we say event "2 in first roll, 5 in second roll", we basically are talking of an event "2 rolls of die". And so our event space consists of 36 elements, each representing a pair as discussed above. But in above formula, we are multiplying probabilities of events which come from  a different event space(containing 6 elements). So basically we are trying to solve problem belonging to some event space using probabilities from different event space.

Am I wrong in my thinking here?

答:
Honestly the simple way is to draw the sample space, that is the following

and immediately you have a clear sight of the elements you need to calculate your probability.

Example 1: sum of the two dice $\leq 7$ and first die =1

Result: [$\frac{6}{36}$]

Example 2: first die 2 and second die 5, as you requested: result $\frac{1}{36}$

Example 3: sum of the two dice $\leq 7$ and at least one "1": Result $\frac{11}{36}$

and so on...any question you have to answer you have only to "visualize" and count the favourable events in the above table and divide it by $36$

PS: the plural of "die" is "dice", I amended you post


答:
I don't quite understand what you mean but I'll try to explain what's going on here.

In the first roll, the probability(2) = 1/6 (assuming a fair die). In the second roll, the probability of getting any number stays the same 1/6. The two rolls are independent - they don't affect each other's outcome.

So we get this:
P(R1 = 2, R2 = Y) = 1/6 Y -> {1,2,3,4,5,6}

Thus, we know that if the second roll is taken in fair circumstances, you are going to have to square the previous probability (1/6)², knowing that the sum of all the possible outcomes in the second roll & P(R1 = 2) is equal to P(R1 = 2)

p.s
If someone is willing to format my formulas using latex you're welcome.
I hope I answered your question.

Title: Understanding probability of independent events

Body:
I am trying to learn probability, and having a hard time to be honest. What I try to do is to break a problem in terms of event space. For eg, event space for rolling a die is {1,2,3,4,5,6}. For rolling two dice it would be 36 elements with pairs from 1,1 to 6,6. Then I try to calculate probability of a given event using simple basic formula of occurrences/total events.

On topic of probability of independent events, suppose I roll 2 dice and want to calculate probability of 2 on first die and 5 on second die. Book I am reading uses "And" rule for independent events:

```<code>P(2 and 5) = P(2) * P(5)
</code>```

I am not sure of this approach. When we say event "2 in first roll, 5 in second roll", we basically are talking of an event "2 rolls of die". And so our event space consists of 36 elements, each representing a pair as discussed above. But in above formula, we are multiplying probabilities of events which come from  a different event space(containing 6 elements). So basically we are trying to solve problem belonging to some event space using probabilities from different event space.

Am I wrong in my thinking here?


问:
Eigen vector of the adjoint operator. Error in this approach?
Reisz Representation Theorem: If $V$ is finite-dimensional and $\phi$ is a linear functional on $V$. Then, there is a unique vector $v \in V~|~\phi(v) = \langle v,u\rangle~\forall~v \in V$

Definition: If $T \in L(V,W) $, then $T^* \in L(W,V)$ represents the adjoint of the linear map $T$. Then, by definition :

$\langle ~Tv, w~\rangle = ~\langle v, T^*w ~\rangle$.

By the Riesz Representation theorem: $T^*w$ is the unique vector in $V$ such that the above equation holds.

This means if for a given $v ~,~\exists ~x \in V ~|~\langle Tv,w~\rangle=~\langle v,x~\rangle$, then $T^*w=x$ as $T^*w$ must be unique by the Riesz Representation theorem.

Am I making a fundamental mistake in the above argument?

As an example, suppose $T \in L(V,V)$ and $T$ has eigenvalue and eigen vector respectively as $\lambda,v$, then : $~\langle Tv,v~\rangle = ~\langle \lambda v,v ~\rangle = ~\langle v, \bar{\lambda}v~\rangle = ~\langle v, T^*v~\rangle$

By the above arguments, this would mean that $v$ is always an eigenvector for $T^*$ which is certainly not true. So, could someone tell me what is the error that I am making?

Another example:  Suppose $v \in V$ is an eigen vector of $V$ and $w \in V$ is arbitrary. Define : $\phi(w) = \langle T v,w \rangle = \langle  \lambda v, w\rangle = \langle  v, \bar{\lambda} w \rangle = \langle v , T^*w \rangle$. Now, $v$ is fixed and $w$ is any arbitrary vector. Therefore, by uniqueness : $T^*w=\bar{\lambda} w ~\forall~w \in V$ which cannot be true. What could be the error here then? Thanks

答:
Suppose $v \in V$ is an eigen vector of $V$ and $w \in V$ is arbitrary. Define : $\phi(w) = \langle T v,w \rangle = \langle  \lambda v, w\rangle = \langle  v, \bar{\lambda} w \rangle = \langle v , T^*w \rangle$. Now, $v$ is fixed and $w$ is any arbitrary vector. Therefore, by uniqueness : $T^*w=\bar{\lambda} w ~\forall~w \in V$ which cannot be true because you are taking $v$ as eigen vector not any random vector in $V$.

Simple fact: you can prove it easily that:
$$<v,w>=<v,w'>$$ for all $v$in $V$ then,
$$w=w'$$ holds!!


答:
There is an analytical answer to this question.

Let us assume an absolute error $\epsilon$, such that

$$\left(1+\dfrac xn\right)^ne^{-x}=1-\epsilon.$$

With $t:=1+\dfrac xn$, we have

$$t^ne^{n(1-t)}=1-\epsilon$$

or

$$-te^{-t}=-\frac{\sqrt[n]{1-\epsilon}}e$$ and

$$t=-W\left(-\frac{\sqrt[n]{1-\epsilon}}e\right),$$

$$x=-n\left(W\left(\frac{\sqrt[n]{1-\epsilon}}e\right)+1\right).$$

Title: How good is the approximation $e^{-x} = \left(1- \frac{x}{n}\right)^n$?

Body: We know that for $n\rightarrow \infty$, the above converges. But how good of an approximation is it for non arbitrarily large $n$ (i.e. $n << \infty$)?


问:
Is every orthogonally diagonalizable matrix invertible?
I've seen a lot of discussion going the other way, but I'm considering the fact that if a matrix A is orthogonally diagonalizable then $A=PDP^{-1}=PDP^{T}$. And because the RHS is invertible so is A?

答:
Take a diagonal matrix $D$ with some zero on the diagonal. Then it is not invertible. Then take whatever orthogonal and invertible matrix you want $P$.

Then $A = PDP^{T}$ is clearly not invertible (since it has the same rank as $D$), but orthogonally diagonalizable.

$0$ is a trivial example.


问:
Test function $\phi$ for which $\int \phi dx = 1$
I'm looking for a test function $\phi$ in $\mathbb{R}^n$ (i.e. $\phi\in C_{\infty}(\mathbb{R}^n)$ for which $[\phi]$ is compact) such that $\int_{\mathbb{R}^n} \phi(x)dx=1$. I have seen multiple examples of test functions, but they all involve exponentials. I could take the easy road and pick such a test function and divide it by its integral over $\mathbb{R}^n$. But this raises the question of integral computation: f.ex. over $\mathbb{R}$, $$ \int_{-1}^1 \exp\left(-\frac1{1-x^2}\right)dx.$$

So,

is there a test function, for which the integral can easily be computed and equals 1 (preferably without exponentials)?

How to compute the integral mentioned above; or generally $\int_{B(a,R)} \exp\left( -\frac1{R^2-|x-a|^2}\right)dx$?

Thanks!

答:
Let $\Phi(x_1,x_2,\dots,x_n)=\Psi(x_1)\Psi(x_2)\cdots \Psi(x_n)$, where $\Psi:\mathbb{R}\to \mathbb{R}$ is given by the standard bump function over $\mathbb{R}$:
$$\Psi(x)=\begin{cases} \exp\left(-\frac{1}{1-x^2}\right),&x\in (-1,1) \\ 0,&\mathrm{otherwise}. \end{cases}$$
Then $\Phi$ is a smooth test function with compact (closed) support with a non-zero integral over $\mathbb{R}^n$, but is not normalized to $1$. To this end, we compute the integral of $\Psi(x)$ over $\mathbb{R}$:
$$\begin{align} \int_{\mathbb{R}} \Psi(x)~dx&=\int_{-1}^1 \exp\left(-\frac{1}{1-x^2}\right)~dx\\&=2\int_{0}^1 \exp\left(-\frac{1}{1-x^2}\right)~dx\\&=2\int_0^{\infty} \exp\left(-\frac{1}{1-\tanh^2(u)}\right)~\operatorname{sech}^2(u)~du\\&=2\int_0^{\infty} \exp\left(-\cosh^2(u)\right)~\operatorname{sech}^2(u)~du\\&=-2\int_0^{\infty} \tanh(u)\cdot (-2\sinh(u)\cosh(u)\exp(-\cosh^2(u))~du\\&=4\int_0^{\infty} \exp(-\cosh^2(u))\sinh^2(u)~du\\&=2e^{-1/2}\int_0^{\infty} \exp\left(-\frac{\cosh(2u)}{2}\right)(\cosh(2u)-1)~du\\&=e^{-1/2}\int_0^{\infty} \exp\left(-\frac{\cosh(v)}{2}\right)(\cosh(v)-1)~dv\\&=e^{-1/2}\left(K_1(1/2)-K_0(1/2)\right)\approx 0.4439938,\end{align}$$
where we have used that for $\Re(x)>0$ the modified Bessel function admits the integral representation
$$K_{\alpha}(x)=\int_0^{\infty} e^{-x\cosh(t)}\cosh(\alpha t)~dt.$$
Note that above we use the substitutions $x=\tanh(u)$ and $v=2u$ and we have used integration by parts in the third line. Hence a smooth test function for $\mathbb{R}^n$ whose integral is $1$ and whose (closed) support is compact is given by
$$\Phi_1(x_1,x_2,\dots,x_n)=\frac{e^{n/2}}{(K_1(1/2)-K_0(1/2))^n}\Psi(x_1)\Psi(x_2)\cdots \Psi(x_n).$$

Edit: After looking at Are there other kinds of bump functions than $e^\frac1{x^2-1}$?, it seems like there is a way to avoid non-elementary functions (if for some reason you dislike modified Bessel functions). Inspired by Robert Israel's answer, with
$$\Psi(x)=\begin{cases} \frac{4\left( {x}^{2}+1 \right)\ {{\rm e}^{{\frac {4x}{{x}^{2}-1}}}}}{\left( \left( {x}^{2}-1 \right)  \left( 1+{{\rm e}^{{\frac {4x}{{x}^{
2}-1}}}} \right)\right)^2},&x\in (-1,1) \\ 0,&\mathrm{otherwise}, \end{cases}$$
one has that $\int_{\mathbb{R}} \Psi(x)~dx=1$ (it has elementary anti-derivative $\frac{1}{1+e^{(4x)/(x^2-1)}}$ on $(-1,1)$), so the function $\Phi(x_1,x_2,\dots,x_n)=\Psi(x_1)\Psi(x_2)\cdots \Psi(x_n)$ should satisfy the properties you want.


Title: an exponent problem

Body:
I need the value of $x$, but I cannot figure out how to deal with ${\sqrt x-1}$ in the following equation.

$x^{\sqrt x-1} =3/2$

Could anyone please help me to figure this out.

Body:
x=9/4 is a solution

$\sqrt(9/4)=3/2$

$3/2-1=1/2$

$9/4^{\sqrt(9/4)-1}=\sqrt(9/4)=3/2$


答:
In some sense, your understanding is correct -- and your confusion is understandable. Let me explain.

If we let $X$ denote the value of the first die, and we let $Y$denote the value of the second die, then we may study $(X,Y)$, which exactly takes values in $\{(1,1), (1,2),\ldots,(6,5),(6,6)\}$.

Independence tells us that
$$
\mathbb{P}(X = x, Y = y) = \mathbb{P}(X = x)\mathbb{P}(Y=y).
$$
Indeed the right hand side involves events from different event spaces, namely the marginal event spaces of $X$ and $Y$. In this case, both are $\{1,2,3,4,5,6\}$.

This shows us that if $X$ and $Y$ are independent, we only need to know how they act on their individual event space, i.e. we need only to assign a probability to each event $(X=x)$ and $(Y=y)$.

Of course, we may relate an event such as $(X=x)$ to the simultaneous event space. Clearly, $(X=x) = (X=x, Y \in \{1,2,3,4,5,6\})$. In particular,
\begin{align*}
\mathbb{P}(X=x) = \sum_{y=1}^6 \mathbb{P}(X=x, Y = y).
\end{align*}
(This holds even if $X$ and $Y$ are not independent.)

I hope this somewhat answers your questions.


问:
$G=D_8$ and $H=\langle r^2 \rangle $ why is $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$
In my homework I have:

$G=D_8$ (Dieder) and $H=\langle r^2 \rangle$

It it said is that $G/H \cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$

Why is that? Why not $\mathbb{Z}/4\mathbb{Z}$?

答:
Just Calculate $G/H$:
Clearly, $$G/H=\{g o H|g \in G \}$$
So, $$G/H=\{1 o <r^2>,r o <r^2>,s o <r^2>,rs o <r^2>\}$$
where $s$ is reflection and $r$ is rotation and $o$ is composition operation of $G$.
Clearly $G/H$ is not cyclic as it does not have any generator but $\mathbb{Z_4}$ is cyclic.


问:
Point $D$ is on line $BO$ such that $O$ is between $B$ and $D$ and $\angle ADC = \angle ABC$.

There is a $\Delta ABC$ with $AB < BC$ , with orthocenter $H$ and circumcenter $O$. Point $D$ is on line $BO$ such that $O$ is between $B$ and $D$ and $\angle ADC = \angle ABC$. The semi-line starting at $H$ and parallel to $BO$ meets the circumscribed circle at points $E$ and $F$ respectively. Prove that $BH = DE$.

What I Tried:  Here's a picture in Geogebra to get an idea of the problem :-

Really Tough. Spent around $1.5$ hours on this problem but no progress, as I am a little weak at Geometry I tried elementary methods like angle-chasing , similarity and so on, but no success. Another problem is that $DABC$ is not a parallelogram as it may seem, so angle-chasing and similarity do not work to any triangles at all till now, so that I could have found something useful.

We basically need to prove that $DEHB$ is an Isosceles Trapezium, and I have no idea. I did not even any reasonable cyclic quadrilaterals to work on and gain some information. I am quite stuck on this problem.

Can anyone help me?

答:
Let $O', H'$ be the circumcenter and the orthocenter of $\triangle ADC$ respectively. Denote, furthermore, by $G$ the orthogonal projection of $O$ onto $CA$.

First of all, observe that the Law of Sines implies that the circumcircle of $\triangle ADC$ has the same radius as the circumcircle of $\triangle ABC$, since $R'=\frac{CA}{\sin\angle CDA}=\frac{CA}{\sin\angle ABC}=R$. It follows that $G$ is the midpoint of $OO'$.

Besides, it is well-known (at least considering the ratios of the Euler line), that $$2OG=BH\implies OO'=BH$$ and since both $OO'$ and $BH$ are perpendicular to $CA$, we have that $OO'HB$ is a parallelogram. Yet this implies that $OB\parallel O'H$, and, hence, $O',E,H$ are collinear. Notice now that, since $OE'\parallel OD$ and $DO'=R'=R=EO$, the quadrilateral $DO'EO$ is an isosceles trapezium; we infer that the diagonals have the same length, i.e. $DE=OO'$. Thus $$DE=OO'=BH$$

Some additional observations. As I noticed while playing with Geogebra, this problem could lead to other nice relationships such as

The Nine-point-Center $N_9'$ of $\triangle ADC$ lies on $BD$.

$H'$ and $E$ are symmetric wrt $BD$.


问:
How to show $I_k(t)=\int_\Omega (u(x,t)-k)_+^2 dx$ is absolutely continuous for any $u\in \mathring W^{1,1}_2(Q_T)$?
$\Omega$ is bounded smooth domain of $\mathbb R^n$.

$Q_T=\Omega\times [0,T]$.

$\overset{\circ}{W}^{1,1}_2(Q_T)$ is the parabolic Sobolve space and is zero at $\partial \Omega \times[0,T]$.

$u_+=\max\{u,0\}$.

How to show for any $k>\sup_\limits{\partial_pQ_T} u$,   (where $\partial_pQ_T=\partial \Omega \times[0,T] \cup\Omega\times \{t=0\}$)
$$
I_k(t)=\int_\Omega (u(x,t)-k)^2_+dx
$$
is  absolutely continuous on $[0,T]$ ?

答:
It is enough to show that $I_k$ is in $W^{1,1}(0,T)$. So multiply by $f’(t)$, where $f\in C^1_c(0,T)$, integrate in $t$ in $(0,T)$ and then use the definition of weak derivative to integrate by parts.Since the function $g(z)=(z-k)_+$ is Lipschitz continuous, you can apply the chain rule, which tells you that $v(x,t)=(u(x,t)-k)_+$ is in $W^{1,2}$ with
$$
\frac{\partial v}{\partial t}(x,t)=\left\{
\begin{array}
[c]{ll}%
\frac{\partial u}{\partial t}(x,t) & \text{if }u(x,t)>k,\\
0 & \text{otherwise.}%
\end{array}
\right.
$$
Hence,
\begin{align*}
\int_{0}^{T}f^{\prime}(t)I_{k}(t)\,dt  & =\int_{0}^{T}\int_{\Omega}f^{\prime
}(t)v^{2}(x,t)\,dxdt\\&=-\int_{0}^{T}\int_{\Omega}f(t)2v(x,t)\frac{\partial
v}{\partial t}(x,t)\,dxdt\\
& =-\int_{0}^{T}\int_{\Omega}f(t)2(u(x,t)-k)_{+}\frac{\partial u}{\partial
t}(x,t)\,dxdt\\
& =-\int_{0}^{T}f(t)\int_{\Omega}2(u(x,t)-k)_{+}\frac{\partial u}{\partial
t}(x,t)\,dxdt.
\end{align*}
This shows that the weak derivative of $I_{k}(t)$ is the function
$$
\omega(t)=\int_{\Omega}2(u(x,t)-k)_{+}\frac{\partial u}{\partial t}(x,t)\,dx,
$$
which is integrable since by Holder's inequality
\begin{align*}
\int_{0}^{T}|\omega(t)|\,dt  & \leq2\int_{0}^{T}\int_{\Omega}(u(x,t)-k)_{+}%
\left\vert \frac{\partial u}{\partial t}(x,t)\right\vert \,dxdt\\
& \leq2\left(  \int_{0}^{T}\int_{\Omega}(u(x,t)-k)_{+}^{2}\,dxdt\right)
^{1/2}\left(  \int_{0}^{T}\int_{\Omega}\left\vert \frac{\partial u}{\partial
t}(x,t)\right\vert ^{2}dxdt\right)  ^{1/2}<\infty.
\end{align*}


问:
All real numbers $(p,q)$ such that $|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$
Find all pairs of real numbers $(p, q)$ such that the inequality
$|\sqrt{1-x^{2}}-p x-q| \leq \frac{\sqrt{2}-1}{2}$ holds for every $x \in[0,1]$

Originally I thought to rephrase it in geometric terms, seeing one degree terms I think about equation of a line.

I also have that $\sqrt{1-x^{2}}-\frac{\sqrt{2}-1}{2} \leq p x+q \leq \sqrt{1-x^{2}}+\frac{\sqrt{2}-1}{2}$

I think that a geometric solution might be possible, please help me to proceed.

Thanks.

答:
Let $f(x) = \sqrt{1-x^{2}}-p x-q$, then the condition becomes $$|f(x)| \leq \frac{\sqrt{2}-1}{2}$$
Thus, the extrema of $f$ should lie in $\left[ -\frac{\sqrt{2}-1}{2}, \frac{\sqrt{2}-1}{2}\right]$. Now, let's see how its extrema look like:
$$f'(x) = \frac{-x}{\sqrt{1-x^2}} - p = 0 $$ After bringing to a common denominator, numerator should be equal to zero: $$ -x - p\sqrt{1-x^2} = 0 \\[2mm] x = r\sqrt{1-x^2}, \ \ (r=-p) \\[2mm] x^2 = r^2(1-x^2) \\[2mm] (1+r^2)x^2 = r^2 \\[2mm] x_0 = \pm \frac{r}{1+r^2} = \mp\frac{p}{\sqrt{1+p^2}}$$
However, since $x \ge 0$, we get $$x_0 = -\frac{p}{\sqrt{1+p^2}} $$
Therefore, $$\begin{align}f(x_0) &= \sqrt{1-\frac{p^2}{1+p^2}} + \frac{p^2}{\sqrt{1+p^2}}-q \\[1mm]&= \frac{1 + p^2}{\sqrt{1+p^2}} -q \\[1mm]&= \sqrt{1+p^2}-q \end{align}$$
Hence, the condition seems to be satisfied for all $p$ and $q$ such that
$$\left| \sqrt{1+p^2} -q \right| \le \frac{\sqrt2 - 1}{2} \tag{1}$$
Also, we need to see the edge cases when $x = 0$ and $x = 1$, which give
$$|f(0)| = \left|1 - q\right| \le \frac{\sqrt{2}-1}{2} \tag{2}$$ and $$|f(1)| = \left|-p - q\right| \le \frac{\sqrt{2}-1}{2} \tag{3}$$ Finally, combining $(1)$, $(2)$ and $(3)$ we get the final answer.

I didn't solve the last three inequalities putting all together, but after graphing I noticed (but still needs to be justified rigorously by solving the inequalities) that only one pair satisfies the statement, namely, $$(p,q) = \left(-1, \frac{\sqrt 2 + 1}{2}\right)$$

Below is the graph with those $p$ and $q$:


问:
Show that a subset $V$ of $X$ is open iff every point $x \in V$ has a neighborhood $V_x \subseteq V$
Show that a subset $V$ of $X$ is open iff every point $x \in V$ has a neighborhood $V_x \subseteq V$. </p>

<p>Any help would be appreciated. I am working in topological spaces right now. </p>

答:
Considering the fact that implication $\Rightarrow$ is clear, you can finish the other implication and therefore your "proof" by realizing that
\begin{equation}
V = \bigcup_{x \in V} \{x\} \subset \bigcup_{V_x \subset V} V_{x} \subset V
\, .\end{equation}
That gives us that V is an union of open sets.


问:
If $n\ge 2$, find $\binom 31 -\binom 42 +\binom 53....(-1)^{n-1} \binom {n+2}{n}$
I have no clue on how to start solving. Can I get a hint to get me started?</p>

Body:
Hint:
This sum is equal to
$$\frac{1}{2} \sum_{k=1}^n (-1)^{k-1} (k+1)(k+2)$$


问:
Distribution of red balls in a picking without replacements
Suppose that there is a urn with $n$ balls, $pn$ red balls and $(1-p)n$ green balls, where $0< p < 1$.

I pick from this urn balls in a sequential order, uniformly without replacement. For $1 \leq  k \leq n$, let $R_k$ be the total number of red balls I have at the $k$'th step. I am interested in the distribution of $R_k$, for every $k$.

My attempt. First, we know that $\Pr[R_1=1] = p$, and that $\Pr[R_n = pn] = 1$. We also know that $$\Pr[R_{k+1} = R_k+1 | R_k] = \frac{pn - R_k}{n-k},$$
since there are $pn-R_k$ red balls left out of $n-k$ (if this value is below $0$ (or above $1$), I assume it is $0$ (or $1$)).
Thus, for some value $m$,
$$\Pr[R_{k+1} = m] = \Pr[R_{k+1} = R_{k}+1 | R_{k} = m-1]\times \Pr[R_{k} = m-1] \\
 + \Pr[R_{k+1} = R_{k} | R_{k} = m]\times \Pr[R_{k} = m], $$

which continues in a recursive manner. I am not sure how to extract from this recursion a closed formula of the distribution of $R_k$, or if it even possible.

答:
The distribution of $R_k$ is hypergeometric and can be found directly without any appeal to recursion.

Picking $k$ balls one by one in a specific order does not essentially differ from picking $k$ balls simultaneously if at stake is the finding of the distribution of $R_k$.

Recursion might come in sight if you are after the distribution of random vector $(R_1,…,R_k)$.


Title: If $n\ge 2$, find $\binom 31 -\binom 42 +\binom 53....(-1)^{n-1} \binom {n+2}{n}$

Body: I have no clue on how to start solving. Can I get a hint to get me started?</p>

Body:
Hint:

$${{r+2}\choose {r}} = {{r+2}\choose {2}} = \frac{(r+2)(r+1)}{2} $$ so the sum is $$\frac 12\sum_1^n (-1)^{r-1} (r+2)(r+1) \\ =\frac 12\sum_0^{\lfloor (n-1)/2 \rfloor}(2k+3)(2k+2) -\frac 12\sum_1^{\lfloor n/2 \rfloor} (2k+2)(2k+1)$$


问:
How do I prove that $\int_{-\infty}^x \frac{\sin(ny)}{\pi y}\text{d}y$ tends towards the Heaviside step function?
I want some (ideally short) proof of the statement
$$\int_{-\infty}^x \frac{\sin(ny)}{\pi y}\text{d}y\,\,\stackrel{n\rightarrow\infty}{\longrightarrow}\,\,H(x)$$
with which I can later proof that $\frac{\sin(nx)}{\pi x}\stackrel{n\rightarrow\infty}{\longrightarrow}\delta_0$. The proofs I found so far (1 2) always assume prior knowledge of something similar to my question, for instance "obvious" properties of $Si(x)$ function.

答: If $\lim_{A,B\to \infty}\int_{-A}^B f(y)dy$ converges then $F(x)=\int_{-\infty}^x f(y)dy$ is well-defined and $F(nx)\to F(\infty) H(x)$ boundedly on $\Bbb{R}$ and locally uniformly on $|x|\ge 1$.


答:
Hint
$$
\int_{-\infty}^x {\sin ny\over \pi y}dy=\int_{-\infty}^{nx} {\sin ny\over \pi ny}dny
=\int_{-\infty}^{nx} {\sin u\over \pi u}du
$$
Now, tend $n\to \infty$ for $x>0$ and $x<0$ and substitute the integral bounds with proper values.

Title: How do I prove that $\int_{-\infty}^x \frac{\sin(ny)}{\pi y}\text{d}y$ tends towards the Heaviside step function?

Body:
I want some (ideally short) proof of the statement
$$\int_{-\infty}^x \frac{\sin(ny)}{\pi y}\text{d}y\,\,\stackrel{n\rightarrow\infty}{\longrightarrow}\,\,H(x)$$
with which I can later proof that $\frac{\sin(nx)}{\pi x}\stackrel{n\rightarrow\infty}{\longrightarrow}\delta_0$. The proofs I found so far (1 2) always assume prior knowledge of something similar to my question, for instance "obvious" properties of $Si(x)$ function.


问:
Why is $r(\theta) = \frac{1}{\| \theta \|}$ here?

Why is $$r(\theta) = \frac{1}{\| \theta \|}$$ for $\theta \in S^{n-1}$, where $S^{n-1}$ is the $n-1$ dimensional unit spherical shell?

This is part of the following extract, which I'm a little lazy to type (so I'm putting a screenshot):

"since $r(\theta)$ is the largest number $r$ for which $r\theta \in K$" - why is this true, and how does it help?

By the way, we are working in $\mathbb{R}^n$, and using spherical polar coordinates.

答:
Earlier in the notes, Prof. Ball states that the notation $|\cdot|$ is reserved for the Euclidean norm. So here he must be talking about something else. It should be the norm $\|\cdot\|_K$; in this specific instance he is talking about the norm given by $K=$ the just mentioned octahedron, i.e. the $\ell^1$ norm $\|x\|=\sum_1^n |x_i|$.

It seems that definitionally, $r(\theta)$ is  the "radius of $K$ in direction $\theta$", namely the largest $r\ge0$ such that $r\theta \in K$. From $K = \{x\in \mathbb R^n : \|x\| \le 1\}$, it should be easy to see that all the points $r(\theta)\theta$ correspond to the boundary of the ball
$$ \partial K = \partial \{x\in \mathbb R^n : \|x\| \le 1\} = \{x\in \mathbb R^n : \|x\|=1\}= \{ r(\theta)\theta : \theta\in S^{n-1}\} $$
Thus, letting $x=r(\theta)\theta$ be a general point of $\partial K$, on taking norms,
$$ 1 = r(\theta)\|\theta\|$$
hence the result.

The context that would be missing from your question without the screenshot is that this is strongly dependent on the choice of $K$. if $\|x\|=|x|$ was indeed the euclidean norm, then the radius would be identically 1, which does indeed correspond to the unit ball in said norm.