问:
Permutations and fixed points
The permutation σ ∈ S2020 has its order = 2019. What is the maximum number of fixed points?

I was thinking that since its order is 2019, the permutation with 2019 different elements is equal to the neutral permutation. So in essence, there is only one point which stays the same, and by that I mean fixed. Is this correct or am I missing something ?

答:
$$2019=3×673$$
We can achieve a permutation of order $2019$ by letting there be a $3$-cycle and a $673$-cycle. Thus $2020-3-673=1344$ points are fixed, and since there is no other factorisation of $2019$ with smaller sum of factors, this is the best we can do.

Title: Permutations and fixed points


问:
How to prove there is only one positive root?
If I want to prove $x^n+x^{n-1}-a=0$ has only one positive root for $a> 0$ and $n \ge 2$
Can I say for $n=2$ : $x^2+x-a=0$ and we know only one of the roots is positive.
Now if we we know there is only one root for $x^n+x^{n-1}-a=0$
now I prove for $x^{n+1}+x^{n}-a=0$
In the end I get to $x=1$ , but I don't think I’m doing it correctly.
Should I use different $a$ s?

Is my way wrong? If it is, could you tell me what I should do?

答:
Your method is hard to understrand and  ts unclear what you are trying to do

Suppose  by contradiction ther is more than one possible positive root say $c,b$ are two of them.WLOG $c>b$ now $$c^n+c^{n-1}=a$$ $$b^n+b^{n-1}=a$$ subtract $$\color{red}{c^n-b^n}+\color{blue}{c^{n-1}-b^{n-1}}=0$$ which is not possible as both colored terms are positive

ContentLicense: CC BY-SA 4.0


问:
Properties of basic set theory
The question is about a set:
$$B=\{a_1,a_2,a_3,...,a_n\} \subset \mathbb R$$
And would like to know how to calculate $B^n$ where $n \in\Bbb N$?</p>

答:
Assuming $B = \{a_1,a_2,a_3,...,a_n\} \subset \mathbb R$, with $|B| = n$,

$B^n$ is the set of all ordered n-tuples of elements of $B$:

The exponent refers to the operation of the Cartesian Product of $B$ with itself, n times:
$$B^n = \underbrace{B \times B \times \cdots \times B}_{\Large\text{n times}} =\{(b_1, b_2,\cdots, b_n) \mid b_i \in B\}.$$

Can you figure out how to calculate: $|B^n|$? There are $n$ choices for position 1, $n$ choices for position 2, ..., $n$ choices for position $n$:

$$|B^n| = \underbrace{n\times n\times \cdots \times n}_{\Large \text{n times}} = n^n$$

ContentLicense: CC BY-SA 3.0

Title: Properties of basic set theory


问:
Closed-form formula for $E_{x}[\max(u^\top x,0)\max(v^\top x ,0)]$ where $u,v$ are fixed vectors in $\mathbb R^d$ and $x$ is uniform on the sphere
Let $x$ be uniformly distributed on the unit-sphere in $\mathbb R^d$ and let $u,v$ be fixed nonparallel vectors in $\mathbb R^d$

Question. Is there a closed-form formula for $f(u,v) := \mathbb E_{x}[\max(u^\top x,0)\max(v^\top x ,0)]$ ?

Due to rotationational symmetry of the distribution of $x$, my guess is that $f(u,v)$ admits a simple expression in terms of the angle between $u$ and $v$.

答:
We can look at normalized vector: $f(u,v) = ||u||\cdot||v||\cdot f\Big(\frac{u}{||u||}, \frac{v}{||v||}\Big)$, so let us assume that $u$ and $v$ are normalized. Then we can write $v = \rho u + \sqrt{1-\rho^2} u_2$ with $\rho = (u^Tv) \in [-1,1]$ and $u_2$ unitary orthogonal to $u$, and then complete $(u,u_2)$ into an orthogonal base of $\mathbb{R}^d$, and look at coordinates $(x_1,...,x_d)$ in that base.

Initial remarks:

the volume $V_k$ of the unit ball $B_k$ in $\mathbb{R}^k$ is $V_k = \frac{\pi^{k/2}}{\Gamma(k/2+1)}$

the density of $(R, \theta)$ s.t. $(R\cos(\theta), R\sin(\theta))$ is uniformly distributed on the unit disk is $\frac{R}{\pi}dRd\theta$

\begin{align*}
f(u,v) & = \displaystyle{\int_{B_d}} \max(x_1, 0)\max\big(\rho x_1 + (1-\rho)x_2, 0\big)\frac{dx_1...dx_d}{V_d} \\
 & = \displaystyle{\int_0^1} \displaystyle{\int_{-\pi}^{\pi}} \max(r\cos(\theta),0)\max(\rho r\cos(\theta) + \sqrt{1-\rho^2}r\sin(\theta),0) \cdot r d\theta \cdot \frac{(1-r^2)^{d/2-1} V_{d-2}}{V_d} dr \\
 & = \displaystyle{\int_{-\pi}^{\pi}} \max(\cos(\theta),0)\max(\rho \cos(\theta) + \sqrt{1-\rho^2}\sin(\theta), 0)d\theta \cdot \displaystyle{\int_0^1} r^3 (1-r^2)^{d/2-1} dr \cdot \frac{d}{2 \pi}\end{align*}

where $(1-r^2)^{(d-2)/2} V_{d-2}$ stems from the integration over $x_3,...,x_d$ with $(x_1,...,x_d) \in B_d$ and $x_1^2+x_2^2=r^2$, and on the last line we use our formula for $V_d$ and $V_{d-2}$.

Let us compute the two integrals. The second one is simpler:

$$\displaystyle{\int_0^1} r^3(1-r^2)^{d/2-1}dr = \Big[-\frac{(1-r^2)^{d/2} \big(2 + dr^2\big)}{d(d+2)}\Big]_0^1 = \frac{2}{d(d+2)}$$

In the first integral, the integrand is non zero iff $\theta \in \big[-\frac{\pi}{2}, \frac{\pi}{2}\big]$ and $\rho \cos(\theta) + \sqrt{1-\rho^2} \sin(\theta)$, so iff $\theta \in \big[0, \frac{\pi}{2}\big]$ or $-\mbox{Arctan}\big(\frac{\rho}{\sqrt{1-\rho^2}}\big) \le \theta \le 0$. Let us denote $\theta_0 = -\mbox{Arctan}\big(\frac{\rho}{\sqrt{1-\rho^2}}\big)$.

Thus the integral is \begin{align*} \displaystyle{\int_{\theta_0}^{\frac{\pi}{2}}} \rho \cos(\theta)^2 + \sqrt{1-\rho^2} \cos(\theta)\sin(\theta)d\theta & = \frac{2\sqrt{1-\rho^2}\cos(\theta_0)^2 + \rho(\pi - 2\theta_0 - \sin(2\theta_0))}{4} \\
 & = \frac{\rho \pi + 2\sqrt{1-\rho^2} + 2\rho \mbox{Arctan}\big(\frac{\rho}{\sqrt{1-\rho^2}}\big)}{4} \end{align*}

To conclude, $$\mbox{with } \rho = \frac{(u^Tv)}{\sqrt{(u^Tu)(v^Tv)}}, \quad f(u,v) = \sqrt{(u^Tu)(v^Tv)} \cdot \frac{\rho \pi + 2\sqrt{1-\rho^2} + 2\rho \mbox{Arctan}\big(\frac{\rho}{\sqrt{1-\rho^2}}\big)}{4(d+2)\pi}.$$

Nota bene: if it can make you any less dubious about the previous computations, I made some Python simulations with $d=2$ and $d=3$ and Monte-Carlo simulations coincide with the theoretical formula.


问:
On the convergence of improper integrals
Question

Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.

My working

It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0\ $.

If $r \neq 0\ $, then

\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{a\to\infty}[r\ln(x + 1) - \frac 3 4 \ln|2x^2 + r|]^{x = a}_{x = 1}
\\[5 mm] & =
\lim\limits_{a\to\infty}[r\ln(a + 1) - \frac 3 4 \ln|2a^2 + r| - r\ln 2 + \frac 3 4 \ln|2 + r|]
\\[5 mm] & =
\lim\limits_{a\to\infty}\left(r\ln\frac {a + 1} 2 + \frac 3 4 \ln\left|\frac {2 + r} {2a^2 + r}\right|\right)
\end{align}

This question just appeared in my final calculus examination this afternoon and I got stuck here ):. We have practiced such questions before but I cannot seem to get this one. I would usually proceed by trying L'Hopital's Rule since we want the limit to exist. I know I probably will not be touching calculus for quite a while from now, but I still want to figure this out. Any intuitions as to how I should proceed will be greatly appreciated :)

答:
By using the comparison test, we know that if we integrate a rational function $\int_a^{\infty} p(x)/q(x)dx$ such that $q(x)\neq 0$ on the interval, then we converge if and only if $\deg(p)-\deg(q)<-1$. However, actually combining and simplifying the rational functions in this problem is tedious algebra, so one can ask if there is a better way. In fact, there is.

Let us write $r/x+1=r/x+a(x)$ where $a(x)$ is some rational function of degree less than $-1$, and let us write $3x/(2x^2+r)=(3/2)/x+b(x)$ where $b(x)$ has degree less than $-1$. We see that if $r=3/2$, we are left with the integral of $a(x)-b(x)$, which will converge. However, for any other choice of $r$, we do not get perfect cancelation and we will be left with something of degree $-1$, so our integral will be divergent.

This allows us to avoid having to integrate anything, or even having to do algebra, simply to compare our integral to the right thing.


问:
Is the function $ f$ differentiable at every point $x$?
Let $ f $ be the function defined as follows: $ f(0) = 1$ and  $f(x)= \frac{\sin x}{x} $ if  $x \neq 0 $. Show that $ f $ is differentiable at every point $x $  ( at $ 0 $ also! ) and evaluate its derivative.

At first I took $f(x)= \frac{\sin x}{x} $ function and wrote that it is continuous function because

$ \lim_{x\to 0^{+}} $  and $ \lim_{x\to 0^{-}}$ both equals to  one.

Now if we discus the graph of  $ f(x) $  when $ x \in ( - \infty ; \infty ) $ we will see that the graph is continuous so the function is differentiable.

$$
\begin{cases} f(x) = \frac{\sin x}{x} & x \neq 0 \\ f(x) = 1 & x = 0  \end{cases}$$

$$
\begin{cases} f^{'}(x) = \frac{x\cos x - \sin x}{x^{2}}  \\ f^{'}(x) = ?  \end{cases}$$
is the derivative of $ f(x) $ when $ x = 0 $  $ \rightarrow $ $ f^{'}(x) = 0 $ ?

答:
It it exists ,
$$f^{\prime}(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\frac{\sin(x+h)}{x+h}-1}{h}=\lim_{h\to0}\frac{\sin(x+h)-(x+h)}{h(x+h)}$$
and in $x=0$ this is
$$=\lim_{h\to 0}\frac{\sin(h)-h}{h^2}=\frac{\not h-\frac{h^3}{3!}+o(h^4)-\not h}{h^2}=-\frac{h}{6}+o(h^2)\rightarrow 0\mbox.$$
This is how i would do it, hope there are no mistakes tho.


问:
No $n$-stair graph has a Hamiltonian path for any $n \ge 3$.
$\textbf{Definition 1:}$ An $n$-stair is the collection of the unit squares bounded by the $x$-axis, $y=x$ and $x=n+1$.

$\textbf{Definition 2:}$ An $n$-stair graph is obtained by replacing each of the unit squares by a vertex where two vertices are adjacent if two unit squares share a common side.

$\textbf{Claim:}$ No $n$-stair graph has a Hamiltonian path for any $n \ge 3$.

I have a kind of visualization of the graph in mind but not getting any clue how to proceed.

答:
Here is a depiction of the $5$-stair graph. Each non-space character is a vertex and there is an edge between and only between adjacent characters:

```<code>    e
   xo
  xoo
 xooo
eoooo
</code>```

Clearly, any Hamiltonian path must start and end at the es, which are only of degree $1$. The path must also pass through the x vertices on the diagonal; since they are degree $2$ their two incident edges must be in the path. When you link up all these "forced" edges, however, you find that a complete path has been formed without any possibility of reaching the other vertices. Thus the $n$-stair graph has no Hamiltonian cycle for $n\ge3$.

Body:
Here is a depiction of the $5$-stair graph. Each non-space character is a vertex and there is an edge between and only between adjacent characters:

    e
   xo
  xoo
 xooo
eoooo

Clearly, any Hamiltonian path must start and end at the es, which are only of degree $1$. The path must also pass through the x vertices on the diagonal; since they are degree $2$ their two incident edges must be in the path. When you link up all these "forced" edges, however, you find that a complete path has been formed without any possibility of reaching the other vertices. Thus the $n$-stair graph has no Hamiltonian cycle for $n\ge3$.


问:
Find the maximum
I would appreciate if somebody could help me with the following problem:

Find the maximum of the function $$f(x,y,z) = x$$ on the curve defined by the equations $F(x,y,z) = G(x,y,z) =0$ with
$$F(x,y,z):= x^2 +y^2 +z^2 -1 \qquad \text{and} \qquad G(x,y,z) :=x^3+y^3 + z^3.$$

答:
Using the method of Lagrange multipliers should yield the following system of equations:

$\begin{eqnarray}
1&=&2ax+3bx^2\\
0&=&2ay+3by^2\\
0&=&2az+3bz^2\\
1&=&x^2+y^2+z^2\\
0&=&x^3+y^3+z^3\\
\end{eqnarray}$

Solving the second and third of these tells you that $y$ is either $0$ or $-2a/3b$ and $z$ is either $0$ or $-2a/3b$. This gives you four cases to work through.

Title: Find the maximum


答:
For the simpler version of what $B^n$ looks like, set up the Cartesian product of $B$'s.

$$\underbrace{B \times B \times \cdots \times B}_{\text{n times}}$$

Then, we have n-tuples consisting on elements from $B$.  As what Stahl just showed to you, we have:

$$B^n = \{(b_1, b_2, \dots, b_n) | \text{for each } b_j \in B \text{ for j} = \{1,2,3,\dots, n\} \}$$

Title: Properties of basic set theory

Body:
The question is about a set:
$$B=\{a_1,a_2,a_3,...,a_n\} \subset \mathbb R$$
And would like to know how to calculate $B^n$ where $n \in\Bbb N$?</p>


问:
When is composition of linear maps an isomorphism
Let $T:V\rightarrow W$ and $L:W\rightarrow U$ be linear maps between finite-dimensional $\mathbb{R}$-vector spaces.  I'm curious to know when $L\circ T:V\rightarrow U$ is an isomorphism.

My hypothesis is that $L\circ T$ is an isomorphism if and only if $Ker(L)^{\perp} = Im(T)$. (By this I mean that $Im(L) \cap Ker(L)={0}$).

Here's what I've gotten far, by this post we know that $L$ must be injective and (arguing dually) we find that $T$ must be surjective.  So, applying the splitting lemma: we write $W\cong V\oplus U$.  Since $T$ is injective and linear then $V\cong Im(T)$.  Now, since $L$ is surjective then if $Im(T)$ intersects $\ker(L)$ non-trivially (ie more than just at $0$) then $Im(L)$ is of strictly lower dimension than $U$; whence it cannot be surjective.  Therefore, $Im(T)\cap \ker(L)={0}$.  The converse direction is clear.

Would my argument also hold if $L\circ T$ is only injective?

答:
The splitting lemma does not apply in this situation. Also for $L \circ T$ to be bijective $L$ must be surjective and $T$ injective.

The following statement is true for all composition of maps.
$L \circ T$ is bijective iff $T$ is injective and $L|_{im T} $ is bijective.
When looking at linear maps this translates to:

$L \circ T$ is bijective iff $T$ is injective, $L$ surjective and $im(T) \cap ker(L) = {0}$


问:
Time transform of Brownian motion
I have the following doubt. In an exercise I was given $(B_t)_{t \geq 0}$ as a standard Brownian motion (i. e. $B_t \sim N(0,t) \ \forall \ t \geq 0$). Now I have to investigate whether or not $B_{e^{2t}}  \sim N(0, e^{2t}) \ \  \forall t \geq 0$. If this is indeed true (may be something related to the invertibility of the exponential function?... how can I show that process has that distribution?), the process defined by $(e^{- t} \cdot B_{e^{2t}})_{t \geq 0}$ is not a standard BM because it lacks $B_0 = 0$, right?

答:
$B_t \sim N(0,t)$ for all $t >0$ so it is legitimate to replace $t$ by $e^{2t}$ and conclude that $B_{e^{2t}} \sim N(0,e^{2t})$. There is no probability theory involved in this implication.

You are right in saying that $(e^{-t}B_{e^{2t}})$ is not a standard BM.

Title: Time transform of Brownian motion


问:
The integral of the log spectral density for a infinite MA process.
Let us consider a MA($\infty$) process:

$y_t= \sum_{j=0}^{\infty} b_j \varepsilon_{t-j}  $

where

$ \sum_{j=0}^{\infty} b_j^2 < \infty  \>\>\>\>\>\> $   (1)

Given $f(\lambda)$ as the spectral density of the process $y_t$, does the square summabilty condition (1) imply that the following equality $\int_{-\pi}^\pi \log f(\lambda) d \lambda =0   $ holds ?

答: As a consequence of Theorem 5.81 in Brockwell and Davids (1986) "Time Series: theory and methods, second edition Springer", the relation $\int_{-\pi}^\pi \log f(\lambda) d \lambda = =0 $ holds for any stationary process $y_t$. In particular, if $y_t$ is long memory such that $ f(\lambda_c) = \infty$ for some $ \lambda_c \in [-\pi, \pi]$, then the previous relation is still true, because the theorem allows for a spectral density defined in $[-\pi, \pi]$ except for a set of zero Lebesgue measure.


问:
How to draw a $(f \circ g)(x)$ function without a rule and with graphs of $f(x)$ and $g(x)$?
Is it possible to draw the composite function $(f \circ g)(x)$ without having a rule for the two functions $f$ and $g$ and only having graphs for these two functions?

For example these two functions

答: Let $x_0$ in the domain of $g$. You do have the point $(x_0, g (x_0))$. For $f (g (x_0))$ to make sense, $g (x_0)$ has to be in the domain of $f$. In this case we clearly can get the point $(g (x_0), f (g (x_0)))$ which is none other than $f\circ g (x_0)$.This way you can have the graph of $f\circ g$ point by point.


答:
By L'Hospital, twice,

$$\lim_{h\to0}\dfrac{\dfrac{\sin(h)}{h}-1}{h}=\lim_{h\to0}\dfrac{\cos(h)-1}{2h}=\lim_{h\to0}\dfrac{\sin(h)}{2}=0.$$

For nonzero angles, you can use formal differentiation and all expressions are defined.

Title: Is the function $ f$ differentiable at every point $x$?

Body:
Let $ f $ be the function defined as follows: $ f(0) = 1$ and  $f(x)= \frac{\sin x}{x} $ if  $x \neq 0 $. Show that $ f $ is differentiable at every point $x $  ( at $ 0 $ also! ) and evaluate its derivative.

At first I took $f(x)= \frac{\sin x}{x} $ function and wrote that it is continuous function because

$ \lim_{x\to 0^{+}} $  and $ \lim_{x\to 0^{-}}$ both equals to  one.

Now if we discus the graph of  $ f(x) $  when $ x \in ( - \infty ; \infty ) $ we will see that the graph is continuous so the function is differentiable.

$$
\begin{cases} f(x) = \frac{\sin x}{x} & x \neq 0 \\ f(x) = 1 & x = 0  \end{cases}$$

$$
\begin{cases} f^{'}(x) = \frac{x\cos x - \sin x}{x^{2}}  \\ f^{'}(x) = ?  \end{cases}$$
is the derivative of $ f(x) $ when $ x = 0 $  $ \rightarrow $ $ f^{'}(x) = 0 $ ?


问:
$\|f*g\|_q\leq \|g\|_q \|f\|_1$ and $\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}}$, $(1/q+1/q^{'}=1)$?
Now I'm reading the Young inequality. It says that if $f \in L^p(R)$, $g \in L^q(R)$, $1\leq p,q\leq \infty$, $1/p+1/q\geq 1$. Then how could we have the following inequalities:

$$\|f*g\|_q\leq \|g\|_q \|f\|_1$$

$$\|f*g\|_\infty\leq \|g\|_q \|f\|_{q^{'}},\quad (1/q+1/q^{'}=1)$$

$f*g$ means the convolution

The two inequalities seems not obvious, the second one is similar with the holder inequality, but actually not.

Could someone give me some hints or helpful links?
Thanks a lot!

答:
I'll give a proof just using Holder's Inequality so that it may easier for you to understand, but as the above comments said the proof for these inequalities is already somewhere else.

Proof for the first: By Holder's Inequality we have
\begin{align*}
\left|\int f(x-y)g(y)dy\right|\le&\int|f(x-y)|^{1/q}|g(y)||f(x-y)|^{1-1/q}dy
\\
\le&\left(\int|f(x-y)||g(y)|^qdy\right)^{1/q}\left(\int|f(x-y)|dy\right)^{1-1/q}
\\
=&\|f\|_1^{1-1/q}\left(\int|f(x-y)||g(y)|^qdy\right)^{1/q},\qquad\forall x.
\end{align*}
Then by Fubini Theorem we have
\begin{align*}
\|f*g\|_q^q\le&\|f\|_1^{q-1}\int\int|f(x-y)||g(y)|^qdydx
\\
=&\|f\|_1^{q-1}\int\left(|g(y)|^q\int|f(x-y)|dx\right)dy
\\
=&\|f\|_1^{q}\|g\|_q^q.
\end{align*}
Taking off $q$ we prove the first one.

The proof for the second one: By Holder's Inequality we have
\begin{align*}
\left|\int f(x-y)g(y)dy\right|\le\left(\int|f(x-y)|^{q'}dy\right)^{1/q'}\left(\int|g(y)|^qdy\right)^{1/q}=\|f\|_{q'}\|g\|_{q},\qquad\forall x.
\end{align*}
By taking the supremum we get the second one.


问:
Ratio in triangle $\Delta ABC$. Find ratio $|CF|:|FP|$.
Let $\Delta ABC$ be triangle and  $P$ and $Q$ points on $AB$ and $AC$ such that $|AP|:|PB|=1:5$ and $|AQ|:|QC|=2:1$. Lines $BQ$ and $CP$ intersect in point $F$. Find ratio $|CF|:|FP|$.

I want to solve this problem without using triangle similarities or some theorems, but only vectors. I tried to write vector $\vec{CF}$ and $\vec{CP}$ using for example $\vec{AB}$ and $\vec{AC}$ that would solve my problem, but can't do it.

答: Use a reference system centered at $O$ and let for simplicity $\overrightarrow{OA}=:\vec{a}, \overrightarrow{OB}=:\vec b, \overrightarrow{OC}=:\vec c,\overrightarrow{OP}=:\vec p$, etc. Notice that, using the given ratios, we have $$\vec p=\frac56\cdot \vec a+\frac16\cdot \vec b, \qquad\vec q=\frac13\cdot \vec a+\frac23\cdot \vec c$$ Since $F$ is collinear with $B,Q$ and $P,C$, consider the real numbers $\lambda, \mu$ satisfying $$\begin{cases}\overrightarrow{CF}=\lambda \cdot \overrightarrow{CP}=\lambda\cdot \left(\frac56\cdot \vec a+\frac16\cdot \vec b-\vec c\right)\\\overrightarrow{BF}=\mu\cdot \overrightarrow{BQ}=\mu\cdot \left (\frac13\cdot \vec a+\frac23\cdot \vec c-\vec b\right)\end{cases}$$ We deduce that $$\begin{align*}\overrightarrow{CF}+\overrightarrow{FB}&=\overrightarrow{CB}\\\iff\lambda\cdot \left(\frac56\cdot \vec a+\frac16\cdot \vec b-\vec c\right)+ \mu\cdot \left (-\frac13\cdot \vec a-\frac23\cdot \vec c+\vec b\right)&=\vec b-\vec c \end{align*}$$ This yields $$\begin{cases}\frac56\cdot \lambda-\frac13\cdot \mu=0\\\frac16\cdot \lambda+\mu=1\\-\lambda -\frac23\mu=-1\end{cases}$$ Solve the system to obtain $\lambda=\frac38$ and $\mu=\frac{15}{16}$. Hence $$\overrightarrow{CF}=\frac38 \cdot \overrightarrow{CP}\implies \lvert CF\rvert:\lvert FP\rvert=3:5$$


问:
Orthogonal random variables - an exercise
Long time ago, I was given the following exercise in a class.

Let $$X_1, X_2, \dots, X_n, \dots,$$
be a sequence of random variables, such  that $\mathbb{E}X_iX_j =0$
for $i\not=j$ and $\mathbb{E}X_i^2 =1$ for all $i$. Let
$$a_1,a_2,\dots, a_N,$$ be a sequence of real numbers. Prove that
for
$$Y_N:= \ \max_{i\le N} \Big|\sum_{k=1}^{i}a_k X_k\Big|,$$
we have
$$\mathbb{E}Y_N^2 \ \le \ O(\log^2N)\cdot\sum_{k=1}^{N}a_k^2.$$
Unfortunately, I never learned how to solve this problem. It looks elementary, but I don't know how to attack it. Does anybody know how to do it? I will be grateful for any insight.

答:
I don't really know either, but i will try to start.

Taking $N$ as you gave it, there exists an index $\overline{i}$ such that

$$\mathbb{E}Y_N^2=\mathbb{E}\bigl{\{}\sum_{k=1}^{\overline{i}}a_kX_k \bigr{\}}^2\le\mathbb{E}\bigl{\{}\sum_{k=1}^{\overline{i}}X_k \bigr{\}}^2\biggl{\{}\sum_{k=1}^{\overline{i}}a_k\biggr{\}}^2\le \mathbb{E}\bigl{\{}\sum_{k=1}^{\overline{i}}X_k \bigr{\}}^2\sum_{k=1}^{N}a_k^2$$
and now it would be convenient to show that
$$\lim_{N\to +\infty}\frac{\mathbb{E}\bigl{\{}\sum_{k=1}^{\overline{i}}X_k \bigr{\}}^2}{log^2(N)}=0$$

Anyways, this is merely a way for you to be inspired, nothing else.


问:
How can I find this limit of $f(n)/n$
First of all, we have arbitrary positive integer $n$ and function $f(n)$. Function is a number of positive dividers of $n$ (for example when $n = 6$, then dividers $- 1, 2, 3, 6$, so $f(n) = 4$. What I need to find is
$\lim_{n\rightarrow\infty} f(n)/n$. I think it is $0$, but don't know how to prove that.

答:
This is known as the divisor function, $d(n)$.
Look it up.

In the Wikipedia article on it,
it is stated with a reference that
$d(n) = o(n^c)$
for any $c > 0$.
More precise results are also given.

Title: How can I find this limit of $f(n)/n$


问:
Completeness property of R
I was reading up on the completeness property of R that sets it apart from Q

It has been stated in 2 ways which are

Every non empty subset of R which is bounded above has a supremum in R

Every non empty subset of R which is bounded below has an infimum in R

Now this is very clear. The problem lies in what these 2 statements are followed by which has been marked in red in the given picture. It says that these 2 properties are equivalent in the sense that one implies the other. How is that even possible?

according to thus language, they r suggesting that 1. implies 2. i.e. if supremum exists than infimum exists too and vice versa which is clearly not the case. Am i interpreting it wrong?

Can someone help urgently plz!

答:
Take a non-empty subset $S$ of $\mathbb{R}$ that is bounded below.

Let $T$ be the set $S$ where every element is multiplied by $(-1).$

Therefore, $T$ is bounded above.

By assumption, $T$ has a supremum.

Therefore, $S$ has an infinum.

Title: Completeness property of R


答:
You don't have to care about the exact value of the integral. Simply use asymptotics.

To see how the integrand functions behaves asymptotically as $x \to \infty$, compute the sum of the two fractions.

$$\frac{r}{x+1} - \frac{3x}{2x^2+r} = \frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)}$$

Now consider two cases:

If $2r-3 = 0$, then asymptotically as $x \to \infty$
$$\frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)} \sim \frac{-3x}{2x^3} = O (1/x^2)$$
hence the integral is convergent.

If $2r-3 \neq 0$, then asymptotically as $x \to \infty$
$$\frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)} \sim \frac{(2r-3)x^2}{2x^3} = C \cdot \frac{1}{x}$$
(where $C \neq 0$ is a constant), hence the integral is divergent.

This means that the integral converges only when $r=3/2$.

Title: On the convergence of improper integrals

Body:
Question

Determine the value of $r$ for which the integral $$\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x$$ is convergent and evaluate the integral for that value of $r$.

My working

It is trivial to see that the integral is divergent for $r = 0\ $, so let $r \neq 0\ $.

If $r \neq 0\ $, then

\begin{align}
\int^{\infty}_1 (\frac r {x + 1} - \frac {3x} {2x^2 + r})\ \mathrm {d}x & =
\lim\limits_{a\to\infty}[r\ln(x + 1) - \frac 3 4 \ln|2x^2 + r|]^{x = a}_{x = 1}
\\[5 mm] & =
\lim\limits_{a\to\infty}[r\ln(a + 1) - \frac 3 4 \ln|2a^2 + r| - r\ln 2 + \frac 3 4 \ln|2 + r|]
\\[5 mm] & =
\lim\limits_{a\to\infty}\left(r\ln\frac {a + 1} 2 + \frac 3 4 \ln\left|\frac {2 + r} {2a^2 + r}\right|\right)
\end{align}

This question just appeared in my final calculus examination this afternoon and I got stuck here ):. We have practiced such questions before but I cannot seem to get this one. I would usually proceed by trying L'Hopital's Rule since we want the limit to exist. I know I probably will not be touching calculus for quite a while from now, but I still want to figure this out. Any intuitions as to how I should proceed will be greatly appreciated :)


答:
You can find in An Introduction to the Theory of Numbers by Hardy that, if $n$ is not a prime, $d(n) \approx \log{n}$ for large $n$ with
$$d(n) = \text{card}\{d \in \mathbb N~|~1\le d \le n~\text{ and }~ d | n\}.$$

Title: How can I find this limit of $f(n)/n$

Body:
First of all, we have arbitrary positive integer $n$ and function $f(n)$. Function is a number of positive dividers of $n$ (for example when $n = 6$, then dividers $- 1, 2, 3, 6$, so $f(n) = 4$. What I need to find is
$\lim_{n\rightarrow\infty} f(n)/n$. I think it is $0$, but don't know how to prove that.


问:
Second Derivative of basic fraction using quotient rule
I know this is a very basic question but I need some help.

I have to find the second derivative of:

$$\frac{1}{3x^2 + 4}$$

I start by using the Quotient Rule and get the first derivative to be:

$$\frac{-6x}{(3x^2 + 4)^2}$$

This I believe to be correct.
Following that I proceed to find the second derivative in the same manner but I get this as my answer:
$$\frac{(54x^4 + 144x^2 +96) - (-36x^3 + 48x)}{(9x^4 +24x^2 +16)^2}$$

This I believe to be correct just not simplified.  However the answer I need to get is:
$$- \frac{6(4 - 9x^2)}{(3x^2 + 4)^3}$$
I do not know what the best way to approach this would be, should I multiply out the denominator and try to cancel?  Could someone point me in the right direction, I want to solve it myself but I need some guidance.

Thanks

答:
The first derivative is correct. Now we want to differentiate $\frac{-6x}{(3x^2+4)^2}$. The main thing to remember is do not "simplify" unless there is good reason to do so.

The derivative of $\frac{-6x}{(3x^2+4)^2}$ is
$$\frac{(3x^2+4)^2 (-6)-(-6x)(6x)(2)(3x^2+4)}{(3x^2+4)^4}.$$
Cancel a $3x^2+4$, and simplify the top.

Remark: I probably would want to take out that ugly $-6$ from the top, which is an invitation to minus sign errors and other errors, and differentiate $\frac{x}{(3x^2+4)^2}$.


问:
Invariant distribution of a Markov chain via $P^t$
For a discrete Markov chain with transition probability matrix $P$, we see that the invariant distribution satisfies

$\pi = \pi P^n\tag{1}$

now suppose I can calculate $\Pi:= \lim_n P^n$ analytically. I know from the theory that $\pi_i$ represents the long-run portion spent in the state $i$.

In the concrete example we have the Ehrenfest two urn model with 3 balls. I was able to calculate the transition probabilities as well as $\Pi$. The value $X_n$ of the Markov Chain denotes the number of balls in urn $1$ at time step $n$. What I want to calculate is the long-run portion of time that the first urn was empty, i.e. $\pi_0$. Now its mentioned that one can use $\Pi$ to calculate $\pi_0$ but I'm not quite sure how. Usually, I would just solve equation $(1)$ above. Any clarification would be appreciated

The transition matrix in this case is given as

$$ \begin{equation*}
P =
\begin{pmatrix}
0 & 1 & 0 & 0 \\
\frac{1}{3} & 0 & \frac{2}{3} & 0\\
0 & \frac{2}{3} & 0 & \frac{1}{3}\\
0 & 0 & 1 & 0\\ \end{pmatrix}
\end{equation*}$$

and $$\Pi = \lim_n P^n = \begin{equation*}
\begin{pmatrix}
0.25 & 0 & 0.75 & 0 \\
0 & 0.75 & 0 & 0.25\\
0.25 & 0 & 0.75 & 0\\
0 & 0.75 & 0 & 0.25\\ \end{pmatrix}
\end{equation*}$$

I know from calculating $(1)$ that $\pi_0=0.125$. But how can I get this via $\Pi$?

答:
The "problem" lies in your matrix $P$. The chain is periodic $(2)$ and thus you are not guaranteed that

$$\lim_n P(X_n=j|X_0=i)=\pi_j \forall i,j $$

You would need to calculate the $\pi_0$ vie $\pi_0=\pi_0 P$ or check that the chain is reversible and use the detailed balanced equation if applicable.


答:
Given a positive divisor $ d$ of $n$, either $d$ or $n/d$ (or both) is no larger than $\sqrt n$. It follows that $f(n)\le2\sqrt n$.

Now you can squeeze...

Title: How can I find this limit of $f(n)/n$


答:
When you perform the quotient rule, it's often easier to not multiply everything out until the end, because there are a lot of cases where you can factor things out and cancel, and if you multiply out first, it will be much harder to see that. Your first derivative is indeed correct, but here's what I'd recommend to get the suggested answer for the second derivative:
\begin{align*}
\frac{d}{dx}\left(\frac{-6x}{\left(3x^2 + 4\right)^2}\right) &= \frac{-6\left(3x^2 + 4\right)^2 - (-6x)(2)\left(3x^2 + 4\right)(6x)}{\left(3x^2 + 4\right)^4}\\
&= \frac{6\left(3x^2 + 4\right)\left(-\left(3x^2 + 4\right) + 12x^2\right)}{\left(3x^2 + 4\right)^4}\\
&= \frac{6\left(9x^2 - 4\right)}{\left(3x^2 + 4\right)^3},
\end{align*}
which is what you wanted.

Title: Second Derivative of basic fraction using quotient rule

Body:
I know this is a very basic question but I need some help.

I have to find the second derivative of:

$$\frac{1}{3x^2 + 4}$$

I start by using the Quotient Rule and get the first derivative to be:

$$\frac{-6x}{(3x^2 + 4)^2}$$

This I believe to be correct.
Following that I proceed to find the second derivative in the same manner but I get this as my answer:
$$\frac{(54x^4 + 144x^2 +96) - (-36x^3 + 48x)}{(9x^4 +24x^2 +16)^2}$$

This I believe to be correct just not simplified.  However the answer I need to get is:
$$- \frac{6(4 - 9x^2)}{(3x^2 + 4)^3}$$
I do not know what the best way to approach this would be, should I multiply out the denominator and try to cancel?  Could someone point me in the right direction, I want to solve it myself but I need some guidance.

Thanks


问:
Interval of convergent of $\sum_n \frac{x^{2^{n}}}{\left( 1-x\right) ^{2^{n+1}}}$
$$ \sum_n \frac{x^{2^{n}}}{\left( 1-x\right) ^{2^{n+1}}} $$

I think I got the answer but I'm not sure. Anyone who verify my answer?
$R=1$ and $x\in \left(0, 2\right) \backslash \left\{ 1\right\}$

答: Your answer is not correct. For  $|\frac x {(1-x)^{2}}| <1$ so the series is convergent.  $|\frac x {(1-x)^{2}}| \geq 1$ it is divergent. Hence the region of convergence is $\mathbb R \setminus [\frac {3-\sqrt 5} 2, \frac {3+\sqrt 5} 2]$.  Radius of convergence is not defined for this series.


问:
Prove that the image of $F$ is a closed set in $\mathbb R^n$
Being $F : \mathbb R^n\to \mathbb R^n$ a continuous function such that for every $x,y\in \mathbb R^n$,
$$\|x − y\| \leq\|F(x) − F(y)\|\,.$$
I'd like to prove that the image of $F$ is a closed set in $\mathbb R^n$.

So far I have tried to use that if $x_n$ is a sequence in $\mathbb R^n$ such that $F(x_n)$ is a Cauchy sequence, $x_n$ must be Cauchy as well, in addition to the fact that any closed set has to contain all the limits of its sequences, but with this approach I haven't been able of find a proof.

Any other ideas on how to solve this proof?

答: A set $C$ in $R^n$ is closed exactly iff all the Cauchy sequences in $C$ converge to an element inside $C$. You can carry a Cauchy sequence in  $F(C) $ to a cauchysequence in $C$ with the preimage. This Cauchy sequence converges in $C$. Using continuity of $F$ we get that the sequence in $F(C) $ converges as well to an element of $F(C) $.


问:
Finding a limit at negative infinity
$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$

Hello,

I know that the answer to this is $- \infty$, but I'm having trouble showing it. Can anyone please point me in the right direction?

Thanks!

Title: Finding a limit at negative infinity

Body:
You can solve it as follows:

$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$
$$=\lim_{x \to -\infty}{x^7(\frac{1}{x^5} + 2)}$$
$$=-\infty(\frac{1}{-\infty} + 2)$$
$$=-\infty(0 + 2)$$
$$=-\infty$$


问:
To show P is a zero polynomial
Suppose that $P$ is a polynomial with integer coefficients that $n$ divides $P(2^n)$ for every positive integer $n$.

Prove that $P$ must be the zero polynomial.

What I did was apply some induction on the expression by considering $$P (x)= a_nx^n+ \cdots +a_0$$ which results in nothing for proving the required result.
Any hints/solution would be appreciated.

答: If a prime $p$ divides $f(2^{mp})$ then it divides $f(2^m)$.

Title: To show P is a zero polynomial


问:
how to apply hint to question involving the pigeonhole principle
The following question is from cut-the-knot.org's page on the pigeonhole principle

Question

Prove that however one selects 55 integers $1 \le x_1 < x_2 < x_3 < ... < x_{55} \le 100$, there will be some two that differ by 9, some two that differ by 10, a pair that differ by 12, and a pair that differ by 13. Surprisingly, there need not be a pair of numbers that differ by 11.

The Hint

Given a run of $2n$ consecutive integers: $a + 1, a + 2, ..., a + 2n - 1, a + 2n$, there are n pairs of numbers that differ by n: $(a+1, a+n+1), (a + 2, a + n + 2), \dots, (a + n, a + 2n)$. Therefore, by the Pigeonhole Principle, if one selects more than n numbers from the set, two are liable to belong to the same pair that differ by $n$.

I understood the hint but no concrete idea as to how to apply it, but here are my current insights:

My Insights

break the set of 100 possible choices into m number of 2n(where $n \in \{9,12,13\}$ ) consecutive numbers and since 55 numbers are to be chosen , show that even if one choses randomly there will be n+1 in one of the m partitions and if so, by the hint there will exist a pair of two numbers that differ by n.

答:
Here is $9$ done explicitly:  Break the set into subsets with a difference of $9$: $\{1,10,19,28,\ldots,100\},\{2,11,20,\ldots 92\},\ldots \{9,18,27,\ldots 99\}$.  Note that there is one subset with $12$ members and eight with $11$ members.  If you want to avoid a pair with a difference of $9$ among your $55$ numbers, you can't pick a pair of neighbors from any set.  That means you can only pick six from within each set, but that gives you only $54$ numbers, so with $55$ you will have a pair with difference $9$.

The reason this fails with $11$ is you get one subset with $10$ members and ten with $9$ members.  You can pick $5$ out of each and avoid neighbors.


问:
If the Riemann Integral of a continuous nonnegative function is 0, then that function is identically 0.
Here's what I'm doing:

Prove that if $f: [a,b] \to [0,+\infty)$ is a continuous nonnegative function with $\int_{a}^{b} f(x) \ dx = 0$, then $f(x) = 0$ for all $x \in [a,b]$.

Proof Attempt:

Suppose that $f(x) \neq 0$ for some $x \in [a,b]$. Let $c$ be one of those points in $[a,b]$ where $f(c) \neq 0$. Let $\epsilon > 0$ be given. Since $f$ is Riemann Integrable, there exists a $\delta > 0$ such that for all partitions $P$ with partition norm $|P| < \delta$ and all associated evaluation sets $T$:

$$|R(f,P,T)-0| = |R(f,P,T)| < \epsilon$$

where $R(f,P,T)$ is shorthand for the Riemann Sum taken over some partition and evaluation set. Now, we know that $R(f,P,T) \geq 0$ and, in fact, if $c \in T$, then $R(f,P,T) > 0$.

So, then, pick $T$ above so that $c \in T$ and let $\epsilon = R(f,P,T)$. Then, we get:

$$R(f,P,T) < R(f,P,T)$$

which is impossible. Hence, $f(x) = 0$ for all $x \in [a,b]$. $\Box$

Does the proof above work? If it doesn't, then why? How can I fix it?

答:
An usual and simple way would be to say, like you did, if there exists $c\in [a,b]$ such that $f(c)>0$ than by continuity, given $\epsilon =f(c)/2$, there exists $\delta >0$ for which for every $x\in (c-\delta,c+\delta)$, $f(x)\ge \frac{f(x)}{2}$. So we have this:
$$\int_a^bf(x)dx \ge \int_{c-\delta}^{c+\delta}f(x)dx\ge \int_{c-\delta}^{c+\delta}\frac{f(c)}{2}dx=\frac{f(c)}{2}2\delta>0\mbox,$$
which goes against our hipoyhesys


问:
Understanding the Leibniz formula for determinants
I am a programmer and trying to write the Leibniz formula for determinants into C++ code, but I am unable to fully understand it.
Can someone walk me through it?

$$\det(A)=\sum_{\sigma \in S_n}\text{sgn}(\sigma)\prod_{i=1}^{n} a_{\sigma(i),i}$$

答: The best method to calculate determinant would be to first perform LU decomposition followed by calculating the determinants of L and U, which are the products of their main diagonals.</p>


问:
Are $aX + bX$ and $bX + cX$ independent if $X$ is normal distribution?
As in the title: $X$ is a Normal random variable with mean $0$ and variance $1$. Let $a, b, c \in \mathbb{R}_+$, $Y = aX + bX$ and $Z = bX + cX$. Are $Y$ and $Z$ independent?

答:
They are not independent, in fact they are "on the opposite side of the spectrum",

i.e. they are fully linearly dependent.

Note that

$$Z = \frac{b+c}{a+b} \cdot Y$$

If you know the value of $Y$ you know the value of $Z$.

So there's no way these can be independent.

Also, I don't think it matters that $X$ is normal, it can be pretty much any random variable.


答: As $x\to -\infty$ we have $$3x^7\le x^2+2x^7\le x^7$$ now squeeze....

Title: Finding a limit at negative infinity

Body:
$$\lim_{x \to -\infty}{x^2 + 2 \cdot x^7}$$

Hello,

I know that the answer to this is $- \infty$, but I'm having trouble showing it. Can anyone please point me in the right direction?

Thanks!


Body:
$$\left(-\frac{6x}{(3x^2+4)^2}\right)'=-\frac{6(3x^2+4)^2-72x^2(3x^2+4)}{(3x^2+4)^4}=$$

$$=-\frac{18x^2+24-72x^2}{(3x^2+4)^3}=-\frac{6(-9x^2+4)}{(3x^2+4)^3}$$


问:
Find sequences $a_n$ that converges to 1 and $b_n$ that converges to $\infty$ such that $a_n^{b_n}$ converges to 5
I stuck in this question for a few hours now.
Can't understand how to find one and all my guesses failed.

The Question :

Find sequence $a_n$ that converges to 1 (not constant sequence).

Find sequence $b_n$ that converges to $\infty$.

such that $a_n^{b_n}$ converges to $5$.

I found such sequences that converges to $2e$ (5.4) but it's not exactly 5..

Thank you very much and have a nice day!

答:
In response to a comment by the proposer about "not using $\ln$": For $n\in \Bbb N$ we know there exists  $m\in \Bbb N$ such that $(1+1/n)^m> 5$ because by the Binomial Theorem, $(1+1/n)^{5n}=1+\binom {5n}{1}(1/n)+\binom {5n}{2}(1/n^2)+...>5.$

So let $b_n$ be the $least$ $m\in \Bbb N$ such that $(1+1/n)^m>5.$ And let $a_n=1+1/n.$

By definition of $b_n$ we have $(1+1/n)^{-1+b_n}\le 5<(1+1/n)^{b_n}.$ Therefore $$0<(a_n)^{b_n}-5\le (a_n)^{b_n}-(a_n)^{-1+b_n}=(a_n)^{-1+b_n}(a_n-1)\le 5(a_n-1)=5/n.$$

So $(a_n)^{b_n}\to 5. $

We cannot have $b_{n+1}<b_n.$ Otherwise $b_{n+1}$ would be an $m\in \Bbb N$ with $m<b_n$ and $(1+1/n)^m>5,$ contrary to the definition of $b_n.$ Because  $5<(1+1/(n+1))^{b_{n+1}}<(1+1/n)^{b_{n+1}}.$

So, since $b_n\le b_{n+1},$ and each $b_n\in\Bbb N,$ we can conclude that $b_n\to\infty$ by showing that there is no largest $b_n$, as follows:   If $n'$ is large enough that $1+1/n'<\sqrt {1+1/n}$ then $(1+1/n')^{-2+2b_n}<(1+1/n)^{-1+b_n}\le 5.$ So we must have $b_{n'}>-2+2b_n.$ This implies  $b_{n'}>b_n$ because $b_n\ge b_1>2$.

Brought to you by the Year 1500. John Napier discovered/invented logarithms in the 1500's.


问:
Doubt in the definition of Group Action.
A group action is defined as a map from $G \times A \to A$, and it follows two properties. Now my basic question is:

How do we know that any arbitrary element $g$ acting on $a \in A$ wil give me an elemnet of $A$? What if $g.a $ is not in the set $A$, then is it not a group action?

I am sorry if the question is very trivial but could someone help me out with the definition of group action.

答:
A group action is first of all a function. So, if $g \cdot a \notin A$ for some $g \in G$ and $a \in A$, then we don't have a function $G \times A \to A$ anymore, let alone a group action.

How do we know that any arbitrary element $g$ acting on $a\in A$ wil give me an element of $A$?

Answer to this question is actually we define an action in a way that it is a function first, then it satisfies the two properties you mentioned. So, if we have a case like $g \cdot a \notin A$ for some $g \in G$ and $a \in A$, we could extend $A$ to a bigger set $B$ so that $g\cdot a \in B$ and it satisfies the group action properties. But note that in this case, we must have an action $G \times B \to B$.


问:
Boundary value of weakly convergence
If $u_m\in \mathring W_2^{1,1}(Q_T)$, and $u_m$ weakly convergence to $u$ in $W_2^{1,1}$,  then whether $u$ is belong to $\mathring W_2^{1,1}$ ?  How to prove it?

ps:  $\Omega\subset \mathbb R^n$, and $Q_T=\Omega\times (0, T)$,

$$
\mathring C^\infty(Q_T)=\{u\in  C^\infty(Q_T) : u \text{ is zero near } \partial\Omega\times (0, T)\}
$$
and  $\mathring W_2^{1,1}(Q_T) $

is closure of  $\mathring C^\infty(Q_T)$  in $W_2^{1,1}(Q_T) $.

答: I do not know your definitions, but typically the space with an "o" on top is a closed subspace of the other space. Closed subspaces are also weakly closed (by an application of Hahn-Banach).</p>

Title: Boundary value of weakly convergence


问:
$\Vert f_n\Vert_p \longrightarrow \Vert f\Vert_p$ implies the convergence of $f_n$ to $f$ in $\textit{L}^p$ with the right hypotesis?
I think this might work, but how i don't know:

Let $p\in (1,+\infty)$ and $\{f_n\}$, $f\in \textit{L}^p(E)$, if we know that:

$\Vert f_n\Vert_p \longrightarrow \Vert f\Vert_p$ as $n\longrightarrow +\infty$;

$\forall$ measurable $F\subseteq E$ with $|F| < +\infty \Longrightarrow \int_Ff_n\longrightarrow \int_Ff$, as $n\rightarrow +\infty$;

is it true that $f_n$ converges to $f$ in $\textit{L}^p(E)$ in norm (strong convergence)?

Any suggestions?

答:
Here is a sketch for a proof:

The sequence is bounded, hence it has a weakly convergent subsequence.

The second bullet gives that the weak limit is $f$.

A subsequence-subsequence argument shows that the entire sequences converges weakly.

Weak convergence and convergence of norms implies strong convergence in $L^p$, $1 < p <\infty$.


问:
Ratio in triangle $\Delta ABC$. Find ratio $|CF|:|FP|$.
Let $\Delta ABC$ be triangle and  $P$ and $Q$ points on $AB$ and $AC$ such that $|AP|:|PB|=1:5$ and $|AQ|:|QC|=2:1$. Lines $BQ$ and $CP$ intersect in point $F$. Find ratio $|CF|:|FP|$.

I want to solve this problem without using triangle similarities or some theorems, but only vectors. I tried to write vector $\vec{CF}$ and $\vec{CP}$ using for example $\vec{AB}$ and $\vec{AC}$ that would solve my problem, but can't do it.

Body:
$\vec{AC} = \vec{b}, \vec{AB} = \vec{c}$

Say, $|\vec{PF}|:|\vec{FC}| = (1-x):x$, $|\vec{BF}|:|\vec{FQ}| = (1-y):y$

$\vec{FC} = x \,\vec{PC} = x(\vec{PA} + \vec{AC}) = x(\vec{b} - \frac{\vec{c}}{6})$

Similarly,

$\vec{FC} = (\vec{FQ} + \vec{QC}) = y \vec{BQ} + \frac{\vec{b}}{3} = y(\vec{BA} + \vec{AQ}) + \frac{\vec{b}}{3} = y(\frac{2\vec{b}}{3} -\vec{c}) + \frac{\vec{b}}{3} \,$

Equating the coefficients of $\vec{b}$ and $\vec{c}$,

$x = \frac{2y+1}{3}, \frac{x}{6} = y$

Solving both, $x = \frac{3}{8}$, so $\, 1-x = \frac{5}{8}$

$CF:FP = 3:5$


问:
Sum of Multiple Series
I don't know how to solve this series summation question. I tried breaking it down, but I'm just kind of lost how to manipulate it to give me the value.

$$f(n,k) = \sum\limits_{s=0}^n\sum\limits_{r=s}^n\sum\limits_{t=0}^s\frac{\binom{n}{r}\binom{r}{s}\binom{s}{t}t(3k^2)^{t/2}I(t)}{s+1}$$

where $I(t) = \begin{cases}1&\text{if }x\equiv 0\pmod{4}\\0&\text{if }x\equiv 1\pmod{4}\\-1&\text{if }x\equiv 2\pmod{4}\\0&\text{if }x\equiv 3\pmod{4}\end{cases}$

The variables (N) and (K) are given to me. However, I want to print the answer modulo 10^9+21. If the answer ends up being P/Q, then print the answer PQ^-1 (mod 10^9+21).

I understand that we're using Binomial Coefficient and that's:

(n k) = [n! / k! * (n - k)!]

and that I can simplify the last part of the formula from (3k^2)^t/2 to be:

t * (3k) ^ t

also, that, I(t) is just I^t = sqrt(-1)

Examples,

f(1, 2) = 0

f(2, 1) = 1000000019

f(5, 3) = 4464

f(3, 1) = -33/2

How,

```<code> I guess you need to work in modulo arithmetic - for that particular example:

 Ex:
 f(3, 1) = [3 = N, 1 = K]

 => x/y = x * y^-1

 => -33 = 1000000021 - 33 = 999999988

 => 2 ^ -1 = 500000011

 => 500000011 * 2 = 1000000022 = 1000000021 + 1 = 1

 => 999999988 * 500000011 = 500000004999999868 =

 => 499999994 * 1000000021 + 499999994 = 499999994
</code>```

Hence the answer of the example f(3, 1) in modulo arithmetic is 499999994

No clue how the answer is coming out to be that.

I want to be able to program this question, but I want to understand the math behind it before I attempt to program this.

答:
For nested summations, you can choose to evaluate the inner summation before the outer summation if you are able to algebraically manipulate things.  Since you are not and you want to program this, it is easier instead perhaps to iterate over the sum of sums, making splitting it into several separate summations to add together.

For instance:

$$\sum\limits_{i=1}^3\sum\limits_{j=1}^i i\cdot j = \sum\limits_{j=1}^1 1\cdot j + \sum\limits_{j=1}^2 2\cdot j + \sum\limits_{j=1}^3 3\cdot j$$

$$=1\cdot (1) + 2\cdot (1+2) + 3\cdot (1+2+3) = 23$$

Alternatively $\sum\limits_{i=1}^3\sum\limits_{j=1}^i i\cdot j = \sum\limits_{i=1}^3 i\cdot i(i+1)/2 = \dots$

As far as pseudocode, this is just nested for loops.  You have something like:

```<code>runningtotal = 0
for i ranging from 1 to 3 {
    for j ranging from 1 to i {
        runningtotal += i * j
    }
}
</code>```

modified for whatever language you are specifically using for the format.

As far as taking $I$ into account, to read this in pseudocode, it may be best to think of this as a switch statement.

```<code>I = function(x) {
    switch (x%4) {
        case 0:
             return 1
             break
        case 2:
             return -1
             break
        default:
             return 0
             break
      }
 }
</code>```

That is, if $x$ is a multiple of $4$, $I$ will give a value of $1$, else if $x$ is a multiple of $2$ which is not a multiple of $4$, then $I$ will give a value of $-1$.  Finally, if $x$ is odd it will give a value of $0$.

As for using this inside of the loop... just define your function/method $I$ before running the loop and have $I$ appear inside of the expression that is added to the runningtotal.

For further example including $I$, this becomes:

$$\sum\limits_{i=1}^3\sum\limits_{j=1}^i I(i\cdot j) = \sum\limits_{j=1}^1 I(1\cdot j) + \sum\limits_{j=1}^2 I(2\cdot j) + \sum\limits_{j=1}^3 I(3\cdot j)$$

$$=I(1\cdot 1) + I(2\cdot 1) + I(2\cdot 2) + I(3\cdot 1)+I(3\cdot 2)+I(3\cdot 3)$$

$$=I(1)+I(2)+I(4)+I(3)+I(6)+I(9) = 0+(-1)+1+0+(-1)+0 = -1$$

Title: Sum of Multiple Series

Body:
For nested summations, you can choose to evaluate the inner summation before the outer summation if you are able to algebraically manipulate things.  Since you are not and you want to program this, it is easier instead perhaps to iterate over the sum of sums, making splitting it into several separate summations to add together.

For instance:

$$\sum\limits_{i=1}^3\sum\limits_{j=1}^i i\cdot j = \sum\limits_{j=1}^1 1\cdot j + \sum\limits_{j=1}^2 2\cdot j + \sum\limits_{j=1}^3 3\cdot j$$

$$=1\cdot (1) + 2\cdot (1+2) + 3\cdot (1+2+3) = 23$$

Alternatively $\sum\limits_{i=1}^3\sum\limits_{j=1}^i i\cdot j = \sum\limits_{i=1}^3 i\cdot i(i+1)/2 = \dots$

As far as pseudocode, this is just nested for loops.  You have something like:

runningtotal = 0
for i ranging from 1 to 3 {
    for j ranging from 1 to i {
        runningtotal += i * j
    }
}

modified for whatever language you are specifically using for the format.

As far as taking $I$ into account, to read this in pseudocode, it may be best to think of this as a switch statement.

I = function(x) {
    switch (x%4) {
        case 0:
             return 1
             break
        case 2:
             return -1
             break
        default:
             return 0
             break
      }
 }

That is, if $x$ is a multiple of $4$, $I$ will give a value of $1$, else if $x$ is a multiple of $2$ which is not a multiple of $4$, then $I$ will give a value of $-1$.  Finally, if $x$ is odd it will give a value of $0$.

As for using this inside of the loop... just define your function/method $I$ before running the loop and have $I$ appear inside of the expression that is added to the runningtotal.

For further example including $I$, this becomes:

$$\sum\limits_{i=1}^3\sum\limits_{j=1}^i I(i\cdot j) = \sum\limits_{j=1}^1 I(1\cdot j) + \sum\limits_{j=1}^2 I(2\cdot j) + \sum\limits_{j=1}^3 I(3\cdot j)$$

$$=I(1\cdot 1) + I(2\cdot 1) + I(2\cdot 2) + I(3\cdot 1)+I(3\cdot 2)+I(3\cdot 3)$$

$$=I(1)+I(2)+I(4)+I(3)+I(6)+I(9) = 0+(-1)+1+0+(-1)+0 = -1$$


问:
Use of uniformity of a limit for proof of continuity when right continuity is known
Consider a function $f: [0,\infty) \rightarrow \mathbb{R}$. Suppose that, for any $t \in (0, \infty)$
$$
\lim_{h \downarrow 0 } \,  \, (  f(t+h) - f(t) )=  O(h),
$$
namely that the function $f$ is right continuous. I read in a book that if O(h) is uniform in $t$, then we can replace, $t=s-h$ and deduce that, for any $s \in (0, \infty)$,
$$
\lim_{h \uparrow 0 }(  f(s) - f(s-h) )=  O(h),
$$
Namely that the function is also left continuous. How us uniformity in t used?

答:
You have that there is $C>0$ and $\delta >0$ s.t. for all $t>0$,
$$|f(t+h)-f(t)|\leq C|h|,\quad |h|\leq \delta .$$

Then if $u=t+h$ and thus $t=u-h$, we get $$|f(u)-f(u-h)|=|f(t+h)-f(t)|\leq C|h|.$$

Taking $h\to 0$ yields the wished result.


问:
Is it true that $\langle u,w\rangle= \langle v,w\rangle$ implies $u = v$?

Let $V$ be a vector space over $\Bbb R$.

Is it true that $\langle u,w\rangle= \langle v,w\rangle$ implies $u = v$? Prove it or give a counter-example.

The above is exactly what my assignment demands. It does not state "for every $w$" belonging to vector space $V$.

I wonder what difference does it make to not state that "for every $v$" in $V$.

I was able to find a counter-example, let $u = (1,-1,0)$, $w = (0,-1,1)$ and $v = (1,1,1)$.

$\langle(1,-1,0), (1,1,1)\rangle = 0$

$\langle (0,-1,1), (1,1,1)\rangle = 0$

but $u \neq w$

Is that correct?

答:
As you've noted, $\langle u, v\rangle = \langle w, v\rangle$ is not sufficient to conclude $u = w$. In fact, what you do get is that $$\langle u - w, v\rangle = 0.$$
That is, $u - w$ is orthogonal to $v$. Now, if you are given that the above is true for all $v \in V$, then you can put $v = u - w$ to get $$\langle u - w, u - w\rangle = 0.$$
By the positive-definiteness of inner product, this implies that $u - w = 0$ or $u = w$, as desired.


问:
What is the problem in my solution $\mathbb{P}[ \{ n,n+1\} ] = C\cdot{11}\cdot{10}^{-n-1}$ find $C$
The answer I got is inccorect and I dont know why

given $\Omega=\mathbb{N}=\{0,1,2,...\}$ with the probability : $$\mathbb{P}[ \{ n,n+1\} ] = C\cdot{11}\cdot{10}^{-n-1}$$
I am asked to find $C$

my attempt

$$P({\Omega})=1=\large\Sigma_{n=0}^{n=\infty}C\cdot\frac{11}{10^{n+1}} $$
$$C\frac{11}{10} \large\Sigma_{n=0}^{n=\infty}\frac{1}{10^{n}}$$
$$C\frac{11}{10} \cdot\frac{1}{\frac{9}{10}}=1$$
$$C=\frac{9}{11}$$

Body:
I think there is a problem with the index $n$, in fact $\Omega=\cup_{n=0}^{+\infty}\{2n,2n+1\}$, so
$$1=P(\Omega)=\sum_{n=0}^{+\infty}C\frac{11}{10^{2n+1}}=C\frac{11}{10}\sum_{n=0}^{+\infty}\frac{1}{100^n}=C\frac{11}{10}\frac{100}{99}=C\frac{10}{9}\Rightarrow C=\frac{9}{10}$$


问:
Convolution, Fourier transforms, and area preservation
I've learned from this answer that the integral of a convolution f * g (i.e. the area of the convolution) yields the product of the integrals of f and g (i.e. a product of areas):

Af *g = Af * Ag

But from this answer I've learned that convolution corresponds to polynomial multiplication in one space and term-wise multiplication in the dual space.

Now I'm trying to bring together three seemingly different types of products:

Multiplication of "areas" (i.e. of sums or integrals)

Multiplication of polynomials (i.e. of corresponding monoms + summation)

Term-by-term multiplication

I'm lacking the intuition how this is all supposedly the same, i.e. with respect to the multiplication of areas, and how the Fourier transformation comes into all this.

Consider a simple example from this answer:

x[n] = {2, 4, 1} and p[n] = {5, 1, 8}

with areas Ax = 2+4+1 = 7 and Ap = 5+1+8 = 14

Product of areas (= "multiplication in real space"):

=> Ax * Ap = 7 * 14 = 98

Term-wise multiplication ("convolution in real space"):

termw[n] = {2⋅5, 4⋅1, 1⋅8} = {10, 4, 8}

=> Atermw = 10 + 4 + 8 = 22

Convolution ("polynomial multiplication in dual space"):

conv[n] = {5⋅2, 5⋅4+1⋅2, 5⋅1+1⋅4+8⋅2, 1⋅1+8⋅4, 8⋅1} = {10, 22, 25, 33, 8}

=> Aconv = 10 + 22 + 25 + 33 + 8 = 98

How is this all the same, yet not the same? Am I misinterpreting the concept of duality?

To make the confusion complete, the Fourier transformation is obviously not area-preserving as can be easily seen from these graphs:

Fourier Transforms

But it does seem to preserve the "multiplicativity of areas" when it comes to convolutions when transforming convolution integrals (areas) between real and dual space (i.e. convolution/deconvolution).

There must be some underlying concept that unifies all this clearly. I'm not a mathematician, but I've read that the mechanics behind convolution (crosscorrelation, autocorrelation) and Fourier transformation is intimately linked to the concept of products (Plancherel's theorem, square-integrable functions etc.). I remember from basic functional analysis the concept of complete orthonormal bases (ONBs) and that a Fourier series "is" a projection of functions onto an ONB of harmonic functions. And somehow, this translates into two different ways of multiplication (term-wise vs. polynomial). From this, convolution appears to be some kind of "generalized product" defined on functions and if we represent functions by harmonic series, Fourier transformation somehow transforms the "mechanics" of the multiplication (term-wise vs. polynomial).

I have a hunch to how Fourier transformation might serve to "simplify" a polynomial multiplication (which scales with the square of the number of monoms) into a term-wise multiplication (which scales linearly with the number of monoms): In a Fourier transformation, a function gets "projected" onto a complete ONB of harmonic functions (represented by complex exponential functions) and integrated. So any harmonic component in "disharmony" with an exponential's frequency will get averaged out. So for the polynomial multiplication, I imagine that all mixed terms are being averaged out leaving only monomic terms.
In other words, we are plugging prior information about the posited "harmonicity" of our functions (i.e. they can be represented by Fourier series) into our multiplicative operation in order to "save some work".
Of course, this is a very imprecise way of describing things - we are not losing any information by applying the Fourier transform as the original functions can be perfectly retrieved via the reverse transform, but somehow we have a representation of functions that lends itself "inherently" for doing multiplicative calculations (products). And it is the nature of this "inherent structure" that I'm asking about here.

I'm sorry if I cannot get any clearer here - but the fact that I cannot is precisely the issue I'm trying to clear up.

答:
In time domain, when everything converges
$$\int_{-\infty}^\infty \int_{-\infty}^\infty f(t-y)g(y)dydt=\int_{-\infty}^\infty  f(t)dt\int_{-\infty}^\infty g(u)du$$
becomes in frequency domain

$$\widehat{f \ast g}(0)=(\hat{f}\hat{g})(0)=\hat{f}(0)\hat{g}(0)$$
It works the same way for the Fourier series and the discrete Fourier transform (with circular convolution for the latter), with polynomials (so that $P(e^{2\pi it})$ is a Fourier series) it gives nothing else than $$P Q(1)=P(1)Q(1)$$ where $P(1)$ is the sum of the coefficients and the coefficients of $PQ$ are the convolution of the coefficients.


问:
How to find a pair of possible solutions (x, y) of this system of equations?
Let $(x, y)$ be a pair of possible solutions of this system of equations:
$x + y = 7$;
$x^2 + y^2 = 25$.

This exercise is all about finding $xy$ product.
there are four possible solutions,

(a) 15

(b) 12

(c) 8

(d) 6

Here's how I've solved it:
I've rewritten the second equation as "$x + y = 5$", since each member was squared.

Now, I've rewritten again this equation as "$y = 5 - x$".
I've substituted and I've got: "$x + 5 - x = 7$", but x cancels out.  Therefore I can say this is not the way to solve it.

I try another one: I've rewritten the first equation as "$y = 7 - x$". Then, I've substituted, and I've got: "$x^2 + (7-x)^2 = 25$"
The last step is: $x^2 = -12$, which isn't solvable using real numbers. So, it's not solvable using this method.

If you know an alternative method, let me know.

Body:
$x^2 + (7 - x)^2 = 25 \implies$

$x^2 + (49 - 14x + x^2) = 25 \implies$

$2x^2 - 14x + 24 = 0 \implies$

$x^2 - 7x + 12 = 0 \implies$

$(x - 3)(x - 4) = 0.$

$x = 3$ or $x = 4$.

$y = 4$ or $y = 3.$


问:
Probability with three parts
Suppose that a town has the population of $100$ and $10$ of them are infected with HIV. Assume that currently Sasha is not infected and that if she meets with a group of one or more people that includes a infected person, then she will get infected.

$(a)$ What is the probability that she will get infected when she meets with one person?

$(b)$ What is the probability that she will get infected when she meets with two persons?

$(c)$ What is the probability that she will get infected when she meets with three persons?

It seems working backwards would be beneficial for solving this problem. Namely, probability that she will not get infected when she meets with $k$ people would be $\frac{C(89,k)}{C(99,k)}$, thus desired answer will be $1 - \frac{C(89,k)}{C(99,k)}$?

答:
Your attempt for part (a) is incorrect, I'll show  you how to start for part (b), you should then be able to finish it and do the other parts.

Leaving out Sasha, there are $10$ infected in the $99$ remaining people.

P(Sasha is not infected meeting $2$ persons) $= \frac{89}{99}\cdot\frac{88}{98}$

So what is P(Sasha is infected) ?

Title: Probability with three parts


问:
Almost sure convergence, proof verification
I have the sample space $\Omega = [0,1]$ and the random variable on that space $X_{n}(\omega)=n^{2}1_{[0,1/n]}(\omega).$ My goal is to prove that this random variable converges almost surely at 0.

My attempt:

From definition I have to prove that $\mathbb{P}(X_{n}\rightarrow 0)=1$

$\mathbb{P}(\left \{ \omega \in \Omega:\left | n^{2}1_{[0,1/n]}(\omega)\right |<\epsilon, \ \ \forall \ n\geq n_{0}  \right \}),$ where $\mathbb{P}$ the Lebesgue measure.

Inside the probability, we have that the inequality must hold for all $n\geq n_{0}$.

$\cap_{n\geq n_{0}}\left \{ \omega \in [0,1/n] \right \}=\cup_{n\geq n_{0}}\left \{ \omega\in [1/n,1] \right \}$

Because we work  with Lebesgue measure we have that

$\lambda(\cup_{n\geq n_{0}}\left \{ \omega\in [1/n,1]\right \})$ and $\left \{ \omega\in [1/n,1]\right \}$ are increasing events as $n$ increases.

Hence, from monotonicity

$\lambda(\cup_{n\geq n_{0}}\left \{ \omega\in [1/n,1]\right \})=1$

答:
I guess that $\mathbb P$ il the Lebesgue measure (otherwise, it's not true in general). Let $\omega \in(0,1]$. Let $N\in \mathbb N$ s.t. $\frac{1}{n}<\omega $ for all $n\geq N$. Then $$X_n(\omega )=0,$$
for all $n\geq N$. Therefore $X_n(\omega )\to 0$ for all $\omega \in (0,1]$, i.e. $X_n\to 0$ a.s.


问:
Closed form solution of Fibonacci-like sequence
Could someone please tell me the closed form solution of the equation below.

$$F(n) = 2F(n-1) + 2F(n-2)$$

$$F(1) = 1$$
$$F(2) = 3$$

Is there any way it can be easily deduced if the closed form solution of Fibonacci is known?

Body:
Define $g(z) = \sum_{n \ge 0} F(n + 1) z^n$, write the recurrence as:
$$
F(n + 3) = 2 F(n + 2) + 2 F(n + 1) \qquad F(1) = 1, F(2) = 3
$$
Multiply the recurrence by $z$, sum over $n \ge 0$ and get:
$$
\frac{g(z) - F(1) - F(2)}{z^2} = 2 \frac{g(z) - F(1)}{z} + 2 g(z)
$$
Solve for $g(z)$:
$$
g(z) = \frac{1 + z}{1 - 2 z - 2 z^2}
     = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 - (1 + \sqrt{3}) z)}
        + \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot \frac{1}{1 + (1 - \sqrt{3})z}
$$
Two geometric series:
$$
T(n+ 1) = \frac{2 + \sqrt{3}}{2 \sqrt{3}} \cdot (1 + \sqrt{3})^n
            + \frac{2 - \sqrt{3}}{2 \sqrt{3}} \cdot (1 - \sqrt{3})^n
$$


问:
Strictly positive element in unital $C^*$ algebra is invertible
Let A be a unital $C^∗$ algebra. $a\in A_+$ is called strictly positive if for all positive linear functional $\phi \in A_+^*$ and $\phi \neq 0$, we have $\phi(a)>0$. Prove that $a\in A_+$ is strictly positive iff $a$ is invertible.

My attempt: if $a\in A_+$ is invertible, then there exists $x\in A$ such that $a=x^*x$ and $a^{-1}=x^{-1}(x^{-1})^*$, Using Cauchy-Schwarz inequality: $\forall \phi \in A_+^*$
$$\phi(a)\phi(a^{-1})=\phi(x^*x)\phi((x^{-1})^{**}(x^{-1})^{*})\ge |\phi(x^*(x^{-1})^*)|^2=|\phi(1)|^2=\|\phi\|^2 > 0 $$
then we have $\phi(a)\neq 0$, then $\phi(a)>0$.

But for other direction, i have no idea. I knew that $a\in A_+$ is called strictly positive also can be definted that $\overline{aAa}=A$, and from this definition i can prove that if $a$ is strictly positive, then $axa$ is invertible, then $a$ is invertible. but i want to prove $a$ is invertible from defintion which i give firstly.

I guess if $\phi(a)>0$, then $1-\phi(a) <1$, and let $\phi$ be some specific function ,then apply functional calculus, but i can't prove that.

Any idea will be appreciated!

答:
EDITED Since I originally just reproved what you already included in your statement.

Suppose $a\in A$ is strictly positive.  Since $A$ is unital, the state space of $A$ is weak$^*$-compact, and it follows that $\varepsilon:=\inf\{\phi(a):\phi\text{ is a state on }A\}>0$. It then follows that $a-\varepsilon\cdot1$ is positive, from which it follows that the spectrum of $a$ is contained in $[\varepsilon,\infty)$, i.e., $0$ is not in the spectrum of $a$, i.e., $a$ is invertible.

Here's another proof of the reverse direction:

If $a\in A$ is positive and invertible, its spectrum is a compact subset of $(0,\infty)$, and thus $a-\varepsilon\cdot 1$ is positive for some $\varepsilon>0$.  Thus if $\phi$ is a non-zero positive linear functional on $A$, we have
$$\phi(a)=\varepsilon\phi(1)+\phi(a-\varepsilon\cdot1)\geq \varepsilon\|\phi\|+0>0.$$


答:
For the other direction, let $a$ be a positive non-invertible element. Then $0\in \text{Spec}(a)$ and there is a unital *-homomorphism $\Phi: C^*(a) \rightarrow C(\text{Spec}(a))$ mapping $a$ to the identity function on $\text{Spec}(a)$. Here $C^*(a)$ is the $C^*$-subalgebra generated by $a$ and $C(\text{Spec}(a))$ the continuous functions on $\text{Spec}(a)$. Define a state $\phi$ on $C^*(a)$ by $\phi(x)=\Phi(x)(0)$.

$\phi$ vanishes on $a$ and we can extend it to all of $A$.

Title: Strictly positive element in unital $C^*$ algebra is invertible

Body:
Let A be a unital $C^∗$ algebra. $a\in A_+$ is called strictly positive if for all positive linear functional $\phi \in A_+^*$ and $\phi \neq 0$, we have $\phi(a)>0$. Prove that $a\in A_+$ is strictly positive iff $a$ is invertible.

My attempt: if $a\in A_+$ is invertible, then there exists $x\in A$ such that $a=x^*x$ and $a^{-1}=x^{-1}(x^{-1})^*$, Using Cauchy-Schwarz inequality: $\forall \phi \in A_+^*$
$$\phi(a)\phi(a^{-1})=\phi(x^*x)\phi((x^{-1})^{**}(x^{-1})^{*})\ge |\phi(x^*(x^{-1})^*)|^2=|\phi(1)|^2=\|\phi\|^2 > 0 $$
then we have $\phi(a)\neq 0$, then $\phi(a)>0$.

But for other direction, i have no idea. I knew that $a\in A_+$ is called strictly positive also can be definted that $\overline{aAa}=A$, and from this definition i can prove that if $a$ is strictly positive, then $axa$ is invertible, then $a$ is invertible. but i want to prove $a$ is invertible from defintion which i give firstly.

I guess if $\phi(a)>0$, then $1-\phi(a) <1$, and let $\phi$ be some specific function ,then apply functional calculus, but i can't prove that.

Any idea will be appreciated!


问:
Find Taylor polynomial of an integral function
I ran into this problem and I'm absolutely clueless. The problem is:

Let $f$ be: $$f(x)=\int_0^{\ln(1+2x)} \frac{e^t}{1+t^2}\;dt$$ Find the Taylor polynomial of order $1$ around $x=0$.

The only idea I've had is that you could write $f(x)= F(ln(1+2x))-F(0)$, and then I could use that $f'(x)=F'(ln(1+2x))-F'(0)$, but whenever I try that I always just get $0$.

Any help is greatly appriciated.

答:
If you have a function of x that's defined as the integral from a contant to x, then the derivative of that function is just the integrand evaluated at x.

So if you can do a substitution inside of the integral so that one boundary value is a constant and the other one is just x, you can easily find the nth derivative of your function f


问:
Dependence of Axioms of Equivalence Relation?
This question is problem 11(a) in chapter 1 in 'Topics in Algebra' by I.N. Herstein.

These are the properties of equivalence relation given in this book.

Prop 1 $a \sim a$

Prop 2 $a \sim b$ implies $b \sim a$.

Prop 3 $(a \sim b$ and $b \sim c)$ imply $a \sim c$.

Statement Property 2 of an equivalence relation states that if $a \sim b$ then $b \sim a$. By property 3, we have the transitivity i.e. $a \sim b$, and $b \sim c$ then $a \sim c$. What is wrong with following proof that property 2 and 3 imply property 1? Let $a \sim b$; then $b \sim a$, whence by property 3 (using $a = c$), $a \sim a$.

I think I can prove this to be wrong. Without proving equivalence relation first, one can not use $a = c$. Right? After all, equality is equivalent to 'equivalence relation' and 'axiom of substitution' are satisfied. If this is right, then I have trouble with the next part of this problem.

Part 2 Can you suggest an alternative of property 1 which will insure us that prop 2 and prop 3 do imply 1?

Can one give such a formulation without using the idea of '=' or otherwise?

EDIT : Italics are my comments. Rest is as it appeared in the book.

Notion of Equality

I have read in Terry 'Analysis 1' in Appendix A.7 published by Hindustan Book Agency that there are four axioms of 'equality'. First 3 are same as equivalence relation where $\sim $ in replaced by $ = $. The fourth one is known as axiom of substitution. Given any two objects $x$ and $y$ of some type, if $ x = y $, then $f(x) = f(y) $ for all functions or operations $f$.

答:
Part 1. Here's an example of a relation that is symmetric and transitive, but not reflexive.

The set is $X=\{1,2,3\}$. The relation is
$$R = \Bigl\{ (2,2),\ (2,3),\ (3,2),\ (3,3)\Bigr\}.$$
Verify that $R$ is symmetric and transitive. Verify that $R$ is not reflexive. Then try to see why the alleged proof fails in this example. Use that to explain where the fallacy in the proof lies. It does not lie in taking $c$ equal to $a$.

Part 2. Think about what the fallacy is in the proof you are given; what extra hypothesis on $\sim$ would make the argument correct? The argument is fallacious because it assumes that something happens, when you have no warrant for asserting it will happen. So try to come up with some hypothesis that will guarantee this happens.


问:
Example that isometry doesn't preserve extrinsic distance
Is there any example that an isometry doesn't preserve the extrinsic distance? That is, I want some example that for some isometry $f:S\to\tilde{S}$ between connected regular surfaces such that it doesn't satisfy this property: $|f(p)-f(q)| = |p-q|$ for all $p,q\in S$.

答:
In a comment I gave a complicated answer (using Nash-Kuiper theorem). But here is a much more simple example. Let $\mathbb{S}^1\subset \mathbb{R}^2$ be the unit circle, and let $\mathcal{E}_{a,b} = \{(x,y) \in \mathbb{R}^2~|~\frac{x^2}{a^2} + \frac{y^2}{b^2}=1 \}$ be an ellipse. Chose $a<b$ such that this ellipse has length $2\pi$, and then is isometric to $\mathbb{S}^1$.

The arc-length parametrization of the ellipse gives an isometry between the circle and the ellipse that does not preserve the extrinsic distance in the plane. One can see this because antipodal points in the circle has constant distance $2$, while in the ellipse, it varies between $2a$ and $2b$.


问:
“Unique” doesn't have a unique meaning
When using words like “unique” and “any”, particularly in technical communication, I sometimes find myself deliberating over which definition and tenor is the most natural, or which alternative phrasing might be clearer even if less succinct or accessible.

Does “Every boy has a unique shirt” mean that

no two boys have the same shirt,

or does it mean that

no two shirts belong to the same boy?

I suppose the former; if so, then does the latter mean “Every shirt belongs to a unique boy”?

答:
Further to MJD's, Dario's and Cheerful Parsnip's comments and answers:

“Every $A$ has a unique $B$” has multiple interpretations:

 $(1)$ Each $A$ has at least one $B$ that no other $A$ has;

 $(2)$ Each $A$ has exactly one $B,$ which no other $A$ has;

 $(3)$ Each $A$ has exactly one $B.$

To illustrate these interpretations, read each of the following
lines (their labels correspond to the interpretation numbers
above) as “Every bin has a unique score” (bins are separated
by indentation and scores by commas):
\begin{gather}6,4\quad 7,5\quad 8,0\quad 9,0\quad \tag{1}\\ 6\quad\quad 7\quad\quad 8\quad\quad 9\quad\quad \tag{1,2,3}\\
7\quad\quad 7\quad\quad 7\quad\quad 7\quad\quad \tag{3}\end{gather}

Interpretation $(1)$ corresponds to the most common definition of ‘unique’ as having no duplicate.

Interpretation $(2)$ corresponds more accurately to “Each $A$ has a distinct $B$”.

Interpretation $(3)$ corresponds to the common technical usage of ‘unique’ to connote having no alternative possibility.

All in all, due to its ambiguity and frequent conflation with the word ‘distinct’, I would just avoid using the word ‘unique’ in the above sense.

(On the other hand, “uniquely determined” isn't ambiguous and corresponds squarely to interpretation $(3)\,).$

It turns out that all the four statements in the Question have different meanings!

$(S_1)\quad$ “Every boy has a unique shirt.”

$(S_2)\quad$ “Every shirt belongs to a unique boy.”

$(S_3)\quad$ “No two boys have the same shirt.”

    Every shirt belongs to at most one boy. $$∀s,b_1,b_2\;\Big(P(s,b_1)∧P(s,b_2)\implies b_1=b_2\Big).$$

$(S_4)\quad$ “No two shirts belong to the same boy.”

    Every boy has at most one shirt. $$∀b,s_1,s_2\;\Big(P(s_1,b)∧P(s_2,b)\implies s_1=s_2\Big).$$

$(S_1)$ and $(S_2)$ are discussed in the previous section, and are clearly inequivalent;

since $(S_3)$ and $(S_4)$ each allows some boy to own no shirt, neither is equivalent to $(S_1);$

since $(S_3)$ and $(S_4)$ each allows some shirt to have no owner, neither is equivalent to $(S_2);$

$(S_3)$ and $(S_4)$ are clearly inequivalent.

Title: “Unique” doesn't have a unique meaning


问:
$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$
Given $a,b,c\geq 0$, then this following is true
$$\min(a,b+c)\leq \min(a,b)+ \min(a,c)$$

I tried to prove it by using some cases, i.e.

Case 1: if $\,\,a\leq b \leq c\,\,$ and $\,\,a\leq c \leq b,\,\,$ it's clear that
\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq a+a=2a
\end{align*}
Case 2: if $\,\,b\leq a\leq c\,\,$ and $\,\,c\leq a \leq b,\,\,$ we have

\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq b+a
\end{align*}
and
\begin{align*}
\min(a,b+c)&\leq \min(a,b)+\min(a,c)\\
a &\leq a+c
\end{align*}
Case 3: $\,\,c\leq b\leq a\,\,$ and $\,\,b\leq c \leq a,\,\,$

I am not sure about this case, because although $a$ is the biggest from $b$ and $c$, it can satisfies $a\leq b+c$ or $b+c\leq a$. Is the third case must be done with two cases again? i.e for $a\leq b+c$ and $b+c\leq a$?

or there is another way to prove $\min(a,b+c)\leq \min(a,b)+ \min(a,c)$ for $a,b,c\geq 0$?

Any help will be appreciated. Thanks.

答:
Alternative proof:

Case 1: $a\ge b+c.$ Then $a\ge b, a\ge c$. $$\min(a,b+c)=b+c =\min(a,b)+ c=\min(a,b)+\min(a,c).$$

Case 2: $a \le b+c.$ Then $\exists x, y \ge 0, a=x+y$, such that $x\le b, y \le c$. (For example $x=\frac{ab}{b+c}, y=\frac{ac}{b+c}.)$

Then
$$ x \le a, x \le b \implies x \le \min(a,b)$$
$$ y \le a, y \le c \implies y \le \min(a,c)$$
Therefore
$$
a=x+y \le \min(a,b)+\min(a,c). \blacksquare
$$


问:
Drawing numbers from a set, quantile
We have a set containing $20$ numbers from $1$ to $20$. Each time we draw only one number, and repeat it $15$ times (without replacement). Let's denote $X-$ the largest drawn number. Find the smallest ${{16} \choose {15}}/{20 \choose 15}$-fractile  of a random variable $X$.

So, we look for such $x_p$ that $F(x_p)\geq {{16} \choose {15}}/{20 \choose 15}.$ If $X$ is maximum of all drawn numbers, then $F_X(t)=\Bbb P(X\leq t)=\Bbb P(X_1\leq t, ...,X_m\leq t)$ and, if all of $X_1,...X_m$ are iid then we have $\Bbb P(X_1\leq t)\cdot...\cdot\Bbb P(X_m\leq t)$. But... how to proceed?

答:
There are $\binom {20} {15}$ ways to choose 15 numbers from the set.

$\binom{16}{15}$ is the number of ways to choose 15 numbers only from $1$ to $16$ inclusive. The maximum of such subset is at most $16$. So the cumulative probability up to and including $t=16$ is the $\binom {16} {15} / \binom {20} {15}$.

Alternatively, consider breaking down to individual probabilitites $\Pr(X=t)$, where $15\le t \le 20$.

To choose a subset with a maximum of $t$, the subset will have 14 elements between $1$ and $t-1$. The number of subsets with a maximum $t$ is $\binom {t-1}{14}$.

Verify that the total number of subset is $\binom {20}{15}$: (the hockey-stick identity)

$$\sum_{t=15}^{20}\binom{t-1}{14} = \sum_{t'=14}^{19}\binom{t'}{14} = \binom {20}{15}$$

And similarly the $\binom {16}{15}$ is the sum

$$\sum_{t=15}^{16}\binom{t-1}{14} = \sum_{t'=14}^{15}\binom{t'}{14} = \binom {16}{15}$$

And so

$$\begin{align*}
\Pr(X \le 16) &= \sum_{t=15}^{16} \Pr(X = t)\\
&= \sum_{t=15}^{16} \frac{\binom{t-1}{14}}{\binom {20}{15}}\\
&= \frac{\binom{16}{15}}{\binom {20}{15}}
\end{align*}$$

Title: Drawing numbers from a set, quantile


问:
understanding $P(X|Y=y)$ and $E(X|Y=y)$ for continuous random variables $X,Y$
I am trying to read and understand a bit about conditional probability and conditional expectation. Especially, I would like to understand a bit more about what $P(X|Y=y)$ or $E(X|Y=y)$ actually mean. So I was reading the dedicated wikipedia article. It is a bit hard since I don't know that much about measure theory. In the computation section it is written that if both $X$ and $Y$ are continuous random variable we have ${\displaystyle \operatorname {E} (X\mid Y=y)=\int _{\mathcal {X}}xf_{X\mid Y}(x\mid y)\,\mathrm {d} x,}$ where $ {\displaystyle f_{X\mid Y}(x\mid y)={\frac {f_{X,Y}(x,y)}{f_{Y}(y)}}}$.

First, what exactly is the left hand side $\operatorname {E} (X\mid Y=y)$? Is it actually $E(X|\sigma(Y=y))$ where $\sigma(Y=y)=\{(Y=y),(Y=y)^c,\Omega,\emptyset\}$ is the sigma algebra generated by the event $(Y=y)$? Then how to prove this equal to $\int_{\mathcal {X}}xf_{X\mid Y}(x\mid y)\,\mathrm {d} x$?

答:
Ignoring measurability issues, you may simply regard the conditional probability $P(\cdot\mid Y=y)$ as a form of density:

\begin{align*}
P(A\mid Y=y)
&= \lim_{h\to0^+} P(A\mid Y\in[y,y+h]) \\
&= \lim_{h\to0^+} \frac{P(A \cap \{Y\in[y,y+h]\})}{P(Y\in[y,y+h])}
\end{align*}

Similarly,

\begin{align*}
E(X\mid Y=y)
&= \lim_{h\to0^+} E(X\mid Y\in[y,y+h]) \\
&= \lim_{h\to0^+} \frac{E(X \cdot \mathbf{1}_{\{Y\in[y,y+h]\}} )}{P(Y\in[y,y+h])}.
\end{align*}

So, if we fix $h > 0$, then by the law of total probability,

\begin{align*}
E[X \mid Y \in [y, y+h]]
& \approx \sum_{k} (kh) P(X \in [kh, (k+1)h) \mid Y \in [y, y+h]) \\
& = \sum_{k} (kh) \frac{P(\{X \in [kh, (k+1)h)\} \cap \{Y \in [y, y+h]\}}{P(Y \in [y, y+h])} \\
& \approx \sum_{k} (kh) \frac{f_{X,Y}(kh, y) h^2}{f_Y(y)h} \\
& \approx \int_{\mathbb{R}} x \frac{f_{X,Y}(x,y)}{f_Y(y)} \, \mathrm{d}x.
\end{align*}

And we expect that the error hidden in the approximate equality $\approx$ will become smaller as $h\to0$, hence obtaining the desired identity.

Of course, all these can be derived rigorously by invoking the measure theory.


问:
Number of possible events is $2^N$ ( $N$ is the number of outcomes )?
A probability space is a random process or experiment with
three components:
–

$Ω,$ the set of possible outcomes $O$

*number of possible outcomes = $|Ω| = N$

*$F$, the set of possible events $E$
      - an event comprises $0$ to
$N$ outcomes

number of possible events
= $| F | = 2^N$

Here I am not able to extract the exact meaning of number of possible outcomes. In case of tossing a coin, we have only $2$ outcomes. So, $N = 2$. So, either we have head or tail. According to the above definition, $|F| = 2^N = 2^2 = 4.$ How come this formula or axiom is valid in probability space ?

答:
$2^N$ is very special. It describes so much and is a key to understanding something truly magical.

An event that can only have $2$ outcomes and more so, can occur equally likely, event conducted in a fair setting. Fair as in pure, without any bias or favouring either of the outcome deliberately! Conducting a trial, Bernulli trial, determining an answer, detecting level of responsiveness.

Fundamentally, during the occurrence of the most simplest event possible, there are only $2$ possible states for this event to be in. At any given time of conducting a measurement the event has not yet occured or it already has.

Consider a sequence, of question -> answer pairs:

Has the event occured? No

Has the event occured? No

Has the event occured? Yes

Has the event occured? Yes

Has the event occured? No

In such case the ratio of number of No's and Yes's observed will be equal

Any sequnce of events with $2$ possible states of outcome has a chance of being observed once in $2^N$ number of trials.

$$\frac12 \times \frac12 \times \ldots = \frac1{2^N}$$

Knowing this it is possible to "game the system".

Here's the simplest thought experiment illustrating the situation if this exploitation occurring. In a way one can use randomness against itself like this:

Decide on the outcome in advance (Red) and conduct the experiment (Throw the ball).

Has the ball stopped on Red? No

Has the ball stopped on Red? No

Has the ball stopped on Red? No

Has the ball stopped on Red? No

Has the ball stopped on Red? Yes

Bet $1$ is $\$1$. Loss. Total loss is $\$1$

Bet $2$ is $\$2$. Loss. Total loss $1+2=3$

Bet $3$ is $\$4$. Loss. Total loss $+4=7$

Bet $4$ is $\$8$. Loss. Total loss $+8=15$

Bet $5$ is $\$16$. Win. Total win is equal to my original $\$16$ used to place bet 5 coming back, plus the win itself. Win equal to the amount I placed on 5th coin toss and $16 additional. Bringing my total win to $32=$16+$16

Lets count our chickens.

Our total loss was $\$15$ the casino took it and that money is not coming back. $-\$15$.
Win is $+\$32$. Assume original $\$16$ were borrowed from a friend to play in casino. So we return his money back $-\$16$ and keep the other earned $\$16$ to ourselves. We've spent $\$15$ out of our pocket to play the coin toss game and ended up with $\$16$ so our net win $-\$15+\$16=\$1$.

Introducing a particular notation O(n)=1 means even if you throw n number of tosses and lose -$X amount and win back +$Y amount you'll end up with -$X+$Y=$1 always at the end.

Next time you throw the coin 4 times and in that circumstance lose 3x but win on the 4th you'll lose -$7 and win +$8 you'll end up now with $2 extra.

Here's what I mean:

Consider a sequence of 5+3 coin tosses:
[Loss Loss Loss Loss Win Loss Loss Win]  or simply [0 0 0 0 1 0 0 1] shows after observing 2 wins you'll end up with $2 extra.

So again O(n)=1 means tossing the coin n times so long as a win at the end has occured will always bring $1 extra.

I keep increasing my bet at the rate of 2 times the previously placed bet so eventually a win is observed and I end up with an extra $1 or an extra unit if your initial bet was not $1 but something else.

The next bet after a win (6th flip or 9th flip) will be back to $1 since we reset the event and start anew.

2^N-1 is also an important formula. It is the total sum, the closed form solution to the SUM(2^N). At any given point in time the total loss is 2^N-1. The sequence of N losses is equal to $2^N-1 dollars.

What if we don't double?

Bet $1$ is $\$1$. Loss. Total loss is $-\$1$

Bet $2$ is $\$3$. Loss. Total loss $-4$

Bet $3$ is $\$7$. Loss. Total loss $-\$11$

Bet $4$ is $\$15$. Loss. Total loss $-26$

Bet $5$ is $\$31$. Win. Total win is $2\times\$31=62$. Net profit $\$-(26+31)+2\times31=+\$5$

Number of flips is $5$. Profit for this batch of flips is $5$.

Bet $6$ is $\$1$. Loss. Total loss is $-1$

Bet $7$ is $\$3$. Loss. Total loss $-4$

Bet $8$ is $\$7$. Win. Total loss is $-11$, total win is $\$14$ so net profit is $-11+14=3$.

Number of flips is $3$. Total profit is $3$ for the second batch of flips.

Total profit is $5+3=8$. Total number of flips is $8$.

Using the introduced notation above $O(n)=n$.

This would mean flip a coin $5$ times and bet $\$100$ on the first flip, the total profit will be $\$500$. Flip another $3$ times, another $\$300$. So long as you can sustain your bets that grow exponentially once a win is observed total number will always be equal to n. So $f(x)=x$.

Now think about $\mathcal{O}(n)=n^2$
Here flip a coin $5$ times and win $\$2500$ plus $\$900$ if you flip another $3$ times.


问:
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?
What is the order of $\bar{2}$ in the multiplicative group $\mathbb Z_{289}^×$?

I know that $289 = 17 \times 17$

so would it be $2^8\equiv 256\bmod17 =1$

and therefore the order of $\bar{2}$ is $8$? I'm not too sure about this

答:
Define the set $H \subset {\displaystyle (\mathbb {Z} /289\mathbb {Z} )^{\times }}$ by

$\tag 1 H = \bigr\{[a + 17m] \,\large \mid \, \normalsize  a \in \{-1,+1\} \text{ and } 0 \le m \lt 17\bigr\}$

It is easy to show that $H$ contains exactly $34$ elements.

Proposition 1: The set $H$ is closed under multiplication.

Proof

Consider,

$\quad (a + 17m)(b+17n) = ab + 17(an +bm) + mn\cdot 17^2$

while dividing $an +bm$ by $17$ to get the non-negative residue. $\quad \blacksquare$

So we can state (see bullet $1$ of this elementary group theory)

Proposition 2: The set $H$ forms a group of order $34$.

Continuing,

Proposition 3: The element $[16]$ generates $H$.

Proof

The order of $[16]$ must divide $34$.

The order of $[16]$ is not equal to $2$. Moreover, by applying the binomial theorem we can write

$\quad 16^{17} = \bigr((-1) + 17\bigr)^{17} = (-1)^{17} + \binom{17}{16}(-1)^{16}\cdot 17^{1} + K\cdot 17^2 \equiv -1 \pmod{289}$

and so the order of $[16]$ must be $34$. $\quad \blacksquare$

There are two methods we can use here to finding the order of $[2]$.

Method 1:

Since $[2]^4 = [16]$ and $[2] \notin H$ the order of $[2]$ is strictly greater than $34$. Also, with this fact and

$\quad [2]^{136} = [16]^{34} = [1]$

we must conclude that the order of $[2]$ is either $68$ or $136$.

Now

$\quad [2]^{68} = [16]^{17} \ne [1]$

and we therefore conclude that the order of $[2]$ is $136$.

Method 2

Since $[2]^1, [2]^2, [2]^3 \notin H$ and $[2]^4 = [16] \in H$ we can employ the group theory found here and conclude that the order of $[2]$ is $4 \times 34 = 136$.


问:
Fourier transform of analytic functions and radius of convergence
Let $\phi$ be a fonction in the Schwartz space $\mathscr S(\mathbb R)$ which is also analytic on the real line and such that the radius of convergence on the real line is bounded below by $\rho>0$. Then, with
$\hat \phi(\xi)=\int e^{-2iπ x\xi}\phi(x) dx$, do we have
$$
\vert\hat \phi(\xi)\vert\le C e^{-2π\rho \vert\xi\vert}\quad?
$$
Is it true for $\phi$ real analytic on $\mathbb R$ which happens to be a temperate distribution? Is there a converse?

答:
You need that $\phi$ is analytic on $|\Im(z)|\le r$,

$L^1$ on horizontal lines and $\lim_{|x|\to \infty}\sup_{|y|\le r} |\phi(x+iy)|=0$
to obtain from the Cauchy integral theorem that $$\int_{-\infty}^\infty e^{2i\pi (x+iy)\xi} \phi(x+iy)dx$$
converges and doesn't depend on $|y|\le r$ so that
$$|\hat{\phi}(\xi)| \le (\|\phi(.+ir)\|_{L^1}+\|\phi(.-ir)\|_{L^1}) e^{-2\pi r |\xi|}$$

Conversely if both $\hat{\phi},\hat{\phi}'$ are $O(e^{-2\pi r|\xi|})$ then $\phi$ satisfies the above conditions for $|\Im(z)|\le r-\epsilon$.

Try with $\phi(x)=e^{-e^{x^2}}$ which is unbounded on the horizontal lines others than the real axis, if $\hat{\phi}$ was $O(e^{-4\pi |\xi|})$ we'd have that $\|\phi(.+i)\|_{\infty}\le \|\hat{\phi}e^{2\pi \xi}\|_{L^1}<\infty$.


问:
Permutations with global limited repetition

What is the number of permutations of $m$ objects in $n$ spaces such that there are exactly $l$ repetitions in the permutation?

for example,
if $m = 5, n = 7, l = 4$

then $$a_1, a_1, a_1, a_3, a_3, a_4,a_4$$
is a legal sequence (4 repetitions, $a_1$ twice, $a_3$ once and $a_4$ once) but

$$a_1, a_2, a_3, a_1, a_3, a_5, a_3$$ is not (only 3 repetitions).
and neither is

$$a_1,a_1,a_1,a_1,a_1,a_1,a_1$$ legal.

The main difference of this problem from other posts such as
Calculating number of permutations given N repeats allowed
is that my number of repeats is global, i.e I require that the number of repetitions of all objects together is exactly l.

答:
You can pick the elements to be in the sequence that are going to be repeated(say $k$ of them) and then consider the repetitions by constructing a sequence $x_i=\text{# times object i repeats},$ so that
$$\sum _{i=1}^kx_i=l,\, x_i\geq 1.$$
Notice that if we call $k$ the ones that are going to be repeated, then $n-(l+k)$ are non repeated elements.
So your expression is
$$\sum _{k=0}^m\binom{m}{k}\binom{m-k}{n-(l+k)}\sum _{x_1+\cdots +x_k=l}\binom{n}{\underbrace{1,\cdots ,1}_{n-(l+k)},x_1+1,\cdots ,x_k+1}.$$
Now, if this can be expressed in easier terms seems a problem for generating functions, the problem being multivariate by nature.


问:
An asymptotic about Integral of Legendre Polynomials
I want to show asymptotics of the following integral involving  Legendre Polynomial:

For $0<t<\theta<\frac\pi2$,
$$\Big|\int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}}  \frac{1-P_n(\cos t)}{t}dt \Big| \leq C \theta^{-\frac12} \cdot \log (n\theta),$$ where $C$ is the constant,  $P_n(x)$ is a Legendre Polynomial, $n$ is a positive integer.

I am trying to use Bernstein's inequality for trigonometric polynomials:

Since $P_n(1)=1$ and $|P_n(\cos t)|\leq 1$, $$|P_n(\cos t)-1|=|P_n(\cos t)-P_n(\cos 0)|\leq nt||P_n ||_{\infty}\leq nt.$$

Then I can only get the below estimate but I have no further idea $$\Big|\int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}}  \frac{1-P_n(\cos t)}{t}dt \Big| \leq  \int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}}  \frac{|1-P_n(\cos t)|}{t}dt \leq  \int_0^\theta \frac{1}{\sqrt{\theta}+\sqrt{\theta-t}}  \frac{nt}{t}dt=O(n).$$

Any suggestions are welcome! Thank you for your help!

答:
First, suppose that $\frac{1}{n} <\theta <\frac{\pi}{2}$. We split the range of integration into two parts: $0<t<\frac{1}{n}$ and $\frac{1}{n}<t<\theta$, respectively. On the first interval, we use the known limiting relation between the Legendre polynomials and the Bessel function of the first kind $J_0$:
\begin{align*}
&\int_0^{1/n} {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - t} }}\frac{{1 - P_n (\cos t)}}{t}dt}  = \int_0^1 {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - s/n} }}\frac{{1 - P_n \left( {\cos \left( {\frac{s}{n}} \right)} \right)}}{s}ds} \\ & \sim \int_0^1 {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - s/n} }}\frac{{1 - J_0 (s)}}{s}ds}  \le K\int_0^1 {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - s/n} }} s ds}  \le \frac{K}{2}\theta ^{ - 1/2}
\end{align*}
with a suitable positive constant $K$ and large $n$. On the remaining interval, we have
\begin{align*}
\left| {\int_{1/n}^\theta  {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} } \right| & \le \theta ^{ - 1/2} \int_{1/n}^\theta  {\left| {\frac{{1 - P_n (\cos t)}}{t}} \right|dt} \\ & \le 2\theta ^{ - 1/2} \int_{1/n}^\theta  \frac{{dt}}{t} = 2\theta ^{ - 1/2} \log (n\theta ) .
\end{align*}
Thus, there is a constant $C >0$ such that
$$
\left| {\int_0^\theta  {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} } \right| \le C \theta ^{ - 1/2} \max(1,\log (n\theta )).
$$
If  $0 <\theta <\frac{1}{n}$, then
\begin{align*}
& \int_0^\theta  {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - t} }}} \frac{{1 - P_n (\cos t)}}{t}dt \le \theta ^{ - 1/2} \int_0^\theta  {\frac{{1 - P_n (\cos t)}}{t}dt}  \\ & \le \theta ^{ - 1/2} \int_0^{1/n} {\frac{{1 - P_n (\cos t)}}{t}dt}   \sim \theta ^{ - 1/2} \int_0^1 {\frac{{1 - J_0 (s)}}{s}ds} \\ & \le K\theta ^{ - 1/2} \int_0^1 {sds}  = \frac{K}{2}\theta ^{ - 1/2} .
\end{align*}
In summary, there is a constant $C >0$ such that
$$
\left| {\int_0^\theta  {\frac{1}{{\sqrt \theta   + \sqrt {\theta  - t} }}\frac{{1 - P_n (\cos t)}}{t}dt} } \right| \le C \theta ^{ - 1/2} \max(1,\log (n\theta )).
$$


问:
I cannot solve the given Probability math. Can anyone help?
In winter of a country, it rains 45% of the days and shine 55% of the days. Mr. X has a barometer which wrongly predicts 5% of the time in rainy days and 10% on shiny days. Mr. X does not carry umbrella if barometer predicts shine. On good shiny morning in winter, his barometer predicts “shine” and he does not carry umbrella. What is the probability that Mr. X will suffer from rain on that day.</p>

答:
Let $R$ be the event that it is rainy and $S$ the event that it is sunny.

Let $X_r$ be the event that it is predicted to be rainy and $X_s$ the event that it is predicted to be sunny.

The problem statement tells you may things.  Make sure you can read the problem statement fully and correctly and be able to translate the statements into mathematics

 $\Pr(R) = 0.45,~\Pr(S)=0.55,~R=S^c,~\Pr(X_r\mid R)=0.95,~\Pr(X_s\mid R)=0.05,~\Pr(X_r\mid S) = 0.1,~\Pr(X_s\mid S)=0.9$

You are asked to find $\Pr(R\mid X_s)$

To do this, use Bayes' Theorem and Total Probability.

Recall that:

$$\Pr(A\mid B) = \dfrac{\Pr(A\cap B)}{\Pr(B)}=\dfrac{\Pr(B\mid A)\Pr(A)}{\Pr(B)}$$

and that

$$\Pr(B) = \Pr(B\cap A)+\Pr(B\cap A^c) =\Pr(B\mid A)\Pr(A)+\Pr(B\mid A^c)\Pr(A^c)$$

At this point, it is just correctly using the identities and plugging in numbers to finish.


问:
A fact about non-homothetic lattices
Let $\mathbb H$ be the upper half-plane, and let $\Gamma:=SL_2(\Bbb Z)$. Let $\tau, \tau' \in \mathbb H$ such that $\Gamma\tau \neq \Gamma\tau'$, and let $L:=\mathbb Z \oplus \tau\mathbb Z$ and $L':=\mathbb Z \oplus \tau'\mathbb Z$. Then $L$ and $L'$ are two compact Riemann surfaces. Why they are not $\Bbb C$-analytically isomorphic? Why they are diffeomorphic? (I can check that the action of $SL_2(\Bbb R)$ is transitive, but I can not see why the transitivity of the action of $SL_2(\Bbb R)$, implies they are diffeomorphic.)

答:
A complex analytic isomorphism $f:\Bbb{C}/L\to \Bbb{C}/L'$ lifts to a complex analytic isomorphism $\Bbb{C\to C}$ thus it has to be of the form $z\to az+b$ ie. $L' = aL$.

Equivalently $dz$ is an holomorphic 1-form on $\Bbb{C}/L$, the other mermorphic 1-forms are $h(z)dz$ with $h$ meromorphic on $\Bbb{C}/L$, thus $h(z)dz$ is an holomorphic 1-form iff $h$ has no poles nor zeros ie. $h$ is constant. $df(z)$ is also an holomorphic 1-form on $\Bbb{C}/L$, thus $df(z)=adz$ for some $a\ne 0$ and hence $L'=aL$.

Title: A fact about non-homothetic lattices


问:
How many ways are there to choose 4 letters out of the following 8 letters without duplicates?
There are eight letters: $A BCD EF AG$

How many ways can I choose four letters without duplicates?

(1) Eliminating the duplicate letter $A$, no of ways is ${7 \choose 4} = 35$

(2) No of ways to choose $4$ out of $8$ letters is ${8 \choose 4} = 70$

$\quad $ No of duplicates is ${(8-2) \choose 2} = 15$

$\quad $  No of ways is $70 - 15 = 55$

Both seem right, I don't know why and which one of them is wrong?
Thank you.

答:
First way should be the correct one. Note that in your second way, when we say $8 \choose 4$, we consider two $A$'s different. But normally, total number of ways of choosing $4$ out of $8$ letters with duplicates should be
$${6 \choose 4}+{6 \choose 3}+{6 \choose 2} = 50$$
where ${6 \choose 4}$ is the number of choices where we don't choose any $A$'s, ${6 \choose 3}$ is the number of choices where we choose one $A$ and ${6 \choose 2}$ is the number of choices where we choose two $A$'s. So when we exclude the cases where we choose two $A$'s, we get the same answer as in your first way, $35$.


问:
Can I prove continuity of a function of two variables in this way?
Common approach in handling functions of two variables is to express this function in polar coordinate system. For example, in the classic example $$f(x,y)=\left\{\begin{array}{lr}\frac{xy}{x^2+y^2} & (x,y)\neq (0,0)\\0 & \text{otherwise}\end{array}\right.$$ say $x=r\cos(\phi)$ and $y=\sin(\phi)$, and then we can write $f(r, \phi)=\frac{r^2\sin(\phi)\cos(\phi)}{r^2(\cos^2(\phi) + \sin^2(\phi))}=\frac{1}{2}\sin(2\phi)$. Now if we let $r\rightarrow 0$, then it and, for example, $\phi=\frac{\pi}{4}$, the $$\lim_{r\rightarrow 0}f(r, \frac{\pi}{4})=\frac{1}{2}\neq 0$$ so $f$ isn't continuous in $(0,0)$.

Now I am given a function $$f(x,y)=(x+y)\exp(xy)$$Since this function is composition of continuous functions, it must be continuous, but I wanted to ask in common: earlier I used polar coordinates to tell that the function isn't continuous, so can I use the same approach to say that function is continuous, that is:
$$\lim_{r\rightarrow0}f(r,\phi)=\lim_{r\rightarrow0}r(\cos(\phi)+\sin(\phi))\exp(r^2\sin(\phi)\cos(\phi))=0$$
so function must be continuous, or is it not enough?

Thank you in advance!

答: Yes, that should be enough. As long as you are sure to take an arbitrary $\phi(r)$ (continuous, of course) for your angle (which will serve to parameterize the path in $\Bbb R^2$ towards the point), you can just take the limit as $r\to 0$ to see if it depends on $\phi(r)$. If not, then there is no dependence on path: i.e. the limit exists, and your function is continuous. Be careful though, sometimes the dependence on the path is subtle (watch out for trig functions in denominators)... However, here it looks like there's no problem: there's nothing to stop that $r$ in front from making everything $0$.</p>


问:
Proving that exponential is growing faster than polynomial
I want to prove this limit:$\lim_{x \to +\infty}(|P(x)|-e^x)=-\infty$.
I can't use Hopital's rule, but only the fact that $a_{n}=(1+\frac{x}{n})^n$ and $b_{n}=(1+\frac{x}{n})^{n+1}$ converge to $e^x$.

答:
For any polynomial,
$$\lim_{x\to\infty}\frac{P(x+1)}{P(x)}=1$$ so there is an $X$ such that

$$\forall x>X:\frac{P(x+1)}{P(x)}<2$$ and $P(x)$ is bounded by a geometric progression of constant ratio $2<e$.


问:
Limit in $\mathbb{Z} / P^{2} \to \mathbb{Z} / P $
Consider the commutative diagram

in Ab, where the morphisms are the canonical ones. Let us denote the top horizontal sequence by A and the bottom one by B. Then the vertical maps induce a morphism of abelian groups $\alpha : Lim A \to Lim B$. I am going to state which of the following statement are correct:

$\alpha$ is bijective

$\alpha$ is injective but not surjective

$\alpha$ is surjective but not injective

$\alpha$ is neither injective nor surjective.

However, I am having some trouble seeing what the limits are of both sequences as I am new to homological algebra. I believe the limit of A is $\mathbb{Z}$ but I am not sure. Is this correct? I am also having some trouble seeing what the limit of B is, but I believe it could simply be 0. Is this correct?

答:
The universal property of limits is the following:

Given a set $\{A_i\}_{i\in\Bbb N}$ of objects, and a diagram $$\cdots \to A_3\to A_2\to A_1$$ (where we say $f_i:A_{i+1}\to A_i$), the limit of this diagram is an object $A$ and a set of morphisms $\{g_i:A\to A_i\}_{i\in \Bbb N}$ such that

$g_i=f_i\circ g_{i+1}$ for all $i\in \Bbb N$

For any other object $A'$ and morphisms $\{g_i':A'\to A_i\}_{i\in \Bbb N}$, there is a unique morphism $h:A'\to A$ such that $g_i'=g_i\circ h$

Now, I could just tell you what the limits here are. They are well-known and often studied objects. But let's work through it.

You say you suspect that the limit of the top horizontal diagram is $\Bbb Z$. Does it fulfill the above? Well, with $g_i=\operatorname{id}$ it immediately fulfills everything but the last point. What about the last point? Well, we see that for any other $A'$ and $\{g_i'\}$, the commuting requirement gives $g_i=g_{i+1}$, and we see that $h=g_1$ and we indeed have that $\Bbb Z$ is the limit.

As for the bottom diagram, that's not quite as straight-forward. We can solve the injectivity without further studies of the limit itself, though. Let $P$ be the limit, and let $g_i':\Bbb Z\to \Bbb Z/p^i$ be the canonical maps (the vertical maps in your figure). By the last point of the universal property, there is a map $h:\Bbb Z\to P$, and this is the map we're interested in.

Let $x,y\in \Bbb Z$ be such that $h(x)=h(y)$. Thus we get $g'_i(x)=g'_i(y)$ for all $i$. But as $i$ grows, at some point we have $|x-y|<p^i$, and $g_i'(x)=g_i'(y)$ therefore gives $x=y$. So $h$ is injective.

For surjectivity, we first need to work a little more. Take an $x\in\Bbb Z$, and consider the sequence $x_i=g'_i(x)$. We have $f_i(x_{i+1})=[x_{i+1}]=x_i$ as elements of $\Bbb Z/p^i$. We can recover $x$ from any such sequence, because it is, in some sense, eventually constant as $i$ grows: The subsets ${g_i'}^{-1}(x_i)$ of the integers have a non-empty intersection, and that is $x$.

However, let's look at the set $A$ of all sequences $\{a_i\in \Bbb Z/p^i\}_{i\in\Bbb N}$ with the property that $f_i(a_{i+1})=a_i$. Some of them arise from integers, but far from all (consider, for instance, $a_i=\frac{p^i-1}{p-1}$ for any $p\neq 2$). And this $A$ injects into the limit, basically by construction. So clearly there are elements in the limit that don't come from $\Bbb Z$, so the map we're after is not surjective.

As a matter of fact, this $A$ is the limit we're after. It's called the $p$-adic integers. If we write numbers in base-$p$, these look like the integers, except we allow infinitely many digits to the left. The integers themselves have finitely many digits, but, say $a_i=\frac{p^i-1}{p-1}$ corresponds to the $p$-adic number $\ldots1111$.