问:
Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.</p>

答: Here's an analogy that may help.  The derivative of $\frac{x}{3}$ is $\frac{1}{3}$, while the derivative of $\frac{3}{x}$ is $-\frac{3}{x^2}$.  The quotient rule gives different results when you swap the numerator and denominator; the same thing happens for powers.</p>


问:
Question about a simple derivative
Let's say we have $y = 10^2$

How much does $y$ increase when $x$ increases by $1$ unit?

We have the form $y=x^2$ and $\dfrac{dy}{dx}=2x \ldots$ here $x=10$, so why isn't the answer $2(10)=20$?

By simple evaluation the answer is $(11^2)-(10^2)=21$

But thinking about it from the perspective of using the general form of the derivative of $y=x^2$, we could answer it by computing $2(10.5)=21$

Finally, my question is:
What is the reasoning behind the $x$ in "$2x$" being $10.5$?

答:
When you say that at $x=10, \frac {dy}{dx}=20$, it means that the gradient at the point $(10,100)=20$.

But, being a parabola, the gradient increases as $x$ increases, so that by the time $x=11$, the gradient at $(11,121)$ is $22$.

The gradient of the straight line joining the two points is $21$. Your simple evaluation is actually calculating that gradient, not the gradient at $x=10$.

It just happens to be the case that for this parabola, the point at $x=10.5$ is the midway point, or average, between $x=10$ and $x=11$ and that the gradient at that point is the average gradient between the two other points.

ContentLicense: CC BY-SA 4.0

Title: Question about a simple derivative


问:
$u(x)$ satisfies the integral equation $u(x) = \int_0^x \sin(u(t))u(t)^p dt$ on $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval
Suppose that $u(x)$ is continuous and satisfies the integral equation

\begin{equation}\label{1.4.1}
        u(x) = \int_0^x \sin(u(t))u(t)^p dt
    \end{equation}

on the interval $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval if $p \geq 0$.

This is what I have:

Since $\sin(u(t))u(t)^p$ is continuous, it follows from the integral definition that $u(x)$ is differentiable. Let us differentiate both sides of equation above with respect to $x$. This yields:

\begin{equation}
    u'(x) = \sin(u(x))u(x)^p
    \end{equation}

This ODE is separabale and becomes:

\begin{equation}
        \frac{1}{\sin(u(x))u(x)^p}du(x) = dx
    \end{equation}

However, this doesn't seem easily solvable so I'm not sure how to show that $u(x) = 0$ from this.

答:
WARNING: As pointed out by Matthew in the comments, this isn't as sound as I wanted it to be and has a few holes - namely continuity for $p<1$ and the interval for which uniqueness is guaranteed. Check out his answer, if he posts one!

I'm going to write the differential equation
$$u'=\sin(u)u^p$$
with the initial value condition $u(0)=0$. Notice that $\frac{\partial}{\partial u}\left(\sin(u)u^p\right)$ and $\sin(u)u^p$ are both continuous on the rectangle $[0,1]\times[a,b]$ for any arbitrary $a,b$. Thus, we can guarantee the uniqueness and existence of our solutions.

$u(x)=0$ is a solution to the differential equation $u'=\sin(u)u^p$. So we can argue that if $u(x)=\int_0^x\sin(u(t))u(t)^pdt$, it must satisfy the differential equation $u'=\sin(u)u^p$ with the initial condition $u(0)=0$. Therefore, $u(x)=0$.

(I hope this is correct; it's been a while since I touched a differential equation)


问:
If symmetric matrices $A$ and $B$ are similar, are they necessarily orthogonally similar
An affirmative answer is claimed on page 3 here (unless I misunderstand),

Lemma:
If two symmetric matrices $A$ and $B$ are similar, then they are orthogonally similar.

but I was unable to find a proof or citation.

To understand this claim I considered the simple case of

$$
A = \left(
\begin{array}{ccc}
 0 & 0 & 1 \\
 0 & 0 & i \\
 1 & i & 0 \\
\end{array}
\right), \qquad
B = \left(
\begin{array}{ccc}
 i \sqrt{2} & 1 & 0 \\
 1 & 0 & 1 \\
 0 & 1 & -i \sqrt{2} \\
\end{array}
\right),
$$
which (by virtue of their Jordan normal form) are both similar to the matrix
$$
J =
\left(
\begin{array}{ccc}
 0 & 1 & 0 \\
 0 & 0 & 1 \\
 0 & 0 & 0 \\
\end{array}
\right).
$$
The aforementioned Lemma then claims that $A$ and $B$ are related by a similarity transformation $A = P B P^{-1}$, such that $P$ is equal to its inverse transpose: $P = P^{-T}$. Following pointers from @chhro I see that
$$
P = \frac{1}{4}\left(
\begin{array}{ccc}
 -i & -3 \sqrt{2} & i \\
 -3 & i \sqrt{2} & 3 \\
 -2 \sqrt{2} & 0 & -2 \sqrt{2} \\
\end{array}
\right)
$$
is a solution. However I would be grateful for reference for the general claim.

答:
As @chhro mentions the answer to the question is yes, and the proof follows from Theorem 20 of Horn and Merino. However, a more direct proof is also provided in the later work of Bukovšek and Omladic:

We assume $A = A^T$, $B = B^T$, $A = P B P^{-1}$ and wish to construct complex orthogonal matrix $O$ ($OO^T = I$) such that $A = O B O^T$.

From the premise we have $A P = P B$, by transposition we have $P^T A = B P^T$, and multiplication by $P$ we have $P P^T A = P B P^T = A P P^T$ hence $A$ and $PP^T$ commute.

Let $R$ be the square root of the complex symmetric invertible matrix $PP^T$: $R^2  = PP^T$, $R$ is symmetric, invertible, and commutes with $A$.

Let $O = R^{-1}P$. $O$ is complex orthogonal $OO^T = R^{-1}P P^T R^{-1} = R^{-1}R^2 R^{-1} = I$. Moreover, $O$ relates $A$ and $B$ in the manner desired: $A O = A R^{-1} P = R^{-1} A P = R^{-1} P B = O B$.


问:
Help needed in calculating the related change
I need help solving an issue of mine.

The question goes like this:

Q) $A$ is a lighthouse and $\overline{BC}$ is the shoreline. It is a right triangle and the lighthouse $A$ is perpendicular to the shoreline. The lighthouse turns once a second so $\dfrac{d\theta}{dt}=2\pi$ radians/second. Let the distance from the lighthouse to shoreline at $B$ be the perpendicular distance $100$ and the light move up from $B$ to $C$ and the distance $\overline{BC}=y$. And finally the hypotenuse is $z$.
Find a relationship between the rate of change of $z$ to $y$ with respect to time.

My solution was to use the Pythagorean theorem
So
$$z^2 = y^2 + 100^2$$
differentiating with respect to time implicity:
$$2z\frac{dz}{dt} = 2y\frac{dy}{dt}$$
so $$\frac{dz}{dt} = \frac{y}{z}\times \frac{dy}{dt}$$
as $\frac{y}{z} = \sin\theta$
$$\frac{dz}{dt} = \sin\theta\frac{dy}{dt}$$
This is the right answer. I tried another approach I do not know why this method does not work.
I tried the following
$$y = z\sin\theta$$
$$\frac{d}{dt}(y) = \frac{d}{dt}(z\sin\theta)$$
From here if I use product rule on the RHS then I do not get the right answer. I do not know why this approach is not working.
I will appreciate any insight as to why this does not work or if I am doing this question wrong.

Thank you and stay safe!!!!

答:
From the first picture (left), you've got $$\begin{split}y^2+100^2&=z^2\\
\frac{dz}{dt}&=\frac{y}{z}\frac{dy}{dt}\end{split}$$

Now suppose you are given the additional information that $\theta$ is $\angle A$ (right). You can indeed find the relationship between dz/dt and dy/dt from this, you just need 2 equations now.

$$\begin{split}\sin\theta&=\frac{y}{z}\\
\cos\theta\frac{d\theta}{dt}&=-\frac{y}{z^2}\frac{dz}{dt}+\frac{1}{z}\frac{dy}{dt}\end{split}$$

and

$$\begin{split}\cos\theta&=\frac{100}{z}\\
-\sin\theta\frac{d\theta}{dt}&=-\frac{100}{z^2}\frac{dz}{dt}\\
\Rightarrow\frac{d\theta}{dt}&=\frac{100}{z^2\sin\theta}\frac{dz}{dt}\end{split}$$

Go back to above

$$\begin{split}\frac{100\cot \theta}{z^2}\frac{dz}{dt}&=-\frac{y}{z^2}\frac{dz}{dt}+\frac{1}{z}\frac{dy}{dt}\\
\left(\frac{100^2}{yz^2}+\frac{y}{z^2}\right)\frac{dz}{dt}&=\frac{1}{z}\frac{dy}{dt}\\
\underbrace{\left(\frac{100^2+y^2}{z^2}\right)}_{=\frac{z^2}{z^2}=1}\frac{dz}{dt}&=\frac{y}{z}\frac{dy}{dt}\end{split}$$


答:
Derivatives are based on infinitesimal changes.  However, you can also formulate those with non-infinitesimal changes.

$$y' = \frac{f(x + h) - f(x)}{h}$$

We normally do $\lim\limits_{h\to 0}$ on this, but there's no reason you have to.  You can actually set $h$ to whatever you want and it still works.  So, we can find the "full-unit derivative" (where $h = 1$) as follows:

$$f(x) = x^2 \\
f'(x) = \frac{(x + h)^2 - (x)^2}{h} \\
f'(x) = \frac{x^2 + 2xh + h^2 - x^2}{h} \\
f'(x) = \frac{2xh + h^2}{h} \\
f'(x) = 2x + h
$$

So, when $h = 1$, then $f'(x) = 2x +1$.  So, at $x = 10$ the amount that $y$ changes for $1$ unit of $x$ is $2x +1 = 2(10) + 1 = 21$.

Technically, this is called a "difference" equation instead of a "full-unit derivative", but, you know, technicalities.

Title: Question about a simple derivative

Body:
Let's say we have $y = 10^2$

How much does $y$ increase when $x$ increases by $1$ unit?

We have the form $y=x^2$ and $\dfrac{dy}{dx}=2x \ldots$ here $x=10$, so why isn't the answer $2(10)=20$?

By simple evaluation the answer is $(11^2)-(10^2)=21$

But thinking about it from the perspective of using the general form of the derivative of $y=x^2$, we could answer it by computing $2(10.5)=21$

Finally, my question is:
What is the reasoning behind the $x$ in "$2x$" being $10.5$?


问:
Double sigma formula with nested index
I am trying to obtain a formula for:

$G = m_1m_2 + m_1m_3 + m_2m_3$ using sigma notation.

I've come up with:
$\sum_{i=1}^{3} \sum_{j=2}^{3} m_im_j = \sum_{i=1}^{3}m_im_2 + m_im_3 = m_1m_2 + m_1m_3 + m_2m_2 + m_2m_3 +m_3m_2 + m_3m_3$. My question is, how do I exclude the values $m_2m_2, m_3m_2, m_3m_3$ in order to obtain $G$? That is, how should I exclude the values where $i \geq j$? Also, is there any way that I can express the above sum "without" the double sigma?

Title: Double sigma formula with nested index

Body: $\sum_{i=1}^2 \sum_{j=i+1}^3 m_im_j$ works.


答:
$$e^x=1+\dfrac{x}{1!}+\dfrac{x^2}{2!}+\dfrac{x^3}{3!} \dots$$

Now I give you freedom to differentiate the RHS w.r.t $x$. You will still get the same RHS.

ContentLicense: CC BY-SA 3.0

Title: Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?

Body: When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.</p>


问:
Does the following necessarily converge to a normal random variable in distribution?
Suppose $X_1, \dots, X_n$ are i.i.d. random variables with $E(X_1) = 0$ and $\operatorname{Var}(X_1) = 1$. Let
$$
S_n = \frac{1}{n}\sum_{i=1}^n\sqrt{i}X_i.
$$
Does $S_n$ converge to a normal random variable?

Originally, I attempted to use Lindeberg CLT to prove this. However, I ran into a wall because I can't figure out a way to check the Lindeberg condition that for $\forall \varepsilon > 0$
$$
\sum_{i=1}^nE(|Y_i|^2\mathbf{1}(|Y_i| > \varepsilon)) \to 0
$$
where $Y_i = \sqrt{i}X_i/s_n$ and $s_n^2 = \sum_{i=1}^ni\operatorname{Var}(X_i) = \frac{n(n+1)}{2}$. If I can prove this, then I can use Slutsky, then we are done. But I have no idea what $X_i$ actually is so I don't know how to verify the condition.

Then I tried using characteristic functions and try to do expansion and approximation. However, I also hit a wall due to the changing index of $i$.

I also tried finding counterexample, but nothing came up.

Can anyone provide some hint? Thank you!

答:
Following up on @StubbornAtom's answer, here is a proof of the Hájek - Šidák CLT.

Claim: Let $(Y_n)$ be IID with zero mean and unit variance. For each $n$ let $c_{n1},\ldots,c_{nn}$ be constants. If
$$\frac{\max_{1\le k\le n}c_{nk}^2}{\sum_{k=1}^n c_{nk}^2}\to 0\qquad\text{as $n\to\infty$,}$$ then $$\frac{\sum_{k=1}^n c_{nk}Y_k}{\left(\sum_{k=1}^n c_{nk}^2\right)^{1/2}}\stackrel d\to N(0,1).$$

Proof: WLOG we may take $\sum_{k=1}^n c_{nk}^2=1$. Put $X_{nk}:=c_{nk}Y_k$. Then each $X_{nk}$ has mean zero. Put $s_n^2:=\operatorname{Var}(\sum_{k=1}^n X_{nk})=\sum_{k=1}^n c_{nk}^2=1$.

Let $\epsilon>0$. For each $n$ and $k$ we have
$$
\begin{aligned}
E(X_{nk}^2 ; |X_{nk}|>\epsilon)&=E(c_{nk}^2 Y_k^2 ; |c_{nk}Y_k|>\epsilon)\\
&=c_{nk}^2 E( Y^2 ; c_{nk}^2Y^2>\epsilon^2)\\
&\le c_{nk}^2 E( Y^2 ; m_nY^2>\epsilon^2),\\
\end{aligned}
$$ where for the sake of brevity we write $m_n:=\max_{1\le k\le n}c_{nk}^2$. Summing over $k$ gives for each $n$
$$\frac1{s_n^2}\sum_{k=1}^nE(X_{nk}^2 ; |X_{nk}|>\epsilon)\le\sum_{k=1}^nc_{nk}^2 E( Y^2 ; m_nY^2>\epsilon^2)= E( Y^2 ; m_n Y^2>\epsilon^2).
$$
Since $m_n\to0$ as $n\to\infty$, this last expectation tends to zero by dominated convergence. This verifies the Lindeberg condition, hence $s_n^{-1}\sum_{k=1}^n X_{nk}$ converges in distribution to a standard normal, and the claim follows.


问:
Are limits of integration defined with respect to the pronumeral or the differential?
Consider the indefinite integral $$\int x^2 \,d(x^2).$$ It evaluates fairly easily to $\tfrac{x^4}{2} + C$. My question is about what happens when we start evaluating definite integrals with respect to these functions. In a specific example, how would $\displaystyle \int_0^2 x^2 \,d(x^2)$ be evaluated?
Would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot\left(x^2\right)^2\right]_0^2 =\frac{4}{2}-0=2 $$ with the limits applying to $x^2$, or would it be: $$\int_0^2 x^2 \,d(x^2)=\left[\frac{1}{2}\cdot x^4\right]_0^2 =\frac{16}{2}-0=8 $$ with the limits applying to $x$.

Hope someone can clarify, thanks in advance!

Body:
There are two ways to interpret $I= \int_0^2 x^2 \,d(x^2):$
$$I=\int_0^2 x^2 2x\,dx={{2x^4}\over 4}\big|_0^2=8$$
and, substituting $u=x^2,$
$$I=\int_0^4 u\,du={u^2\over 2}\big|_0^4=8.$$


答: Without a double sigma, you can write $\sum_{i=1,j>i}^{i<3,j\le3}m_im_j=m_1m_2+m_1m_3+m_2m_3$.

Title: Double sigma formula with nested index

Body:
I am trying to obtain a formula for:

$G = m_1m_2 + m_1m_3 + m_2m_3$ using sigma notation.

I've come up with:
$\sum_{i=1}^{3} \sum_{j=2}^{3} m_im_j = \sum_{i=1}^{3}m_im_2 + m_im_3 = m_1m_2 + m_1m_3 + m_2m_2 + m_2m_3 +m_3m_2 + m_3m_3$. My question is, how do I exclude the values $m_2m_2, m_3m_2, m_3m_3$ in order to obtain $G$? That is, how should I exclude the values where $i \geq j$? Also, is there any way that I can express the above sum "without" the double sigma?


问:
longest increasing subsequence with at most k deletion
Given a number array of length n and a number k, what is the longest non-decreasing contiguous subarray if we allow at most k deletions in our original array?

example:
n, k = 5, 2
array = 5 2 1 3 4

answer = 2 3 4
with deleting 1 from index 2.

答:
You can use DP to solve this problem.

subproblem: $dp(i)$ = length of non-decreasing subsequence ending at index i

recurrence: $dp(j) = max_{i < j, A[i] <= A[j]} ${dp(i)}$ + 1$ where . This is because the subsequence ending at index j is going to be extended by some subsequence ending at a previous index of i. E.g. 1,2,3,...i,..j. So we maximum over all position ending before j s.t. $A[i] <= A[j]$. After we find it, we add 1 additional element, which is jth element.

To find the solution, we have to take $max_{0 <= j <= n}$  ${dp(j)}$ and the sum of 'decreasing' element should less than k. This is because j can be anywhere in the array and we also need to consider k.

The time complexity is $O(n^{2})$ because we have to solve n subproblem and each of them takes O(n) time.


问:
Are there non-continuable functions that become continuable when raised to some power?
Let $X$ be a complex algebraic variety (integral, separated scheme of finite type over $\mathbb C$) and $U\hookrightarrow X$ an open subvariety. I will say that $f\in\mathscr O_X(U)$ is continuable if there exists some $g\in \mathscr O_X(X)$ with $g|_U=f$. My question is simply: Can it happen that $f$ is not continuable, but there exists a $k\in\mathbb N^+$ with $f^k$ continuable?

I was debating with a friend who thinks it is possible but could not give an example, while I thought that it is impossible but wasn't able to give a satisfying proof. So, I'd be really glad if you could provide one of the two (that is, a simple example of such a phenomenon or a proof that it can not happen).

答:
This answer is a generalization of Martin's answer.

Claim: If $X$ is a normal complex algebraic variety, then any function $f\in \mathcal{O}_X(U)$ with $f^k$ continuable must be continuable too.

Proof:

Notice first that wlog we can choose $U:=\{x\in X:f\in\mathcal{O}_{X,x}\}$ to be the maximal open subset of $X$ such that $f|_U$ is regular. If $U\neq X$, by algebraic Hartog theorem $X\setminus U$ is of purely codimension $1$, which means $X\setminus U$ is a finite union of distinct prime Weil divisors $D_i$.

Since $f^k$ is regular on $X$, we have $\operatorname{ord}_{D_i}(f^k)\geq 0$, hence $\operatorname{ord}_{D_i}(f)=\operatorname{ord}_{D_i}(f^k)/k\geq 0$, which means $f\in \mathcal{O}_{X,\zeta_i}$, where $\zeta_i$ is the generic point of $D_i$. Therefore, $\zeta_i \in U\cap D_i=\emptyset$, we get a contradiction.

Edit: Actually here is a much simpler proof:

Suppose $X$ is covered by affine open subsets $U_j$, then $\mathcal{O}_X(U_j)$ is integrally closed by the normality of $X$. Since $f$ is regular at the generic point of $X$, $f$ belongs to the quotient field of $\mathcal{O}_X(U_j)$ for every $j$. Since $\mathcal{O}_X(U_j)$ is integrally closed and $f^k$ is regular on every $U_j$, so $f$ is also regular on every $U_j$, hence $f$ is regular on $X$.


问:
Formula for number of pairs that can be made from $n$ items

If I have
$$\begin{align}
1 \text{ item}\phantom{s} &= \phantom{1}0 \text{ pair}\phantom{s} \\
2 \text{ items} &= \phantom{1}1 \text{ pair}\phantom{s} \\
3 \text{ items} &= \phantom{1}3 \text{ pairs} \\
4 \text{ items} &= \phantom{1}6 \text{ pairs} \\
5 \text{ items} &= 10 \text{ pairs} \\
6 \text{ items} &= 15 \text{ pairs}
\end{align}$$
and so on. How about if I have $n$ items? How many pairs can be made?

Thank you.

答:
We could also phrase this question "how many ways are there to pick two elements from a set of size $n$". We can answer this using the mathematics of combinations. In general, if we want to choose $k$ elements from a set of size $n$, there are
$$
\begin{pmatrix}
n \\
k
\end{pmatrix} := \frac{n!}{k!(n-k)!}
$$
ways of doing so. In the case of $k=2$, we have
$$
\textrm{# of pairs} = \begin{pmatrix}
n \\
2
\end{pmatrix} = \frac{n!}{2!(n-2)!}=\frac{n(n-1)}{2}.
$$

You may also recognise $\frac{n(n-1)}{2}$ as the sum of the first $(n-1)$ integers. Can you think where this relationship might come from?


问:
Asymptotic of incomplete Gamma function, $\Gamma(n,n/a)$
Let $a>1$. I need to approximate (for n large)
$$
\Gamma(n,n/a) \approx f(n)
$$

Approximations I found are for the case which the second argument is fixed. Is there any simple formula for this asymptotic?

答:
$\Gamma(n, n/a)$ and $\Gamma(n)$ will become very close (in the sense that their ratio converges to $1$) as $n$ grow. The following computation provides a more precise error bound for their ratio:

I will write $\alpha = 1/a$ so that $0 < \alpha < 1$. Then

$$ \Gamma(n, \alpha n) = \Gamma(n) \biggl[ 1 - \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \biggr]. $$

Since $t \mapsto t^{n-1}e^{-t}$ is increasing on $0 \leq t \leq n-1$, for large $n$ we have

$$ \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \leq (\alpha n)^n e^{-\alpha n}. $$

So by the Stirling's approximation,

$$ \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t
\leq \frac{(\alpha n)^n e^{-\alpha n}}{\sqrt{2\pi} \, n^{n-\frac{1}{2}} e^{-n}}
= \frac{1}{\sqrt{2\pi}} n^{1/2} (\alpha e^{1-\alpha})^n .
$$

Similarly, by noting that

$$ \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t
\geq \int_{\alpha n-1}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t
\geq (\alpha n - 1)^{n-1} e^{-\alpha n+1}, $$

we have

$$ \frac{1}{\Gamma(n)} \int_{0}^{\alpha n} t^{n-1}e^{-t} \, \mathrm{d}t \geq c_{\alpha} n^{-1/2} (\alpha e^{1-\alpha})^{n} $$

for some constant $c_{\alpha} \in (0, 1)$ depending only on $\alpha$. This tells that

$$ \Gamma(n, \alpha n) = \Gamma(n) \left(1 - n^{\Theta(1)}(\alpha e^{1-\alpha})^n \right), $$

where $\Theta(1)$ represents a bounded sequence in $n$.

Remarks. A less precise asymptotic formula $\Gamma(n, \alpha n) = \Gamma(n)(1 - o(1))$ can be obtained by applying the central limit theorem to

$$ \frac{\Gamma(n, \alpha n)}{\Gamma(n)} = \mathbf{P}(\tau_1 + \cdots + \tau_n \geq \alpha n), $$

where $\tau_k$'s are independent $\operatorname{Exp}(1)$ variables.


问:
Monotone Convergence Theorem with given a1 and a2
I have a question regarding the Monotone Convergence Theorem which goes like this,

a1=2

a2=3

an+1=$\frac{a_n +a_{n-1}}{(n+1)n}$       for n>=2

Does the sequence an converge?

So i tried this by taking a look at values a3 a4 a5 using the formula and it does look like a decreasing function which is bounded by 0. Then i tried to show that the function was decreasing by taking the difference of an+1 - an but this is where i got stuck. I'm not sure how to show this function is decreasing and bounded by 0. What should be my take on this question?

答:
Assume $0\le a_{n-2}\le a_{n-1} \le a_n$ (which you actually have from $a_3,a_4,a_5$). You then have $a_{n+1} \ge 0$ and
\begin{align*}
a_{n+1} - a_n
&= \frac{a_n + a_{n-1}}{(n+1)n} - \frac{a_{n-1} + a_{n-2}}{n(n-1)}\\
&\le \frac{a_n + a_{n-1}}{(n+1)n} - \frac{a_{n-1} + a_{n-2}}{n(n\color{red}{+}1)}\\
&\le \frac{a_n - a_{n-1}}{(n+1)n} + \frac{a_{n-1} - a_{n-2}}{n(n+1)}\\
&\le 0.
\end{align*}
Due to the Induction Principle, this sequence is positive and decreases after $n=3$. Therefore, it converges. Moreover, the limit is given by
$$0\le l = \lim_{n\to\infty}a_{n+1} = \lim_{n\to\infty}\frac{a_n+a_{n-1}}{(n+1)n} \le \lim_{n\to\infty}\frac{2a_3}{(n+1)n} = 0,$$
ie. $l=0$.


问:
Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?

答:
HINT: Use AM-GM inequality for real positive $a_i$s (where $1\le i\le n$)

$$\frac{\sum_{1\le r\le n}  a_i}n\ge \sqrt[n]{\prod_{ 1\le r\le n}a_i}$$


问:
Parametric Equation for the intersection of a curve
I am trying to find the parametric equation for the cylinder x^2+z^2=4 and the plane through the points (-1,3,0), (1,2,0), and (0,2,2).

I obtained 2x + 4y + z = 10. As the equation of the plane through the points, but I am having trouble finding the parametric equations for the ellipse.

I know the intersection of the cylinder and the plane is an ellipse!

答:
To find parametric equation of the intersection curve of cylinder $x^2+z^2=4$ and plane $2x+4y+z = 10$, we can use polar coordinates.

$x = 2 \cos\theta, z = 2 \sin\theta$

$y = \displaystyle \frac{10-2x-z}{4} = \frac{5-2\cos\theta-\sin\theta}{2}$

So parametric equation of the intersection curve is

$(2\cos\theta, \frac{5-2\cos\theta-\sin\theta}{2}, 2 \sin\theta), \ 0 \leq \theta \leq 2\pi$.


问:
Finding the CDF of a random variable
Your statistics teacher challenges you to write a mathematics paper, which depending on your time spent, will earn you a cash reward. You're given the paper and a candle stands on the table in front of you. The life span $X$, in minutes, of the candle is a continuous random variable with a uniform distribution on $(0, 60)$. You must leave after $\frac{1}{2}$ an hour has been spent writing the paper or as soon as the candle burns out, whichever of these happens first. The teacher gives you a cash reward $M$ equal to half the amount of time the candle was lit. Find the cumulative distribution function (cdf) of $M$ and determine the expected value $E[M]$.

My attempt:

By inspection $M = \frac{X}{2}$.

I think cdf of $M$ is of the form:

$$F(m)= \left\{
  \begin{array}{lr}
      0 & \mbox{if } x < 0 \\
      2x = m &  \mbox{if } 0 \leq x < 60 \\
      1 & \mbox{if } 60 \leq x
      \end{array}
\right.$$

As for $E(M)$, we have

$$E(M) = \int_{0}^{60} m \cdot F'(m) \,dm = \int_{0}^{60} m \,dm = 1800$$

Is this correct? My work above seems lacking and I am not sure if I did it correctly. Any assistance is much appreciated.

答:
If the lifespan of the candle is $X\sim \mathscr {Unif}(0, 60)$ then your reward is $M=\frac 1 2 \min(30, X)=15\cdot \mathbb I(60>X>30)+\frac 1 2 X\cdot \mathbb I(0<X<30)$. I means it is a 1 if the criteria inside is satisfied and a 0 if it is not satisfied. So this is a convenient way of writing a binary outcome. It just means if X is between 30 and 60, M is 15, and if X is between 0 and 30, M is X/2.

The expected value of this is $$\begin{align}E(M)&=15\int_{-\infty}^\infty\frac{1}{60}\Bbb I(30<x<60)dx+\frac 1 2\int_{-\infty}^\infty\frac{x}{60}\Bbb I(0<x<30)dx\\[1ex]&=15\int_{30}^{60}\frac {1}{60}dx+\frac{1}{2}\int_{0}^{30}\frac{x}{60}dx\\[1ex]&=\frac{15}{2}+\frac{1}{240}(30)^2\\[1ex]&=\frac{45}{4}\\[1ex]&=11.25\end{align}$$

Or, if you don't want to think about "I", you see that M is 15 with probability 1/2 and 1/2X with probability 1/2. Then $E(M)~{=15\, P(M=15)+\int_0^{30} (1/2)xf(x)dx\\=7.5+\int_0^{30}\frac 1 2x\frac{1}{60}dx\\~\vdots}$

M has pmf $P(M=15)=\frac 1 2$ and a pdf of $f(M)=\frac 1{30}$ for 0<M<15 and $f(M)=0$ otherwise.

Half of $M$'s probability mass is uniformly distributed over $[0..15)$ and half of the probability mass occurs at $15$.

Thus it has cdf

$$F(m)=\begin{cases}0&\text{if }& ~~~~~~~m<0\\\frac{m}{30}&\text{if }&0\le m<15\\1&\text{if }&15\le m\end{cases}$$

Title: Finding the CDF of a random variable


问:
Find $\int_0^4(g\circ f\circ g)(x)\mathrm{d}x$ where $f(x)=\sqrt[3]{x+\sqrt{x^2+1/27}}+\sqrt[3]{x-\sqrt{x^2+1/27}}$, $g(x)=x^3+x+1$

Let $$f(x)=\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}+\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}$$
and
$$g(x)=x^3+x+1$$
then, find $$\int_0^4(g\circ f\circ g)(x) \mathrm dx$$

My attempt:

Let $\displaystyle h(x)=\sqrt{x^2+\frac{1}{27}}$
$$(g\circ f)(x)=2x+3((2x)(x^2-[h(x)]^2))^{1/3}+(x+h(x))^{1/3}+(x-h(x))^{1/3}+1$$

Is finding $(g\circ f \circ g)(x)$ in term of $x$ necessary? Because it is quite a work to do.

答:
This is a nice problem.

$$f(x)=\underbrace{\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}}_{a}+\underbrace{\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}}_b$$

Now note that from the identity $(a+b)^3=a^3+b^3+3ab(a+b)$, and using $ab=-\dfrac{1}3$,

we have:

$$(f(x))^3=2x-f(x) \implies (f(x))^3+f(x)=2x $$

And from $g(x)=x^3+x+1$ , we have $$g(f(x))=(f(x))^3+f(x)+1=2x+1$$

Therefore $$g(f(g(x))=2g(x)+1=2x^3+2x+3$$

So all you have got is $$\int_0^4 (2x^3+2x+3)\mathrm{d}x$$


问:
Existence of field injections between $\mathbb{R}(x)$ and $\mathbb{C}$.
This came with a set of questions:

Find a field injection between the following fields or prove that such injection doesn't exist:

$\mathbb{Q}[i]\subset\mathbb{C}\to\mathbb{R}$

$\mathbb{Q}(x)\to\mathbb{C}$

$\mathbb{R}(x)\to\mathbb{C}$

$\mathbb{Q}[\alpha]\subset\mathbb{C}\to\mathbb{R}$, where$\alpha=2^{\frac{1}{3}}e^{2\pi i/3}$

I have worked with the finite case before, (for example $\mathbb{F}_{29}\to \mathbb{F}_{57}$), in which case I could use the cardinality of the image to show that no such field injection exists. But I don't think I could continue with this strategy when everything is infinite.

I think I have tackled all other parts, my proof goes something like this:

Suppose $\phi:\mathbb{Q}[i]\to\mathbb{R}$ is an injection. Suppose $\phi(i)=r\in \mathbb{R}$. Then $\phi(-1)=\phi(i\cdot i)=\phi(i)\cdot\phi(i)=r^2$. But since $$\phi(0)=0=\phi(1)+\phi(-1)=1+\phi(-1)$$,  $\phi(-1)=-1=r^2$, and this is a contradiction since there is no such real number whose square is $-1$.

For the second part, I think $\phi:\mathbb{Q}(x)\to\mathbb{R}:p(x)\to p(e)$ would be an injection, based off the fact that $e$ is transcendental over $\mathbb{Q}$. But I have no idea about the third part. I only have a vague feeling that there might be some argument based on cardinality.

For the last part,the map $\phi:\mathbb{Q}[\alpha]\to\mathbb{R}:(a_0+a_1\alpha+a_2\alpha^2)\to(a_0+a_12^\frac{1}{3}+a_22^\frac{2}{3})$ seems to be valid.Many thanks to @reuns for pointing this out.

I also found this question, which seems to be more or less relate：https://math.stackexchange.com/questions/935877/how-to-prove-that-the-evaluation-map-is-a-ring-homomorphism

Any help is appreciated, thank you in advance.

答:
Part $(3)$ is a bit subtle.

There's one key difference here between $\mathbb{R}$ and $\mathbb{Q}$. From a field perspective, $\mathbb{Q}$ is very "explicit:" if $F$ is a field, there is only ever at most one field homomorphism $\mathbb{Q}\rightarrow F$. This is basically because every element of $\mathbb{Q}$ is "definable" in a particularly nice way, namely by combining the additive and multiplicative units ($0$ and $1$) via addition, multiplication, subtraction, and division. Since all of that structure is preserved by field homomorphisms, this gives the uniqueness above.

By contrast, $\mathbb{R}$ does not have this property. In particular, a homomorphism $h:\mathbb{R}(x)\rightarrow\mathbb{C}$ is allowed to do terrible things to the "$\mathbb{R}$-part" of $\mathbb{R}(x)$: it need not be the case that $h$ restricted to $\mathbb{R}$ is just the inclusion map $\mathbb{R}\rightarrow\mathbb{C}$.

At this point it's a good exercise to prove that there is no field embedding $h:\mathbb{R}(x)\rightarrow\mathbb{C}$ which restricts to the inclusion on $\mathbb{R}$ - that is, such that $h(r)=r$ for all $r\in\mathbb{R}\subset\mathbb{R}(x)$. Basically, think about what happens if we try to whip up such an $h$ which sends $x\in\mathbb{R}(x)$ to $a+bi\in\mathbb{C}$.

So when we look at $\mathbb{R}(x)\rightarrow\mathbb{C}$ we're in the realm of very "flexible" objects. In fact, the fields in question are now pathologically flexible - for example, the field of complex numbers has $2^{2^{\aleph_0}}$-many nontrivial automorphisms.

Here, now, is a rather indirect proof that there is a field homomorphism from $\mathbb{R}(x)$ to $\mathbb{C}$:

Every field has an algebraic closure. Let $F$ be an algebraic closure of $\mathbb{R}(x)$ and $i:\mathbb{R}(x)\rightarrow F$ be a field homomorphism. Then $F$ is an algebraically closed field of characteristic $0$ (immediate) of transcendence degree $2^{\aleph_0}$ (by thinking about cardinality). But up to isomorphism, there is only one algebraically closed field of characteristic $0$ and transcendence degree $\kappa$ for any $\kappa$ - and in the particular case of $\kappa=2^{\aleph_0}$, it's just $\mathbb{C}$ itself.

The bolded fact above is not trivial. It's an extension of the more familiar result that any two vector spaces of the same dimension are isomorphic - here, "algebraically closed field of characteristic $0$" (or indeed any fixed characteristic) replaces "vector space" and "transcendence degree" replaces "dimension" - but more work needs to be done to prove it.

That's not even the worst bit, though. The above argument uses the axiom of choice in two ways: to build an algebraic closure of $\mathbb{R}(x)$, and to build transcendence bases of $F$ and $\mathbb{C}$ in order to get an isomorphism between them. This is not in general something we can avoid, and I am not certain whether $\mathsf{ZF}$ (= set theory without the axiom of choice) alone can prove that there is a field homomorphism from $\mathbb{R}(x)$ to $\mathbb{C}$.


问:
Rearrange this expression for lambda and substitute it back into the expression
I was asked to rearrange this expression for $\lambda$ and substitute it back into the expression for $\mathbf{r}'$ in order to get $\mathbf{r}'$ in terms of $\mathbf{r}$:

$\mathbf{r}.\hat{\mathbf{e}}_3 + \lambda s_3 = 0$  (where $s_3 = \hat{\mathbf{s}}.\hat{\mathbf{e}}_3$)

I worked out $\lambda = - \mathbf{r}.\hat{\mathbf{e}}_3/s_3$

I then had to substitute the solved $\lambda$ into the following expression:

$\mathbf{r}' = \mathbf{r} + \lambda \hat{\mathbf{s}}$

Which is solved by doing the following:

$\mathbf{r}' = r + \hat{\mathbf{s}} (- \mathbf{r}.\hat{\mathbf{e}}_3/s_3)$

$\mathbf{r}' = r - \hat{\mathbf{s}} (\mathbf{r}.\hat{\mathbf{e}}_3)/s_3$

I get confused at the last step. Why does the minus get swapped out and $\mathbf r + \hat{\mathbf{s}}$ become $\mathbf r - \hat{\mathbf{s}}$?

答:
You are not reading the expressions correctly. There is no $r + \hat{\mathbf{s}}$ or $r - \hat{\mathbf{s}}$. In the original expression, the $\hat{\mathbf{s}}$ is multiplied by $\lambda$, then the result is added to $r$.  So after the substitution, it becomes

$$ \mathbf{r}' = r + \left(\hat{\mathbf{s}} (- \mathbf{r}.\hat{\mathbf{e}}_3/s_3) \right)$$

I inserted extra parentheses to emphasize where the confusion may be. The above is then equal to

$$ \mathbf{r}' = r - \left(\hat{\mathbf{s}} (\mathbf{r}.\hat{\mathbf{e}}_3/s_3) \right)$$


问:
Problem with a partial fraction decomposition
I don't know why, but for some reason I cannot solve the following partial fraction decomposition no matter how much I try.

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

When decomposing that to $\frac{1}{(v-1)^2(v+1)^2} = \frac{A_1}{v-1} + \frac{A_2}{(v-1)^2} + \frac{A_3}{v+1} + \frac{A_4}{(v+1)^2}$ I did figure out that $A_2$  and $A_4$ are $\frac{1}{4}$, but I cannot for the love of me figure out $A_1$ and $A_3$.

Sorry if this sounds like a stupid question, but I'm really stuck here and don't know how to proceed.

Body:
$$\frac{1}{(v-1)^2(v+1)^2} = \frac{A_1}{v-1} + \frac{A_2}{(v-1)^2} + \frac{A_3}{v+1} + \frac{A_4}{(v+1)^2}$$

This isn't the short way, but I enjoy doing it this way because I enjoy watching it get simpler as I go along.

$$1 = (v-1)(v+1)^2A_1 + (v+1)^2A_2 + (v-1)^2(v+1)A_3 + (v-1)^2A_4$$

Let $v = -1$, and you get

$$1 = 4A_4 \implies A_4 = \frac 14$$

$$4 = 4(v-1)(v+1)^2A_1 + 4(v+1)^2A_2 + 4(v-1)^2(v+1)A_3 + (v-1)^2$$

$$-(v^2-2v-3) = 4(v-1)(v+1)^2A_1 + 4(v+1)^2A_2 + 4(v-1)^2(v+1)A_3$$

$$-v+3 = 4(v-1)(v+1)A_1 + 4(v+1)A_2 + 4(v-1)^2A_3$$

Let $v=-1$, again, and you get

$$4 = 16A_3 \implies A_3 = \frac 14$$

$$-v+3 = 4(v-1)(v+1)A_1 + 4(v+1)A_2 + (v-1)^2A_3$$

$$-(v^2-v-2) = 4(v-1)(v+1)A_1 + 4(v+1)A_2$$

$$-(v-2) = 4(v-1)A_1 + 4A_2$$

Let $v=1$, and you get

$$1 = 4A_2 \implies A_2 = \frac 14$$

$$-v+2 = 4(v-1)A_1 +1$$

$$-v+1 = 4(v-1)A_1 \implies A_1 = -\frac 14$$


答:
By the AM-GM inequality, we have
$$\frac{x_1/x_n+x_2/x_{n-1}+\cdots+x_n/x_1}{n}\geq\sqrt[n]{\frac{x_1}{x_n}\frac{x_2}{x_{n-1}}\cdots\frac{x_n}{x_1}}=\sqrt[n]{1}=1.$$</p>

Title: Show $\frac{x_1}{x_n} + \frac{x_2}{x_{n-1}} + \frac{x_3}{x_{n-2}} + \dots + \frac{x_n}{x_1} \geq n$

Body:
I was recently asked this question which stumped me.

How can you show $\dfrac{x_1}{x_n} + \dfrac{x_2}{x_{n-1}} + \dfrac{x_3}{x_{n-2}} + \dots + \dfrac{x_n}{x_1} \geq n$ for any positive reals $x_1, x_2, \dots, x_n$?


Title: Find $\int_0^4(g\circ f\circ g)(x)\mathrm{d}x$ where $f(x)=\sqrt[3]{x+\sqrt{x^2+1/27}}+\sqrt[3]{x-\sqrt{x^2+1/27}}$, $g(x)=x^3+x+1$

Body:
Let $$f(x)=\sqrt[3]{x+\sqrt{x^2+\frac{1}{27}}}+\sqrt[3]{x-\sqrt{x^2+\frac{1}{27}}}$$
and
$$g(x)=x^3+x+1$$
then, find $$\int_0^4(g\circ f\circ g)(x) \mathrm dx$$

My attempt:

Let $\displaystyle h(x)=\sqrt{x^2+\frac{1}{27}}$
$$(g\circ f)(x)=2x+3((2x)(x^2-[h(x)]^2))^{1/3}+(x+h(x))^{1/3}+(x-h(x))^{1/3}+1$$

Is finding $(g\circ f \circ g)(x)$ in term of $x$ necessary? Because it is quite a work to do.

Body:
Note that $f^{-1}(x) = \dfrac{x^3+x}{2}\\$

$\therefore g(x) = 2f^{-1}(x)+1\\$

$\therefore g(f(g(x))) = 2f^{-1}(f(g(x)))+1\\$

and I'm sure you know $f^{-1}(f(x)) = x \\$

$\therefore g(f(g(x))) = 2g(x)+1\\$

$$\therefore \int_0^4g(f(g(x)))dx = \int_0^4(2g(x)+1)dx = \int_0^4(2x^3+2x+3)dx\\$$


问:
$u(x)$ satisfies the integral equation $u(x) = \int_0^x \sin(u(t))u(t)^p dt$ on $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval
Suppose that $u(x)$ is continuous and satisfies the integral equation

\begin{equation}\label{1.4.1}
        u(x) = \int_0^x \sin(u(t))u(t)^p dt
    \end{equation}

on the interval $0 \leq x \leq 1$. show that $u(x) = 0$ on this interval if $p \geq 0$.

This is what I have:

Since $\sin(u(t))u(t)^p$ is continuous, it follows from the integral definition that $u(x)$ is differentiable. Let us differentiate both sides of equation above with respect to $x$. This yields:

\begin{equation}
    u'(x) = \sin(u(x))u(x)^p
    \end{equation}

This ODE is separabale and becomes:

\begin{equation}
        \frac{1}{\sin(u(x))u(x)^p}du(x) = dx
    \end{equation}

However, this doesn't seem easily solvable so I'm not sure how to show that $u(x) = 0$ from this.

答: Assume there exists $t\in(0,1)$ such that $u(t)>0$. Because $u$ is continuous and $u(0)=0$, we can identify $a,b\in [0,1]$ such that $u(a)=0,a<t<b,$ and $u>0$ on $(a,b)$. Since $\sin\big(u(t)\big)$ is also continuous, we can find $c\in (a,b)$ such that $\sin \big(u(t)\big)>0$ on $(a,c)$. Now consider the IVP: $$\frac{dy}{dx}=y^p\sin(y):x\in(a,c)$$ $$y(a)=0$$ We know $u$ is a solution to this IVP, so for $x_1,x_2\in (a,c)$ we have $$\int_{u(x_1)}^{u(x_2)}\frac{dy}{y^p\sin(y)}=\int_{x_1}^{x_2}dx=x_2-x_1$$ Take $x_1 \longrightarrow a^{+}$ to get $$\lim_{x_1 \longrightarrow a^{+}}\Bigg[\int_{u(x_1)}^{u(x_2)}\frac{dy}{y^p\sin(y)}\Bigg]=x_2-a$$ This reduces to $$\lim_{h \longrightarrow 0^{+}}\Bigg[\int_{h}^{u(x_2)}\frac{dy}{y^p\sin(y)}\Bigg]=x_2-a$$ However, if $p\geq 0$ the limit on the left side doesn't exist. This is by a limit comparison test to $\lim_{h\longrightarrow 0^{+}}\Big[\int_h^{u(x_2)}\frac{dy}{y^{p+1}}\Big]$. You will arrive at a similar contradiction if we assume $u(t)<0$.


问:
Integral of $\int\ln(a\sin(2x)+b)\cos(x)dx$
Mathematica is able to solve the indefinite integral

$$\int\ln(a\sin(2x)+b)\cos(x)dx$$

analytically for $b>a>0$, but the resulting $\arctan$ terms are too complicated for the next steps I need to perform. I assume these $\arctan$ terms are the result of Weierstrass substitution, so I am wondering if this integral can be evaluated in a different way, maybe by substitution, so there are no complicated $\arctan$ terms in the result?

Thanks!

答:
It looks much better if you write
$$\int\log(a\sin(2x)+b)\cos(x)\,dx=$$ $$\int\log(\sin(2x)+(c^2-1))\cos(x)\,dx+\log(a)\int \cos(x)\,dx$$ with $c^2-1=\frac b a$ specifying $c>1$ and using the FullSimplify command.

Simplifying the arctangents and the logarithms, you should end for the first integral with
$$-\sqrt{c^2-2} \tan ^{-1}\left(\frac{\sqrt{c^2-2} (\sin (x)+\cos (x)+1)}{2-c^2+\sin(x)+\cos (x)}\right)+$$
$$\frac c 2 \log \left(-\frac{c+\sin (x)-\cos (x)}{c-\sin (x)+\cos (x)}\right)+\sin (x) \left(\log \left(c^2+\sin (2 x)-1\right)-2\right)$$


问:
Solving a simple quadratic in modular arithmetic

I've been trying to solve this congruence> $$x^2 \equiv 9\pmod{13}$$

I keep coming back to $x =3$ but it seems to be wrong, any help on what's the correct technique to find the answer?

答:
Because $13$ is a prime number. Then $\Bbb Z_{13}$ is a finite field. That means that

$x^2 \equiv 0 \pmod{13}$ has exactly two solutions $x \in \{3, -3\} = \{3, 10\}$


问:
Largest prime that remains prime when any one of its digits is deleted
Recently I came across this beautiful question: Deleting any digit yields a prime... is there a name for this?

The question talks about primes that remain prime when any one of their digits is deleted (zeros allowed). The answers there claim that there is only a finite number of such primes. So naturally I started wondering what is the largest such prime?

According to A051362 the largest known prime with this property is 9977770001777. Can we find a larger one?

I searched for primes in the form $aa \ldots aabb \ldots bb$, where $a$ and $b$ are digits in $\{1,3,7,9\}$ and there are up to $n \leq 400$ of each one. But I couldn't find any solutions other than the trivial small ones.

答:
Primes very much behave like random numbers once you erase the basic properties and you start looking at very large numbers. So your question is if probability of a large number having a property P is $\frac{1}{\ln(n)}$ and we erase one digit, what is the probability of a new number having the same property P.

The question here is quite obvious, and that is which digit you have chosen to delete. Probability that a number $m-k10^n$ has a property P again, directly correlates with $10^n$. So you are asking if $m$ is prime what is the probability of $m-k10^n$ being prime.

Again the average gap between primes is $\ln(n)$ meaning that if you take digits up to $k\log_{10}(\ln(n))$ meaning first few (with some constant k that we can find heuristically), even for a very large number, the rest will keep the probability of $\frac{1}{\ln(\frac{n}{10})}$ for being a prime, regardless which digit you have deleted.

For the first few digits the situation is very complicated and very likely not to be solved any time soon regarding the precise distribution of prime gaps on such a small scale.

So being a prime number, $p$, and deleting a little bit larger digit does not differ much from guessing any prime number that has one digit less. Therefore you can still expect that a new number will be prime with the probability of

$$\frac{1}{\ln(\frac{p}{10})}$$

which is quite large.

Now if you want it for each digit you come up with the probability of the order of

$$\left ( \frac{1}{\ln(\frac{p}{10})} \right)^{\log_{10}(p)-k\log_{10}(\ln(p))} $$

which is the expression of the form

$$\frac{1}{(d-1)^d}$$

where $d$ is the number of digits, explaining why it will be notoriously difficult to verify the existence of the prime with such property with a very large number of digits just by listing them.

The option here is, then, to search for numbers in a specific form so we eliminate quite some number of possible factors when we delete a digit.


答:
$M=\tfrac 12\min(X,30)$ where $X\sim\mathcal U[0..60]$

Half of the probability mass of $M$ will be distributed uniformly over $[0..15)$ while half will be massed at the point of $15$.

Therefore $F_{\small M}(m)= (m/30)\mathbf 1_{m\in[0..15)}+\mathbf 1_{15\leqslant m}$

$\therefore \mathsf E(M) ~{=\int_0^\infty 1-F_{\small M}(m)\,\mathrm d m\\= \int_0^{15} 1-m/30\,\mathrm d m\\=15-15^2/60\\=15(3/4)\\=11.25}$

Title: Finding the CDF of a random variable

Body:
Your statistics teacher challenges you to write a mathematics paper, which depending on your time spent, will earn you a cash reward. You're given the paper and a candle stands on the table in front of you. The life span $X$, in minutes, of the candle is a continuous random variable with a uniform distribution on $(0, 60)$. You must leave after $\frac{1}{2}$ an hour has been spent writing the paper or as soon as the candle burns out, whichever of these happens first. The teacher gives you a cash reward $M$ equal to half the amount of time the candle was lit. Find the cumulative distribution function (cdf) of $M$ and determine the expected value $E[M]$.

My attempt:

By inspection $M = \frac{X}{2}$.

I think cdf of $M$ is of the form:

$$F(m)= \left\{
  \begin{array}{lr}
      0 & \mbox{if } x < 0 \\
      2x = m &  \mbox{if } 0 \leq x < 60 \\
      1 & \mbox{if } 60 \leq x
      \end{array}
\right.$$

As for $E(M)$, we have

$$E(M) = \int_{0}^{60} m \cdot F'(m) \,dm = \int_{0}^{60} m \,dm = 1800$$

Is this correct? My work above seems lacking and I am not sure if I did it correctly. Any assistance is much appreciated.


问:
Intuition for plane graphs
In the graph theory section of 2-Dimensional Categories by Johnson and Yau, they define a plane graph as a graph $G$ together with a topological embedding $m_G:|G|\to\mathbb{C}$ of the geometric realization of $G$ into the complex plane. This is supposed to fix a drawing of the graph, where (unless I'm mistaken) we want edges to be allowed to cross each-other as needed to draw the graph the way we want to draw it, even if these crossings aren't at vertices. For posterity:

Definition. A graph $G$ is a triplet $(V_G,E_G,\psi_G)$ where $V_G$ is a finite set called the set of vertices, $E_G$ is a finite set called the set of edges, and $\psi_G:E_G\to V_G\times V_G$ is a function called the incidence function. The geometric realization of $G$, denoted $|G|$, is the topological quotient $$|G|=\Big[\big(\coprod_{v\in V_G}\{v\}\big)\coprod\big(\coprod_{c\in E_G}[0,1]_e\big)\Big]\big/\sim$$ where $\{v\}$ is a one point space indexed by a vertex $v\in V_G$, $[0,1]_e$ is a copy of the unit interval indexed by an edge $e\in E_G$, and the identification $\sim$ is generated by $$u\sim0\in[0,1]_e\ni1\sim v\ \ \ \ \text{if}\ \ \ \ \psi_G(e)=(u,v).$$

This is where I'm confused -- the geometric realization has distinct copies of the unit interval for each edge of $G$, only glued together at endpoints where edges share vertices.  If $m_G$ is a topological embedding then it's injective, so we aren't allowed to send two non-endpoints on two separate unit intervals to the same point in $\mathbb{C}$. That is, for $x,y\in(0,1)$ and edges $e,e'\in G$ we have that $$e\neq e'\implies m_G(e,x)\neq m_G(e',y)$$ if $m_G$ is an embedding. But this prevents drawings where edges overlap at non-vertex points, and this would prevent us from even having a geometric realization for non-planar graphs using this definition, correct?

If the above nitpicking is correct, are the authors implicitly restricting attention to planar graphs because all pasting diagrams have planar underlying graphs? Are there any diagrams in (bi-) categories, pasting or otherwise, whose underlying graphs are not planar?

答:
Yes, this definition restricts to planar graphs.  I'm not sure exactly what their motivation for doing so is (I haven't read most of the paper), but I imagine it is because their theorems are about how the diagrams they are considering interact with the topology of the plane.  For instance, from a quick skim, Theorem 3.3.7 seems to be saying there is a coherent way to compose 2-morphisms by juxtaposing appropriate plane graphs geometrically, corresponding to "gluing together" faces of a plane graph to form larger faces.  This is specifically using the geometric configuration of the faces to tell you what 2-morphisms you are composing together, and so it doesn't make sense if you don't have a plane graph with well-defined faces.  Another way to think about it is that what's going on is that these diagrams don't really correspond to graphs at all, but instead to 2-dimensional polyhedral complexes, and they are looking at the special case that those polyhedral complexes are embedded in the plane.  The graphs are merely the 1-skeletons of these 2-dimensional diagrams.

It's certainly not true that an arbitrary diagram in a category is always a planar graph.  For instance, if you take a 5-element linear order and draw all its non-identity morphisms, that forms a diagram whose graph is $K_5$, which is non-planar.

Title: Intuition for plane graphs


问:
Why isn't the derivative of $e^x$ equal to $xe^{(x-1)}$?
When we take a derivative of a function where the power rule applies, e.g. $x^3$, we multiply the function by the exponent and subtract the current exponent by one, receiving $3x^2$. Using this method, why is it that the derivative for $e^x$ equal to itself as opposed to $xe^{x-1}$? I understand how the actual derivative is derived (through natural logs), but why can't we use the original method to differentiate? Furthermore, why does the power rule even work? Thank you all in advance for all of your help.</p>

答:
The rule $\frac d{dx}x^k=kx^{k-1}$ for $k\geq1$ integer is a straightforward application of the basic case $\frac d{dx}x=1$ and the rule
$\frac d{dx}(fg)=f\frac d{dx}g+g\frac d{dx}f$ reiterated $k-1$ times.

The problem with $e^x$ is that it doesn't admit an expression as product of simpler functions.


问:
Can segment sphere intersection be view as point cylinder intersection
If we have a sphere with origin at $p$ and radius $r$, and a segment with endpoints $a$ and $b$. And we want to check if they intersect. Would it be correct to assume that we could also view this as checking whether a point $p$ lies inside a cylinder defined by endpoints $a$, $b$ and radius $r$?

答:
Consider this example:

\begin{align}
p &= (0,0,0),\\
r &= 1,\\
a &= (0.8,0,0),\\
b &= (2,0,0).
\end{align}

Here $a$ is inside the sphere and $b$ is outside, so there is definitely an intersection, but $p$ is not inside the cylinder.

But if you add a half-sphere of radius $r$ onto each end of the cylinder,
then $p$ is inside that combined region if and only if the segment $ab$ intersects the interior of the sphere.


问:
convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinates
Context

This is an interim problem related to a Green's function solution for a boundary-value problem in the cylindrical coordinate system.

Question

How do I convert $(x-x')^2 + (y-y')^2 + (z-z')^2$ to cylindrical coordinate system?

My worked solution is given in the answer below.

答:
$$(x-x')^2 + (y-y')^2 + (z-z')^2
=r^2 + {r'}^2 -2\,r\,{r}' \,\cos\left(\theta - \theta'\right) + (z-z')^2$$

My attempt

I know that to tranform from Cartesian to cylindrical requires the following three equations
\begin{align}
       r &= \sqrt{x^2 + y^2} \\
  \theta &= \arctan{\left(\frac{y}{x}\right)} \\
       z &= z \quad.
\end{align}

Directly, the question reduces to how to covert the following to cylindrical system.
$$(x-x')^2 + (y-y')^2.$$
Its natural to begin with $r$ and $r'$.
\begin{align}
(x-x')^2 + (y-y')^2
&=
 x^2 - 2\,x\,x' +{x'}^2 +  y^2 - 2\,y\,{y'} +{y'}^2
\\
&=
 x^2 +  y^2 +{x'}^2 +{y'}^2 - 2\,x\,x'  - 2\,y\,{y'}
\\
&=
r^2 +{r'}^2 - 2\,x\,x'  - 2\,y\,{y'}
\end{align}
Now the problem is to convert $ - 2\,x\,x'  - 2\,y\,{y'}   $ to cylindrical coordinate system.

The next part of the answer required trial and error, and some good fortune.
\begin{align*}
r\,r' \,\cos\left(\theta - \theta'\right)
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{y}{x}} - \arctan{\frac{y'}{x'}}\right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{\frac{y}{x} - \frac{y'}{x'}}{1+ \frac{y}{x} + \frac{y'}{x'}}} \right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{\frac{y}{x} - \frac{y'}{x'}}{1+ \frac{y\,y'}{x\,x'}  }} \right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\cos\left(\arctan{\frac{ y\, x'  -  y'\,x  }{x\,x'+  y\,y'   }} \right)
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{1}{\sqrt{1+\left(\frac{ y\, x'  -  y'\,x  }{x\,x'+  y\,y'   }\right)^2}}
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+  y\,y' }{\sqrt{(x\,x'+  y\,y' )^2+\left(  y\, x'  -  y'\,x   \right)^2}}
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+  y\,y' }{\sqrt{\left(x\,x'\right)^2+      2\,x\,x' \,  y\,y' + \left( y\,y' \right)^2+\left(  y\, x' \right)^2   -2\, y\, x'\,y'\,x     + \left( y'\,x    \right)^2}}
\\
&=
\sqrt{\left(x^2 + y^2\right)\,\left( {x '}^2 + {y'}^2\right)}\,\frac{x\,x'+  y\,y' }{\sqrt{\left(x\,x'\right)^2+        \left( y\,y' \right)^2+\left(  y\, x' \right)^2         + \left( y'\,x    \right)^2}}
\\
&=
x\,x'+  y\,y'
\end{align*}
At this point the nut is cracked, and I rebuild the solution as follows:
\begin{align}
 - 2\,x\,x'  - 2\,y\,{y'}
&= -2\,r\,r' \,\cos\left(\theta - \theta'\right)
\\
r^2 + {r'}^2 - 2\,x\,x'  - 2\,y\,{y'}
&=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right)
\\
(x-x')^2 + (y-y')^2
&=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right)
\\
(x-x')^2 + (y-y')^2 + (z-z')^2
&=r^2 + {r'}^2 -2\,r\,r' \,\cos\left(\theta - \theta'\right) + (z-z')^2
\end{align}


问:
Smooth curves and Line Integrals
Let $f:R^3->R$ be a continuous, and let $C_{1},C_{2}$ be two smooth curves joining two points $p,q$. Then prove that line integrals of $f$ with respect to $C_{1}$ and $C_{2}$ are equal.

My progress: To be honest, I am quite unsure whether this statement is true or not. I have tried to find counter examples, but in each time, I was finding equality for each two points with different curves. I would be glad if anyone could help me to solve this problem.

答: Take $f(x,y,z)=1$ and let $C_1$ and $C_2$ be two different curves of different lengths. Then the line integral of $f$ along $C_1$ equals the length of the curve $C_1$. Similarly, the line integral of $f$ along $C_2$ equals the length of the curve $C_2$. Thus, these integrals must be different.

Title: Smooth curves and Line Integrals


问:
Can someone explain how to get the minimal polynomial for this matrix?
This is example 8.45 from Sheldon Axler's Linear Algebra Done Right. $T$
is the operator on $\mathbb{C}^5$ whose matrix with respect to the standard basis is:\begin{matrix}
0 & 0 & 0 & 0 & -3\\
1 & 0 & 0 & 0 & 6\\
0&1&0&0&0\\
0&0&1&0&0\\
0&0&0&1&0
\end{matrix}
Find the minimal polynomial of $T$.
The solution is supposed to be $z^5-6z+3$ but I don't understand the steps to get to this.

答:
Let $e_n$ be the column vector with all $0$'s except for a $1$ in the $n$-th column from the top, for $n=1,2,3,4,5$. Then $Te_1=e_2$, $Te_2=e_3$, $Te_3=e_4$, $Te_4=e_5$, and $Te_5=-3e_1+6e_2$. Therefore,
$$
                T^5e_1=-3e_1+6Te_1 \\
                (T^5-6T+3I)e_1=0.
$$
By applying powers of $T$, you can conclude that $(T^5-6T+3I)e_k=0$ for $k=1,2,3,4,5$. $p(\lambda)=\lambda^5-6\lambda+3$ is the minimal polynomial of $T$.


问:
How do I solve $32x \equiv 12 \pmod {82}$?
I am able to solve simpler linear congruences, for example $3x \equiv 2 \pmod 5$. What I would do in this case is use that $0 \equiv 10 \pmod 5$ and then utilising a theorem: $3x \equiv 12 \pmod 5$. Then I can divide by $3$ leaving me $x \equiv 4 \: \left( \mathrm{mod} \: {\frac{5}{\mathrm{GCD}(5,3)}} \right) \quad \Longleftrightarrow x \equiv 4 \pmod{5}$ which means the solution is $x = 4k + 5$ where $k \in \mathbb{Z}$.

But I cannot apply the same method to this congruence:
$$ 32x \equiv 12 \pmod {82} $$
This is how far I got:
$$ 8x \equiv 3 \: \left( \mathrm{mod} \: \frac{82}{\mathrm{GCD}(82, 4)} \right) $$
$$ \Updownarrow $$
$$ 8x \equiv 3 \pmod {41} $$

What could I do next? Please provide solutions without the Euclidean algorithm.

EDIT:

What I found later is that I can say that
$$ 0 \equiv 205 \pmod {41} $$
And then I can add it to the congruence in question and divide by $8$.
So I guess my question is essentially 'How can I find a number that is a multiple of $41$ (the modulus) and which, if added to $3$ gives a number that is divisible by $8$?'

I reckon the Euclidean algorithm is something which gives an answer to these kinds of questions?!

答:
\begin{align}
   32x &\equiv 12 \pmod {82} &\text{Divide by $2$}\\
   16x &\equiv 6 \pmod {41}
       &\text{Divide by $2$. ($2^{-1}$ exists since $\gcd(2,41)=1$)}\\
   8x &\equiv 3 \pmod {41} &\text{$3 \equiv 3+41 \pmod{41}$}\\
   8x &\equiv 44 \pmod {41} &\text{divide by $4$}\\
   2x &\equiv 11 \pmod {41} &\text{$11 \equiv 11+41 \pmod{41}$}\\
   2x &\equiv 52 \pmod {41} &\text{Divide by $2$} \\
   x &\equiv 26 \pmod {41}
\end{align}


问:
3D software like GeoGebra
Does it exist a free interactive geometry software, like GeoGebra, which works for 3D geometry? I would be able to draw spheres, great circles, and so on.</p>

答:
Geomview is fairly flexible, and allows you to work in Euclidean, spherical or hyperbolic geometry.  Depending on which operating system you use it might be easier or harder to get it installed.  It made it into the Ubuntu repositories recently, so if you're running Ubuntu it takes little effort to install.

That said, it's not the easiest software to use -- quite a ways away from the interface of GeoGebra. More often than not, if I need a high-quality rendered image of something 3-dimensional, I code a PovRay script and render it.

Blender is open-source and interactive.  It appears to be quite flexible -- it's used to produce models for 3d video games, for example.  I haven't really got comfortable using it, though.

ContentLicense: CC BY-SA 2.5

Title: 3D software like GeoGebra


问:
What is the area of the portion of 1/8 of an sphere cut off by two parallel planes?
So the problem that I'm trying to solve is as follows:

Assume 1/8 of a sphere with radius $r$ whose center is at the origin (for example the 1/8 which is in $R^{+}$). Now two parallel planes are intersecting with this portion where their distance is $h$ (Note:$h$ < $r$) and one of the planes passes through the origin.What is the area cut off by these two planes on the 1/8 of the sphere?
You may assume anything that you think is required to calculate this area, as given, my suggestion would be the angles at which the parallel planes cross xyz planes.

Obviously I'm not interested in trivial cases for example when the plane which passes through origin is one of xy, xz or zy planes.

I hope I explained everything clearly. I wish I could draw a picture for this but I don't how. Let me know if you need more clarification. Any hint or help about how to find this area is highly appreciated.

答:
This first draft gives the main components of what might be a "near solution". As the edits take place later on, hopefully a convincing argument will (and can) be made. This attempt at a solution has not yet been reviewed that much, so it might end up not being that near to a solution.

Future edits might include a yet-to-be-written computer program that would use the results of this solution to calculate the required surface area given a set of input parameters.

Future edits might also come with graphs (if not a real-time picture generator) and use latex or whatever it is that is used at math.stackoverflow.com to improve appearance.

Critique is greatly welcomed. Proofs below are adhoc and can be improved.

Problem:

Determine the area of surface, S, in 3 space: S overlaps a sphere of radius R centered at the origin. S is further limited to the intersection of this sphere with the following: the xy, xz, and yz Cartezian planes (so S lies entirely within the (+,+,+) octant), as well as two other planes, parallel to each other, separated by a distance h, and where one of the planes, P0, goes through the origin. Depending on the orientation of P0, the other parallel bounding plane, P1, might lie "below" or "above" it. P0 does not overlap the xy, xz, or yz planes. A full solution would include suitable constraints on h and on any other variables so that two such parallel planes touching S will exist.

The solution does not attempt to find a reduced form for various terms. Please comment if you notice simplifications that were not carried out.

Solution:

Because we have to deal with both a spherical entity and a planar one, the chosen coordinate system is a right-hand Cartezian coordinate system in 3 dimensions. [This may not be ideal.]

P0 is defined with arbitrary constants, A, B, and C:
Ax + By + Cz = 0

P1 is defined:
A'x + B'y + C'z = D'

Since P1 is parallel to P0,
A' = A, B' = B, C' = C. [not proven]

The first significant step:

To solve for D' [and there may be a much better way to do this], we observe that P1 is tangent to a sphere of radius h centered at the origin. [Proof of existence of this tangent point is left for a later edit.]

We solve for the tangent point, labeled (x,y,z) below in order to keep clutter down (instead of (x0, y0, z0)).

First, let's calculate some partial derivatives of the sphere (shown using d notation for now) dz/dy dz/dx and dy/dx [todo: use instead dz/dy, dy/dx, and dx/dz to (perhaps) preserve symmetry in solution of D']

For x^2 + y^2 + z^2 = h^2,

dy/dx = -x/(sqrt(h^2-x^2-z^2))

dz/dy = -y/(sqrt(h^2-x^2-y^2))

dz/dx = -x/(sqrt(h^2-x^2-y^2))

These linear changes in the given directions will match the corresponding "slopes" of P1.

For Ax + By + Cz + D',

dy/dx = -A/B

dz/dy = -B/C

dz/dx = -A/C

From this we can find (x,y,z). After equating corresponding pairs of these 6 equantions, removing the denominators, and squaring:

(B^2)(x^2) = A^2(h^2-x^2-z^2)

(C^2)(x^2) = A^2(h^2-x^2-y^2)

(C^2)(y^2) = B^2(h^2-x^2-y^2)

Moving terms around, it becomes clearer that we have a system of 3 equations linear in x^2, y^2, and z^2

(A^2+B^2)x^2 + 0 + (A^2)z^2 = (A^2)(h^2)

(A^2+C^2)x^2 + (A^2)y^2 + 0 = (A^2)(h^2)

(B^2)x^2 + (C^2+B^2)y^2 + 0 = (B^2)(h^2)

Solving using Cramer's rule [details in later edit].. and note the assymetry [todo: did I make a mistake above or is this simply a consequence of breaking symmetry in not using dx/dz?]

x^2 = ( (h^2)(A^2)(C^2) ) / ( (A^2+C^2)(C^2+B^2) )

y^2 = ( (h^2)(B^2) ) / (C^2+B^2)

z^2 = ( (h^2) ((A^2)(B^2)+(C^4)) ) / ( (A^2+C^2)(C^2+B^2) )

When we take the square roots, we will have to determine whether + or - is required. [todo: not sure yet precisely which + or - is taken for each of x, y, and z, but one set corresponds to a plane "below" P0 and another to one "above". We may or may not have to figure out + or - this way]

For emphasis, let's use x0, y0, and z0:

x0 = +- ( (h)(A)(C) ) / (sqrt( (A^2+C^2)(C^2+B^2) ))

y0 = +- ( (h)(B) ) / (sqrt( (C^2+B^2) ))

z0 = +- ( (h) sqrt((A^2)(B^2)+(C^4)) ) / (sqrt( (A^2+C^2)(C^2+B^2) ))

Knowing the tangent point, we can calculate D'

D' = A(x0) + B(y0) + C(z0)

From here on out, we'll just use D' (or D) but knowing that it has been resolved in terms of A, B, C, and h.

The second significant step:

We'll need to find where the parallel planes intersect the radius R (+,+,+) sphere.

But first we will create the strips and summation that will approximate the surface area.

We will choose the strips approximating S as follows. Each strip is approximately the intersection of a very thin ring (along a minor circle) of the sphere's intersection with S. These rings are parallel to the yz plane (so along the x axis). A ring's area can be seen as the difference in area of the side of two similar cones with one's height being H + delta x and the other's being H. Each such strip (ring) will correspond to a pair of cones that would be tangent at the center x value of that strip. Adjacent strips share a side. A similar construction can be done inside the sphere where the strips are sections of secant planes that lie entirely within the sphere.

Note, the sum of the external strips may not form an upper bound to the surface area of S. It is assumed for the moment that as the strip's width (delta x) goes to 0, the sum of their areas approach the area of S arbitrarily close and give us our answer. This construction and argument needs to be redone. A later edit may set everything up carefully and try to prove this final step as a Riemann integral. The integral (and the series) may not be reduced fully, but the series that comprises the Riemann sum can be directly encoded into a computer algorithm to produce an answer hopefully arbitrarily close to the exact result.

My very limited experience is that if we develop a Riemann sum that for infinitesimals has each component area approximate the exact area of (flat) planar region tangent to the surface piece being approximated, then the formalism that leads to the integral will be possible to carry out (whether we carry it out in our proof or not). Note the following problem.

continuing for the moment...

The endpoints of each strip are defined by where the two parallel planes intersect the sphere.

Rather than calculating the cones' areas and subtracting [todo: reconsider such a cone calculation and compare]. We will assume [without proof for the moment] that the area of each such flat and circular strip is equal to the area of a rectangular strip with

width = delta s

height = (r)(alpha1 - alpha0)

We see the latter because

height = minor circle arc length

= 2(pi)(r) ( (alpha1-alpha0)/(2(pi)) )

We also note that delta s does not equal delta x as the ring strips are slanted at an angle that matches the slope of the sphere at that minor circle corresponding to x. The precise slopes depend on y as well, but there is significant symmetry here (see reference to cone shape). So the delta s is the hypotenuse of a right triangle with delta x as one of the sides and delta r as the other.

These alphas, in radians, are the angle values taken from the center of the implied minor circle (ie, angle formed at x axis by rays that go towards the sphere surface). An alpha value of zero would correspond to the point where each ray overlap and lie on the xy plane. When one ray is on the xy plane and the other on the xz plane, we have pi/2. Remember, these points that define one of the rays that form a pair, one each for alpha1 and alpha0, for a given strip (ie, at a given x), come from the intersection, respectively, of P1 and P0 (at that x) with the sphere. [These two values of alpha will be calculated in the third section.]

Why do we select the strips this way (ie, slices of the sphere parallel to the yz plane)? Because the length and width of each such strip can easily be calculated. Each such length is a section of the perimeter of a minor circle of the sphere.. so we can avoid doing a second limit/sum to calculate their lengths since we know the lengths already for circles. We can also express both the width and the length of each strip in terms of x. This leaves the entire integral (ie, the limit of the sum of strips) expressing the area of S as a function of x only (not y or z).

The radius of the minor circle:
r = sqrt(R^2-x^2)

The width:
(delta s)^2 = (delta r)^2+(delta x)^2

delta r (a constant for each ring which equals the radial distance from the x axis to the longer end of the strip - the distance from x axis to the shorter end) is equal to delta z when we look at where those rings intersect the xz plane.

(delta s)^2 = (delta z)^2+(delta x)^2

On this xz plane, we see that

(delta z) / (delta x) = dz/dx

From the equation of the sphere, with y=0, z=sqrt(R^2-x^2)

dz/dx = x / (sqrt(R^2-x^2))

delta z = x (delta x) / (sqrt(R^2-x^2))

So

(delta s)^2 = (x^2(delta x)^2)/(R^2-x^2) + (delta x)^2

= (delta x)^2 ( (x^2+1)/(R^2-x^2) )

delta s = (delta x) (sqrt( (x^2+1)/(R^2-x^2) ))

Thus the area of the rectangle that equals the area of the corresponding circular ring strip:

strip area = [delta s] [r] [alpha1 - alpha0]

= [ (delta x) (sqrt( (x^2+1)/(R^2-x^2) )) ] [sqrt(R^2-x^2)] [alpha1 - alpha0]

The total area of all strips is SUM x=0 to R of all of these strips above.

This moves to the integral, x=0 to x=R, of [(sqrt( (x^2+1)/(R^2-x^2) )) ] [sqrt(R^2-x^2)] [alpha1 - alpha0], as delta x goes to zero.

The third  significant step:

alpha0 and alpha1 will now be calculated as functions of x.

The steps will be added in a later edit, but here is how we solve this:

We take the eqn of the sphere and the equation of each plane. P0 solves for alpha0 and P1 solves for alpha1. The steps of these two mimic each other.

Combining these two eqns, we get a quadratic in z and a quadratic in y (which mimic each other), where x is treated as a constant within these quadratics. We solve these quadratics.

Then,

alpha0 = a bifurcation of two parts: { arctan (solution to z quadratic / solution to y quadratic), iff solution to z quadratic > 0 and solution to y quadratic > 0; 0, otherwise }

If everything was done properly, we'll be able to show that the solutions to y and z quadratic can be complex for values of A, B, C, and x where the P0 does not intersect the sphere at that x, but the solution will otherwise have one positive value and one negative value. Only the positive is useful within the +,+,+ octant. [x will be squared within each of these quadratic solution pairs, meaning one solution corresponds to negative x values and one to positive x values.]

The h value and whether we look at the plane "above" or the plane "below" P0 plays a role here as well (ie, for some h, A, B, and C, we might have to consider "above" plane or else "below" plane, labeled P1 in each case but differentiated by +- value of x0,y0,z0 calculated and discussed earlier).

alpha1 is similar (calculated using P1's eqns that has D') but instead of "0, otherwise" we get "pi/2, otherwise"

Now, let's just list the y solutions for alpha1 (with details left for later edit). The y solutions for alpha0 can be found by replacing D with 0. Also, the z solutions are identical to the corresponding y solns but have B and C reversed.

y solns used to calculate alpha1 = [-b+-sqrt(b^2-4ac)] / 2a

a = 1 + (B^2/C^2)
b = (2Bx/C) (Ax/C-D)
c = x^2 - R^2 + (Ax/C-D)^2

...

Well, lots of details to be cleaned up (and maybe changed entirely) and to be filled in, but supposedly we found the integral expression that would yield the surface area desired. This integral expression depends on constants A, B, C, h, and R and also on the integration variable x. Computer algorithm could be used to approximate this surface area by relying on the Riemann sum described and using a small step delta x. Hopefully such a computer program will also be provided in a future edit.


问:
Basics in Complex Analysis
I am learning complex analysis and having doubts in the basics as follows:-

I read that the derivative of a complex valued function $f(z)$
behaves exactly in the similar fashion as if it were a real valued
function $f(x)$.I would like to know why that happens. If i guess
that it is because the definition of differentiability in both the
worlds, the complex as well as the real is $'identically'$ similar .
If there's more to it ,please help me here.

The integral of a complex-valued function is defined along a contour as the sum of the product of the function value at $z$ times the displacement $dz$ .What does this represent if it's not area under the curve ? $$\int_\gamma f(z)\,dz .$$

Following from above ,in complex integrations , we essentially multiply two complex numbers (i.e. $f(z)$ and $dz$) , simplify them to a single number and then integrate. How is this action inside an integral is justifiable ?

In Cauchy's theorem of analytic functions in a simple closed curve , is it necessary to have
function $f(z)$ being $'extending ...smoothly'$ towards the contour. Is it necessary as well as suffficient if the function is only analytic on and inside the contour and outside the contour it can behave in a rather funny way.Can that happen ?

If there's some text that people here can share on complex analysis , that would be really helpful.

Thanks for helping.

答:
I think the source of the problem is thinking about the integral primarily as "the area under a curve".  That's one application of integration, but it isn't what integration is.  Integration is the infinite sum of infinitely small values.  That's why complex integration works just like real integration.  $f(z)\,dz$ is just an infinitely small value that gets summed.  It works for multivariable equations just like single-variable equations.  It is literally just an infinite sum of infinitely small values.

The reason why, in the reals, it represents an area under a curve, is that if you divide the curve $f(x)$ into infinitely small rectangles, each rectangle will be $f(x)$ tall and $dx$ wide.  So, $\int f(x)\,dx$ is just the sum of all those areas.  But the function that the integration itself is performing is not area, but summation.  It doesn't matter to the integral that $f(x)\,dx$ represents an area or anything else.  Therefore, since addition works the same in the complex numbers as it does in the reals, integration also works the same, because it is just a summation.

If you realize that integration is just the infinite sum of infinitely small values, then I think that answers all of your questions, at least implicitly.

Title: Basics in Complex Analysis


问:
How to prove $N_{L/K }(\mathfrak{P})=\mathfrak{p}^{f_{\mathfrak{P}/\mathfrak{p}}}$?
Let $L/K$ be a finite extension of number fields and the prime $\mathfrak{P}$ lies above the prime$(\mathfrak{p})$ of $K.$ To prove $N_{L/K }(\mathfrak{P})=\mathfrak{p}^{f_{\mathfrak{P}/\mathfrak{p}}}$, I tried using ideal norm arguments, but I was not successful.
Any help is much appreciated.

答:
Let $L/K/\Bbb{Q}$ be a tower of finite extensions and $\mathfrak{P}\subset O_L$ a maximal ideal. Let $N_{L/K}(\mathfrak{P})$ be the $O_K$ ideal generated by the norms of all elements $\in \mathfrak{P}$.

$N_{L/K}(\mathfrak{P})$ is a power of $\mathfrak{p}=\mathfrak{P}\cap O_K$: this is because for each $a\in O_K-\mathfrak{p}$ there is $b\in \mathfrak{p}$ such that  $(a,b)=O_K$ thus $(a,N_{L/K}(\mathfrak{P}))\supset (a,b^{[L:K]})=O_K$ and hence $N_{L/K}(\mathfrak{P})$ is comaximal with $(a)$.

From our knowledge of Dedekind domains this implies that $N_{L/K}(\mathfrak{P})=\mathfrak{p}^r$ for some $r$.

$N_{L/\Bbb{Q}}$ is defined the same way, but we also know that

$$N_{L/\Bbb{Q}}(\mathfrak{P})=|O_L/\mathfrak{P}|\Bbb{Z}=p^{f(\mathfrak{P}/p)}\Bbb{Z},\qquad N_{L/\Bbb{Q}}(\mathfrak{P})=N_{K/\Bbb{Q}}(\mathfrak{p}^r)=p^{r\ f(\mathfrak{p}/p)}\Bbb{Z}$$

Whence $$r = \frac{f(\mathfrak{P}/p)}{f(\mathfrak{p}/p)}=f(\mathfrak{P/p})$$
where $f(\mathfrak{P/p})$ means $[O_L/\mathfrak{P}:O_K/\mathfrak{p}]$.

I think it is a bit non obvious that $N_{L/\Bbb{Q}}(\mathfrak{P})=N_{K/\Bbb{Q}}(\mathfrak{p}^r)$:

by definition $N_{L/\Bbb{Q}}(\mathfrak{P})\subset N_{K/\Bbb{Q}}(\mathfrak{p}^r)$, for the reverse inclusion I would take $c,d\in \mathfrak{P}$ such that $(c)=\mathfrak{P} I,(d)=\mathfrak{P} J$, $(N_{L/\Bbb{Q}}(I),N_{L/\Bbb{Q}}(J))=(1)$, therefore $\mathfrak{p}^r=(N_{L/K}(c),N_{L/K}(d)), N_{K/\Bbb{Q}}(\mathfrak{p}^r)=(N_{L/\Bbb{Q}}(c),N_{L/\Bbb{Q}}(d))$.

$N_{L/\Bbb{Q}}(\mathfrak{P})=|O_L/\mathfrak{P}|\Bbb{Z}$ follows from the same idea.


问:
Show that $\lim\limits_{x\to0}\frac{e^x-1}{x}=1$

Show that $\displaystyle\lim_{x\to0}\frac{e^x-1}{x}=1$

Letting $y=e^x-1\implies e^x=y+1\implies x=\log(y+1)$ the evaluation is easy.

But I can't understand how to express given function as a composition of two functions so that the following rule can be used limit operation can be done.

Let $A\subset\mathbb R,f:A\to\mathbb R,g:D\to\mathbb R$ such that $f(A)\subset D.$ Let $c$ be a limit point of $A$ and $\lim_{x\to c}f(x)=l.$ If $l\in D$ and $g$ is cont at $l$ then $\lim_{x\to c}(gf)(x)=g(l)$ and if $l\notin D$ but a limit point of $D$ then $\lim_{x\to c}(gf)(x)=\lim_{y\to l}g(y).$

答:
Take $f:\mathbb R-\{0\}\to\mathbb R:x\mapsto e^x-1\\g:(-1,\infty)-\{0\}\to\mathbb R:x\mapsto\dfrac{x}{\log(x+1)}.$

Note $\forall~x\in\mathbb R,~e^x>0\text{ i.e. } e^x-1>-1\text{ and }e^x=1\iff x=0.\text{ So image}(f)\subset\text{domain}(g).$ Now $gf:\mathbb R-\{0\}\to\mathbb R:x\mapsto \dfrac{e^x-1}{x}.$

The given limit is $\displaystyle\lim_{x\to0}(gf)(x).$ $e^x-1$ being continuous at $0,\displaystyle\lim_{x\to0}f(x)=0,$ a limit point of the domain$(g)$ and $\displaystyle\lim_{x\to0}\frac{\log(1+x)}{x}=1(\neq 0)\implies\displaystyle\lim_{x\to0}\frac{x}{\log(1+x)}=1.$ Hence the given limit is $\displaystyle\lim_{x\to0}g(x)=1.$


问:
Could this Beta density related integral be simplified
I encounter the following integral, where $F^{-1}(u)$ is the quantile function for $\mathcal{N}(0,1)$ distribution: for $k \in\mathbb{N}$:

$$\int\limits_{1/2}^1 F^{-1}(u) u^k(1-u)^k\,\mathrm{d}u$$

Could this be simplified to some easier form as a function of $k$? Or if this could not be derived to an explicit form, as $k\rightarrow\infty$, the integrand becoming small, could we say something about its decreasing order in terms of $k$.

答:
If
$$F^{-1}(u)=\sqrt{2}\, \text{erf}^{-1}(2 u-1)$$
$$I(k)=\int\limits_{\frac12}^1 F^{-1}(u) u^k(1-u)^k\,du$$ does not show any analytical form.

Computing it and making a quick and dirty linear regression
$$\log\big[I(k)\big]=a+ b\,k$$ for $0 \leq k \leq 50$ gives with $R^2=0.999919$
$$\begin{array}{clclclclc}
 \text{} & \text{Estimate} & \text{Standard Error} & \text{Confidence Interval}
   \\
 a & -2.72809 & 0.11301 & \{-2.95532,-2.50086\} \\
 b & -1.44222 & 0.00389 & \{-1.45005,-1.43439\} \\
\end{array}$$

Edit

One thing we can try is to let $u=\frac {x+1} 2$ to make
$$I(k)=2^{-2 k-\frac{1}{2}}\int_0^1  \text{erf}^{-1}(x) \left(1-x^2\right)^k\,dx$$ and use the series expansion
$$\text{erf}^{-1}(x)=\sum_{n=0}^\infty \frac{c_n}{2 n+1} \Big[\sqrt{\frac{\pi }{2}} x\Big]^{2n+1}$$ where
$$c_0=1 \quad \text{and}\quad c_n=\sum _{m=0}^{n-1} \frac{c_m \,c_{n-m-1}}{(m+1) (2 m+1)}$$
$$I(k)=2^{-2 k-\frac{1}{2}}\sum_{n=0}^\infty \frac{2^{-n-\frac{3}{2}} \pi ^{n+\frac{1}{2}}  \Gamma (2 k+1) \Gamma
   (n+1)}{(2 n+1) \Gamma (2 k+n+2)}c_n$$ Using a few terms, the asymptotics is
$$I(k)=\frac{\sqrt{\pi } }{k}\,2^{-2 k-\frac{5}{2}}\exp\Big[-\frac{12-\pi }{12 k} \Big]$$

For $k=5$, this gives $I_5=0.00005280$ while numerical integration gives $0.00005326$.


问:
How to write that the only integers that have a multiplicative inverse are $1$ and $-1$, in symbolic form
Could it be something like for $(1,-1) \in \mathbb{Z}$, there exists $y \in \mathbb{Z}$, such that $xy = 1$?

答:
For any (unital) ring $R$, the set of units $R^\times$ is the group consisting of
$$
\{ x \in R \mid \exists\, y \in R \text{ s.t. } xy = 1 \}.
$$

Your (true) claim that the only units in the integers could be written:
$$
\{ x \in \mathbb{Z} \mid \exists\, y \in \mathbb{Z} \text{ s.t. } xy = 1 \}
= \{ -1, +1 \}
$$


问:
Show that $\dim V = \dim W$
$V$ and $W$ are subspaces of $\mathbb{C}^m$ and $\mathbb{C}^n$ respectively. $m, n$ be positive integers. Let $A$ be a $(n,m)$ matrix and $B$ be a $(m,n)$ matrix assume that $\lambda \neq 0$

$V= \{x\in \mathbb{C}^m |$ for a positive integer $k$ , $(BA-\lambda I_m)^k x=0$ holds}

$W= \{y\in \mathbb{C}^n |$ for a positive integer $l$ , $(AB-\lambda I_n)^l y=0$ holds}

Show that $\dim V= \dim W$

I have proved $BA$ and $AB$ have the same eigenvalue $\lambda$ but then how do we involve these statements, any hints would be appreciated.

答:
I'm editing the post after johnny10 pointing out my mistake in the previous post.

If we denote for fix $k$:

$V_k= \{x\in \mathbb{C}^m |  (BA-\lambda I_m)^k x=0 \}$
and
$W_k= \{x\in \mathbb{C}^n |  (AB-\lambda I_n)^k x=0 \}$
then $\dim V_k =\dim W_k$.

For the proof, I just extend the answer that johnny10 gave above (but the comment doesn't allow me to type too many things so I make a post here):

The following is the same as that given by Christiaan Hattingh in the post with adapted size:

From $\begin{bmatrix} I_n & -A \\ 0 & I_m\end{bmatrix}\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix}\begin{bmatrix} I_n & A \\ 0 & I_m\end{bmatrix} = \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$, we see that $\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} \text{ and } \begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix}$ are similar so they have same Jordan form.

In particular, $\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda  I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix} - \lambda  I_{m+n}\right )^k$ for all $k$.

On the other hand, we have $\text{rank}\left (\begin{bmatrix} AB & 0 \\ B & 0\end{bmatrix} - \lambda  I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} (AB-\lambda I_n)^k & 0 \\ D & (-1)^k \lambda^k I_m\end{bmatrix} \right ) = \text{rank}(AB - \lambda I_n)^k +m$
and
$\text{rank}\left (\begin{bmatrix} 0 & 0 \\ B & BA\end{bmatrix} - \lambda  I_{m+n}\right )^k = \text{rank}\left (\begin{bmatrix} (-1)^k \lambda^k I_m & 0 \\ D' & (-1)^k (BA-\lambda I_m)^k\end{bmatrix} \right ) = \text{rank}(BA - \lambda I_m)^k +n$

Therefore, $n-\text{rank}(AB - \lambda I_n)^k=m-\text{rank}(BA - \lambda I_m)^k$ for all $k$ and it implies the statement above.

Hence, $\dim V =\dim W $ since in general $V$ and $W$ are just $V_k$ and $W_k$ for $k$ large (we can take $k$ to be the algebraic multiplicities).

Title: Show that $\dim V = \dim W$


问:
Intuitive Understanding of Independence of 3 Events
I understand what it means to say that an even E is independent of an event F, it means that knowing that F has occurred does not change the probability that E will occur. Algebraically, it implies:

$$P(E) = P\left(E \mid F\right)   \stackrel{\text{def}}{=} \frac{P\left(E \cap F\right)}{P\left(F\right)} \Leftrightarrow P\left(E \cap F \right) = P\left(E\right)  P\left(F\right)$$

Now I'm trying to tackle what it means for events E, F, and G to be independent, the definition stated in the book says:

$$\begin{gathered}
  P\left(E \cap F \cap G\right) = P\left(E\right) P\left(F\right) P\left(G\right)\\
  P\left(E \cap F\right) = P\left(E\right) P\left(F\right) \\
  P\left(E \cap G\right) = P\left(E\right) P\left(G\right) \\
  P\left(F \cap G\right) = P\left(F\right) P\left(G\right)\end{gathered}$$

Based on my previous understanding the bottom three lines would be implying that E is independent of F, E is independent of G, and that F is independent of G. But I'm not sure what the first line is saying. The way I've made sense of it so far is that:

$$ P\left(E \cap F \cap G\right)= P\left( E \cap \left( F \cap G  \right) \right) \stackrel{?}{=} P\left(E \right) P\left(F \cap G\right) \stackrel{\alpha}{=} P\left(E\right) P\left(F\right)P\left(G\right)$$

(Where $\alpha$ comes from the fact that E and F were independent.)

In order for the equality with the question mark to hold, I think we would need $E$ to be independent from $F \cap G$.

I am a little confused by their definition and tried to figure it out algebraically, but that's what I've gotten up to.

I am hoping someone can:

Give me an intuitive understanding of what it means for 3 events to be independent

Help me connect this understanding to the definition they have provided

Bonus: Help me take that understanding to the independence of $n$ events

答:
For three events to be independent, knowledge of the others should not affect the probability of each event.  For pairwise independence, knowing one other event tells you nothing. But knowing both other events might shift the odds.
For example: draw 3 overlapping circles. Put a 1 in the 3 way overlap. Put 7 in each of the 2-way overlaps that are not in the center.  Put a 1 in each of the regions belonging to exactly one set.  Put a 7 in the complement of the union. The total is 32; each set contains 16, each 2-way overlap contains 8, so for example $A$ and $B$ are independent since $P(A) = \frac{1}{2}, P(B) = \frac{1}{2},$ and $P(A\cap B) = \frac{1}{4}$. Likewise for the other pairs, so we have full two way independence. But 3-way independence fails: $\frac{1}{8} = P(A)P(B)P(C) ≠ P(A\cap B \cap C) = \frac{1}{32}$.

Now what does this mean?  Suppose I know $A$ is true. $P(C)$ is unaffected, since half of the elements of $A$ are in $C$.

Now uppose I know $A$ is true and $B$ is true. This makes $P(C)$ drop from $\frac{1}{2}$ to $\frac{1}{32}$.  Knowing one of the other truth values told you nothing, but knowing both of them does give you information that changes the probability.


答:
You are confusing things. If I define $f : \mathbb{R} \to \mathbb{R}$ by $f(x)=x^n$ this is very different from defining $g: \mathbb{R} \to \mathbb{R}$ by $g(x)= a^x$, note that in the first one the exponent is not varying and on the other function the variable appears on the exponent.

For the first function the derivative is just $f'(x) = nx^{n-1}$, for the second one things are different. First it turns out that first you need to define what it means to raise something to a real number (notice that the usual definition doesn't work, what would mean multiplying a number by itself $\pi$ times?), in that case for reasons that I won't explain here we define this function as:

$$a^x = e^{x\ln a}$$

In that case, if we know how to differentiate $e^x$ (and usually when we construct this, we already know), we'll have the following:

$$(\ln \circ g)(x)=x \ln a$$

Now the chain rule gives:

$$\ln'(g(x))g'(x)=\ln a$$

However $\ln'(x) = 1/x$ because of the construction of $\ln$ and $g(x)=a^x$ so tha we have:

$$\frac{1}{a^x}g'(x)=\ln a \Longrightarrow g'(x) = a^x \ln a$$

Notice that there was a crucial appeal to the definition $a^x = e^{x \ln a}$. To know why we define things this way look at Spivak's Calculus, there's an entire chapter devoted to all the constructions about logs and exponentials.


问:
Poisson distribution with an interval
Customers arrive at the shopping mall according to a Poisson process with $\lambda = 3$ and the process begins at $t = 0$. Let $X$ be the number of arrivals in the interval $[2, 4]$. (i) Find $P(X \leq 4|X \geq 2)$. (ii) Let $Y$ be the time of the $10$th arrival. Find the values of $m$ and $n$ if $P(Y > 8) = \sum_{j = 0}^{m} \frac{n^j \cdot e^{-n}}{j!}$ . (iii) Let $W$ be the time of the third arrival. Find $E[\frac{1}{W^2}] $.

My attempt:

(i) $X$ follows a Poisson distribution with $\lambda = 3 *(4 - 2) = 6$. Hence

$P(X \leq 4|X \geq 2) = \frac{P(2 \leq X \leq 4)}{1 -P(X < 2)} = \frac{P(X = 2) + P(X = 3) + P(X = 4)}{1 - P(X = 0) - P(X = 1)} \approx 0.27243228$

(ii) Consider the following argument:

Let $T$ be the time of the $k$th arrival where $k \in \mathbb{Z}^+$. If $S$ is the number of arrivals in $[0, s]$ then $S$ is Poisson distributed with $s\lambda$. So

$P(T \leq t) = P(S \geq k) = 1 - P(S < k) = 1 - \sum_{j = 0}^{k - 1} \frac{e^{-t\lambda}(t\lambda)^j}{j!}$

By a similar argument, if $X$ is the number of arrivals in $[2, 4]$ then this is equivalent to the number of arrivals in $[0, 2]$.

$P(Y > 8) = 1 - P(Y \leq 8) = P(X < 10) = \sum_{j = 0}^{9} \frac{e^{-8 \cdot 6}(8 \cdot 6)^j}{j!}$

So $m = 9, n = 48$.

(iii) No idea how to do this one.

For the ones I attempted, are they correct? The second part seems incorrect and I have no clue about the third part. Any assistance much appreciated.

答:
Solutions to the first two parts look correct to me.

For the third part, if $W_n$ denotes the time of the $n^\mathrm{th}$ arrival, then
$W_n$ is distributed according to $\mathrm{Gamma}(n,\lambda)$. Hence the pdf of $W_n$ is given by
\begin{align*}
f_{W_n}(t)=\frac{\lambda\exp(-\lambda t)(\lambda t)^{n-1}}{(n-1)!}
\end{align*}
Since we want to know about $W_3$, its pdf is given by
\begin{align*}
f(t)=\frac{\lambda^3\exp(-\lambda t)t^{2}}{2}
\end{align*}
Then,
\begin{align*}
\mathbb{E}\left(\frac{1}{W_3^2}\right)&=\int_{0}^\infty dt\frac{1}{t^2}f(t)\\
&=\frac{\lambda^3}{2}\int_{0}^\infty dte^{-\lambda t}\\
&=\frac{\lambda^2}{2}
\end{align*}

Title: Poisson distribution with an interval


问:
Showing that $\frac{(j+a)!(j-a)!}{a^2}$, for $a=1, \ldots, j$, reaches its minimum at $a\approx \sqrt{j}$
Let $j$ be a positive integer, and $a=1,\dots,j$. From numerical result, I get that $$\frac{(j+a)!(j-a)!}{a^2}$$ reaches its minimum at $a\approx\sqrt{j}$. But how to prove this?

More general form: $\alpha \neq 2$ case

答:
Approximation of the factorial

Observe that
$$\log N!=\sum_{n=1}^{N}\log n\approx \int\limits_{1}^{N}\log x\,\mathrm{d}x = N\log N-N+1\approx N\log N-N.$$
Raising both sides to $e$ yields
$$N!\approx\left(\frac{N}{e}\right)^N.$$
One may even use Stirlings formula but at the end we'll see that this approximation is more than enough.

Replacing

If we replace the factorial in your formula with this approximation we have
$$\frac{1}{a^2}\left(\frac{j+a}{e}\right)^{j+a}\left(\frac{j-a}{e}\right)^{j-a}=\frac{1}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}.$$
The domain of this function is $\{x\in\mathbb{R}\mid j>a.\}$

Setting up the equation

In order to find the minimum we have to compute the derivative of this function and set it equal to zero
$$-\frac{2}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}+\frac{1}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}(\log(j-a)+1)+\frac{2}{a^2}e^{-2j}\left(j+a\right)^{j+a}\left(j-a\right)^{j-a}(\log(j+a)+1)=0.$$
Solving the equation

Write the left hand side as a single fraction. $$\frac{1}{a^2}e^{-2j}(j+a)^{j+a}(j-a)^{j-a}(\log(j+a)+\log(j-a))=0$$

Multiply both sides by $a^2e^{2j}(j-a)^a$. $$(j+a)^{j+a}(j-a)^{j}(\log(j+a)+\log(j-a))=0$$

Now we have $3$ different possibilities:

P1: $(j-a)^j=0$. Violates $j>a$.

P2: $(j+a)^{j+a}=0$. Base has to be $0$ but $0^0$ is undefined.

P3: $\log(j+a)+\log(j-a)=0$. Is solvable.

Solving P3

Combine logarithms
$$\log((j+a)(j-a))=0 \Rightarrow (j+a)(j-a)=1.$$
Expand out the terms and add $a^2$ on both sides to find the minimum.
$$j^2=a^2+1\Rightarrow j=\sqrt{a^2+1}.$$
The equation above does not have $2$ solutions since our domain was $\{x\in\mathbb{R}\mid j>a\}$ (=no negative solution). Equivalently
$$a=\sqrt{j^2-1}.$$
This is as good as $a=j$, since for big values of $j$ we may even say $j>6$, the difference is very small. For $j\to\infty$ there is no difference.


问:
Data structure to find GetMean in O(log n) time
I need ideas for designing a data structure that does insert, delete and getmean(a,b)  all in

O(log n) time. getmean(a,b) is the arithmetic mean of all number x in [a,b)

My Thoughts -

Generally, insert and delete operations could be done in O(log n) time if we store data in

Balanced Search Trees like AVL Trees. But to solve getmean(a,b) in O(log n) time we need to store

some additional information. For calculating mean we could do the following:

recursively do an in-depth traversal. If the current element then we don't need to traverse

the left sub-tree. If current element > b then we don't need to traverse the right-subtree.

If the current element is between a and b then we include that element in our set.

But this approach needs some modification. For consider the case where all the elements in the

collection are in [a,b). Then this method will traverse all the items in the collection.

Resulting in O(n) operation.

答: I think you can get $O(\log n)$ by making every node store the number of nodes and the sum of all beneath it.  Inserting/deleting one node changes $\log n$ nodes above it, and a tree-rotation at balancing time only requires recomputation of one nodes from its children (the other node just copy what was already there), so it doesn't hurt our $O(\log n)$ addition/deletion.  Now to find mean you locate the smallest and largest node from $[a,b)$, find their closest common ancestor, and remove from the sum at that ancestor record at most $2\log n$ disjoint subtrees that doesn't appear in the reduced copy of the tree only featuring $[a,b)$ nodes, so $O(\log n)$ to get the sum and the number of elements, hence the mean.


问:
Why does the Galton-Watson Process dies out a.s. when the mean is less than 1?
The following image describes a Galton-Watson process.

Let $\xi_i^n$, $i,n\geq1$, be i.i.d. nonnegative integer-valued random variables. Define a sequence $Z_n$, $n\geq0$ by $Z_0=1$ and
$$Z_{n+1}=\left\{\begin{array}{lll}
\xi_1^{n+1}+\cdots+\xi_{Z_n}^{n+1} & \text{if} & Z_n>0,\\
0 & \text{if} & Z_n=0.
\end{array}\right.$$

This is the Galton-Watson Process where the $Z_n$ term is supposed to represent the number of individuals in the $n^{th}$ generation.

If $\mu:=E[\xi_i^m]<1$ then $Z_n$ is $0$ a.s. for $n$ large enough. However, I do not understand the proof which goes as follows:

$\frac{Z_n}{\mu^n}$ is a martingale. (This is fine.)

Since $E[\frac{Z_n}{\mu^n}]=E[X_0]=1$ this implies that $E[Z^n]=\mu^n$. Also $Z_n1_{\{Z_n>0\}}\geq 1_{\{Z_n>0\}}$ which implies that:$$P(Z_n>0)\leq E[Z_n;Z_n>0]\leq E[Z_n]=\mu^n\rightarrow 0.$$

The proof ends there. However, I don't see why this shows that $Z_n=0$ for large enough $n$. It just shows that for large enough $n$, $P(Z_n>0)$ is really small but it might not be zero which is what we want since almost surely requires $P(Z_n=0)=1$ for large enough $n$.

Any ideas how one might complete the proof?

答: $Z_n=0$ implies $Z_{n+1}=0$. Let $E_n$ be the event $Z_n=0$. Then $P(Z_n=0 \text  { for some } n)=P(\bigcup_n E_n)=\lim_{n \to \infty} P(E_n)$ since $E_n$'s are increasing. Hence,$P(Z_n=0 \text  { for some } n)=1$.


问:
Prove that in a $32$ team games, there must exist $2$ teams with same number of wins,losses and ties.

Suppose that we have $32$ teams in a tournament, we divide them into $8$ groups, where each group has $4$ teams, in each group, every team will play two games with the rest $3$ teams and each game can be win/loss/tie, now I want to prove that there must exist two teams having same number of wins,losses and ties.

答:
Hint:

Suppose for a particular team, there are $a$ wins, $b$ losses and $c$ ties, then we have $a+b+c=6$ (Why?).

Count the number of solutions to the above equation and compare them to the number of teams we have. This is what is known as pigeonhole principle.


问:
Density function of Beta distribution
Let $a, b, \alpha, β$ be real numbers such that $a  0$ and $β> 0$. If the random variable $X$ is distributed beta with parameters $α$ and $β$, find the density function of the random variable $Y: = (b - a) X + a $.

This is my solution:
I know that the density function is
$$
f_X(x)=\frac{1}{B(\alpha,\beta)}x^{\alpha-1}(1-x)^{\beta-1}
$$
with $B(\alpha,\beta)=\displaystyle\int_0^{1}x^{\alpha-1}(1-x)^{\beta-1} dx$.

And by a theorem (i don't know the name of the theorem)
$$
f_Y(y)= \left\{ \begin{array}{lcc}
             f_X(h^{-1}(x))\left|\frac{\partial(h^{-1}(x))}{\partial y}\right| &   if  & y=h(x) \\
             \\ 0 &  if & y\neq h(x)
             \end{array}
   \right.
$$
So as $h(x)=(b-a)x+a$, we have $h^{-1}(y)=\frac{y-a}{b-a}=x$. Then $\frac{\partial(h^{-1}(x))}{\partial y}=\frac{\partial}{\partial y}\left(\frac{y-a}{b-a}\right)=\frac{1}{b-a}$.

Finally i have $$f_Y(y)= \left\{ \begin{array}{lcc}
            \frac{1}{B(\alpha,\beta)}\left(\frac{y-a}{b-a}\right)^{\alpha-1}\left(1-\frac{y-a}{b-a}\right)^{\beta-1} \left(\frac{1}{b-a}\right)&   if  & y=h(x) \\
             \\ 0 &  if & y\neq h(x)
             \end{array}
   \right.
$$
is okey my solution? thanks for the help

答:
I would say that since $X\in(0, 1)$ and $Y=(b-a)X+a$, then the image of $(0,1)$ is $(a,b)$. These are the values that the random variable $Y$ can take. So instead of saying $y=h(x)$ you could put the actual bounds of Y like this:

$$g(Y)=\begin{cases}\frac{(Y-a)^{\alpha-1}(b-Y)^{\beta-1}}{B(\alpha,\beta)(b-a)^{\alpha+\beta-1}}&\text {if } a<Y<b\\
0&\text{otherwise}\end{cases}$$

Otherwise it looks fine.

Title: Density function of Beta distribution


问:
Delta function and $\sum_{t}\exp\{ i k t\} $
I am interested in evaluating the following, which appears in an integrand,
$$
\sum_{t=0}^{\infty}e^{ikt}
$$
where $k$ is real. Using the following relation
$$
\sum_{t=-\infty}^{\infty}e^{ikt} = 2\pi\delta(k),
$$
the real part of it is

\begin{eqnarray}
\Re\left[\sum_{t=0}^{\infty}e^{ikt}\right]&=&\frac{1}{2}+\frac{1}{2}\Re\left[\sum_{t=-\infty}^{\infty}e^{ik}\right]\\&=&\frac{1}{2}+\pi\delta(k).
\end{eqnarray}

Note the addition of $\frac{1}{2}$ term! This $\frac{1}{2}$ term does not agree with the following derivation using $\epsilon$ trick!

\begin{eqnarray}
\lim_{\epsilon\rightarrow0_{+}}\lim_{T\rightarrow\infty}\sum_{t=0}^{T}e^{ikt}e^{-\epsilon t}&=&\lim_{\epsilon\rightarrow0_{+}}\lim_{T\rightarrow\infty}\frac{e^{\epsilon}-e^{ik\left(T+1\right)}e^{-\epsilon T}}{e^{\epsilon}-e^{ik}}\\
&=&\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}
\end{eqnarray}

which becomes, when $k \rightarrow 0$,

\begin{eqnarray}
\lim_{k\rightarrow0,}\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}&=&\lim_{k\rightarrow0,}\lim_{\epsilon\rightarrow0_{+}}\frac{1}{\epsilon-ik}\\
&=&\lim_{k\rightarrow0}\lim_{\epsilon\rightarrow0_{+}}\frac{i}{k+i\epsilon}\\&=&\lim_{k\rightarrow0}\left[\pi\delta\left(k\right)+i\mathcal{P}\frac{1}{k}\right]
\end{eqnarray}

where Sokhotsky's formula is used.

Now when $k\nrightarrow0$,
$$
\lim_{\epsilon\rightarrow0_{+}}\frac{1}{e^{\epsilon}-e^{ik}}=\frac{1}{1-e^{ik}}
$$

Therefore, for all $k$,
$$
\sum_{t=0}^{\infty}e^{ikt}=\pi\delta\left(k\right)+\mathcal{P}\frac{1}{1-e^{ik}}
$$

Note that the real part of it does not have $\frac{1}{2}$ addition term, instead has $\Re\mathcal{P}\frac{1}{1-e^{ik}}$.

What am I doing wrong?

Building on Svyatoslav's answer which in turn supported by vitamin d, the conclusion is the following:
$$
\sum_{t=0}^{\infty}e^{ikt} = \frac{1}{2} + \pi \delta(k) + \frac{i}{2} \mathcal{P} \cot (k/2)
$$

答:
In addition to Svyatoslav's answer

In order to prove that $$\delta_\varepsilon(k) =\frac{1}{\pi}\frac{\varepsilon}{\varepsilon^2+k^2}$$ is a nascent delta function we have to consider it's definition:

A sequence $(\delta_\alpha)_{\alpha\in\mathbb{N}}$ of integrable functions $\delta_\alpha\in L^1(\mathbb{R}^n)$ is called a nacent delta function if

for all $x\in\mathbb{R}$ and $\alpha\in\mathbb{N}$, $(\delta_\alpha)_{\alpha\in\mathbb{N}}\ge0,$

for all $\alpha\in\mathbb{N}$, $$\int_{\mathbb{R}^n}\delta_{\alpha}(x)\,\mathrm{d}x=1,$$

for all $\varepsilon$, $$\lim_{\alpha \to \infty} \int_{\mathbb{R}^n \setminus B_\varepsilon (0)} \delta_{\alpha}(x) \mathrm{d}x = 0.$$

It can easily be shown that the first condition is satisfied. I will prove condition 2. and I will leave 3. to to you. A hint: A sphere in $1$ dimension is a line. If you still have questions, ask in the comments.

We want to prove
$$\int\limits_{-\infty}^{\infty}\frac{1}{\pi}\frac{\varepsilon}{\varepsilon^2+k^2}\,\mathrm{d}k=1.$$
Substitute $x=k/\varepsilon$. We get
$$\frac{1}{\pi}\int\limits_{-\infty}^{\infty}\frac{\varepsilon^2}{\varepsilon^2+k^2\varepsilon^2}\,\mathrm{d}k = \frac{1}{\pi}\int\limits_{-\infty}^{\infty}\frac{1}{1+x^2}\,\mathrm{d}x.$$
This is a well known integral, the $\arctan$. Applying the limits
$$\frac{1}{\pi}\left[\arctan x\right]_{-\infty}^\infty = \frac{\pi}{\pi} = 1.$$

If you prove the third condition right, we can conclude that
$$\lim_{\varepsilon\to0}\frac{1}{\pi}\frac{\varepsilon}{\varepsilon^2+k^2} = \delta(k).$$