问:
Combination calculation with reducing set size
My statistics aren't too great, so I'm struggle to work out the result of the following situation.

Say you have 5 sets of 5 possible options (25 options total); and you select 1 option from each set. Each time you select 1 option from a set, that set is removed from the next round of possible options; leaving 4 sets of 5 possible options. Again, pick another option, leaving 3 sets of 5 possible options.

So the selection process is to pick 5 options, one from each set; and the total amount of options reduces by 5 on each round of selection.

Eg.

```<code>Set 1
Option 1, Option 2, Option 3, Option 4, Option 5

Set 2
Option 6, Option 7, Option 8, Option 9, Option 10

Set 3
Option 11, Option 12, Option 13, Option 14, Option 15

Set 4
Option 16, Option 17, Option 18, Option 19, Option 20

Set 5
Option 21, Option 22, Option 23, Option 24, Option 25
</code>```

So you pick Option 1 first, that leaves Sets 2-5 (20 options remaining)

Then you pick Option 6, that leaves Sets 3-5 (15 options remaining)

You pick Option 11, that leaves Sets 4-5 (10 options remaining)

You pick Option 16, that leaves Set 5 (5 options remaining)

You pick Option 21, there are no items left

The order that the items are selected in is not important - but how many possible combinations does it mean you could select?

The very basic maths of

```<code>25 * 20 * 15 * 10 * 5 = 375,000
</code>```

Wouldn't factor in out-of-order repetition (which we don't care about).

So how many combinations could there be?

答: You don't need to worry about the person choosing from the second set first, then the first, since in the end, there is exactly one option chosen from each set. Thus you can view the choices being made in order: first one option is chosen from 1-5, then one in 6-10, and so on, so that you get $5^5=3125$ possibilities.</p>


问:
Confusion with lottery probabilities.

Problem

I tried to compute some probabilities for a lottery (6/49 to be specific) and found a discrepancy between two ways to compute the same probability, so I must have done something wrong. But I do not see at which point my thoughts diverge from reality.

The lottery works as follows. There are $N$($=49$) numbers to choose from ($1,\dots,N$) and a correct solution consisting of $n$($=6$) of those numbers all different from one another. The player tries to guess the correct solution by also choosing $n$ distinct numbers. The order of the numbers does not matter.

Mathematically, the solution space $\mathcal L$ consists of all subsets of $\{1,\dots,N\}$ with cardinality $n$, and has $|\mathcal L|=\binom Nn$ elements. If $\ell_0\in\mathcal L$ is the correct solution, then $C(\ell)=|\ell\cap\ell_0|$ is the number of correctly guessed numbers.

I tried to find formulas for $P(C=k)$ and $P(C\geq k)$, but the formulas I found do not satisfy $P(C\geq k)=\sum_{j=k,\dots,n}P(C=j)$ even though $\{C\geq k\}=\bigcup_{j=k,\dots,n}\{C=j\}$ is a disjoint union.

My approach

To find $P(C=k)=|\{C=k\}|\big/|\mathcal L|$ I need to count all sets $\ell\in\mathcal L$ where $k$ numbers are correct (i.e. belong to $\ell_0$) and where the other $n-k$ numbers of $\ell$ are not correct (i.e. belong to $\{1,\dots,N\}\setminus\ell_0$). I count $\binom nk$ possibilities for the former condition and $\binom{N-n}{n-k}$ possibilities for the latter condition, giving me $$P(C=k)=\frac{\displaystyle\binom nk\binom{N-n}{n-k}}{\displaystyle\binom Nn}.$$

Similarly, to find $P(C\geq k)=|\{C\geq k\}|\big/|\mathcal L|$ I need to count all sets $\ell\in\mathcal L$ where $k$ numbers are correct (i.e. belong to $\ell_0$) and where the other $n-k$ numbers of $\ell$ are arbitrary (i.e. can be any of the remaining $N-k$ numbers, regardless of whether they are correct or not). I count $\binom nk$ possibilities for the former condition and $\binom{N-k}{n-k}$ possibilities for the latter condition, giving me $$P(C\geq k)=\frac{\displaystyle\binom nk\binom{N-k}{n-k}}{\displaystyle\binom Nn}$$

Error

If I compute $P(C\geq k)$ using the formulas above for $(N,n,k)=(49,6,3)$ I get

$$
P(C\geq k)
=\frac{50}{2,303}
\color{red}\neq
\frac{4,654}{249,711}
=\frac{8,815}{499,422}+\frac{645}{665,896}+\frac{43}{2,330,636}+\frac{1}{13,983,816}
=\sum_{j=k,\dots,n}P(C=j),
$$

which means I miscounted $\{C\geq k\}$ and $\bigcup_{j=k,\dots,n}\{C=j\}$ by about 43,000 elements. I strongly suspect that my formula for $P(C\geq k)$ is wrong. Any insight is appreciated.

答:
For solutions $\ell$ that has $j$ matching numbers, your $P(C\ge k)$ formula is overcounting $\binom jk$ times as many as it should be.

For example $(N, n, k) = (49, 6, 3)$ and if the winning solution is $\ell_0 = \{1,\ldots, 6\}$, then $\ell$ like $\{1,2,3,4,48,49\}$ are counted multiple times, because the $1,...,4$ are counted both among the $k$ numbers $\in \ell_0$ and among the $n-k$ arbitrary numbers.

$$\begin{align*}
\sum_{j=k}^n P(C=j)&\ne\sum_{j=k}^n \binom{j}{k}P(C=j)\\
&= \sum_{j=k}^n \binom{j}{k}\cdot\frac{\dbinom nj\dbinom{N-n}{n-j}}{\dbinom Nn}\\
&= \frac{1}{\dbinom Nn}\sum_{j=k}^n\dbinom nj\dbinom jk\dbinom{N-n}{n-j}\\
&= \frac{1}{\dbinom Nn}\sum_{j=k}^n\dbinom nk\dbinom {n-k}{j-k}\dbinom{N-n}{n-j}&\dbinom nj\dbinom jk = \dbinom nk\dbinom {n-k}{j-k}\\
&= \frac{\dbinom nk}{\dbinom Nn}\sum_{j=k}^n\dbinom {n-k}{j-k}\dbinom{N-n}{n-j}\\
&= \frac{\dbinom nk}{\dbinom Nn}\sum_{j'=0}^{n-k}\dbinom {n-k}{j'}\dbinom{N-n}{n-k-j'}&j' = j-k\\
&= \frac{\dbinom nk}{\dbinom Nn}\dbinom{N-k}{n-k} &\text{by Vandermonde's identity}
\end{align*}$$

ContentLicense: CC BY-SA 4.0

Title: Confusion with lottery probabilities.


问:
A practice ACT problem involving parametric equations
On a recent ACT practice problem that I did, I came across this problem:

Given the parametric equations $x=\sec\theta$ and $y=\tan\theta$, which of the following is the graph of the points $(x,y)$?

(A)$\quad$circle
(B)$\quad$ellipse
(C)$\quad$parabola
(D)$\quad$hyperbola
(E)$\quad$None of these

I have plotted this on Desmos (an online graphing tool) and determined that the answer is (D), but I am not too sure how to arrive at this answer. Note that this is a problem that was found on ACT, a test meant to measure high school students' readiness for college, meaning that the solution shouldn't be too complicated, and it shouldn't take too long either. How can I solve this problem in a reasonable amount of time(say, 60 to 90 seconds), and if you want to, could you also provide some tips for problems similar to this problem?

I don't know if this is a feasible way or not, but my thought process while doing this problem was to plug in arbitrary values for $\theta$ then plotting the resulting points to obtain the answer.

答:
To fill out the details of Matthew's suggestion: note that, since $x = \sec \theta$ and $y = \tan \theta$, then perhaps some identity relating secant and tangent is worth exploring. We have the identity

$$\sec^2 \theta = \tan^2 \theta + 1$$

In terms of $x,y$, that identity is

$$x^2 = y^2 + 1$$

That is,

$$x^2 - y^2 = 1$$

This describes a hyperbola, centered at the origin, with transverse/conjugate axis lengths of $1/2$, which opens towards left/right half-planes.

(Though in the case of a multiple-choice question like this, you could just note "oh hey I have $x^2 - y^2 = 1$ and thus it's a hyperbola" and that will be sufficient.)


问:
Let $f: [1,7]→ \mathbb R$ be a twice differentiable function such that $f(7) = 3f(3)−2f(1)$. Prove that there is $c∈(1,7)$ such that $f{′′}(c) = 0$
Let $f: [1,7]→ \mathbb R$ be a twice differentiable function such that $f(7) = 3f(3)−2f(1)$. Prove that there is $c∈(1,7)$ such that $f{′′}(c) = 0$

Not certain what to do with this question. Am I supposed to come up with a function that holds this property?

Body:
$$f(7)=3f(3)-2f(1)\\\iff f(7)-f(3)=2(f(3)-f(1))\\\iff \frac{f(7)-f(3)}{7-3}=\frac{f(3)-f(1)}{3-1}$$

By LMVT and the above equality, there exists $c_1\in [1,3]$ and $c_2\in [3,7]$ such that $f'(c_1)=f'(c_2)$

Apply Rolle's theorem on $f'$ on $[c_1,c_2]$


答:
If $f$ is twice differentiable on $[1,7]$, then it must be differentiable on $[1,3]$ and $[3,7]$. Applying the mean value theorem to each interval then implies that there exists an $x_1\in(1,3)$ and an $x_2\in (3,7)$ such that

$$f'(x_1)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$$

and

$$f'(x_2)=\frac{f(7)-f(3)}{7-3}=\frac{f(7)-f(3)}{4}$$

But $f(7)=3f(3)-2f(1)$ implies that $f(7)-f(3)=2f(3)-2f(1)$, so

$$f'(x_2)=\frac{f(7)-f(3)}{4}=\frac{2f(3)-2f(1)}{4}=\frac{f(3)-f(1)}{2}=\frac{f(3)-f(1)}{3-1}=f'(x_1)$$

We thus have $f'(x_1)=f'(x_2)$, and since $f$ is twice differentiable on $[x_1,x_2]\subset[1,7]$, it must be the case that $f'$ is differentiable on $[x_1,x_2]$. We can now apply Rolle's theorem to $f'$ on the interval $[x_1,x_2]$ to deduce that there is a $c\in(x_1,x_2)\subset(1,7) $ where $f''(c)=0$. And that's what we wanted!

Body:
Let $f: [1,7]→ \mathbb R$ be a twice differentiable function such that $f(7) = 3f(3)−2f(1)$. Prove that there is $c∈(1,7)$ such that $f{′′}(c) = 0$

Not certain what to do with this question. Am I supposed to come up with a function that holds this property?


问:
Does the dilogarithm function (which is multi-valued) have a single-valued inverse?
The $p$-logarithm is defined for $|z|<1$ by

$$\text{Li}_p(z)=\sum_{n=1}^\infty\frac{z^n}{n^p}$$

and defined elsewhere in $\mathbb C$ by analytic continuation, though it may be multi-valued, depending on the path of continuation.

For an integer $p<0$, the $p$-logarithm is a rational function, whose inverse is an algebraic function, which is multi-valued. For example,

$$\text{Li}_{-1}(z)=\frac{z}{(1-z)^2},\qquad\text{Li}_{-1}^{-1}(z)=\frac{2z+1\pm\sqrt{4z+1}}{2z}.$$

For $p=0$ we have

$$\text{Li}_0(z)=\frac{z}{1-z},\qquad\text{Li}_0^{-1}(z)=\frac{z}{1+z}=-\text{Li}_0(-z),$$

and for $p=1$ we have a form of the ordinary logarithm, whose inverse is an entire function:

$$\text{Li}_1(z)=-\ln(1-z),\qquad\text{Li}_1^{-1}(z)=1-e^{-z}.$$

So, for $p<0$ the $p$-logarithm is single-valued while its inverse is multi-valued, for $p=0$ the $p$-logarithm and its inverse are both single-valued, and for some $p>0$ the $p$-logarithm is multi-valued while its inverse is single-valued.

This isn't much reason to expect the "pattern" to hold for all integers $p>0$. Nevertheless, I'd like to know if it does hold. There's a lot of information on polylogarithms on Wikipedia, but I didn't see any obvious answer to this simple question: Is $\text{Li}_p^{-1}$ single-valued? Equivalently, is $\text{Li}_p$ injective?

("Injective" is usually defined for single-valued functions. Here it means, if $z_1\neq z_2$, then the sets $\text{Li}_p(z_1)$ and $\text{Li}_p(z_2)$ don't intersect.)

We may focus on the case $p=2$.

Let's consider the values of $\text{Li}_2(z)$ near the branch point $z=1$, or equivalently of $\text{Li}_2(1-z)$ near $z=0$, noting that $\text{Li}_2(1)=\pi^2/6$:

$$\frac{\pi^2}{6}-\text{Li}_2(1-z)=\text{Li}_2(z)+\ln(z)\ln(1-z)$$

$$=\left(z+\frac{z^2}{4}+\frac{z^3}{9}+\cdots\right)-\ln(z)\left(z+\frac{z^2}{2}+\frac{z^3}{3}+\cdots\right)$$

(throw away the higher powers of $z$ since $z\approx0$)

$$\approx z-z\ln(z)=z\big(1-\ln(z)\big)$$

(use $\ln(z)\approx-\infty$)

$$\approx z\big({-\ln(z)}\big)=-z\ln(z).$$

Now let $z=e^u$ where $\text{Re}(u)\approx-\infty$, so that $\ln(z)=u$ (as one of the possible values), and

$$\frac{\pi^2}{6}-\text{Li}_2(1-e^u)\approx-ue^u.$$

This latter function is certainly not injective. For any $a\in\mathbb R$, there is a sequence of points $u_k\in\mathbb C$ with $\text{Re}(u_k)<a$, along a curve which is almost a vertical line, all having the same value of $u_ke^{u_k}$, and all having different values of $z_k=e^{u_k}$. I can see this by graphing and analyzing the level curves of $|ue^u|$ and $\arg(ue^u)$: if $u=x+yi$, these curves have the form $y=\pm\sqrt{C^2e^{-2x}-x^2}$ and $x=y\cot(D-y)$, respectively.

Can we use this to prove that $\text{Li}_2$ is not injective? Or is something lost in these approximations? What other methods can we use?

答:
$\text{Li}_2'(z)=\frac{-\log(1-z)}{z}$ doesn't vanish so any curve $\gamma:0\to ?$ along which (the continuation of) $\text{Li}_2(z)$ is analytic gives a curve $\text{Li}_2^{-1}(\gamma):0\to ?$ along which $\text{Li}_2^{-1}$ is analytic.

Next $\text{Li}_2'(z)=\frac{-\log(1-z)}{z}$ shows that $$\text{Li}_2(1-z)=\frac{\pi^2}{6}-\text{Li}_2(z)-\log(z)\log(1-z)$$
for $z\in (0,1)$ and it is meant the principal branch of each term.

Continuing analytically by starting at $1/2$ and rotating one time around $z=0$ we get a new branch

$$\text{Li}_2(1-z)^{new\ branch}=\frac{\pi^2}{6}-\text{Li}_2(z)-(\log(z)+2i\pi)\log(1-z)$$
analytic for $\Im(z)>0,|z|<1$.

This branch has a zero near $0.08+0.18 i$, and since $\text{Li}_2(0)=0$ too it means that  $\text{Li}_2^{-1}$ is multivalued, ie. there are some curves $\Gamma:0\to 0$ along which $\text{Li}_2^{-1}$ is analytic but not the same at its departure and arrival.

(it is the argument principle which proves that there is a zero on this plot)


问:
Proving a property of exponential function applied to linear operators
I'm having trouble proving a statement regarding the properties of the exponential function for linear bounded maps.More specifically, given a Banach space $X$ and a linear, bounded map $B:X \to X$ I want to show

$$
\lim_{n \to \infty}(I+\frac{1}{n} B)^n=e^B \tag{1} \\
\lim_{n \to \infty}(I-\frac{1}{n} B^n)^{-n}=e^B
$$

where $e^B=\sum_{k=0}^{\infty} \frac{B^k}{k!}$ and $I$ is the identity map. By the Binomial theorem I have

$$
(I+\frac{1}{n} B)^n=\sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} B^k.
$$

My idea was to show that

$$
\lim_{n \to \infty} ||\sum_{k=0}^{n} \binom{n}{k} \frac{1}{n^k} B^n-e^B||=0.
$$

Since $||B|| \in \mathbb{R}$ I think I have to  exploit the fact that (1) holds for real numbers. However, I do not see how to proceed.

Body:
\begin{align*}
  %%
\Big\|\sum_{k=0}^{n} \binom{n}{k}& \frac{1}{n^k} B^k-e^B\Big\|  =
  %%
\Big\|\sum_{k=0}^{n} (1-\frac1n)(1-\frac2n)\cdot(1-\frac{k-1}{n})
  %%
 \frac{B^k}{k!} -e^B\Big\|\\
  %%
& \leq \sum_{k=0}^{n} \big|(1-\frac1n)(1-\frac2n)\cdot
  %%
(1-\frac{k-1}{n})-1\big| \frac{\|B\|^k}{k!}+
  %%
\sum_{k=n+1}^\infty\frac{\|B\|^k}{k!}\rightarrow 0
  %%
\text{ as } n\rightarrow\infty.
  %%
\end{align*}


问:
probability that a random line segment parallel to the hypot. of a triangle with legs 3 and 4 will inclose an area of at least half
Sorry for the unclear title. It was difficult to explain the problem in a concise way in 150 characters.

A right triangle has the legs 3 and 4 units, respectively. Find the probability that
a line segment, drawn at random parallel to the hypotenuse and contained entirely in the
triangle, will divide the triangle so that the area between the line and the vertex opposite
the hypotenuse will equal at least half the area of the triangle. Make a sketch of the triangle.

I am completely stuck on this question. First of all I am assuming that the legs of length 3 and 4 are the ones containing the right angle but I am not sure if this is even right. If this is the case then a line segment drawn at random will divide the area of the triangle if it intersects the two legs (call them $a$ and $b$) at exactly $\frac{1}{\sqrt{2}}$ of their length. This way both the height and width of the smaller triangle will be divided by $\frac{1}{\sqrt{2}}$ and thus the new area will be $\frac{ab}{2}$ which is half of $ab$, the original area. How do I extend this idea to answer the question fully? Is the probability just $1-\frac{1}{\sqrt{2}}$? Thanks in advance!

答: Having drawn a line parallel to the hypotenuse, the smaller triangle will be proportional to the larger triangle.  We will assume that 'at random' means the value of the proportionality constant is chosen uniformly between $0$ and $1$.  As you've noted, if it is anything larger than $\frac{\sqrt2}{2}$, the area will be at least half of the original area.  Hence, the desired probability is indeed $1-\frac{\sqrt{2}}{2}$.</p>

ContentLicense: CC BY-SA 3.0


问:
The proof of the kernel and cokernel formula of a certain map.
If I have the following map:

$\phi_s: \mathbb Z_{p^a} \rightarrow \Bbb Z_{p^b}$ defined by multiplication by $kn/d$ where $n = p^b$ and $d = \gcd(p^a, p^b)$ and $k= p^s$ and $0 \leq k < d$ and I know that if $a\geq b$ and we are not in the case of the zero homomorphism i.e. we took $s = b -1$, we will have:

1- ker $\phi_{s} = Z_{p^{a-s+1}}.$

2- coker  $\phi_s = Z_{p^{a-s}}.$

How can I start proving the above results about the kernel and cokernel, should I use the properties of SES or what? could anyone help me please?

答:
Unfortunately, the formula you have is not right. Consider the case where $a=b$ and $s=0$. Then $n=d$ and $k=1$, so the map you've defined is the identity map and hence has trivial kernel and cokernel, but your formula would give its kernel to be $\mathbb{Z}/p^{a+1}$ and its cokernel to be $\mathbb{Z}/p^a$.

If you don't mind, I'll use the notation $\phi_s$ for what you refer to by $k\Lambda$. So $\phi_s$ takes $\overline{1}\in\mathbb{Z}/p^a$ to $\overline{p^{s+b}/d}\in\mathbb{Z}/p^b$, where $d=\gcd(p^a,p^b)$. Note that the only positive divisors of $p^a$ and $p^b$ are powers of $p$, so in fact $d=p^{\min\{a,b\}}$, and we can write $\phi_s(\overline{1})=\overline{p^{s+b-\min\{a,b\}}}$. To see that $\phi_s$ is a well-defined map, you need to show that $p^a\phi_s(\overline{1})=\overline{0}\in\mathbb{Z}/p^b$. This amounts to showing that $p^ap^{s+b-{\min\{a,b\}}}$ is divisible by $p^b$ in $\mathbb{Z}$, ie that $a+s+b-\min\{a,b\}\geqslant b$. Since $s\geqslant 0$, this is clear.

Now, $\overline{n}\in\ker\phi_s$ if and only if $\overline{0}=\phi_s(\overline{n})=\overline{n{p^{s+b-\min\{a,b\}}}}\in\mathbb{Z}/p^b$, which is in turn true if and only if $np^{s+b-\min\{a,b\}}$ is divisible by $p^b$ in $\mathbb{Z}$. If $p^k$ is the highest power of $p$ dividing $n$, then this is true if and only if $k+s+b-\min\{a,b\}\geqslant b$, ie if and only if $k+s\geqslant\min\{a,b\}$. You are assuming that $\phi_s$ is not the zero map, so we know $s<\min\{a,b\}$. Thus a generator for $\ker\phi_s$ is given by $\overline{p^{\min\{a,b\}-s}}$. The order of $\overline{p^{\min\{a,b\}-s}}$ is $p^{a-(\min\{a,b\}-s)}=p^{a+s-\min\{a,b\}}$, and subgroups of cyclic groups are cyclic, so $\fbox{$\ker\phi_s\cong\mathbb{Z}/p^{a+s-\min\{a,b\}}$}$. In particular, in the case you brought up, where $b\leqslant a$ and $s=b-1$, we have $\ker\phi_s\cong\mathbb{Z}/p^{a-1}$.

Now, the cokernel of $\phi_s$ is the quotient of $\mathbb{Z}/p^b$ by the image of $\phi_s$. Quotients of cyclic groups are cyclic, and cyclic groups are uniquely determined by their cardinalities, so the cokernel of $\phi_s$ will be determined by its size. Now, by the first isomorphism theorem, the image of $\phi_s$ is isomorphic to the quotient $\frac{\mathbb{Z}/p^a}{\ker\phi_a}$. Since $|\mathbb{Z}/p^a|=p^a$, and $|\ker\phi_s|=p^{a+s-\min\{a,b\}}$ by the computation above, we have $$|\operatorname{im}\phi_s|=\frac{p^a}{p^{a+s-\min\{a,b\}}}=p^{\min\{a,b\}-s}.$$ (Again, you are assuming $s<\min\{a,b\}$, so this number makes sense.) Since $|\mathbb{Z}/p^b|=p^b$, we can hence compute $$|\operatorname{coker}(\phi_s)|=\frac{|\mathbb{Z}/p^b|}{|\operatorname{im}\phi_s|}=\frac{p^b}{p^{\min\{a,b\}-s}}=p^{b+s-\min\{a,b\}},$$ and so $\fbox{$\operatorname{coker}\phi_s\cong\mathbb{Z}/p^{b+s-\min\{a,b\}}$}$. In particular, in the case you brought up, where $b\leqslant a$ and $s=b-1$, we have $\operatorname{coker}\phi_s\cong\mathbb{Z}/p^{b-1}$.


问:
Show that two groups that are isomorphic to a third group are isomorphic to each other
Prove that if G is isomorphic to K and H is isomorphic to K then G and H are isomorphic to each other.

My strategy thus far has been to show that we can use composition of the respective isomorphisms to map from G to K then K to H. Would it be conclusive to show that this process can be inverted thus proving G and H are isomorphic?

答: Isomorphism is transitive, since the composition of two isomorphisms is an isomorphism.  This is utterly straight forward.  In fact, it's not hard to see that isomorphism defines an equivalence relation.</p>


问:
Is this problem solvable as stated, or is there an error?
I asked this problem, and one of the users suggests the problem is ill posed.

Here is the statement of the problem, copied exactly from the homework.

Is there really an error in the problem, which makes it unable to be solved?

The problem is as follows:

Suppose $W : [-1,1] \to \Bbb R$ is a continuous function. Show that

$$f(t) = \begin{cases}
 W(t) & \text{if } t \in [-1,0] \\
 1+W(t) & \text{if } t \in (0,1] \end{cases}$$

is integrable on $[-1,1]$.

答:
Hint: Recall that, for a function $f : [a,b] \to \Bbb R$, the quantity

$$\int_a^b f(x)dx$$

exists if $f$ is continuous on $[a,b]$ (save for most finitely many discontinuities on $[a,b]$).

With this in mind, I believe the question is perfectly well-framed and can be solved using the information given.

An example of this would be the function on $[0,1]$ given by

$$f(x) = \begin{cases}
1 & x \in [0,1/2) \cup (1/2,1] \\
2 & x = 1/2
\end{cases}$$

which has $\int_0^1 f(x)dx = 1$. Similarly, for a function on $[0,100]$ defined by

$$f(x) = \begin{cases}
x & x \text{ is not an integer} \\
-x^2 & x \text{ is an integer}
\end{cases}$$

has $\int_0^{100} f(x)dx = \int_0^{100} x \, dx$. This becomes clear if you think of things in terms of signed area: how much area would a single point contribute to the whole integral?


问:
Hölder Inequality
I am wondering how I get
$$
 \frac{a_{1}^{k}}{b_{1}}+\frac{a_{2}^{k}}{b_{2}}+\cdots+\frac{a_{n}^{k}}{b_{n}}\geq\frac{\left(a_{1}+\cdots+a_{n}\right)^{k}}{n^{k-2}\cdot\left(b_{1}+\cdots+b_{n}\right)}.
$$
from the Hölder inequality

$$ \sum_{i =1}^{n}a_{i}b_{i}\leq\left(\sum_{i=1}^{n}a_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i =1}^{n}b_{i}^{q}\right)^{\frac{1}{q}}. $$

I was reading through AoPS and I am struggling to see how the first was obtained from the second.

答:
First we apply

$$\sum_{i =1}^{n}x_{i}y_{i}\leq\left(\sum_{i=1}^{n}x_{i}^{p}\right)^{\frac{1}{p}}\left(\sum_{i =1}^{n}y_{i}^{q}\right)^{\frac{1}{q}}$$

with $p=k$, $q=k/(k-1)$, $x_i=a_i/b_i^{1/k}$, $y_i=b_i^{1/k}$ ,

to get

$$\sum_i a_i \leq \left(a_i^k/b_i \right)^{1/k} \left( b_i^{1/(k-1)} \right)^{(k-1)/k} \; ,$$

which is equivalent to

$$\left( \sum_i a_i \right)^k \leq \left( \sum_i a_i^k/b_i \right) \left( \sum_i b_i^{1/(k-1)} \right)^{k-1} \; .$$

By concavity of $x \mapsto x^{1/(k-1)}$ (I guess $k \geq 2$), we also have that

$$1/n \sum_i b_i^{1/(k-1)} \leq \left( \sum_i b_i/n \right)^{1/(k-1)}$$

which combined with the preceding inequality, gives the desired result.

Title: Hölder Inequality


问:
why measure theory for probability?
I studied elementary probability theory. For that, density functions were enough. What is a practical necessity to develop measure theory? What is a problem that cannot be solved using elementary density functions?</p>

答: Measure theory goes beyond probability theory. It generalizes our notion of length, area and volume.</p>

Title: why measure theory for probability?


问:
How to construct a simple test to signficance level of $5\%$
I am stuck at the following homework question:

You visit a fortune teller. You test his abilities by tossing a fair coin 100 times. The fortune teller manages to predict the outcome of 60 tosses correctly. Construct a statistical test for the hypothesis
$$H_0: \text{The fortune teller is a fraud.}$$
Does your test reject the hypothesis with a significance level of $5\%$?

I would say that the obvious idea would be to define the critical area $K$ of the test as follows:

$$K := \bigg\{(x_1,\ldots,x_n) \in \{0,1\}^n \ \mid \  \sum_{i=1}^n x_i \ge k \bigg\}$$

But how do I know how to choose $k$ to reach the required significance level? Could you help me?

答:
You can run a binomial test considering under the null hypothesis (the fortune teller is a fraud) what is the probability to get by pure guessing $60$ or more correct answers. The proportion expected (the population proportion, so to speak) is $0.5:$

$$\Pr(\text{counts} \geq 60)=\displaystyle \sum_{n=60}^{100} {100\choose n} \,0.5^n\, 0.5^{100-n}$$

or

```<code>> pbinom(59,100,.5,lower.tail=F)
[1] 0.02844397
</code>```

... or given that the sample is large, we can use the normal approximation to test if the sample proportion is different from a population proportion - the equivalent to the $Z$-test for sample means. Under the null hypothesis (random guesses), the probability that the proportion of predictions is $60$ when the expected proportion is $0.5$ is

$$Z=\frac{.6-.5}{\sqrt{\frac{.5(1-.5)}{100}}}=2$$

which is greater than

```<code>> qnorm(.95)
[1] 1.644854
</code>```

Hopefully, though, and despite the evidence, you would use common sense, and not reject the null: the fortune teller is indeed a fraud.

Body:
You can run a binomial test considering under the null hypothesis (the fortune teller is a fraud) what is the probability to get by pure guessing $60$ or more correct answers. The proportion expected (the population proportion, so to speak) is $0.5:$

$$\Pr(\text{counts} \geq 60)=\displaystyle \sum_{n=60}^{100} {100\choose n} \,0.5^n\, 0.5^{100-n}$$

or

> pbinom(59,100,.5,lower.tail=F)
[1] 0.02844397

... or given that the sample is large, we can use the normal approximation to test if the sample proportion is different from a population proportion - the equivalent to the $Z$-test for sample means. Under the null hypothesis (random guesses), the probability that the proportion of predictions is $60$ when the expected proportion is $0.5$ is

$$Z=\frac{.6-.5}{\sqrt{\frac{.5(1-.5)}{100}}}=2$$

which is greater than

> qnorm(.95)
[1] 1.644854

Hopefully, though, and despite the evidence, you would use common sense, and not reject the null: the fortune teller is indeed a fraud.


问:
Let $a_1/b_1,\ldots,a_n/b_n$ be rational numbers in lowest terms. If $M={\rm lcm}(b_1,\ldots,b_n)$, prove $\gcd(Ma_1/b_1,\ldots,Ma_n/b_n)=1$.

Problem. Let $a_1/b_1,\ldots,a_n/b_n$ be rational numbers in lowest terms. If $M={\rm lcm}(b_1,\ldots,b_n)$, prove that the gcd of $Ma_1/b_1,\ldots,Ma_n/b_n$ is $1$.

I think this statement is false. For example, consider $$\frac{2}{3}\quad{\rm and}\quad \frac{2}{9}.$$ Both of them are in the lowest form, but $M={\rm lcm}(3,9)=9$ where $$9\cdot\frac{2}{3}=6\quad{\rm and}\quad 9\cdot\frac{2}{9}=2$$ are not relatively prime.

Nonetheless, if I add the assumption that $\gcd(a_1,\ldots,a_n)=1$, will the assertion be true? I tried writing $1$ as a linear combination of those numbers, but it seems not working.

答:
With your added assumption

$$\gcd(a_1, \, \ldots \, , a_n) = 1 \tag{1}\label{eq1A}$$

then the statement of

$$\gcd\left(\frac{Ma_1}{b_1}, \, \ldots \, , \frac{Ma_n}{b_n}\right) = d \tag{2}\label{eq2A}$$

having $d = 1$ is true. To prove this, assume instead that $d \gt 1$ so there's a prime $p \mid d$. There are $2$ basic cases to consider.

First, there's at least one $1 \le i \le n$ where $p \mid \frac{M}{b_i}$. This means there's one or more $1 \le j \le n$ where $p \mid b_j$, so there's also one or more indices where the exponent of $p$ in the prime factorization of $b_j$ is a maximum, call it $m$. Choose any one of these indices, call it $k$. The definition of $\operatorname{lcm}$ means that $M$'s exponent of $p$ in its prime factorization is also the maximum among all of the $b_i$, so it's the same as that of $b_k$, i.e., it's $m$ for both, meaning that $p \not\mid \frac{M}{b_k}$. As such, due to $d$ being the $\gcd$ in \eqref{eq2A}, this requires that $p \mid a_k$ instead. However, since $p \mid b_k$, this means $\frac{a_k}{b_k}$ is not in lowest terms, which is not allowed.

Thus, the second case must apply, i.e., there are no $i$ where $p \mid \frac{M}{b_i}$. This means $p \mid a_i$ for all $1 \le i \le n$ instead. However, this contradicts your added assumption in \eqref{eq1A}.

These contradictions in the $2$ possible cases mean the original assumption of $d \gt 1$ must be false, i.e., it must be that $d = 1$ in \eqref{eq2A} instead.


问:
Algorithmic simplification of expressions with $\arctan$, e.g. $2\arctan\frac{5-\sqrt2}{5+\sqrt2}=3\arctan\frac{1}{2\sqrt2}$.
This is a problem I encountered in another (unrelated) question.

I got an expression which is a bit ugly, $2\arctan\frac{5-\sqrt2}{5+\sqrt2}$, but when changing the method I got another prettier expression, $3\arctan\frac{1}{2\sqrt2}$.

Since they are both equal to the same integral, they must be equal as well (and we can easily prove that by doing some simple calculation). But what makes it interesting is, a lot of common mathematics software cannot do this kind of simplification (I tried SageMath and WolframAlpha, they both did nothing).

So let's do it by hand to see what happened.

\begin{align}
2\arctan\frac{5-\sqrt2}{5+\sqrt2} &= \arctan\frac{23}{10\sqrt2}\tag{automatic}\\
&= \arg\left(10\sqrt2+23i\right)\tag{automatic}\\
&= \arg\left(2\sqrt2+i\right)^3\tag{miracle}\\
&= 3\arctan\frac{1}{2\sqrt2}.\tag{automatic}
\end{align}

(In the general case, we should also count how many $\pi$ should be added at the end.)

As you can see, the key step here is the factorization $10\sqrt2+23i=\left(2\sqrt2+i\right)^3$. This is a factorization in $\mathbb Z[\sqrt2,i]$, which might be difficult even for computers (I'm not sure).

What's more, normally there's not just $\sqrt2$ that appears, e.g.
$$\arctan\frac{\sqrt5-77\sqrt2}{7+11\sqrt{10}}=3\arctan\sqrt5+5\arctan\sqrt2-3\pi.$$

So here is my question:

Is there an algorithmic way to do this kind of simplification?

Let me explain a little where this problem comes from.

I got an integral, for $a>0$:

$$\int_0^\infty e^{-at}\left(\operatorname{erf}\sqrt t\right)^3\,dt=\frac{4}{\pi}\frac1{a\sqrt{a+1}}\arctan\frac{1-b}{1+b},$$

where $b=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}}$.

I wanted to simplify this expression but failed. The example mentioned before is just the case $a=1$.

I'd be glad if anyone can simplify this expression for any $a>0$, and I'll take it as an acceptable answer since I think this expression is already complicated enough.

I added my answer to the integral problem only.

$$\int_0^\infty e^{-at}\left(\operatorname{erf}\sqrt t\right)^3\,dt=\frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).$$

And it didn't give us a general method to deal with all cases.

答:
Just an answer to what I edited after.

For $a>0$, let

$$b(a)=\frac{a}{a+4}\sqrt{\frac{a+3}{a+1}},$$
$$f(a) = \arctan\frac{1-b(a)}{1+b(a)}.$$

Then,

$$f'(a) = -\frac{3}{2 \, {\left(a + 2\right)} {\left(a + 1\right)}}\sqrt{\frac{a+1}{a+3}},$$
$$\int f'(a)\,da =3\arctan\sqrt{\frac{a+3}{a+1}} + Const,$$

and here we have $Const = -\frac34\pi$. Thus, for $a>0$,

$$\int_0^\infty e^{-at}\left(\operatorname{erf}\sqrt t\right)^3\,dt=\frac{12}{\pi}\frac1{a\sqrt{a+1}}\left(\arctan\sqrt{\frac{a+3}{a+1}}-\frac\pi4\right).$$


问:
Let $f: [0,6] \rightarrow \mathbb R$ be a three times differentiable function such that $f(0)=f(1.4)=f(3.9)=f(5.2)$.
Let $f: [0,6] \rightarrow \mathbb R$ be a three times differentiable function such that $f(0)=f(1.4)=f(3.9)=f(5.2)$. Prove that there is c $\epsilon$ (0,6) such that $f^{'''}(c)=0$.

Now using Rolle Theorem I know that from (0,5.2) I have that $f^{'}(x)=0$. But I am struggling with what to do from (5.2,6). I figure I can use the fact that we are looking for $f^{'''}(c)=0$, but I don't know what to do.

答: Suppose by contradiction $f''(c)\ne 0$ for any $c\in (0,6)$.In that case suppose $f''(x)>0$ for all $x\in(0,6)$.This means the function is concave upwards and can only hit the $x$ axis at atmost $2$ points  which is not the case.Similarly in the case when $f''(x)<0$ for all $x\in (0,6)$ it will be concave downwards and meet the x axis at atmost $2$ points.Again  a contradiction


问:
Exact sequence of 4 abelian groups, 3 of them being free
Let $G$ be a finitely generated abelian group that fits in an exact sequence of the form

$$0 \to \mathbb{Z} \to \mathbb{Z}^n \stackrel{f}{\to} G \stackrel{g}{\to} \mathbb{Z}^m \to 0$$

for some $n \geq 1$ and $m \geq 0$. What can we say about $G$?

答:
Since $\mathbb{Z}^m$ is free, the surjection $G\to \mathbb{Z}^m$ splits. Thus, $G$ is a direct product of $\mathrm{Im}(f)$ and $\mathbb{Z}^m$, So the only question is what is $\mathrm{Im}(f)$.

Every subgroup of $\mathbb{Z}^n$ is free, and moreover, given any subgroup $H$ of $\mathbb{Z}^n$, either $H$ is trivial or there exists a basis $x_1,\ldots,x_n$ of $\mathbb{Z}^n$, an integer $r$, $1\leq r\leq n$, and positive integers $d_1,\ldots,d_r$ such that $d_1|\cdots|d_r$ and $H$ is (freely) generated by $d_1x_1,\ldots,d_rx_r$.

Here, $r$ must be equal to $1$, so there exists an integer $d\gt 0$ such that $\mathbb{Z}^n/H\cong (\mathbb{Z}/d\mathbb{Z})\oplus \mathbb{Z}^{n-1}$. Thus,
$$G \cong \frac{\mathbb{Z}}{d\mathbb{Z}}\oplus \mathbb{Z}^{n+m-1},$$
for some $d\geq 1$.


Title: Integrate by parts: $\int \ln (2x + 1) \, dx$

Body:
$$\eqalign{
  & \int \ln (2x + 1) \, dx  \cr
  & u = \ln (2x + 1)  \cr
  & v = x  \cr
  & {du \over dx} = {2 \over 2x + 1}  \cr
  & {dv \over dx} = 1  \cr
  & \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
  &  = x\ln (2x + 1) - \int 1 - {1 \over {2x + 1}}   \cr
  &  = x\ln (2x + 1) - (x - {1 \over 2}\ln |2x + 1|)  \cr
  &  = x\ln (2x + 1) + \ln |(2x + 1)^{1 \over 2}| - x + C  \cr
  &  = x\ln (2x + 1)^{3 \over 2} - x + C \cr} $$

The answer $ = {1 \over 2}(2x + 1)\ln (2x + 1) - x + C$

Where did I go wrong?

Thanks!

Body:
$$ = x\ln (2x + 1) + \ln |{(2x + 1)^{{1 \over 2}}}| - x + C  $$

$$ = \ln(2x + 1)^x +  \ln(2x + 1)^\dfrac 12 - x + C $$

$$= \ln({(2x + 1)^x \cdot(2x + 1)^\dfrac 12}) - x + C $$
$$= \ln{(2x + 1)^\dfrac{2x+1}{2} } - x + C $$
$$= \dfrac {1}{2}\cdot (2x+1) \ln{(2x + 1)} - x + C $$


问:
Show subset of $\ell^2$ is compact
How does one go about showing if the following subsets of $\ell^2$ are compact or not:

$$\{a\in \ell^2 : \|a\|_2=1\}$$

$$\{a\in \ell^2 : |a_k|\leq \frac{1}{k} \forall k\in\mathbb{N}\}$$

I thought that a subset being comapct meant that every sequence $\{x_n\}$ in the subset contains a subsequence that converges to an element in the metric space. But for example the first sequence when $a$ belongs to $\ell^2$ and it's norm is less than one isn't it 'trivial' that the sequence would converge to an element of $\ell^2$? For the second sequence I suppose that $k=0$ which would make the norm tend to infinity so that it's not compact in $\ell^2$?

答:
As Emmanuel has shown your first set is not compact.

We shall show that the second set, which we shall call $A$, is sequentially compact, thus compact. Let $(x_n)\subset A$. We first note that for every $k\in \mathbb N$ we can find an $x^k\in A$ and a subsequence $(x_{n}^k)\subset (x_n)$ such that $|\pi_ix^k-\pi_ix^k_n|\to 0$ for all $i=1,\dots,k$, where $\pi_i$ is just the projection onto the $i$-th coordinate. Note that superscripts are just used to denote a label. Furthermore we can make this choice inductively, so that for each $j<k\in \mathbb N$ we have $\pi_ix^j=\pi_ix^k$ for all $i=1,\dots,j$, and $(x^k_n)\subset (x^j_n)$. Now let us define $x\in A$ by $\pi_i x=\pi_i x^i$ for all $i\in \mathbb N$. We further define a subsequence $(x_{n_j})\subset (x_n)$ by setting $x_{n_j}=x_j^j$  for all $j\in\mathbb N$ (this is what is often referred to as a diagonalisation argument).

Finally, we claim that $x_{n_j}\to x$ in $\ell^2$. Fix $\varepsilon>0$. By definition of $A$ there must exist some $N\in \mathbb N$ such that
$$\sum_{k=N}^\infty|\pi_kx_{n_j}-\pi_kx|^2<\varepsilon$$
for all $j\in\mathbb N$.
Furthermore, by our choice of $x_{n_j}$ and $x$ there must exist a $J\in\mathbb N$ so
$$\sum_{i=1}^{N-1}|\pi_ix_{n_j}-\pi_ix|^2<\varepsilon$$for all $j>J$. Putting these two facts together proves the claim.

Title: Show subset of $\ell^2$ is compact


问:
Evaluating $\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} dx$
This question is not a homework question, but I just searched it on WolframAlpha and just want to know how it's done. WolframAlpha lists the following:

$$\int_{-1}^{0} \frac{1}{\sqrt{1+x^3}} \mathrm dx = \frac{\sqrt{\pi}\Gamma (\frac{4}{3})}{\Gamma (\frac{5}{6})}$$

How can this be evaluated? I'm not too sure where to start.

答:
Substituting $v=-x$ and $u=v^3$ gives
$$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}=\int_0^1\frac{dv}{\sqrt{1-v^3}}=\frac{1}{3}\int_0^1 u^{-\frac{2}{3}}(1-u)^{-\frac{1}{2}}du=\frac{1}{3}\beta\left(\frac{1}{3},\frac{1}{2}\right)=\frac{\Gamma\left(\frac{1}{3}\right)\Gamma\left(\frac{1}{2}\right)}{3\Gamma\left(\frac{5}{6}\right)} $$
Notice that $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ and $\Gamma\left(\frac{4}{3}\right)=\frac{1}{3}\Gamma\left(\frac{1}{3}\right)$, we then get
$$ \int_{-1}^0\frac{dx}{\sqrt{1+x^3}}dx=\frac{\sqrt{\pi}\Gamma\left(\frac{4}{3}\right)}{\Gamma\left(\frac{5}{6}\right)} $$


问:
Is there a closed form for $\int_0^1\,_3F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)dx$?

Is there a closed form evaluation for the integral $$J=\int_0^1 {_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)dx?$$

Context:
I have been investigating integrals of the form
$$e_{p,q}^{n,m}\left({\begin{array}ca_1,..., a_p\\b_1,...,b_q\end{array}}\right)=\int_0^1x^n\left[{_p}F_q\left({\begin{array}ca_1,..., a_p\\b_1,...,b_q\end{array}};x\right)\right]^mdx.$$
Obviously there is no reason to expect a general closed form, but I have found the following:
$$E_1=e_{2,1}^{1,2}(\tfrac12,1;2)=12-16\ln2,\tag1$$
and $$E_2=e_{2,1}^{1,3}(\tfrac13,\tfrac23;\tfrac32)=\frac{27}{32}.\tag2$$
I found these through applying the Lagrange inversion theorem to the functions $x^2-x$ and $x^3-x$, respectively. The proofs are below.

Theorem. We have the explicit evaluation
$$\int_0^1x\left[{_2}F_1(\tfrac12,1;2;x)\right]^2dx=12-16\ln2.\tag{1'}$$
Proof. Using the Lagrange inversion theorem, the function $g(x)$, satisfying $$g(x)^2-g(x)=x,$$
is given by the hypergeometric series $g(x)=-x\,{_2}F_1(\tfrac12,1;2;-4x)$, for $x\in[-1/4,\infty)$. Thus, the function $F(x)={_2}F_1(\tfrac12,1;2;x)$ satisfies $$xF(x)^2=4(F(x)-1),$$
for $x\in(-\infty,1]$. Thus $$E_1=\int_0^1xF(x)^2dx=-4+4\int_0^1F(x)dx.$$
Then using $$_2F_1(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(b)\Gamma(c-b)}\int_0^1\frac{t^{b-1}(1-t)^{c-b-1}}{(1-zt)^a}dt,$$
we have $$F(x)=\frac2\pi\int_0^1\sqrt{\frac{1-t}{t}}\frac{dt}{1-xt}=\frac4\pi\int_0^\infty\frac{t^2dt}{(t^2+1)(t^2+1-x)}=\frac{2}{1+\sqrt{1-x}}.$$
It is then not too difficult to show that $$\int_0^1F(x)dx=\int_0^1\frac{2dx}{1+\sqrt{1-x}}=4-4\ln2,$$
which gives $(1')$ and thus $(1)$.   $\square$

Theorem. We have the explicit evaluation
$$\int_0^1 x\left[{_2}F_1(\tfrac13,\tfrac23;\tfrac32;x)\right]^3dx=\frac{27}{32}.\tag{2'}$$
Proof. The Lagrange inversion theorem gives $g(x)^3-g(x)=x$, for $$g(x)=-x{_2}F_1(\tfrac13,\tfrac23;\tfrac32;\tfrac{27}{4}x^2),\qquad |x|<\frac{2}{3\sqrt3}.$$
Setting $F(x)={_2}F_1(\tfrac13,\tfrac23;\tfrac32;x)$, we have $$4xF(x)^3=27(F(x)-1),$$
and thus
$$E_2=\int_0^1xF(x)^3dx=\frac{27}{4}\left(-1+\int_0^1F(x)dx\right).$$
Then from here and here, we have
$$\int_0^1F(x)dx=\left(\frac32\right)^2\left(\frac{\Gamma(\tfrac12)\Gamma(\tfrac32)}{\Gamma(\tfrac56)\Gamma(\tfrac76)}-1\right)=\frac{9}{8},$$
which is equivalent to $(2)$ and $(2')$.   $\square$

Here is my work on the current problem.

As you may have guessed, we use the Lagrange inversion theorem to see that $g(x)^4-g(x)=x$, where
$$g(x)=-x{_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;-\tfrac{4^4}{3^3}x^3).$$
Setting $F(x)={_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)$, we have
$$xF(x)^4=\frac{4^4}{3^3}(F(x)-1),$$
so that
$$e_{1,4}^{3,2}(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43)=\int_0^1xF(x)^4dx=\frac{4^4}{3^3}(J-1),$$
where $J$ is the integral in the title. It may or may not help, but we can use integral representations of $_pF_q$ to get
$$J=\int_0^1 F(x)dx=\frac{\Gamma(\tfrac43)\Gamma(\tfrac23)}{\pi\Gamma(\tfrac7{12})}\int_0^1\int_0^1\int_0^1\frac{dtdxdz}{x^{1/2}(1-x)^{1/2}t^{1/4}(1-t)^{5/12}(1-txz)^{1/4}}.$$
According to Desmos, the value of $J$ is roughly $J\approx 1.08494289471$, but for some reason I can't get wolfram alpha to get me anything better.
Is there any way to evaluate $J$? Thanks :)

Body:
I did it! We see that $F(x)$ is of the form
$$\,{_{k+1}}F_{k}\left(\tfrac{1}{k+1},\tfrac{2}{k+1},...,\tfrac{k}{k+1};\tfrac{2}{k},\tfrac{3}{k},...,\tfrac{k-1}{k},\tfrac{k+1}{k};x\right),$$
so we have
$$F\left(-r_3\cdot(m(m^3-1))^3\right)=\frac{1}{1-m^3},$$
where $r_3=4^4/3^3$. The above is from eq. (25) here, with $k=3$. Thus, we may write $x=-r_3m^3(m^3-1)^3$, and we have
$$J=\int_0^1F(x)dx=\int_{x=0}^{x=1}\frac{1}{1-m^3}\frac{dx}{dm}dm=3r_3\int_0^\alpha m^2(m^3-1)(4m^3-1)dm,$$
where $\alpha$ is the real root of $r_3\alpha^3(\alpha^3-1)^3+1=0$. Thus,
$$J=\frac{2^7}{3^4}\alpha^3(8\alpha^6-15\alpha^3+6).$$
However, we have $f(\alpha)^3=-1/r_3$, so that $\alpha=g(-r_3^{-1/3})=r_3^{-1/3}F(1)$. But  according to W|A, we have $F(1)=4/3$, so that $\alpha=\frac{1}{\sqrt[3]{4}}$. Thus

$$J=\frac{88}{81},$$ and thus $$e_{3,2}^{1,4}(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43)=\frac{1792}{2187}.$$


问:
Relation between meagre sets
Let $(X, \tau)$ a topological space and $(Y,\tau_y)$ a topological subspace. If $A \subset Y$ is meagre in $Y$, then $A$ is meagre in $X$.

My attempt of solution:

$A$ meagre in $Y$ $\implies$ $A \subset \bigcup_n \{F_n^y;\,\, F_n^y\,\,\, \text{closed in}\,\,\, Y\,\,\,\text{and}\,\,\, \text{int}_y(F_n^y)=\emptyset\}.$

So we have that
$A \subset \bigcup_n \{\overline{F_n^y};\,\, F_n^y\,\,\, \text{closed in}\,\,\, Y\,\,\,\text{and}\,\,\, \text{int}_y(F_n^y)=\emptyset\}$ because $\overline{F_n^y}^y = F_n^y=\overline{F_n^y} \cap Y.$

It is easy to see too that $\text{int}(F_n^y)=\emptyset$ because $\text{int}_y(F_n^y)=\emptyset.$

I need help to prove that $\text{int}(\overline{F_n^y})=\emptyset$

Any help, hint is very appreciated.

答: Let $U=\operatorname{int}\operatorname{cl}F_n^y$, and suppose that $U\ne\varnothing$. Let $x\in U$. Then $x\in\operatorname{cl}F_n^y$, and $U$ is an open nbhd of $x$, so $U\cap F_n^y\ne\varnothing$. But then $U\cap F_n^y$ is a non-empty relatively open subset of $F_n^y$ in $Y$, contradicting the hypothesis that $\operatorname{int}_YF_n^y=\varnothing$.

Title: Relation between meagre sets


问:
How to use Green's Theorem to find area bounded by two curves?
Just starting out for calculus but are you able to help explain how to use Green's Theorem. How to fins the area bounded by the curves:

$x = a\cos^3{t}, y = a\sin^3{t}, (0 \leq t \leq 2\pi)$

How do I use the theorem for this context?

答:
You can take in Green's formula $P=0, Q=x$ or/and $P=-y, Q=0$ and obtain
$$S=\int_\limits{\partial G^+}xdy=-\int_\limits{\partial G^+}ydx=\frac{1}{2}\int_\limits{\partial G^+}(xdy-ydx)$$
Can you finish from here?


问:
Let $X$ and $Y$ be independent exponential random variables with means $\beta_1$ and $\beta_2$ . What is the probability distribution of $Z=|X-Y|$?
We know that $X\sim\text{Exp}(\beta_1)$ and $Y\sim\text{Exp}(\beta_2)$ with the exponential pdf given by

$F_X(x) = \frac{1}{\beta_1}e^{-x/\beta_1}$.

My attempt was to try

$P(Z < z) = P(X < Y+z) - P(X < Y-z)$,

and got it to

$P(X < Y + z) = 1 - \frac{\beta_1 e^{-z/\beta_1}}{\beta_1 + \beta_2}$,

but couldn't work out the other probability since I got that the integral diverges. Any help moving forward is appreciated =)

答:
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$\ds{\bbox[5px,#ffd]{}}$

\begin{align}
\on{P}_{Z}\pars{z} & \equiv \bbox[5px,#ffd]{\int_{0}^{\infty}{\expo{-x/\beta_{1}} \over \beta_{1}}
\int_{0}^{\infty}{\expo{-y/\beta_{2}} \over \beta_{2}}\,
\delta\pars{z - \verts{x - y}}\dd y\,\dd x}
\\[5mm] & =
{\bracks{z > 0} \over \beta_{1}\beta_{2}}\int_{0}^{\infty}\expo{-x/\beta_{1}}
\int_{0}^{\infty}\expo{-y/\beta_{2}}\ \times
\\[2mm] &\
\phantom{= {\bracks{z > 0} \over \beta_{1}\beta_{2}}}\braces{\delta\pars{y - \bracks{x - z}} +
\delta\pars{y - \bracks{x + z}}}\dd y\,\dd x
\\[5mm] & =
{\bracks{z > 0} \over \beta_{1}\beta_{2}}\int_{0}^{\infty}\expo{-x/\beta_{1}}
\,\expo{-\pars{x - z}/\beta_{2}}\,\,\bracks{x - z > 0}\dd x
\\[2mm] &
+ \,\,{\bracks{z > 0} \over \beta_{1}\beta_{2}}\int_{0}^{\infty}\expo{-x/\beta_{1}}
\,\expo{-\pars{x + z}/\beta_{2}}\,\,\bracks{x + z > 0}\dd x
\\[5mm] & =
{\bracks{z > 0} \over \beta_{1}\beta_{2}}\expo{z/\beta_{2}}
\int_{z}^{\infty}\expo{-x/\beta}\,\dd x
\\[2mm] & +
\,\,{\bracks{z > 0} \over \beta_{1}\beta_{2}}\expo{-z/\beta_{2}}
\int_{0}^{\infty}\expo{-x/\beta}\,\dd x
\end{align}
where $\ds{{1 \over \beta} \equiv {1 \over \beta_{1}} +
{1 \over \beta_{2}}}$.

Then,
\begin{align}
\on{P}_{Z}\pars{z} & =
{\bracks{z > 0} \over \beta_{1}\beta_{2}}
\pars{\expo{z/\beta_{2}}\beta\expo{-z/\beta} +
\expo{-z/\beta_{2}}\beta}
\\[5mm] & =
{\bracks{z > 0} \over \beta_{1} + \beta_{2}}
\bracks{\expo{-z/\beta_{1}} + \expo{-z/\beta_{2}}}
\end{align}
\begin{align} \mbox{} \\
&\bbx{\on{P}_{Z}\pars{z} =
\bracks{z > 0}
{\expo{-z/\beta_{1}} + \expo{-z/\beta_{2}} \over
\beta_{1} + \beta_{2}}} \\ &
\end{align}


问:
Joint probability $P(X>0, Z>-X)$ for $X,Z \overset{iid}{\sim} N(0,1)$

Let $X \sim N(0,1)$ and $Z \sim N(0,1)$ be independent normal random variables. Find the probability $P(X>0,Z>-X)$.

This is apparently an elementary calculation but I struggle to understand it. It is given that $X$ and $Z$ are independent but the limit of $Z$ depends on $X$ which is the source of my confusion.

First of all, could we write
$P(X>0,Z>-X)=P(X>0)P(Z>-X)$? If yes, then I can simply proceed as
$$P(X>0,Z>-X) = \int_{0}^{\infty}\phi(x)\,dx \int_{-x}^{\infty}\phi(s)\,ds.$$ But something looks odd. Is this the right direction?

Thanks.

答: As I wrote in my comments, the fact that $X$ and $Z$ are iid $N(0,1)$ random variables tells us that the joint density of $(X,Z)$ (namely $f_{XZ}$) factors as $$f_{XZ}(x,z)=f_{X}(x)f_{Z}(z)=\frac{1}{2\pi}\exp\bigg({-\frac{x^2+z^2}{2}}\bigg)$$ In order to evaluate an expression like $P(X>0,Z>-X)$ you need to integrate your joint density $f_{XZ}(x,z)$ over the region $\mathcal{R}$ in the $(x,z)-$plane given by $$\mathcal{R}=\{(x,z)\in \mathbb{R}^2\big|x>0,z>-x\}$$ This means $$P(X>0,Z>-X)=\int_{0}^{\infty}\int_{-x}^{\infty}f_{XZ}(x,z)dzdx=\frac{3}{8}$$ On the other hand, if we wanted to compute $P(Z>-X)$ we would have to integrate our joint density $f_{XZ}$(x,y) over the region $$\mathcal{S}=\{(x,z)\in\mathbb{R}^2\big|z>-x\}$$ giving us $$P(Z>-X)=\int_{-\infty}^{\infty}\int_{-x}^{\infty}f_{XZ}(x,z)dzdx=\frac{1}{2}$$ Note that $P(X>0)=\frac{1}{2}$ and so $$P(X>0,Z>-X)=\frac{3}{8}\neq \frac{1}{4}=P(X>0)P(Z>-X)$$


问:
A problem about $SS=G?$
Suppose $G$ is a Group, $S$ is a subset of $G$. Given that $\exists a,aS\cap S\neq \emptyset, aS\cup S=G$, prove or give a counterexample: $\forall g\in G,\exists x,y\in S,s.t.xy=g\ (i.e.\; SS=G)$

I can prove the proposition is true when G is a abelian group, the prove is as follow:

wts. $\forall g'\in G, g'S^{-1}\cap S\neq \emptyset $
, here $S^{-1}=\{s^{-1}: s\in S\}$

Anti-evidence method:
if $g'S^{-1}\cap S=\emptyset $
,then $ag'S^{-1}\cap aS=\emptyset$

given that $S\cap aS\ne \emptyset$ and $S\cup aS=G$, so $ag'S^{-1}\subsetneqq S$

proposition 2: if $pS^{-1}\subset S$, then $pS^{-1}=S$

prove: Anti-evidence method: if there is $s$ in S but not in $pS^{-1}$

then $ps^{-1} \in pS^{-1}\subset S$, $p(ps^{-1})^{-1} \in pS^{-1}$

$G$ is a abelian group, so $p(ps^{-1})^{-1}=s\in pS^{-1}$, contradictory.

So the proposition is proved.

so $ag'S^{-1}\subsetneqq S$ leads to a contradiction.

then SS=G is proved.

I wonder whether the proposition holds when G is not a abelian group. Can anyone help me?(thanks)

答:
I post this question on my school's BBS, and finally got the answer. Thank you for all your help.

The proof process is as follows:

$S \cap gS\neq \emptyset,S\cup gS=G$

$\Leftrightarrow S\cap g^{-1}S\neq \emptyset,S\cup g^{-1}S=G$

therefore the "$g$" in the following proposition can be all replaced by "$g^{-1}$".

Proposition 1. If $a\notin S(a\in G)$, then $ga,g^{-1}a\in S$

Proof. Given that $S\cup gS=G$, if $a\notin S$, then $a\in gS$, $g^{-1}a\in S$; similarly $a\in g^{-1}S,ga\in S$.

Proposition 2. $S\cap Sg\neq \emptyset$

Proof by contradiction. Suppose that $S\cap Sg=\emptyset$,

$S\cap Sg=\emptyset \Rightarrow Sg\subset g^{-1}S$（deduced by $S\cup g^{-1}S=G$）

$\Leftrightarrow gSg\subset S\Rightarrow gSg\cap Sg=\emptyset$（deduced by $S\cap Sg=\emptyset$）

$\Leftrightarrow gS\cap S\notin \emptyset$, and this lead to a contradiction.

Back to the main result:

Conclusion. $SS=G$.

Proof by contradiction. suppose that there is $h\notin SS(h\in G)$, then $\forall s\in S, sg,sg^{-1}\notin S$.

Proof. $\forall s\in S$, $h=s(s^{-1}h)=(hs^{-1})s$, so $s^{-1}h,hs^{-1}\notin S$

Let $a=s^{-1}h$, using proposition 1, we can conclude that:

$a=s^{-1}h\notin S \Rightarrow gs^{-1}h\in S\Rightarrow h(gs^{-1}h)^{-1}\notin S \Leftrightarrow sg^{-1}\notin S$. Similarly, $sg\notin S$.

$\forall s\in S,sg\notin S \Rightarrow Sg\cap S = \emptyset$, and this lead to a contradiction with proposition 2!

Q.E.D.

Title: A problem about $SS=G?$


问:
Probability to get 13 cards such that none of the 13 cards have the same number
There are $52$ cards in a deck is distributed equally to $4$ people. What is the probability for one of 4 people to get $13$ cards such that none of them have the same value?

Example:

{2♥, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, Ace♥} is correct

{3♠, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, Ace♥} is incorrect

Note: my teacher said result $\frac{4^{13}}{^{52}C_{13}} \approx 0.0001057$ is incorrect and the order of 13 cards on player's hand does not matter.

My programming experiment:

I know a bit about Python programming, so I did a test, my test specific was:

A deck is distributed equally to $4$ people

Then I check if any player satisfies 13 cards and none of them have the same number $\Rightarrow$ this is a successful experiment and if none of 4 people matches the condition so this is an unsuccessful experiment.

I do this test 100 000 times, add up all the successful experiments and divide it by 100000, the probability always falls between 0.0003 and 0.0005

答:
You should get $13$ different values. First card has $52$ choices, second has $48$ choices, etc down to $4$. So multiply all multiples of $4$ from $4$ to $52$, and divide by the total, which is ${52} \choose 13$.

When you simplify you should get $\frac{4^{13}\times 13!}{52 \choose 13}$


答:
If the question asks for the probability that a particular person's hand doesn't have two cards of same value, then:

Out of the 13 cards in your hand, the first can be any of the 52 cards.

For the second one, it can be any of 48 cards (excluding the first card and 3 other cards of other suits with the same value)

For the third one, it can be any of 44 cards (excluding the first two cards and the cards with same value but different suits)

Continuing similarly, there are $52\times 48\times\dots\times 8\times 4=4^{13}\times 13!$ favorable configurations out of the $52!$ configurations. This results in the answer you gave, ie, $\dfrac{4^{13}\times 13!}{52!}=\dfrac{4^{13}}{\binom{52}{13}}$, so I think your teacher referred to the latter case in the problem.

If the probability is that no person gets two cards of same value, then count configurations where each person gets a hand with all value cards but different suits, so $4!^{13}$ configurations out of a total of $\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}$ configurations (total number of possible deals) because for each value card, the suits in the 4 player hands can be permuted in $4!$ ways for each of the 13 valued cards.

Title: Probability to get 13 cards such that none of the 13 cards have the same number

Body:
There are $52$ cards in a deck is distributed equally to $4$ people. What is the probability for one of 4 people to get $13$ cards such that none of them have the same value?

Example:

{2♥, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, Ace♥} is correct

{3♠, 3♥, 4♥, 5♥, 6♥, 7♥, 8♥, 9♥, 10♥, J♥, Q♥, K♥, Ace♥} is incorrect

Note: my teacher said result $\frac{4^{13}}{^{52}C_{13}} \approx 0.0001057$ is incorrect and the order of 13 cards on player's hand does not matter.

My programming experiment:

I know a bit about Python programming, so I did a test, my test specific was:

A deck is distributed equally to $4$ people

Then I check if any player satisfies 13 cards and none of them have the same number $\Rightarrow$ this is a successful experiment and if none of 4 people matches the condition so this is an unsuccessful experiment.

I do this test 100 000 times, add up all the successful experiments and divide it by 100000, the probability always falls between 0.0003 and 0.0005


问:
Request for crazy integrals
I'm a sucker for exotic integrals like the one evaluated in this post. I don't really know why, but I just can't get enough of the amazing closed forms that some are able to come up with.

So, what are your favorite exotic integral identities, and how do you prove them?

Title: Request for crazy integrals

Body:
We have
$$\int_0^1x\left[{_2}F_1(\tfrac12,1;2;x)\right]^2dx=12-16\ln2,$$
$$\int_0^1 x\left[{_2}F_1(\tfrac13,\tfrac23;\tfrac32;x)\right]^3dx=\frac{27}{32},$$
$$\int_0^1x\left[{_3}F_2(\tfrac14,\tfrac12,\tfrac34;\tfrac23,\tfrac43;x)\right]^4dx=\frac{1792}{2187},$$
and
$$\int_0^1\int_0^1\int_0^1\frac{dtdxdz}{x^{1/2}(1-x)^{1/2}t^{1/4}(1-t)^{5/12}(1-txz)^{1/4}}=\frac{2^{9/4}\cdot11\sqrt{\pi}}{3^{19/8}}\frac{\Gamma(\tfrac13)}{\Gamma(\tfrac14)}\sqrt{\sqrt{3}-1},$$ all from here.

Another cool one:

$$\int_0^\infty\left\{x-\log(2\sinh x)\right\}\cos^2(x)dx=\frac18\left[\frac{\pi^2}{3}+\pi\coth\pi-1\right]$$


答:
I am sure there are more tidy ways to do this but as an alternative...

Why not do $\int \ln(2x + 1)dx$ using:

$v^\prime = 1 \Rightarrow v = x$ and $u = \ln(2x+1)\Rightarrow u^\prime=\frac{2}{2x+1}$

Therefore,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{2x}{2x+1}dx$$

Then make the substitution $u=2x + 1$ to yield,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \int \frac{1}{2}\frac{u-1}{u}du$$

Thus,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}u - ln(u) + c$$

Then,

$$\int \ln(2x + 1)dx = x\ln(2x+1) - \frac{1}{2}(2x-1) - \ln(2x-1) + c$$

$$= \ln(2x-1)(x-1) - \frac{1}{2}(2x-1) + c$$


答:
You need to count the number of ways in which each person can receive one card of each denomination. Each denomination can be distributed to the $4$ players in $4!$ different ways, so there are $4!^{13}$ different deals that give each player one card of each denomination. There are

$$\binom{52}{13}\binom{39}{13}\binom{26}{13}\binom{13}{13}=\binom{52}{13,13,13,13}=\frac{52!}{13!^4}$$

possible deals, so the desired probability is

$$\frac{4!^{13}}{\frac{52!}{13!^4}}=\frac{4!^413!^4}{52!}\,,$$

which is a bit more than $6\times 10^{-24}$.

Added: This is the probability that each player receives one card of each denomination. Subsequent edits make it unlikely that this is the intended interpretation of the question. If what is desired is the probability that at least one player receives a card of each denomination, or that exactly one player does so, the calculation is significantly messier.


答: Imagine arranging the cards in 13 rows and 4 columns. The first row contains all four aces, the second row contains all four kings, the third row contains all four queens, etc. Assign the first column of cards to person #1, the second column of card to person #2, the third column of cards to person #3, and the fourth column of card to Person #4. Since are $4!$ ways to arrange the cards in each row, the total number of such arrangements is $(4!)^{13}$ making the desired probability $\frac{(4!)^{13}}{{52 \choose 13}{39 \choose 13}{26 \choose 13}{13 \choose 13}}$


问:
$\lim_{r\to\infty} \int_0^\infty rx/(exp(x) + r^2x)dx = 0$
I've got a feeling I need to use the Dominated Convergence Theorem, but I'm honestly not sure how to even begin tackling this.

I realize the limit of the integrand will ultimately be $0$ as the denominator will be dominated by n, but I'm not sure how to continue.

答:
Let $f_n(x)$ be the integrand of your problem, so if we were to differentiate $f_n(x)$, we discover that

$$
f_n'(x)={ne^x+n^3x-nxe^x-n^3\over(e^x+n^2x)^2}
$$

and by some further manipulation we see that this function attains maximum at $x=1$, indicating

$$
0\le f_n(x)\le{n\over e+n^2}
$$

Thus we have

$$
\lim_{n\to\infty}\sup_{x\ge0}|f_n(x)-0|=\lim_{n\to\infty}{n\over e+n^2}=0
$$

Therefore, $f_n(x)$ converges uniformly to zero on $[0,\infty)$, which justified the interchanging:

$$
\lim_{n\to\infty}\int_0^\infty f_n(x)\mathrm dx=\int_0^\infty\lim_{n\to\infty}f_n(x)\mathrm dx=0
$$


问:
Limit of $\sqrt[10]{n^{10} + 8n^9} - n$ as $n \rightarrow \infty$ using two "standard" limits
I must use the two standard limits $\lim\limits_{n \rightarrow \infty} \frac{e^{\alpha_n}-1}{\alpha_n} = 1$ and $\lim\limits_{n \rightarrow \infty} \frac{\ln(1+\beta_n)}{\beta_n} = 1$ if  $\alpha_n, \beta_n \rightarrow 0$ so multiplying by conjugate approach won't work.

My attempt:

$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n) = \lim\limits_{n \rightarrow \infty} e^{\ln(\sqrt[10]{n^{10} + 8n^9} - n)} = e^{\lim\limits_{n \rightarrow \infty} (\ln(\sqrt[10]{n^{10} + 8n^9} - n))}$

I am not sure how to proceed further. Any help/hints will be appreciated.

答:
$$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n)=\lim\limits_{n \rightarrow \infty}n \frac{\sqrt[10]{1+\frac{8}{n}}-1}{\frac{8}{n}}\cdot\frac{8}{n}=\frac{8}{10}$$
Using, for $x \to 0$
$$\frac{e^{\alpha \ln (1+x)}-1}{\alpha \ln (1+x)}\cdot \frac{\alpha \ln (1+x)}{x}=\frac{e^{\alpha \ln (1+x)}-1}{x}=\frac{(1+x)^\alpha-1}{x}\to \alpha$$


答:
Here is a cute variant. Let $u=\ln(2x+1)$ and let $dv=dx$. Then $du=\frac{2}{2x+1}$ and (this is the cute part) we can take $v=x+\frac{1}{2}$. It follows that
$$\int \ln(2x+1)\,dx=\left(x+\frac{1}{2}\right)\ln(2x+1)-\int \left(x+\frac{1}{2}\right)\frac{2}{2x+1}\,dx.$$
But the remaining integrand is just $1$! It follows that our integral is
$$\left(x+\frac{1}{2}\right)\ln(2x+1) -x+C.$$


答:
First, we have

$$\sqrt[10]{n^{10}+8n^9}-n=e^{\log(\sqrt[10]{n^{10}+8n^{9}}) }-n\ne e^{\log(\sqrt[10]{n^{10}+8n^{9}}-n) }$$

Second, $\lim_{n\to\infty}\frac{e^{\alpha_n}-1}{\alpha_n}=1$ and $\lim_{n\to\infty}\frac{\log(1+\beta_n)}{\beta_n}=1$ for $\alpha_n\to 0$ and $\beta_n\to 0$, respectively.

So, how shall we proceed?

Let's first write $\sqrt[10]{n^{10}+8n^9}-n$ as

$$\begin{align}
\sqrt[10]{n^{10}+8n^9}-n&=n\left(\sqrt[10]{1+\frac8n}-1\right)\\\\
&=n\left(e^{\frac1{10}\log\left(1+\frac8n\right)}-1\right)\\\\
&=n\left(\frac{e^{\frac1{10}\log\left(1+\frac8n\right)}-1}{\frac1{10}\log\left(1+\frac8n\right)}\right)\times \left(\frac{\frac1{10}\log\left(1+\frac8n\right)}{\frac8n}\right)\times\left(\frac8n\right)\\\\
\end{align}$$

Can you finish now?

Title: Limit of $\sqrt[10]{n^{10} + 8n^9} - n$ as $n \rightarrow \infty$ using two "standard" limits

Body:
I must use the two standard limits $\lim\limits_{n \rightarrow \infty} \frac{e^{\alpha_n}-1}{\alpha_n} = 1$ and $\lim\limits_{n \rightarrow \infty} \frac{\ln(1+\beta_n)}{\beta_n} = 1$ if  $\alpha_n, \beta_n \rightarrow 0$ so multiplying by conjugate approach won't work.

My attempt:

$\lim\limits_{n \rightarrow \infty} (\sqrt[10]{n^{10} + 8n^9} - n) = \lim\limits_{n \rightarrow \infty} e^{\ln(\sqrt[10]{n^{10} + 8n^9} - n)} = e^{\lim\limits_{n \rightarrow \infty} (\ln(\sqrt[10]{n^{10} + 8n^9} - n))}$

I am not sure how to proceed further. Any help/hints will be appreciated.


答:
Calculation based on first card is free.  Get probability all remaining cards have to be different.  Second card is $\frac{48}{51}$ third card is $\frac{44}{50}$ summary is $\frac{12!\times 4^{12}\times 39!}{51!}=0.000105681$

This is the probability for one particular player getting it right.  Since there are four players the question seems to be at least one gets the desired result.  That is approximately $4$ times as much, ignoring dependency of the hands.

............
exact calculation.
Let $P_1$ be the result of the above calculation.  $P_2= \frac{12!\times3^{12}\times26!}{38!}=0.0001962866$ and  $P_3=\frac{12!\times2^{12}\times13!}{25!}= 0.0007876469$

The probability at least one good hand $=P_1$, the probability of at least two good hands is $Q=P_1P_2$, and the probability of all four hands are good is $R=P_1P_2P_3$.

One or more good hands occurs four times each includes two or more three times, while each two or more includes one all.  Adding it all up and removing duplicates the final probability is $4P_1-6Q+3R$.


问:
Prove $1^r+2^r+.....\equiv 0\pmod p$ , given that p is odd prime and $0<r<p-1$
An exercise reads as follows:
If $p$ is an odd prime and $0<r<p-1$ then prove that $s=1^r+2^r+...(p-1)^r\equiv 0\mod p$

I know one clever way to prove it is to use a primitive root $g$ and write the given sum as: $g^r+g^{2r}+g^{3r} +...+g^{(p-1)r}$ which is a geometric process easy to handle and deduce the result.

But I am looking for a more primitive way right now (suppose I know nothing about primitive roots). My thought is: the polynomial $$f=x^{p-1}-1$$ has $p-1$ roots distinct roots in the field $\Bbb{Z_p}$ so one can also write $$f= (x-1)(x-2)(x-3)...(x-(p-1))$$. By inspection of the coefficients of the 2 equal polynomials we get

$t_1=1+2+3+...+(p-1)\equiv0 \mod p$

$t_2=1\cdot 2+1\cdot 3+...+(p-2)\cdot( p-1)\equiv 0 \mod p$

$t_3=1\cdot 2\cdot3+1\cdot 2\cdot4+...+(p-3)\cdot (p-2)\cdot(p-1)\equiv 0\mod p$

and so on ....

now suppose $r=2$ then $s=t^2_1 -2t_2$, where we deduce that $p$ divides $s$

But how in general to compute $s$ by the $t_i$ mentioned above? Is there an algorithm for that? Thanks a lot.

答:
There is another way to do this, I am not sure if this type of approach is what you had in mind:

Claim 1: Let $H$ be a subgroup [under multiplication] of $(\mathbb{F}_p)^{\times}$ that satisfies $|H| \ge 2$. Then $\sum_{a \in H} a \equiv_p 0$.

Indeed, we first note the following: For any $a_0 \in H$ the sets $\{a; a \in H\}$ and $\{a_0a; a \in H\}$ are the same. Thus for any $a_0 \in H; a_0 \not = 1$; we note the following:

$$a_0\sum_{a \in H} a  = \sum_{a \in H} a_0a \equiv_p \sum_{a' \in H} a' =\sum_{a \in H} a.$$

However, as $a_0 \not =1$ the above implies that $\sum_{a\in H} a$ must be 0 mod $p$ and so Claim 1 follows. $\surd$

Now for each positive integer $r$, the set of $r$-th powers of $(\mathbb{F}_p)^{\times}$ form a subgroup $H_r$ of $(\mathbb{F}_p)^{\times}$, and for all $a \in H_r$ the number of solutions to $x^r=a$ is the same $\frac{p-1}{|H_r|} < p-1$ for all $r<p-1$ [because $(\mathbb{F}_p)^{\times}$ is cyclic and so has an element of order $p-1$]. Thus $|H_r| >1$ and so

$$\sum_{x \in \mathbb{F}_p} x^r = 0^r + \frac{p-1}{|H_r|}  \sum_{a \in H_r} a, $$

is indeed divisible by $p$  by Claim 1.


问:
Derivative of conjugate transpose of matrix
Building off of my previous question, I am trying to derive the normal equations for the least squares problem:

$$
\min_W \|WX - Y\|_2 \\
W \in \mathbb{C}^{N \times N} \quad X, Y \in \mathbb{C}^{N \times M}
$$

The intuitive way of viewing this problem is that I am trying to predict a vector $y$ (of length $N$) from a corresponding $x$ vector using a matrix $W$, and to estimate $W$ I have multiple ($M$, to be precise) realizations of $x$ and $y$ packed into matrices $X$ and $Y$.

I am trying to define this in terms of a least-squares problem and derive the normal equations myself, but I'm running into issues in taking the derivative.  To spell it out explicitly, I can re-state the above equation as:

$$
\min_W (WX - Y)^H (WX - Y) \\
= \min_W (X^H W^H - Y^H) (WX - Y) \\
= \min_W (X^H W^H W X - Y^H W X - X^H W^H Y + Y^H Y)
$$

Now typically I would take the derivative with respect to $W$, set it equal to zero, and solve for $W$.  However, my matrix calculus is rusty and everything I know is basically summed up on this webpage, where it explicitly states:

Note that the Hermitian transpose is not used because complex conjugates are not analytic.

Now, because of Michael C. Grant's answer I feel there must be some way of doing this, but I am at a loss as to how. Thank you all in advance!

答:
In the first version of my answer I overlooked the fact that you defined the $L_2$ error in a way that is not suitable for matrices. For vectors it would have been OK, but if your complex error is $E=WX-Y$, then $E^HE$ is not the error that you want to minimize, because $E^HE$ is a matrix (if $E$ were a vector, then $E^HE$ would be a scalar and you could minimize it). The appropriate definition of the (scalar) error measure is given by the Frobenius norm

$$\epsilon=\|E\|_F^2=\|WX - Y\|_F^2=\\
=\text{trace}\left ( E^HE\right)=\text{trace}\left ( (WX-Y)^H(WX-Y)\right)=\\
=\text{trace}\left ( X^HW^HWX - X^HW^HY - Y^HWX +Y^HY\right)$$

Now we can apply the trick of taking the conjugate complex derivative (reference) w.r.t. $W^H$:

$$\frac{\partial\epsilon}{\partial W^H} = WXX^H - YX^H$$

Setting the derivative to zero gives the solution

$$W=YX^H(XX^H)^{-1}$$

assuming $(XX^H)^{-1}$ exists.

EDIT: just an additional simple example to show how the trick with the conjugate derivative works when solving for an unknown complex vector (instead of a matrix). In this case we get a well-known result:

Assume we have an overdetermined system of complex linear equations

$$Ax=b$$

where $A$ is $m\times n$, $m>n$, $x$ is $n\times 1$, and $b$ is $m\times 1$. Minimizing the squared error

$$\epsilon=(Ax-b)^H(Ax-b)=x^HA^HAx - x^HA^Hb-b^HAx+b^Hb$$

by taking the derivative w.r.t. $x^H$ and equating it with zero gives

$$A^HAx-A^Hb=0$$

which leads to the well-known Moore-Penrose pseudoinverse:

$$x=(A^HA)^{-1}A^Hb$$

EDIT 2: Let's have a look at two little examples showing that it is correct to regard a complex variable $z=z_R+iz_I$ as constant when taking the derivative w.r.t. $z^*$. The conjugate derivative is defined as (ref.)

$$\frac{\partial f}{\partial z^*}=\frac{1}{2}\left [\frac{\partial f}{\partial z_R} +i\frac{\partial f}{\partial z_I} \right]$$

For $f=z$ we get

$$\frac{\partial f}{\partial z^*}=
\frac{1}{2}\left [\frac{\partial (z_R+iz_I)}{\partial z_R} +i\frac{\partial (z_R+iz_I)}{\partial z_I} \right]=\frac{1}{2}(1+i^2)=0$$

For $f=zz^*$ we have

$$\frac{\partial f}{\partial z^*}=
\frac{1}{2}\left [\frac{\partial (z_R^2+z_I^2)}{\partial z_R} +i\frac{\partial (z_R^2+z_I^2)}{\partial z_I} \right]=\frac{1}{2}(2z_R+2iz_I)=z$$


问:
Calculate the number of different ways of tiling 'special' rooms [Read Full Que. Below]
It's a really long question so please bear with me. I will write my approach in the comments since the question is already so big.

You have to tile a room that is two units wide and N units long.  You have with you two types of tiles: a rectangle that is one unit wide and two units long, and an L-shaped tile covering three square units.  Here are pictures of the two types of tiles.

You are allowed to rotate the tiles when you lay them.  You have an infinite supplyof both types of tiles.  However, your architect is from Nauru and is a believer in thePolynesian system of tiling calledBatlar  Guntuwhich forbids any corner where fourtiles meet.For instance, here are some ways to tile a 2×4 room.  The first three are consistentwith theBatlar Guntusystem while the fourth tiling is not

Your  task  is  to  calculate  the  number  of  different  ways  a  room  can  be  tiled  withoutviolating the principles ofBatlar Guntu.  For instance, a 2×3 room can be tiled in 5different ways, as follows:

Calculate the number of different ways of tiling rooms of the following sizes:

a) 2 x 8

b) 2 x 10

c) 2 x 12

答:
Any such tiling will be built out of these block patterns:

with the rules

There are no adjacent $B$ blocks (the only way to have 4 corners together)

$E$ can only be followed by $F$ or $G$

$F$ can only follow $E$ or $F$ and only be followed by $F$ or $G$

$G$ can only follow $E$ or $F$

$H$ can only be followed by $I$ or $J$

$I$ can only follow $H$ or $I$ and only be followed by $I$ or $J$

$J$ can only follow $H$ or $I$

From this you can build some recursion formulas:. Let $a_n$ be the number of tilings for a $2\times n$ room, and let $b_n$ be the number of tilings that do not end with a $B$ block. Setting $a_n =b_n = 0$ for $n < 0$, and $a_0 = b_0 = 1$, for each $n$ there will be

$a_{n-1}$ tilings ending in an $A$ block.

$b_{n-2}$ tilings ending in a $B$ block.

$a_{n-3}$ tilings ending in a $C$ block, and similarly for $D$.

$a_{n-4}$ tilings ending in $EG$, and similarly for $HJ$.

$a_{n-4-2k}$ tilings ending in $EF^kG$, and similarly for $HI^kJ$.

So $$a_n = a_{n-1} + b_{n-2} + 2a_{n-3} + 2\sum_{k \le (n-4)/2} a_{n-4-2k}\\b_n = a_n - b_{n-2}$$

The $b_{n-2}$ term can be replaced by a sum of $a_k$ values to get a recursion in one dimension. Taking the difference between $a_n$ and $a_{n-2}$ results in a simpler recursion formula, having only a summation for the replaced $b_{n-2}$ term. The same trick can be applied a second time to remove this remaining sum, leaving a linear recursion formula in $a_n$ of fixed length. I'll leave the details of it to you.


问:
Find the angle in the drawing

In the below drawing, we are given:
Triangle ABC is right. D is the midpoint of AB. Angle $ACD = φ$ and $BCE = φ$.
Angle $EAB = φ$.
$ED=α$ and $CD=4a$.

We are looking for angle φ (to be solved by using Geometry, not trigonometry).

What I have tried so far:

Since angles EAB and ECB are both φ, and AB is vertical to BC, then their other sides must also be vertical, so AEC is right.

Therefore the quadrilateral AEBC is inscribed in a semicircle (which I have drawn to see if I get any clue out of it) and AC is its diameter.

If G is the center of the circle, then EG and BG are equal to the circle radius.
Also angle $BDC = φ + BAC = EAC$ so triangles EAC and BDC are similar.

But I don't know how to use $ED=a$ and how to combine all this, to calculate angle φ.
By the way, I have found it in Geogebra to be ~19 degrees.

答:
Alright, I constructed the geometry and verified that the angle is 19 degrees, and I can't tell you how to calculate this angle without trigonometry, but I can tell you how this structure is built, and what the lengths $ \alpha $ and $ 4 \alpha $ reveal. Buckle up!

$$ \widehat {ACE} = \widehat {DCB} $$
Therefore the right triangles $ ACE $ and $ DCB $ are similar. As a result:
$$ \frac{AC}{DC} = \frac{AE}{DB} \qquad(1)$$
Now recall that $ CD $ is the median of $ AB $, so
$$ AD = DB $$
And replacing in (1) we will have:
$$ \frac{AC}{DC} = \frac{AE}{AD} \qquad(2)$$
Now consider the triangles $ACD$ and $EAD$. They have an equal angle ($\varphi$) and equal proportions of the lengths of the sides of that angle, as shown in (2). Therefore these two triangles are similar. And here is an interesting result:
$$ \frac{DC}{AD} = \frac{AD}{DE} $$
Or
$$ AD^2 = DC.DE $$
Now, recalling that $DE = \alpha$ and $DC = 4\alpha = 4DE$  we have:
$$ AD^2 = 4DE^2 $$
$$ AD = 2DE \qquad(3) $$
Et Voila!
$$ DC = 4DE = 2AD = 2DB \qquad(4) $$
Look at the right triangle $DBC$ : side $DB$ is half the hypothenuse $CD$ , which means angle $\widehat{DCB}$ must be 30 degrees.

So, to construct the angle $\varphi$ , all we need to do is draw the right triangle $DBC$ , extend the side $BD$ by equal length from point $D$ to find point $A$ , and connect $A$ to $C$ . The angle $\widehat{ACD}$ will be our $\varphi$ .

Title: Find the angle in the drawing


问:
Norm of a polynomial $N_{E/K}(f)(X)$ with $E/K$ a Galois extension
I am quite new to Galois Theory and tried to give a go at this exercise:

Let $E/K$ be a finite Galois extension with $G=\text{Gal}(E/K)$. For each polynomial $f(X)\in E[X]$ we define:
$$N_{E/K}(f)(X)=\prod_{\sigma \in G} f^{\sigma}(X)$$

Check that $N_{E/K}(f)(X) \in K[X]$.

Prove that if $N_{E/K}(f)(X)$ is irreducible in $K[X]$, then $f(X)$ is irreducible in $E[X]$. Is the converse true?

Prove that if $N_{E/K}(f)(X)$ is irreducible in $K[X]$ then the coefficients of $f$ generate $E$ over $K$.

Let $f(X)\in E[X]$ be an irreducible polynomial and let $\alpha$ be a root of this polynomial in an algebraic closure $\bar{E}$. Prove that the normal closure of the extension $E(\alpha)/K$ in $\bar{E}$ is the splitting field of the polynomial $N_{E/K}(f)(X)$.

I thought I understood the theory, but I don't know where to begin. For (1), I'd like to see that the coefficients of the polynomial are in $K$, but how do I continue? I know $\sigma$ fixes $K$ but $f(X)\in E(X)$. For the others, I lay the definitions and try to go for a contradiction but nothing comes from it.

I've looked all over the reference books and the internet and can only find similar definitions regarding elements of $E$, not polynomials. Any help, indications or references would be greatly appreciated.

答:
$\sigma(f(X))$ means applying $\sigma$ to the coefficients, usually denoted $f^\sigma(X)$ (well KCd prefers the notation $(\sigma f)(X)$) and the subring of $E[X]$ fixed by $Gal(E/K)$ is $K[X]$.

is obvious

for $E/F/K$ then $F=E$ iff $F$ is not fixed by any $\sigma$.

the roots of $f^\sigma$ are $K$-conjugates to the roots of $f$.


问:
Can a closed (non-trivial) subspace of an incomplete vector space be complete?
While thinking about the statement:

A subspace of a complete vector space is closed if and only if it's complete.

I was trying to drop the first "complete" and see what gets broken.

And my question is:

Can a closed (proper) subspace of an incomplete vector space be complete?

If the answer is yes, I'd be happy to see some relatively simple example (over $\mathbb R$), since I haven't had functional analysis yet.

If the answer is no, please give an example of an incomplete but closed subspace of an incomplete vector space over $\mathbb R$. For this case, I was thinking about $C^1(\mathbb R)$ and $f_n=|x|^{1+\frac{1}{n}}$ yet I couldn't think of any closed (proper) subspace with $f_n$.

Thanks a million!

答:
Yes, it can. Just take an incomplete space $V$ and a complete one $W$; then $V$ can be identified with a closed subspace of $V\oplus W$ with the product topology (that is normed if $V$ and $W$ are); this subspace of $V\oplus W$ is not complete, but closed, because it is $\{(v,w):w=0\}$.

Another example: take a noncomplete normed space (for instance the continuous functions on $[0,1]$ under the $L^1$-norm). Any finite dimensional subspace is complete.


问:
non-homogenous poisson process
buses arrive at a bus stop according to a non-homogeneous Poisson process with the rate
function $$\mu(t) = 1 + t$$
I have some idea on how to do this but I don't know if I'm correct.

Q1: Assuming the first bus arrives at bus stop at time t = 2, what is the probability that one has to wait more than 3 time units for the second bus to arrive?

So if the first bus arrive at t = 2, then is the probability of one has to wait more than 3 time units from the second bus to arrive just $$P(N([2,5]) = 0)?$$

Q2: Assuming that 2 buses has arrived at the bus stop at t = 6, find the probability that the first bus arrived at the bus stop at t = 2?

So $$N([0,6]) = 2$$,  and $$N(]0,2]) = 1,$$
$$P(N([0,2]) = 1|N([0,6]) = 2) = \frac{P(N([0,2]) = 1, P(N([2,6]) = 1)}{P(N([0,6]) = 2)}$$
Is this correct?

答: Let $t_0,t_1,\ldots ,t_n\in[2,5]$ such that $2=t_0<t_1<\ldots<t_{n-1}<t_n=5$. For each $1\leq i \leq n$ take $\Delta t_i=t_i-t_{i-1}$ and let $X_i$ denote the number of arrivals that occur in the time interval $[t_{i-1},t_i)$. Notice how $X_i$ is approximately $\text{Poisson}\Big(\mu(t_i)\Delta t_i \Big)$ and that $X_1, \ldots ,X_n$ are independent. Hence $$P\Big(\text{No arrivals in} [2,5]\Big) \approx \prod_{i=1}^nP(X_i=0)=\exp{\bigg(-\sum_{i=1}^n\mu(t_i)\Delta t_i\bigg)}$$ Taking $n \longrightarrow \infty $ gives us $$P\Big(\text{No arrivals in} [2,5]\Big)=e^{-\int_2^5\mu(t)dt}=\frac{1}{e^{13}\sqrt{e}}$$ In fact, if $N\Big([a,b]\Big)$ denotes the number of arrivals in $[a,b]$, then it turns out that $$N\Big([a,b]\Big) \sim \text{Poisson}\bigg(\int_a^b\mu(t)dt\bigg)$$

Title: non-homogenous poisson process


问:
Null space matrix
Once a matrix is put into RREF, it can be re-arranged as R=[I F], the identity variables followed by free variables. The null space matrix is that which, multiplying R, gives the zero vector. In the MIT open courseware lecture, Dr. Strang says this can be expressed as $N=\begin{bmatrix} -F \cr I \end{bmatrix}$, making RN=0. I understand how this works in theory, but in practice I'm having difficulty setting up N once I figure out R. He says the columns of N are the special solutions. I think I'm missing a conceptual intuition.
For example, after Gauss-Jordan Elimination, I have the matrix R:

$\begin{bmatrix} 1\text{ }3\text{ }0\text{ }0\text{ }3\cr 0\text{ }0\text{ }1\text{ }0\text{ }9\cr0\text{ }0\text{ }0\text{ }1\text{ -}4 \end{bmatrix}$

If I rearrange columns so the pivot columns are followed by the free columns, what does N look like?

Title: Null space matrix

Body:
$$R = \left[\begin{array}{ccc|cc} 1 & 0 & 0 & 3 & 3\\ 0 & 1 & 0 & 0 & 9 \\ 0 & 0 & 1 & 0 & -4 \end{array}\right]$$

$$N = \left[\begin{array}{cc}-3 & -3 \\ 0 & -9 \\ 0 & 4\\ \hline  1 & 0 \\ 0 & 1\end{array}\right]$$


Body:
$$(\sqrt[10]{n^{10} + 8n^9} - n)=n\left(\sqrt[10]{1+\frac{8}{n}}-1\right) $$
$$\sqrt[10]{1+\frac{8}{n}}=\sum_{k=0}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}=1+\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^k}$$
$$n\left(\sqrt[10]{1+\frac{8}{n}}-1\right)=\sum_{k=1}^\infty 8^k \binom{\frac{1}{10}}{k}\frac 1{n^{k-1}}=\frac{4}{5}-\frac{72}{25 n}+\cdots $$


问:
Consecutive Integers Pythagorean Triplets
I have a question for which I was not able to find an answer online. I was wondering how many Pythagorean Triplets we have found till now which consists of three consecutive integers like $(3,4,5)$.

答:
For slightly simpler numbers, let $x$ be the middle number then you have to solve
$$
\begin{split}
(x-1)^2 + x^2 &= (x+1)^2 \\
x^2 &= (x+1)^2 - (x-1)^2 = 2x\cdot2
\end{split}
$$
which yields $x=0$ and $x=4$. Can you finish?


问:
Compute the Eigenvectors & Show A is diagonalizable
$A =
        \begin{bmatrix}
        1&2&1 \\
        0&1&0 \\
        1&3&1 \\
        \end{bmatrix}
$

I computed the eigenvalues:

$λ_ 1 = 1$

$λ_ 2 = 0$

$λ_ 3 = 2$

The corresponding eigenvalues I computed are:

$X_{λ1} =
        \begin{bmatrix}
        x_1 \\
        (-1/2) x_3  \\
        x_3 \\
        \end{bmatrix}
$ Substitute $x_1$ as 1, $x_3$ as 2: $X_{λ1} =
        \begin{bmatrix}
        x1 \\
        -1  \\
        2 \\
        \end{bmatrix}
$

$X_{λ2} =
        \begin{bmatrix}
        -x_3 \\
        0  \\
        x_3 \\
        \end{bmatrix}
$ Substitute $x_3$ as 1: $X_{λ2} =
        \begin{bmatrix}
        -1 \\
        0  \\
        1 \\
        \end{bmatrix}
$

$X_{λ3} =
        \begin{bmatrix}
        x_3 \\
        0  \\
        x_3 \\
        \end{bmatrix}
$ Substitute $x_3$ as 1: $X_{λ3} =
        \begin{bmatrix}
        1 \\
        0  \\
        1 \\
        \end{bmatrix}
$

To show A is diagonalizable:

$S =
        \begin{bmatrix}
        1&-1&1 \\
        -1&0&0  \\
        2&1&1 \\
        \end{bmatrix}
$

$S^{-1}AS = D$

AS = SD

After matrix multiplication, I got

$
        \begin{bmatrix}
        1&0&2 \\
        -1&0&0  \\
        0&0&2 \\
        \end{bmatrix}
$ = $
        \begin{bmatrix}
        1&0&2 \\
        -1&0&0  \\
        2&0&2 \\
        \end{bmatrix}
$

I know I made a mistake somewhere because AS does not equal SD. I can't find my mistake. Any help would be appreciated.

答: Just to remove this question from the "Unanswered" queue, it's clear from the comments that Christina's problem has been solved.</p>


问:
Is this proof of a ring being Noetherian correct or we should prove that it is left Noetherian and then right Noetherian and why?
Here is the question I want to answer:

Let $\varphi : R \rightarrow S$ be a surjective homomorphism of commutative rings. Show that, if $R$ is Noetherian, then $S$ is Noetherian.

Here is a trial to the solution:

Let $\varphi : R \rightarrow S$ be a surjective homomorphism of commutative rings and assume that $R$ is Noetherian. We want to show that $S$ is Noetherian.

Now, since we know the following definition of a Noetherian ring : a commutative ring $S$ is Noetherian iff it satisfies the $ACC$ condition. And since we know that a commutative ring $S$ has the ascending chain condition $(ACC),$ if, for every ascending chain of ideals $$I_1 \subset I_2 \subset I_3 \subset \dots $$ in $S,$ there is an $N > 0,$ so that $I_n = I_{n+1}$ for $n \geq N.$

So, let $I_1 \subset I_2 \subset I_3 \subset \dots $ be an ascending chain of ideals in $S,$ Now since we have a surjective ring homomorphism, then the preimage of the ideal $I_k$ in $S$ (i.e. $f^{-1}(I_k)$) is an ideal of $R$(to be proved at the end). And similarly for all other ideals. So we get the following ascending (we will prove later why it is ascending) chain of ideals in $R,$ $$f^{-1}(I_1) \subset f^{-1}(I_2) \subset f^{-1}(I_3) \subset \dots \subset f^{-1}(I_k) \subset \dots $$

Now, since by assumption, $R$ is Noetherian, so its ascending chain of ideals terminate. That is $\exists N \in \mathbb Z$ such that  $f^{-1}(I_n) = f^{-1}(I_{n+1})$ for all $n \geq N.$

Now since $f$ is surjective, we have $$f (f^{-1}(I_k)) = I_k, \quad  \forall k > 0.$$

Therefore, it follows that we have $$ I_n = I_{n+1}, \quad \forall n \geq N.$$
So, each ascending chain of ideals of $S$ terminates, as we have taken an arbitrary ascending chain of $S$ and so $S$ is Noetherian as required.

My questions are:

1-Is this proof correct? or we should prove that $S$ is left Noetherian and then right Noetherian? If so, why?

2- why we are sure that we get an ascending chain of ideals in $R$?

答:
As has been pointed out in the comments to the question itself, and is in fact easily seen, the hypothesis that $R$ and $S$ be commutative implies that every ideal in either ring is two-sided, so we needn't separately consider left or right ideals.

It strikes me that the essential fact here is the

Assertion that for surjective homomorphisms

$\varphi:R \to S \tag 1$

the image $\varphi(I)$ of an ideal

$I \subset R \tag 2$

is in fact an ideal in $S$.

It should be noted that if we lift the hypothesis that $\varphi$ is surjective, we still have that

$I = \varphi^{-1}(J) \subset R \tag 3$

is an ideal for any ideal

$J \subset S; \tag 4$

for if

$i_1, i_2 \in I, \tag 5$

we have

$\varphi(i_1), \varphi(i_2) \in J, \tag 7$

whence

$\varphi(i_1 - i_2) = \varphi(i_1) - \varphi(i_2) \in J; \tag 8$

thus

$i_1 - i_2 \in I; \tag 9$

also, for

$r \in R, \tag{10}$

$\varphi(ri_1) = \varphi(r) \varphi(i_1) \in J, \tag{11}$

since $J$ is an ideal; from this,

$r i_1 \in I; \tag{12}$

(9) and (12) together show that $I$ is an ideal in $R$.

The present Assertion is thus in a sense a logical complement of the above result, for it allows the affirmation that $I$ is a ideal if and only if $J$ is in the event of surjective $\varphi$.

Now once again assuming that $\varphi$ is surjective, we let $I$ be an ideal in $R$ and let

$J = \varphi(I) \tag{13}$

as a set.  Then for

$j_1, j_2 \in J \tag{14}$

we may find

$i_1, i_2 \in I \tag{15}$

such that

$\varphi(i_1) = j_1, \; \varphi(i_2) = j_2; \tag{16}$

$j_1 - j_2 = \varphi(i_1) - \varphi(i_2) = \varphi(i_1 - i_2) \in J, \tag{17}$

and if

$s \in S, \tag{18}$

using the surjectivity of $\varphi$ we have some

$r \in R \tag{19}$

with

$\varphi(r) = s; \tag{20}$

and now

$sj_1 = \varphi(r) \varphi(i_1) = \varphi(ri_1) \in J, \tag{21}$

since

$ri_1 \in I; \tag{22}$

(17) and (21) in concert show that $J$ is in fact an ideal in $S$, proving the Assertion.

Thus we see that, in the event $\varphi$ is surjective, $I$ is an ideal in $R$ if and only if $J = \varphi(I)$ is an ideal in $S$.

It is a short step from this to the requested result, for if $R$ is Noetherian and $J_k$ is an ascending sequence of ideals in $S$, that is

$J_i \subset J_{i + 1}, \tag{23}$

we can form the corresponding sequence of ideals

$I_i = \varphi^{-1}(J_i) \subset R; \tag{24}$

we know the $I_i$ are ideals in $R$ by virtue of what has just been proven in the above.  Now since $R$ is Noetherian, the sequence $I_i$ stabilizes at some point, whence

$\exists n \in \Bbb N, \; I_j = I_n \; \text{for} \; j \ge n; \tag{25}$

from elementary set theory it follows that

$J_j = \varphi(\varphi^{-1}(J_j)) = \varphi(I_j) = \varphi(I_n) = J_n \; \text{for} \; j \ge n, \tag{26}$

which shows the sequence of ideals $J_j$ stabilizes in $S$ and thus that $S$, like $R$, is a Noetherian ring.

Note Added in Edit, Friday 11 December 2020 10:22 PM PST:  A few final remarks meant to address our OP Confusion's two closing questions.  Yes, the proof given in the text of the question itself appears to be quite correct; indeed, it is for all intents and purposes the same as mine, and I'm pretty damn sure that is OK, so . . . I'm pretty damn sure Confusion's proof is OK as well.  There is no need to consider right and left Noetherian rings separately: indeed, since the rings in question are commutative, there is not distinction 'twixt right, left, and two-sided ideals.  As for Confusion's second question, that is, how do we know the sequence of ideals (24) (in the notation of my answer) is in fact ascending, note that

$r \in I_i = \varphi^{-1}(J_i) \tag{27}$

implies

$\varphi(r) \in J_i \subset J_{i + 1}, \tag{28}$

whence

$r \in \varphi^{-1}(J_{i + 1}) = I_{i + 1}, \tag{29}$

which shows that

$I_i \subset I_{i + 1}, \tag{30}$

that is, the ideals $I_i$ comprise an ascending chain in $R$.  End of Note.


问:
Joint probability $P(X>0, Z>-X)$ for $X,Z \overset{iid}{\sim} N(0,1)$

Let $X \sim N(0,1)$ and $Z \sim N(0,1)$ be independent normal random variables. Find the probability $P(X>0,Z>-X)$.

This is apparently an elementary calculation but I struggle to understand it. It is given that $X$ and $Z$ are independent but the limit of $Z$ depends on $X$ which is the source of my confusion.

First of all, could we write
$P(X>0,Z>-X)=P(X>0)P(Z>-X)$? If yes, then I can simply proceed as
$$P(X>0,Z>-X) = \int_{0}^{\infty}\phi(x)\,dx \int_{-x}^{\infty}\phi(s)\,ds.$$ But something looks odd. Is this the right direction?

Thanks.

答:
Let $X,Y$ be independent and identically distributed symmetric continuous RVs.

Let $p=P( \underset{=A}{\underbrace{X>0,X+Y>0}})=P(\underset{=B}{\underbrace{X>0,X+Y>0}})$.

Then $A\cup B = \{X+Y >0\}$ and $A\cap B = \{X>0,Y>0\}$, so that $P(A\cup B) = \frac 12$  as $X+Y$ is a continuous symmetric RV, and $P(A \cap B) =  P( X>0)P(Y>0) = \frac 14$.

Use this to get

$$\frac 12 = P(A \cup B)= P(A) + P(B) - P(A\cap B) = 2 p - \frac 14,$$

giving $p=\frac38$.


问:
Runge Kutta Fehlberg goes crazy when using adaptive h in python
So basically what i'm trying to do here is try to define a function that integrates for runge kutta. The most "physical" aspects of it i have already controlled for (it's not a problem of the values of the functions , or the physical parameters (initial conditions, etc), since i have controlled for those and they don't change the result) .
The problem i'm solvin is an ODE of second order (harmonic oscilator),

```<code>x''-μ(1-x^2 )x'+x=0 + CI x(0)=2, x'(0)=0, for t in [0,25], and μ=4
</code>```

which i have separated into two first-order ODES;

```<code>μ*(1-x**2)*v-x = v'
v = x'
</code>```

and we have inital conditions x0,v0,t0 and a interval on which to integrate t0,tf, and a step size h with which to integrate.
ε is my error tolerance.
The error I have is that when i activate the adaptivetoggle, so that i should use an adaptive h, x just goes to a value and v goes to 0, instead of oscilating like a harmonic oscilator should. I suspect the problem is only about that bit of code, because when i deactivate the adaptive toggle , everything runs just fine.
I am not sure of what is happening, my values should be oscilating, and small, and instead the error just goes to 0 ( it shouldn't, i think ) , v does too and x tends to something instead of oscilating.

The code i'm running is:

```<code>def rk452D(v, f, x0, v0, t0, tf, ε, h, adaptivetoggle):

x = x0; t = t0
valuesx = []
valuesv = []
while t<tf :
#We define the runge kutta functions on which to iterate while t is in [t0,tf], and they are of the kind (t,x,v)
    f1v = f(t,               x,                                                                                v0                                                                                )
    f1  = v(t,               x,                                                                                v0                                                                                )

    f2v = f(t + (1/4)*h,     x + (1/4)*f1,                                                                         v0 + (1/4)*f1v                                                                           )
    f2  = v(t + (1/4)*h,     x + (1/4)*f1,                                                                         v0 + (1/4)*f1v                                                                           )

    f3v = f(t + (3/8)*h,     x + (3/32)*f1      + (9/32)*f2  ,                                                     v0 + (3/32)*f1v      + (9/32)*f2v                                                        )
    f3  = v(t + (3/8)*h,     x + (3/32)*f1      + (9/32)*f2  ,                                                     v0 + (3/32)*f1v      + (9/32)*f2v                                                        )

    f4v = f(t + (12/13)*h,   x + (1932/2197)*f1 - (7200/2197)*f2 + (7296/2197)*f3,                                 v0 + (1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v                                 )
    f4  = v(t + (12/13)*h,   x + (1932/2197)*f1 - (7200/2197)*f2 + (7296/2197)*f3,                                 v0 + (1932/2197)*f1v - (7200/2197)*f2v + (7296/2197)*f3v                                 )

    f5v = f(t + h,           x + (439/216)*f1   - 8*f2           + (3680/513)*f3  - (845/4104)*f4,                 v0 + (439/216)*f1v   - 8*f2v           + (3680/513)*f3v  - (845/4104)*f4v                )
    f5  = v(t + h,           x + (439/216)*f1   - 8*f2           + (3680/513)*f3  - (845/4104)*f4,                 v0 + (439/216)*f1v   - 8*f2v           + (3680/513)*f3v  - (845/4104)*f4v                )

    f6v = f(t + h/2,         x - (8/27)*f1      + 2*f2           - (3544/2565)*f3 + (1859/4104)*f4 - (11/40)*f5,   v0 - (8/27)*f1v      + 2*f2v           - (3544/2565)*f3v + (1859/4104)*f4v - (11/40)*f5v )
    f6  = v(t + h/2,         x - (8/27)*f1      + 2*f2           - (3544/2565)*f3 + (1859/4104)*f4 - (11/40)*f5,   v0 - (8/27)*f1v      + 2*f2v           - (3544/2565)*f3v + (1859/4104)*f4v - (11/40)*f5v )

   #Now we calculate the positions and velocities, for the fourth order runge kutta aproximation. Commented we have the fifth order approxiation results, which we use to estimate the error ( Error = abs(order5approximation-order4approximation)
    x4 = x + h*((25/216)*f1  + (1408/2565)*f3   + (2197/4104)*f4     -(1/5)*f5              )
    #x5 = x + h*((16/135)*f1  + (6656/12825)*f3  + (28561/56430)*f4   -(9/50)*f5  +(2/55)*f6 )

    v4 = v0 + h*((25/216)*f1v + (1408/2565)*f3v  + (2197/4104)*f4v    -(1/5)*f5v             )
    #v5 = v0 + h*((16/135)*f1v + (6656/12825)*f3v + (28561/56430)*f4v  -(9/50)*f5v +(2/55)*f6v)

  #If we want to use an adaptive h for our calculations,
    if adaptivetoggle == True :
       #We calculate error in x using fs, eror in v using fvs, and we take the smaller hnew of the two
       Errorx = abs((1/360)*f1  - (128/4275)*f3  - (2197/75240)*f4  + (1/50)*f5  + (2/55)*f6  )
       Errorv = abs((1/360)*f1v - (128/4275)*f3v - (2197/75240)*f4v + (1/50)*f5v + (2/55)*f6v )

       hnewx  = 0.9*h*((ε)/Errorx)**(0.25)
       hnewv  = 0.9*h*((ε)/Errorv)**(0.25)

       hnew = min(hnewx,hnewv)

       print(hnew)
       #if hnewx < hnewv:
        #   hnew = hnewx
       #if hnewx > hnewv:
         #  hnew = hnewv

       #After calculating hnew, we compare it to the integration step h we have used, to decide if we can keep our calculation or we have to repeat it using hnew.

       if hnew >= h :
          #we increment the loop,and take hnew for the next loop
          t += h
          h = hnew
          x = x4
          v0 = v4
          valuesx.append(x4)
          valuesv.append(v4)

       elif hnew < h:
          h  = hnew #we don't increment t , the loop variable, when we repeat the integration step using the new h integration step

    else :#if we don't want an adaptive h ( this works just fine)
       valuesx.append(x4)
       valuesv.append(v4)
       x = x4
       v0 = v4
       t+=h #increment the loop
return valuesx, valuesv
#Then we implement the function
#We define the two functions ( of the ODEs) to feed the RK solver
def f(t,x,v):
    return μ*(1-x**2)*v-x
def v(t,x,v):
    return v

#we feed it the parameters
μ  = 4; x0 = 2;  v0 = 0; t0 = 0; tf = 25; h  = 1; ε  = 10**-3
adaptivetoggle = True

solution = rk452D(v, f, x0, v0, t0, tf, ε, h, adaptivetoggle)
print(solution)
</code>```

答:
The code doesn't run as it is now. <s> I think your formula for the error is wrong. Judging by the $4$th root you took, the tolerance $\varepsilon$ is meant to be for error per time since error per time is $O(h^4)$ for RK4. So you should use Error/h instead of Error in your formula for hnew. </s> Also your criteria for when to accept the approximation is harsh (it may not accept an approximation even if it was within the tolerance). You could just check whether Error <= $\varepsilon$.

Also you could take the maximum of the errors before computing the hnew rather than computing two hnew and taking a minimum. This error would be the $l_{\infty}$ norm of (RK4 - RK5)/h, and the hnew formula is based on the fact that the error is $O(h^4)$.


问:
Finite abelian $2$-group

If $G$ is a finite abelian group such that $o(x)=2$ for all $x \neq e$ and $|G|=2^n$ for some $n\in\mathbb N$, prove that $G \cong \mathbb{Z}_2\times\cdots\times\mathbb{Z}_2$ ($n$ factors).

Any help is appreciated. I asked this question before and was suggested to try induction but I haven't been able to write down the inductive proof properly. If anybody can help that will be really nice. Thanks!

答:
If you write $G$ in additive notation, it's fairly easy to check that you can make it into a vector space over the field $\mathbb{F}_2$ with two elements, where

$$
0g=0,\quad 1g=g
$$

Then, take a basis of $G$ as vector space: then $G$ becomes the direct sum of $n$ copies of $\mathbb{F}_2$, which, with respect to addition is none else than $\mathbb{Z}/2\mathbb{Z}$.

This can be of course generalizable to abelian groups $G$ such that $x^p=e$ for any $x\in G$, where $p$ is prime.

Title: Finite abelian $2$-group


问:
Finding the limit as x approaches 1 of $\sum \frac {(-1)^{n+1}}{n} x^n$
I want to find $\displaystyle \lim_{x \rightarrow 1^-} \sum_{n \in \mathbb N} \frac {(-1)^{n+1}}{n} x^n$

What I'm guessing I'm supposed to do is use a certain property that states that if $\displaystyle \sum_{n \in \mathbb N} \frac {(-1)^{n+1}}{n} x^n$ converges for some $x$ in $[0,1]$ and the series of the derivatives $\displaystyle \sum_{n \in \mathbb N} (-1)^{n+1} x^{n-1}$ converges uniformly in $[0,1]$, then the original series converges uniformly and thus I can exchange the limit on $x$ and the limit on $n$, giving

$$\displaystyle \lim_{x \rightarrow 1^-} \lim_{N \rightarrow \infty} \sum_{n = 1}^N \frac {(-1)^{n+1}}{n} x^n = \lim_{N \rightarrow \infty} \lim_{x \rightarrow 1^-}  \sum_{n = 1}^N \frac {(-1)^{n+1}}{n} x^n$$

Which would result in the alternate Harmonic series (multiplied by $-1$).

The problem, however, is that while it is true that the derivatives converge uniformly on $[0,1)$, they don't on $[0,1]$ since when x = 1 the partial sums for the derivatives are $1,0,1,0 \dots$

However, the statement that the derivatives need to converge uniformly on $[0,1]$ is what's making my head spin. The original functions are defined on $[0,1]$ so shouldn't the derivatives be defined on $(0,1)$ ?. If not, what does it mean for the derivatives to converge uniformly on $[0,1]$ ?

答: Let $f(x)= \sum\limits_{n=1}^{\infty} {(-1)^{n-1} x^{n-1}} $. This is geomertic seris and the sum is $\frac 1 {1+x}$. Now $\int_0^{x} f(t) dt=\sum\limits_{n=1}^{\infty} \frac {(-1)^{n-1} x^{n}} n$. Since $(-1)^{n-1}=(-1)^{n+1}$ the given sum is $\int_0^{x} f(t) dt=\ln (1+t)|_0^{x}=\ln (1+x)$. As $x \to 1$ this tends to $\ln 2$.


问:
Why are sine and cosine identities always true (and thus sentences) even though they have free variables?
In mathematics, we often see statements such as $\sin ^2x + \cos^2 x = 1$.

Notice that this is not a sentence, as the variable $x$ is free. But we all agree that this statement is true, given the usual interpretations of the symbols. How can we square this with the claim that sentences are the formulas that can be either true or false?

答:
In practice, a statement such as

$$
\sin^2 x + \cos^2 x = 1
$$

serves as a convenient abbreviation for

$$
\forall x \, \sin^2 x + \cos^2 x = 1.
$$

In other words, free variables like $x$ are governed by implicit universal quantifiers. Most of the time, writing out all quantifiers increases  clutter and hurts readability, so it is avoided.


问:
Determining end coordinates of line with the specified length and angle?
I have the point $(x_1, y_1)$, the angle $0 \leq a < 360^o$ and the length $l > 0$.

How do I determine the end point $(x_2, y_2)$ if there is a line between $(x_1, y_1)$ and $(x_2, y_2)$ of length $l$ and with angle $a$?

答: Knowing the length $l$ and the angle $a$ gives you a right triangle with base parallel to the $x$-axis and the hypotenuse is the line itself. From this you can calculate the length of the legs, using trigonometry: denote the length of the leg parallel to the $x$-axis by $X$, and the other one by $Y$. Then, by the law of sine you get: $\frac{X}{\cos a}=\frac{Y}{\sin  a}=\frac{l}{\sin \frac{\pi}{2}}=l$. From here you can find $X,Y$. The the end-point will be $(x_2,y_2)=(x_1+X,y_1+Y)$.</p>