问:
Does the matrix product $X^T X$ have a special meaning?
I have come across this specific matrix product several times, lately in the context of stochastic linear models where it is an integral part of the Least Squares Estimator (LSE). Often times in linear algebra there is some beautiful intuition hidden behind recurring formulae and since I don't see the one behind this one I'm asking for help.

Is there a geometric interpretation or special meaning for the matrix product

$$X^T X$$

for a matrix $X \in \mathbb{R}^{n \times p}$ with $p \leq n$?

答: When we find the least squares solution $\hat{x}$ to a system $Ax = b$, we are finding the point $p = A\hat{x}$ of the column space that is closest to $b$. Geometrically, this is given by projecting $b$ onto the column space of $A$. The error $e = b - p$ is thus orthogonal to the column space. The orthogonal complement of the column space of $A$ is the nullspace of $A^T$, so this means we want $A^Te = 0$. But $e = b - p = b - A\hat{x}$, so $A^Te = 0$ is equivalent to $A^T(b - A\hat{x}) = 0$, or $A^TA\hat{x} = A^Tb$. This gives us the normal equations, with the appearance of the matrix $A^TA$.


问:
How to find nonhomogeneous solutions of the following PDE: $b^2d_{,xx}+d-1=0$?
I have a question about the solutions for the following PDE.
$$
b^2d_{,xx}+d-1=0, \text{where}\quad x,b\in \mathbb{R} \quad\text{and} \quad b>0
$$

It is easy to verify that $d=1$ is a homogeneous solution and $d(x)=1-\sin \left(\frac{|x|}{b}\right)$ is also a solution which is nonhomogeneous. The problem is when I use COMSOL Multiphysics to solve this equation, I can only get the homogeneous solution and by no means the nonhomogeneous one. I can see that the nonhomogeneous solution is not continuous which may be the reason why I can't get it. Is there any way to get this solution through a numerical way? Thanks!

答:
$$b^2d_{,xx}+d-1=0$$
Rewrite it as:
$$b^2(d-1)_{,xx}+(d-1)=0$$
$$b^2w_{,xx}+w=0$$
Where $w=d-1$. Then the DE is homogeneous.

The characteristic polynomial is $(b \ne 0)$:
$$b^2r^2+1=0 \implies (br-i)(br+i)=0$$
$$\implies r=\pm \dfrac i b$$
The solution is :
$$w(x)=c_1 \cos (x/b)+c_2 \sin(x/b)$$
And
$$d(x)=1+w(x)$$
$$d(x)=1+c_1 \cos (x/b)+c_2 \sin(x/b)$$

ContentLicense: CC BY-SA 4.0


问:
I do not understand this particular step in a proof using the Well Ordering Principle
Below is a proof using the Well Ordering Principle. I get lost starting at $(13)$...

$$
\begin{aligned}
P(c): &:=c^{3} \leq 3^{c} \\
& \equiv c^{3} \leq 3(c-1)^{3} \\
& \equiv c \leq \sqrt[3]{3} \times(c-1)
\end{aligned}
$$

I don't understand how we get $c^{3} \leq 3(c-1)^{3}$ from $c^{3} \leq 3^{c}$.. why does the righthand side of the inequality change like that?

答:
There's so much notational and organizational sloppiness to the given proof that it doesn't really convey a valid argument. A more clear, complete version would be something like:

Let $C=\{n\in \Bbb N:n^3>3^n\}$. We wish to show that $C$ is empty.

We proceed by contradition: suppose that $C$ is nonempty. Then by the well-ordering principle, $C$ has a smallest element $c$. Note the following:

Since $c\in C$, we have $c^3>3^c$.

Since $c$ is the smallest element of $C$, we know that $c-1\notin C$, so $(c-1)^3\le 3^{c-1}$.

We can verify directly that $n^3\le3^n$ for $n\in\{0,1,2,3\}$, so we must have $c\ge4$.

Combining these observations, we have
$\left(\frac43\right)^3=(1+\frac1{4-1})^3\ge(1+\frac1{c-1})^3=(\frac c{c-1})^3=\frac {c^3}{(c-1)^3}>\frac{3^c}{3^{c-1}}=3$,
i.e. $64>81$, which is false. Therefore our assumption that $C$ is nonempty must be false. In other words, every $n\in\Bbb N$ satisfies $n\le3^{n/3}$.


问:
What's the maximum and minimum of $\sin Z$ and $\cos Z$ where $Z$ is a complex number?
I see that a lot of people try solving something  like $\sin Z=2$. So that breaks the the maximum of $\sin x$. Now I wonder if there is a maximum and minimum of $\sin Z$ and $\cos Z$ if $Z$ is a complex number.

答:
Note that $\sin z=sin(x+iy)=\sin x \cosh y+i\cos x\sinh y$

Taking absolute value to have the meaning of maximum and minimum, $|\sin z|=\sqrt{\sin^2 x\cosh^2 y+\cos^2 x\sinh^2 y}=\sqrt{\sin^2 x+\sinh^2  y}=\sqrt{\sin^2 x+(\frac{e^y-e^{-y}}{2})^2}$

which is unbounded along the imaginary axis in both directions. Similarly you can proceed for $\cos z$. The minimum value for both the functions can be easily seen to be zero at origin for $|\sin z|$ and at $z=\pi/2$ for $\cos z$.


问:
Linear independence of partial derivatives
Consider the general solution $y(x,c_1,c_2)$ of a 2nd-order linear homogeneous ODE, where $c_1$ and $c_2$ are two parameters and $y$ is assumed to be smooth in $x$, $c_1$, and $c_2$. Take the partial derivatives
$$
\frac{\partial y}{\partial c_1}(x,c_1,c_2), \quad \frac{\partial y}{\partial c_2}(x,c_1,c_2).
$$
Now fix some $(c_1,c_2) = (k_1,k_2)$. Is is true that the functions of $x$ defined by
$$u_1(x) = \frac{\partial y}{\partial c_1}(x,k_1,k_2), \quad u_2(x) = \frac{\partial y}{\partial c_2}(x,k_1,k_2)$$
are linearly independent? If not, what extra assumptions are necessary? Does it depend on the choice of $(k_1,k_2)$, and if so, in what way? The final two questions are extra; I'd really love a proof (even an outline) or counterexample for the first question.</p>

答:
Unless I'm mistaken, the partial derivatives are always linearly independent.

The general solution of a second order linear homogeneous ODE $y'' + ay' + by = 0$ is either $y = c_1e^{r_1x} + c_2e^{r_2x}$, if $r_1$ and $r_2$ are the two roots of the characteristic polynomial $r^2 + ar + b$, or $y = c_1e^{rx} + c_2xe^{rx}$ if r is a double root of the characteristic polynomial.

In the first case, $\frac {\partial} {\partial c_1}c_1e^{r_1x} + c_2e^{r_2x} = e^{r_1x}$, and $\frac {\partial} {\partial c_2}c_1e^{r_1x} + c_2e^{r_2x} = e^{r_2x}$, which are linearly independent because $r_1$ and $r_2$ are distinct, and in the second case, $\frac{\partial}{\partial c_1} y = c_1e^{rx} + c_2xe^{rx} = e^{rx}$, and $\frac{\partial}{\partial c_2} y = c_1e^{rx} + c_2xe^{rx} = xe^{rx}$, which are again linearly independent.

Note that, since the solutions are linear in $c_1$ and $c_2$, these partial derivatives do not depend on $c_1$ and $c_2$.

Edit: As has been pointed out several times, this only works for constant-coefficient equations. Here is a more general answer:

First we have to notice that any given differential equation won't necessarily have just one "general solution," so the partial derivatives $\frac{\partial y}{\partial c_1}$ and $\frac{\partial y}{\partial c_2}$ won't be well defined (for example, consider the equation $y'' + y = 0$, whose general solution can be written as either $y = c_1cos(x) + c_2sin(x)$ or as $y = c_1e^{ix} + c_2e^{-ix}$). However, for any general solution with a reasonably standard form, the result does hold.

A common form of the general solution of a homogeneous linear second order ODE involves first finding two linearly independent solutions $\phi_1$ and $\phi_2$, and setting $y(x,c_1,c_2) = c_1\phi_1(x) + c_2\phi_2(x)$. The partial derivatives with respect to $c_1$ and $c_2$ are then $\phi_1$ and $\phi_2$, which are linearly independent.

ContentLicense: CC BY-SA 3.0


问:
Proving the convergence of $\log(n+2)/\log(n+1)$
Currently trying to prove that  $\log(n+2)/\log(n+1)$ converges to $1$ with analysis tools.

if I can prove that $\forall \varepsilon >0$, $\exists N$ such that
$$\left| \log(n+2)/\log(n+1)-1|<\varepsilon \right|$$
whenever $n\geq N$.

I tried using the integral definition of the log, along with $1=\log(n+1)/\log(n+1)$
$$
\left| \frac{\int^{n+2}_1\frac{1}{u}-\int^{n+1}_{1}\frac{1}{u}}{\int^{n+1}_1\frac{1}{u}}\right|\
$$
which I suppose simplifies to
$$
\left| \frac{\int^{n+2}_{n+1}\frac{1}{u}}{\int^{n+1}_1\frac{1}{u}}\right|\
$$
but I'm not sure where to go from here or even if this is the right direction. Or is proving that the sequence is cauchy is a better approach?

答: You want to prove the fraction you give converges to $0$.  Note that the numerator is less than $\frac 1{n+1}$ because the integrand is less than $\frac 1{n+1}$ and the interval is $1$.  Then note the denominator (for $n \gt 3$) is greater than $1$


答:
Hint:

your idea can be made to work, we just need to estimate your two integrals.

Since $\frac{1}{u}$ is decreasing, your top integral

$$
\int_{n+1}^{n+2} \frac{1}{u}\ du
\leq \left ( (n+2) - (n+1) \right )\frac{1}{n+1}
= \frac{1}{n+1}
$$

Can you find a way to upper bound $$\frac{1}{\int_1^{n+1} \frac{1}{u}\ du}?$$

Can you turn these upper bounds into an upper bound for your fraction, and show it can be made $< \epsilon$?

I hope this helps ^_^

Title: Proving the convergence of $\log(n+2)/\log(n+1)$

Body:
Currently trying to prove that  $\log(n+2)/\log(n+1)$ converges to $1$ with analysis tools.

if I can prove that $\forall \varepsilon >0$, $\exists N$ such that
$$\left| \log(n+2)/\log(n+1)-1|<\varepsilon \right|$$
whenever $n\geq N$.

I tried using the integral definition of the log, along with $1=\log(n+1)/\log(n+1)$
$$
\left| \frac{\int^{n+2}_1\frac{1}{u}-\int^{n+1}_{1}\frac{1}{u}}{\int^{n+1}_1\frac{1}{u}}\right|\
$$
which I suppose simplifies to
$$
\left| \frac{\int^{n+2}_{n+1}\frac{1}{u}}{\int^{n+1}_1\frac{1}{u}}\right|\
$$
but I'm not sure where to go from here or even if this is the right direction. Or is proving that the sequence is cauchy is a better approach?


问:
Probability of drawing at least J cards out of I cards of interest from a deck of N cards within D draws with replacement
Given a regular deck of 52 playing cards, if you draw from the deck 10 times with replacement, what is the probability that you have drawn at least 3 different aces (any of the 4 aces) at least once each? And how does this generalize to N playing cards, D draws, I many cards that we are interested in, and J many of the cards we are interested in that we must draw at least once.

I came across a simpler version of this problem while working on a hobby project trying to figure out some statistics related to a Pokémon-themed Discord bot gacha game, but I was intrigued with this problem after I found a few unanswered Math StackExchange posts about this variant. (1, 2) Unfortunately those posts didn't get any attention so I figured I would make a post with the additional information that I found.

Formalization

I was able to formalize the problem as $f(n_1, n_2, k_1, k_2, p_1)$, where:

$n_1$ is the number of cards drawn

$n_2$ is the number of cards that we are interested in

$k_1$ is the number of times at least that we want to draw a card we are interested in

$k_2$ is the number of cards at least that we are interested in that we want to draw at least $k_1$ times

$p_1$  is the probability of drawing a specific card

My attempts so far

Binomial distribution of binomial distributions

I had some initial attempts to solve it by modeling a binomial distribution where the events were obtaining at least 1 of a card of interest in $n_1$ draws (that also being a binomial distribution). But I validated that against some simulation code that I wrote, and the formulation came up with different results, which I think were due to the events being not independent (drawing one particular card in a set of draws effects the probability of drawing at least one of another particular card on the same set of draws).

I think that approach should give an upper bound on the answer for the problem. It models drawing at least $k_1$ copies of a card we are interested in by using a binomial distribution. It then (incorrectly) assumes that we can model that for each card we are interested in drawing as a binomial distribution (not true, since those probabilities are not independent).

But I think the lack of independence is always destructive in this case (drawing a particular card prevents you from have drawing any other card in that same draw action), so the actual probability should be upper bounded by this formula. But I'm not 100% sure on that.

$$
  p_2 = \sum_{k=k_1}^{n_1} \binom{n_1}{k} * p_1^{k} * (1 - p_1)^{n_1 - k}
  $$
$$
  \sum_{k=k_2}^{n_2} \binom{n_2}{k} * p_2^{k} * (1 - p_2)^{n_2 - k}
  $$

Multinomial distribution

After looking around on the internet for similar problems or approaches that might work I learned about multinomial distributions. I think you can try modelling the card drawing with replacement as a multinomial distribution and then sum the results of the probability distribution function with the different input sets that make up all the possible winning sets of draws, where there is probably a trick to group those sets into categories of equal probability.

This is my current best guess as to a framework for a solution that might work, but I haven't worked out the details on it and tested it against the simulation yet.

Simpler variant

Before I came across this variant of the problem, in my hobby project I came across the simpler case where $k_2 = n_2$ and $k_1 = 1$, the "collect all of the Pokémon at least once" problem. Unfortunately I haven't solved this question yet either.

Starting with solving this simpler variant might be useful in finding a solution for the more general case. But if we do find a solution for the general case, it should work for this specific case as well.

Simulation code

I posted the simulation code I wrote on a GitHub gist at the link below. You can run it by providing $n_1$, $n_2$, $k_1$, $k_2$, and the number of total cards in the deck. You can also optionally provide an RNG seed and a number of iterations (number of draws of $n_1$ cards to simulate).

```<code>$ python sim.py simulate 10 4 1 3 52 --seed 42
Namespace(command='simulate', iterations=100000, k_1=1, k_2=3, n_1=10, n_2=4, seed=42, total=52)
100%|████████████████████████████████| 100000/100000 [00:01<00:00, 95673.52it/s]
estimated_probability = 0.01499
</code>```

https://gist.github.com/ExcaliburZero/dfc6ed8b3c7e69f6acbb556844d70add

The script also includes a mode to run the "binomial distribution of binomial distributions" approach that I think upper bounds the result.

```<code>$ python mathoverflow_answer.py calculate 10 4 1 3 52
p_1 = 0.019230769230769232
p_exactly = 0.16147234355824405
p_2 = 0.17649104785295508
result = 0.019079344697167273
</code>```

Internet research

While working on this problem I did some searching online to try to find similar problems or potential approaches for solutions. Here are some links I found the seemed potentially relevant.

Math StackExchange posts with the same question that went unanswered. (1, 2)

Resources related to cumulative distribution functions of multinomial distributions.

A page from the book "Chemical Process Dynamics and Controls", which includes a section 2.2 that gives a cumulative distribution function for multinomial distributions. (3)

A 1981 paper on a representation for a multinomial cumulative distribution function, which I don't really understand yet. (4)

Conclusion

Above are my thoughts on and current attempts at solving this problem. My current best lead is to try something with multinomial distributions, but I haven't been able to solidify and test that approach yet.

Does anyone have thoughts on how to go about solving this problem? Does this approach seem reasonable? Are there better approaches that yield a solution to this problem?

答:
This can be expressed as a recursive formula.

First, let's look only at $k_1=1$. If $n_1>0$ and $n_2 \geq k_2>0$, then

$$ f(n_1,n_2,1,k_2,p) = n_2 p \cdot f(n_1-1, n_2-1, 1, k_2-1,p) + (1-n_2 p) f(n_1-1, n_2, 1, k_2, p) $$

This is because if the first draw is a wanted card ($n_2 p$), then the conditional probability is the same as for an instance with one less draw ($n_1-1$), one less card type still "interesting" ($n_2-1$), and one less success needed to reach the target ($k_2-1$). Or if the first draw is not a wanted card ($1-n_2 p$), then the conditional probability is the same as for an instance with one less draw ($n_1-1$) but everything else staying the same.

We have initial conditions,

$$ \begin{align*}
 f(n_1, n_2, 1, 0, p) &= 1 \\
f(0, n_2, 1, k_2, p) &= 0 & \mathrm{if}~k_2>0 \end{align*}$$

since wanting to collect no more cards ($k_2=0$) means the player has already succeeded, and otherwise having no draws left ($n_1=0$) means they've failed.

From the recursion relation, we can see that each set of $f$ values with $n_2-k_2$ constant is an independent problem. We can also consider $p$ constant in the recursion problem. So let's call $m = n_2-k_2$, the maximum number of interesting card types we'd be okay with missing out on. Then there are just two variables to recurse on to get $f(n_1, n_2, 1, n_2-m, p)$.

A computer program can evaluate this kind of recursion fairly efficiently and with whatever accuracy is needed, including exact rational ratios. For $f(10, 4, 1, 3, 1/52)$, this would start like

$$ \begin{align*}
f(1, 2, 1, 1, 1/52) &= \frac{2}{52} \cdot 1 \\
f(2, 2, 1, 1, 1/52) &= \frac{2}{52} \cdot 1 + \frac{50}{52} \cdot \frac{2}{52} = \frac{204}{52^2} \\
f(2, 3, 1, 2, 1/52) &= \frac{3}{52} \cdot \frac{2}{52} = \frac{6}{52^2} \\
f(3, 2, 1, 1, 1/52) &= \frac{2}{52} \cdot 1 + \frac{50}{52} \cdot \frac{204}{52^2} \\
f(3, 3, 1, 2, 1/52) &= \frac{3}{52} \cdot \frac{204}{52^2} + \frac{49}{52} \cdot \frac{6}{52^2} \\
f(3, 4, 1, 3, 1/52) &= \frac{4}{52} \cdot \frac{6}{52^2}
\end{align*} $$

For $k_1 > 1$, things get trickier. A recursion will need more than the existing $f$ in five variables, since it can matter whether the player is looking for two more identical cards, or one more of each of two card types. Instead, we could introduce a sequence of counter variables $w_1, \ldots w_{n_2}$, representing for each interesting card type the number of cards the player still needs to reach a "success" for that card type: $g(n_1, k_2, p, w_1, \ldots, w_{n_2})$ is the probability that a player will get at least $k_2$ successes within the next $n_1$ draws, where $p$ is the probability of drawing any particular card and $w_i$ more draws of card $i$ counts as a success.

This relates to the original $f$ by taking all $w_i = k_1$. That is,

$$ f(n_1, n_2, k_1, k_2, p) = g(n_1, k_2, p, \underbrace{k_1, k_1, \ldots, k_1}_{n_2 \mathrm{~times}}) $$

Note that the value of $g$ does not change when the values of $w_i$ are permuted. This should drastically cut back on the number of calculations needed when applying the recursion, whether by hand or by computer. It would probably be helpful to keep the $w_i$ in non-descending or non-ascending order.

For initial conditions, again $k_2=0$ is a success and $k_2>0, n_1=0$ is a failure. The recursion is

$$ g(n_1, k_2, p, w_1, \ldots, w_{n_2}) =
 p \sum_{w_i = 1}
 g(n_1-1, k_2-1, p, w_1, \ldots, w_{i-1}, 0, w_{i+1}, \ldots, w_{n_2}) \\
{}+ p \sum_{w_i > 1}
 g(n_1-1, k_2, p, w_1, \ldots, w_{i-1}, w_i - 1, w_{i+1}, \ldots, w_{n_2}) \\
{}+ \left(1 - p \sum_{i=1}^{n_2} w_i\right)
 g(n_1-1, k_2, p, w_1, \ldots, w_{n_2}) $$


答:
Why not to use $$a_n=\frac{\log (n+2)}{\log (n+1)}$$ and compose Taylor series to get
$$a_n=1+\frac{1}{n \log (n)}-\frac{3 \log (n)+2}{2 n^2 \log ^2(n)}+O\left(\frac{1}{[n \log(n)]^3}\right)$$


问:
Existence for the vectors with the same length and isotropic position.
Suppose that $v_1,\ldots,v_m\in\mathbb{R}^n$ satisfying $$\sum_{i=1}^m v_iv_i^T=I_n$$ and $$\|v_i\|^2=\frac{n}{m}$$ for each $i$.

Do the vectors $\{v_i\}_{1\leq i\leq m}$ always exist? If not, what's the condition for the existence?

Note that $[v_1,\ldots,v_m]$ can be viewed as a submatrix extracted $n$ rows from an orthogonal matrix, and each columns have the same length.

答:
As noticed by Reinhard, a necessary and sufficient condition is $m\geq n$. Below you can read a simple existencial proof of this fact.

Consider $B_{m\times m}=\begin{pmatrix}(1-\frac{n}{m})Id_{n\times n} & 0_{n\times (m-n)}\\
0_{(m-n)\times n} & -\frac{n}{m}Id_{(m-n)\times (m-n)}\end{pmatrix}$.

Notice that $B$ is a traceless real symmetric matrix. Hence there is an orthogonal matrix $U_{m\times m}$ such that $UBU^t=C$ and $C_{ii}=0$ for $i=1,\ldots,m$.

Define $D=\frac{n}{m}Id_{m\times m}+C$.

Notice that $Spec(D)= \stackrel{n}{\overbrace{1,\ldots,1}},\stackrel{m-n}{\overbrace{0,\ldots,0}}$ and $D_{ii}=\frac{n}{m}$ for $i=1,\ldots,m$.

Since $D_{m\times m}$ is a positive semidefinite symmetric matrix of rank $n$, there exists $R_{m\times n}$ such that $D=RR^t$.

Next, $R^tR=Id_{n\times n}$, since $Spec(R^tR)=\stackrel{n}{\overbrace{1,\ldots,1}}$ and $R^tR$ is a real symmetric matrix of order $n$.

Let $R^t=(v_1,\ldots,v_m)$, where $v_i$ is column $i$ of $R^t$.

Then $Id_{n\times n}=R^tR=\sum_{i=1}^mv_iv_i^t$.

Finally, $v_i^tv_i=D_{ii}=\frac{n}{m}$, for $i=1,\ldots,m$. $\square$

Edit: Let us prove that every traceless real symmetric  matrix $S_{m\times m}$ is orthogonally equivalent to a symmetric matrix with zeros in the diagonal.

Let $v_1,\ldots,v_m$ be an orthonormal basis of eigenvectors of $S$. Define $v=\frac{v_1+\ldots+v_m}{\sqrt{m}}$.

Notice that $v^tSv=\frac{tr(S)}{m}=0$. Let $U_{m\times m}$ be any orthogonal matrix such that the first column of $U$ is $v$. So $U^tSU_{11}=0$.

Now, since $U^tSU_{11}=0$ and $U^tSU$ is traceless, the submatrix $A_{m-1\times m-1}$ occupying the last $m-1$ rows and columns of $U^tSU$  is also a traceless real symmetric matrix.

By induction on $m$, there is an orthogonal matrix $V_{m-1\times m-1}$ such that $V^tAV$ is symmetric with zeros in the diagonal.

So $W=\begin{pmatrix}1 & 0_{1\times m-1}\\
0_{m-1\times 1} & V_{m-1\times m-1}\end{pmatrix}$ is orthogonal and $W^tU^tSUW$ is symmetric with zeros in the diagonal.

Finally, $UW$ is orthogonal too.


问:
Finding Radon-Nikodym derivative given integral

Let $\lambda$ be the Gaussian measure on $\mathbb{R}$ defined by
$$\lambda(E):=\int_{E}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx.$$
Let $\operatorname{m}$ be Lebesgue measure on $\mathbb{R}$. What is the Radon-Nikodym derivative of $\operatorname{m}$ with respect to $\lambda$?

I've already show that $\operatorname{m}$ is absolutely continuous with respect to $\lambda$, written
$\operatorname{m}\ll\lambda$. As $f:=1/\sqrt{2\pi}e^{-x^2/2}$ is non-negative and both $\operatorname{m}$ and $\lambda$ are $\sigma$-finite measures on $\mathbb{R}$, I take it the Radon-Nikodym derivative of $\lambda$ with respect to $\operatorname{m}$ is

$$\frac{d\lambda}{d\operatorname{m}} = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\tag{1}.$$

Is this correct, and if so how do I then get the Radon-Nikodym derivative of $\operatorname{m}$ with respect to $\lambda$?

答:
The notation for the Radon-Nikodym derivative is well chosen. That is,

$$\frac{dm}{d\lambda} = \frac{1}{\frac{d\lambda}{dm}}$$

whenever we have both $\lambda \ll m$ and $m \ll \lambda$. Of course, nothing is special about lebesgue measure $m$ in this theorem. You can find this as Corollary 3.10 in Folland.

In your case, if $\frac{d\lambda}{dm} = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}$, then we would see

$$\frac{dm}{d\lambda} = \sqrt{2\pi}e^{x^2/2}$$

I hope this helps ^_^


答: The key is to show that the equality extends to $$\int_Eg\,d\lambda=\int_Egf\,dm.$$ Then if $g=dm/d\lambda$, you get $$\int_E(1-gf)\,dm=0$$ for all measurable $E$. Conclude that $fg=1$ a.e.

Title: Finding Radon-Nikodym derivative given integral

Body:
Let $\lambda$ be the Gaussian measure on $\mathbb{R}$ defined by
$$\lambda(E):=\int_{E}\frac{1}{\sqrt{2\pi}}e^{-x^2/2}dx.$$
Let $\operatorname{m}$ be Lebesgue measure on $\mathbb{R}$. What is the Radon-Nikodym derivative of $\operatorname{m}$ with respect to $\lambda$?

I've already show that $\operatorname{m}$ is absolutely continuous with respect to $\lambda$, written
$\operatorname{m}\ll\lambda$. As $f:=1/\sqrt{2\pi}e^{-x^2/2}$ is non-negative and both $\operatorname{m}$ and $\lambda$ are $\sigma$-finite measures on $\mathbb{R}$, I take it the Radon-Nikodym derivative of $\lambda$ with respect to $\operatorname{m}$ is

$$\frac{d\lambda}{d\operatorname{m}} = \frac{1}{\sqrt{2\pi}}e^{-x^2/2}\tag{1}.$$

Is this correct, and if so how do I then get the Radon-Nikodym derivative of $\operatorname{m}$ with respect to $\lambda$?


问:
Evaluating $\int_{0}^{1} \frac{x^{2n+2}}{x^2-2x+2}\,{\rm d}x$
$$\int_{0}^{1} \frac{x^{2n+2}}{x^2-2x+2} \,{\rm d}x \tag{1}$$

This integral came up as a result of another integral I was trying to evaluate. Just for recreational purposes. Is there some closed form for this, even in terms of special functions? In general, is it possible to evaluate integrals of the form:

$$\int_{0}^{1} \frac{P(x,x^n)}{Q(x)} \,{\rm d}x \tag{2}$$

with $\text{deg}(P) > \deg(Q)$? For example:

$$\int_{0}^{1} \frac{x^n}{1+x^2} \,{\rm d}x \tag{3}$$

Can be written in terms of the digamma function. If I'm not mistaken, $(1)$ does not follow from $(3)$, because if you try to complete the square in the denominator of $(1)$, you'd end up having to perform a $u$-sub with $u = x-1$ and then a $(u+1)^{2n+2}$ in the numerator. This ruins things. Can someone give some help?

答:
Like most things in life, this is a hypergeometric integral:

$$
\fbox{$\frac{(1+i) \left(\, _2F_1\left(1,2 n+3;2
   (n+2);\frac{1}{2}-\frac{i}{2}\right)-i \,
   _2F_1\left(1,2 n+3;2
   (n+2);\frac{1}{2}+\frac{i}{2}\right)\right)}{8
   n+12}\text{ if }\Re(n)>-\frac{3}{2}$}
$$
The way to get it there is to transform $[0, 1]$ to $[0, \infty]$ by a linear fractional, and then it looks like a standard hypergeometric integral (I did not do that, but just asked mathematica).


问:
Holomorph of Cyclic $p$-groups
Assume that $G$ is a cyclic $p$-group (or up to isomorphism more explicitly $\mathbb{Z}_{p^r}$). Do you know if there does exist any explicit way to describe the holomorph of the above group? Moreover, do you know if we can say something about the representations of these groups? (modular, non-modular)</p>

答:
My representation theory is almost nada.

But the holomorph is, ims, the canonical semi-direct product  $G\rtimes\rm_\varphi  {Aut}G $, with the action  being left multiplication.

So we get  $\mathbb Z_{p^r}\rtimes_\varphi\mathbb Z_{p^r-p^{r-1}} $, unless  $p=2$.

Title: Holomorph of Cyclic $p$-groups


问:
Proof of uniqueness of the bounded linear transformation extended in the Bounded Linear Transformation theorem
B.L.T Theorem (from Reed/Simon):  Suppose $T$ is a bounded linear transformation from a normed linear space $\langle V_1, \|\cdot\|\rangle$ to a complete normed linear space $\langle V_2, \|\cdot\|\rangle$.  Then $T$ can be uniquely extended to a bounded linear transformation (with the same bound), $\widetilde{T}$, from the completion of $V_1$ to $\langle V_2, \|\cdot\|\rangle$.

The proof for $\widetilde{T}$ being bounded was given and very straightforward, and proving that it was linear was pretty simple as well.  I have been having issues with proving that $\widetilde{T}$ is unique, however, despite it probably being easy.  I tried supposing towards a contradiction and using that $V_1$ is dense in its completion, $\tilde{V_1}$, the extending transformations must agree on $V_1$, the extending transformations must both have the same bound as $T$, and since this implies that the extending transformations must have different bounds to be different themselves, this proves that $\widetilde{T}$ is unique. I don't think that this reasoning is right since I think that a transformation can act differently on some subset of $\tilde{V_1}\setminus{V_1}$ without changing the bound, and my only other idea was to use that both of these spaces are $T_1$, so if a sequence converges, it must converge to a unique point, and since $T$ is bounded and therefore continuous, and any extensions must be bounded and therefore continuous, we'll have sequences being mapped to their limits, and since these limits are unique, we will only yield one extension $\widetilde{T}$ that works for all of $\tilde{V_1}$.  Again, I think that this reasoning is missing something.

Any insights, whether it be a nudge in the right direction or full proofs, are very welcome!  Thanks in advance.

答:
Let $\{x_n\}$ be a Cauchy-Sequence in $V_1$. Then $\{T(x_n)\}$ is a Cauchy-Sequence in $V_2$. (This follows from the boundedness of $T$)

Then the natural way to extend $T$ is using the limit of Cauchy-Sequence in $V_2$. (which is possible by completeness of $V_2$)

For the proof of uniqueness, suppose there are two extensions $T_1$ and $T_2$,
and use the fact that $T_1-T_2$ is bounded linear. (hence continuous)

Apply to the Cauchy-Sequence  $\{x_n\}$. Then for $x^*=\lim x_n \in \bar{V_1}$, $(T_1-T_2)(x_n) = 0$ for all $n$. Thus $T_1(x^*) = T_2(x^*)$ by continuity.


问:
Finding the sum of possible values
$14$ students have taken part in a math olympiad. Every problem in that olympuad is solved by exactly $7$ students. Furthermore, the first $13$ students have all solved $10$ problems each. If the $14$th student has solved exactly  $k$ problems, then what is the sum of the allpossible values of $k$?

I have managed the diophantine equation: 130+$k$=7$x$ where $x$ is the total number of questions in the question paper....but i have been stuck determining the sum of the values of $k$.

答:
Suppose the 14th student - say, Robin - did very poorly, and managed to solve some minimum number, while most of the solutions were found by the other students. Then the “$7x$” of your equation would be only a little bigger than $130$.

Now suppose in contrast that Robin did very well, so the other students were only contributing the other six answers to most of the questions. What’s the biggest set of questions that they could have given six answers to? Note that they have to supply full sets of six answers - you can’t have a case where Robin gives more than one answer to a question, so “spare” answers from the others will make sets of seven without Robin’s help.

These two limits and the multiple of $7$ rule give you the possible values of $k$.

Title: Finding the sum of possible values


问:
Prove that $\text{dim}(W\cap V^{\perp})=1$ (Linear Algebra)
Let $V$, $W$ vector subspaces of $\mathbb{R}^n$, such that: $$V\leq W\hspace{1cm}\textrm{and}\hspace{1cm}\textrm{dim }W=1+\textrm{dim }V.$$
How can i prove that:$$\textrm{dim }(W\cap V^{\perp})=1.$$Where $V^{\perp}=\{x \in \mathbb{R}^n: \langle x,v \rangle=0\hspace{0.5cm}\textrm{for all }v \in V\}$, and $\langle x,v \rangle$ is the usual inner product in $\mathbb{R}^n$.

答: Write $m=\dim V$, and choose an orthonormal basis $w_1,\dots,w_m,w_{m+1}$ for $W$ such that $w_1,\dots,w_m$ is a basis for $V$. Then, as $\langle w_i,w_{m+1} \rangle = 0$ for $i=1,\dots,m$, we see that $w_{m+1} \in V^\perp$. On the other hand, if $w \in W \cap V^\perp$, write $w = a_1w_1 + \cdots + a_mw_m + a_{m+1}w_{m+1}$ for some $a_1,\dots,a_m,a_{m+1} \in \mathbb R$, and then note that, for $j=1,\dots,m$, $$0 = \langle w,w_j \rangle = \sum_{i=1}^{m+1} a_i \langle w_i,w_j \rangle = \sum_{i=1}^m a_i \delta_{ij} = a_j$$ so $w = a_{m+1}w_{m+1}$.


问:
Definable well-ordered class relations
Suppose we a working with an axiomatic set theory such as $\bf ZFC$. Formulae with 2 free variables define class relations. "Small" class relations can be turned into sets of ordered pairs, the others are proper classes. Some formulae define relations that are well-orders. A "small" such relation can be turned into a well-ordered set that is order-isomorphic to some ordinal $\alpha\in\bf Ord$. Proper class relations also be can well-orders, e.g. a formula $\phi(x,y)$ saying that $x$ and $y$ are both ordinals and $x\in y$ defines the usual well-order on the class of ordinals $\bf Ord$.

Now, we can also write a formula that well-orders all ordinals except $\bf 0$ in the usual way, but considers $\bf 0$ as larger than all of them. Apparently, it gives us an order type that is "longer" than $\bf Ord$, and can be denoted as $\bf Ord+\bf1$. In a similar way, formulae having even "longer" order types such as $\bf Ord+\bf Ord$, etc. can be written. These order types are not sets existing in the universe, but rather, informal equivalence classes of formulae defining well-ordered class relations under (provable) order isomorphism. But, still, we can think of them as very long "ordinals" that occur when we count beyond all set-sized ordinals and even beyond the proper-class-sized "ordinal" $\bf Ord$. I'm interested in how far those "ordinals" can extend, and whether there is some systematic or formal way to study them, or develop ordinal arithmetic on them. Perhaps, a different axiomatic theory, rather than $\bf ZFC$, would be more suitable for that?

答:
A good way in my opinion to study this phenomenon is to switch from $V$ to transitive set-sized models of $\mathsf{ZFC}$ (making appropriate existence assumptions of course). Specifically, suppose $M$ is a transitive model of $\mathsf{ZFC}$. Then we can talk unproblematically about the supremum $S_M$ of the order-types of $M$-definable well-orderings; this supremum will always be $>\mathsf{Ord}^M$ (which is just $\mathsf{Ord}\cap M$ of course) by precisely the example you give.

Especially when $M$ is countable (if $M$ is uncountable we need to perform some circumlocutions) there is a natural computability-theoretic ordinal to compare with $S_M$, namely the supremum $O_M$ of the ordinals computable from an isomorphic copy of $M$ with domain $\omega$. This $O_M$ "sees" that $M$ is countable, so on the face of things doesn't seem to be connected to $M$ very well at all. However, it turns out that this is a rather important thing to consider, so it's natural to compare it to $S_M$. We always have $O_M\ge S_M$, and for $M$ "sufficiently closed" this inequality is strict; however, equality can occur.

If we really want to talk about class relations on $V$ rather than "set-ifying" things, we need to look at some class theory. The standard examples are $\mathsf{NBG}$ (which is a conservative extension of $\mathsf{ZFC}$) and $\mathsf{MK}$ (which isn't). The latter is more appropriate for things related to class-length recursions, which suggests that it's the right framework here. There is definitely some work done on class theories and their relation to "long" ordinals; I think Joel David Hamkins has some good papers on the topic.

Title: Definable well-ordered class relations


问:
Conditional Probability for mutually exclusive events - What is the probability $P(X_j = 1 \ \vert \ \text{Only one of } X_i=1)$
Suppose I have $n$ random variables $X_1, X_2, X_3, ..., X_n$ such that $X_i =\begin{cases} 1 & p_i \\[5pt] 0 & 1-p_i \end{cases}$.

I would like to find the probability that
$$P(X_j = 1 \ \vert \ \text{Only one of } X_i=1)$$

This is an abstraction of a likelihood function$^1$ I am trying to find, but you can also think of it a bit like winning a lottery with only one prize, but everyone has different chances of winning.

My source tells me it's $$\frac{p_j}{\sum_{i=1}^n p_i} \tag{1}$$

but I can't quite understand how they arrived at this.

I feel like it should be more like $$\frac{\frac{p_j}{1-p_j}}{\sum_{i=1}^n \frac{p_i}{1-p_i}} \tag{2}$$

however the result is quite a famous one (Cox's Partial Likelihood function) so I'm pretty sure the answer should be $(1)$.

Is there some interpretation of the fractions $\frac{p_{j}}{1-p_j}$? Are they some sort of rescaled probabilities?

The actual context uses conditional densities $f_{X_i \vert X_i \ge x} (x)$, not probabilities. Might that fix the matter?

1 Cox regression, partial likelihood function, Equation $(12)$

答: $p/1-p$ is called the odds ratio. Your intuition is correct, working with densities changes things a little bit. The “equivalent” of the odds ratio with a continuum of results is called the hazard rate, and Eq (12) of Cox’s paper is just (2) adapted to a continuum of states. That $\lambda$ is not a probability itself but rather the hazard rate, which is the derivative of the log of the probability function.


问:
Numbers in a list which are perfect squares and perfect cubes of numbers
How many numbers in the list $$1,2,3,...,2001$$ are perfect squares and perfect cubes of whole numbers?

My progress: Well I do know $$1,4,9,16,25,36,...$$ are perfect squares and $$1,8,27,64,...$$
are perfect cubes but I can't manage to get a formula/pattern to determine how many there are before 2001 without actually counting them.

Body:
56 perfect squares and cubes before 2001.

Squares:1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169,
196, 225, 256, 289,
324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784,
841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369,
1444, 1521, 1600, 1681, 1764, 1849, and 1936.

Cubes: 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, and 1728.

44 squares and 12 cubes.

Numbers with both perfect squares and cubes in common : 1, (1^2 and 1^3) 64, (8^2 and 4^3) and 729 (27^2 and 9^3)

3 total. 1, 64, 729


问:
Expected value and variance of kth order statistic given maximum value
Let X(1) < X(2) < X(3) < X(4) < X(5) be the order statistics corresponding to a random sample of size 5 from a uniform distribution on [0, θ], where θ ∈ (0, ∞).
Prove that the variance of E[2X(3)| X(5)] <= variance of 2X(3)

Intuitively I understand that given a value, there is more information about X(3) and less variance. But how to mathematically prove it?

答: Hint: Since $2 X_{(3)}$ and $X_{(5)}$ are random variables in the same probability space , you may use the Law of Total Variance.


问:
Geometric view of isomorphism between sets
My studying of Commutative Algebra lead me to the following question (though not a strictly Commutative Algebra question) :

What is a (or the) geometric intuition of the isomorphism
$$\mathbb{R}[x,y]/(x^2-y,x-x^3+xy)\simeq\mathbb{R}$$
So $\mathbb{R}[x,y]$ is the ring of polynomials with variables $x,y$ and $(x^2-y,x-x^3+xy)$ is an ideal of $\mathbb{R}[x,y]$.

I would like to see an awnser for the spesific question for I have searched for 'geometric approaches' of isomorphisms in classics like the Atiyah MacDonald 'Introduction to Commutative Algebra' and in Algebraic Geometry material and I cannot figure out the idea. Really tough, I would like to know your mind considering geometric approaches of isomorphisms 'in general' (perhaps by giving an example that you have in mind, like the one above and of course, the more examples, the better). Also, any sources for further studying or links to previously related posts are welcome.

答:
The short answer: the curves $x^2 - y = 0$ and $x-x^3+xy = 0$ intersect in $1$ point, namely $(0,0)$. Correspondingly, the ring $\mathbb{R}[x,y]/(x^2 - y, x-x^3+xy)$ is isomorphic to $\mathbb{R}$ via the map sending $x \mapsto 0$ and $y \mapsto 0$. Note that

$\mathbb{R}$ has Krull dimension $0$, which reflects that the intersection (a point) is a $0$-dimensional variety; and

$\mathbb{R}$ is a $1$-dimensional $\mathbb{R}$-vector space, corresponding to the number of intersection points of the curves. (In general, the dimension is the number of points counted with multiplicity.)

The longer answer: Let $f_1 = y - x^2$ and $f_2 = x-x^3+xy = x(y - (x^2 - 1))$. To simplify things in the beginning, let’s work over $\mathbb{C}$, which is algebraically closed. The vanishing locus of $\{f_1, f_2\}$ (or equivalently of the ideal $(f_1, f_2)$) is the set $\newcommand{\V}{\mathbb{V}} \newcommand{\C}{\mathbb{C}} \mathbb{V}(f_1,f_2)$ of points $(x_0, y_0) \in \C^2$ such that
$$
f_1(x_0,y_0) = f_2(x_0,y_0) = 0 \, .
$$

Given a point $(x_0, y_0) \in \mathbb{C}^2$, then $(x_0, y_0) \in \V(f_1,f_2) \iff (f_1, f_2) \subseteq (x - x_0, y - y_0)$; see here for a proof. By the correspondence theorem for ideals in a quotient, this means that $(x - x_0, y - y_0)/(f_1, f_2)$ is an ideal, and in fact a maximal ideal of the coordinate ring $\frac{\mathbb{C}[x,y]}{(f_1, f_2)}$. Moreover, all such maximal ideals of $\frac{\mathbb{C}[x,y]}{(f_1, f_2)}$ are of this form by the Nullstellensatz. Thus we get an algebro-geometric dictionary in which the points of the algebraic set given by $f_1 = f_2 = 0$ are in bijection with the maximal ideals of the coordinate ring $\frac{\mathbb{C}[x,y]}{(f_1, f_2)}$.

Over a non-algebraically closed field like $\mathbb{R}$, things are a bit more complicated. Maximal ideals of the coordinate ring $\frac{\mathbb{R}[x,y]}{(f_1, f_2)}$ don’t correspond to $\mathbb{R}$-points of $\V(f_1,f_2)$, but instead to Galois orbits of $\mathbb{C}$-points. For instance, since there are no real solutions to the equation $x^2 + 1 = 0$, the maximal ideal $(x^2 + 1)$ of $\mathbb{R}[x]$ doesn’t correspond to any real point. Instead it corresponds to the Galois orbit $\{i, -i\}$ of complex solutions to the equation, since $x^2 + 1 = (x-i)(x+i)$.

Fortunately in your example things are simple. The vanishing locus of $f_1 = x^2 - y$ is the parabola $y = x^2$, and the vanishing locus of $f_2 = x(y - (x^2 - 1))$ is the union of the $y$-axis and the parabola $y = x^2 - 1$. As you can see in the plot below, the intersection of these two algebraic sets is simply the point $(0,0)$.

$\hspace{2.5cm}$

Algebraically, one can see this by noting that $(f_1, f_2) = (x,y)$, since
$$
x = f_2 + x f_1 \qquad \text{and} \qquad y = f_1 + x^2 = f_2 + x(f_2 + x f_1) \, .
$$
Thus
\begin{align*}
\frac{\mathbb{R}[x,y]}{(f_1, f_2)} = \frac{\mathbb{R}[x,y]}{(x,y)} \cong \mathbb{R}
\end{align*}
where the last isomorphism takes $x \mapsto 0, y \mapsto 0$.

For similar connections between algebra and geometry, see this post and this post. For more on the algebro-geometric dictionary and how geometric properties are reflected in the corresponding algebraic structures, see this post.


答:
Squeezing works quite well here:

\begin{eqnarray*}1
& \leq & \frac{\log (n+2)}{\log (n+1)} \\
& = & \frac{\log (n+1 + 1)}{\log (n+1)} \\
& = & \frac{\log \left((n+1)\left(1+\frac 1{n+1} \right)\right)}{\log (n+1)} \\
& = & \frac{\log (n+1)+ \log \left(1+\frac 1{n+1} \right)}{\log (n+1)} \\
& = & 1+ \frac{\log \left(1+\frac 1{n+1} \right)}{\log (n+1)} \\
& \leq & 1+ \frac{\log 2}{\log (n+1)} \\
\end{eqnarray*}
Now, send $n\to \infty$.


问:
Are there any equation that could produce a monotonic smooth step function by parameter
I want to write a mathematic formular that, given any number of monotonic arbitrary point, it will produce a monotonic smooth step function

Such as a figure below, I give it 2 point (the intersect point of red line and black curve). It will plot a graph as a function curve that pass all given points with derivative 0, so between each point will be like S curve

I want a function that was not being step, being a single function that could inverse and differentiate/integrate. So it could be use in forward and reverse calculation

Are there any function that fit this requirement?

答:
I have found a simple solution. Which is just keep adding logistic function and shifting offset of the x. The result is nearly smooth step as I wish

This need more tweak to have each term join together at the same tangent point and manipulate variable to have each step start and end at the point I expect. But for now I think this is what satisfied me the most

https://www.desmos.com/calculator/hzynrk3tu1


问:
Maximising exponent of a Gaussian random variable
I am trying to solve the maximisation problem defined by:

$$\max_x \mathbb{E}_t \Big[ -e^{-\gamma (\omega + x (Y_{t+1} - Y_t))} \Big]   $$

where I am told that $Y_{t+1} - Y_t \sim N(0,\sigma_Y ).$ It should be straightforward to verify that
$$ x^* = \frac{\mathbb{E}(Y_{t+1} - Y_t)}{\gamma \sigma_Y}. $$

I cannot seem to back out this simple expression though, I must be making a simple algebraic mistake somewhere. I am trying to use the fact that for a lognormal we have $\mathbb{E}_t\Big[e^{(Y_{t+1}-Y_t)}\Big]=e^{\sigma_Y^2/s}.$ This hasn't helped though.

答:
If we just get rid of the clutter, your question boils down to

$$\min_x \mathbb{E}_t \Big[ e^{-\gamma x Z} \Big] = \min_x e^{\frac{1}{2}\gamma^2 x^2 \sigma_Y^2}$$

The minimum is attained for $x^*=0$.


问:
Condition for a rigid body in equilibrium on a horizontal surface
The following question is from a past paper from Further Mathematics and it has been bothering me immensely I have spent hours cracking at a way to make sense of it, the question is in two parts

 Part 1:

An object consists of a uniform solid circular cone, of vertical height 4r and radius 3r, and a uniform solid cylinder, of height 4r and radius 3r. The circular base of the cone and one of the circular faces of the cylinder are joined together so that they coincide. The cone and the cylinder are made of the same material.

 Find the distance of the centre of mass of the object from the end of the cylinder that is not attached
to the cone.

I first roughly sketched the object as such

Then finding the $\bar{x}$ of each part seperately:

For the cone I used the standard result of the centre of mass being $\frac{1}{4}r$ away from the base:

$$\bar{x}_{Cone}=\frac{1}{4}.4r + 4r=5r$$

For the cylinder I derived $\bar{x}$ via integration:

$$y=3r$$

$$\bar{x}_{cylinder}=\frac{\int_0^{4r} xdV}{\int_0^{4r}dV}$$

$$\because dV=y^2 \pi dx $$

$$\implies \bar{x}_{cylinder}=\frac{\int_0^{4r} 9r^2 \pi x dx}{\int_0^{4r}9r^2 \pi dx}$$

$$\implies \bar{x}_{cylinder}=\frac{9r^2 \pi \int_0^{4r} x dx}{9r^2 \pi\int_0^{4r} 1 dx}$$

$$\implies \bar{x}_{cylinder}=\frac{\frac{1}{2}\left[x^2 \right]^{4r}_0}{\left[x \right]^{4r}_0}$$

$$\therefore  \bar{x}_{cylinder}=2r $$

Then by taking the weighted average of both objects I arrived at the correct value for the distance of the centre of gravity from the base of the cylinder which was

$$\bar{x}=\frac{11}{4}r$$

However the next part has had me at a complete loss for the better part of the day, it states that:

Show that the object can rest in equilibrium with the curved surface of the cone in contact with a horizontal surface.

I tried coming up with a rough sketch for this also

However I do not understand how to tackle this question at all, all I know is that for a body to be at equilibrium on such a surface, the centre of mass must pass through the point of suspension, however clearly this does not give an insight on how to answer this.

The condition for this question given in the marking scheme is that

Can someone explain what this means and why this is the case?

答:
As you correctly found out, the center of gravity $G$ is on the axis of the cone and cylinder ($BF$ in the diagram) and $\frac{11r}{4}$ from $F$. So from the vertex of the cone $B$, it is at $\frac{21r}{4}$ distance $(\gt BI = 4r)$.

We know $G$ is between $F$ and $I$. For the solid to be in equilibrium with the curved surface of the cone touching a horizontal surface (ray $AB$), we need to show that $G$ is directly above a point on segment $AB$, in other words we need to show that $G$ is between $H$ and $I$.

$\small AB = \sqrt{AI^2 + BI^2} = 5r$

If $\small \angle ABH = \theta$,

$\displaystyle \small \cos \theta = \frac{BI}{AB} = \frac{4}{5}$.

Now in $\triangle ABH, \displaystyle \small \cos \theta = \frac{4}{5} = \frac{AB}{BH} = \frac{5r}{BH}$

$\implies \displaystyle \small BH = \frac{25r}{4} \gt \frac{21r}{4} = BG$

This proves that $G$ is between $H$ and $I$.


Title: Evaluating $\int_{0}^{1} \frac{x^{2n+2}}{x^2-2x+2}\,{\rm d}x$

Body:
$$\int_{0}^{1} \frac{x^{2n+2}}{x^2-2x+2} \,{\rm d}x \tag{1}$$

This integral came up as a result of another integral I was trying to evaluate. Just for recreational purposes. Is there some closed form for this, even in terms of special functions? In general, is it possible to evaluate integrals of the form:

$$\int_{0}^{1} \frac{P(x,x^n)}{Q(x)} \,{\rm d}x \tag{2}$$

with $\text{deg}(P) > \deg(Q)$? For example:

$$\int_{0}^{1} \frac{x^n}{1+x^2} \,{\rm d}x \tag{3}$$

Can be written in terms of the digamma function. If I'm not mistaken, $(1)$ does not follow from $(3)$, because if you try to complete the square in the denominator of $(1)$, you'd end up having to perform a $u$-sub with $u = x-1$ and then a $(u+1)^{2n+2}$ in the numerator. This ruins things. Can someone give some help?

Body:
$$I=\int \frac{x^{2n+2}}{x^2-2x+2}\, dx=\int x^{2n}\frac{x^2}{(x-(1+i)) (x-(1-i))}\, dx$$
$$ \frac{x^2}{(x-(1+i)) (x-(1-i))}=1+\frac{1}{x-(1-i)}+\frac{1}{x-(1+i)}$$ So, we face the problem of
$$I_a=\int \frac{x^{2n}}{x-a}\,dx$$ where $a$ is a complex number.
$$I_a=-\frac{x^{2 n+1} }{a (2 n+1)}\, _2F_1\left(1,2 n+1;2 n+2;\frac{x}{a}\right)$$
$$J_a=\int_0^1 \frac{x^{2n}}{x-a}\,dx=-\frac{\, _2F_1\left(1,2 n+1;2 n+2;\frac{1}{a}\right)}{(2  n+1)a}$$ As a total
$$K_n=\int_0^1 \frac{x^{2n+2}}{x^2-2x+2}\, dx=
\frac 1{2n+1}-$$ $$\frac{(1-i) \left(\, _2F_1\left(1,2 n+1;2n+2;\frac{1-i}{2}\right)+i
   \, _2F_1\left(1,2 n+1;2n+2;\frac{1+i}{2}\right)\right)}{2(2 n+1)}$$

Now, depending on the parity of $n$, we have two kinds of results.
$$K_{2m}=(-1) ^m\alpha_m+(-4)^m \,\log(2)$$
$$K_{2m+1}=(-1) ^m\beta_m+(-4)^m\,\pi$$
$$\left(
\begin{array}{ccc}
m & \alpha_m & \beta_m \\
0 & 1 & \frac{10}{3} \\
 1 & \frac{79}{30} & \frac{436}{35} \\
 2 & \frac{14087}{1260} & \frac{174434}{3465} \\
 3 & \frac{3990557}{90090} & \frac{1810840}{9009} \\
 4 & \frac{217474889}{1225224} & \frac{260047122}{323323} \\
 5 & \frac{10326377909}{14549535} & \frac{71767882892}{22309287} \\
 6 & \frac{19001942777579}{6692786100} & \frac{64592068202966}{5019589575} \\
 7 & \frac{3306285643032971}{291136195350} &
   \frac{232272319634479216}{4512611027925}
\end{array}
\right)$$


问:
For a regular Borel measure with infinite support, how to find a decreasing sequence of positive measure open sets?
Let $K$ be a compact Hausdorff space and $\mu$ be a regular Borel measure on $K$ with infinite support. How to find a decreasing sequence of open sets $\{V_{n}\}$ such that $\mu(V_{n}) > 0$ and $\lim_{n \rightarrow \infty} \mu(V_{n}) = 0$?

If $(X, \tau)$ is a topological Hausdorff space and $\mu$ is a measure on $X$, then definition of support of a measure is $\{x \in X: x \in N_{x} \in \tau \implies \mu(N_{x})>0\}$.

I am sure that the regularity of $\mu$ plays a role in finding the above decreasing sequence, but I don't know how. Please provide hints.

答:
Consider $\{x\in supp (\mu): \mu (\{x\}) >0\}$. If this is infinite then
there exist $x_n$'s with $\mu (\{x_n\}) >0$ for all $n$ and $\sum \mu (\{x_n\}<\infty$. Hence, $\mu (\{x_n\})  \to 0$. By going to a subsequence we may suppose $\mu (\{x_n\}) <\frac 1 {2^{n}}$.  Now there exist open sets $U_n$ with  $\mu(U_n)<\mu (\{x_n\})+\frac 1 {2^{n}}$ and $x_n \in U_n$. Note that $\mu(U_n) >0$ for all $n$ because $x_n$ belongs to the support of $\mu$.  Take $V_n=\bigcup_{k\geq n} U_k$.

The case where $\{x\in supp (\mu): \mu (\{x\}) >0\}$ is a finite set is easier. Here there exists at least one point $x$ in the support with $\mu (\{x\})=0$. Can you finish?


问:
How to show $e_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}$ is standard orthogonal basis of $L^2[0,2\pi]$?
I find the conclusion in a book, but without proof. It say that  $e_n(t)=\frac{1}{\sqrt{2\pi}}e^{int}$ is standard orthogonal basis of $L^2[0,2\pi]$. I can prove that
$$
\lVert e_n \rVert =1
$$
and
$$
(e_n,e_m)=\delta_{nm}
$$
But I don't know how to show it is a basis, namely, for any $u\in L^2[0,2\pi]$, the $u$ can be presented as
$$
u=\sum\limits_{i=1}^\infty  (u, e_i) e_i.
$$
I know it is equal to prove the Parseval equation, namely
$$
\lVert u \rVert^2=\sum\limits_{i=1}^\infty |(x,e_i)|^2.
$$
Or equal to prove
$$
\{e_n\}^\bot=\{\theta\}.
$$
But no one of them I can prove.

答:
Assume that the orthogonal complement is non-zero, containing $f$.

Think to it as a $2\pi$-periodic function. The shifts of $f$ are in the orthogonal complement too. For the right choice of $\phi$ continuous we get that $g(t) = \int_0^{2\pi} f(t-u)\phi(u)du$ is non-zero, continuous, and orthogonal to every $e^{int}$.

But if $g(a)\ne 0$ then for $m$ large enough $\int_0^{2\pi} g(t) \left(\frac{1+\cos(t-a)}2\right)^mdt$ can't be $0$, contradicting that $g$ is orthogonal to the exponentials.


问:
Are Rational Power of e is transcendental?
It is well known to all of us that the rational powers of $e$ are irrational numbers. Many of the proofs proving this use a similar approach as proving $e$ irrational using Niven's Polynomials. Is it true that rational powers of $e$ are also transcendental numbers using proofs similar(as proved by Hermite) for proving $e$ transcendental?

How to measure the irrationality measure of those rational powers of $e$?

答: In field theory language: We have $\mathbb{Q}(e^{m/n})\subseteq\mathbb{Q}(e^{1/n})$ so WLOG we may consider $e^{1/n}$. Observe $\mathbb{Q}(e^{1/n})/\mathbb{Q}(e)$ is finite and $\mathbb{Q}(e)/\mathbb{Q}$ is infinite, so $\mathbb{Q}(e^{1/n})/\mathbb{Q}$ is infinite: $e^{1/n}$ is transcendental.


问:
Is there a specific name for parallel line segments that are the same length and aligned?
I'm trying to name an algorithm that identifies line segments with the following properties.

They are parallel.

They have the same length.

Their endpoints are aligned along the perpendicular axis.

In other words, if you connected them, they would form a rectangle.

Is there a specific name for parallel lines with these properties?

Edit: I've taken to calling lines with this property "co-rectangular"

答:
It is a particular case of "equipollent line segments".

See https://en.wikipedia.org/wiki/Equipollence_(geometry).


问:
Probability of going to school A
I commute to teach at one of two schools everyday by train.
I live at the terminal station where every 30 min there is a train leaving for school A, every 60  min there is a train leaving for school B. I will jump on the earliest train available at the station and teach at that school.

My  question is: What is the probability I teach at school A?   Since the frequency of the train leaving for school A is twice compared to that of the train leave for school B, my guess is that the probability is $\frac{2}{3}$. However, I am not sure if my intuition is correct. Can anyone help me out?

答:
I suppose it is a sort of puzzle, which makes all sorts of (mostly unwarranted) simplifying assumptions. So I assume

$\bullet$ Train A arrives/departs punctually and instantaneously at $8:00, 8:30, 9:00,..$

$\bullet$ Train B arrives/ departs punctually and instantaneously at $8:15$, $9:15$, etc.

$\bullet$ I arrive with equal probability at any time  between $8:00$ and $9:00$ and catch whichever train arrives next.

So if  I arrive between $8:00$ and $8:15$, I catch train $B$ and if I arrive between $8:15$ and $8:30$, it is train $A$
If I arrive between $8:30$ and $9:00$ I will only get train $A$, so  overall probability of catching train $A$ and hence going to school $A = \frac12\frac12 + 1\cdot\frac12 = \frac34$

Title: Probability of going to school A


问:
Partial sum of ${A \choose i} {B\choose n-i}$, when $B=-1$?
It's easy to see that
$$ \sum_i {A\choose i} {B\choose n-i} = {A+B\choose n} $$
since when we choose $n$ things out of $A+B$, some ($i$ of them) are in the $A$ and the rest are in the $B$.

Is there any reasonable formula for
$$ \sum_{i< I} {A\choose i} {B\choose n-i} = {A+B\choose n}, $$
i.e. we have a bound on how many of them are from the $A$ side?

Is the $B=1$ case any easier? (That being my real question.)

EDIT: I totally misasked this question, and have hopefully fixed it here: Partial sum over $M$, of ${m+j \choose M} {1-M \choose m+i-M}$?

答:
For B=1 this is rather easy:

${1 \choose j}$ is $1$ if $j$ is $0$ or $1$, and is $0$ for any other value of $j$.  So your first expression becomes $ {A\choose n-1} {1\choose 1} + {A\choose n} {1\choose 0} = {A+1 \choose n} $
or as Fabian says, Pascal's rule

$$ {A\choose n-1} + {A\choose n} = {A+1 \choose n}. $$

Your second expression depends on $I$.  If $I \le n-1$ then it gives a sum of 0; if $I = n$ and $i \lt I$ it give a sum of  ${A\choose n-1}$, and if $I \gt n$ it gives a sum of ${A+1 \choose n}$.


问:
How did they prove this big o inequality
I was reading my textbook when I came across this example

I was wondering how they got 35n^3 since we have log(n)?

Body:
In fact, $\color{red}{\log n<n}$ for all $n>1$, so $20n^3+10n\log n+5<20n^3+10n^2+5$, and $n^2<n^3,1<n^3$, so $10n^2<10n^3$, $5<5n^3$, so $20n^3+10n\log n+5<20n^3+10n^2+5<20n^3+10^3+5n^3=35n^3$.

Given $f(x)=x-\log x$, then $f^{'}(x)=1-\frac{1}{x}$, so when $x>1$, $f^{'}(x)>0$, so $f(x)$ increase on $(1,\infty)$, and $f(x)>f(1)=1$ for $x\in (1,\infty)$. As for the case in $\mathbb{N}$, we always have $n\geq1$, so $\color{red}{n>\log n}$.


问:
Is $\exists a \in A. \forall b \in B \Rightarrow P(a, b)$ equivalent to $\forall c \in B. \exists d \in A \Rightarrow P(c, d)$?
Given 2 statements:

$$\exists a \in A. \forall b \in B \Rightarrow P(a, b)$$
and
$$\forall c \in B. \exists d \in A \Rightarrow P(d, c)$$

where lowercase letters are elements(or items);
A and B are sets that define the elements; P is a relationship between the two items in the bracket.

Are they equivalent to each other?
And Why?

答:
No.

It is not true that there exists a woman on Earth (a in A) such that for every man on Earth (b in B), the woman is his mother [P(a,b)].

However, it is true that for every man on Earth (c in B) there exists a woman on Earth (d in A) such that the woman is his mother [P(c,d)].


问:
Explanation of higher order derivative from Lawrence Perko book
enter image description hereenter image description here
In the Lawrence Perko differential equations and dynamical systems book Page no 338, Theorem $1$, he introduced the notion $D^2f(x_0,\mu_0)(v,v)$ and $D^3f(x_0,\mu_0)(v,v,v)$. He also directly wrote the values of these quantities in subsequent examples. I have understood the notation $Df(x_0,\mu_0)$ but could not understand the higher orders how this vector $v$ applied? Can someone please explain it in detail

答:
$D$ is the differentiation operator in $x$ direction. Thus
\begin{align}
Df(x,\mu)(v)&=\sum_i\frac{∂ f}{∂ x_i}(x,μ)\,v_i\\
D^2f(x,\mu)(v,v)&=\sum_{i,j}\frac{∂^2 f}{∂ x_i∂ x_j}(x,μ)\,v_iv_j\\
D^3f(x,\mu)(v,v,v)&=\sum_{i,j,k}\frac{∂^3 f}{∂ x_i∂ x_j∂ x_k}(x,μ)\,v_iv_jv_k\\
\end{align}
etc.


问:
Topology of $(S^1 \times S^1)/C_n$
Prepare two $S^1$. Let a cyclic group $C_n$ of order $n$ acts on each $S^1$ by rotation. The actions may be distinct and may not be faithful. What is the topology of the quotient space $(S^1 \times S^1)/C_n$?

I think this is again $S^1 \times S^1$ as topological space. Am I right?

答: This questions is a duplicate of What is $(S^1\times S^1)/C_{n}$ topologically?. You may be interested to read Hatcher's section on lens space where he did this in more detail. </p>

Title: Topology of $(S^1 \times S^1)/C_n$


问:
Let X and Y be jointly distributed random variables having the joint probability density function
Let $X$ and $Y$ be jointly distributed random
variables having the joint probability
density function
[
f(x,y) = \begin{cases}
            \frac{1}{\pi}, &\text{if}\quad x^2 + y^2 \leq 1\\
             0, &\text{otherwise}\\
            \end{cases}

Then $P(Y > |X|)$ is

I tried writing integration for this but I am getting wrong. I tried writing cdf for this function as well but that did not help. How should I proceed further?

答:
$(X,Y)$ is uniformly distributed on the unit disk (the disk centered in the origin with radius 1), thus geometrically you immediately get that

$$\mathbb{P}(Y>|X|)=\frac{1}{4}$$

No calculations are needed, just a geometric observation because the favourable area is $\frac{1}{4}$ of the total disk


问:
Composite of Homeomorphisms is Homeomorphism
In this ProofWiki article, Composite of Homeomorphisms is Homeomorphism

By definition of homeomorphism, $f$ and $g$ are both bijections.

From Composite of Bijections is Bijection it follows that $g∘f$ is also
a bijection.

Similarly, also by definition of homeomorphism, $f$ and $g$ are both
continuous mappings.

From Composite of Continuous Mappings is Continuous it follows that
$g∘f$ is also a continuous mapping.

Hence the result, from definition of homeomorphism.

I feel like the proof is not complete, as by the definition on ProofWiki of a Homeomorphism

Definition 1
$f$ is a homeomorphism if and only if both $f$ and $f^{−1}$ are
continuous.

I think the proof should also mention that $(g \circ f)^{-1}$ is also continuous as $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ and each $f^{-1}$ and $g^{-1}$ are continuous by the definition of homeomorphism. And still, composite of continuous mapping is continuous.

Hence, $(g \circ f)^{-1} = f^{-1} \circ g^{-1}$ is continuous.

Is my extra step needed in the proof? Or it is implied somehow in the proof itself?

Any feedback is appreciated!

答:
You're correct to say that specifying continuity of the inverse is necessary. It is not implied by the given proof. For instance, we can find two continuous bijections that are not homeomorphism, and such that their composition isn't a homeomorphism either. Hence, continuity of the inverse cannot be assumed from the given proof.

Let $\mathbb R_0$ be $\mathbb R$ as a set with the discrete topology, let $\mathbb R_1$ be $\mathbb R$ with the usual metric topology, and let $\mathbb R_2$ be $\mathbb R$ with the indiscrete topology, i.e. the only open sets are the empty set and the whole space. Then the identity maps $\mathbb R_0 \longrightarrow \mathbb R_1 \longrightarrow \mathbb R_2$ are continuous bijections which are not homeomorphisms, as is their composition.


答:
Here's how I think about it. $\mathbb{R}[x, y]$ is the ring of all polynomial functions on $\mathbb{R}^2$ under pointwise addition and multiplication. The "$\mathbb{R}$" on the right-hand side of your isomorphism is not some abstract version of the real field. It's actually all polynomial functions whose domain is the single point $(0, 0) \in \mathbb{R}$.

Thought of in these terms, the restriction from $\mathbb{R}^2$ to $\{(0, 0)\}$ corresponds to the obvious surjective map $\mathbb{R}[x, y] \to \mathbb{R}$ given by $x \mapsto 0, y \mapsto 0$ that just picks off the constant term. This is a ring homomorphism because function arithmetic is defined pointwise, which restriction clearly respects.

As for the ideal $(x^2-y, x-x^3+xy)$, passing to the quotient $\mathbb{R}[x, y]/(x^2-y, x-x^3+xy)$ literally says we want to consider all polynomial functions on $\mathbb{R}^2$ with the additional relations that $x^2-y = 0$ and $x-x^3+xy = 0$ but no others except those implied by these. The claimed isomorphism is a concrete geometric way to think of what this abstract quotient "looks like": if it's true, it says that any two polynomials in $\mathbb{R}[x, y]$ with the same constant term can be shown to be equal if you allow yourself to write $x^2-y=0$ and $x-x^3+xy=0$ in intermediate steps, and that there is no additional "collapsing" beyond this.

To prove it, first off the two relations respect restriction to $\{(0, 0)\}$ since $0^2 - 0 = 0$ and $0-0^3+0\cdot0 = 0$, so restriction continues to give a surjective map $\mathbb{R}[x, y]/(x^2-y, x-x^3+xy) \to \mathbb{R}$. It is natural to ask if this map is injective as well. Geometrically, that's equivalent to asking: having imposed the two abstract relations above and no more, can I have two different polynomial functions which have the same value at $(0, 0)$?

Before answering that question, let's consider what would happen if $x^2-y$ and $x-x^3+xy$ were also both zero at some point $(a, b)$ different from $(0, 0)$. Then we could pick polynomials $p(x, y), q(x, y)$ such that $p(0, 0) = q(0, 0)$ yet $p(a, b) \neq q(a, b)$. Using the restriction to $\{(a, b)\}$ rather than $\{(0, 0)\}$, that would force $p \neq q$ in the quotient even though their constant terms are the same, so the map would not be injective. Hence more common zeros keep the quotient bigger. However, it's easy to see there are no common zeros here beyond $(0, 0)$, so this obstruction doesn't appear in this case.

It's inevitable at this point that we'll have to start playing with algebra. For instance, we get the same obvious surjection $\mathbb{R}[x, y]/(x^2, y^2) \to \mathbb{R}$ by restricting to $\{(0, 0)\}$, but the quotient is larger. (The map which takes the partial derivative with respect to $x$ and then evaluates at $(0, 0)$ is well defined.)

Ok, let's play. We've got $y=x^2$ and $xy=x^3-x$. If we could divide by $x$ the second would be $y=x^2-1$ and then we'd have imposed $0=-1$ which would result in the zero ring and everything would collapse completely. But we can't divide by $x$, roughly since that might mean dividing by $0$. But if we multiply $y=x^2$ by $x$ we can do almost the same thing: $xy = x^3$ and $xy = x^3-x$. Together this says $x=0$, but since $y=x^2$ we get $y=0$ too. Thus in the quotient $p(x, y) = p(0, 0) + x(\cdots) + y(\cdots) = p(0, 0)$ always, and our map is indeed injective, hence bijective, hence an isomorphism.

Title: Geometric view of isomorphism between sets

Body:
My studying of Commutative Algebra lead me to the following question (though not a strictly Commutative Algebra question) :

What is a (or the) geometric intuition of the isomorphism
$$\mathbb{R}[x,y]/(x^2-y,x-x^3+xy)\simeq\mathbb{R}$$
So $\mathbb{R}[x,y]$ is the ring of polynomials with variables $x,y$ and $(x^2-y,x-x^3+xy)$ is an ideal of $\mathbb{R}[x,y]$.

I would like to see an awnser for the spesific question for I have searched for 'geometric approaches' of isomorphisms in classics like the Atiyah MacDonald 'Introduction to Commutative Algebra' and in Algebraic Geometry material and I cannot figure out the idea. Really tough, I would like to know your mind considering geometric approaches of isomorphisms 'in general' (perhaps by giving an example that you have in mind, like the one above and of course, the more examples, the better). Also, any sources for further studying or links to previously related posts are welcome.


问:
Two questions about Euler's number $e$
I am on derivatives at the moment and I just bumped into this number $e$, "Euler's number" . I am told that this number is special especially when I take the derivative of $e^x$ , because its slope of any point is 1. Also it is an irrational ($2.71828\ldots$) number that never ends, like $\pi$.

So I have two questions, I can't understand

What is so special about this fact that it's slope is always 1?

Where do we humans use this number that is so useful, how did Mr Euler come up with this number?

and how come this number is a constant? where can we find this number in nature?

答:
You don't take the derivative of a constant. You could, but it's zero.

What you should be talking about is the exponential function, $ e^x $ commonly denoted by $ \exp(\cdot ) $. Its derivative at any point is equal to its value, i.e. $ \frac{d}{dx} e^x \mid_{x = a} = e^a $. That is to say, the slope of the function is equal to its value for all values of $ x $.

As for how to arrive at it, it depends entirely on definition. There are numerous ways to define $ e $, the exponential function, or the natural logarithm. One common definition is to define $$ \ln x := \int\limits_1^x \frac{1}{t} \ dt $$ From here, you can define $ e $ as the sole positive real such that $ \ln x = 1 $ and arrive at it numerically.

Another common definition is $ e = \lim\limits_{n \to \infty}\left(1 + \frac{1}{n}\right)^n $, although in my opinion it is easier to derive properties from the former definition.

Title: Two questions about Euler's number $e$


问:
Analytic Solution to the Differential equation $(1-t^2)x'' + tx' - x = 0$.
I am looking to solve the following problem:

Consider the differential equation,

$$(1-t^2)x'' + tx' - x = 0$$

(a) Find a power series expansion about t = 0 for a general solution to the differential equation. Your answer should include a general formula for the coefficients.

(b) Solve the IVP $$(1-t^2)x'' + tx' - x = 0, x(0) = -1, x'(0) = 2$$

I have come to a dead end on this problem myself after my work and solution shown below;

$$x(t) =  \sum_{n=0}^\infty a_nt^n$$
$$x'(t) =  \sum_{n=1}^\infty a_nnt^{n-1}n$$
$$x''(t) =  \sum_{n=2}^\infty a_nn(n-1)t^{n-2}$$

Subsituting into the differential equation and simplifying gave:

$$\sum_{n=0}^\infty [a_{n+2}(n+2)(n+1) - a_nn(n-1)]t^n = 0$$

Since $t^n > 0$, we must have

$$a_{n+2}(n+2)(n+1) - a_nn(n-1) = 0 $$

This gives the recursive sequence:

$$a_{n+2} = \frac{a_nn(n-1)}{(n+2)(n+1)}$$

I am stuck here and am not sure how to proceed with solving the rest of the problem. I am looking for a solution to the initially stated problem so that I can compare my work to it and learn from it. As a note, I suspect my formula for the recursive sequence is incorrect since it terminates, although I am not sure.

Body:
The coefficient iteration is
$$
t^n:\, 0=(n+2)(n+1)a_{n+2}-n(n-1)a_n+na_n-a_n=(n+1)(n+2)a_{n+2}-(n-1)^2a_n
$$
For the first powers this gives
\begin{align}
2a_2&=a_0\\
6a_3&=0\\
12a_4&=a_2\\
20a_5&=4a_3=0\\
30a_6&=9a_4
\end{align}
so the general solution starts as
$$
x(t)=x'(0)t+x(0)\left(1+\frac12t^2+\frac1{24}t^4+\frac1{80}t^6+...\right)
$$


问:
On continuity of the Gateaux derivative of p-Laplacian operator
Let $\Omega\subset \mathbb{R}^n, N\geq3$, be an open set. For $p\in(1,+\infty)$, define a functional $J:W_0^{1,p}(\Omega)\rightarrow\mathbb{R}$ by

$J(u)=\int_\Omega |\nabla u|^p\,dx.$

Then $J$ is differentiable in $W_0^{1,p}(\Omega)$ and

$J'(u)v=p\int_\Omega |\nabla u|^{p-2}\nabla u\cdot\nabla v\,dx.$

My question is about how to prove the Gateaux derivative $J_G'(u)$ is continuous. Indeed, for any sequence $u_k\rightarrow u$, I can prove that there exists a subsequence $u_{k_j}$ satisfying $J_G'(u_{k_j})\rightarrow J_G'(u)$ in [$W^{1,p}(\Omega)$]', but I can't prove it for the original sequence.

Thanks for any help!

答:
Use the following convergence principle:

A sequence is converging to some $x$ if and only if every subsequence contains a subsequence that converges to $x$.


问:
Ordinary Power series
How do I find an  exponential generating function for $$\{p(n)/n!\}_{1}^{\infty}$$ where $$ p \in \mathbb{Z}^{+}[x] $$I see a method to find an opsgf for the same sequence by applying $x \frac{d}{dx}$ operator to $e^x$ but I am not able to figure out a good way for egf. Help would be appreciated.
Thank You.

答:
With the modified Bessel function $I_0$ we have

$$ \begin{array}{ll} \displaystyle\sum_{n=0}^\infty p(n)\frac{x^n}{n!^2} & \displaystyle = \sum_{n=0}^\infty \Big(\sum_{k=0}^d a_kn^k\Big)\frac{x^n}{n!^2} \\[5pt] & \displaystyle = \Big[\sum_{k=0}^d a_k\Big(x\frac{d}{dx}\Big)^k\Big]\sum_{n=0}^{\infty} \frac{x^n}{n!^2} \\[5pt] & = \displaystyle p\Big(x\tfrac{d}{dx}\Big) I_0(2\sqrt{x}). \end{array} $$

Title: Ordinary Power series


问:
On proving $\partial_\theta^n\int_0^1t^{\theta-1}\,\mathrm dt=\int_0^1\partial_\theta^nt^{\theta-1}\,\mathrm dt$
Let $n=1,2\dots$ and $\theta\in(0,\infty)$. I want to prove
$$
\partial_\theta^n\int_0^1t^{\theta-1}\,\mathrm dt=\int_0^1\partial_x^nt^{\theta-1}\,\mathrm dt.
$$
I was using this document as a reference for understanding the conditions under which integration and differentiation can be interchanged. Using the notation in the linked paper: $\Theta=(0,\infty)$, $X=[0,1]$, and $g:\Theta\times X\to \Bbb R$ with $g(\theta,x)=x^{\theta-1}$.

Clearly,
$$
\int_0^1|g(\theta,x)|\,\mathrm dx<\infty,\quad\forall\theta\in\Theta.
$$
Furthermore, its easy to show that $|\partial_\theta^ng(\theta,x)|=(-\log x)^{n}x^{a-1}$ and
$$
\int_0^1|\partial_\theta^ng(\theta,x)|\,\mathrm dx<\infty,\quad\forall\theta\in\Theta.
$$
If I am understanding correctly, all that is left to show is uniform local integrability. If so, I have to show for all $\theta_0\in\Theta$ that there is an $\epsilon>0$ such that
$$
\sup_{\theta\in(\theta_0-\epsilon,\theta_0+\epsilon)}|\partial_\theta^ng(\theta,x)|\leq h(x),
$$
where $\int_0^1h(x)\,\mathrm dx<\infty$. But how do I find the dominating function $h$? It is true that
$$
\sup_{\theta\in(\theta_0-\epsilon,\theta_0+\epsilon)}|\partial_\theta^ng(\theta,x)|\leq (-\log x)^n x^{\theta_0-\epsilon-1},
$$
which is integrable if $\theta_0>\epsilon$. Since I could choose $\epsilon=\theta_0/2$ it would seem I can always find a dominating function that is integrbale, however, this dominating function is not independent of $\theta$.

How to I prove the uniform local uniform integrability condition here to finish the proof?

答:
Notice that
$$\partial^k_{\theta} (x^{\theta-1}) = x^{\theta-1} \log^k x.$$

Let $\theta_0 > 0.$
Then, for $\theta \in [\theta_{0},\infty)$ we have
$$|x^{\theta-1}\log^k x| \leq x^{\theta_0-1} |\log^k x|$$
for all $x \in (0,1),$
and the right hand side is integrable, independent of $\theta$.

By the criterion above, the function $\theta \mapsto \int_0^1 x^{\theta-1} d x$ is $k$ times differentiable in $(\theta_0, \infty)$ with $k$-th derivative $\int_0^1 x^{\theta-1} \log^k x \,d x.$
Since $\theta_0 > 0$ was arbitrary, it follows that the function is $k$ times differentiable in $(0,\infty).$


问:
Why is the answer to this $2$ dice and a coin probability question not $\frac{1}{2}$?
I was trying to solve the following homework probability question which has the following setup:

We have $2$ dice: $A$ and $B$. Die $A$ has $4$ red faces and $2$ white faces, whereas die $B$ has $2$ red faces and $4$ white faces. On each turn, a fair coin is tossed. If the coin lands on heads then die $A$ is thrown, but if the coin lands on tails then die $B$ is thrown. After this the turn ends, and on the next turn the coin is once again tossed to determine the die (i.e. we throw the coin and the corresponding die on each turn).

From this game I'm asked to answer $2$ questions:

Show that the probability of obtaining a red face on any $n$-th throw is $\frac{1}{2}$.

If the first $2$ consecutive die throws result in red faces, what is the probability that the third throw is also red?

To answer part $1$ I used the law of total probability. Denoting obtaining a red face on the $n$-th die throw as $P(R_n)$ I get

\begin{align*}
P(R_n) &= P(R_n \vert A) P(A) + P(R_n \vert B) P( B) \\
& = \left(\frac{4}{6}\right)\left(\frac{1}{2}\right) + \left(\frac{2}{6}\right)\left(\frac{1}{2}\right)\\
& = \frac{1}{2}
\end{align*}

But on question $2$ is where I started running into trouble. Using the same notation, what I want to calculate is $P(R_3 \vert R_2 R_1)$. And recalling that for events $E_1$ and $E_2$ we can say that
$$
P(E_2 \vert E_1) = \frac{P(E_2E_1)}{P(E_1)}
$$
I get that
$$
P(R_3 \vert R_2 R_1) = \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)} \tag{1}
$$
and from here I obtain 2 different solutions using $2$ distinct methods:

Answer 1

Using that on the first part of the question we showed that $P(R_n) = \frac{1}{2}$, and noticing that the die throws are independent since what I threw before does not affect how I throw the next coin toss or how I roll the next die, from equation $(1)$ I get
\begin{align*}
P(R_3 \vert R_2 R_1) &= \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)}\\
&= \frac{P(R_3) P(R_2) P(R_1)}{P(R_2) P(R_1)}\\
& = P(R_3) = P(R_n) = \frac{1}{2}
\end{align*}

Answer 2

Using the law of total probability on $(1)$ I get
\begin{align*}
P(R_3 \vert R_2 R_1) & = \frac{P(R_1 R_2  R_3 \vert A)P(A)+ P(R_1  R_2  R_3 \vert B)P(B)}{P(R_1  R_2 \vert A ) P(A)+ P(R_1  R_2 \vert B) P(B)}\\
&= \frac{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+ \left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+\left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}\\
& = \frac{3}{5}
\end{align*}

To me both of the previous answers seem to be following coherent logic, but since I didn't get the same answer I knew one of them was wrong. I decided to write a program to simulate the game and I found out that the correct solution was $P(R_3 \vert R_2 R_1) = \frac{3}{5}$. But even though I verified this answer to be correct I couldn't seem to understand what part of my analysis is wrong on Answer 1.

So my question is, why is
$$
P(R_3 \vert R_2 R_1) \neq \frac{1}{2}\quad ?
$$

答:
The flaw in your second answer is that it is not necessarily the same die that is thrown after each coin toss.  The way you have written the law of total probability is acceptable in your first answer, but not in the second, because it is only in the second answer that you conditioned on the events $A$ and $B$, which represent the outcomes of the coin toss.  Consequently, the result is incorrect because it corresponds to a model in which the coin is tossed once, and then the corresponding die is rolled three times.

First, let us do the calculation the proper way.  We want $$\Pr[R_3 \mid R_1, R_2] = \frac{\Pr[R_1, R_2, R_3]}{\Pr[R_1, R_2]}$$  as you wrote above.  Now we must condition on all possible outcomes of the coin tosses, of which there are eight:
$$\begin{align}
\Pr[R_1, R_2, R_3] &= \Pr[R_1, R_2, R_3 \mid A_1, A_2, A_3]\Pr[A_1, A_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, A_2, B_3]\Pr[A_1, A_2, B_3] \\
&+ \Pr[R_1, R_2, R_3 \mid A_1, B_2, A_3]\Pr[A_1, B_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid A_1, B_2, B_3]\Pr[A_1, B_2, B_3] \\
&+\Pr[R_1, R_2, R_3 \mid B_1, A_2, A_3]\Pr[B_1, A_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid B_1, A_2, B_3]\Pr[B_1, A_2, B_3] \\
&+\Pr[R_1, R_2, R_3 \mid B_1, B_2, A_3]\Pr[B_1, B_2, A_3] \\ &+ \Pr[R_1, R_2, R_3 \mid B_1, B_2, B_3]\Pr[B_1, B_2, B_3] \\
\end{align}$$
and since each of the $2^3 = 8$ triplets of ordered coin tosses has equal probability of $1/8$ of occurring,
$$\Pr[R_1, R_2, R_3] = \tfrac{1}{8}\left((\tfrac{2}{3})^3 + 3(\tfrac{2}{3})^2(\tfrac{1}{3}) + 3(\tfrac{2}{3})(\tfrac{1}{3})^2 + (\tfrac{1}{3})^3\right)
= \tfrac{1}{8}.$$  A similar (but simpler) calculation for the denominator yields $1/4$, and the result follows.

Of course, none of this is necessary; it is only shown here to illustrate how the calculation would be done if it were to be done along the lines of your second answer.


答:
With the  second approach:

To have first two die throws resulting in red, there are 2x2 paths: AA,AB,BA, BB $\to$

$P(R1\cap R2)=(\frac{1}{2})^2(\frac{2}{3})^2$+2$(\frac{1}{2})^2\frac{2}{3}\frac{1}{3}$+$(\frac{1}{2})^2(\frac{1}{3})^2$ =$\frac{1}{4}$

To have three  die throws resulting in red, there are 2x2x2 paths:AAA,BBB,AAB,ABA,BAA,BBA,BAB,ABB $\to$

$P(R1\cap R2\cap R3)= (\frac{1}{2})^3(\frac{2}{3})^3$+3$(\frac{1}{2})^3(\frac{2}{3})^2\frac{1}{3}$+3$(\frac{1}{2})^3(\frac{1}{3})^2\frac{2}{3}$+ $(\frac{1}{2})^3(\frac{1}{3})^3$ =$\frac{1}{8}$

$\frac{P(R1\cap R2\cap R3)}{P(R1\cap R2)}$ =$\frac{1}{2}$

In this way, you get consistent result compared to first approacch.

Title: Why is the answer to this $2$ dice and a coin probability question not $\frac{1}{2}$?

Body:
I was trying to solve the following homework probability question which has the following setup:

We have $2$ dice: $A$ and $B$. Die $A$ has $4$ red faces and $2$ white faces, whereas die $B$ has $2$ red faces and $4$ white faces. On each turn, a fair coin is tossed. If the coin lands on heads then die $A$ is thrown, but if the coin lands on tails then die $B$ is thrown. After this the turn ends, and on the next turn the coin is once again tossed to determine the die (i.e. we throw the coin and the corresponding die on each turn).

From this game I'm asked to answer $2$ questions:

Show that the probability of obtaining a red face on any $n$-th throw is $\frac{1}{2}$.

If the first $2$ consecutive die throws result in red faces, what is the probability that the third throw is also red?

To answer part $1$ I used the law of total probability. Denoting obtaining a red face on the $n$-th die throw as $P(R_n)$ I get

\begin{align*}
P(R_n) &= P(R_n \vert A) P(A) + P(R_n \vert B) P( B) \\
& = \left(\frac{4}{6}\right)\left(\frac{1}{2}\right) + \left(\frac{2}{6}\right)\left(\frac{1}{2}\right)\\
& = \frac{1}{2}
\end{align*}

But on question $2$ is where I started running into trouble. Using the same notation, what I want to calculate is $P(R_3 \vert R_2 R_1)$. And recalling that for events $E_1$ and $E_2$ we can say that
$$
P(E_2 \vert E_1) = \frac{P(E_2E_1)}{P(E_1)}
$$
I get that
$$
P(R_3 \vert R_2 R_1) = \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)} \tag{1}
$$
and from here I obtain 2 different solutions using $2$ distinct methods:

Answer 1

Using that on the first part of the question we showed that $P(R_n) = \frac{1}{2}$, and noticing that the die throws are independent since what I threw before does not affect how I throw the next coin toss or how I roll the next die, from equation $(1)$ I get
\begin{align*}
P(R_3 \vert R_2 R_1) &= \frac{P(R_3 R_2 R_1)}{P(R_2 R_1)}\\
&= \frac{P(R_3) P(R_2) P(R_1)}{P(R_2) P(R_1)}\\
& = P(R_3) = P(R_n) = \frac{1}{2}
\end{align*}

Answer 2

Using the law of total probability on $(1)$ I get
\begin{align*}
P(R_3 \vert R_2 R_1) & = \frac{P(R_1 R_2  R_3 \vert A)P(A)+ P(R_1  R_2  R_3 \vert B)P(B)}{P(R_1  R_2 \vert A ) P(A)+ P(R_1  R_2 \vert B) P(B)}\\
&= \frac{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+ \left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}{\left[\left(\frac{2}{3}\right)\left(\frac{2}{3}\right)\right]\frac{1}{2}+\left[\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)\right]\frac{1}{2}}\\
& = \frac{3}{5}
\end{align*}

To me both of the previous answers seem to be following coherent logic, but since I didn't get the same answer I knew one of them was wrong. I decided to write a program to simulate the game and I found out that the correct solution was $P(R_3 \vert R_2 R_1) = \frac{3}{5}$. But even though I verified this answer to be correct I couldn't seem to understand what part of my analysis is wrong on Answer 1.

So my question is, why is
$$
P(R_3 \vert R_2 R_1) \neq \frac{1}{2}\quad ?
$$


问:
prove/disprove: the functions $f_n(x)=\cos(x+n)+\ln\left(1+\frac{\sin^2(n^nx)}{\sqrt{n+2}}\right)$ are uniformly equicontinuous

Prove/Disprove that the sequence of functions $(f_n)$ from $\Bbb R$ to $\Bbb R$ defined by $$f_n(x)=\cos(x+n)+\ln\left(1+\frac{\sin^2(n^nx)}{\sqrt{n+2}}\right)$$
  are uniformly equicontinuous. That is, prove/disprove that for all $\epsilon>0$ there is a $\delta>0$ such that $$|x-t|<\delta, n\in\Bbb N\Rightarrow |f_n(x)-f_n(t)|<\epsilon.$$

Thoughts/Context. It is my gut feeling that $f_n$ are uniformly equicontinuous. I have a not-very-rigorous "proof" of this. For $n\in\Bbb N$ and $x\in\Bbb R$, $\sin^2(n^n x)$ is between $0$ and $1$, so $$0\le \frac{\sin^2(n^nx)}{\sqrt{n+2}}\le \frac{1}{\sqrt{n+2}}<1.$$
Then from the squeeze theorem $a_n(x)=\frac{\sin^2(n^nx)}{\sqrt{n+2}}\to 0$ as $n\to\infty$, regardless of the value of $x$. Hence $\ln(1+a_n(x))\to0$ as $n\to \infty$ as well. So for large $n$, $f_n$ is practically $\cos(x+n)$. I know that the set of functions $$\mathcal{E}=\left\{f\in C^1(\Bbb R,\Bbb R): |f'(x)|\le1\right\}$$
is uniformly equicontinuous. Thus the subset $$\{\cos(x+n):n\in\Bbb N\}\subset \mathcal{E}$$
is also uniformly equicontinuous. Of course this does not necessarily imply that functions that behave like $\cos(x+n)$ (in the aforementioned way that $f_n$ does) are also uniformly equicontinuous.

Unfortunately, $\{f_n(x):n\in\Bbb N\}$ does not seem to be a subset of $\mathcal{E}$, so I can't really take the easy way out here.

Is there some way to rigorize my argument, or is there a better way to go about things? Thanks.

答:
Legit approach? If not, flaw in logic/conceptual understanding pointer would be appreciated. Think my notation is fine, but first post.

Let $\epsilon\gt0$ be given and fix $n\in\mathbb{N}$. Let $s,t\in\mathbb{R}$. First, note that for
$$f_n(x)=\cos(n+x)+\log\biggl(1+\frac{1}{\sqrt{n+2}}\sin^2(n^nx)\biggl),$$
the fact that (1) the logarithm is nonnegative for all relevant $n,x$ coupled with (2) the fact that $0\leq\sin^2(z)\leq1$ for any real input $z$ and (3) number sense regarding $\frac{1}{\sqrt{n+2}}$ with $n\in\mathbb{N}$ allows us to write
$$\cos(n+x)\leq f_n(x)\leq \cos(n+x)+\log(1+\frac{1}{\sqrt{3}}).$$
Note that this inequality holds for all $n\in\mathbb{N}$ and $x\in\mathbb{R}$. Thus, write
$$\cos(n+s)\leq f_n(s)\leq \cos(n+s)+\log(1+\frac{1}{\sqrt{3}})$$
and
$$\cos(n+t)\leq f_n(t)\leq \cos(n+t)+\log(1+\frac{1}{\sqrt{3}}).$$
Subtraction and simplification yields
$$\cos(n+s)-\cos(n+t)\leq f_n(s)-f_n(t)\leq \cos(n+s)-\cos(n+t).$$
No longer regarding the LHS, we note that
$$\left\lvert f_n(s)-f_n(t)\right\rvert\leq \left\lvert\cos(n+s)-\cos(n+t)\right\rvert.$$
In what follows, I will make use of angle sum formula, reorganization of terms, triangle inequality, absolute value properties, and the Mean Value Theorem, in that order:
$$\begin{equation}\begin{aligned}
\left\lvert f_n(s)-f_n(t)\right\rvert
  &\leq \left\lvert\cos(n+s)-\cos(n+t)\right\rvert \\
  &= \left\lvert\cos(n)\cos(s)-\sin(n)\sin(s)-(\cos(n)cos(t)-\sin(n)\sin(t))\right\rvert \\
  &= \left\lvert\cos(n)(\cos(s)-\cos(t))-\sin(n)(\sin(s)-\sin(t))\right\rvert \\
  &\leq \left\lvert\cos(n)(\cos(s)-\cos(t))\right\rvert+\left\lvert-\sin(n)(\sin(s)-\sin(t))\right\rvert \\
  &\leq \left\lvert\cos(s)-\cos(t)\right\rvert+\left\lvert\sin(s)-\sin(t)\right\rvert \\
  &\leq \left\lvert s-t \right\rvert+\left\lvert s-t \right\rvert \\
\end{aligned}\end{equation}
$$
So let $\delta=\frac{\epsilon}{2}$. Then from the work above we see that if $\left\lvert s-t \right\rvert\lt\frac{\epsilon}{2},$ we have:
$$f_n(s)-f_n(t)\leq\left\lvert f_n(s)-f_n(t)\right\rvert \leq \left\lvert s-t \right\rvert+\left\lvert s-t \right\rvert \lt \epsilon. \blacksquare$$

If there's a big error it's like 0205 so I'm tired.


问:
If G is a graph such that $\Delta(G) = 3$ then $\chi'(G) \le 4$
$\chi'(G)$ being the chromatic index of $G$, $\chi(G)$ its chromatic number and  $\Delta(G)$ its maximum degree.

Here's the solution that I'm given, but I don't understand it:

Apply Brooks' theorem to the line graph of G.

I see how $\chi'(G) = \chi(L(G))$.

But a graph G with $\Delta(G) = 3$ can obviously have a line graph such that $\Delta(L(G)) > 3$, take for example:

Additionally the line graph could turn out to be an odd cycle...

Any idea where to go from there?

答:
You don't need to worry about odd cycles, because they can be colored with three colors and you are allowed four.

You need to justify three things:

Show that the maximum possible degree is $4$ (not so bad, I think).

Show that thine graph cannot be the complete graph on five vertices, i.e. that you will not get a clique (you can check by going backwards if you like).

You don't care if you get an odd cycle.

The you can use Brook's theorem in all its glory.

For further information, this result is called Vizing's Theorem.


问:
Relation between the order of a set and the number of orbits

How can I show that if $G$ is a finite group that acts on a set $X$ and $m$ denotes the number of orbits that
$$m \ge \frac{|X|}{|G|}\:$$

Is it enough to intuitively say that $\sum|\mbox{Orb}_{G}(x)|=|X|$, thus $m=|X|$ in the case that the order of the group $G$ is one, otherwise it must be greater?

答:
The definition of the orbit of $x\in X$ is: $O_x:=\{g\cdot x, g\in G\}$, and hence $|O_x|\le|G|$ for every $x\in X$. Now, the set of the orbits $\mathcal{O}:=\{O_x,x\in X\}$ forms a partition of $X$; therefore:

$$|X|=\sum_{O_x\in\mathcal{O}}|O_x|\le\sum_{O_x\in\mathcal{O}}|G|=m|G|$$

whence $m\ge\frac{|X|}{|G|}$.

Alternatively. By Burnside's lemma, $m=\frac{1}{|G|}\sum_{g\in G}|\operatorname{Fix}(g)|$, where $\operatorname{Fix}(g):=\{x\in X\mid g\cdot x=x\}$. Since $e\in G$ fixes every $x\in X$, $|\operatorname{Fix}(e)|=|X|$ and then:

\begin{alignat}{1}
m &= \frac{1}{|G|}\sum_{g\in G}|\operatorname{Fix}(g)| \\
&= \frac{1}{|G|}\bigl(|\operatorname{Fix}(e)|+\sum_{g\in G\setminus\{e\}}|\operatorname{Fix}(g)|\bigr) \\
&= \frac{1}{|G|}\bigl(|X|+\sum_{g\in G\setminus\{e\}}|\operatorname{Fix}(g)|\bigr) \\
&= \frac{|X|}{|G|}\bigl(1+\frac{1}{|X|}\sum_{g\in G\setminus\{e\}}|\operatorname{Fix}(g)|\bigr) \\
&\ge\frac{|X|}{|G|}
\end{alignat}


问:
Evaluating $\int_{0}^{1} \frac{x^{2n+2}}{x^2-2x+2}\,{\rm d}x$
$$\int_{0}^{1} \frac{x^{2n+2}}{x^2-2x+2} \,{\rm d}x \tag{1}$$

This integral came up as a result of another integral I was trying to evaluate. Just for recreational purposes. Is there some closed form for this, even in terms of special functions? In general, is it possible to evaluate integrals of the form:

$$\int_{0}^{1} \frac{P(x,x^n)}{Q(x)} \,{\rm d}x \tag{2}$$

with $\text{deg}(P) > \deg(Q)$? For example:

$$\int_{0}^{1} \frac{x^n}{1+x^2} \,{\rm d}x \tag{3}$$

Can be written in terms of the digamma function. If I'm not mistaken, $(1)$ does not follow from $(3)$, because if you try to complete the square in the denominator of $(1)$, you'd end up having to perform a $u$-sub with $u = x-1$ and then a $(u+1)^{2n+2}$ in the numerator. This ruins things. Can someone give some help?

答:
Let's first try to find a simple expression for $ I_{k} $, defined for all $ k\in\mathbb{N} $, as follows : \begin{aligned}I_{k}=\int_{0}^{1}\frac{x^{k}}{x^{2}+1}\,\mathrm{d}x\end{aligned}

Let $ k\in\mathbb{N} $. Using the geometric formula for partial sums, we have : $$ \frac{\left(-1\right)^{k}}{1+x^{2}}+\sum_{j=0}^{k-1}{\left(-1\right)^{k-1-j}x^{2j}}=\frac{x^{2k}}{1+x^{2}} $$

Thus : $$ I_{2k}=\left(-1\right)^{k}\frac{\pi}{4}+\sum_{j=0}^{k-1}{\frac{\left(-1\right)^{k-1-j}}{1+2j}} $$

And : $$ I_{2k+1}=\left(-1\right)^{k}\frac{\ln{2}}{2}-\sum_{j=1}^{k}{\frac{\left(-1\right)^{k-j+1}}{2j}} $$

Now let $ n\in\mathbb{N} $, let's get back to the original integral : \begin{aligned} \int_{0}^{1}{\frac{x^{2n+2}}{x^{2}-2x+2}\,\mathrm{d}x}&=\int_{0}^{1}{\frac{x^{2n+2}}{\left(1-x\right)^{2}+1}\,\mathrm{d}x}\\ &=\int_{0}^{1}{\frac{\left(1-x\right)^{2n+2}}{x^{2}+1}\,\mathrm{d}x}\\ &=\sum_{k=0}^{2n+2}{\left(-1\right)^{k}\binom{2n+2}{k}I_{k}} \\ &=\sum_{k=0}^{n+1}{\binom{2n+2}{2k}I_{2k}}-\sum_{k=0}^{n}{\binom{2n+2}{2k+1}I_{2k+1}}\end{aligned}

We have all the formulas we need, we can continue for further possible simplifications, but I think the final formula would seem a little ugly.


问:
Lathe Bed Differential Wear Tool Offset Calculation
This is probably an odd one here,

I'm trying to calculate the linear offset that a tool needs to have to meet a circle of any radius r with an arbitrary offset y and arbitrary angle ϴ. The tool holder "height" from the bed surface is fixed and can be taken as a constant, but is also arbitrary and will vary lathe to lathe. I am trying to "map" this offset on my own, but would like to have a general formula to calculate the offset programmatically from measured values to incorporate into a LinuxCNC component so other users with this issue can measure and compensate their own machines.

I've linked an example with a green circle showing the distance between the circle and "tool" I'm trying to calculate, note this distance should be along theta and not along the origin axis. All of the dimensions are exaggerated for effect, in reality the difference will likely be within a few thousandths of an inch. Note that I have assumed that as the bed wears the left edge remains in the same plane. This is the edge of the lathe carriage that is "guided," the other is free to move laterally.

I've racked my brain and just can't come up with anything to describe how this gap interacts with the variables. CAD has been no help, either. I'd really appreciate it if someone could help me out with this.

答: With Glarbo's explanation spurring me along I have arrived at what I believe to be the correct solution. The height error varies as a function of the slope of the bed and the wear offset at x=0. This forms a line, y=mx+b. If I substitute this into the equation for a circle where r is the radius that is intended to be cut x^2 + y^2 = r^2, I can simplify and arrive at x=(-mb+sqrt(r^2 + m^2r^2 - b^2)/m^2+1 and x=-((mb+sqrt(r^2+m^2r^2 - b^2))/m^2+1) in general. This gives me intercepts of the line and circle and thus one of these I "should" be cutting at in order to meet the circle. The linear offset is thus r-x. From this linear offset, the "true" offset (distance the x axis needs to move along the sloped line) would be offset=(linear offset)/cos(m). Now all I need to figure out is which x value to ignore.  The illustration shows in green the offset that is calculated through this method. I am also aware that below a radius value where the tool can intercept the work, this will fail as there are no intercepts. I should be able to use the discriminant described here to check for the condition that the tool will not intersect the work and simply disable compensation for those cases.


问:
Find the sum of the First $50$ Natural Numbers starting from $11$. Is it from $11-50$ or $11-60$?
This a simple question yet confusing for me, I found the answer as 1220 by taking sum from $11$ to $50$, by inferring the question as first 50 natural numbers {$1,2,3,4,5,6,7,8,9,10,11,...49,50$}

and sum starting from 11 which means sum $= 11+12+...+49+50$.
But my friend says otherwise, he took from 11 to 60 and says answer as 1775. Can anyone explain how to infer the word first in the context of maths.

答:
The question says to calculate the sum of 50 natural numbers from 11 , it doesn't mean that to calculate the sum of natural numbers from 11 to 50 . So there is an high school equation to calculate this its is
Sn = n/2 (2f+ (n -1)d)
S - sum of natural numbers
n - no. Of terms ( eg: in this situations we are going to calculate the
sum of 50 terms)
f - First term ( eg : here 11 is the first term because we are adding the
natural numbers starting frm 11)
(n - 1) - we have to subtract 1 from the no. Of terms .( eg: here no. Of
terms we are going to add is 50 , so we need to subtract 1
from it (50 - 1) and we get 49 )

d   - it is the commen difference between terms ( here the first term is
11 and 2nd term will be 12 . So difference between terms is 12-
11=1)
Therefore ans;
S50 = 50 /2( 2×11+49×1)
=25(22+49)
= 25 × 71
= 1775


问:
Prove That $\sum_{i=1}^{\infty}\left(\frac{1}{P_i^2}\left(1-\sum_{k=1}^{i-1}\frac{1}{P_k^2}\right)\right)=1-\frac{6}{\pi^2}$
($P_i$ denotes the ith prime)

I recently derived this question while finding ways to express powers of natural numbers using primes.

My proof is long and tedious to work with and only works for powers of -2.

Are there similar variants to this question in history and is there a more generalised result (Im sure this is related to the Riemann Zeta Function).And if you could ,please give a better proof)

Also i checked in desmos for the first 1000 sums and it seems to be correct

Thanks a lot ☺☺☺

答:
I actually think that the question is wrong. This is what I think the question is supposed to be:

Prove that $\Sigma_{i=1}^\infty \frac{1}{P_i^2} \left(1 - \Sigma_{k=1}^{i-1} \frac{1}{P_k^2} \left(1 - \Sigma_{j=1}^{k-1} \frac{1}{P_j^2}\left(...\right) \right)\right) = 1 - \frac{6}{\pi^2}$ where $P_k$ is the kth prime.

I do have a solution to this. Multiply the RHS by $\frac{\pi^2}{6}$, and the LHS by $\left(
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} .... \right)$, which are both the same thing.

Now the RHS is just $\frac{\pi^2}{6} - 1$, which is just $\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} ...$

I will expand the LHS and prove that the terms match that on the RHS.

Observe that the LHS is just $\left(
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} .... \right)\left(\Sigma \frac{1}{p_i^2} - \Sigma \frac{1}{p_i^2 p_k^2} + \Sigma \frac{1}{p_i^2 p_k^2 p_j^2} - ....\right)$ where the signs keep swapping every time and the summations are taken over all possibilities of the product of primes.

Now expand this product of infinite sums and consider the coefficient of a number $\frac{1}{n^2}$, for all $n \ge 2$, if we can prove that the coefficient is $1$, then we are done since then this will be the same as the RHS.

If $n$ is a prime, then obviously, it will appear only once in the expansion and so the coefficient is $1$.

Now, suppose $n = p_1^{\alpha_1}p_2^{\alpha_2}....p_k^{\alpha_k}$. Then, you can check that the coefficient becomes

$ 1 + \left(\binom{k}{1} - \binom{k}{2} + \binom{k}{3} ..... + (-1)^{k-1}\binom{k}{k-1}\right)$, where the second part is well known to be just $0$. So, the coefficient is always $1$ and so the proof is done

The above two solutions also work perfectly, but only when the question is changed to this. In N-N's solution, the formula for $T_n$ was incorrect as by PIE (Principle of inclusion and Exclusion), there are going to be a lot more terms as you would have overcounted and undercounted some numbers. For example, you counted the multiples of $5$ excluding mutiples of $2$ and $3$ but forgot to again add back multiples of $6$ as you had subtracted them twice.

Hope this helps :)


问:
Limit as $x\to \infty,$ of $\;\frac{(1+\frac{1}{x})^{x}}{x}$
I'm having some trouble evaluating the $\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}$ in the context of real analysis. Denote this sequence by $(a_k)$

If so, then I need to prove that for all $\varepsilon>0$

$$\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right| <\varepsilon$$

We know that $\lim_{x\to\infty}1/x=0$, however $|1/x|\leq\left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}\right|$, so I'm not sure this information is useful. I considered the sequential characterization of continuity, so if the sequence above converges to 0, then $f((a_k))$ converges to $f(0)$, but there is also not an obvious choice of $f$. Any suggestions into how to evaluate the limit?

答:
We will show that
$$\lim_{x\to\infty}\frac{\left(1+\frac{1}{x}\right)^{x}}{x}=0.$$
We let $\varepsilon >0$ be given. We need to find a $\delta >0$ such that $|x-0|<\delta$ implies ${\displaystyle \left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right|<\varepsilon }$.

In this post, a user proved that $\left(1+\frac{1}{x}\right)^{x}<3.$ Also for every $x>1$, $1/x<x$. Choose $\delta=\varepsilon/3$.
$${\displaystyle \left|\frac{\left(1+\frac{1}{x}\right)^{x}}{x}-0\right|<\frac{3}{x}<3x<3\delta<\varepsilon }.$$


问:
If $A\subseteq B$, then is $\langle A \rangle \le \langle B \rangle$?
For instance, if we have a group $\langle x_1, x_2 \rangle$, then is $\langle x_1 \rangle$ a subgroup of this? I would think that this is not true since every element in $\langle x_1, x_2 \rangle$ can be represented as $x_1^{\epsilon_1}x_2^{\epsilon_2}$ for $\epsilon_i \in \{-1, +1\}$ and $x_1^a \in \langle x_1 \rangle$ for any $a>1$ cannot necessarily be represented in this form. However, I am not sure if this is a valid argument.

答: Let $G$ be the group that you are working with. The group $\langle A\rangle$ is the intersection of all subgroups of $G$ containing $A$. Since $A\subset B$, $\langle B\rangle$ is one such subgroup, and therefore $\langle A\rangle$ is a subgroup of $\langle B\rangle$.


问:
Usual conditions - filtration
$(\Omega, \mathcal{F}, P)$ - probability space. What is the definition of filtration whch satisfy usual conditions? I know that it must be right-continuous and $\mathcal{N}\subset \mathcal{F_0}$, but what is the form of $\mathcal{N}$? Is it $\mathcal{N}=\{X\subset \Omega:X\subset Y\in\mathcal{F}, P(Y)=0\}$ or $\mathcal{N}=\{A\in \mathcal{F}:P(A)=0\}$?

答: A filtered probability space $(\Omega, \mathcal{F}, (\mathcal{F}_t)_{t\geq0}, P)$ is said to satisfy the usual conditions if $(\mathcal{F}_t)_{t\geq0}$ is right-continuous and $\mathcal{N}\subset \mathcal{F}_0$ with $\mathcal{N}=\{X\subset \Omega|\exists Y\in\mathcal{F}: X\subset Y  \land P(Y)=0\}$.

Title: Usual conditions - filtration