答:
You're right that there's an identity that will be very helpful here: it's the functional equation for the Gamma function,
$$
\Gamma(z+1) = z\Gamma(z),
$$
which is valid wherever $\Gamma$ is defined. (Indeed it can be used to define $\Gamma(z)$ wherever $\Gamma(z+1)$ is defined, which is even stronger.)

And you're also right about integrating by parts, in that this functional equation is proved precisely by integrating by parts.

Title: Tips for proving: $\Gamma(n-a)/\Gamma(-a)=(-1)^na(a-1)(a-2)...(a-n+1),~$ for $~n = 1, 2, ...$?

Body:
I am trying to prove how $\Gamma(n-a)/\Gamma(-a)=(-1)^na(a-1)(a-2)...(a-n+1),~$ for $~n = 1, 2, ...$? So I believe that the numerator is $\Gamma(n-a)=(n-a-1)!=((n-a)!)/(a-n)$, and I proved this by using an analogy to the recurrence relation, and checked it with Wolfram Alpha. However, I do not know what to do with the denominator. The lecture notes state, $1/\Gamma(-n)=0$, where n = 0, 1, 2... However, the domain of "a" is not specified in the homework (the problem is given just as I have stated it in the title), so I am assuming that "a" can be any number, complex, negative, noninteger, etc.. So now I don't know where to start.

One route I was thinking of is using integration by parts and using a pattern that may come up to solve the integral (since simply integration by parts will go on endless without recognizing a pattern). In class today, he implied that the answer should be relatively simple to get, so it makes me think that integration by parts is not what he intended. Another route I was wondering about is if there is a relevant identity/formula for the Gamma function that I can use that may help me by analogy? Any starting point would be great, and especially in generic terms/concepts so that I can apply those tips to any similar problem (rather than an answer to the question I have).

This is my first time posting so hopefully I included all relevant information.


问:
Forward Kolmogorov Equation and Normal Distribution
I read through my textbook and find one interesting statement. The normal pdf function $\frac{1}{\sqrt[]{2\pi\sigma^2}\ t}\exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$ satisfies the forward Kolmogorov equation. I find the forward Kolmogorov equation and try to prove it. But I'm confused about how to deal with the derivatives, especially the $\frac{\partial P}{\partial t}$ since I never see the t in other normal pdf. Could someone give me some ideas about how to go forward? Thank you very much.

$$\frac{1}{2}\sigma^2\frac{\partial^2 P}{\partial x^2} - \mu\frac{\partial P}{\partial x}=\frac{\partial P}{\partial t}$$

答:
Ok, the problem isn't hard if you know what those partials mean, it's just a huge mess haha  To take the partial with respect to $t$ (or any variable) you just treat the other variables like constants.  For example, if $P = xt^2$ then $\partial P/\partial t = 2xt$.  So to compute the partial $\partial P / \partial t$ first use the product rule and take the derivative of the left summand:

$$\partial P / \partial t = \frac{\partial}{\partial t}\left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right) =$$

$$= \left( \frac{-1}{t^2 \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right)$$

Now to take the derivative on the right, we use the chain rule for $(e^{f(t)})' = f'(t)e^{f(t)}$:

$$\frac{\partial}{\partial t} \left( \exp\big(-\frac{(x-\mu t)^2}{2\sigma^2t}\big) \right) = \frac{\partial}{\partial t} \left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right) =$$

$$ = - \left( \frac{-4\mu \sigma^2 t(x-\mu t) - 2\sigma^2(x-\mu t)^2}{4 \sigma^4 t^2}\right) \cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right) =$$

$$ = \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t^2}\cdot \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t} \right)$$

Where the first equality is just the well-known quotient rule for derivatives, and the second equality is just distributing the minus sign and reducing.  I'd double-check these calculations if I were you, I just woke up.  Putting these together and writing $A = \exp\left(-\frac{(x-\mu t)^2}{2\sigma^2t}\right)$, we get the atrocious:

$$\partial P/\partial t = \left( \frac{-1}{t^2 \cdot \sqrt{2\pi\sigma^2}}\right) \cdot A + \left( \frac{1}{t \cdot \sqrt{2\pi\sigma^2}}\right) \cdot \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t^2} \cdot A =$$

$$ =\frac{A}{t^2 \cdot \sqrt{2 \pi \sigma^2}} \cdot \left( \frac{2\mu t(x-\mu t)+(x-\mu t)^2}{2 \sigma^2 t} - 1\right) =$$

$$= \frac{A}{t^2 \cdot \sqrt{2 \pi \sigma^2}} \cdot \left( \frac{x^2 - \mu^2t^2}{2 \sigma^2 t} - 1\right)$$

Now you need to check that the functional equation is satisfied, taking the partials wrt $x$.  Hopefully it just falls right out - which it basically always does when something like that is true.  But without some heavy theorems I see no way to prove this without doing the monster calculations.  The partial wrt $x$ is a lot simpler, thankfully.

ContentLicense: CC BY-SA 4.0


问:
Not rejecting null hypothesis not equivalent to accepting it
As title states, why is not rejecting null hypothesis not equivalent to accepting it? I can't find any information about a 'state' other than accepting or rejecting null hypothesis.</p>

答:
There are three states an hypothesis can be in: accepted, rejected, or not concluded.

Before we do an experiment and analyze the data, we can't conclude anything about the null hypothesis. Then after we do the experiment and the statistics, if the results are far away from what the null hypothesis would imply, we say we reject the null hypothesis.

However, if the analysis reveals that the result was rather close to what the null hypothesis predicts, then we can't reject it. However, this does not mean we immediately accept it. It could be that the null hypothesis is wrong, but that the true solution is too close to the null hypothesis for our experiment to distinguish.

This is how failing to reject the null hypothesis does not mean accepting it. It just means we can't conclude.


问:
Using the trapezoid rule show that $\sum_{k=1}^nk = \frac{n(n+1)}{2}, \forall n \in \mathbb{N}$
$\sum_{k=1}^nk = \frac{n(n+1)}{2}, \forall n \in \mathbb{N}$

I got as far as

$$k = w_kf(x_k) \\
\sum_{k = 1}^n k = \int_1^n w_kf(x_k)dx = \\
\frac{(n-1)}{2(n-1)}[f_o+2(f_1+f_2+f_3+...+f_{n-1}) +f_n] \\
f(x_k) = k \Rightarrow \\
\frac{1}{2}[(n+1)+2(\sum_{k=2}^{n-1}k)]$$

I got stumped on this part so I improvised

$$\frac{1}{2}[(n+1)+2(\frac{(n-1)n}{2}-1)] = \\
\frac{1}{2}[(n+1)+(n-1)n-2] = \\
\frac{(n-1)(n+1)}{2}$$

I have a gut feeling that on $\frac{(n-1)}{2(n-1)}$ it should be $\frac{(n)}{2(n-1)}$ but the formula states it's $\frac{(b-a)}{2N}$ where N is the number of spaces between the points so I don't know.

Help?

答: You can write the sum as the area of a trapezium that splits into $n$ trapezia, but that's not the usual meaning of "the trapezoid rule". Since $k$ is the area of a width-$2$ trapezium whose parallel sides have lengths $k\pm\tfrac12$, $\sum_{k=1}^nk$ is the area of a width-$2n$ trapezium whose parallel sides have lengths $n+\tfrac12,\,\tfrac12$.


问:
Trying to understand and derive the binomial distribution
I have been trying to derive the binomial distribution through reasoning and logic however I am a bit stuck and in need of possible hints on how to proceed or where I have gone wrong.

My reasoning is as follows:

Let $$n=\text{total number of trials}$$
$$p=\text{probability that a trial ends in success}$$

Then the probability for the case where in n trials there is only one success is

$$p(1-p)^{n-1}$$

And the probability for the case where in n trials there are two successes is

$$p^2(1-p)^{n-2}$$

More generally, the probability that in $n$ trials there are $k$ successes is

$$p^{k}(1-p)^{n-k}$$

As you can see I am far from reaching my goal, any helpful hints or corrections would be appreciated.

答: Very close! In the probability that in n trials there are k successes is ${n \choose k}p^k (1-p)^{n-k}$ because you need to choose which of the n trials are k successes. $p^k(1-p)^{n-k}$ is the probability of success of a single configuration of k successes in n trials. For example, if you have n=5, k=2, this is the probability of S F F S F (with S=success, F=failure). Another possibility is F S F S F. So you have to multiply your probability with the total number of possibilities, which is $n\choose k$.


问:
Every compact subset of Baire space $\omega^\omega$ has empty interior.
Here is the well-known proposition, I assumed that $K$ is compact subset of $\omega^\omega$ and it contains an open Set $N_s$, so $N_s$ is clopen and thus it is compact. If I can show $N_s$ can not be covered by a finite subcover, the solution is done, but I don’t know how to start, can you please help me?

答:
It helps to use a basis. Basic open sets in Baire Space look like:
$$\mathcal{U} = \{m_1\} \times \{m_2\} \times \ldots \times \{m_n\} \times \Bbb{N} \times \Bbb{N} \times \ldots \subseteq \omega^\omega,$$
for some finite sequence $m_1, \ldots, m_n \in \Bbb{N}$. If $K$ has non-empty interior, then it must contain a set of this form. These sets are indeed clopen, so by your logic, we just need to show $\mathcal{U}$ is not compact. We will find an open cover, without a finite subcover.

This is not too hard to do. Essentially, we just take open sets
$$\mathcal{U}_k = \{m_1\} \times \{m_2\} \times \ldots \times\{m_n\} \times \{k\} \times \Bbb{N} \times \Bbb{N} \times \ldots,$$
where $k \in \Bbb{N}$.

Points in $\mathcal{U}$ are precisely the natural sequences which begin with $m_1, \ldots, m_n$. If we take such a sequence $(x_k) \in \mathcal{U}$, then $x_{n+1}$ must have some value, in which case, $(x_k) \in \mathcal{U}_{x_{n+1}}$. That is,
$$\mathcal{U} \subseteq \bigcup_{k \in \Bbb{N}} \mathcal{U}_k.$$
That is, we have an (infinite) open cover. Does it have a finite subcover? No; in fact it doesn't even have a proper subcover. If we remove $\mathcal{U}_k$ from the union, then suddenly we are not covering the sequence
$$(m_1, m_2, \ldots, m_n, k, k, k, \ldots) \in \mathcal{U}.$$
So, every single $\mathcal{U}_k$ is necessary to cover $\mathcal{U}$, and hence there cannot be a finite subcover.


问:
Intersection of perpendicular chords: is this true for a sphere?
It is true that in a circle with radius $R$, if the intersection of any two perpendicular chords divides one chord into lengths $a$ and $b$ and divides the other chord into lengths $c$ and $d$, then
$$a^2 + b^2 + c^2 + d^2 = 4R^2.$$

Does this equation hold for a sphere with radius $R$, and three mutually perpendicular chords:
$$a^2 + b^2 + c^2 + d^2 + e^2 + f^2 = 6R^2?$$

答:
HINTS:

For a circle we have constraints

$$ a b = c d = e ( 2 R- e ), ...,....; $$

where e is minimum distance to circle periphery.

Similarly for a sphere

$$ ab=cd = x ( 2p-x), ....,...; $$
plus two more such pair wise relations where $p,q,r $ could be radii of three mutually orthogonal circles contained in such perpendicular  planes and $ x,y,z $ could be their corresponding minimum distances.


问:
Some alternating sum of integer part of $\frac{kb}{1722}$.
For every integer $k$ coprime to 1722, how can one compute the sum

$\sigma(k)= \sum_{1\leq b\leq 1722, (1,b)=1722} \lfloor\frac{kb}{1722}\rfloor (-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\rfloor-\lfloor \frac{b}{7}\rfloor-\lfloor \frac{b}{41}\rfloor}$?

(Here the sum is taken over integers $b$ between $1$ and 1722 coprime to 1722.)

An approach would be to consider a linear function $\tau(k)$ and compute $\sigma(k)-\tau(k)$ where

\begin{align*}\tau(k) &= \sum_{1\leq b\leq 1722, (1,b)=1722} \frac{kb}{1722}(-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\rfloor-\lfloor \frac{b}{7}\rfloor-\lfloor \frac{b}{41}\rfloor}\\
&=\frac{k}{1722} \sum_{1\leq b\leq 1722, (1,b)=1722}b\cdot (-1)^{b-\lfloor \frac{b}{2}\rfloor-\lfloor \frac{b}{3}\rfloor-\lfloor \frac{b}{7}\rfloor-\lfloor \frac{b}{41}\rfloor} = -\frac{80\cdot 1724k}{1722}
\end{align*}

It is not hard to see that $\sigma(k)-\tau(k)$ is bounded. I used Python to compute $\tau(k)$.

I would like to know how to compute both $\sigma(k)$ and $\tau(k)$ by hand.

Body:
You can use $\tau(k)$ to approximate $\sigma(k)$.

Your calculation of $\tau(k)$ was not correct. This is the correct function:
$$\frac{k}{1722} \sum _b b (-1)^{-\left\lfloor \frac{b}{2}\right\rfloor -\left\lfloor \frac{b}{3}\right\rfloor -\left\lfloor \frac{b}{7}\right\rfloor -\left\lfloor \frac{b}{41}\right\rfloor +b}=-\frac{68960}{861}k$$

And this is a table to show how good is the approximation
$$
\begin{array}{r|r|r}
k & \sigma(k) & \tau(k)\\
\hline
 100 & -8008 & -8009.29 \\
 200 & -16016 & -16018.6 \\
 300 & -24032 & -24027.9 \\
 400 & -32036 & -32037.2 \\
 500 & -40044 & -40046.5 \\
 600 & -48060 & -48055.7 \\
 700 & -56052 & -56065. \\
 800 & -64076 & -64074.3 \\
 900 & -72088 & -72083.6 \\
 1000 & -80092 & -80092.9 \\
 2000 & -160184 & -160186. \\
 3000 & -240276 & -240279. \\
 4000 & -320372 & -320372. \\
 5000 & -400464 & -400465. \\
 6000 & -480548 & -480557. \\
 7000 & -560648 & -560650. \\
 8000 & -640744 & -640743. \\
 9000 & -720832 & -720836. \\
 10000 & -800928 & -800929. \\
\end{array}
$$


问:
Probability of lines in a pentagon intersecting internally or at vertices.
I'm studying maths as a hobby.

The points A,B,C,D & E are the vertices of a regular pentagon. All possible lines joining the pairs of these points are drawn. If two of these lines are chosen at random, what is the probability that their point of intersection is (a) inside the pentagon, (b) one of the points A,B,C,D,E?

I start out by saying the number of ways of choosing 2 lines from 10 is $\binom{10}{2} = 45$

Then I started getting confused. I can see there are 5 internal points of intersection, which I mark in red, so that would give me $\frac{5}{45} = \frac{1}{9}$ probability of choosing lines which meet internally. And this is indeed the answer my text book gives. But I'm not sure if this is just coincidence.

As for (b), I'm not sure how to proceed. The book gives the answer as $\frac{2}{3}$

答:
We either have side-side (S-S) intersection, side-diagonal (S-D) intersection or diagonal-diagonal (D-D) intersection.

For S-S intersection, if you take any side, out of $4$ possible intersections,

i) it intersects with $2$ adjoining sides on pentagon $(\frac{1}{2})$

ii) it intersects with $2$ of them externally $(\frac{1}{2})$

For S-D intersection, if you take any side, out of $5$ possible intersections,

i) it intersects with $4$ diagonals on pentagon $(\frac{4}{5})$

ii) it never intersects with $1$ which is parallel to it $(\frac{1}{5})$

For D-D intersection, if you take any diagonal, out of $4$ possible intersections,

i) it intersects with $2$ diagonals at an interior point $(\frac{1}{2})$

ii) it intersects with other $2$ diagonals on the pentagon. $(\frac{1}{2})$

Now, probability of choosing two sides $ = \displaystyle \frac{5 \choose 2}{10 \choose 2} = \frac{2}{9}$

Probability of choosing two diagonals is the same $\frac{2}{9}$.

Probability of choosing a side and a diagonal $ \displaystyle = \frac{{5 \choose 1}^2}{10 \choose 2} = \frac{5}{9}$

So as you can see that only two diagonals intersect internally and so the probability for the first is $ = \frac{1}{2} \times \frac{2}{9} = \frac{1}{9}$.

Probability for the second is $ \displaystyle = \frac{2}{9} \times \frac{1}{2} + \frac{2}{9} \times \frac{1}{2} + \frac{5}{9} \times \frac{4}{5} = \frac{2}{3}$.


问:
Finding solutions to $[x]^2 + 2[x]-3x=0$
Related question, I was attempting the question in an alternative way I wrote $ \{ x \} = x-[x]$, then after subbing that into the equation I got the following quadratic :

$$ [x]^2 + 2[x]-3x=0$$
I rearrange this as:

$$ [x]^2 + 2 [x]  = 3x$$

Adding one:

$$ ( [x] +1)^2 = 3x+1$$

Or,

$$ [x] + 1 = \pm \sqrt{3x+1}$$

Now, I use property to take in $1$ to the box function:

$$ [1+x] = \pm \sqrt{3x+1}$$

Now , I substitute $ x+1 = u$, this leads to:

$$ [u] = \pm \sqrt{3u-2}$$

The bound on $u$ is given as $ u \geq \frac23$, it is clear that the $ [u]= - \sqrt{3u-2}$ has no solutions because left side is always positive for $u \geq \frac23$ , hence can't be equal to negative number of RHS. This leads to a single inequality:

$$ [u] = \sqrt{3u-2}$$

This leads for checking when $3u-2$ is perfect square:

$$ 3u-2 = 0$$

$$ 3u-2=1 \implies u=1$$
$$ 3u -2 = 4 \implies u=2$$
$$ 3u-2 =9 \implies u= \frac{11}{2}$$

Seems like there are no solutions for $ u >2$ , so how do I proof that this is the case?

答:
Using $\{x\}$ is advantageous. The original equation transforms to
$$[x]^2-[x]=3\{ x\}$$

LHS is an integer $\Rightarrow$ $\{x\}=0,1/3,2/3$.
So $x$ is $n$ or $n+1/3$ or $n+2/3$ for some integer $n$.

Substituting these in original equation gives
$$n^2-n=0,1,2$$

which can be easily solved. Now there is no need to check anything. Only correct solutions will be obtained.


问:
Computing a difficult integral
This is actually a follow-up question to what I posted here: Computing double integral for expected value. I think that the result in the previous post may be helpful in computing this integral.

The result in the previous post was:

$$\frac{\int_{\mathbb{R}^{2}}\frac{p^2}{2} e^{-\beta (V(x) + p^2/2)}\mathop{dx}\mathop{dp}}{\int_{\mathbb{R}^{2}} e^{-\beta (V(x) + p^2/2)} \mathop{dx}\mathop{dp}} = (k_BT)^{-1},$$

where $\beta = (k_BT)^{-1}$. So we can use this result if it makes the following integral easier.

I would now like to compute the following:

$$\frac{\int_{\mathbb{R}^{6n}} \frac{p^2}{2} e^{-\beta(V(x) + \frac{1}{2} \sum_{i = 1}^{3n} p_i^2)} \mathop{dx dp}}{\int_{\mathbb{R}^{6n}} e^{-\beta(V(x) + \frac{1}{2}\sum_{i = 1}^{3n} p_i^2)} \mathop{dx dp}},$$

where $V$ is some unknown function of $x$. The answer should be $3nk_BT/2 = \frac{3n}{2\beta}$, but I'm really not sure about how to show this result.

Any help is appreciated

答:
By the same factoring trick we can say that the $x$ integral cancels again and we are left with

$$\frac{\sum_{i=1}^{3n}\int_\Bbb{R} \frac{p_i^2}{2}\exp\left(-\frac{\beta p_i^2}{2}\right)dp_i \Pi_{j\neq i} \int_\Bbb{R} \exp\left(-\frac{\beta p_j^2}{2}\right)dp_j}{\left(\int_\Bbb{R} \exp\left(-\frac{\beta p^2}{2}\right)dp\right)^{3n}}$$

$$ = \frac{3n \int_\Bbb{R} \frac{ p^2}{2}\exp\left(-\frac{\beta p^2}{2}\right)dp \cdot \left(\int_\Bbb{R} \exp\left(-\frac{\beta p^2}{2}\right)dp\right)^{3n-1}}{\left(\int_\Bbb{R} \exp\left(-\frac{\beta p^2}{2}\right)dp\right)^{3n}}$$

$$= \frac{3n \int_\Bbb{R} \frac{p^2}{2}\exp\left(-\frac{\beta p^2}{2}\right)dp}{\int_\Bbb{R} \exp\left(-\frac{\beta p^2}{2}\right)dp} = \frac{3n}{2\beta}$$

since all of the terms in the sum were identical.

Title: Computing a difficult integral


问:
If $u \in C$, $|\int u \phi_{ij}| \leq 2 \|h\|_p \|\phi\|_{p'} \forall \phi \in C_c^\infty$, then $\|u_{ij}\|_p \leq 2\|h\|_p$?
Let $\Omega$ be a domain in $\mathbb R^N$, $u \in C(\Omega)$ and $h \in L^p(\Omega)$. Suppose that
$$
\left|\int_\Omega u \varphi_{ij} \ dx  \right| \leq 2\|h\|_p \|\varphi\|_{p'} \quad \forall \varphi \in C_c^\infty(\Omega).
$$

Is it true that $u_{ij} \in L^p(\Omega)$ and $\|u_{ij}\|_p \leq 2\|h\|_p$? If so, how to prove it?

Context: from what I understand, Caffarelli and Cabré use this fact in the proof of Proposition 1.1 in their book Fully Nonlinear Elliptic Equations. I understand that $u_{ij}$ is to be interpreted as distributional derivatives, but I am unable to prove the inequality.

Any hints will be the most appreciated.

Thanks in advance.

答:
Define $L : C_c^\infty(\Omega) \to \mathbf R$ by
$$L\phi = \int_\Omega u \phi_{ij} \, dx.$$ By hypothesis you have
$$|L\phi| \le M \|\phi\|_{p'},\quad \phi \in C_c^\infty(\Omega).$$ In particular $L$ is bounded (in the $L^{p'}$ norm) on $C_c^\infty(\Omega)$.  Since $C_c^\infty(\Omega)$ is dense in $L^{p'}(\Omega)$ you can extend $L$ by continuity to all of $L^{p'}(\Omega)$. If you denote the extension by $\tilde L$ then
$$ |\tilde Lg| \le M \|g\|_{p'},\quad g \in L^{p'}(\Omega).$$
Now apply Riesz representation theorem (you need to insist $1 < p \le \infty$) to find a function $f$ satisfying $\|f\|_p \le M$ and
$$\tilde L g = \int_\Omega fg \, dx, \quad g \in L^{p'}(\Omega).$$
In particular, if $\phi \in C_c^\infty(\Omega)$ then
$$ \int_\Omega u \phi_{ij} \, dx = L\phi = \tilde L\phi = \int_\Omega f \phi \, dx.$$
Consequently $f = u_{ij}$ in the distributional sense and $\|u_{ij}\|_p \le M$.


问:
If I know the function is continuous, is it okay to calculate the value of an unknown constant like this?
I have the function

$f(x) =
\begin{cases}
x+3, & x\in[-2,1]\\\\
(x-2)^2 + k, & x\in(1,3]
\end{cases}$

where $k \in \mathbb{R}$.

If I know the function is continuous, would it be wrong to calculate the value like this:

$ x+3 = (x-2)^2 + k $

when x = 1 as the function is continuous

$ 1+3 = 1 - 4 + 4 + k$

$ 4 = 1 + k$

$3 = k$

I know this may not work if the function is not continuous, but if it is continuous it should work every time. Right?

答: Yes, it will always work. But you can get to the same conclusion with less computations  if you note that $(1-2)^2=(-1)^2=1$. There is no need to use the fact that $(1-2)^2=1^2-2\times2+2^2$.


问:
Showing identity with the integers modulo
Given the set $S_m=\{[\frac{k(k+1)}{2}]_m\}$ for $k\in\mathbb{N}$.
Where $[x]_m$ denotes the equivalence class (modm).

I am trying to show that: $S_m=\mathbb{Z}_m$ if and only if $m=2^s$ for some $s \in\mathbb{N}$.

A hint is also given that for each prime $p>2$ there is a $b\in\mathbb{Z}_p$ such that the equation $x^2\equiv b\pmod{p}$ has no solutions.

I've tried using the condition for what m has to equal in order for the identity to hold but I'm quite lost in how to proceed really, any help would be greatly appreciated.

答:
If $m=2^s$

For $k$ and $l$  residues modulo $2^{s+1}$ suppose $$\frac{k(k+1)}{2}\equiv \frac{l(l+1)}{2}\pmod {2^s}.$$

Then $k(k+1)\equiv l(l+1)\pmod {2^{s+1}}$

i.e. $(k-l)(k+l+1)\equiv 0\pmod {2^{s+1}}$.

Now one of the two factors on the LHS is odd and so the other factor has to be divisible by $2^{s+1}$. So equality occurs if and only if $k+l+1\equiv 0\pmod {2^{s+1}}$. Hence everyone of the $2^{s+1}$ values for $\frac{k(k+1)}{2}$ is taken precisely twice and so, by the pigeon-hole principle, $\frac{k(k+1)}{2}$ takes every value modulo $2^s$.

If $m$ has an odd prime factor $p$

Let $b$ be an element such that $x^2\equiv b\pmod p$ cannot be solved.

Suppose that $\frac{k(k+1)}{2}\equiv \frac{b-1}{8}\pmod p$ could be solved. >!Then $(2k+1)^2\equiv b \pmod p$, a contradiction.


问:
Challenging Integrals for High School Students
I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

Integration by parts, inspection, substitution, partial fraction decomposition

Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions

I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.

Thank you for your suggestions.

答:
Here are a couple of calculus questions concerning hyperbolic functions:

Just as $\sin$ can be defined as the unique function $f:\mathbb{R} \mapsto \mathbb{R}$ satisfying
\begin{align}
f''(x) &= \color{blue}{-}f(x) \\
f'(x) &= 1 \\
f(x) &= 0 \, ,
\end{align}
$\sinh$ can be defined as the unique function $f:\mathbb{R} \mapsto \mathbb{R}$ satisfying
\begin{align}
f''(x) &= \color{red}{+}f(x) \\
f'(x) &= 1 \\
f(x) &= 0 \, .
\end{align}
Derive the exponential form of hyperbolic sine by solving the above equation.

The classical way to define $\sinh$ is via the 'unit hyperbola':

If the region OPR has an area of $t/2$, then the $x$- and $y$-coordinates are, respectively, $\cosh t$ and $\sinh t$. (If you're interested in the details of how hyperbolic functions can be defined using a hyperbola, then see here.) If however you define $\sinh$ and $\cosh$ using exponentials, then it is possible to prove that the shaded area is equal to $t/2$, given that $OQ = \cosh t$ and $PQ = \sinh t$. Prove this using integration. You may need to use hyperbolic identities to simplify your answer.

The graph of $\cosh$ has a curious property. For any interval $[a,b]$, the area under the graph is equal to the length of the arc connecting the points $(a,\cosh a)$ and $(b,\cosh b)$. Prove this using the formula
$$
\text{arc length} = \int_{a}^{b}\sqrt{1+\left(f'(x)\right)^2} \, dx \, ,
$$
where $y=f(x)$ is the graph of the function in question. (Here, $f=\cosh$.) And if you're interested in seeing why the arc length formula works, then see here.


答: Suppose $K$ is compact. It follows that all $p_n[K]$ are also compact in $\omega$, which is a discrete space so every projection of $K$ is finite ( the only kind of compact subset of a discrete space) An open subset always has infinitely many projections that are $\omega$, by the form of standard basic open sets. From this the statement follows.

Title: Every compact subset of Baire space $\omega^\omega$ has empty interior.

Body: Here is the well-known proposition, I assumed that $K$ is compact subset of $\omega^\omega$ and it contains an open Set $N_s$, so $N_s$ is clopen and thus it is compact. If I can show $N_s$ can not be covered by a finite subcover, the solution is done, but I don’t know how to start, can you please help me?


问:
Proving strong continuity of semigroup in $L^2(\mu)$ ($\mu$ invariant measure) from non-cadlag Markov process
I am trying to derive the definition of a Markov semigroup from Section 1.1 Bakry, Gentil, Ledoux Analysis and Geometry of Markov Diffusion Operators from a Markov process. I am unsure about my proof of strong continuity of the semigroup on $L^2(\mu)$ for an invariant measure $\mu$. Here is my work:

Suppose we are given a Markov process $X_t, t \geq 0$ on a Polish space $\mathcal X$. The semigroup is defined as

\begin{equation}
        \mathbf{P}_{t}f(x) = \mathbb E [f(X_t)|X_0=x].
\end{equation}
on bounded measurable functions $\mathcal M_b (\mathcal X)$.

The semigroup property, mass conservation $\mathbf{P}_{t}  1 = 1$ and positivity preservation  $f \geq 0 \Rightarrow \mathbf{P}_{t} f \geq 0$ are straightforward to show from the definition.

Also, these properties allow to show a Jensen's inequality for the semigroup, for convex functions $\Phi$
\begin{equation}
    \mathbf{P}_{t}(\Phi(f)) \geq \Phi\left(\mathbf{P}_{t} f\right)
\end{equation}

A $\sigma$-finite measure $\mu$ is invariant under the semigroup if
\begin{equation*}
  \int_\mathcal X \mathbf{P}_{t}f \: d \mu = \int_\mathcal X f \: d \mu, \quad \forall f \in \mathcal M_b (\mathcal X).
\end{equation*}

With this, we can extend the semigroup to a contraction of $L^p(\mu), 1 \leq p\leq\infty$

\begin{align*}
   \| f \|_{L^p(\mu)}^p= \int_\mathcal X| f|^p \:d \mu =\int_\mathcal X \mathbf{P}_{t} |f|^p \:d \mu \geq \int_\mathcal X |\mathbf{P}_{t} f|^p \:d \mu  = \|\mathbf{P}_{t} f \|_{L^p(\mu)}^p
\end{align*}

Finally, for any bounded, continuous function $f \in \mathcal C_b(\mathcal X), 1 \leq p< \infty$,

\begin{equation*}
\int_\mathcal X |P_t f(x)- f(x)|^p \: d \mu(x) \to 0 \quad \text{as}\quad  t \to 0
\end{equation*}
by the DCT since $|P_t f- f|^p(x) \leq 2^{p-1}(\|P_t f\|_\infty +\|f\|_\infty)\leq 2^{p}\|f\|_\infty< \infty$. Since $C_b(\mathcal X)$ is dense in $L^p(\mathcal X, \mu)$ we are done.

Now this argument with the DCT seems only to work for finite measures. For $\sigma$-finite measures, should I split the state-space into finite regions and apply the argument?

Also I am not sure what licenses me to say that $P_t f(x) \to f(x)$ as $t \to 0$ for $\mu$-a.e. $x$ and bounded continuous $f$. This works when the process is right continuous but I am not sure about the general case.

答:
Added note: As far as I can tell, BGL asserts strong continuity in $L^2$ of the semigroup of a right-continuous Markov process that is symmetric with respect to an invariant measure.

All the arguments that I know of need enough functions $f\in L^1(\mu)\cap L^\infty(\mu)$ such that $t\mapsto P_tf(x)$ is right continuous for $\mu$-a.e. $x$. If the process has right-continuous paths, if the state space is nice (locally compact, second countable), and if $\mu$ is a Radon measure, then there are plenty of bounded continuous  functions $f$ in $L^1(\mu)$. More generally, if $\mu$ is $\sigma$-finite and the process is a "right Markov process" (essentially, right continuous and strong Markov) then the set of bounded integrable $f$ such that $t\mapsto P_tf(x)$ is right continuous for $\mu$-a.e. $x$, is dense in $L^2(\mu)$.


问:
Integral equations for research in quantum mechanics
In my research, I have reached a point where I need to find a nonnegative function which satisfies the following properties:

It's an even function:

$$f(-\theta) = f(\theta)$$

It's normalised as:

$$\int^\pi_0f(\theta)d\theta = \frac{\pi}{2}$$

For any $\phi$ in the interval $[0,\pi]$,the following property is satisfied:

$$\int^\pi_0[f(\theta)+f(\theta - \phi)-2f(\theta)f(\theta - \phi)]d\theta = \pi\sin^2\frac{\phi}{2}$$

I would be very much in your debt if you help me find such a function.

答:
This falls into the problem of computing convolutional square roots, modulo the flaws already outlined by Cameron Williams.

A more sensible problem would be; compute a function $f$ on the circle (that is, a $2\pi$ periodic function) that satisfies $f(\theta)=f(-\theta)$ and such that
$$\tag{1}
f\ast f (\phi) = \pi \cos^2\frac \phi 2.$$
Here
$$
f\ast f(\phi):= \int_{-\pi}^\pi f(\phi-\theta)f(\theta)\,d\theta.$$
The normalisation condition is already encoded in (1); indeed, evaluating (1) at $\phi=0$,
$$
f\ast f(0)= 2\int_0^\pi f^2(\theta)\, d\theta = \pi.$$

Convolution square roots have been discussed before on Math.StackExchange. (Here there is an answer of Julián Aguirre (who, unfortunately, recently died). In the comments to that question, I have linked a related problem, answered by Terry Tao. All of these links, however, describe the problem on $\mathbb R$. I doubt they can be useful to the OP).

OwnerDisplayName: Arthur


问:
differentiating covariance
$$\sum_{i=1}^n x_{i}^2C_{ii} + \sum_{i=1}^n \sum_{j=1 \atop{j\neq i}}^n x_{i}x_{j}C_{ij}$$
I am trying to differentiate the above expression with respect to $x_i$. I did partial differentiation to get $$2x_i C_{ii} + \sum_{j=1 \atop{j\neq i}}^n x_{j}C_{ij}$$. However the answer is given as $$2x_i C_{ii} + 2\sum_{j=1 \atop{j\neq i}}^n x_{j}C_{ij}$$.

Could I get assistance with the regards to the extra 2 before the summation and why it should be there?

答:
Note that

$$\frac{\partial x_p}{\partial x_i} = \delta_{ip} = \begin{cases}1, & i=p\\0,& i \neq p \end{cases},$$

and by relabeling the first index,

$$S= \underset{j \neq i}{\sum_{i=1}^n \sum_{j=1 }^n} x_ix_{j}C_{ij} = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_kx_{j}C_{kj}$$

Hence,

$$\frac{\partial S}{\partial x_i} =\frac{\partial}{\partial x_i}\underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_kx_{j}C_{kj} = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} \frac{\partial (x_k x_{j })}{\partial x_i}C_{kj} \\ = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_k\frac{\partial  x_{j }}{\partial x_i}C_{kj}+  \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_j\frac{\partial  x_{k }}{\partial x_i}C_{kj} = \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_k\delta_{ij}C_{kj}+  \underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_j\delta_{ik}C_{kj} $$

For the second sum on the RHS, since $\delta_{ik} = 0$ if $k \neq i$, we have

$$\underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_j\delta_{ik}C_{kj} = \underset{j \neq i}{\sum_{j=1 }^n}x_jC_{ij} $$

For the first sum on the RHS, since $\delta_{ij} = 0$ if $j \neq i$, we have

$$\underset{j \neq k}{\sum_{k=1}^n \sum_{j=1 }^n} x_k\delta_{ij}C_{kj}= \underset{k \neq j}{\sum_{j=1}^n \sum_{k=1 }^n} x_k\delta_{ij}C_{kj} = \underset{k \neq i}{\sum_{k=1}^n } x_kC_{ki} =\underset{j \neq i}{\sum_{j=1}^n } x_jC_{ji}$$

Thus,

$$\tag{*}\frac{\partial S}{\partial x_i} = \underset{j \neq i}{\sum_{j=1 }^n}x_jC_{ij} + \underset{j \neq i}{\sum_{j=1}^n } x_jC_{ji}$$

A covariance matrix is symmetric.  Thus $C_{ij} = C_{ji}$ and (*) becomes

$$\frac{\partial S}{\partial x_i} = 2\underset{j \neq i}{\sum_{j=1 }^n}x_jC_{ij}$$

Title: differentiating covariance


问:
Closed subset with derived set non closed
I'm looking for a topologic space $X$ and a $A\subset X$ closed such that $A'$ is not closed. Where $A'$ is the derived set of $A$. I'm trying with a $T_0$ space, like Sierpinski topology but I couldn't find the example.

If you can help me with any ideas it would be of great help.

答: You can't find an example because the derived set of a closed set is always closed. You can find the proof in the answer to this question.</p>


问:
Give syntax proof of $\forall x (P(x) \lor Q(x))$ , $\forall y. \neg P(y)$ $\vdash$ $\forall x.Q(x)$
So the question is in the title already. I have to prove the following using first-order logic but I am failing to do so. Please someone help me understand how do I do it?

$\forall x (P(x) \lor Q(x))$ , $\forall y. \neg P(y)$ $\vdash$ $\forall x.Q(x)$

答:
So as for the first rule of natural deduction you can only use the elimination of the universal quantifier; can you go on from there?

it is just like PL until (and not including) the last necessary rule.


问:
Finding bases for a matrix's four subspaces without computing the matrix
enter image description here

I have attached what A is in a picture. I'm wondering how I can find the bases of A without computing A (i.e. by just looking at the LU factorisation of A, how can we work out the bases for the column, row, null and left nullspace?)

So far I think rank of A is 3, because rank of L and U are both 3. And because L and U both have 3 linearly independent columns, A (=LU) has 3 independent columns as well? Which leads to dimension of column space of A being 3?

I also know that the dimension of the nullspace of A is 4-3 = 1. But how do I work out some general bases of the 4 subspaces?

Thank you for your time!

答: It turns out $A=LU$ has rank $3$. To see this, we'll show that $R(A)=\mathbb{R}^3$. Choose $b\in \mathbb{R}^3$ arbitrary. Note $Ax=b$ has a solution iff $\Big[U\Big|L^{-1}b\Big]$ is consistent. The latter augmented system is consisent since $U$ has $3$ pivot columns, so $A$ has rank $3$; any basis for $\mathbb{R}^3$ will suffice as a basis for $R(A)$. Next, observe how $Ax=0$ iff $Ux=0$ so $N(A)=N(U)$. You can find $N(A^T)$ using the fact $\Big[R(A)\Big]^{\perp}=N(A^T)$. Lastly, to find $R(A^T)$, use the facts that $$\text{rank}(A^T)=\text{rank}(A)=3$$ $$\text{rank}(U^T)=3$$ $$R(A^T)\subseteq R(U^T)$$ to conclude $R(A^T)=R(U^T)$.


问:
Inequality Involving Supremum of Expectation
Let $X, Y$ be two random variables and $f$ be a continuous function. If for any realization $X=x$, we have
$$
\mathbb{E}[f(x,Y)] \leq \sup_{x} \mathbb{E}[f(x,Y)]
$$
I was wondering if we can conclude that $\mathbb{E}[f(X,Y)] \leq \sup_x \mathbb{E}[f(x,Y)]?$

I guess it is fine, by something like substitution law for conditional expectation, but not pretty sure...Any comment is appreciated. Thank you.

答:
First of all, note that the assumption about particular realisations is vacuously true ($\mathbb{E}[f(X(\omega_0), Y)] \leq \sup_{\omega \in \Omega} \mathbb{E}[f(X(\omega), Y)]$ trivially- I have assumed $X$ and $Y$ are defined on a probability space $(\Omega, \mathcal{B}, \mathbb{P})$).

We will choose $X = Y$ to be Rademacher variables, i.e. $\mathbb{P}(X = 1) = \mathbb{P}(X = -1) = \frac{1}{2},$ supported on exactly these two points and $f: \mathbb{R}^2 \rightarrow \mathbb{R},$ $f(x, y) := xy,$ which is clearly continuous. Note that $\mathbb{E}[f(X, Y)] = \mathbb{E}[X^2] = 1.$ However, all the possible realisations for $X$ are $1$ and $-1$ and we have that $\mathbb{E}[f(1, Y)] = \mathbb{E}[f(-1, Y)] = \mathbb{E}[X] = 0,$ so it is not true that $\mathbb{E}[f(X, Y)] \leq \sup_{\omega \in \Omega} \mathbb{E}[f(X(\omega), Y)].$


问:
Expected Value of Logit
If the only information that we know is that the following holds

$$\mathbb{E}[Y]=\frac{a}{a+b}$$

Is it possible for

$$\mathbb{E}[logit(Y)]=logit\Bigl(\frac{a}{a+b}\Bigr)$$

to hold?? Is there a property that might make it possible to hold?

答: Clearly, if $Y$ is constant, your property will hold, but if $Y$ is a general random variable, this is not true. So in general you seem to be asking under which conditions it is possible to interchange integration and a continuous function.

Title: Expected Value of Logit


问:
Solving $e^{-x} + x e^{-x} + \frac12x^2e^{-x} + \frac16x^3 e^{-x}=0.99$
Equation:

$$e^{-x} + x e^{-x} + \frac12 x^2 e^{-x} + \frac16x^3e^{-x}=0.99$$

I tried simplifying it and got it down to

$$e^{-x}\left(6+6x+3x^2+x^3\right)=5.94$$

although I am not sure how I would solve the $(6+6x+3x^2+x^3)$ portion.

Thanks to Wolfram Alpha I know the roots (solution for $x$), so I am really looking for guidance on how you would approach this equation and how far you can simplify it before having to use a calculator.

答: To solve $0.99=\frac{1+x+x^2/2+x^3/6}{1+x+x^2/2+x^3/6+x^4/24+\cdots}\in1-x^4/24+o(x^4)$, work numerically (e.g. by Newton–Raphson), starting from $x\approx0.24^{1/4}\approx0.70$. We quickly get a root it sounds like you already know, $0.82$. But I doubt a non-numerical option is available.


问:
indefinite integral of $e^{\frac{-x^2}{1+x}}$
I'm wondering if there's some simple form of the integral: $\int_0^X e^{\frac{-x^2}{1+x}} dx$

For context, I'm fitting to a distribution (of the velocities of local stars) as shown below and it's well approximated by $p(v) \propto e^\frac{-v^2}{1+|v|}$. AFAIK this is not a common probability distribution, but it seems a useful one, especially if the CDF can be written in simple functions.
histogram of observed stars in red, and the proposed pdf in green

答:
It doesn't look like there is an elementary integral, but we can find asymptotic forms.

For small $X$

Since $\frac{-x^2}{1+x}=1-x-\frac{1}{1+x}$, I will consider the integral

$$
I(b)=\int\limits_0^b dx \  e^{ -x-\frac{1}{1+x}}
$$

Which differs from the original by a multiplicative constant $e$, and investigate $b \to 0$. Integrating by parts

$$
I(b)=\int\limits_0^b dx \ (1+x)^2 e^{-x} \left[ \frac{d}{dx} e^{-\frac{1}{1+x}}\right]
$$

$$
I(b)=(1+x)^2  e^{ -x-\frac{1}{1+x}} \Bigg\vert_0^b - \int\limits_0^b dx \ (1-x^2) e^{ -x-\frac{1}{1+x}}
$$

The boundary term may be readily evaluated. Within the integral on the right we recognize $I(b)$

$$
2 I(b)=(1+b)^2 e^{ -b-\frac{1}{1+b}}-e^{-1}+\int\limits_0^bx^2e^{ -x-\frac{1}{1+x}}
$$

For $b \to 0$, the integral on the right is $O(b^3)$ and we have

$$
I(b)>>\frac{1}{2} \int\limits_0^bx^2e^{ -x-\frac{1}{1+x}}  \ \ , \ \ b \to 0
$$

Therefore

$$
I(b) \sim \frac{1}{2} \left( (1+b)^2 e^{ -b-\frac{1}{1+b}}-e^{-1} \right) \ \ , \ \ b \to 0
$$

Simplifying, I find the leading order term

$$
I(b) \sim \frac{b}{e} \ \ , \ \ b \to 0
$$

In principle, repeated integration by parts could produce the asymptotic series. Here is a plot of the first term versus the numerical integration for small $b$:

For large $X$

I'll use the substitution in the comments: $t=x+1$ to write

$$
e^{-2} \int\limits_0^X \ dx \exp \left(-\frac{x^2}{1+x} \right) =\int\limits_1^a dt \ \exp(-t-t^{-1})
$$

And we're investigating $a \to \infty$. Let $f(t)=\exp(-t-t^{-1})$, rewrite the integral

$$
\int\limits_0^\infty dt \ f(t) = \int\limits_0^1 dt \ f(t) + \int\limits_1^a dt \ f(t) + \int\limits_a^\infty dt \ f(t)
$$

The integral on the left is exactly $2K_1(2)$, where $K$ is a modified Bessel function of the second kind. The first integral on the right is also constant, independent of $a$. We are left with the simpler problem of studying

$$
I(a)=\int\limits_a^\infty dt \ e^{-t-t^{-1}}
$$

For large $a$. Integrate by parts

$$
I(a)=\int\limits_a^\infty dt \ e^{-t^{-1}} \frac{d}{dt} \left[ -e^{-t} \right]
$$

$$
I(a)=-e^{-t-t^{-1}} \Big\vert_a^\infty + \int\limits_a^\infty dt \ \frac{1}{t^2} e^{-t-t^{-1}}
$$

The integral on the right differs from $I(a)$ by a factor of $1/t^2$, thus we have

$$
I(a)>> \int\limits_a^\infty dt \ \frac{1}{t^2} e^{-t-t^{-1}} \ \ , \ \ a \to \infty
$$

Which leads to

$$
I(a) \sim e^{-a} \ \ , \ \ a \to \infty
$$

Finally,

$$
\int\limits_1^a dt \ \exp(-t-t^{-1}) \sim 2K_1(2) - C - e^{-a} \ \ , \ \ a \to \infty
$$

Where $C=\int_0^1 dt \ f(t)$, and can be found numerically to be about $0.072$; probably there is a nice way to estimate it, but I don't see it right now. Here is a plot of the approximation versus the exact numerical integral:

EDIT: In terms of the incomplete Bessel functions, defined as

$$
K_\nu(x,y)=\int\limits_1^\infty dt \ t^{-\nu-1} \exp(-xt -y/t)
$$

We have for $I(a)$, by changing variables $u = t/a$

$$
I(a)=a \int\limits_1^\infty du \  e^{-au-a^{-1} u^{-1}} = a K_{-1}(a,a^{-1})
$$

And for $C$, by changing variables $u= 1/t$

$$
C= \int\limits_1^\infty du \ u^{-2} e^{-u- u^{-1}} =  K_{1}(1,1)
$$

Thus your original integral, with $a=X+1$ may be written

$$
\int\limits_0^X \ dx \exp \left(-\frac{x^2}{1+x} \right) = e^2 \left[ 2K_1(2) - K_{1}(1,1) - a K_{-1}(a,a^{-1}) \right]
$$


问:
Extensions with the same degree are $K-$isomorphic?
Suppose we have two simple algebraic extensions $K(a)$, $K(b)$ over a field $K$

(I am interested in case $\mathrm{char}(K)=0$ but I guess there's no difference)

If they have the same (finite) degree over $K$, then they are always $K-$isomorphic?

答:
If $[K(a):K]=[K(b):K]$, then $\dim_K K(a)=\dim_K K(b)<\infty$ by definition and hence they are isomorphic as $K$-vector spaces by elementary linear algebra. That is, there is an invertible $K$-linear map $K(a)\to K(b)$. If we want this map to be a $K$-isomorphism (i.e. fixing $K$) we can write down explict bases for $K(a)$ and $K(b)$

$$\mathfrak B_1=\{1,a,\dots,a^{n-1}\},\quad\mathfrak B_2=\{1,b,\dots,b^{n-1}\}$$

where $[K(a):K]=n=[K(b):K]$ and force $1\mapsto1$. By linearity this ensures that $K$ is fixed. This also works in the absense of a power basis as we can freely insert $1$ into any given basis.

It is, indeed, not necessary that $\operatorname{char} K=0$ for this to work.


答:
This one's really nice.

$$\int\frac{1}{x^7-x}\text{ }dx$$

There's a clever trick that can save you from a tedious partial fraction decomposition. Can you find it?

Title: Challenging Integrals for High School Students

Body:
I am now in my last year of high school. We have covered all the techniques useful for indefinite integration that are included in our Maths and Further Maths courses. This includes:

Integration by parts, inspection, substitution, partial fraction decomposition

Integration of regular and inverse trigonometric, regular and inverse hyperbolic, exponential, logarithmic, polynomial functions

I would like to have some challenging integrals to attack that are possible for me to solve at my current level of knowledge. By challenging, I mean integrals similar to the ones in this document. They were generally enjoyable and very satisfying to solve. If you have an integral that you think I could do that is more challenging than those in the aforementioned link, so much the better.

Thank you for your suggestions.


问:
What alternatives are there to the generalized continuum hypothesis?
Since the generalized continuum hypothesis is independent of ZFC, we can adopt its negation as an axiom. However, this would not be a very "nice" axiom, since all it does is assert that there is at least one instance in which GCH fails.

My question is what axioms have been studied that prove the negation of the GCH, but has many more consequences than the negation of GCH. In particular, I am looking for ones that mathematicians would consider appealing axioms for mathematical foundations, not merely any statement that implies the negation of the GCH.

As an example, I would consider the Axiom of determinacy a "nice" alternative to the Axiom of choice.

答:
One thing to keep in mind are models where the GCH holds up to a point and then "fails." Gitik says something about the argument in "No bound for the first fixed point" requiring the GCH below the FFP but then there is no special bound on the FFP's powerset. Otherwise, I don't know if there's a generalization from $2^{\aleph_0} = 2^{\aleph_1} = \aleph_2$ but...

Now here it says that:

... Woodin showed (again assuming large cardinals) that it is possible that for all κ, 2κ = κ++.

But the context is singular cardinals so I'm not sure "all $\kappa$" means all cardinals whatsoever, or singular ones. Still, whatever the proof actually is, it constitutes some kind of at least partial alternative to the GCH.


问:
Is this a special kind of exponential equation? How can I solve it?
I like looking through random equations online, and can't figure out what kind of equation this is:

$$ 2^x + 2^y = 2^z$$

Seems like a multivariate exponential equation, but the results on the internet are all basic. I'm trying to solve 'Find all x, y, z such that $$ x+y+z = 100 $$
$$ 2^x + 2^y = 2^z$$'

Would appreciate help on this question I thought of while studying exponential equations. I tried using logarithms and laws of exponents but they didn't really help.

答:
Assuming we are looking for integer solutions, let $m$ be the smallest of $x,y,z$. Then $2^{m+1}$ would divide the other two terms and therefore the $2^m$ term as well (which is absurd) unless two of $x,y,z$  equal $m$.

We then have $2^m + 2^m = 2^{m+1}$ and so $m+m+(m+1)=100$ and $m=33$.


问:
Homotopy equivalent chains and I take the tensor product of both with the same complex?
This is my first post here, so forgive me for any potential mistakes.
So i have two chain complexes that are chain homotopy equivalent, say $C_1$ and $C_2$. Now if i take the tensor product of both of them with the same complex, as in $C_1 \otimes F[t,t^{-1}]$ and $C_2 \otimes F[t,t^{-1}]$ where the field is just $\mathbb{Z}_2$, is there a universal property or theorem that will give me a chain homotopy equivalence between the new complexes?
Intuitively, I have thought about just considering a map of the form $\mathcal{H} \otimes id$ where $\mathcal{H}$ is the chain homotopy equivalence before, but I failed to show why this should work.
Thanks a lot for any help!</p>

答:
I think you had the right ideia about the problem. Let's formalize that:

Let $f: C_1\rightarrow C_2$ and $g: C_2\rightarrow C_1$, such that $fg\sim id_{C_2}$ and $gf\sim id_{C_1}$. Let $id_F$ be the identity of $F[t, t^{-1}]$ and  consider the maps $$f\otimes id_F: C_1\otimes F[t,t^{-1}] \rightarrow C_2\otimes F[t, t^{-1}]$$ $$g\otimes id_F: C_2\otimes F[t,t^{-1}] \rightarrow C_1\otimes F[t, t^{-1}]$$
Since $fg\sim id_{C_2}$ there exists $h$ such that $fg-id_{C_2}=hd+dh$, so $$(f\otimes id_F)(g\otimes id_F)-id_{C_2}\otimes id_F= (fg-id_{C_2})\otimes id_F= (hd+dh)\otimes id_F = (h\otimes id_F)(d\otimes id_F) + (d\otimes id_F)(h\otimes id_F)$$

The same argument holds to show that $(g\otimes id_F)(f\otimes id_F)\sim id_{C_1}\otimes id_F$.

ContentLicense: CC BY-SA 3.0


问:
Prove that a symmetric group $S_n$ has a cyclic subgroup of order $n$.
Prove that a symmetric group $S_n$ has a cyclic subgroup of order $n$.

I can see it for a symmetric group $S_3$ and $S_4$ but how can I generalize it for any number $n.$

答: Hint: The element $$(1,2,3,\dots, n)$$ has order $n$.


问:
Visualising Plaques of Foliations
So I just recently started to study foliation theory and I realized I'm visualizing plaques of foliations the wrong way.

Given a manifold $M$, a foliation $\mathcal{F}$ of the manifold and a local chart $(U,\varphi)$ of $\mathcal{F}$ such that $\varphi(U)=U_1 \times U_2 \subset \mathbb{R}^n \times \mathbb{R}^{m-n}$, my definition of a plaque ($\alpha$) is a set of the form $\varphi^{-1}(U_1 \times \{c\}), \enspace c \in U_2$.

In my head, I imagine a plaque simply being the intersection of a leaf $F$ of the foliation and my open set $U$ of $M$.

For some time this made sense to me until I came across the observation that, in the topology of the leaf $F$, one could have $F$ intercepting the domain $U$ of a chart $(U,\varphi)\in \mathcal{F}$ in a sequence of plaques $(\alpha_n)_{n\in\mathbb{N}}$, which accumulate in a plaque $\alpha \subset F$.

For me, this is not coherent with the image I had of a plaque because, given a domain $U$ and a leaf $F$, I could only have one plaque of $F$ in $U$.

What is the problem here? I would appreciate any help. Please be kind since I have just recently started to study this.

答: With a tiny change, you can convert your incorrect intuition into a correct intuition, like this: A plaque is a connected component of the intersection of a leaf $F$ and the open set U.

Title: Visualising Plaques of Foliations


问:
Why is $ab + bc + c\bar{a} = ab + c\bar{a}$ true in binary?
I was simplifying the equation of a logic gates problem and I realized that ab + bc + cā and ab + cā followed the same truth table which is the following:

```
-----------------
| A | B | C | S |
-----------------
| 0 | 0 | 0 | 0 |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 0 | 0 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 0 | 0 |
| 1 | 0 | 1 | 0 |
| 1 | 1 | 0 | 1 |
| 1 | 1 | 1 | 1 |
-----------------
```

The problem is that I haven't found any way to get to the simplified expression from the main one by applying boolean algebra procedures. Any idea? Thanks.

答:
Another way to illustrate this equivalence is a Karnaugh-Veitch map:

The "green" term $b c$ can be eliminated, because it is covered by the disjunction of $c a'$ and $a b$.


问:
Circular segment using polar coordinates
Imagine, you'd like to integrate  $f(x,y)=x^2-y$ over this region:

How would you parameterize this circular segment? My try: $$\Phi : [1,2]\times [-\pi /3, \pi /3]\to \mathbb{R} \, (r,\varphi)\mapsto (x,y)=(r\cos(\varphi),r\sin(\varphi))$$ is this correct?

答:
HINT

The integral you are interested in is given by:
\begin{align*}
I = \int_{\arctan(-\sqrt{3})}^{\arctan(\sqrt{3})}\int_{\sec(\theta)}^{2}rf(r\cos(\theta),r\sin(\theta))\mathrm{d}r\mathrm{d}\theta
\end{align*}

Can you take it from here?


问:
Projecting Sphere to Rectangle/Square using Mercerator Projection
I am learning some projection technique where we can project a 3d object like globe to a 2d. I have the 3d coordinates of points on the surface of sphere same as globe. Here is a reference where a globe can be projected on a plane using Stereo-graphic projection (https://en.wikipedia.org/wiki/Stereographic_projection) but this projection project the globe to a circle but i am trying to project sphere to a rectangle the same way as the globe is projected on a small paper (https://en.wikipedia.org/wiki/Mercator_projection).</p>

答: Both Stereographic and Mercator projections map the globe to unbounded regions. However, the Lambert cylindrical equal area projection does map the entire globe to a rectangle.


问:
Proof related to $C([a,b],\mathbb{R}^n)=\{ f:[a,b]\to \mathbb R^n:f=(f_1\ldots f_n)/f \text{ is continuous on }[a,b]\}$ where $[a,b]\subset \mathbb R$

Question

Let $[a,b] \subset \mathbb{R}$,

We define the set $C([a,b], \mathbb{R}^n) = \{ f : [a,b] \rightarrow \mathbb{R}^n:f=(f_1\ldots f_n)/f \text{ is continuous on } [a,b]\}$

Prove that $C([a,b], \mathbb{R}^n)$ forms a linear space in relation to the estimation operations of two functions in $C([a,b], \mathbb{R}^n)$, respectively of multiplication with real scalars with functions from $C([a,b], \mathbb{R}^n)$.

My Attempt

I know that we have to prove the $10$ axioms of vector spaces, I just don't know how to take the functions, I mean in what form to take them, so that I can prove those axioms in $\mathbb{R}^n.$ Please, I just need a hint on how to take those two functions $f$ and $g$ in what form to prove those $10$ axioms.

答:
Some of the axioms are readily verified and don't require much attention, seeing as they are more or less pretty obvious. Perhaps the more sensible ones would be showing that the addition of two continuous functions is again continuous, as well as that any scalar multiple of a continuous function is indeed continuous, so I will prove these for you here.

Let $f, g \in C([a, b], \mathbb{R}^n)$ and $x \in [a, b]$ fixed. Let $\varepsilon > 0$ and $\lambda \in \mathbb{R} \setminus \{0\}$ be fixed (that the function which is identically zero is continuous is immediate). Since $f$ and $g$ are continuous at $x,$ there exist $\delta_\varepsilon^f, \delta_\varepsilon^g > 0$ such that $\Vert f(x) - f(y) \Vert < \min \{\frac{\varepsilon}{2}, \frac{\varepsilon}{\vert \lambda \vert} \} > \Vert g(x) - g(y) \Vert$ whenever $\vert x - y \vert < \delta_\varepsilon^f$ and $\vert x - y \vert < \delta_\varepsilon^g,$ respectively, where I think it's pretty clear how they correspond to each other ($\Vert \cdot \Vert$ denotes the standard euclidean norm on $\mathbb{R}^n$). It follows that for all $y \in [a, b]$ with $\vert x - y \vert < \min \{\delta_\varepsilon^f, \delta_\varepsilon^g\},$ we have that $$\Vert f(x) + g(x) - f(y) - g(y) \Vert \leq \Vert f(x) - f(y) \Vert + \Vert g(x) - g(y) \Vert \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon$$ and $$\Vert \lambda f(x) - \lambda f(y) \Vert = \vert \lambda \vert \Vert f(x) - f(y) \Vert \leq \vert \lambda \vert \frac{\varepsilon}{\vert \lambda \vert} = \varepsilon,$$ which proves continuity of $f + g$ and $\lambda f.$


问:
Determine the truth value with proof
I've been stuck on these two propositions for a while especially how to prove their truth value.
Note that $x$ and $y$ belong to $\mathbb{N}$.
Any idea how to solve this??

$\forall x ~ \forall y ~~ (x \ne 1) \to (xy \ne y)$

$\forall x ~ \forall y ~ \exists z ~~ (x > y) \to (x - y = z)$

答:
As you define $\mathbb N$ to include $0$, the first one is false:
Setting $y = 0$ and say, $x = 10$ we find that $x \neq 0$ but $0 = xy = y = 0$.

The second statement holds basically by definition; it is hard to "prove" this without also giving some actual definition of the $<$ relation (which is usually done by saying that $x > y$ if there exists some positive $z$ such that $x = y + z$, or equivalently, $x - y = z$).

Title: Determine the truth value with proof


问:
ode formation when constant appears with arc length.
How to form ode and solve for

$$ s( 4-s) = y^2+5 $$

where $s$ is arc length of a curve in x-y plane ? The constant 5 disappears when derivating.

答:
Add an arbitrary arc length $\lambda s $

$$ s^2-4 s + y^2 + 5 + \lambda s =0$$

Differentiate  w.r.t. arc s, $ \sin \phi =\frac{dy}{ds} $ the curvature $ \phi'$

$$ 2s-4 +2y \sin \phi + \lambda =0,\,$$
Differentiate again,

$$ 2+ 2( \sin^2\phi +y \cos \phi \;\phi^{'})=0 $$

simplifies to

$$ \phi^{'}(s)=\dfrac{\cos \phi-1/\cos \phi}{y}$$

which can be numerically integrated using suitably chosen IC. Product of two arc segments depends on $t$ as required.


问:
A hard geometry problem on circles
I found this problem on a website and I couldn't do anything. Do you have any ideas, hints?

Edit: If I say $$\frac { { \partial }^{ 2 }f }{ \partial { a }^{ 2 } } +\frac { { \partial  }^{ 2 }f }{ \partial { b }^{ 2 } } =0$$

am I wrong?

答:
Nothing to do with harmonic functions :)

This is a version of the classical Problem of Apollonius, for which there are both geometric and algebraic solutions. See http://en.wikipedia.org/wiki/Problem_of_Apollonius#Ten_combinations_of_points.2C_circles.2C_and_lines and/or http://mathworld.wolfram.com/ApolloniusProblem.html.
For the algebraic approach, you can use the chord lengths $a$ and $b$ to write down equations of the two lines (say, through the fixed point $(-1,0)$ on the unit circle).

Edit #3:
OK, it turns out one has to be a bit more careful than I was. Given $a<b$, there are two configurations (up to congruence). So, at long last, we have
$f(a,b)=$
$$ \frac{a^2 (2+\sqrt{4-b^2})+a b
   (\sqrt{4-a^2}-\sqrt{4-b^2})+(\sqrt{4-a^2}-2)(4
   \sqrt{4-b^2}+8-b^2)}{(a+b)^2}
$$
when the chords are on the same side of the diameter through the common point of the chords, and

$$\frac{a^2(2-\sqrt{4-b^2})+ab(\sqrt{4-a^2}+\sqrt{4-b^2})-(\sqrt{4-a^2}-2) (b^2+4\sqrt{4-b^2}-8)}{(a+b)^2}$$
when they are on opposite sides of the diameter.

I thank @Rahul Narain for insisting I get it right :) It is reassuring that these formulas do agree when $b=2$.

Still no harmonic functions.

Title: A hard geometry problem on circles


问:
Schur Theorem implies Finite Schur Theorem
$\text{If } \mathbb{N}=C_{1} \cup C_{2} \cup \ldots \cup C_{r} \text{ is a finite partition, then for some } j \in\{1,2, \ldots, r\}, \text{ there exist } a, b, c \in C_{j}  \text{ with } a+b=c.$

$\Rightarrow \text { Given } r \in \mathbb{N}, \text { there exists a positive integer } N=
N(r) \text { such that if }\{1,2, \ldots, N\}=C_{1} \cup C_{2} \cup \ldots \cup C_{r}, \text { is a finite partition, then for some } j \in\{1,2, \ldots, r\}, C_{j} \text { contains } a, b, c \text { with } a+b=c $

I am trying to understand the following proof to the given statement.

Suppose not. Then for each $N \in \mathbb{N},$ there is a coloring of $\{1,2, \ldots, N\}$ with $r$ colors that does not contain any monochromatic Schur triple. Use any of the $r$ colors to color the integers $\{N+1, N+$ $2, \ldots\},$ thus extending each coloring of $\{1,2, \ldots, N\}$ to a coloring of $\mathbb{N}$ with $r$ colors. Thus for each $N \in \mathbb{N}$, we have a corresponding coloring of $\mathbb{N}$. (*) Since only finitely many colors were used, by the Pigeonhole Principle there are infinitely many $N \in \mathbb{N}$ such that the corresponding coloring uses the same color for the integer $1 .$ Consider only this subsequence of colorings and note that we are still left with an infinite set of colorings. We can now repeat this process in this subsequence. By passing to a subsequence, if necessary, we can assume that all the colorings use the same color for the integer 2 . Inductively, we can pass to a subsequence of the original list of colorings that fixes each $n \in \mathbb{N}$ with a single color. Taking the limit along this subsequence, we obtain a coloring of $\mathbb{N}$ that does not contain a monochromatic Schur triple, a contradiction.

I am stuck at (*). How does the author use the Pigeonhole Principle? What exactly is the process of taking sub-sequences?

There is a very similar question here which was helpful to me in proving what I want but I want to understand this proof too.

答:
For each $N\in\Bbb N$ we have a coloring $\mathscr{C}_N$ of $\Bbb N$ that uses $r$ colors and does not have any monochromatic Schur triple in the set $\{1,2,\ldots,N\}$. Let $c_N$ be the color given to the number $1$ by the coloring $\mathscr{C}_N$. Since each of the colorings $\mathscr{C}_N$ uses the same $r$ colors, there are only $r$ possibilities for the colors $c_N$. But there’s a $c_N$ for every $N\in\Bbb N$, so at least one of those $r$ colors must be used for infinitely many of the $c_N$. Thus, there are an increasing sequence $\langle N_k:k\in\Bbb N\rangle$ of positive integers and a color $c(1)$ such that $c_{N_k}=c(1)$ for every $k\in\Bbb N$. That is, we have an infinite subsequence $\langle\mathscr{C}_{N_k}:k\in\Bbb N\rangle$ of the original sequence $\langle\mathscr{C}_N:N\in\Bbb N\rangle$ of colorings with the property that every coloring $\mathscr{C}_{N_k}$ in this subsequence colors $1$ with the color $c(1)$.

Now start with this subsequence $\langle\mathscr{C}_{N_k}:k\in\Bbb N\rangle$ and do the same thing. Every coloring $\mathscr{C}_{N_k}$ gives $2$ one of the same $r$ colors, so there must be an infinite subsequence of it, $\langle\mathscr{C}_{N_{k_i}}:i\in\Bbb N\rangle$, consisting of colorings that all give $2$ the same color, say $c(2)$. We continue repeating this process. At stage $n+1$ we start with an infinite sequence of colorings that all color $1$ with color $c(1)$, $2$ with color $c(2)$, and so on $n$ with color $c(n)$. We repeat the argument of the first paragraph to conclude that an infinite subsequence of this sequence must all color $n+1$ with the same color, which we’ll call $c(n+1)$. This recursive construction can be carried out to produce a color $c(n)$ for each $n\in\Bbb N$.

Let $\mathscr{C}$ be the coloring defined by the colors $c(n)$ for $n\in\Bbb N$: we color $1$ with $c(1)$, $2$ with $c(2)$, and so on for all $n\in\Bbb N$. Suppose that $\mathscr{C}$ had a monochromatic Schur triple $\langle u,v,w\rangle$, with $u+v=w$. At stage $w$ in the construction we had an infinite subsequence of the original sequence of colorings such that every coloring in the subsequence gave every positive integer $n<w$ the color $c(n)$, and we found an infinite subsequence $\sigma$ of that consisting of colorings that gave $w$ the color $c(w)$. The triple $\langle u,v,w\rangle$ is monochromatic, so $c(u)=c(v)=c(w)$. Since $\sigma$ is an infinite subsequence of $\langle\mathscr{C}_N:N\in\Bbb N\rangle$, there is an $N\in\Bbb N$ such that $N>w$ and $\mathscr{C}_N$ is one of the sequences in the subsequence $\sigma$. And now we have our contradiction: $\mathscr{C}_N$ has no monochromatic triple in the set $\{1,2,\ldots,N\}$, but $\langle u,v,w\rangle$ is a monochromatic triple in that set.

This contradiction shows that $\mathscr{C}$ cannot have a monochromatic Schur triple, and that contradicts the Schur theorem. This final contradiction then shows that no such sequence as $\langle\mathscr{C}_N:N\in\Bbb N\rangle$ can exist, and therefore there must be some $N\in\Bbb N$ such that any $r$-coloring of $\{1,2,\ldots,N\}$ has a monochromatic Schur triple.


问:
Complex function with property: $f(z+w) = f(z) + f(w)$.
If $f$ is a complex vauled fuction which in analytic in the complex plane and  satisfies,

$ f(z+w) = f(z) + f(w) ,    z, w  $ are complex  numbers

Then  show that  $ f(z) = cz$     where $c$ is a constant..

I have tried to break the function in real and imaginary part. But failed to solve.
Any help is highly appreciated .

答:
Substituting into the equation
$$
f(z+w)=f(z)+f(w)\tag1
$$
$z=w=0$ one obtains
$$
f(0)=0.\tag2
$$

Since the function is analytic in the whole complex plane its derivative exists everywhere and is constant:

$$
f'(z)=\lim_{\Delta z\to0}\frac{f(z+\Delta z)-f(z)}{\Delta z}\stackrel{(1)}=\lim_{\Delta z\to0}\frac{f(\Delta z)}{\Delta z}=c,\tag3
$$
which after combining with (2) gives
$$
f(z)=cz.
$$


问:
Are mobile 2FA verification PINs completely random?
In my experience, when I have to use some 6 digit pin to log in to, say, my internet banking, or some other service which requires 2FA; the PIN itself always seems "catchy".

Example PINs include:

232323

888666

120130

112211

228229

Is this my brain trying to find patterns, or is their generation somehow done in such away which prefers "catchy" PINs (to make them easy to remember when typing them in)? It feels like I never get a truly "random feeling" PIN like 703926 (say).

答:
Is this my brain trying to find patterns [...]?

Yes. The numbers generated internally are actually fully random 31-bit integers which then get reduced $\bmod 10^6$ yielding a bias of less than $2^{10}$ to small values (the bias gets slightly worse for 7 and 8 digits).

Details

The actual details are in RFC 6238 (TOTP) which esstially just invokes HOTP (RFC 4226) with a conversion of the current time in seconds (since epoch) into the number of time steps (since epoch). HOTP then essentially computes $\operatorname{HMAC-SHA1}(K,C)$ using a fixed shared key between you and the service and using the counter value $C$ as derived from the time. This yields a 20 byte random value (assuming HMAC-SHA1 is a PRF which is a pretty weak and still true assumption). From these 20 random bytes 31 random bits are extracted in an non-straightforward (and security-wise useless) way. Then these 31 bits are treated as a normal 32-bit integer by your processor (with the top bit cleared to avoid wonkiness in the modular reduction) and reduced modulo $10^6$ yielding the displayed number.

Applying a PRF keyed with a random key to a not before queried value yields a freshly random value (if not the PRF assumption would be broken). Extracting the 31 bits is (hopefully) done in a way that preserves as much entropy as possible (i.e. 31 bit). Reducing a 31-bit value $\bmod 10^d$ then introduces a small bias towards the low $2^{31}\bmod 10^d$ values by about $10^d/2^{31}$ which is small enough for all standardized values of $d\in\{6,7,8\}$ - especially given the fact that we only have 31 bits to work with here.


问:
How to find a function given parameters?
A continuous function defined on $[-1,1]$ with $f(-1) = -1$ and $f(1) = 1$ such that $\left|f(x)\right |\geq \frac{1}{2}$ for all $x \in [-1,1]$

How can I find a function that fits this criteria?

答:
$f $ is continuous at $ [-1,1] $ and $ f(-1)f(1)<0$, So by IVT,

$$\exists c\in(-1,1) \;:\; |f(c)|=0<\frac 12$$

The function you look for cannot exist.


问:
Is the order relation to $<$ on the positive rational numbers (fractions) well-founded?
I am pretty new with well-founded relations and the concept, so i have sort of gave a brief reason nothing to in depth would i be correct with what i have said any help and explanations are greatly appreciated.

This is what i have said i believe the relation to not be well-founded am i correct?

the order relation to < strictly less than on the positive rational
numbers is not a well-founded relation as it has an infinite
descending chain, for example this infinite descending chain ...
$< a_n < ...< a_4 < a_3< a_2 <a_1 <a_0$ where $a_0 = \frac{9}{10}$ then ($... < an < ... <\frac{1}{5} < \frac{1}{4}< \frac{1}{3} <\frac{1}{2} <\frac{9}{10}$ ) this would be an infinite descending chain.

答:
It could be stated more clearly, but your reasoning is correct. There’s no need to use $0.99$; you could simply let $a_n=\frac1{n+1}$ for $n\in\Bbb N$ and observe that $\langle a_n:n\in\Bbb N\rangle$ is an infinite strictly descending chain in $\Bbb Q^+$. Another simple example is to let $a_n=2^{-n}$ for $n\in\Bbb N$.

(By the way, the usual term for $<$ is strictly less than.)


问:
How does permuting rows and columns change the eigenvectors of a matrix?
Let $A\in\mathbb{R}^{n\times n}$ be any matrix and let $P\in\mathbb{R}^{n\times n}$ be a permutation matrix.

Left multiplication of $A$  by $P$ corresponds to permuting the rows of $P$. On the other hand, right multiplication of $A$  by $P$ corresponds to permuting the columns of $P$. However, I'm not sure how these operations affect the eigenvalues and eigenvectors.

In other words, how are the eigenvalues and eigenvectors of $A$, $PA$, and $AP$ related or not?

答:
Well, we first note that the eigenvalues do change, just consider a quick example:
$$ A=\begin{pmatrix}-1 &0\\0&1\end{pmatrix} \Rightarrow \lambda_{1,2}=\pm1, \text{ while } PA=\begin{pmatrix}0&1\\-1&0\end{pmatrix} \Rightarrow \lambda_{1,2}=\pm i$$
However, the product of all the eingenvalues (counting multiplicity) will only change its signal when applying a permutation (or will not change at all if you have a even permutation). That's true because the determinant of a matrix is equal to the product of its eigenvalues.


答:
A binary relation $R$ on a class $X$ is said to be well founded if every non empty subset $S \subseteq X$ has a minimal element.

Obviously, $\lt$ is not well founded on $\mathbb Q$ as $\mathbb Q$ itself has no minimal element: any $q$ can’t be a minimal element of $\mathbb Q$ as $q-1 \lt q$.

In your question, you used the infinite descending chain characterization of well founded that requires a form of axiom of choice.

Title: Is the order relation to $<$ on the positive rational numbers (fractions) well-founded?

Body:
I am pretty new with well-founded relations and the concept, so i have sort of gave a brief reason nothing to in depth would i be correct with what i have said any help and explanations are greatly appreciated.

This is what i have said i believe the relation to not be well-founded am i correct?

the order relation to < strictly less than on the positive rational
numbers is not a well-founded relation as it has an infinite
descending chain, for example this infinite descending chain ...
$< a_n < ...< a_4 < a_3< a_2 <a_1 <a_0$ where $a_0 = \frac{9}{10}$ then ($... < an < ... <\frac{1}{5} < \frac{1}{4}< \frac{1}{3} <\frac{1}{2} <\frac{9}{10}$ ) this would be an infinite descending chain.


问:
how to find $t$ and $r$ of Archimedean spiral with known spiral length?
I have a generic archimedean spiral $r=a\cdot\frac{t}{\pi}$ and have a spiral length $s$.

by knowing the formula for the spiral, and knowing that the spiral has no rotation applied, and knowing the spiral length how can $r$ and $t$ be calculated?

答:
EDIT

Derivate wrt polar angle t.

$$r= at/\pi,\quad r'= a/\pi\;$$

Arc length

$$ L= \int_0^ T \sqrt{ r^2+r^{'2}} dt =(a/\pi)\int_0^T \sqrt{1+t^2}dt$$ which is evaluated using  hyperbolic functions.

$$ L= (  a/\pi)\frac12 (T\sqrt{1+T^2} +ArcSinh T )$$

This result helps to numerically find $T$ when $L$ is given (no need r anymore) with an iterator.


问:
How to show $[f^{-1}][f]\cong e_x?$
$\pi_1(X,x_0)$ is  a group with respect to the product $[f][g]=[f.g]$

satisfy three properties

$1. $Associative

$2.$ Identity :$[c]$ where $c$ is the constant loop i,e $c(s)=x_0$ for any $s$

$3.$Inverse:The inverse  of $[f]$ will $[\bar f]$ , where  $\bar f(s)= f(1-s)$

My question : How to show the inverse  properties in $\pi_1(X,x_0)?$

My attempt :   our motive  $[f][f^{-1}] \cong e_x $ and $[f^{-1}][f]\cong e_x$

First case : to show $[f][f^{-1}] \cong e_x $

Proof : $[f]\cdot [f^{-1}](t)= \begin{cases}
      f(2t)& 0\le t\le 1/2 \\
      f(2(1-t) & 1/2\le t\le 1
   \end{cases}$

Now we can construct a homotophy between  $[f][f^{-1}]$ and $e_x$ i,e given by

$H : I \times I  \to X$ is given by

$H(t,s)= \begin{cases}
      f(2t(1-s)& 0\le t\le 1/2 \\
      f\left((2-2t)(1-s))\right) & 1/2\le t\le 1
   \end{cases}$

$H(t,0)= [f][f^{-1}](t)$  and since $f$ is a loop so  $H(t,1)= f(0) =e_x(t) $

where  $e_x$ denote the constant path $ e_x:[0,1] \to X$  defined by  $e_x(t)=x$ for all $t\in [0,1]$

This implies that  $H$ is homotopy relative $\{0,1\}$

Therefore $[f][f^{-1}] \cong e_x $

My doubts : my doubt is  in converse direction that

How to show  $[f^{-1}][f]\cong e_x?$

答:
Proof : $[f^{-1}]\cdot [f](t)= \begin{cases}
      f^{-1}(2t)& 0\le t\le 1/2 \\
      f(2t-1) & 1/2\le t\le 1
   \end{cases}$

Note: $f^{-1}(2t)= f(1-2t) $

$[f^{-1}]\cdot [f](t)= \begin{cases}
      f(1-2t)& 0\le t\le 1/2 \\
      f(2t-1) & 1/2\le t\le 1
   \end{cases}$

Now we can construct a homotophy between  $[f^{-1}][f]$ and $e_x$ i,e given by

$H : I \times I  \to X$ is given by

$H(t,s)= \begin{cases}
      f((1-2st)& 0\le t\le 1/2 \\
      f\left((2t-1)s+(1-s))\right) & 1/2\le t\le 1
   \end{cases}$

since $f$ is a loop so  $H(t,0)= f(1) =e_x(t) $

so,$H(t,0)= f(1)=e_x(t)$  and

$H(t,1)=[f^{-1}][f]$

where  $e_x$ denote the constant path $ e_x:[0,1] \to X$  defined by  $e_x(t)=x$ for all $t\in [0,1]$

This implies that  $H$ is homotopy relative $\{0,1\}$

Therefore $[f^{-1}][f] \cong e_x $

Title: How to show $[f^{-1}][f]\cong e_x?$


问:
$p$-subgroup of Sylow and Normalizer
Let $G$ be a finite group, $p$ a prime that divides $|G|$, and $P$ a $p$-subgroup of Sylow of $G$. Show that $P$ is the unique $p$-subgroup of Sylow that is in $N_G(P)$.

I tried this by assuming there is another $p$-subgroup of Sylow called $Q$. So by the second Sylow's theorem, $Pg=gQ$; but I don't see something to justify that $P=Q$. Any help would be appreciated.

答: If $Q$ is another $p$-Sylow subgroup of $N_G(P)$ then indeed by the Sylow theorems we have $Q=gPg^{-1}$ for some $g\in N_G(P)$. But $P$ is a normal subgroup of $N_G(P)$, and so $gPg^{-1}=P$.

Title: $p$-subgroup of Sylow and Normalizer


答: Hint:  By one of Sylow's theorems, all the Sylow $p$-subgroups are conjugate.

Title: $p$-subgroup of Sylow and Normalizer

Body:
Let $G$ be a finite group, $p$ a prime that divides $|G|$, and $P$ a $p$-subgroup of Sylow of $G$. Show that $P$ is the unique $p$-subgroup of Sylow that is in $N_G(P)$.

I tried this by assuming there is another $p$-subgroup of Sylow called $Q$. So by the second Sylow's theorem, $Pg=gQ$; but I don't see something to justify that $P=Q$. Any help would be appreciated.


问:
Query about functions
Suppose that we have a function $f:(\mathbb{R^2},d_2) \to (\mathbb{R^2},d_2) $ where $d_2$ is the Euclidean metric. A function in this case looks like what exactly. Since we are in $\mathbb{R^2}$ the function could look like $f(x,y)=(x-1,y+2)$ right? If this is the case then what would be meant by $d_2(f(x),f(y))$. Maybe someone could clarify why I'm confused, or if I am just wrong in general about how I am thinking about this.

答:
You are confused because you’re using $x,y$ to denote two different things:

Coordinates $(x,y)$ of a vector $X \in \mathbb R^2$.

Vectors of $\mathbb R^2$ when you write $d(f(x),f(y))$.

This is not an issue usually, except when you’re working on both concepts at the same time. What you’re doing.

Denote $X_1=(x_1,y_1), X_2=(x_2,y_2)$ two vectors will avoid the ambiguity you have.

$d_2(f(X_1),f(X_2))$ means the Euclidean distance between the images of $X_1,X_2$ under $f$.

The map  $f$ is independent from the distance $d_2$ itself. What is not independent from the metric, $d_2$ in your example are some properties of the map $f$. Continuity for example. For such properties, the metric is crucial as a map may be continuous for a metric but not related to another one.

Title: Query about functions


问:
Proof of a theorem on Plateau's problem in "Elementary Differential Geometry" by Andrew Presley
I'm studying the proof of this theorem on Plateau's problem in the textbook "Elementary Differential Geometry" by Andrew Presley, but I don't understand, being more specific, why $\beta^0=\gamma^0=0$ in the final part. For first, let me enunciate the theorem:

Theorem: with the above notation, assumme that the surface $\varphi^{\tau}$ vanishes along the boundary curve $\pi$. then
\begin{align}
(\mathcal{A}(0))' = -2\int_{\operatorname{int}(\pi)} H(EG - F^2)^{1/2}\alpha du dv, \label{1}
\end{align}
where $H$ is the mean curvature of $\sigma$, $E$, $F$ and $G$ are the coefficients of its first fundamental form, and $\alpha = \varphi\cdot N$, where $N$ is the standar unit normal of $\sigma$.

And this is the proof:

I think that reason is related with the following fact, $\varphi^{\tau}(u,v) = 0$ when $(u,v)$ is a point on the curve $\pi$ where $\varphi^{\tau}= (\sigma^{\tau})^{\prime}$, but i didn't get it to, any help will be really appreciate.

答:
Your suspicion is correct.

The theorem assumes that the surface variation $\varphi^\tau(u,v) = \mathbf{0}$ for every point $(u,v)$ on the boundary curve $\pi$.  Now, the equation
$$\varphi^\tau = \alpha^\tau \mathbf{N}^\tau + \beta^\tau\sigma^\tau_u + \gamma^\tau \sigma^\tau_v$$
holds for all points $(u,v)$ in the interior region $\text{Int}(\pi)$ and on the boundary curve $\pi$.  In particular, if $(u,v)$ is on the boundary, then:
$$\mathbf{0} = \alpha^\tau(u,v) \mathbf{N}^\tau + \beta^\tau(u,v) \sigma^\tau_u + \gamma^\tau(u,v) \sigma^\tau_v.$$
Since $\{\mathbf{N}^\tau, \sigma^\tau_u, \sigma^\tau_v\}$ is linearly independent, it follows that for every point $(u,v)$ on the boundary curve $\pi$:
$$\alpha^\tau(u,v) = 0, \ \ \ \beta^\tau(u,v) = 0, \ \ \ \gamma^\tau(u,v) = 0.$$
Setting $\tau = 0$, we arrive at the claim.


答:
The presence of the distances in the definition of $f$ clarifies which topologies we are to use when talking about stuff like continuity for example. What is meant by $d_2(f(x, y), f(z, w))$ is exactly the only thing that can be meant (note that $f$ has two inputs, whilst in your original question it had only one- which is not necessarily a mistake if you also mention $x, y \in \mathbb{R}^2,$ for example). In your particular example, we would have $$d_2(f(x,y), f(z, w)) = \Vert (x - 1, y + 2) - (z - 1, w + 2) \Vert = \Vert (x - z, y - w) \Vert = \sqrt{(x - z)^2 + (y - w)^2} = d_2((x, y), (z, w))$$ for instance.

What you should take away from all of this is that sometimes we are not working with the standard topology on $\mathbb{R}^n$ which we all know and love and the topologies actually dictate what continuity means for functions between those two spaces. :)

Title: Query about functions

Body: Suppose that we have a function $f:(\mathbb{R^2},d_2) \to (\mathbb{R^2},d_2) $ where $d_2$ is the Euclidean metric. A function in this case looks like what exactly. Since we are in $\mathbb{R^2}$ the function could look like $f(x,y)=(x-1,y+2)$ right? If this is the case then what would be meant by $d_2(f(x),f(y))$. Maybe someone could clarify why I'm confused, or if I am just wrong in general about how I am thinking about this.


问:
What is $\;\int xe^{-x^2} \,dx\;?$
What is $$\int xe^{-x^2} dx\quad?$$

I used substitution to rewrite it as $$\int -\dfrac{1}{2}e^u\, du$$ but this is too hard for me to evaluate. When I used wolfram alpha for $\int e^{-x^2} dx$ I got a weird answer involving a so called error function and pi and such (I'm guessing it has something to do with Euler's identity, but I'm fairly certain this is above my textbook's level).

答:
Your substitution was spot on. You put $$u = -x^2 \implies du = -2x\;dx \implies x\,dx = -\frac 12\, du$$

And having substituted, you obtained $$\int -\dfrac{1}{2}e^u\, du$$
Great work. But I think you gave up too early!: $$-\frac 12\int e^u \, du = -\frac 12 e^u + C,\tag{1}$$ and recall, $u = f(x), \;du = f'(x)\,dx$, so $(1)$ is equivalent to $$-\frac 12 \int e^{f(x)}\,f'(x)\,dx = -\frac 12 e^{f(x)} + C$$

So we can integrate as follows, and then back-substitute:
 $$\int -\dfrac{1}{2}e^u\, du = -\frac 12 \int e^u \,du = -\frac 12 e^u + C = -\frac 12 e^{-x^2} + C\tag{2}$$

Now, to remove all doubts, simply differentiate the result given by $(2)$: $$\frac{d}{dx}\left(-\frac 12 e^{-x^2} + C\right) = -\frac 12(-2x)e^{-x^2} = xe^{-x^2}$$
which is what you set out to integrate: $\color{blue}{\bf x}e^{-x^2} \neq e^{-x^2}$

Title: What is $\;\int xe^{-x^2} \,dx\;?$


答:
The Euclidean norm in $\mathbb{R}^{2}$ is defined as follows.

Let $x = (x_{1},x_{2})\in\mathbb{R}^{2}$ and $y = (y_{1},y_{2})\in\mathbb{R}^{2}$. Then we have that
\begin{align*}
d_{2}(x,y) = \sqrt{(x_{1} - y_{1})^{2} + (x_{2} - y_{2})^{2}}
\end{align*}

If we have two functions $f(x,y) = (f_{1}(x,y),f_{2}(x,y))$ and $g(x,y) = (g_{1}(x,y),g_{2}(x,y))$, then
\begin{align*}
d_{2}(f(x,y),g(x,y)) = \sqrt{(f_{1}(x,y) - g_{1}(x,y))^{2} + (f_{2}(x,y) - g_{2}(x,y))^{2}}
\end{align*}

More generally, we can define the Euclidean metric in $\mathbb{R}^{n}$ as follows.

Let $x = (x_{1},x_{2},\ldots,x_{n})$ and $y = (y_{1},y_{2},\ldots,y_{n})$. Then we have that
\begin{align*}
d_{2}(x,y) = \sqrt{(x_{1}-y_{1})^{2} + (x_{2}-y_{2})^{2} + \ldots + (x_{n}-y_{n})^{2}}
\end{align*}

More generally, we can talk about metric spaces.

Given a set $X$, we say the function $d_{X}:X\times X\to\mathbb{R}_{\geq 0}$ is a metric iff given $x\in X$, $y\in X$ and $z\in  X$, we have

(a) $d_{X}(x,y) \geq 0$ and $d_{X}(x,y) = 0$ iff $x = y$.

(b) $d_{X}(x,y) = d_{X}(y,x)$ for every $x$ and $y$.

(c) $d_{X}(x,z) \leq d_{X}(x,y) + d_{X}(y,z)$ for every $x$, $y$ and $z$.

In the case of $\mathbb{R}^{n}$, there are two examples whose presentation is standard.

More precisely, given $x\in\mathbb{R}^{n}$ and $y\in\mathbb{R}^{n}$, one has that
\begin{align*}
d_{1}(x,y) = |x_{1} - y_{1}| + |x_{2} - y_{2}| + \ldots + |x_{n} - y_{n}|
\end{align*}
which is known as the taxi-cab metric.

Another example is given below:

\begin{align*}
d_{\infty}(x,y) = \max\{|x_{1} - y_{1}|, |x_{2}  - y_{2}|,\ldots, |x_{n} - y_{n}|\}
\end{align*}

Besides such examples, there are plenty of metric spaces.

If you still want to go further, you can study Topology, which generalizes the ideas previously presented.

Hopefully this helps!

Title: Query about functions


问:
Circular segment using polar coordinates
Imagine, you'd like to integrate  $f(x,y)=x^2-y$ over this region:

How would you parameterize this circular segment? My try: $$\Phi : [1,2]\times [-\pi /3, \pi /3]\to \mathbb{R} \, (r,\varphi)\mapsto (x,y)=(r\cos(\varphi),r\sin(\varphi))$$ is this correct?

答: By symmetry, the $y$ term doesn't contribute, so we just want$$2\int_0^{\pi/3}d\theta\cos^2\theta\int_{\sec\theta}^2r^3dr=\int_0^{\pi/3}d\theta(4+4\cos2\theta-\tfrac12\sec^2\theta)=4\pi/3+\sqrt{3}/2.$$


问:
Show a Subspace of regular space is regular

I use this definition .Proofs of this theorem
I saw are  based on a similar def,  but with functions.
Source:
(https://www.math.tamu.edu/~tomzz/math636/ass7.pdf)

Attempt

Let (X$,\mathcal{T}$) be a regular topology.
Let $\mathcal{T}_S$={O$\cap$ S :O$\in\tau$}

Suppose X is a regular and $S\subseteq X$,S closed and any point  x$\in A$ and V$\subset A$

Let x$\in S\setminus A$, then $S\setminus A$ is open

Then there exists U open set in X s.t U $\cap S$ =V

Clearly $ x\notin$ U else x$\in V$

Since X is regular  (S$\setminus A )\cap S=\emptyset$

答:
Do the proof more systematically. Don’t needlessly  repeat definitions.

Let $(X,\tau)$ be a regular space and let $S \subseteq X$ be a subset in the subspace topology.

Let $x \in S$ and let $C \subseteq S$ be closed in $S$ such that $x \notin C$.

By standard facts about the subspace topology, there is a closed subset $C’$ of $X$ such that $$C = C’ \cap S$$

It’s clear that $x \notin C’$ as well, so by regularity of $X$ there are open sets $U$ and $V$ of $X$ such that $x \in U$, $C’ \subseteq V$ and $U \cap V = \emptyset$.

Then $U’ = U \cap S$ is open in $S$ and contains $x$, $V’=V \cap S$ is open in $S$ and $$C = C’ \cap S \subseteq V \cap S = V’$$

Of course $U’ \cap V’ \subseteq U \cap V = \emptyset$ so these sets show that $S$ is regular.


问:
ANOVA parameter estimation
I cannot find a formula for estimating coefficients ($a$ and $b$) of ANOVA model given by: $$\operatorname{weight} = a \cdot \operatorname{sex} + b + \epsilon$$
where $\operatorname{sex}=1$ means male and $\operatorname{sex}=0$ means female. The following table is provided

\begin{array}{|c|c|c|c|}
\hline
\text{sex}& \text{amount} & \text{mean weight} & \text{std} \\ \hline
 \text{male} & 94 & 31.7 &2.62\\ \hline
\text{female}& 83 & 25.23&2.00\\ \hline
\end{array}

答:
The following process works generally if your independent variable is binary and you can assume the usual assumptions for a linear regression.

Let $y = \text{weight}$ and $x = \text{sex}$. If you take an expectation of both sides conditioned on sex, you have:

$$ E[y | x = 0] = a*0 + b + E[\varepsilon | x = 0] , $$
$$ E[y | x = 1] = a*1 + b + E[\varepsilon | x = 1] . $$

If you assume $ E[\varepsilon | x ] = 0$ (exogeneity assumption of linear regression), then you have

$$ b = E[y | x = 0] , $$
$$ a = E[y | x = 1] - E[y | x = 0] . $$

Then, by the law of large numbers, a good estimator for the expectations is the relevant sample means:

$$ \hat{b} = \bar{y}_{x=0} , $$
$$ \hat{a} = \bar{y}_{x=1} - \bar{y}_{x=0} . $$

In other words, you can use the average weight of women as an estimator of $b$ and the difference in the average weight of men and women as an estimator of $a$.

Title: ANOVA parameter estimation


问:
Prove $|B_i \Delta B_j|\le 2$ without coding theory or other overkills
$A=\{0,1,2,3,4,5,6,7,8,9\}$, and $B_1, B_2,... B_{100}$ are arbitrary subsets of $A$.

Prove that there exist at least two subsets $B_i$ and $B_j$ such as $|B_i \Delta B_j|\le2$ without using coding theory

Note: this question is asked in Prove that cardinality of the symmetric difference of subsets less than 3, but the accepted solution is an overkill. The challenge is to revisit the problem and come up with something more elegant

Attempt

I started with proof by contradiction aiming at pigeon hole at some point. I tried to split all subsets of $A$ by number of elements $(0, 1,..., 10)$ and find the max number of subsets from each group. The goal is to show that from all $11$ groups the sum of max numbers of subsets is less than $100$ in this case. But I got stuck

答:
For each $i\in \{1,\dots,100\}$, let
$$
S_i=\Big\{B_i,B_i\Delta\{0\},B_i\Delta\{1\},\dots,B_i\Delta\{8\},B_i\Delta\{9\}\Big\}
$$
That is, $S_i$ is the set of sets whose symmetric difference with $B_i$ has zero or one elements, or the set of sets which can be obtained from $B_i$ by adding/removing at most one element.

If it were true that $|B_i\Delta B_j|> 2$ for each $i,j$, then the sets $S_i$ would be pairwise disjoint. Indeed, if $C\in S_i\cap S_j$, then you could obtain $B_i$ from $B_j$ by adding/removing at most one element from $B_j$ to get $C$, then adding/removing at most one element from $C$ to get $B_i$, implying $|B_i\Delta B_j|\le 2$.

But you cannot have $100$ pairwise disjoint subsets of the power set of $\{0,1,\dots,9\}$ with $11$ elements each, because $100\cdot 11> 2^{10}$. Contradiction!


答:
$\;\;\;\;\;$ Firstly, note that if $P$ is a Sylow $p$-subgroup of $G$ then $P$ is also a Sylow $p$-subgroup of any $G$ subgroup containing $P$.

$\;\;\;\;\;$ By definition, $P$ is a normal subgroup of its $\it{normalizer}$ $$\mathcal N(P):=\{g\in G:gP=Pg\}=\{g\in G:P=gPg^{-1}\}$$ Equivalently, $P$ itself is the only conjugate of $P$ in $\mathcal N(P)$. Sylow gaurantees that the $\mathcal N(P)$ Sylow subgroups are conjugates and thus $P$ is the unique Sylow $p$-subgroup of $\mathcal N(P)$.

Title: $p$-subgroup of Sylow and Normalizer


问:
Kleene Star over a formal Language containing Unions
I am a little bit confused, how the following language should be understood or further more, how the Kleenestar is interpreted in some ways:

$ ( \{0\} \cup \{1\}^*)^*$

I think the language looks like $\{0,1,11,111,1111....\}$.

So does the Outer-Kleene Star nothing? Or do I understand something wrong here?

Thanks in Advance

答:
$\{0\}\cup\{1\}^*=\{\epsilon,0,1,11,111,1111,\ldots\}$: it contains $0$, the empty string (since that is in $\{1\}^*$), and all non-empty finite strings of ones. When you take the star of this, you get all finite concatenations of these words. Many of those concatenations are just strings of ones, so they’re already present, as is the empty string. However, you also get all possible concatenations of $0$ and $1$, which means that you get the entire set $\{0,1\}^*$ of all possible finite sequences of zeroes and ones.

$$\begin{align*}
\big\{\{0\}\cup\{1\}^*\big\}^*&=\{0,1\}^*\\
&=\{\epsilon,0,1,00,01,10,11,000,001,\ldots\}
\end{align*}$$