问:
Can $\int_{2}^{\infty} [\zeta(x)-1] dx $ be evaluated?
I wonder whether an exact evaluation of $$\int_{2}^{\infty} [\zeta(x)-1] dx $$ can be obtained. Its sum analogue is already known: $$\sum_{n=2}^{\infty} [\zeta(n)-1] = 1. $$
Presumably, finding the indefinite integral is pretty difficult or even impossible, but I suspect the definite version could be found. So far, I have only found results on the mean values of the Riemann zeta function, but nothing on the integral mentioned above.

The two expressions are related via the Euler-Maclaurin summation formula, but thus far I have not been able to use it to calculate the integral.

答:
A try, not a final answer yet, I will update and have to check and correct stuff.

We need to find: $$\sum_{c=1}^{\infty} \frac{1}{(c+1)^2\ln(c+1)}$$

We know from the product series of ln(c+1)/c
Found similair to the product formula/refined stirling numbers found here
How to find a power series representation for a divergent product?

$$\ln\left({\ln(1+c)\over c}\right)=\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x! n}c^n (-1)^n dx$$

$$ \frac{1}{(c+1)\ln(c+1)} =1/c+\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x!}c^{n-1} (-1)^n dx$$

$$ \frac{1}{(c+1)^2\ln(c+1)} =\frac{1}{c(c+1)}+\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x!}\frac{c^{n-1}}{c+1} (-1)^n dx$$
$$=\frac{1}{c(c+1)}+\sum_{k=0}^{\infty}\sum_{n=1}^{\infty}\int_{-1}^0 \frac{(n+x)!}{n! x!}c^{n-1}c^{k} (-1)^{n+k} dx$$

Sum it diagonally under n+k=p

$$=\frac{1}{c(c+1)}+\sum_{p=1}^{\infty}\sum_{n=1}^{p}\int_{-1}^0 \frac{(n+x)!}{n! x!}c^{p-1} (-1)^{p} dx$$

$$=\frac{1}{c(c+1)}+\sum_{p=1}^{\infty}\int_{-1}^0 \bigg(\frac{(p+x+1)!}{p! (x+1)!}-1\bigg)c^{p-1} (-1)^{p} dx$$

$$=\frac{1}{c(c+1)}+\sum_{p=1}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg)c^{p-1} (-1)^{p} dx$$

We can now extend all it's powers over all p, in order to regularize it over c.
$$=1+\sum_{p \in \mathbb{Z}}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p} dx$$

If p goes to limit 0, it's $-\gamma$
If p<0 it's 0.

Now we have now over all positive integers, which we need to regularize.

$$=1-\gamma+\sum_{p \in \mathbb{N}}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p} dx$$

Unfortunally I don't know the answer yet. I've to do some rewriting.

To check
Because regularised sum over the negative zeta's (e.g. postive numbers) will be roughly equal to it's first few terms, before it diverges, we can actually see it moves around the value to be found.

We know numerically the answer is ~0.6055. Thuse

0.6055~=+1-0,577+1/4-11/144+1181/720/120-~0.009= between 0.61 and 0.60 so it's somewhat okay, especially when refined a little by taking the second order average of the alternating value's.

Alternative derivation:

$$\sum_{n=1}^{\infty} \frac{1}{(n+1)^2\ln(n+1)}=\sum_{n=1}^{\infty}\frac{1}{(n+1)n\ln(n+1)}-\frac{1}{(n+1)^2n\ln(n+1)}=$$$$\sum_{n=1}^{\infty}\sum_{j=1}^{\infty} \frac{(-ln(n+1))^{j-1}}{j!n(n+1)} $$
$$1+\sum_{n=1}^{\infty}\sum_{j=2}^{\infty} \frac{(-ln(n+1))^{j-1}}{(j)!n(n+1)} $$

$\frac{(-ln(n+1))^{j-1}}{j!n(n+1)}$ when extended over n writen as polynomal are the stirling numbers of the first $-S(n,j)/j/(n-1)!$
So you get

$$\sum_{j=2}^{\infty}\sum_{n=1}^{\infty} \frac{-S(n,j) \zeta(j-n)}{j*(n-1)!} $$

Or if use them unisgned

$$\sum_{j=2}^{\infty}\sum_{n=1}^{\infty} )\frac{(-1)^{n+1}S(n,j) \zeta(j-n)}{j*(n-1)!} $$

Which is equal to the integral/factorial above when writen out. However the difference is with this formula you need to extend it beyond n<1.

Basically this:

$$\sum_{p =1}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p}/c^p dx=\sum_{j=2}^{\infty}\sum_{n=1}^{\infty} \frac{-S_1(n,j) \zeta(n-j)}{j*(n-1)!}  $$
$$ \sum_{n=1}^{\infty}\sum_{j=2}^{\infty} \frac{(-ln(n+1))^{j-1}}{(j)!n(n+1)}=\sum_{j=2}^{\infty}\sum_{n\in \mathbb{Z}}^{\infty} \frac{-S_1(n,j) \zeta(n-j)}{j*(n-1)!} =$$

Proof for the negative side here
$$\sum_{n=-\infty}^{0} \frac{-S_1(n,j) \zeta(n-j)}{j*(n-1)!}  =(-1)^{j+1}\zeta(j)/j$$
$$\sum_{j=2}^{\infty}(-1)^{j+1}\zeta(j)/j=-\gamma$$

Continuation/some more rewriting: You can also rewrite the previous found integral, as those coefficients turned out to be common:

$$=1-\gamma+\sum_{p=1}^{\infty}\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}-1\bigg) \zeta(1-p) (-1)^{p} dx$$

If W_i(n) are the bernoulli polynomals of the second kind then:
$$\int_{0}^1 \bigg(\frac{(p+x)!}{p! (x)!}\bigg)=W_p(p)$$
Turns out there are Gregory $G_i$ coefficients which are related to them.

$$W_p(p)-1=\sum_{i=1}^{p*}\frac{p!}{(p-i)!(i)!} G_{i}$$

So if we rewrite our to calculate sum part:
$$\sum_{p=1}^{\infty} \sum_{i=1}^{n*}\frac{p!}{(p-i)!(i)!} \zeta(1-p) (-1)^{p} G_{i} =$$

Remove the first term because I think it might help
$$-\zeta(0)(G_1)+\sum_{p=2}^{\infty} \sum_{i=1}^{n*}\frac{p!}{(p-i)!(i)!} \zeta(1-p)  G_{i} =$$
$$1-\gamma+1/4+\sum_{p=2}^{\infty} \sum_{i=1}^{n*}\frac{p!}{(p-i)!(i)!} \zeta(1-p)  G_{i} =$$

We know that:
$$n!=\bigg(\frac{n}{e}\bigg)^{n}\big(\sqrt{2 n\pi}\big) \exp\bigg(-\sum_{j=1}^{\infty}\frac{\zeta(-j)}{j(n)^{j}}\bigg)$$

$$\frac{d}{dn} \ln(n!)=\ln(n)+\frac{1}{2n}+\bigg(\sum_{j=1}^{\infty}\frac{\zeta(-j)}{(n)^{j+1}}\bigg)$$

I've haven't often used the polgamma yet, but it's nice to introduce it:
$\frac{d^s}{dn}\ln(n!)=\psi_{s-1}(n+1)$

$$\psi_{0}(n+1)=\ln(n)+\frac{1}{2n}+\bigg(\sum_{j=2}^{\infty}\frac{\zeta(1-j)}{(n)^{j}}\bigg)$$

m=1/n
$$\frac{d^s}{d^sn}\bigg(\sum_{j=2}^{\infty}\zeta(1-j)(n)^{j}\bigg)= \bigg(\sum_{j=2}^{\infty} \frac{j!}{(j-s)!}\zeta(1-j)(m)^{j-s}\bigg)$$

$$\frac{d^s}{d^sn}\bigg(\psi_{0}(1+1/m)+\ln(m)-m/2 \bigg) = \bigg(\sum_{j=2}^{\infty} \frac{j!}{(j-s)!}\zeta(1-j)(m)^{j-s}\bigg)$$

$$\frac{d^s}{d^sn}\psi_{0}(1+1/m)=(-1)^s\sum_{k=0}^{s-1}\frac{\frac{s!}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!}\psi_{s-k}(1+1/m)}{m^{2s-k}}$$

With m goes to 1

$$\bigg(\sum_{j=2}^{\infty} \frac{j!}{(j-s)!}\frac{\zeta(1-j)}{}\bigg)=\bigg((-1)^s\sum_{k=0}^{*s-1}\frac{s!}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!}\psi_{s-k}(2)\bigg)+\bigg|\frac{d^s}{d^sn} ln(m)-m/2 \bigg|_{m=1}$$

$$\bigg|\frac{d^s}{d^sn} ln(m)\bigg|_{m=1}=(-1)^{s+1}(s-1)!$$

the derivative of -m/2 is only relevant if s=1, which gives -1/2.
The other coefficents are $\gamma$ and $-1/4$

$$1-\gamma+1/4+\sum_{p=2}^{\infty} \sum_{s=1}^{p*}\frac{p!}{(p-s)!(s)!} \zeta(1-p)  G_{s} =$$

$$1+ \sum_{s=1}^{\infty}\sum_{k=0}^{*s-1}(-1)^sG_{s}\bigg(\frac{1}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!}\psi_{s-k}(2)\bigg)  =$$

$$1+ \sum_{s=1}^{\infty}\sum_{k=0}^{*s-1}(-1)^sG_{s}\bigg(\frac{1}{(s-k)!k!}\frac{(s-1)!}{(s-k-1)!} (s-k)!((-1)^{1+s-k}\zeta(1+s-k)-1)\bigg)  =$$

$$1+ \sum_{s=1}^{\infty}\sum_{k=0}^{*s-1}(-1)^{1+k}G_{s}\bigg(\frac{1}{k!}\frac{(s-1)!}{(s-k-1)!} (\zeta(1+s-k)-1)\bigg)  =$$

$$1+ \sum_{s=1}^{\infty}\sum_{j=1}^{s}(-1)^{1+s-j}G_{s}\bigg(\frac{1}{(s-j)!}\frac{(s-1)!}{(j-1)!} (\zeta(1+j)-1)\bigg)  =$$

The -1 part is 0 when summe over j, unless s=1 and at s=1, $G_1=1/2$

$$ \sum_{s=1}^{\infty}\sum_{j=1}^{s}(-1)^{1+s-j}G_{s}\bigg(\frac{1}{(s-j)!}\frac{(s-1)!}{(j-1)!} (-1)\bigg)  =1/2$$

So now we have rewriten it to:
$$3/2+ \sum_{s=1}^{\infty}\sum_{j=1}^{s}(-1)^{1+s-j}G_{s}\bigg(\frac{1}{(s-j)!}\frac{(s-1)!}{(j-1)!} \zeta(1+j)\bigg)$$

Probably could been derived much easier but I thought it was cool. Obviously there must be some rule for how and when you can rewrite a positive power function into an negative one. I think it got something to do with asymtotics, but maybe someone else could explain it, as i am just starting to figure it.


问:
If $(X_n)_{n\in \mathbb{N}}$ is a martingale s.t. $\sup_n E[|X_n|]\leq M < \infty$, then $\sum_{n\geq 2}(X_n-X_{n-1})^2<\infty$ almost surely.
I need to prove the above statement. The hint provided was to consider the stopping time $T_l=\inf\{n\in \mathbb{N}||X_n(w)|\geq l\}$ where $l\in \mathbb{N}$ and then show that $E[\sum_{n=2}^K(X_n-X_{n-1})^21_{T_l>K}]$ is uniformly bounded in $K$.

I have 2 issues here:

$E[\sum_{n=2}^K(X_n-X_{n-1})^21_{T_l>K}]$ is uniformly bounded: I know that if $T_l>K$ this implies that $|X_n-X_{n-1}|^21_{T_l>K}\leq (2l)^2 $.

But this just shows that $E[\sum_{n=2}^K(X_n-X_{n-1})^21_{T_l>K}]\leq K(2l)^2$ which is not uniform in K. I feel we have to use the fact that $(X_n)$ converges to a finite limit since it is $L_1$-bounded.

How does the hint then prove the original statement? I am stuck here. I think we have have to use Lebesgue Dominated Convergence Theorem but I don't see how.

Any ideas? I have been stuck for some time now. I would also appreciate another alternative method if anyone has one.

答:
First suppose all $X_n$ are square-integrable. We have \begin{align}
    &E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>K}\right)\notag\\
    \leqslant& E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>n}\right)\notag\\
    =& E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>n-1}\right)-E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l=n}\right)
\end{align}
(Note that the term after $\leqslant$ may not exist if we do not assume $X_n$ is square-integrable.)
The first term is \begin{align*}
    & E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>n-1}\right)\\
    =&E\left(E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2}|\mathcal{F}_{n-1}\right) 1_{T_l>n-1}\right)\\
    =&\sum_{n=2}^{K}E\left[\left(E\left(X_n^2|\mathcal{F}_{n-1}\right)+X_{n-1}^2-2E\left(X_nX_{n-1}|\mathcal{F}_{n-1}\right)\right)1_{T_l>n-1}\right]\\
    =&\sum_{n=2}^{K}E\left((X_n^2-X_{n-1}^2) 1_{T_l>n-1}\right)\\
    =&\sum_{n=2}^{K}E\left(X_n^21_{T_l=n}+X_n^21_{T_l>n}-X_{n-1}^21_{T_l>n-1}\right)\\
    =&E\left(X_n^21_{T_l>n}\right)-E\left(X_1^21_{T_l>1}\right)+\sum_{n=2}^{K}E\left(X_n^21_{T_l=n}\right)
\end{align*}
The second term is \begin{align*}
    &E\left(\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l=n}\right)\\
    =&\sum_{n=2}^{K}E\left(X_n^21_{T_l=n}\right)+\sum_{n=2}^{K}\left(E(X_{n-1}^21_{T_l=n})-2E(X_{n-1}X_n1_{T_l=n})\right)
\end{align*}
So (1) becomes\begin{align*}
    E\left(X_n^21_{T_l>n}\right)-E\left(X_1^21_{T_l>1}\right)-\sum_{n=2}^{K}E(X_{n-1}^21_{T_l=n})+2\sum_{n=2}^{K}E(X_{n-1}X_n1_{T_l=n})
\end{align*}
By definition of $T_l$, the first two terms are both bounded by $l^2$. The second term is bounded by $$l^2\sum_{n=2}^{K}P(1_{T_l=n})\leqslant l^2$$
For the last term \begin{align*}
    &\left|2\sum_{n=2}^{K}E(X_{n-1}X_n1_{T_l=n})\right|\\
    \leqslant&2\sum_{n=2}^{K}E(|X_{n-1}X_n|1_{T_l=n})\\
    \leqslant&2l\sum_{n=2}^{K}E(|X_n|1_{T_l=n})\\
    \leqslant&2lM\sum_{n=2}^{K}P(T_l=n)\\
    \leqslant&2lM
\end{align*}
Therefore, (1) is bounded by $3l^2+2lM$, which is uniform in $K$.
For general $\{X_i\}$ (not necessarily square integrable), take a number $H>l$. Then $\{X_{n\wedge T_H}\}$ is a square integrable martingale, so $$E\left(\sum_{n=2}^{K}\left(X_{n\wedge T_H}-X_{(n-1)\wedge T_H}\right)^{2} 1_{T_l>K}\right)$$ is bounded by $3l^2+2lM$. Notice that $$\sum_{n=2}^{K}\left(X_{n}-X_{n-1}\right)^{2} 1_{T_l>K}=\sum_{n=2}^{K}\left(X_{n\wedge T_H}-X_{(n-1)\wedge T_H}\right)^{2} 1_{T_l>K}$$ since $H>l$. This concludes the proof of the statement in the hint.

Denote
$$A_l=\bigcap_{n=2}^{\infty}\{\omega:T_l(\omega)>n\}=\{\omega:|X_n(\omega)|<l,\,\forall n\}$$
Let $K$ goes to infinity, we have
$$E\left(\sum_{n\geqslant 2}(X_n-X_{n-1})^21_{A_l}\right)<+\infty$$
due to the statement provided by the hint. So $$\sum_{n\geqslant 2}(X_n-X_{n-1})^21_{A_l}<+\infty \qquad a.e.$$ for every $l>0$. We then have $$\sum_{n\geqslant 2}(X_n-X_{n-1})^21_{A_{\infty}}<+\infty \qquad a.e.$$ where $$A_{\infty}=\bigcup_{l=1}^{\infty} A_l=\{\omega:\sup_n |X_n(\omega)|<\infty\}$$ Note that $X_n$ converges a.e. since it is $L_1$-bounded. So $P(A_{\infty})=1$.

ContentLicense: CC BY-SA 4.0


问:
Supremum of a subharmonic function
Le  be an open ball of $\mathbb{R}^2$, :$\mathbb{R}^2→\mathbb{R}$ a function and $\Delta$ the Laplacian operator. $m_1$ is a local maximum of $f$ in $D$.

Recall that $\forall h \in \mathbb{R}^2, d^2_{m_1}(h,h)\leq 0$ where $d^2_{m_1}$ is the second-order differential of $f$ at $m_1$.

Show that if $\Delta f >0$ on , then $\forall m \in D$, $$f(m) < \sup_{m' \in \partial D} f(m').$$

答:
I think I have found the answer.

Since $d^2_{m_1}f(h,h) \leq 0$, the Hessian matrix of $f$ is negative in a local interior maximum of $f$ in $D$. Therefore,

$$\frac{\partial^2 f}{\partial x^2} \leq 0 \text{  and  } \frac{\partial^2 f}{\partial y^2} \leq 0$$

So at a local interior maximum: $$\Delta f \leq 0$$

More generally, for all interior maximum of $f$ in $D$,  $$\Delta f \leq 0$$

Hence, if $\Delta f > 0$, then $f$ can only reach its maximum can't reach its maximum on an interior point  of $D$.

Finally,

$$\forall m \in D, f(m) < \sup_{m \in \partial D} f(m')$$

Title: Supremum of a subharmonic function


问:
How do you calculate variables in vector addition?
I'm currently completing Specialist Maths Unit 1, Topic 2 "Vectors in the plane", and I've come across a question I can't solve.

Four vectors are given. The first two of these vectors, when added, lead to the same point as the last two vectors when added. However, the magnitude of one of the first two vectors isn't given, and the magnitude of one of the last two vectors isn't given.

The first two vectors are (10.5, $\angle225^\text{o}$) and ($\alpha, \angle315^\text{o}$).

The last two vectors are (14, $\angle330^\text{o}$) and ($\beta, \angle240^\text{o}$).

I tried converting them to Cartesian form:

a = |a|cos(θ)î + |a|sin(θ)ĵ

a = 10.5 × cos(225°)î + 10.5 × sin(225°)ĵ

a = –7.42î – 7.42ĵ (2 d. p.)

b = |α|cos(θ)î + |α|sin(θ)ĵ

b = α × cos(315°)î + α × sin(315°)ĵ

b = 0.71αî – 0.71αĵ (2 d. p.)

c = |c|cos(θ)î + |c|sin(θ)ĵ

c = 14 × cos(330°)î + 14 × sin(330°)ĵ

c = 12.12î – 7ĵ (2 d. p.)

d = |d|cos(θ)î + |d|sin(θ)ĵ

d = β × cos(240°)î + β × sin(240°)ĵ

d = –0.5βî – 0.87βĵ (2 d. p.)

And using substitution to calculate α and β:

–7.42î – 7.42ĵ + 0.71αî – 0.71αĵ = 12.12î – 7ĵ – 0.5βî – 0.87βĵ

–7.42î + 0.71αî = 12.12î – 0.5βî

0.71α = 12.12 – 0.5β + 7.42

0.71α = 19.54 – 0.5β

α = (19.54 – 0.5β) / 0.71

12.12î – 0.5βî = –7.42î + 0.71αî

12.12î – 0.5βî = –7.42î + 0.71((19.54 – 0.5)/0.71)î

12.12 – 0.5β = –7.42 + 19.54 – 0.5β

–0.5β = –7.42 + 19.54 – 0.5β

–0.5β = 12.12 – 0.5β

β = (12.12 – 0.5β)/–0.5

β = 24.24 + β

2β = 24.24

β = 24.24/2

β = 12.12

–7.42î + 0.71αî = 12.12î – 0.5βî

–7.42 + 0.71α = 12.12 – 0.5(12.12)

–7.42 + 0.71α = 12.12 – 0.5(12.12)

–7.42 + 0.71α = 12.12 – 6.06

0.71α = 12.12 – 6.06 + 7.42

0.71α = 13.48

α = 13.48/0.71

α = 18.99

I then tried to solve the vectors using the discovered variables.

–7.42î – 7.42ĵ + 0.71αî – 0.71αĵ = 12.12î – 7ĵ – 0.5βî – 0.87βĵ

–7.42î – 7.42ĵ + 0.71(18.99)î – 0.71(18.99)ĵ = 12.12î – 7ĵ – 0.5(12.12)î – 0.87(12.12)ĵ

–7.42î – 7.42ĵ + 13.48î – 13.48ĵ = 12.12î – 7ĵ – 6.06î – 10.5ĵ

(–7.42 + 13.48)î + (–7.42 – 13.48)ĵ = (12.12 – 6.06)î + (–7 – 10.5)ĵ

6.06î – 20.9ĵ ≠ 6.06î – 17.5ĵ

As you can see, the result is incorrect.

Is this the correct method for solving this problem and I simply made a mistake in the process? Or should a different method be used for determining an answer?

答:
You have two linear simultaneous equations to solve:

$$10 \sin(225^\circ)+\alpha \sin(315^\circ) = 14 \sin(330^\circ) + \beta \sin(240^\circ)$$
$$10 \cos(225^\circ)+\alpha \cos(315^\circ) = 14 \cos(330^\circ) + \beta \cos(240^\circ)$$

which we can write (multiplying by $- 2$) as

$$10 \sqrt2+\alpha \sqrt{2} = 14 + \beta \sqrt{3}$$
$$10 \sqrt2-\alpha \sqrt{2} = -14 \sqrt{3} + \beta$$

so $\beta = 10 \sqrt2-\alpha \sqrt{2} +14 \sqrt{3}$ and thus

$$\alpha =\frac{ 56 + 10 \sqrt6 -10 \sqrt2}{\sqrt6 + \sqrt2}\approx 17.173 $$
$$\beta =\frac{ 40 + 14 \sqrt6 -14 \sqrt2}{\sqrt6 + \sqrt2} \approx 14.104$$


问:
showing to be extreme subset (might use Hahn decomposition Theorem)
I am studying Functional analysis by myself and stumbled this question and am completely at a loss.

We want to show that $\{ f \in L^1 [0,1 ] : ||f|| =1 \}$ is an extreme subset of $\{ \mu \in C[0,1]^{\ast} : ||\mu||=1$.

In the hint it says to use Hahn decomposition theorem but I am not sure how to go about this type of problem. Any help would be appreciated.

Thnx.

答:
Suppose that $f\in L^1[0,1]$ is non-negative with $\|f\|_1=1$. If $f=t\mu_1+(1-t)\mu_2$ for measures with $\|\mu_j\|=1$, this means that for every $g\in C[0,1]$
$$\tag{1}
\int_0^1 gf\,dm=t\int_0^1 g\,d\mu_1+(1-t)\int_0^1 g\,d\mu_2.
$$
Taking $g=1$, we get $1=t\mu_1(1)+(1-t)\mu_2(1)$. As $|\mu_j(1)|\leq\|\mu_j\|=1$, we get $\mu_j(1)=1$, $j=1,2$. This forces $\mu_1,\mu_2$ to be positive (since a unital functional of norm one is positive).

Using dominated convergence and the fact that continuous funtions are dense in $L^1$, we can push (1) to the case where $g\in L^\infty[0,1]$ (in particular, $g$ can be any bounded measurable function).

Taking characteristic functions, we get
$$
\int_\Delta\,f\,dm=t\mu_1(\Delta)+(1-t)\mu_2(\Delta).
$$
This shows that $\mu_1,\mu_2$ are absolutely continuous with respect to $f\,dm$. So there exist Radon-Nikodym derivatives $h_1,h_2\geq0$ such that
$$
\mu_j(\Delta)=\int_\Delta\,h_jf\,dm
$$
So now we have
$$
\int gfdm=t\int gh_1fdm+(1-t)\int gh_2fdm=\int gf(th_1+(1-t)h_2)\,dm
$$
for every bounded measurable function $g$. So $f(th_1+(1-t)h_2)=f$ a.e., from where we conclude that $th_1+(1-t)h_2=1$ a.e. (their value on the set where $f=0$ is irrelevant and we can make it zero). As $h_1,h_2\geq0$ (from $\mu_1,\mu_2\geq0$), we deduce that $h_1,h_2\leq1$ a.e., i.e. they are in $L^\infty[0,1]$. Then $h_jf\in L^1[0,1]$ and $\mu_j=h_jf\,dm$.

Now consider a general $f$ (i.e. not necessarily positive) with $\|f\|_1=1$. Write its Hahn Decomposition as $f=f^+-f^-$. This means that there exists $E\subset[0,1]$ with $1_E\,f=f^+$, $-1_{E^c}f=f^-$.
If $f\,dm=t\mu_1+(1-t)\mu_2$, then
$$
f^+\,dm=t\mu_1(E\cap\cdot)+(1-t)\mu_2(E\cap\cdot),\ \ f^-\,dm=t(-\mu_1(E^c\cap\cdot))+(1-t)(-\mu_2(E^c\cap\cdot))
$$
Now we can apply the first part of the proof to normalized versions of each of $f^+,f^-$ (so that they have $\|\cdot\|_1=1$) and we get that $\mu_j=(h_j^+-h_j^-)f\,dm$, $j=1,2$. So the unit sphere is extreme.

ContentLicense: CC BY-SA 3.0


问:
Using bijective correspondence between functions from $\mathbb{R} \rightarrow\{0,1\}$ and power set of $\mathbb{R}$.
I am having some confusion with the notion of interpreting the topological space $2^\mathbb{R}$ the set of all functions from $\mathbb{R} \rightarrow \{0,1\}$ as equivalent to the set of all subsets of $\mathbb{R}$.The book I am using says to interpret each function as associated with the subset of $\mathbb{R}$ containing the point $a \in \mathbb{R}$ if $f(a)=1$ and not containing $a$ if $f(a)=0$. For some reason I am then interpreting the space $2^ \mathbb{R}$ with the product topology to be equivalent with the discrete topology on $\mathbb{R}$. Is this correct?

For example I am looking at the following question:In the product space $2^\mathbb{R}$, what is the closure of the set $Z$ consisting of all elements of $2^\mathbb{R}$ that are $0$ on every rational coordinate but may be $0$ or $1$ on any irrational coordinate?Equivalently, thinking of $2^\mathbb{R}$ as subsets of $\mathbb{R}$ what is the closure of the set $Z$ consisting of all subsets of $\mathbb{R}$ that do not contain any rational?

An answer that I first came up with was that the closure of $Z$ should be $Z$ itself, because every subset of $\mathbb{R}$ in the discrete topology is closed and therefore equal to its closure. However, I know that the problem is not that trivial. Now that I think about this, is my interpretation the notion that the functions themselves are the elements of the product topology on $2^\mathbb{R}$, not the topological space? Instead should I think about the product topology?

For any point $x$ in $2^\mathbb{R}$ and open set containing $x$, I know that there is a basis element of $2^\mathbb{R}$ with all but finitely many factors equal to $\{0,1\}$ containing $x$.Then should this basis element intersect a point in $Z$, necessarily?

I would appreciate if someone could clear up my confusion on interpreting the set $\mathbb{R}$ as equivalent to $\mathbb{R}$ with the discrete topology, and lead me in the right direction in solving this problem.

答:
In $\{0,1\}^{\Bbb R}$ the reals are just an index set, or the domain for the set of functions. We're not implying or putting any topology on it, we're just discussing the set of all functions on $\Bbb R$ with codomain $\{0,1\}$. If we have such a function $f$  we "identify" it with the subset $A=f^{-1}[\{1\}$ of $\Bbb R$. This way, we've "encoded" all subsets with a unique function and every such function encodes a subset. The set $\{0,1\}$ does have the discrete topology, but then the product topology on $\{0,1\}^{\Bbb R}$ is very far from discrete; it has no isolated points at all.

Let $A \subseteq \{0,1\}^{\Bbb R}$ be all functions such that $f(x)=0$ for all $x \in \Bbb Q$, which indeed corresponds to all subsets of the reals that contain no rational under the aforementioned correspondence.

My claim is that this set is closed: Let $g \notin A$. This means that $g(q)=1$ for some $q \in \Bbb Q$. But then $\pi_q^{-1}[\{1\}]$ is subbasic open in $\{0,1\}^{\Bbb R}$, contains $g$ and is disjoint from $A$, as all members in it by definition also have value $1$ at $q$ (i.e. contain $q$) and so do contain a rational. We can see the basic open sets in this space (when we view them as subsets of $\Bbb R$ as "contain-and-miss" sets: A basic set is of the form (own notation here) $[F;G]:= \{A \subseteq \Bbb R: F \subseteq A; A \cap G = \emptyset\}$ where $F,G \subseteq \Bbb R$ are finite. (In the above proof I used $[\{q\}; \emptyset]$, really). This fits into a more general pattern of other topologies on spaces of subsets (like the Vietoris topology, the Pixley-Roy topology and many others).


问:
Continued fraction $[0;0,0,0,\ldots] = \pm 1$
I would like to represent each $x\in\mathbb{R}_{>0}$ as a unique continued fraction. We represent each $x$ with $[a_{0};a_{1},a_{2},\ldots]$ as usual for the continued fraction. for irrational numbers, we know the sequence is infinite. For rational numbers we have a finite sequence. To have a unique representation in this notation I would like to extend the sequence with zero's. For example, $\frac{415}{93} = [4;2,6,7]$ would be extended to $[4;2,6,6,0,0,\ldots]$. We compute $$0+\frac{1}{0+\frac{1}{0+\ddots}},$$

$x_{n+1}=0+\frac{1}{x_{n}}\Rightarrow x^{2}=1\Rightarrow x=\pm 1.$

Would this be a unique representation for $x\in\mathbb{R}_{>0}$, or would this give a problem since the continued fraction of $[0;0,0,\ldots]$ is also $-1$. Possible way to work around this problem?

答:
The value of the continued fraction is, by definition, $\lim r_n$ where the $r_n$ are the convergents
$$
r_0 = 0,\\
r_1 = 0 + \frac{1}{0},\\
r_2 = 0 + \frac{1}{0+\frac{1}{0}},\\
\vdots\\
r_{n+1} = 0+\frac{1}{r_n}\\
\vdots
$$
Because of the zeros in the denominator, we cannot do this in the real numbers. So let's do it in the Riemann sphere (a.k.a. projective line).  Then
$$
r_0 = 0\\
r_1 = 0+\frac{1}{0} = \infty\\
r_2 = 0+\frac{1}{\infty} = 0\\
r_3 = 0+\frac{1}{0} = \infty
$$
alternating between $0$ and $\infty$.  So, in the Riemann sphere, this sequence does not converge.

In the conventional theory of continued fractions with real entries, there is a theorem that if $b_k \ge c$ for some $c>0$, then
$$
b_0 + \frac{1}{b_1+\frac{1}{\displaystyle b_2+\ddots}}
\tag1$$
converges.  Even if the $b_k$ are positive but converge to $0$ we could get divergence for the continued fraction $(1)$.


问:
Residue field of complete number field is finite
Suppose $K_\mathfrak{p}$ is the completion of a number field (i.e. the field of fractions of the $\mathfrak{p}$-adic completion of the ring of integers of $K$). Is there a direct way of showing that its residue field is finite?

I think this would follow fairly easy from $K_\mathfrak{p}$ being a finite extension of $\mathbb{Q}_\mathfrak{p}$ but to establish this seems to require an excursion into valuation extensions, relative norms, ramification etc.

This may well be fine but I wondered if there was something simpler?

答:
We adopt the following notations:

$\mathcal{O}_K$ is the ring of algebraic integers in $K$ and  $\mathfrak{p}\subset \mathcal{O}_K$ is a prime of $K$ with $\mathcal{O}_\nu\subset K$ the corresponding valuation ring.

$(p)=\mathfrak{p}\cap\mathbb{Z}$ with $\mathbb{Z}_{(p)}$ the $p$-adic valuation ring of $\mathbb{Q}$. Since $\nu_{\mathfrak{p}}(p)=1$, $p$ is a uniformiser for $\mathcal{O}_\nu$.

The residue field of $K$ is $k_\nu=\mathcal{O}_\nu/\mathfrak{m}_\nu$ where $\mathfrak{m}_\nu=p\mathcal{O}_\nu$.

Write $\mathcal{O}_\mathfrak{p}=\varprojlim \mathcal{O}_\nu/p^n\mathcal{O}_\nu$ for the valuation ring of $K_\mathfrak{p}$ and $k_\mathfrak{p}=\mathcal{O}_\mathfrak{p}/\mathfrak{m}_\mathfrak{p}$ for the residue field with $\mathfrak{m}_\mathfrak{p}$ the maximal ideal in $\mathcal{O}_\mathfrak{p}$.

First we show that $k_\mathfrak{p}\simeq k_\nu$:

Suppose $a:(a_n)\in \mathcal{O}_\mathfrak{p}$ with $|a|=|a_m|\;\forall\; m\ge N$. We then have a homomorphism
$$
\begin{array}{rcl}
\phi\colon \mathcal{O}_\mathfrak{p}&\rightarrow &\mathcal{O}_\nu\\
a&\mapsto&a_N
\end{array}
$$
which is surjective since $\mathcal{O}_\mathfrak{p}$ contains the constant sequences from $\mathcal{O}_\nu$.

We can extend to a surjective map to the quotient:
$$
\phi\colon \mathcal{O}_\mathfrak{p}\rightarrow \mathcal{O}_\nu\rightarrow \mathcal{O}_\nu/ \mathfrak{m}_\nu
$$
with $\ker\phi=\{a\in\mathcal{O}_\mathfrak{p}:|a|<1\}=\mathfrak{m}_\mathfrak{p}$.

Next we show that $k_\nu$ has a subfield $\mathbb{F}_p$:

We have a composite ring homomorphism
$$
\psi\colon\mathbb{Z}\hookrightarrow \mathcal{O}_K\hookrightarrow \mathcal{O}_{K,\mathfrak{p}}=\mathcal{O}_\nu\rightarrow \mathcal{O}_\nu/ p\mathcal{O}_\nu
$$
and $\psi(\mathbb{Z})=\mathbb{Z}/ (p)\simeq\mathbb{F}_p$.

Finally, we show $[k_\nu\colon \mathbb{F}_p]<\infty$:

For $a\in \mathcal{O}_\nu$ write $\bar{a}$ for its image in $k_\nu$. Suppose $\bar{a}_1,\dots,\bar{a}_n$ are linearly independent over $\mathbb{F}_p$. We will show that $a_1,\dots,a_n\in \mathcal{O}_\nu\subset K$ are then linearly independent over $\mathbb{Q}$:

Suppose not and we have
$$
r_1 a_1+\cdots+r_na_n=0,\quad r_i\in\mathbb{Q}.
$$
Suppose wlog that $r_1$ is the non-zero coefficient with minimum $p$-adic valuation. Then if $r_i\ne 0$ we have $\nu_p(r_i/ r_1) \ge 0$. So dividing through by $r_1$, clearing denominators and reducing $\mod p$ we obtain
$$
g_1\bar{a}_1+g_2 \bar{a}_2+\cdots g_n\bar{a}_n=0, \quad g_i\in \mathbb{F}_p, g_1\ne 0.
$$
which would be a contradiction.

So $a_1,\dots,a_n$ are linearly independent over $\mathbb{Q}$ and thus $n\le [K:\mathbb{Q}]$ which is finite.

i.e. $[k_\nu:\mathbb{F}_p]<\infty$ and $k_\mathfrak{p}\simeq k_\nu$ is finite ( and of characteristic $p$).


问:
Minizing the total lenght of 2 pipelines, Optimization question
A pump house is to be placed at some point $X$ along a river, A pipe from point $A$ and a pipe from point $B$ will then be connected to the point $X$. How far should $X$ be away from $M$, so that the total length of the pipes $\overline{AX}$ and $\overline{BX}$ are minimised?

I just need help with setting up the function.

答:
Here's what I did.

Hyputenuse of triangle AMX= sqrtroot of 4+×^2

Hyputenuse of triangle BNX= sqrtroot of 1+(5-x)^2

Sqrt root 4+x^2 + sqrtroot 1+(5-x)

I differentiate that and got (×/sqrroot 4+x^2) - (-x+5/sqrtroot x^2 - 10x +26)

I graphed the function and got 3.333
I then made a sign diagram which clarified that It was a local minimum


问:
$\mathbb{Z}:= \mathbb{N} \cup (-\mathbb{N}) \cup \{0\}$, $a,b \in \mathbb{Z}$, when $a \le b$?
Let $\mathbb{Z}:= \mathbb{N} \cup (-\mathbb{N}) \cup \{0\}$,  and $a,b \in \mathbb{Z}$, when $a \le b$?
Thanks in advance!!</p>

答: $$a\le b$$ if and only if $$b-a\in \mathbb{N}\cup \{0\}$$</p>


答:
$AM+XB$ is minimum when reflexing $B$ in $D$ wrt $MB$ we connect $AD$. The point $X$ where $AD$ intersects $MN$ is the minimum required. Because any other point $P$ is such that $AP+PD>AD$ for the triangular inequality.

$\triangle ACD$ similar $\triangle AMX$

$MX:AM=CD:AC$

$MX=\frac{AM\cdot CD}{AC}=\frac{10}{3}$km

Title: Minizing the total lenght of 2 pipelines, Optimization question

Body:
A pump house is to be placed at some point $X$ along a river, A pipe from point $A$ and a pipe from point $B$ will then be connected to the point $X$. How far should $X$ be away from $M$, so that the total length of the pipes $\overline{AX}$ and $\overline{BX}$ are minimised?

I just need help with setting up the function.


问:
¿Can a subset $\overline{S}$ of the accumulation points of a set S be greater than S itself? in $R^{n}$
Let $S \subseteq \mathbb{R}^{n}$ and $\overline{S}$ the set of all accumulation points of $S$

Then if $ S \subseteq \overline{S}$.

can $\overline{S}$ be greater than $S$?

答: $S =\Bbb Q$ is a mere countable set, while $\overline{S}=\Bbb R$ is uncountable. Most sets are actually smaller then their set of accumulation points.


问:
Existence of a one-dimensional representation
I am working on the following exercise in relation to representation theory and I have some questions.

Let $G$ be a group with center $Z$. Let $\pi$ be an irreducible complex representation of G on a finite dimensional complex vector space $V$. Prove that for each $z \in Z$, $\pi (z)$ is an endomorphism of $V$ that is for all $g \in G$ we have that $\pi (g) \pi (z) = \pi (z) \pi (g)$. Conclude that there exists a one-dimensional representation $\chi_{\pi} : Z \to \mathbb{C}^{\times}$ of $Z$ such that $\pi (z) = \chi_{\pi} (z)\operatorname{id}_V$ for all $z \in Z$

My idea for an answer is: By definition we have that the center $Z$ of $G$ is given as $Z:= \lbrace z \in G : zg=gz,\,\forall g \in G \rbrace$. Let $g \in G$ and $z \in Z$ and let $\pi$ be a complex and irreducible representation. Then we obtain that

\begin{align*}
\pi (g) \pi (z) &= \pi (gz) \\
&= \pi (zg) \\
&= \pi (z) \pi (g)
\end{align*}

Where it has been used that $\pi$ is a representation (so $\pi (g) \pi (z) = \pi (gz)$) and that $z \in Z$. We have now shown that $\pi (z)$ is an endomorphism. Is this a correct approach or is there more to it which I am missing?

Additionally I am not sure how to conclude the second part of the question.

答:
Yes, your approach to the first part if fine.

If $z\in Z$, consider the kernel of $\pi(z)$. It is stable with respect to the action $\pi$ of $G$. But that's an irreducible representation. Therefore, $\ker\pi(z)=\{0\}$ (in which case $\pi(z)$ is injective) or $\ker\pi(z)$ is $V$ (in which case $\pi(z)$ is the null function). Since $V$ is a finite-dimensional complex vector space, $\pi(z)$ has an eigenvalue $\lambda$. But $\pi(z)-\lambda\operatorname{Id}$ commutes with every $\pi(g)$. So, by the same argument as before, $\ker\bigl(\pi(z)-\lambda\operatorname{Id}\bigr)=0$ or $\ker\bigl(\pi(z)-\lambda\operatorname{Id}\bigr)=V$. But it cannot be $\{0\}$, since $\lambda$ is an eigenvalue, and therefore $\pi(z)=\lambda\operatorname{Id}$.

So, this proves that, for each $z\in Z$, there is some scalar $\lambda_z$ such that $\pi(z)=\lambda_z\operatorname{Id}$. But $\pi(z)$ is invertible, and therefore $\lambda_z\ne0$. So, take $\chi_\pi(z)=\lambda_z$.


问:
Weak$^*$ continous functionals
Let $X$ be a Banach space (not necessarily reflexive) and let $X^{*}$ denote the continuous dual of $X$. Let $\psi:X \to X^{**}$ denote the canonical embedding of $X$ into its double dual $X^{**}$. We know that the weak$^*$ topology on $X^*$ is induced by the family of continuous linear functionals $\{\psi(x): x\in X\}$.

From this, can we say that every weak$^*$ continuous linear functional on $X^*$ is of the form $\psi(x)$ for some $x\in X$? If it is, I would like to know an explanation of the answer. If not then a counter example would help. I am a beginner on this domain and know very little. So, I do not know, if this question is trivial.

Thanks in advance.

答: Let $F$ be a continuous linear functional on $(X^{*}, weak*)$. There exists  a basic neighborhood of $0$ in weak* topology, say $\{f\in X^{*}: |f(x_i)| <r_i, 1 \leq i \leq N\}$ such that $|F(f)| <1$ whenever $f$ lies in this neighborhood. In particular, this implies that if $f(x_i)=0$ for each $i$ then ($nf$ is in the neghborhod for each $n \in \mathbb N$ so) $|F(nf)| <1$ for all $n$. This proves that $\bigcap_i ker (F_{x_i}) \subseteq ker F$, where $F_x(f)\equiv f(x)$. An elementary lemma in linear algebra now says that $F$ is a linear combination of $F_{x_1},F_{x_2},,...,F_{x_N}$  so $F=F_x$ for some $x$ of the form $\sum c_ix_i$and this is exactly what we are trying to prove.

Title: Weak$^*$ continous functionals


问:
Find $\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$
Find the limit of $$\lim_{x\to 0} \frac{1+\sin x-\cos x}{1+\sin(px)-\cos(px)}$$ without l'Hospital.

I tried expanding $1-\cos(x)$ to get something this kind $$\lim_{x\to 0}\frac{\sin x}{x}$$ but couldn't solve it. The answer is $1/p$.

答:
With some basic asymptotic analysis;

First, observe that $\:1-\cos x+\sin x=2\sin^2 \frac x2$.
Now, $\sin^2\frac x2=o(\sin x)$ near $0$, and therefore
$$1-\cos x+\sin x\sim_0\sin x \sim_0x.$$
Similarly, $\;1-\cos px+\sin px\sim_0\sim px\sim_0px$, whence
$$\frac{1-\cos x+\sin x}{1-\cos px+\sin px}\sim_0\frac x{px}=\frac1p.$$


问:
Best definition for distributive property of multiplication
I am teaching math for kids in Perú, however, I find so many different definitions for the distributive property of multiplication. Do you have a solid definition in words with a book reference?

Thanks.

答:
The distributive property of multiplication is that $a(x+y)=ax+ay$. If I were teaching children about this, then I would be reluctant to give this definition right off the bat. Instead, I would use examples. Why is it to so easy to calculate $2$ million plus $7$ million? Both of those numbers are very large, and yet we can compute the answer in a heartbeat. $2$ million plus $7$ million equals $9$ million for the same reason that $2$ apples plus $7$ apples equals $9$ apples. It's because we can add the numbers $7$ and $2$ together, and then give the units. In other words,
$$
7 \text{ million} + 2 \text{ million} = (7+2) \text{ million}= 9 \text{ million} \, .
$$
The reason that we can do this is because of the distributive property, which says that $ax+ay=a(x+y)$. In the above example, $x=7$, $y=2$, and $a=\text{one million}$. There are other examples, too: since $7 \times 12 = 84$, and $9 \times 12 = 108$,
$$
16 \times 12 = (9+7) \times 12 = 12(9+7)=12(9)+12(7)=108+84=192 \, .
$$
More complicated rules, such as the 'FOIL' method are a direct consequence of the distributive property:
$$
(a+b)(x+y)=\underbrace{ax}_{\text{First}}+\underbrace{ay}_{\text{Outside}} + \underbrace{bx}_{\text{Inside}}+\underbrace{by}_{\text{Last}}
$$
This method works because if we let $k=x+y$, then
$$
(a+b)k=k(a+b)=ak+bk=a(x+y)+b(x+y)=ax+ay+bx+by \, .
$$
Teaching students the FOIL method without even mentioning this reason is ill-advised in my opinion.


问:
Differential equation based on initial values
I have a differential equation that looks like:

$$y'=y(4-y)$$
I have identified the critical points to be $0$ and $4$.
I need to identify the behavior of the equation as $t \rightarrow \infty$ and this behavior it seems depends on the initial value.

Now depending on the differential equation I can say that

If $y > 4$, then $y$ is decreasing

If $0<y<4$, then $y$ is increasing

If $y<0$, then $y$ is decreasing

And I infer this based on the slope values calculated from the $y'$ equation.

So based on this I was able to conclude that as $t \rightarrow \infty$, $y$ converges to $4$ and and diverges from $0$. But how do I identify this behavior from initial values? I feel like I'm missing something here.

答:
To go along with the other nice answer (+1), we can do some visual analyses using direction field and phase line plots.

A direction field plot is shown that highlights three initial conditions (pink, green, red). One is outside an equilibrium, one is inside the region between the two equilibria and one is outside the other equilibrium.

Notice the two equilibrium points and the directions that solutions go towards or away from them or trapped between them.

We can also draw a phase line to see this behavior (the open circle is unstable and the solid is stable). If you pick a point below zero, you go to minus infinity, if you pick a point between zero four, you go to four and if you pick a point above four, you go to four.

We can also solve this DEQ and do analyses

$$y(t) = \dfrac{4 e^{4 t}}{e^{4 t}+c}$$


问:
Poisson distribution where $\ \lambda,\ $ the average rate of events, varies with time or space?
A Poisson distribution models tells us the number of times an event occurs in an interval of time or space, but $\ \lambda\ $ is a constant, so this rate is fixed.

What if the rate changes with time or space, e.g. instead of $X \sim Po(10),\ $ what if we had, for example, $X \sim Po(\ln(t+1))),\ t>0,\ t \text{ is time in minutes since math.se website's launch?}\ $

This could represent many other real-world things, but for this example let $X$ be the random variable: "number of questions asked each hour on math.se (e.g. in the first few years of the website's launch)".

Is is known how to do calculations like finding $P(1300\leq X\leq1400) $ in the first week of the website's launch?

[Note the I am just making up the numbers of course, but I have tried to do it in a way so that the probability is not obviously close to $1$ or $0$.]

I think maybe we would need to do some sort of mean of a function or integral or double integral but I'm not sure what or how or if there is any literature on a $\ \lambda\ $ that varies.

答:
If the varying parameter at time $t$ is $\lambda(t)$ then the number between times $t_1$ and $t_2$ will be distributed $\mathrm{Po}(\mu)$ where $\mu = \int\limits_{t_1}^{t_2} \lambda(t) \, dt$.

Then the probability that this number is from $a$ through to $b$ is then $\sum\limits_{n=a}^b e^{-\mu}\frac{\mu^n}{n!}$,

and if this calculation is too difficult then a normal approximation may be $\Phi\left(\dfrac{b+\frac12-\mu}{\sqrt{\mu}}\right)-\Phi\left(\dfrac{a-\frac12-\mu}{\sqrt{\mu}}\right)$ and you can drop the second term if $a=0$.

You will need to ensure consistency in the time units: your example mixes minutes and weeks.


问:
How do I solve $x^3-x=12y+6$
This is what I have so far:

$$x^3-x = 12y + 6$$
$$x(x+1)(x-1) = 2(6y+3)$$
The RHS of the equation is even, so therefore so must the LHS. Given that the three numbers on the LHS are consecutive,
then we know $x$ is even and $(x-1)$ and $(x+1)$ are odd.

Given that they're three consecutive numbers their gcd is 1, making them coprime.

That's as far as I got. Please note that both $x$ and $y$ are positive integers.

答:
Note that necessarily $x\equiv2\pmod{4}$ because
$$x(x-1)(x+1)=12y+6\equiv2\pmod{4},$$
and conversely if $x=4z+2$ then
$$x^3-x=4(16z^3+24z^2+5z)+6,$$
where a quick check shows that $16z^3+24z^2+5z\equiv0\pmod{3}$ for all $z$.

Title: How do I solve $x^3-x=12y+6$


问:
Higgs bundle and spectral curve
I have trouble understanding the following part in E. Wittens paper "More On Gauge Theory And Geometric Langlands":

The context is the following: $(E, \varphi)$ is a Higgs bundle with structure group $SU(2)$ on the Riemann surface $C$. We look at the eigenvalue problem $\varphi \phi =y \phi$ where $\phi \in E$ and $y$ is a holomorphic 1-form.

For $N=2$ this is equivalent to $y^2-1/2 Trace(\varphi^2)=0$. For this case, if the eigenvalues are not distinct, it means that $y=0$ and so $Tr(\varphi^2)=0$.

1)My problem starts when it says "As we require $Tr(\varphi^2)$ to have simple zeros...". Why can we require that?

2)Also, if we assume this, why does $\varphi $ have to be of the form ? We know the following:

$Tr(\varphi)=0$ since $G=SU(2)$.

$Tr(\varphi^2)=0$ for $z=0$.

$Tr(\varphi^2)$ has simple zeros

Now let

$\varphi$=
$\begin{pmatrix}  a & b\\
c & -a \end{pmatrix}$

Then $\varphi^2= \begin{pmatrix}  a^2+bc & 0\\ 0 & a^2+bc \end{pmatrix}$
and we conclude that $(2a^2+2bc)(0)=0$ and that it is a simple zero.
Can we already follow from that that $\varphi$ has to be of the form in the paper?

3)What does it mean that we regard $y$ as a local parameter? I mean, $y$ is arbitrary, why can we just replace $z$ by $y^2$? This seems so random to me.

Thanks in advance for any answer, I understand that this is a question which is maybe hard to answer because you first need to understand the context here so I really appreciate any help!

答:
I realize this is nearly a year-old and you've maybe/probably found an answer by now. Nevertheless, I will have a go at answering your questions.

According to BNR's paper, the Jacobian criterion implies that the spectral curve defined by $y^n + a_1y^{n-1} + ... + a_n = 0$ with $a_i \in H^0(X, K_X^i)$ is smooth iff at every multiple zero of the section $a_n$, the section $a_{n-1}$ is non-zero.

Since $\phi$ is traceless, $\det(y - \phi) = y^2  + \det(\phi)$. Since $a_1 = tr(\phi)$, $\det(\phi)$ must have simple zeroes since the spectral curve is smooth. Since $\det(\phi) = - \frac{1}{2}tr(\phi^2)$, it follows that $tr(\phi^2)$ must have simple zeroes.

The ramification points of the spectral curve are given by the divisor $tr(\phi^2)$. Now, assume that $tr(\phi^2) = 0$ at $z = 0$, which makes $z = 0$ a ramification point. Then locally, the spectral 2-cover looks like $y \mapsto y^2 = z$. Sections of line bundles, including $K_X$, are of the form $s(z) = f_0(z) + yf_1(z)$.

By the spectral correspondence, the Higgs field is conjugate to the "multiplication-by-$y$" endomorphism. The latter sends $s(z)$ to $ys(z) = yf_0(z) + y^2 f_1(z) = yf_0(z) + zf_1(z).$

Taking $f_0(z)$, $f_1(z)$ to be local bases, we have that $\phi$ is conjugate to
$\begin{pmatrix}
0 & z \\
1 & 0
\end{pmatrix}.$ One thing that is lacking about this answer is that I am not sure how $tr(\phi^2)$ having simple zeroes help.

Perhaps the above answers this question. We replace $z$ with $y^2$ because locally at $z = 0$, $tr(\phi^2) = 0$ and the spectral curve looks like the double cover $y \mapsto y^2 = z$.

Title: Higgs bundle and spectral curve


问:
Do $C_b(X)$ and $L^{\infty}(X)$ have the same duals?
I was reading the following theorem where $C_b(X)$ are the continuous and bounded function on $X$.

Theorem : Consider a normal topological space $X$ and a positive linear functional $\varphi\colon C_b(X)\to\mathbb{R}$. There exists a unique finitely additive set function $\mu$ defined on the algebra generated by the open sets in $X$ such that $$\varphi(f)=\int_X f\,d\mu,\quad f\in C_b(X).$$

So the dual of the $C_b(X)$ would be the finitely additive measure. But isn't this also the dual of $L^{\infty}(X)$? If it is the case, how can we explain that they have the same dual?

答:
They are not the same!

$C_b(X)'$ is the finitely additive measures defined on the algebra generated by opens,

$L^\infty(X)'$ is the finitely additive measures defined on the $\sigma$-algebra generated by opens.


问:
Computing integral $\int_0^1 |x-a| \, {\rm d} x$
I have the following question for a seminar and cannot find the relevant information to solve it.

Solve the integral for all $a \in \mathbb R$ $$I(a)=\int_0^1 |x-a| \, {\rm d} x$$

I understand that I have to treat the absolute value of the function $f(x)=|x-a|$ depending on the when $x$ is less or more then $a$, if $a$ had a defined value I could have solved this, but now im stuck because I dont know how to relate its unspecified value to find a solution for the integral.

Any help is appreciated that'll point me in the right direction.

答:
Substitute $u=x-a$. This will give us $\mathrm{d}x=\mathrm{d}u$ and
$$\int\limits_{-a}^{1-a}\lvert u\rvert\,\mathrm{d}u.$$
Now integrate by parts. Choose $f'=1$ and $g=\lvert u \rvert$:
$$\int\limits_{-a}^{1-a}\lvert u\rvert\,\mathrm{d}u = u\lvert u\rvert-\int\limits_{-a}^{1-a}\lvert u\rvert\,\mathrm{d}u.$$
These two integrals are equal. That means that
$$\int\limits_{-a}^{1-a}\lvert u\rvert\,\mathrm{d}u=\frac{u\lvert u\rvert}{2}\Bigg\lvert^{1-a}_{-a}.$$
Now just plug in in the values for $u$ and you'll get the solution for $a\in\mathbb{R}$:
$$\frac{(1-a)\lvert 1-a\rvert+a\lvert a\rvert }{2}$$


答:
welcome to MSE.
$$\int |x-a|dx=\dfrac{\left(x-a\right)\left|x-a\right|}{2}+C$$ also you can rewrite as
$$I(a)=\int_0^1 |x-a|dx=\dfrac{a\left|a\right|}{2}-\dfrac{\left|a-1\right|a-\left|a-1\right|}{2}$$

Title: Computing integral $\int_0^1 |x-a| \, {\rm d} x$

Body:
I have the following question for a seminar and cannot find the relevant information to solve it.

Solve the integral for all $a \in \mathbb R$ $$I(a)=\int_0^1 |x-a| \, {\rm d} x$$

I understand that I have to treat the absolute value of the function $f(x)=|x-a|$ depending on the when $x$ is less or more then $a$, if $a$ had a defined value I could have solved this, but now im stuck because I dont know how to relate its unspecified value to find a solution for the integral.

Any help is appreciated that'll point me in the right direction.


问:
Loewy decomposition of differential operators
The paper by Fritz Schwarz, "Loewy decomposition of linear differential equations", contains the following lemma, which I try to prove in order to understand the algorithm which Schwarz describes and which goes back to Loewy:

Lemma: Determining the right irreducible factors of an ordinary operator up to order three with rational function coeffictents amounts to finding rational solutions of Riccati equations. A second order operator (with $D=\frac{d}{dx}$)

$D^2+AD+B, \; A,B \in \mathbb{Q}(x), \qquad \qquad (1)$

has a right factor $D+a, \; a \in \mathbb{Q}(x)$, if $a$ is a rational solution of

$a'-a^2+Aa-B=0. \qquad \qquad (2)$

A third order operator

$D^3+AD^2+BD+C, \; A,B,C \in \mathbb{Q}(x), \qquad \qquad (3)$

has a right factor $D+a, \; a \in \mathbb{Q}(x)$, if $a$ is a rational solution of

$a''-3aa'+a^3+A(a'-a^2)+Ba-C =0. \qquad \qquad$ (4)

It has a right factor $D^2+bD+c, \, b,c \in \mathbb{Q}(x)$, if b is a rational solution of

$ b''-3bb'+b^3+2A(b'-b^2)+(A'+A^2+B)b-B'-AB+C=0. \qquad \qquad $ (6)

Then $c=-(b'-b^2+Ab-B). \qquad (7) \qquad$  End of the lemma.

The author of the paper takes the proof to be obvious when dividing the given operator by the right factor and requiring that the division be exact, but I don't really see that for the second and the third case.

What we want to do with the help of this lemma is decompose a differential operator $L$ in order to solve the corresponding differential equation $Ly=0$, with y being the unknown function of x and A,B,C,a,b,c being rational functions of x. I understand that the operator product for differential operators is noncommutative and obeys the rule

$Da=aD+a'. \qquad \qquad (5)$

So in order to proof the lemma's first statement, I do the following: Polynomial division (in D) of (1) by $D+a$ gives me $D+A-a$, plus some rest $a^2-Aa+B$. Ignoring the rest for the moment, I build and simplify the operator product, paying attention to the above rule (5):

$(D+A-a)(D+a)= D^2+AD-aD+Da+Aa-a^2 = D^2+AD+a'+Aa-a^2 $,

and what follows is (2), if the division is to be exact. Now the same approach for proving the second statement. Polynomial division of (3) by $D+a$ gives $D^2+D(A-a)-Aa+a^2+B$ plus some rest. Again we build the product, multiply it out and try to simplify acording to (5):

\begin{align}
&(D^2+DA-Da-Aa+a^2+B)(D+a)
\\ &= D^3+DAD-DaD-AaD+a^2D+BD +D^2a+DAa-Da^2-Aa^2+a^3+Ba
\\ &= D^3+((AD+A')D)-(D(Da-a'))-(DAa-(Aa)')+(Da^2-(a^2)')+BD+ \cdots
\\ &= D^3+ (AD^2+A'D)-(D^2a-Da')-(DAa-A'a-Aa')+(Da^2-2aa')+BD+ \cdots
\\ &= D^3+AD^2+BD+A'D+Da'+A'a+Aa'-2aa'-Aa^2+a^3+Ba.
\end{align}

But this does not bring me to (4). In particular, I don't know how to get rid of the terms $A'D$ and $Da'$, where D still appears. What am I doing wrong? Or is my whole approach not valid?

答:
Now I found the necessary trick:

$A'D+Da'=A'D+a'D+a''$,

and in order to make the two terms with D disappear, we find the condition $A'=-a'$. This substituted in the equation gives (4). So the rules employed were correct for proving this part, too.

Now the third part: Polynomial division gives $D+A-b$ plus some rest. We multiply and simplify

\begin{align}
&(D+A-b)(D^2+bD+c)
\\ &= D^3+AD^2-bD^2+DbD+AbD-b^2D+Dc+Ac-bc
\\ &= D^3+AD^2-bD^2+(bD^2+b'D)+AbD-b^2D+(cD+c')+Ac-bc
\\ &= D^3+AD^2+(\underbrace{b'+Ab-b^2+c}_{=B})D+\underbrace{c'+Ac-bc}_{=C}.
\end{align}

So we get two equations. The first one is (7). We get (6) by differentiating (7), substituting therein c' by the second equation and then again c by (7).


问:
Is there a good reason to use covariance and not correlation?
Correlation is a normalization of covariance by the standard deviation of each variable.

So is there a good reason (and example) when we should (and have) to use covariance and not correlation ?

答:
Covariances are generally more easy to handle mathematically.
If $X$ and $Y$ are random variables with expectation zero,
$$
Cov(X,Y)=E[XY]
$$
Whereas:
$$
Corr(X,Y)=\frac{E[XY]}{\sqrt{E[X^2]}\sqrt{E[Y^2]}}
$$
The covariance can be seen as an inner product (dot product) on the space of random variables with finite variance and expectation zero.
Consequentially, it has many nice mathematical properties, and is important to study such random variables geometrically.
An important example is that $X\mapsto Cov(X,Y)$ is linear, but $X\mapsto Corr(X,Y)$ is not linear.


问:
Why is $0^0$ undefined and how would we graph this?
So I saw exponents like $3^0$ and $4^0$, etc which are all equal to $1$.

And then soon I see that $0^0$ is not defined.

I checked the graph of $x^0$

Then I decided to make some observations.

We go like this:

$0.9^{0.9} = 0.909532576$

$0.8^{0.8} = 0.836511642$

$0.7^{0.7} = 0.779055912$

$0.6^{0.6} = 0.73602192281$

$0.5^{0.5} = 0.70710678118$

$0.4^{0.4} = 0.69314484315$

And then comes the moment:

$0.3^{0.3} = 0.69684530193$

And then, so on the values increase, and then finally the calculator breaks at $0.00000000000001^{0.000000000000001}$ and gives $1$.

I'd like to understand why $0.3^{0.3}$ is greater than $0.4^{0.4}$. And why the values go on increasing as we go $<0.3$.

So we get $\lim_{x\to 0} x^{x} = 1$.

Another amazing graph:

This thing has really intrigued me, and I wonder if there are more ways to prove that $0^0$ is undefined. Sharing more methods for the same would be highly appreciated!

答:
I don't entirely agree that $0^0$ "is not defined". By definition of "define", you can define something to be whatever you want... some definitions turn out to be more helpful than others, though. There are lots of things in mathematics that are defined in some contexts and not in others. For instance, you will often have heard that something like $\frac{1}{0}$ is undefined, and that it would be absurd to assign it any value. However, in many areas of mathematics, it makes perfect sense to "invent" some thing $\infty$ that equals precisely $\frac{1}{0}$. When we do this, we have to be precise about what properties our new thing has, so as to avoid making logical errors.

This Wikipedia page for $0^0$ tells us that "The most common possibilities are 1 or leaving the expression undefined". As you have demonstrated in the question, $1$ does indeed seem the most reasonable choice if we do want to define it.

For the sake of completeness, I will also include an informal proof that the limit of $x^x$ as $x$ tends to $0$ from above is indeed $1$. We will use standard properties of the natural logarithm $\log$. For any positive real $x$, we have
$$
x^x = e^{\log(x^x)} = e^{x\log x},
$$
and therefore it suffices to show that $x\log x \to 0$ as $x\to 0^+$. To see this, consider the sequence $(\frac{1}{n})_{n\geq 1}$. We have
$$
\frac{1}{n}\log(\frac{1}{n}) = -\frac{1}{n}\log(n),
$$
which converges to $0$ as $n\to \infty$ (I won't prove this but it follows from elementary properties of logarithms, and you can probably Google it). Therefore, since $x\log x$ is continuous for $x>0$ and $\frac{1}{n}\to 0$ as $n\to \infty$, we have $x\log x \to 0$ as $x \to 0^+$ as required.


问:
Working backwards with Determinants.
The determinant of A is -2. Find $\det(3A^TA^3)$. You may leave your answer as a product of integers.

What I did was this:

$$\det(3A^TA^3)$$

$$\det(3A^T)\det(A^3)$$

$$3\det(A^T)(\det(A))^3$$

This is the part where I became confused;

$$3\det(A^T)(-2)^3 $$

That's as far as I got, and looking at the solution what I was doing was correct, however the solution for it does this;

$$\det(3A^T)\det(A)^{3}$$
$$(3^3)\det(A)\det(A)^{3}$$
$$(27)(-2)^4$$
$$(27)(16)$$
$$432$$

Now the part that confuses me about this question is , where does the other power of three come from in the line, and what happened to the transpose?

$$(3^3)\det(A)\det(A)^{3}$$

答: Hint: $\det(A) = \det(A^{T})$ (can you see why this is?); furthermore, for an $n \times n$ matrix, $\det(kA) = k^{n}\det(A)$, where $k$ is a constant. </p>

Title: Working backwards with Determinants.


答:
I'd like to understand why $0.3^{0.3}$ is greater than $0.4^{0.4}$.
And why the values go on increasing as we go $<0.3$.

As mentioned in the comments you can set the derivative equal to zero to obtain the extreme point. Let $f(x)=x^x, >0$

$\ln(f(x))=x\cdot \ln(x)$

Differentiating. On the left side we use the chain rule.

$\frac1{f(x)}\cdot f'(x)=\ln(x)+1$

$f'(x)=f(x)\cdot (\ln(x)+1)$

$f'(x)=x^x\cdot (\ln(x)+1)$

Setting the derivative equal to zero.

$=x^x\cdot (\ln(x)+1)=0\Rightarrow \ln(x)+1=0$

$\ln(x)=-1\Rightarrow x_E=\frac1e$. And you see that $0.3<\frac1e<0.4$

It remains to show that this extreme point is a (relative) minimum. So we need the second derivative. Using the product rule and we get

$f''(x)=x^x\cdot \left(\frac1x+(\ln(x)+1)^2\right)$

$f''\left(\frac1e\right)>0 \Rightarrow x_E=\frac1e$ is a (relative) minimum.

Title: Why is $0^0$ undefined and how would we graph this?

Body:
So I saw exponents like $3^0$ and $4^0$, etc which are all equal to $1$.

And then soon I see that $0^0$ is not defined.

I checked the graph of $x^0$

Then I decided to make some observations.

We go like this:

$0.9^{0.9} = 0.909532576$

$0.8^{0.8} = 0.836511642$

$0.7^{0.7} = 0.779055912$

$0.6^{0.6} = 0.73602192281$

$0.5^{0.5} = 0.70710678118$

$0.4^{0.4} = 0.69314484315$

And then comes the moment:

$0.3^{0.3} = 0.69684530193$

And then, so on the values increase, and then finally the calculator breaks at $0.00000000000001^{0.000000000000001}$ and gives $1$.

I'd like to understand why $0.3^{0.3}$ is greater than $0.4^{0.4}$. And why the values go on increasing as we go $<0.3$.

So we get $\lim_{x\to 0} x^{x} = 1$.

Another amazing graph:

This thing has really intrigued me, and I wonder if there are more ways to prove that $0^0$ is undefined. Sharing more methods for the same would be highly appreciated!


问:
$f(x) = \int\limits_{a(x)}^{b(x)} \sin{\sqrt t}\,\mathrm{d}t,$ compute $f^{'}(0)$.
Let $a(x)=\frac{\pi^2}{4} + \cos(3x+\frac{\pi}{2})$ and $b(x)=\frac{25 \pi^2}{4} + 2x^2$. $$f(x) = \int\limits_{a(x)}^{b(x)} \sin{\sqrt t}\,\mathrm{d}t,$$ compute $f^{'}(0)$.

My thoughts:

To evaluate $$\int \sin \sqrt x \, dx $$, let $u = \sqrt x, x = u^2, dx = 2u \, du$

$$\int \sin \sqrt x \, dx  = \int \sin u 2u \, du = 2 \int u d (-\cos u) \, du =
2 \left( -u\cos u + \int \cos u \, du \right) = 2 \left( -u\cos u + \sin u  + C \right)  $$

But it doesn't make the final computation much easier..

答: More generally, if $f$ has an antiderivative $F$,$$\frac{d}{dx}\int_{a(x)}^{b(x)}f(t)dt=\frac{d}{dx}[F(b(x))-F(a(x))]=f(b(x))b^\prime(x)-f(a(x))a^\prime(x),$$so we don't need to know $F$ itself, i.e. you don't need to integrate $f$. (Rather nicely, you don't need to differentiate $f$ either.) Now take$$a=\frac{\pi^2}{4}-\sin(3x),\,b=\frac{25\pi^2}{4}+2x^2,\,f=\sin\sqrt{t}.$$I'll leave that to you.


问:
Maximizing the Lagrangian with respect to vector - help solve
I need help solving this following equation, a Lagrangian problem that I encountered during my studies in principal component analysis (PCA).

One should maximize the variance with respect to the first principal component

Following is the case:
We know that $W = {\bf v}^T_1 {\bf Sv}_1 $ (eq 1) and that $\bf v_1$ is an orthonormal basis.

Therefore $||{{\bf v}_1 }||^2 = {\bf v}_1^T{\bf v}_1 = 1$ (eq 2).
The maximization of $W$ under this constraint can be done by introducing the Lagrangian multiplier λ and maximizing the Lagrangian:

$\mathcal{L} = W + \lambda (1-||{\bf v}_1||^2) = {\bf v}^T_1 ( {\bf S}-\lambda {\bf I}){\bf v}_1 + \lambda$ (eq 3)

So expanding $W$ using eq. 1, and the laws of distributivity we end up with eq 3.

Next, taking the derivative with respect to vector and $\lambda$ we obtain:

$$\frac{\partial}{\partial \lambda} \mathcal{L} = 1-{\bf v}_1^T{\bf v}_1  = 0 $$

$$\nabla_{{\bf V}}{_1}\mathcal{L} = ({\bf S} - \lambda {\bf I}){\bf v}_1  = 0 $$

This can later be shown to be an eigenvalue problem. Can anyone explain to me why the derivatives look like that? I can't seem to figure it out.

Best regards :)

答:
I believe I found a solution, to whom this may be interesting.

First partial differentiation is straight forward, $\lambda$  is a constan. I found a slightly different approach, remember the inner product is = 1.

We know the fact that:$$ \lambda\frac{\partial}{\partial {\bf v}_1^T}  ({\bf v}_1^T{\bf v}_1) = 2\lambda {\bf v}_1$$

The second one, uses the fact that
$$\frac{\partial}{\partial {\bf v}_1} ({\bf v}_1^T({\bf S}-\lambda {I\bf }){\bf v}_1) = 2({\bf S}-\lambda {\bf I}) {\bf v}_1 $$

Hence the partial differentiation yields:

$$ \nabla _{{\bf v}_1}\mathcal{L} = 2({\bf S}-\lambda {\bf I}) {\bf v}_1  - 2\lambda{\bf v}_1
$$

The optimization problem then becomes:
$$ \frac{\partial \mathcal{L}}{\partial{{\bf v}_1}} = ({\bf S}-\lambda {\bf I}) {\bf v}_1  - \lambda{\bf v}_1 = 0$$

Normalization of ${\bf v}_1$ and rewriting of the optimization problem yields the eigenvalue problem:
$$ {\bf S v}_1 = \lambda {\bf v}_1 $$


问:
Show that f is differentiable at $(x, y) = (0, 0)$ for the given function
I know that this question has been asked before for the given function, but I still have specific questions regarding this problem that hasn't been addressed and I would like to explain my specific reasoning as well.

We are given the following function:
$$f(x,y) = \begin{cases}(x^2+y^2)\sin(\frac{1}{\sqrt{x^2+y^2}})&\text{ if }(x,y)\not =(0,0)\\0 &\text{ if }(x,y)=(0,0)\end{cases}$$

Show that $f$ is differentiable at $(x, y) = (0, 0)$ and determine the derivative.

In general, we know the following:

If the partial derivatives $f_x$ and $f_y$ exist near $(a, b)$ and are continuous
at $(a, b)$, then $f$ is differentiable at $(a, b)$.

If   $\lim_{(x, y) \to (a, b)} f(x, y) =  f(a, b)$, then $f$ is continuous at $(a, b)$.

If $f$ is differentiable at $(a, b)$, then $f$ is continuous at $(a, b)$.

I know the following about the function:

The partial derivatives $f_x$ and $f_y$ exist at every $(x,y)$, where $f_x(0, 0) = 0$ and $f_y(0, 0) = 0$.

The partial derivatives are not continuous at $(0, 0)$, since $\lim_{(x, y) \to (0, 0)} f_x(x, y)$ does not exist.

We can conclude the following:

We know that the partial derivatives are not continuous at $(0, 0)$, but this doesn't mean that $f$ is not differentiable at $(0, 0)$, since the reverse of the first general statement is not true.

So how can we show that $f$ is differentiable at $(0, 0)$, specifically using the following hint:
$-(x^2+y^2) \leq f(x, y) \leq x^2 + y^2$ for all $(x, y$)?
Checking whether $f$ is continuous at $(a, b)$ isn't enough right? Because if $f$ is continuous at $(a, b)$, then this doesn't mean that $f$ is differentiable at $(a, b)$ (since the reverse of the third general statement isn't true).

答: Since $f_x(0,0)=f_y(0,0)=0$, if $f$ is differentiable at $(0,0)$, then $f'(0,0)$ can only be the null function. And it is the null function if and only if$$\lim_{(x,y)\to(0,0)}\frac{f(x,y)}{\sqrt{x^2+y^2}}=0.$$And this is true because you have$$-(x^2+y^2)\leqslant f(x,y)\leqslant x^2+y^2$$and therefore$$-\sqrt{x^2+y^2}\leqslant\frac{f(x,y)}{\sqrt{x^2+y^2}}\leqslant\sqrt{x^2+y^2}$$and so all you have to do is to apply the squeeze theorem.


问:
Evaluating $\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1}$

I am seeking to evaluate the sum $$S=\sum_{r \in \mathbb{N}} (n-2r+1)^2 \binom{n}{2r-1} \\ =(n-1)^2 \binom{n}{1}+(n-3)^2 \binom{n}{3}+(n-5)^2\binom{n}{5}+\cdots$$

I re-wrote the sum as $$S=(n+1)^2\sum_{r \in \mathbb{N}} \binom{n}{2r-1}-4(n+1)\sum_{r\in\mathbb{N}}r\binom{n}{2r-1}+4\sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$$

It is clear from the binomial expansion of $(1+x)^n$, (by plugging $x=\pm1$ ) we get the value of the sum $\displaystyle \sum_{r\in\mathbb{N}}\binom{n}{2r-1}$ as $2^{n-1}$.

Now, after this I can find the value of the sums $\displaystyle \sum_{r\in\mathbb{N}}r\binom{n}{2r-1}$ and  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$ by considering the binomial expansion of $\displaystyle \dfrac{(1+x)^n-(1-x)^n}{2}$ and substituting $x$ for $\sqrt{x}$ and then differentiating, and by manipulations, I will be able to find the desired sums.

However, this method gets very lengthy, especially when doing differentiation for finding  $\displaystyle \sum_{r\in\mathbb{N}}r^2\binom{n}{2r-1}$.

I am hoping to see some other different approaches which don't involve so much calculation and are easy to understand.

The final closed form is $n(n+1)2^{n-3}$.

Body:
Hint:

Let

$$(n-2r+1)^2=a+b(2r-1)+c(2r-1)(2r-2)$$

Set $2r-1=0\implies a=n^2$

$2r-2=0\implies a+b=(n-1)^2\implies b=?$

Compare the coefficients of $r^2,c=1$

Now $$(2r-1)(2r-2)\cdot\binom n{2r-1}=n(n-1)\binom{n-2}{2r-3}$$


问:
The real number reached by the probability sequence
All numbers are expressed in base $2$.

Start from $0.1_2$

If we reach $r(n) < 1$ after $n$ steps the probability of having $1$ at $n+1^{th}$ decimal position is then $r(n)$.

For example, it is equally probable to have $0.10_2$, $0.11_2$ and then we have two branches, first where it is equally probable to have $0.100_2$ and $0.101_2$ and the other branch where it is more probable to have $0.111_2$.

What is the expected final real value when $n$ tends to infinity? Can you express it in some closed form?

Expressing the problem in the simplest term would be

$$e(1)=\frac{1}{2}$$
$$e(n+1)=e(n)+\frac{1}{2^{n+1}}e(n)$$

However, can we apply this recurrence blindly over the expected value? And then if so, can we find some closed form of the recurrence? (I can't find anything shorter than q-Pochhammer.)

The approximation gives $r(\infty) \approx 0.794$

答:
Work recursively.  Let $E_n$ denote the expected value of the $n-$ digit entry.  Of course $E_1=\frac 12$.  Let $E$ denote $\lim_{n\to \infty}E_n$, so $E$ is the answer we seek.

We have $$E_n=E_{n-1}\times \left(E_{n-1}+\frac 1{2^n}\right)+(1-E_{n-1})\times \left(E_{n-1}\right)\implies E_n=E_{n-1}\times \left(1+\frac 1{2^n}\right)$$

In the limit we have $$E=\frac 12\times \prod_{n=2}^{\infty}\left(1+\frac 1{2^n}\right)$$

That can be evaluated using the Pochhammer Symbol and we get $$\boxed {E\approx 0.794744}$$

To justify the recursion, observe that $$E_n=\sum_w p(w)w$$ where $w$ spans the $n-$digit binary decimals and of course $p(w)$ is the probability of observing $w$.  Of course we can write $w=w_{n-1}X$ where $w_{n-1}$ is a word of length $n-1$ and $X$ is $1$ or $0$ with probability $w_{n-1}, 1-w_{n-1}$ respectively. Splitting the sum into two parts, according to whether $X=1,0$ we see that
$$E_n=\sum_{w_{n-1}} p(w_{n-1})w_{n-1} \left(w_{n-1}+ \frac 1{2^n}\right)+\sum_{w_{n-1}} (1-w_{n-1})p(w_{n-1}) w_{n-1}=E_{n-1}\times \left(1+\frac 1{2^n}\right)$$as desired.


问:
Spruce Budworm bifurcation diagram
The ODE that appears in the Spruce Budworm Population problem is the following:
$$\frac{dN}{dt}=r_PN\left(1-\frac{N}{K}\right) -p(N) \quad \quad \mbox{where} \quad \quad p(N)=\frac{BN^2}{A^2+N^2}$$ where $A$, $B$ are constants.
After some variable changes (you can check them on pages $4$ and $5$ of the document liked above) we get to the following ODE
$$\frac{du}{dτ} = ru \left(1 −\frac{u}{q}\right)−\frac{u^2}{1+u^2}$$
The bifurcation diagram, where we plot the parametric curves $r=\frac{2u^3}{(1+u^2)^2}$, $q=\frac{2u^3}{u^2-1}$ is this one. I want to know how can I find the coordinates of the point $A$ in the graph linked. I already know their values but I want to know a method to manually find their values. Thank you for your help

答: The cusp is only possible when neither $r$ nor $q$ provides a local coordinates, i.e., $r'(u)=q'(u)=0$.  You just have to solve it and get $u=0,\pm\sqrt3$ and you must go back and chck indeed that is a cusp, not a problem of silly parametrization like $y=x=t^2$ at $t=0$.

Title: Spruce Budworm bifurcation diagram


问:
Solve for integer values of $x,y,z$: $\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3$
Solve for integer values of $x,y,z$;
$$\frac{xy}{z} + \frac{xz}{y} + \frac{yz}{x} = 3.$$

My attempt:

Note that all $x, y, z$ are non-zero, otherwise a denominator would be zero.

Mutliplying by $xyz$ gives:
$$
x^2y^2+x^2z^2+y^2z^2 = 3xyz
$$

Since (from 1.) the left hand side (LHS) is strictly positive, then $xyz \gt 0$

Exploring the LHS of the equation from 1.:
$$
2x^2y^2+2x^2z^2+2y^2z^2 = (xy-xz)^2+(xy-yz)^2+(xz-yz)^2+2xyz(x+y+z)
$$
therefore
$$
x^2y^2+x^2z^2+y^2z^2 \geq xyz(x+y+z)
$$

Replacing LHS of the last inequation by $3xyz$ (according to 2.) gives:
$$3xyz \geq xyz(x+y+z)$$  and since (from 3.) $xyz$ is positive, we can divide both  sides by $xyz$, obtaining: $$x+y+z \leq 3$$

$xyz$ can be positive in two cases:

Case I.    $x, y, z$ are all positive. In this case $x \geq 1$, $y \geq 1$, $z \geq 1$,
so
$$
x+y+z \geq 3
$$
but according to 5. this must be an equation. Therefore all inequations for $x$, $y$ and $z$ must turn into equations, so $x=y=z=1$.

Case II.   One of $x, y, z$ is positive, two others are negative. Without loss of generality we can assume $x \gt 0$, $y \lt 0$,  $z \lt 0$

...

答:
If $(x,y,z)$ is a solution then so is $(|x|,|y|,|z|)$, so without loss of generality $x,y,z>0$, and after rearranging we may assume without loss of generality that $x\geq y\geq z$. Then $\tfrac xy,\tfrac xz,\tfrac yz\geq1$, and hence
$$3=\frac{xy}{z}+\frac{xz}{y}+\frac{yz}{x}> x\frac yz\geq x,$$
and similarly $y,z<3$, so $x,y,z\in\{1,2\}$. This leaves just four triplets $(x,y,z)$ to check.


问:
Find all $n$ such that $2^n+1$ and $2^n-1$ are primes

Find all $n\in \mathbb N$ such that $2^n+1$ and $2^n-1$ are both primes .

My Attempt:

let $p=2^n+1$ and $q=2^n-1$

I will start with a claim.

claim:
$n=2\implies p=5, q=3$ is the only solution, because otherwise $2^n+1$ or $2^n-1$ would be divisible by $5$ or $3$.

if $3\mid 2^n+1\iff 2^n+1\equiv 0 \pmod 3 \implies 2^{n-1}\equiv 1 \pmod 3$.

But i don't know how to prove this claim.

答: You are kinda on the right track. Note that one of $2^n-1, 2^n, 2^n+1$ is divisible by $3$, and one of them is never divisible by $3$. Hence if $2^n-1 > 3$, one of $2^n-1, 2^n+1$ is composite.


问:
Uniform convergence and Riemann integral; from compact $0<t<T<+\infty$ to $(0,+\infty)$ [Hint]
I want to know an idea about how to start with this problem:

Let be $g, f_n, \forall n \in \mathbb{N}$ well-defined functions in $(0, \infty)$, and Riemann integrable in every compact subset $[t, T]$ where $0 < t < T < \infty$. In addition, $f_n$ converges uniformly to a function $f$ and $|f_n| \leq g$ in every compact subset of $[0, \infty]$. If we have that:

$\int_0^{\infty}g(x)dx < \infty$

Proove that $\lim_{n \rightarrow \infty} \int_0^{\infty}f_n(x)dx = \int_0^{\infty}f(x)dx$

I am supposed to solve this problem without using directly any theorem about Measure Theory (like the Dominated Convergence Theorem, which has similarities with this exercise), only using facts about convergence of functions.

My problem is that I can't find a way to start. I know that if $f_n$ are Riemann integrable in a certain subset and they converge uniformly to a function $f$ in t his subset then f is Riemann integrable in this subset and I can put the limit inside or outside the integral. The problem is that I know it for every interval $[t,T]$ but they are talking about $(0, \infty)$, and this generalization is non-trivial and here the function $g$ has to help me but I don't know how to use the bound that they give me.

Thanks :)

答:
Consider the difference $\int f_n(x)dx-\int f(x)$. You want to try and make a claim that $|f_n(x)-f(x)|<\epsilon$ for all $x$ when $n$ is big enough but you don't quite have this. Instead, use the fact that $g$ is integrable. This means that $\int_c^\infty g(x)dx$ can be made arbitrarely small as $c$ goes to infinity. So now write:

$$\left|\int_0^\infty f_n(x)dx-\int_0^\infty f(x)dx\right|\leq \int_0^c|f_n(x)-f(x)|dx+2\int_c^\infty g(x)dx$$

where you can now use the fact that $|f_n(x)-f(x)|<\epsilon$ on $[0,c]$ for large enough $n$.


问:
Finding the area of shaded region
Consider a circle with the equation $x^2 + y^2 = 1$. Now find the area of the shaded region which we denote by $u$.

image of circle:

I'm trying to approach this question and would like support on my integral calculation. The difficulty I'm having is with substitution.

For example, given that:

$$u = (OM)(MP)+2\int^{1}_{x} \sqrt{1-x^2} \space dx$$

I should be getting:

$u = xy + [x\sqrt{1-x^2}-\cos^{-1}x]^{1}_{x}$

Although, I currently lack the technical skills to get this answer.

My approach:

I thought that I could substitute back in $x$ for $\sin(x)$ to get $\sqrt{1-\sin^2(x)}$, then when $u = \sin(x) \implies du = \cos(x)$, so that I get:

$$2\int^{1}_{x} \sqrt{1-\sin^2(x)}\cdot \cos(x) \space dx$$

However, this does not lead me close to the answer.

I have also tried inputting this equation into symbolab, and get:

$\arcsin \left(x\right)+\frac{1}{2}\sin \left(2\arcsin \left(x\right)\right)+2C$

Which is also wrong. What might be the approach towards this?

In Progress:

I've currently rearranged my substitution and it looks promising:

$u = 1-\sin^2 (x) \implies du = -cos(x)dx \implies-arcos(x)du = dx$

I cannot manage with this approach neither, as I'll have to integrate the square root.

Looking at the equation, I'll need something like:

$secant(x) -arcos(x)$ to convert the secant into $x\sqrt{x^2-1}$

答:
You may integrate by parts directly

\begin{align}
\int_{x}^{1}\sqrt{1-t^2}dt
& =\int_{x}^{1}\frac{\sqrt{1-t^2}}{2t}d(t^2)
= -\frac12 x\sqrt{1-x^2}+\frac12 \cos^{-1}x
\end{align}

which, with $y=\sqrt{1-x^2}$, leads to the area
$$u =xy + 2 \int_{x}^{1}\sqrt{1-t^2}dt =\cos^{-1}x$$

Title: Finding the area of shaded region


问:
Can you rearrange this simple Hadamard & Matrix product?
I have the following matrix operation:

${((AB^T) \odot C)D}$

where juxtaposition denotes the matrix product and $\odot$ the Hadamard product.

A, B & D are all matrices of shape (3, 2) and C is a matrix of shape (3, 3). Is it possible to rearrange/decompose the operation, such as to avoid calculating the matrix product of A and B?
This doesn't work, but something like
${(A \odot C)(B^TD)}$ ?

答:
$\def\B{\Big}\def\L{\left}\def\R{\right}\def\D{\operatorname{Diag}}$Let $\{a_k,b_k\}$ denote the $k^{th}$ columns of $\{A,B\}$, respectively.

Construct diagonal matrices from these column vectors
$$\eqalign{
{\cal A}_k &= \D(a_k) \qquad {\cal B}_k = \D(b_k) \\
}$$
Then
$$\eqalign{
AB^T &= \sum_{k=1}^2 a_kb_k^T \\
}$$
and the product in question can be expanded as a sum
$$\eqalign{
P &= \B(AB^T\odot C\B)D \\
 &= \L(\sum_{k=1}^2 a_kb_k^T\odot C\R)D \\
 &= \sum_{k=1}^2 {\cal A}_kC{\cal B}_kD \\
 &= {\cal A}_1C{\cal B}_1D + {\cal A}_2C{\cal B}_2D \\
}$$
which also eliminates the Hadamard product.


问:
Using Newton's Method to find eigenvalues of a matrix $A \in \mathbb{R}^{n \times n}$
Let $A \in \mathbb{R}^{n \times n}$ and suppose $\exists$ a nonsingular $V \in \mathbb{R}^{n \times n}$ and a diagonal $D \in \mathbb{R}^{n \times n}$ such that $A = VDV^{-1}$. Define the map $F: \mathbb{R}^n \times \mathbb{R} \rightarrow \mathbb{R}^n \times \mathbb{R}$ by $$F(v,\lambda) = \left[\begin{array}{c}
Av - \lambda v \\
\frac{1}{2}v^T v - \frac{1}{2}
\end{array}\right].$$ Then finding an eigenpair of $A$ with an eigenvector of unit length can be viewed as a root finding problem. One part of the question I am working on asks to show that the Jacobian $F'(v_*,\lambda_*)$ is nonsingular if $\lambda_*$ is a simple eigenvalue.

Attempt: Let $(v_*,\lambda_*)$ be an eigenpair of $A$ where $\lambda_*$ is a simple eigenvalue. Suppose the Jacobian $$F'(v_*,\lambda_*) = \left[\begin{array}{cc}
A - \lambda_* I & -v_* \\
v_*^T & 0
\end{array}\right]$$ is singular. Then there exists a nonzero vector $(x,\mu) \in \mathbb{R}^n \times \mathbb{R}$ such that $$\left[\begin{array}{cc}
A - \lambda_* I & -v_* \\
v_*^T & 0
\end{array}\right]\left[\begin{array}{c}
x \\
\mu
\end{array}\right] = 0.$$ That is, \begin{align*}
Ax - \lambda_*x & = \mu v_* \\
v_*^Tx & = 0.
\end{align*} I am not sure how to derive a contradiction from here. Any help would be sincerely appreciated!

答:
There is a different solution, that doesn't require $ A $ to be diagonalisable.

Let $ x $ be an eigenvector to the simple eigenvalue $ \lambda $, so $ x\neq 0 $.

We assume $ F'(x,\lambda) $ being singular, then there exists $ \begin{pmatrix}
v\\\mu
\end{pmatrix} \in \mathbb(R)^{n+1}\backslash\{0\} $ with
$$ \begin{pmatrix}
A-\lambda I & -x\\
x^T&0
\end{pmatrix}
\begin{pmatrix}
v\\\mu
\end{pmatrix}
=
\begin{pmatrix}
0\\0
\end{pmatrix}$$
which is leading to the following equations:
\begin{align}
(A-\lambda I )v-\mu x&=0\\
x^T v&=0
\end{align}
We will now look to the following cases:

1.) If $ \mu=0 $ then $ v\neq 0 $.
From $ (1) $ it follows that $ Av=\lambda v $, which implies $ v\in E(A,\lambda) $, where $ E(a,\lambda) $ denotes the eigenspace of the eigenvalue $ \lambda $.
With $ x,v\neq 0 $ the orthogonality of $ x $ and $ v $ implied by $(2)$ shows, that $ x $ and $ v $ are linearly independent.
Both being Elements of $ E(A,\lambda) $ implies that the geometric multiplicity of $ \lambda $ is at least 2, which is a contradiction to $ \lambda $ being a simple eigenvalue, because the  geometric multiplicity can never exceed the algebraic multiplicity.

2.) Now, let $ \mu \neq 0 $. Because $ x $ is an eigenvector $ x\neq 0 $.
That leads to $ \mu x \neq 0 $, which implies $ v\neq 0 $ because otherwise $ (1) $ will give us the contradiction
$$ 0=(A-\lambda I)v=\mu x \neq 0. $$
As said before, $ v\neq 0 $ implies (with $ (2) $) that $ x $ and $ v $ are linear independent.
From $ (1) $ we get
$$ (A-\lambda I)^2 v=(A-\lambda I) \mu x =0$$
which means that $ v $ is an element of the generalized eigenspace $ H(A,\lambda) $.
From the linear independence of $ x $ and $ v $ it follows that $ \dim H(A,\lambda)\geq 2>1 $.
Because $ \dim H(A,\lambda)$ is the algebraic multiplicity of an eigenvalue this is a contradiction, to $ \lambda $ being a simple eigenvalue.

So $ F'(x,\lambda) $ must have been regular.


答:
Alternative method:

The area of the shaded region is $\frac{1}{2}r^2\theta,\ $ where $r$ is the radius of the circle and $\theta\ $ is the angle in radians that the region makes at the centre of the circle $O$.

$r = 1,\ $ therefore the area is $\frac{1}{2}\times1^2\times \theta = \frac{1}{2}\theta.$

Now, $\angle POM = \frac{\theta}{2},\ $ so we can express $\frac{\theta}{2}\ $ in terms of the $x-$coordinate of $M = (M_x,M_y)$, $M_x$ say. Looking at the triangle $OPM,\ $ we know that $OP=1,\ $ therefore $\frac{\theta}{2} = \cos^{-1}\left(\frac{M_x}{1}\right) = \cos^{-1}(M_x) \implies \theta = 2\cos^{-1}(M_x).$

So area of shaded region $= \frac{1}{2} \times 2\cos^{-1}(M_x) = \cos^{-1}(M_x).$

Title: Finding the area of shaded region

Body:
Consider a circle with the equation $x^2 + y^2 = 1$. Now find the area of the shaded region which we denote by $u$.

image of circle:

I'm trying to approach this question and would like support on my integral calculation. The difficulty I'm having is with substitution.

For example, given that:

$$u = (OM)(MP)+2\int^{1}_{x} \sqrt{1-x^2} \space dx$$

I should be getting:

$u = xy + [x\sqrt{1-x^2}-\cos^{-1}x]^{1}_{x}$

Although, I currently lack the technical skills to get this answer.

My approach:

I thought that I could substitute back in $x$ for $\sin(x)$ to get $\sqrt{1-\sin^2(x)}$, then when $u = \sin(x) \implies du = \cos(x)$, so that I get:

$$2\int^{1}_{x} \sqrt{1-\sin^2(x)}\cdot \cos(x) \space dx$$

However, this does not lead me close to the answer.

I have also tried inputting this equation into symbolab, and get:

$\arcsin \left(x\right)+\frac{1}{2}\sin \left(2\arcsin \left(x\right)\right)+2C$

Which is also wrong. What might be the approach towards this?

In Progress:

I've currently rearranged my substitution and it looks promising:

$u = 1-\sin^2 (x) \implies du = -cos(x)dx \implies-arcos(x)du = dx$

I cannot manage with this approach neither, as I'll have to integrate the square root.

Looking at the equation, I'll need something like:

$secant(x) -arcos(x)$ to convert the secant into $x\sqrt{x^2-1}$


问:
Two step percentage discount how to arrive to x
Let's say we have a value of $100$ to which we apply a $20$% which arrives at $80$.
Now let's apply a two step percentage decrease so $100 - 10$% - $x$% $= 80$. To get there $x$ would be roughly $11,1$%. Can anyone help how the actual formula should look as for the moment I can only pinpoint that by trial and error?

答:
Start with a number $n$ (=100). We want to reduce it by $k$% (=20%). Call this method 1. Instead, we want to go by method 2 where:

Reduce by $\frac{k}{2}$%.

Then reduce the resultant by $x$%.

We want to find x such that the results of method 1 and method 2 are same.

$n - \frac{k}{100}n = n - \frac{k/2}{100}n - \frac{x}{100}\left[n-\frac{k/2}{100}n\right]$

Simplifying, we get:

$x = \frac{k/2}{1-\frac{k/2}{100}}$

Plugging in $k$ = 20 gives $x$ = 11.111.


问:
H0w have group theory and fractal geometry been combined?
Has there been a significant tie made between group theory and fractal geometry? What are some ways that they have been tied together?
I've been inspired to ask this question by this image of a free group.

答: See the book Indra's pearls by Mumford, Series and Wright for the wonders of Kleinian Groups.</p>


答:
(I assume that $(x,y)$ is in the first open quadrant).

The area of a circular sector with angle $\alpha$ is

$${\frak A}=\alpha/2$$

(in particular, if $\alpha=2\pi$, we get an area equal to $\pi$ in conformity with formula $\pi R^2$ for $R=1$).

As, in triangle $OMP$,

$$\alpha/2=\operatorname{atan}\dfrac{MP}{MO}=\operatorname{atan}\dfrac{y}{x}$$

As a consequence:

$${\frak A}\text{=}\operatorname{atan}\dfrac{y}{x}$$

Title: Finding the area of shaded region


问:
Left coset and right coset of subgroup $H=\langle(234)\rangle$ in alternating group $A_4$
My homework question is:

Partition $G=A_4$ into left cosets of the subgroup $H=\langle (234)\rangle$

but I am not sure how to start with.

I know that $$A_4= \{(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)\},$$ am I right?

Also, the left coset of H is $xH=\{xh:h\in H\}$.

答:
Since $H=\{e,(234),(243)\}$, we get that the four right cosets are
$$\begin{align}
H, &\\
H(12)(34)&=\{(12)(34),(132),(142)\}, \\
H(13)(24)&=\{(13)(24),(143),(123)\},\\
H(14)(23)&=\{(14)(23),(124),(134)\},
\end{align}$$

just by multiplying through by the elements of $V_4$.  (I know that will get me the four cosets since $V_4\cap H=\{e\}$.)

Similarly for the left cosets.


问:
Why does an exponential function eventually get bigger than a quadratic
I have seen the answer to this question and this one.

My $7$th grade son has this question on his homework:

How do you know an exponential expression will eventually be larger than any quadratic expression?

I can explain to him for any particular example such as $3^x$ vs. $10 x^2$ that he can just try different integer values of $x$ until he finds one, e.g. $x=6$. But, how can a $7$th grader understand that it will always be true, even $1.0001^x$ will eventually by greater than $1000 x^2$? They obviously do not know the Binomial Theorem, derivatives, Taylor series, L'Hopital's rule, Limits, etc,

Note: that is the way the problem is stated, it does not say that the base of the exponential expression has to be greater than $1$. Although for base between $0$ and $1$, it is still true that there exists some $x$ where the exponential is larger than the quadratic, the phrase "eventually" makes it sound like there is some $M$ where it is larger for all $x>M$. So, I don't like the way the question is written.

答:
Personally I'd be hesitant to pose such a question to a $7$-th grader. For some this is a big but instructive stretch of the mind, for others the stretch is too big. So my first advice would be

Don't try to force your son to 'get' this.

Reservations aside, here's how I'd approach it: Together you can explore the question

How many steps does it take for such a function to double?

An exponential function grows by $c\%$ at every step. So after some fixed number of steps the output doubles; it isn't difficult to find that it has certainly doubled after $\tfrac{100}{c}$ steps.

This is of course clearer with an example; the output of the function $f(x)=1.02^x$ grows by $2\%$ every time $x$ increases by $1$. So if $x$ increases by $50$ then $f$ increases by at least $100\%$. That is to say $f$ more than doubles every $50$ steps. This is true regardless of the value of $x$! In the same way, every exponential function has some fixed number of steps required to double.

Conclusion 1: An exponential function has a 'doubling number'; every time you increase $x$ by this number, the function doubles.

How many steps do you need for the function $f(x)=Cx^2$ to double, for some constant $C$? This boils down to finding $y$ such that $f(y)=2f(x)$, i.e.
$$Cy^2=2Cx^2,$$
and a bit of algebra then shows that $y=\sqrt{2}x$. So for small $x$ the function doubles quickly, but it takes longer and longer to double. For example, for $x=5$ you find $y\approx7$, so it takes only about $2$ steps for $f$ to double. But from $x=100$ it already takes roughly $41$ steps to double, and from $x=500$ it takes well over $200$ steps!

Conclusion 2: A quadratic function takes longer and longer to double as $x$ increases. The number of steps required to double increases without bound.

So eventually a given quadratic function will take longer to double than a given exponential function. From this point on the exponential function grows faster than the quadratic function, as it doubles faster. And so eventually the exponential function will overtake the quadratic function.


问:
Working backwards with Determinants.
The determinant of A is -2. Find $\det(3A^TA^3)$. You may leave your answer as a product of integers.

What I did was this:

$$\det(3A^TA^3)$$

$$\det(3A^T)\det(A^3)$$

$$3\det(A^T)(\det(A))^3$$

This is the part where I became confused;

$$3\det(A^T)(-2)^3 $$

That's as far as I got, and looking at the solution what I was doing was correct, however the solution for it does this;

$$\det(3A^T)\det(A)^{3}$$
$$(3^3)\det(A)\det(A)^{3}$$
$$(27)(-2)^4$$
$$(27)(16)$$
$$432$$

Now the part that confuses me about this question is , where does the other power of three come from in the line, and what happened to the transpose?

$$(3^3)\det(A)\det(A)^{3}$$

答:
As an answer to your first question, consider the following two facts:

The determinant of a matrix is equal to the determinant of its transpose.

If $A$ is a square matrix, and $B$ is obtained from $A$ by multiplying one row (or column) by a constant $c$, then $\det(B)=c\det(A)$.

Notice that from fact number 2 that an $n\times n$ matrix $A$ satisfies the equation:

$$\det(cA)=c^n\det(A)$$

Title: Working backwards with Determinants.


问:
How to find the number of times a weather station registered a certain temperature?
The problem is as follows:

The temperature in the city of Daegu on March 1st, 2020 is given by:

$15+5\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)$ in celcius, where
$t\in [0,24]$. Assume $t$ is the time elapsed in hours from midnight.
Using this information. Find how many times the temperature of the
city was measured to be $17.5$ celcius.

What I've attempted to do in order to solve this problem isn't much. What I arrived was this:

$$15+5\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=17.5$$

Then this is reduced to:

$$\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=\frac{1}{2}$$

But now what? I could use the inverse sine function as:

$$\frac{\pi t}{12}=\sin ^{-1}\left(\frac{1}{2}\right)-\frac{\pi}{2}$$

But from looking at this effort, I don't seem this would help me to get the value of $t$, needless to say. Assuming that I could get that value. How can I get how many times that temperature was attained?. Therefore I need assistance in this problem.

答:
First note that
$$15+5\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=17.5$$
$$5\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=\frac52$$
$$\sin\left(\frac{\pi t}{12}+\frac{\pi}{2}\right)=\frac12$$
$$\cos\left(\frac{\pi t}{12}\right)=\frac12$$
Which yields
$$\frac{\pi t_1}{12}=\frac\pi3+2\pi k,\ \mbox{for}\ k\in\mathbb Z$$
$$\frac{\pi t_2}{12}=\frac{5\pi}{3}+2\pi k,\ \mbox{for}\ k\in\mathbb Z$$
Solving for $t_1$ and $t_2$, we have
$$t_1=24k+4,\ \mbox{for}\ k\in\mathbb Z$$
$$t_2=24k+20,\ \mbox{for}\ k\in\mathbb Z$$
The only scenario in which $t_i\in[0, 24]$ is when $k=0$. So now we have
$$0\le t_1=4\le 24$$
$$0\le t_2=20\le 24$$
Therefore on March 1st, 2020, there were two occurrences in which the city of Daegu attained a temperature of $17.5^\circ \mathrm C$.


问:
Computing integral $\int_0^1 |x-a| \, {\rm d} x$
I have the following question for a seminar and cannot find the relevant information to solve it.

Solve the integral for all $a \in \mathbb R$ $$I(a)=\int_0^1 |x-a| \, {\rm d} x$$

I understand that I have to treat the absolute value of the function $f(x)=|x-a|$ depending on the when $x$ is less or more then $a$, if $a$ had a defined value I could have solved this, but now im stuck because I dont know how to relate its unspecified value to find a solution for the integral.

Any help is appreciated that'll point me in the right direction.

答:
Compute the integral for $a < 0$, $a \in [0,1]$ and $a > 1$. The solution is

$$I (a) = \begin{cases} \frac12 - a & \text{ if } a < 0\\\\ \frac12 - a + a^2 & \text{ if } a \in [0,1]\\\\ a - \frac12 & \text{ if } a > 1\end{cases}$$


问:
How many elements associated (in the same conjugacy class) with a $\sigma \in S_{3n}$
Let $\sigma = (123)...(3n-2, 3n-1, 3n) \in S_{3n}$

The question is how many elements associated (in the same conjugacy class) with a $\sigma \in S_{3n}$

As I understood we can see how transposition affects our permutation, and then say that every permutation we want to find is product of transpositions.

It worked for $\sigma'=(12...k)$, but I don't know what else I can do here.

答:
It is a fact that two elements in a symmetric group are in the same conjugacy class if and only if they have the same cycle structure.

So all you have to do is compute how many elements of the form $(***)(***)\dots(***)$ the group $S_{3n}$ contains.

This can be done with purely combinatorial methods: let $N$ be this number, then

$$ N = 2^{n}*\frac{1}{n!}{3n \choose 3}{{3n-3}\choose 3}\dots {6 \choose 3}{3 \choose 3} = \frac{2^n}{n!} \prod_{i=0}^{n-1} {{3n-3i}\choose 3}$$

where we multiply by $2$ each factor because given a fixed set of three elements there are two possible distinct cycles that can occur (namely $(123)$ and $(132)$), and we divide by $n!$ since disjoint cycles commute and can be written in any order (namely $(123)(456)=(456)(123)$ )

(thinking about it, there are probably better formulas that use multinomial coefficients: suggestions welcome!)


问:
Existence of continuous function satisfying a condition
After hearing about "locally constant". This question popped up in my mind.

Is there a continuous non-constant function $f: \mathbb{R} \to \mathbb{R}$ which is locally constant everywhere? That is given any $r \in \mathbb{R}$, there exists $\epsilon >0$ such that $f(B_{\epsilon}(r))=\{f(r)\}$. In words, given any $r \in \mathbb{R}$, there is a neighborhood of $r$ where such that the restriction of the function $f$ is constant.

I tried to solve this problem by assuming that $x \neq y$ and then showing that $f(x)=f(y)$, but it is possible that we cannot do that in finite number of steps, and even in countably infinite number of steps. then how can I solve this?

答:
The closest function that comes to mind is the Cantor-Function.
It is continuous, non-constant and its derivative is 0 almost everywhere, in the sense that the derivative doesnt vanish on a lebesgue zero set $A\subset [0,1]$. I hope i didnt miss anything wrong when writing this and could help a little.


问:
Riemann integrable function which is discontinuous at an uncountable number of points on every open set?

Let $n$ be some positive integer. Does there exist a Riemann integrable function
$$f: [0,1]^n \to \mathbb{R} $$
with the following property: for every non-empty open subset of $U \subset (0,1)^n$, the set
$$ \{x \in U \ \mid \ f \text{ is discontinuous at } x \} $$
is uncountable?

I don't really know how to approach this. I know that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere.

I tried to see if, for instance, I can somehow construct a function like this, such as the characteristic function on the Cantor set in $[0,1]$ (which does not satisfy the above property) and extend it to $[0,1]^n$, but I did not succeed.

答:
Here is an example for $n=1$:

It is known that if $S$ is a  $F_\sigma$ set (a countable union of closed sets) in $\Bbb R$, then there is a function $f:\Bbb R \rightarrow\Bbb R$ whose set of discontinuities is $S$. See this post for a reference to a proof.

So, it suffices to find an $F_\sigma$ in $\Bbb R$ of measure zero with the property that its intersection with any non-degenerate open interval is uncountable.  Towards this end, for each pair of distinct rationals, construct a Cantor set between them; then take the (countable) union over all such sets. It is easily verified that this gives what's required.

Also, see Dave Renfro's comments above.


答: If $f:X\to Y$ is a function that is locally constant at every point and $X$ is connected then $f$ is (globally) constant. So the answer is no with domain $\Bbb R$ in the usual topology. If we give it the discrete metric topology then every function on $\Bbb R$ is locally constant trivially. A complete proof for the connected domain case can be found here.

Title: Existence of continuous function satisfying a condition

Body:
After hearing about "locally constant". This question popped up in my mind.

Is there a continuous non-constant function $f: \mathbb{R} \to \mathbb{R}$ which is locally constant everywhere? That is given any $r \in \mathbb{R}$, there exists $\epsilon >0$ such that $f(B_{\epsilon}(r))=\{f(r)\}$. In words, given any $r \in \mathbb{R}$, there is a neighborhood of $r$ where such that the restriction of the function $f$ is constant.

I tried to solve this problem by assuming that $x \neq y$ and then showing that $f(x)=f(y)$, but it is possible that we cannot do that in finite number of steps, and even in countably infinite number of steps. then how can I solve this?


问:
Elementary Row Matrices
Let $A$ =

$$
\begin{align}
\begin{bmatrix}
-4 & 3\\
1 & 0
\end{bmatrix}
\end{align}
$$

Find $2 \times 2$ elementary matrices $E_1$,$E_2$,$E_3$ such that $A$ = $E_1 E_2 E_3$

I figured out the operations which need to be performed which are;

$E_1$ = $R_2 \leftrightarrow R_1$

$E_2$ = $R_2$ = $R_2$ + $4R_1$

$E_3$ = $R_2$ * $\frac{1}{3}$

My question is how would I go about writing the elementary matrices? The solution says that they are;

$E_1$ =
$
\begin{align}
\begin{bmatrix}
1 & -4\\
0 & 1
\end{bmatrix}
\end{align}
$
$E_2$ =
$
\begin{align}
\begin{bmatrix}
3 & 0\\
0 & 1
\end{bmatrix}
\end{align}
$
$E_3$ =
$
\begin{align}
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix}
\end{align}
$

答: Hint: what do elementary matrices correspond to? Can you some how form a correspondence between the row operations you used to reduce the matrix and elementary matrices? In other words, the elementary matrices are related to how $R_{1}$ and $R_{2}$ are manipulated in each row reduction step. </p>

Title: Elementary Row Matrices


问:
Is there an open $S$-immersion from $S$ to $\mathbb A_S^n$
Let $S$ be a scheme, then is there an open $S$-immersion from $S$ to $\mathbb A_S^n$ for some $n\in\mathbb{N}$?

答:
I’m assuming that by “open $S$-immersion” you mean “a $S$- immersion which is an open map”. Let $n \geq 1$ and let $s \in S$ be a point. Consider $i: S \rightarrow \mathbb{A}^n_S$ an open $S$-immersion. Let $G \subset \mathbb{A}^n_S$ be its image, it is an open subset. Then $G \cap \mathbb{A}^n_{k(s)}$ is open in $\mathbb{A}^n_{k(s)}$ and is a point, a contradiction.

It is possible to find a scheme $S$ which is isomorphic to $\mathbb{A}^1_S$, though not over $S$. You can for instance take $S$ to be the spectrum of $\mathbb{Z}[x_1,\ldots,x_n,\ldots]$.


问:
2 numbers are chosen at random from set $\{1,2,3..5n\}$ (without replacement). Find the probability that $n_1^4-n_2^4$ is divisible by 5
All the numbers can be written in the form of $5k,5k+1,5k+2,5k+3,5k+4$

Also the given expression can be written as
$$(n_1-n_2)(n_1+n_2)(n_1^2 + n_2^2)$$

Case 1

Let $n_1-n_2$ be a multiple of 5

Which implies that both numbers have to be of the from $5k$, ie. $n$ possible numbers

Case 2

Let $n_1+n_2$ be a multiple of 5

So $(n_1,n_2)$ can be $(5k,5k), (5k+1, 5k+4), (5k+4, 5k+1), (5k+2,5k+3), (5k+3, 5k+2)$

With $n$ cases for each ordered pair

Case 3

Let $n_1^2 + n_2^2$ be a multiple of 5

So here cases can be 5k and 5k only (IMO)

The cases with $(5k,5k)$ are repeated 3 times, so count that as one case only

So
$$P=\frac {\binom n2 + 2\times \binom n2 +2\times \binom n2}{\binom {5n}{2}}$$

But that doesn’t match with the given answer $\frac{17n-5}{5(5n-1)}$

Where am I going wrong?

Body:
Idea:  separate into two sets $$S_1=\{5,10,15,..,5n\}\\S_2=\{1,2,3,4,6,7,8,9,11,...,5n-1\}$$now note that dividing by $5$ for each number (to the power of 4)has two remainders
$$a=5k+0\to r_1=\underbrace{a^4}_{mod 5}=0\\
a=5k+1\to r_2=\underbrace{a^4}_{mod 5}=1^4=1\\
a=5k+2\to r_3=\underbrace{a^4}_{mod 5}=2^4=16(mod \ 5)=1\\
a=5k+3\to r_4=\underbrace{a^4}_{mod 5}=3^4=81(mod \ 5)=1\\
a=5k+4\to r_5=\underbrace{a^4}_{mod 5}=2^4=256(mod \ 5)=1\\$$ so $n_1^4,n_2^4$ can choose from $S_1$ and $S_2$ such that $5|n_1^4-n_2^4$ now
$$P=\frac { \binom n2 + \binom {4n}2}{\binom {5n}{2}}$$


问:
Why we can write $\frac{\partial \psi}{\partial x}+ \frac{\partial \psi}{\partial y}\frac{dy}{dx}$ as $\frac{d\psi(x,y)}{dx}$?
I'm reading Boyce/Diprima/Meade's book, in this example:

It's not clear to me why we can write $\frac{\partial \psi}{\partial x}+ \frac{\partial \psi}{\partial y}\frac{dy}{dx}$ as $\frac{d\psi(x,y)}{dx}$, I think we can write $\frac{d \psi(x,y)}{d y}\frac{dy}{dx}$ as $\frac{d \psi(x,y)}{d y}$ due to the chain rule, so we get:

$$\frac{d \psi(x,y)}{dx}+ \frac{d \psi(x,y)}{d y}$$

I computed this and obtained the original differential equation, so I might be on the right path. But I still don't get the part where we write it as

$$\frac{d\psi(x,y)}{dx}$$

Can you help me?

答:
The trick here is to separate places in a function (i.e., which argument is which) and the names of variables occupying those places. Suppose, for instance, that we define
$$
f(x, y) = x^3y
$$
and then ask "What's the derivative of $f(y, x)$ with respect to $x$?" Using the chain rule, one might be tempted to say that it's
$$
\frac{\partial f} {\partial x} \frac{dy}{dx} + \frac{\partial f} {\partial y} \frac{dx}{dx},
$$
and untangling that would be a complete mess.

So suppose we agree that if $f$ is a function of two arguments, we'll denote its derivative with respect to the first as $D_1 f$, and similarly for the second. Then the answer becomes clearer, esp. if we add arguments back in place. It's
$$
D_1f (y, x) \frac{dy}{dx} + D_2f (y,x) \frac{dx}{dx},
$$
and assuming that $y$ and $x$ are independent variables here, the first term is zero, so the overall result is just $D_2f (y,x) \frac{dx}{dx}.$

In your case, $x$ and $y$ are used to represent both places in the function $\psi$ and something more --- $y$ is supposed to actually be a function of the variable $x$. With that in mind, let's rewrite

$$
\frac{\partial \psi}{\partial x}+ \frac{\partial \psi}{\partial y}\frac{dy}{dx}
$$

as
$$
D_1 \psi (x, y(x)) + D_2 \psi (x, y(x)) \frac{dy(x)}{dx}.
$$
The is evidently (by the chain rule) the derivative of the function
$$
g(x) = \psi(x, y(x))
$$
with respect to its single argument. It is this function, $g$, that B & D are referring to when at the end they write
$$
\frac{d\psi}{dx}.
$$

It still baffles me that anyone (including me) can make sense of stuff like this, but we all eventually do.


答:
Since $5$ is a prime number
$$
n^4=1\pmod 5
$$
due to Fermat's theorem for all $n\ne 0\pmod5$.

Thus the probability in question is this of the event that the two chosen numbers either both divisible or indivisible by 5:
$$
p=\frac45\cdot\frac{4n-1}{5n-1}+\frac15\cdot\frac{n-1}{5n-1}.
$$

Title: 2 numbers are chosen at random from set $\{1,2,3..5n\}$ (without replacement). Find the probability that $n_1^4-n_2^4$ is divisible by 5

Body:
All the numbers can be written in the form of $5k,5k+1,5k+2,5k+3,5k+4$

Also the given expression can be written as
$$(n_1-n_2)(n_1+n_2)(n_1^2 + n_2^2)$$

Case 1

Let $n_1-n_2$ be a multiple of 5

Which implies that both numbers have to be of the from $5k$, ie. $n$ possible numbers

Case 2

Let $n_1+n_2$ be a multiple of 5

So $(n_1,n_2)$ can be $(5k,5k), (5k+1, 5k+4), (5k+4, 5k+1), (5k+2,5k+3), (5k+3, 5k+2)$

With $n$ cases for each ordered pair

Case 3

Let $n_1^2 + n_2^2$ be a multiple of 5

So here cases can be 5k and 5k only (IMO)

The cases with $(5k,5k)$ are repeated 3 times, so count that as one case only

So
$$P=\frac {\binom n2 + 2\times \binom n2 +2\times \binom n2}{\binom {5n}{2}}$$

But that doesn’t match with the given answer $\frac{17n-5}{5(5n-1)}$

Where am I going wrong?


答:
Take $r$ in the image of $f$ and consider $A=\{x\in\mathbb{R}:f(x)\ne r\}$, which is open in $\mathbb{R}$. Let $x\in\mathbb{R}$ such that $f(x)=r$; then there is a neighborhood $U$ of $x$ such that $f(y)=r$, for every $y\in U$. Therefore also $B=\{x\in\mathbb{R}:f(x)=r\}$ is open. Since $\mathbb{R}$ is connected, the only possibility is that $A=\emptyset$, because $B\ne\emptyset$.

So the only case when the function is locally constant is when it is constant.

More generally, if $f\colon X\to Y$ is a locally constant continuous function between topological spaces (metric spaces, if you prefer) is constant in every connected component of the domain.