问:
Sample mean converges to 1/2 with error less than 0.01 and probability 0.99
The sample mean random variable of $N$ IID random variables with $X_i$ ~ $U(0, 1)$ will converge to 1/2. How many random variables need to be averaged before we can assert that the approximate probability of an error of not more than 0.01 in magnitude is 0.99?

I setup the problem by using the weak law of large numbers like so:
$$
\sigma = \frac{1}{12} \space \mu=\frac{1}{2}
$$
$$
P[|S_N-1/2|<0.01] = 1-\frac{(\frac{1}{12})^2}{N(0.01)^2}=0.99
$$
Solving for $N$, I get $6945$ but apparently the answer is supposed to be $5529$. What am I doing wrong?

答:
The sample mean $$\bar X = \frac{1}{n} \sum_{i=1}^n X_i$$ is approximately normally distributed with mean $\mu = 1/2$, and variance $$\sigma^2 = \frac{\operatorname{Var}[X_i]}{n} = \frac{1}{12n}.$$  Therefore, the criterion we must use is $$0.99 \le \Pr\left[-0.01 \le \bar X - \mu \le 0.01\right].$$  But since $$\frac{\bar X - \mu}{\sigma}$$ is a standard normal random variable with mean $0$ and variance $1$, we have $$0.99 \le \Pr\left[ -\frac{\sqrt{12n}}{100} \le Z \le \frac{\sqrt{12n}}{100}\right] = 1 - 2 \Phi\left(-\frac{\sqrt{12n}}{100}\right)$$ where $\Phi$ is the cumulative distribution function for a standard normal variable.  This leads to the equation $$n \ge \frac{2500}{3} \left(\Phi^{-1}(1/200)\right)^2.$$  When we look up the quantile function, we have $$\Phi^{-1}(1/200) \approx -2.57583,$$ which gives us $n \ge 5529.08$.

Bearing in mind that this is an approximation since the exact distribution of $\bar X$ is a Bates distribution, it is important to note that the large sample size $n$ means that the approximation is fairly precise.  Usually, we would round up the sample size $n$ even though $5529.08$ is closer to $5529$ than to $5530$.  However, an exact calculation yields the following table:

$$\begin{array}{c|c}
n & \Pr[|\bar X - 1/2| \le 1/100] \\
\hline
5528 & 0.98999516849251412362\ldots \\
5529 & 0.99000190666215766225\ldots \\
5530 & 0.99000864018063913425\ldots
\end{array}$$

which shows that $n = 5529$ is the correct answer.

The above calculation may be performed using a computer via the formula

$$\Pr[|\bar X - 1/2| \le 1/100] = 2 F(n, \tfrac{51}{100} n) - 1$$ where $$F(n,x) = \frac{1}{n!} \sum_{k=0}^{\lfloor x \rfloor} (-1)^k \binom{n}{k} (x-k)^n.$$


问:
Graph theory and degree of vertex
A given tree with 9 vertices has precisely five vertices of degree 1, and precisely two vertices of degree 2. The remaining two vertices have degrees $a$ and $b$. Given that $a \le b$, determine $a$ and $b$.

答:
For $d \ge 1$, let $n_d$ be the number of vertices of degree $d$.  Then
\begin{align}
\sum_d n_d &= 9 \\
\sum_d d n_d &= 2(9-1)
\end{align}
Substituting $n_1=5$ and $n_2=2$ yields
$$1\cdot 5 + 2\cdot 2 + a + b = 16,$$
so
$$a + b = 7.$$
Now what does $3 \le a \le b$ imply?

ContentLicense: CC BY-SA 4.0

Title: Graph theory and degree of vertex


问:
Verify an approximation computed on an integer-arithmetic system
Suppose a system which supports only integer-division, and is bounded by the range $[0,M]$.

Given positive integers $x$, $n$, $d$ and $K$, let the following be:

$y = \min(n,x)$

$z = \max(n,x)$

$p = \lfloor\frac{z}{K}\rfloor$

$q = \lfloor\frac{d}{K}\rfloor$

I know that:

$yp\leq M$

$xn>M$ and therefore cannot be computed directly

How can I determine whether or not $\lfloor\frac{yp}{q}\rfloor\leq\lfloor\frac{xn}{d}\rfloor$?

Thank you!

答:
Well in general, a floating-point division can be done with integer functions. Here is a numerical example:

125 / 13 = 9 <-

125 Mod 13 = 8

80 / 13 = 6 <-

80 Mod 13 = 2

20 / 13 = 1 <-

20 Mod 13 = 7

70 / 13 = 5 <-

for 9.615 as 9.62 .


问:
Possibly Möbius Inversion Formula Application
EDIT: I believe I've figured it out! Feel free to take a look in case I've made a mistake.

Problem: Let $n$ and $d$ be positive integers and $m, b \in R$, some ring.  If $F(n) = \sum_{d|n}f(d)$ and $\sum_{d|n}g(d) = mF(n)+b$, then $g(n) = mf(n) + b \iota(n)$.

Proof:

$$\sum_{d|n}g(d) = mF(n)+b$$
$$\Rightarrow$$
$$g(n) = \sum_{d|n}\mu(d)(mF(\frac{n}{d})+b)=m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d)$$
But our other assumption is, $F(n) = \sum_{d|n}f(d) \Rightarrow f(n)=\sum_{d|n}\mu(d)F(\frac{n}{d})$.
Then,
$$m\sum_{d|n}\mu(d)F(\frac{n}{d})+b\sum_{d|n}\mu(d) = mf(n) + b\sum_{d|n}\mu(d)$$
$$\Rightarrow$$
$$g(n) = mf(n) + b\sum_{d|n}\mu(d)$$

But $\iota (n) = 1$ if $n=1$ and $\iota (n) = 0$ if $n>1$, so $\iota(n)$ and $\sum_{d|n}\mu(d)$ have the same property. So we conclude,
$$g(n) = mf(n) + b\iota(n)$$

答:
From the comments I will try to put you in the track:

You correctly have tried to use the Mobius inversion formula getting that
$$f(n)=\sum _{d|n}\mu (n/d)F(d),$$
so, if you try to compute $\sum _{d|n}\mu (n/d)(mF(d)+b),$ and you show that $\iota (n)=\sum _{d|n}\mu (d),$ Hint: Show it for a power of prime and use multiplicativity, then
$$\sum _{d|n}\mu (n/d)(mF(d)+b)=m\sum _{d|n}\mu (n/d)F(d)+\sum _{d|n}\mu (n/d)b=mf(n)+b\iota(n),$$
concluding, again by Mobius, that this is $g(n).$


问:
How to solve $\int \sqrt{1+\sin x}\, dx$?
It's easy to get this:
$$\int \sqrt{1+\sin x}\, dx \\= \int \sqrt{ \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}} + 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\,\, dx  \\ = \int \left | \sin{\frac{x}{2}} + \cos{\frac{x}{2}} \right |\, dx \\= \sqrt{2} \int \left | \sin{\left ( \frac{x}{2} + \frac{\pi}{4} \right )} \right |\, dx$$

So, in fact, my question is how to solve the integrate $\int \left | \sin x \right| dx $.
Or how to deal with the integrate when have absolute in it?

答:
Hint: $|\sin x|=(-1)^n \sin x$ for $\pi n \leqslant x < \pi (n+1)$, where $n=0, \pm 1,\pm2, \cdots$.

To keep antiderivative continuous in $x=\pi (n+1)$ you need to solve
$$\left((-1)^{n+1}\cos x +C_n\right) \Big|_{x=\pi (n+1)}=\left((-1)^{n+2}\cos x +C_{n+1}\right)\Big|_{x=\pi (n+1)}$$
taking some $C=C_0$.


问:
Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$.
I'm trying to do this by induction.  It works for $n=1$ because we get $1 \le 2 - 1 = 1$.  Now suppose for some natural $k \ge 1$, we have $\sum_{i=1}^k  \le 2 - 1/k$.  I must show $\sum_{i=1}^{k+1} \le 2 - 1/(k+1)$.  I started out from the hypothesis and added $1/(k+1)^2$ to both sides of the inequality.  I get

\begin{align*}
  1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{(k+1)^2}
\end{align*}

and it's not obvious what to do from there.  Any hints?

答:
Hint: prove that LHS "grows" slower than RHS

 Prove that $$\frac{1}{n^2} < (2-\frac 1n) - (2-\frac{1}{n-1}), n>1$$


问:
How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral?
So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)

I was hoping someone out there will have a super smart elegant solution to this!

Base step: test when n = 1 ...

Assume true for $n = 1$
ie . $12^k - 4^k - 3^k +1 = 6M$, where $m$ is an integer

RTP: also true for $n = k+1$
ie.  $12^{k+1} - 4^{k+1} - 3^{k+1} +1 = 6N$ where $N$ is an integer

LHS: $12^{k+1} - 4^{k+1} - 3^{k+1} +1$

$= 12( 4^k + 3^k  - 1 + 6M) - 4^{k}(4) - 3^{k}(3) +1$ (from assumption)

$= 6(12M) + 12(4^k) + 12(3^k) -12 -4^{k}(4) - 3^{k}(3) +1$

Here is where I break down and go around in circles.

答:
Work mod 3, and put aside induction for now. Note that  $4 \equiv_3 1$, and so for each integer $n$:

$$4^n \equiv_3 1^n = 1.$$

Thus $12^n -4^n -3^n +1 \equiv_3 (0 -1 -0 -1) = 0$.

So this quantity is divible by 3 for each positive integer $n$.

Lets now get back to mod 6 though. So $12^n -4^n -3^n +1$ is divisible by 3. But it is also even so it is divisible by 6 as well.


问:
What is the Ramanujan summation for the series $\sqrt[n]{2}$
A Ramanujan summation is a

technique invented by the mathematician Srinivasa Ramanujan for assigning a value to divergent infinite series

In my case, I'm interested in assigning a value to the divergent series

$$\sum_{n=1}^\infty f(n) \ \ \ \ \ \ \ \text{where}\ \ \ \ f(n)=\sqrt[n]{2}$$

According to the Wikipedia page (and my understanding), the Ramanujan summation is

$$\sum_{n=1}^\mathfrak{R} f(n)=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N f(n)-\int_{1}^N f(t)dt\Bigg]$$

Thus

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg]$$

Taking the antiderivative

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(\ln2\Big(\text{li}\ 2-\text{Ei}\frac{\ln2}{N}\Big)+N\sqrt[N]{2}-2\Bigg)\Bigg]$$

Moving some constants outside the limit

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=2-\ln2\cdot\text{li}\ 2+\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\Bigg(N\sqrt[N]{2}-\ln2\cdot\text{Ei}\frac{\ln2}{N}\Bigg)\Bigg]$$

It's at this point I'm unsure of how to proceed. I'm not terribly confident what the limit converges to. From my computational estimates up to $N=10^8$, I find that

$$\sum_{n=1}^\mathfrak{R} \sqrt[n]{2}\approx1.6$$

But due to floating point errors or slow convergence, it deviates substantially enough for me to not be confident about any more digits.

I'd like to know if this converges at all, and if it does, is there a (reasonably) closed form / relation to other constants?

答:
I've not yet seen that definition of Ramanujan-summation as in your first formula.

But for testing I've tried your third formula
$$ \sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg] \tag 3
$$
implemented in Pari/GP for high precision.

First, using W|A I've got the following expression for the integral, setting $ß=\log(2)$
$$ \int \exp( \frac ßt) dt = t\exp(\fracßt) - ß \text{Ei}(\fracßt) + const \tag {4.1}
$$
and
$$ \begin{align}I(N)&= \int_{t=1}^{N} \exp( \frac ßt) dt
\\ &= (N\exp(\fracßN) - ß \text{Ei}(\fracßN))&-(1\exp(\fracß1) - ß \text{Ei}(\fracß1))
\\I(N) &=N\exp(\fracßN) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß))  \end{align} \tag {4.2}$$
With this I come for some not too large N in the near of your found value:
$$K(N) = \sum_{k=1}^N \exp( \fracßk) \tag{4.2}$$
and $$S(N) = K(N) - I(N) \tag {4.3}$$

```<code>  N       S(N)
   100  1.60585777814
  1000  1.60273245539
 10000  1.60242047744
100000, 1.60238928520
</code>```

It seems to converge, and to the value that you gave in your OP. But doing the serial summation in $K(N)$ to even higher N (to get more accuracy) is time-consuming, so I reconstruct this series such that Pari/GP can calculate this easier when $N$ is in the billions... This is the rewriting as a double series, where the expression for each $\sqrt[k] 2$ is expanded in the exponential-series on $\fracßk$

$$ \begin{array} {}
  \text{lhs} & =\text{rhs1} &+ \text{rhs2} \\ \hline
 \exp(ß/1) & = 1+ß/1 & +ß^2/1^2/2! &+ß^3/1^3/3! & + \cdots \\
 \exp(ß/2) & = 1+ß/2 & +ß^2/2^2/2! &+ß^3/2^3/3! & + \cdots \\
 \exp(ß/3) & = 1+ß/3 & +ß^2/3^2/2! &+ß^3/3^3/3! & + \cdots \\
 \vdots & \vdots \\
 \exp(ß/N) & = 1+ß/N & +ß^2/N^2/2! &+ß^3/N^3/3! & + \cdots \\
 \vdots & \vdots \\
\end{array} \tag 5$$
Looking at the column-sums we see, that the first two columns $\text{rhs1}$ give divergent series, but the following columns $\text{rhs2}$ are convergent.
So we operate in two parts: evaluate the RHS2 for the limit $N \to \infty$ immediately by the sum of zetas
$$ \text{RHS2}(\infty) = ß^2/2! \zeta(2) + ß^3/3! \zeta(3) + ß^4/4! \zeta(4) + ... \approx 0.473841903568 \tag {5.1}
$$
Since the columnsums in RHS1 diverge we reformulate the columnsums up to the N'th partial sum in immediate values in terms of $N$ and of harmonic-numbers $H(N)=\psi(1+N)+\gamma$ (or H(N)=psi(1+N)+Euler in Pari/GP): $  \text{RHS1}(N)=N+ ß \cdot H(N)$ so
$$ \begin{array} {}K(N) &= N + ß \cdot H(N) &+ 0.473841903568\\
   I(N) &= N\exp(\fracßN) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß)) \\
   S(N) &= K(N)&- I(N) \end{array} \tag {6}$$

giving convergence to the value

```<code>  N    S(N)           difference S(N_{k+1})-S(N_k)
 1000: 1.60297258955
10^5 : 1.60239168746 -0.000580902092269
10^6 : 1.60238640626 -0.00000528119790414
10^12: 1.60238581946 -0.000000586799480429
10^24: 1.60238581946 -5.86800097238 E-13
10^48: 1.60238581946 -5.86800097239 E-25
</code>```

So I think, your own approximation went into the right way.

P.s. In eq.6 the divergence can be reduced by cancellation; the exponential in $I(N)$ be expanded to $N(1+ß/N) + O(1/N)=N + ß + O(1/N)$ and the divergence in $N$ cancels with that $N$ in $K(N)$. Next, the expression $\text{harm}(N)$ can be rewritten as $\log(N)+\gamma$ for $N \to \infty$ and be put together with the $\text{Ei()}$ - expression in $I(N)$ getting $$ \lim_{N\to \infty} \text{Ei}(ß/N) + \log(N) = \gamma + \log(\log(2)) + O(1/N)$$ leading to the shortened version of eq.6
$$ \begin{array} {}
   \lim_{N \to \infty} S(N)
&= & ß \cdot (\log(N)+\gamma) &+ 0.473841903568\\
&&- (  ß + O(1/N) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß))) \\
&=&  ß\gamma -ß+2 -ß\text{Ei}(ß) &+ 0.473841903568\\
&&+  ß (\text{Ei}(\fracßN)+\log(N) ) \\
&=&  2+ß(2\gamma + \log(ß) -1 -\text{Ei}(ß)) &+ 0.473841903568\\
&=&  1.60238581946
 \end{array} \tag {6a}$$ where in the second-last line the divergence $N$ has been cancelled and a constant expression emerged.

Body:
I've not yet seen that definition of Ramanujan-summation as in your first formula.

But for testing I've tried your third formula
$$ \sum_{n=1}^\mathfrak{R} \sqrt[n]{2}=\lim_{N\to\infty}\Bigg[\sum_{n=1}^N \sqrt[n]{2}-\int_{1}^N \sqrt[t]{2}dt\Bigg] \tag 3
$$
implemented in Pari/GP for high precision.

First, using W|A I've got the following expression for the integral, setting $ß=\log(2)$
$$ \int \exp( \frac ßt) dt = t\exp(\fracßt) - ß \text{Ei}(\fracßt) + const \tag {4.1}
$$
and
$$ \begin{align}I(N)&= \int_{t=1}^{N} \exp( \frac ßt) dt
\\ &= (N\exp(\fracßN) - ß \text{Ei}(\fracßN))&-(1\exp(\fracß1) - ß \text{Ei}(\fracß1))
\\I(N) &=N\exp(\fracßN) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß))  \end{align} \tag {4.2}$$
With this I come for some not too large N in the near of your found value:
$$K(N) = \sum_{k=1}^N \exp( \fracßk) \tag{4.2}$$
and $$S(N) = K(N) - I(N) \tag {4.3}$$

  N       S(N)
   100  1.60585777814
  1000  1.60273245539
 10000  1.60242047744
100000, 1.60238928520

It seems to converge, and to the value that you gave in your OP. But doing the serial summation in $K(N)$ to even higher N (to get more accuracy) is time-consuming, so I reconstruct this series such that Pari/GP can calculate this easier when $N$ is in the billions... This is the rewriting as a double series, where the expression for each $\sqrt[k] 2$ is expanded in the exponential-series on $\fracßk$

$$ \begin{array} {}
  \text{lhs} & =\text{rhs1} &+ \text{rhs2} \\ \hline
 \exp(ß/1) & = 1+ß/1 & +ß^2/1^2/2! &+ß^3/1^3/3! & + \cdots \\
 \exp(ß/2) & = 1+ß/2 & +ß^2/2^2/2! &+ß^3/2^3/3! & + \cdots \\
 \exp(ß/3) & = 1+ß/3 & +ß^2/3^2/2! &+ß^3/3^3/3! & + \cdots \\
 \vdots & \vdots \\
 \exp(ß/N) & = 1+ß/N & +ß^2/N^2/2! &+ß^3/N^3/3! & + \cdots \\
 \vdots & \vdots \\
\end{array} \tag 5$$
Looking at the column-sums we see, that the first two columns $\text{rhs1}$ give divergent series, but the following columns $\text{rhs2}$ are convergent.
So we operate in two parts: evaluate the RHS2 for the limit $N \to \infty$ immediately by the sum of zetas
$$ \text{RHS2}(\infty) = ß^2/2! \zeta(2) + ß^3/3! \zeta(3) + ß^4/4! \zeta(4) + ... \approx 0.473841903568 \tag {5.1}
$$
Since the columnsums in RHS1 diverge we reformulate the columnsums up to the N'th partial sum in immediate values in terms of $N$ and of harmonic-numbers $H(N)=\psi(1+N)+\gamma$ (or H(N)=psi(1+N)+Euler in Pari/GP): $  \text{RHS1}(N)=N+ ß \cdot H(N)$ so
$$ \begin{array} {}K(N) &= N + ß \cdot H(N) &+ 0.473841903568\\
   I(N) &= N\exp(\fracßN) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß)) \\
   S(N) &= K(N)&- I(N) \end{array} \tag {6}$$

giving convergence to the value

  N    S(N)           difference S(N_{k+1})-S(N_k)
 1000: 1.60297258955
10^5 : 1.60239168746 -0.000580902092269
10^6 : 1.60238640626 -0.00000528119790414
10^12: 1.60238581946 -0.000000586799480429
10^24: 1.60238581946 -5.86800097238 E-13
10^48: 1.60238581946 -5.86800097239 E-25

So I think, your own approximation went into the right way.

P.s. In eq.6 the divergence can be reduced by cancellation; the exponential in $I(N)$ be expanded to $N(1+ß/N) + O(1/N)=N + ß + O(1/N)$ and the divergence in $N$ cancels with that $N$ in $K(N)$. Next, the expression $\text{harm}(N)$ can be rewritten as $\log(N)+\gamma$ for $N \to \infty$ and be put together with the $\text{Ei()}$ - expression in $I(N)$ getting $$ \lim_{N\to \infty} \text{Ei}(ß/N) + \log(N) = \gamma + \log(\log(2)) + O(1/N)$$ leading to the shortened version of eq.6
$$ \begin{array} {}
   \lim_{N \to \infty} S(N)
&= & ß \cdot (\log(N)+\gamma) &+ 0.473841903568\\
&&- (  ß + O(1/N) -2 &- ß (\text{Ei}(\fracßN)- \text{Ei}(ß))) \\
&=&  ß\gamma -ß+2 -ß\text{Ei}(ß) &+ 0.473841903568\\
&&+  ß (\text{Ei}(\fracßN)+\log(N) ) \\
&=&  2+ß(2\gamma + \log(ß) -1 -\text{Ei}(ß)) &+ 0.473841903568\\
&=&  1.60238581946
 \end{array} \tag {6a}$$ where in the second-last line the divergence $N$ has been cancelled and a constant expression emerged.


问:
I have a question about the identity matrix $I$ being similar to only $I$ (linear algebra, hoffman kunze)
$Recall$ square matrices $A$,$B$ are $similar$ if there exists a invertible matrix $P$ such that $A$=$P^{-1}BP$

$Recall$ Let $T$ is a linear operator on a finite-dimensional vector space $V$ and let $B,B'$ be ordered bases for $V$. then if $[T]_B$,$[T]_{B'}$ are the matrices of $T$ in ordered bases $B,B'$ respectively, these two matrices are similar.

$My$ $Idea$

When $T$ is the identity operator on $V=R^2$ and $B$ is the standard basis {$e_1$,$e_2$} for $R^2$ and $B'$={$e_2,e_1$}, $[T]_B$ is the identity matrix $I$ and $[T]_{B'}$=$$\begin{pmatrix}
0&1\\
1&0
\end{pmatrix}.$$

By the second $Recall$, these two matrices are similar. But they can not be similar ! Because if some matrix $A$ is similar to the identity matrix $I$ then $A$=$I$ by the first $Recall$

$Question$

Am I wrong about the definition of ordered basis?  I don't know where the error is. I got this question while I was looking at the Exercise 11 of sec5.3 in Linear hoffman kunze. If this error is true , the determinant value $detT$ for linear operator $T$ is not unique.

答: Since $Te_2 = e_2 = 1e_2+0e_1$ and $Te_1 = e_1 = 0e_2+1e_1$, then $[T]_{B'}$ is also the identity matrix. Indeed, for any ordered basis $C$ for $\mathbb R^2$ we have that $[T]_C$ is the identity matrix.


答:
As zkutch has given how to calculate with absolute value here is another approach

To avoid confusion its better to do this we have $$\int \frac{\cos x}{\cos x}\sqrt{1+\sin x}dx=\int \frac{\cos x dx}{\sqrt{1-\sin x}}=-2\sqrt{1-\sin x}+C$$
Here in above integral i had put $1-\sin x=t$ which gave answer easily

Title: How to solve $\int \sqrt{1+\sin x}\, dx$?

Body:
It's easy to get this:
$$\int \sqrt{1+\sin x}\, dx \\= \int \sqrt{ \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}} + 2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}\,\, dx  \\ = \int \left | \sin{\frac{x}{2}} + \cos{\frac{x}{2}} \right |\, dx \\= \sqrt{2} \int \left | \sin{\left ( \frac{x}{2} + \frac{\pi}{4} \right )} \right |\, dx$$

So, in fact, my question is how to solve the integrate $\int \left | \sin x \right| dx $.
Or how to deal with the integrate when have absolute in it?


问:
What is the double sum of:
$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \frac{ \sin[ka(m-n)]}{(m-n)} , m \neq n $

where $k$ and $a$ are constants.

How to treat this double sum?

答: Hint: Use the fact that $$\frac{\sin[(m-n)ka]}{m-n} = \frac 1 2\int_{-ka}^{ka}e^{i(m-n)t}dt$$

Title: What is the double sum of:


答:
Hint: To prove the claim that $3\mid4^n-1$, notice that

$$4^{k+1}-1=4(4^k-1)+3$$

Title: How do prove $12^n - 4^n - 3^n +1$ is divisible by 6 using mathematical induction, where n is integral?

Body:
So this question is very challenging because normally the bases of the exponents are the same. There are too many different bases for me to successfully subtitue in the assumption (when $n=k$)

I was hoping someone out there will have a super smart elegant solution to this!

Base step: test when n = 1 ...

Assume true for $n = 1$
ie . $12^k - 4^k - 3^k +1 = 6M$, where $m$ is an integer

RTP: also true for $n = k+1$
ie.  $12^{k+1} - 4^{k+1} - 3^{k+1} +1 = 6N$ where $N$ is an integer

LHS: $12^{k+1} - 4^{k+1} - 3^{k+1} +1$

$= 12( 4^k + 3^k  - 1 + 6M) - 4^{k}(4) - 3^{k}(3) +1$ (from assumption)

$= 6(12M) + 12(4^k) + 12(3^k) -12 -4^{k}(4) - 3^{k}(3) +1$

Here is where I break down and go around in circles.


问:
Show that the following set is countable
Show that the following set is countable:  $\{f ∶ \mathbb{N} \to \mathbb{N} ∶ \forall$ m, n $\in \mathbb{N}$ m $ \leq$ n $\implies$ $f (n) \leq f (m)\}.$

答:
HINT: Let $F$ be the set of functions in question. Don’t try to find an injection from $F$ to $\Bbb N$ directly; use various facts about countable sets that you already know after you prove the following results.

Show that every $f\in F$ is eventually constant, meaning that there are an $n_f\in\Bbb N$ and a $k_f\in\Bbb N$ such that $f(n)=k_f$ for all $n\ge n_f$.

Show that for each $\langle\ell,k,n\rangle$ there are only finitely many $f\in F$ such that $f(0)=\ell$, $k_f=k$, and $n_f=n$. (My $\Bbb N$ includes $0$; if yours starts with $1$, replace $f(0)$ with $f(1)$.)


问:
Is $GL_2(\mathbb{C})\cong \hat{\mathbb{C}}\times \mathbb{C}\times \mathbb{C}^*\times \mathbb{C}^*$?
Is $GL_2(\mathbb{C})\cong \hat{\mathbb{C}}\times \mathbb{C}\times \mathbb{C}^*\times \mathbb{C}^*$ ?

I found that $GL_2(\mathbb{C})/Z_2(\mathbb{C})\cong Mob$, where $Mob$ is the set of Möbius transformations, $Z_2(\mathbb{C})$ is the centre of $GL_2(\mathbb{C})$.
Therefore, $GL_2(\mathbb{C})\cong Z_2(\mathbb{C})\times Mob\cong Z_2(\mathbb{C})\times \left\{f:f=\dfrac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)},z_i\neq z_j, z_i\in\hat{\mathbb{C}}\right\}$
$\cong Z_2(\mathbb{C})\times \hat{\mathbb{C}}\times \mathbb{C}\times \mathbb{C}^*\cong\hat{\mathbb{C}}\times\mathbb{C}\times\mathbb{C}^*\times\mathbb{C}^*$

What if I changed $GL_2(\mathbb{C})$ into $GL_n(\mathbb{C})$, where $n\in\mathbb{N}^*$?

Could there be more general way to prove the case that $n=2$?

$\hat{\mathbb{C}}$ means $\mathbb{C}\cup\{\infty\}$.

答:
A group $G$ is an internal direct product $H\times K$ if every $g\in G$ is uniquely expressible as $g=hk$ and every $h\in H$ and $k\in K$ commute. Any direct product of abelian groups is abelian, so no nonabelian group can be a direct product of abelian groups. A weaker condition is if $G$ is an internal knit product $G=HK$ of two subgroups $H$ and $K$; this happens when every $g\in G$ is uniquely expressible as $g=hk$ for some $h\in H$ and $k\in K$, but with no assumption that $h$s and $k$s commute.

The QR factorization shows $\mathrm{GL}_n\mathbb{C}$ is a knit product of $\mathrm{U}(n)$ and the group $B$ of upper triangular matrices with real positive diagonal entries. This means $\mathrm{GL}_n\mathbb{C}\simeq \mathrm{U}(n)\times B$ is true topologically. Moreover, since positive reals are homeomorphic to all reals (via logarithms), $B\simeq\mathbb{R}^n\times\mathbb{C}^{\binom{n}{2}}\simeq\mathbb{R}^{n^2}$ topologically.

This is also related to the Iwasawa decomposition of a Lie group.

The unitary group $\mathrm{U}(n)$ is not so simple - it's not (topologically) a direct product, but rather a kind of twisted version of a direct product, called a fiber bundle. For illustration, while a band (a cylinder) is a direct product $S^1\times I$ (where $I=[0,1]$), a Mobius band $M$ (a twisted band) is a fiber bundle $I\to M\to S^1$. For $\mathrm{U}(n)$, if we have it act on the unit sphere $S^{2n-1}\subseteq\mathbb{C}^n$, the action is transitive with stabilizer $\mathrm{U}(n-1)$ (of the last standard coordinate basis vector $e_n$) so the orbit-stabilizer theorem says $\mathrm{U}(n)$ is a fiber bundle

$$ \mathrm{U}(n-1)\to\mathrm{U}(n)\to S^{2n-1}. $$

As $\mathrm{U}(1)=S^1$, this means $\mathrm{U}(n)$ is an extremely twisted version of $S^1\times S^3\times\cdots\times S^{2n-1}$. For $n=2$ though, it actually is a knit product $\mathrm{U}(2)=\mathrm{SU}(2)\mathrm{U}(1)$, where we embed $\mathrm{U}(1)\hookrightarrow\mathrm{U}(2)$ in the upper left or lower right corner (with $1$ in the other), and $\mathrm{SU}(2)=S^3$, so $\mathrm{U}(2)=S^3\times S^1$ is a bona fide topological direct product, which means $\mathrm{GL}_2\mathbb{C}\simeq S^3\times S^1\times\mathbb{R}^4$ topologically


问:
Proving the quadratic formula (for dummies)
I have looked at this question, and also at this one, but I don't understand how the quadratic formula can change from $ax^2+bx+c=0$ to $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$. I am not particularly good at maths, so can someone prove the quadratic formula in a simple way, with no complicated words? All help appreciated.</p>

ContentLicense: CC BY-SA 3.0

Body:
Look at each step here:
$$
\begin{align*}
a x^2 + b x + c
  &= 0 \\
a \left( x^2 +  \frac{b}{a} x \right) + c
  &= 0 \\
a \left( x^2 + \frac{b}{a} x + \frac{b^2}{4 a^2} \right) - \frac{b^2}{4 a} + c
  &= 0 \\
a \left( x + \frac{b}{2 a} \right)^2
  &= \frac{b^2}{4 a} - c \\
\left( x + \frac{b}{2 a} \right)^2
  &= \frac{b^2 - 4 a c}{4 a^2} \\
x + \frac{b}{2 a}
  &= \frac{\pm\sqrt{b^2 - 4 a c}}{2 a} \\
x &= \frac{-b \pm\sqrt{b^2 - 4 a c}}{2 a}
\end{align*}
$$


问:
Determine recurrence and transience on an infinite state space
Consider a Markov chain with state space $S = {0, 1, 2, . . .}$ and transition probabilities $\{p_{ij}\}$ given by
\begin{equation*}
\left\{
\begin{array}{rl}
p_{i,i-1} = 1, & \forall i \geq 1,\\ p_{0,i} = q_i & \forall i \geq 1.
\end{array} \right. \end{equation*}
with $\sum_{i=1}^{\infty} q_i = 1$ and $q_i > 0 \ \forall i \geq 1$. Determine which states are recurrent.

Intuitively, I think all states are recurrent, but my proof gives me the opposite answer: all states are transient...Anyway, here is my approach:

For the state $0$: $p_{00}^{(n)} = q_{n-1}p_{n-1,n-2}p_{n-2,n-3}...p_{1,0} = q_{n-1} \cdot (n-1) \cdot 1 = q_{n-1}$ for $n=2,3,4,...$. Thus, $\sum_{n=1}^{\infty} p_{00}^{(n)} = \sum_{n=2}^{\infty} p_{00}^{(n)} = \sum_{n=2}^{\infty} q_{n-1} = \sum_{n=1}^{\infty} q_{n} = 1 < \infty.$ This shows that the state $0$ is transient. Further, it is easy to prove that the Markov chain is indeed irreducible, which then implies that all states are transient.

What is wrong with my argument? Any help is greatly appreciated!

答:
A Markov chain of this form is recurrent. For a state $i>0$, define
$$
\tau_i = \inf\{n>0:X_n=i\mid X_0=i\},
$$
and for positive integers $m$ define
$$
\sigma_m = \inf\{n>0: X_n =0\}.
$$
Then
$$
\mathbb P(\tau_i<\infty) = \mathbb P\left(\bigcup_{m=1}^\infty \{X_{\sigma_m+1}\geqslant i\}\right) = 1,
$$
as by independence,
$$
\mathbb P\left(\bigcap_{m=1}^\infty \{X_{\sigma_m+1}<i\} \right) = \lim_{m\to\infty} \left(\sum_{j=1}^{i-1} q_j\right)^m = 0.
$$
Now, whether the chain is positive or null recurrent is less clear; I believe it depends on the $q_i$, but I am not sure of an example where e.g. $\mathbb E[\tau_1]=\infty$.


问:
Greatest integer function of this limit.
The limit :
$$\lim_{x\rightarrow 0}\left[\dfrac{\tan x}{x}\right]$$

Which somehow tends to 1, Which is the part I can understand graphically but not intuitively

Plugging values for tanx/x for $10^{-3}$,I get 0.017.

By plugging even smaller values I get values still close to 0.017 so the greatest integer function must tend to zero right?

Can someone point out my flawed logic?

答:
You are getting the wrong limit because you are calculating $\tan x$ as if $x$ is measured in degrees, instead of radians.  Note $$0.017 \approx \frac{\pi}{180}.$$

If you calculate it in radians, for small $x$, we have $$\frac{\tan x}{x} \approx 1.$$


答:
We are going to replace $4^n=12^n-3^n+1-6M$

$\begin{align}12^{n+1}-4^{n+1}-3^{n+1}+1
&=12.12^n-4.4^n-3.3^n+1\\
&=12.12^n-4(12^n-3^n+1-6M)-3.3^n+1\\
&=8.12^n+3^n-3+24M
\end{align}$

$12^n$ and $24M$ are obviously divisible by $6$

Notice $3^n-3=3\times\underbrace{(3^{n-1}-1)}_{\text{even}}$ is also divisible by $6$ so you can finish your induction step.


问:
Are (semi)-simple Lie algebras not solvable?
Let $L\ne \{0\}$ be a non-abelian simple Lie algebra, then the only ideals are $L$ and $\{0\}$. We know that $L'$ is the smallest ideal of $L$ such that $L/L'$ is abelian. If $L'=\{0\}$, then $L/L'=L$, but $L$ is non-abelian. So $L'=L$ and hence $L^{(k)}=L\ne \{0\}$ for all $k$, i.e. $L$ is not solvable. So simplicity implies non-solvability.

My question is, can we extend this idea to semi-simple Lie algebras? In other words, is it true that semi-simplicity implies non-solvability?

My definition for semi-simple: $L$ has no non-zero solvable ideals.

答:
As noted in the comments, the answer I give may not coincide with the definitions/assumptions of the book/lecture course you are following.

Suppose $L$ is a non-abelian semi-simple Lie algebra. Then by definition of semi-simplicity (or it is a theorem depending on the approach you take) $L=\bigoplus_{i=1}^n L_i$ for some simple Lie algebras. And because $L$ is non-abelian at least one of the $L_i$ has to be non-abelian.

Because the argument is not really different for bigger $n$, let me for simplicity of notation assume that $L=L_1\oplus L_2$ with $L_1$ non-abelian (and we don't know whether $L_2$ is abelian or not). This means that $L=L_1\oplus L_2$ as vector spaces and additionally $[l_1,l_2]=[l_2,l_1]=0$ for all $l_1\in L_1$ and $l_2\in L_2$.

Now, by definition $L'=[L,L]=[L_1\oplus L_2,L_1\oplus L_2]=[L_1,L_1]+[L_1,L_2]+[L_2,L_1]+[L_2,L_2]$. The last equality follows as the Lie bracket is bilinear. But since we have a direct sum decomposition of Lie algebras, we have $[L_1,L_2]=[L_2,L_1]=0$. Hence $L'=L_1'\oplus L_2'$. Now since $L_1'$ is non-abelian simple, we have that $L_1'=L_1$. Thus $L^{(k)}=L_1\oplus L_2^{(k)}$. In particular, $L$ is not solvable.


答:
Every state is recurrent because the chain is irreducible and $0$ is recurrent; $0$ is recurrent because no matter where you go from $0$ you will go to some state $i$ after which you will return to $0$ in $i$ additional steps. It follows from irreducibility that the whole chain is also recurrent.

Intuitively, without thinking about abstract Markov chain theory, you can explain the recurrence of the entire chain as follows. Beginning at $i$, you will return to zero, and then you will get a trial with probability $\sum_{j=i}^\infty q_j$ to reach $i$ at some point in the next excursion. If you fail, then you return to zero and get another independent trial with the same success probability. Thus you get infinitely many independent trials with a fixed success probability so there will be a success almost surely.

That being said, this chain can definitely be null recurrent. Indeed the expected return time to $0$ is $\sum_i (i+1) q_i$ (one jump to leave, $i$ to come back), which can easily be infinite for instance if $q_i=\frac{1}{\zeta(\alpha)} \frac{1}{i^\alpha}$ and $1<\alpha \leq 2$.

Title: Determine recurrence and transience on an infinite state space

Body:
Consider a Markov chain with state space $S = {0, 1, 2, . . .}$ and transition probabilities $\{p_{ij}\}$ given by
\begin{equation*}
\left\{
\begin{array}{rl}
p_{i,i-1} = 1, & \forall i \geq 1,\\ p_{0,i} = q_i & \forall i \geq 1.
\end{array} \right. \end{equation*}
with $\sum_{i=1}^{\infty} q_i = 1$ and $q_i > 0 \ \forall i \geq 1$. Determine which states are recurrent.

Intuitively, I think all states are recurrent, but my proof gives me the opposite answer: all states are transient...Anyway, here is my approach:

For the state $0$: $p_{00}^{(n)} = q_{n-1}p_{n-1,n-2}p_{n-2,n-3}...p_{1,0} = q_{n-1} \cdot (n-1) \cdot 1 = q_{n-1}$ for $n=2,3,4,...$. Thus, $\sum_{n=1}^{\infty} p_{00}^{(n)} = \sum_{n=2}^{\infty} p_{00}^{(n)} = \sum_{n=2}^{\infty} q_{n-1} = \sum_{n=1}^{\infty} q_{n} = 1 < \infty.$ This shows that the state $0$ is transient. Further, it is easy to prove that the Markov chain is indeed irreducible, which then implies that all states are transient.

What is wrong with my argument? Any help is greatly appreciated!


Body:
Suppose $6 | 12^{n}-4^{n}-3^{n}+1$ for some $n$.

Then $6| 12(12^{n}-4^{n}-3^{n}+1) = 12^{n+1}-12\cdot4^{n}-12\cdot3^{n}+12=$

$12^{n+1}-3\cdot4^{n+1}-4\cdot3^{n+1}+12.$

Since $3^{m} \equiv 3 \pmod 6$, and $4^{m} \equiv 4 \pmod 6$ for every $m$, we have that
$2\cdot4^{n+1} + 3\cdot3^{n+1}-11 \equiv2+3+1\equiv0\pmod 6.$

Hence, $6|(12^{n+1}-3\cdot4^{n+1}-4\cdot3^{n+1}+12)+(2\cdot4^{n+1} + 3\cdot3^{n+1}-11) = 12^{n+1}-4^{n+1}-3^{n+1}+1$.


答:
$$
{1\over(k+1)^2}<{1\over k(k+1)}={k+1-k\over k(k+1)}=\frac1k-{1\over k+1}
$$

As a result, we have

$$
\sum_{i=1}^{k+1}{1\over i^2}\le2-\frac1k+\frac1k-{1\over k+1}=2-{1\over k+1}
$$

Hence, by the principle of mathematical induction, we conclude that the proposition holds for all positive integer $n$.

Title: Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$.

Body:
I'm trying to do this by induction.  It works for $n=1$ because we get $1 \le 2 - 1 = 1$.  Now suppose for some natural $k \ge 1$, we have $\sum_{i=1}^k  \le 2 - 1/k$.  I must show $\sum_{i=1}^{k+1} \le 2 - 1/(k+1)$.  I started out from the hypothesis and added $1/(k+1)^2$ to both sides of the inequality.  I get

\begin{align*}
  1 + \frac{1}{4} + \frac{1}{9} + ... + \frac{1}{k^2} + \frac{1}{(k+1)^2} \le 2 - \frac{1}{k} + \frac{1}{(k+1)^2}
\end{align*}

and it's not obvious what to do from there.  Any hints?


问:
Simple 2-connected graph with diameter 2 and 2n-5 edges
For each $n \ge 5$ find a simple 2-connected graph with $n$ vertices and diameter 2, and with $2n-5$ edges.

I had this as homework, and gave the following solution, which is obviously wrong:

We first construct the graph $(A_1 \cup A_1)*A_{n-4}$ where $K_1$ is the graph with one vertex and $A_{n-4}$ is the graph with $n-4$ vertices and no edges.

The resulting graph has $n-4+2=n-2$ vertices and $2(n-4)=2n-8$ edges, so we need two more vertices and three edges.

My solution was to link them to two central vertices: if $v,u$ the new ones and $x,y$ two random central ones we add the edges $(x,u),(u,v),(v,y)$.

This, while it makes the graph have the desired number of edges, makes the diameter 3.

Choosing the graph $(A_1 \cup A_1)*A_{n-2}$ gives $2n-4$ edges with diameter 2.

Is my approach correct? What am i missing?

答:
I started with the case n=5, the graph C5 is 2-connected and has size m=2(5)-5 = 5.
Let w be a vertex in C5 (deg(w)=2), I added a new vertex v and joined it to w and to one of the two vertices that are not adjacent to w. This new graph is still 2-connected, has diameter 2 and size m=2(6)-5=7. If you continue this process (adding again a new vertex and join it to w and to a vertex nos adyacent to w in the original C5) you will always obtain a graph with diam=2 and m=2(n)-5.</p>


问:
Question about Probabilities
If my team wins 50% of the time when I wear my lucky tie and 90% when I wear my yellow sox, what is the probability of them winning if I wear both.  explain how you got to your response, please</p>

答:
The probability of them winning remains the same, since an individual's choice of clothes does not influence the results of sporting events (unless you're bribing the players).

On a more serious note, you can't calculate this without some more information, since:
$$P(C \ |\  A \cap B) $$
Is not necessarily related to $P(C\ | \ A)$ and $P(C\ | \ B)$, which is the information you provided.

Title: Question about Probabilities


答:
This is what i got.
If there is any error, let me know, please.

Title: What is the double sum of:

Body:
$\sum\limits_{n=0}^\infty \sum\limits_{m=0}^\infty \frac{ \sin[ka(m-n)]}{(m-n)} , m \neq n $

where $k$ and $a$ are constants.

How to treat this double sum?


问:
Show that: $\int_{0}^{1}f\left(x\right)dx\ .\ \int_{0}^{1}g\left(x\right)dx\ \le\ \int_{0}^{1}f\left(x\right)g\left(x\right)dx$
Given f and g defined in $ [0,1] $ continuous and increasing in that interval.
Show that: $\int_{0}^{1}f\left(x\right)dx\ .\ \int_{0}^{1}g\left(x\right)dx\ \le\ \int_{0}^{1}f\left(x\right)g\left(x\right)dx$

I've looked everywhere about the properties of integrals but I haven't found this one. Why?

答:
Since $f, g$ are continuous, there exists $M>0$ such that $|f(x)|, |g(x)| \leq M, \forall x \in [0, 1]$. Consider $$\int_0^1 f(x) dx \int_0^1 g(x) dx \leq \int_0^1 f(x) g(x) dx \iff \int_0^1 (f(x)-M) \int_0^1 (g(x)+M) dx \leq \int_0^1 (f(x)-M)(g(x)+M) dx$$ we can assume that $f \leq 0$ and $g \geq 0$.

Let $H(x) := \int_0^x f(t)g(t) dt - \int_0^x f(t) dt \int_0^x g(t) dt$. Then $H'(x) = f(x)g(x) - f(x)\int_0^x g(t)dt - g(x)\int_0^x f(t)dt = (f(x) - \int_0^x f(t) dt)(g(x) - \int_0^x g(t) dt) - \int_0^xf(t) dt \int_0^x g(t) dt \geq 0, \forall x \in (0, 1)$.

Hence, the inequality holds.


问:
Iff statement for Modulo
I have a question that I've been stuck on for a long time. I wanted to see if friends at Stack Exchange can help me:

Prove that for any odd prime $p$, the congruence $x^{2}+1\equiv0(mod p)$ has a solution if and only if $x^{2}+1\equiv0 (mod p^{2})$ has a solution. So far the forwards proof is giving me some trouble. I have the following down:

Let p be an odd prime. Let n be an integer such that $n^{2}+1\equiv0(mod p)$. This means that there exists an integer k such that $n^{2}+1=pk$. We will show that there exists an integer x such that $x^{2}+1\equiv0(mod p^{2})$.

Let $x=qp+r$ for some integers q and r:

$$(qp+r)^{2}+1\equiv0 (mod p^{2})$$

Then $$2qpr+r^{2}\equiv0(mod p^{2})$$

Setting $r=n$, we get $2qpn+n^{2}+1\equiv0(mod p^{2})$.

This is as far as I have gotten. I'm stuck from this point on: can someone tell me what my next steps should be to complete the forwards proof? Thank you.

Note: this is not really a "proof verification" question since the proof is not fully complete, but I feel that I'm close enough.

答:
The obvious thing to do is apply a Hensel’s-Lemma type of argument to the situation, improving what you have by adding a suitable multiple of $p$ to get something that’s good modulo $p^2$. This is the very sound strategy that motivated you.

But let me show you a sneaky trick that is special to this situation, because the $p$-adic root of your equation $X^2+1$ is a root of unity, namely the fourth root of unity $i$ (or $-i$, naturally).

Your hypothesis is that $x^2+1\equiv0\pmod p$, and I’m about to show you that $(x^p)^2+1\equiv0\pmod{p^2}$. (Check it out for $x=2$, $p=5$.)

The hypothesis can be rewritten $x^2=pr-1$ for some integer $r$. Now,
$$
(x^p)^2=(x^2)^p=(pr-1)^p\,,
$$
and you just expand the right-hand end of this using Binomial Theorem, and see that the result is in the form $p^2R-1$ for a certain integer $R$, and that says that $(x^p)^2\equiv-1\pmod{p^2}$, as desired.

The really wonderful thing about this is that if you take the number you just got, namely $x^p$, and raise it to the $p$-th power, you get something that is an even better root of $X^2+1$.

Title: Iff statement for Modulo


问:
Tangent space of statistical manifold (information geometry)
I am currently reading http://www.danielwagenaar.net/papers/98-Wage2.pdf, where at page 8 it creates a bijection between $T_pM$ and a vector space which basis is $ \partial_\theta l(\theta)$, where $l(\theta)=\ln (p_\theta (x))$ and $p_{\theta}$ is a probability distribution.

My only problem with this is: Why are the derivatives of $l(\theta)$ linearly independent and thus form a vector space? I have been trying to show it using the usual definition of linearly independency but I can not come up with a contradiction.

What is the intuition behind this?

Thanks!

答: $\partial_\theta\ell(\theta)/d\theta^\mu$ is something called a directional derivative. It tells you the rate at which $\ell(\theta)$ changes when you travel in direction $\theta^\mu$. The directional derivative also points in that direction, so $\theta^\mu$ and $\partial_\theta\ell(\theta)/d\theta^\mu$ point in the same direction. Therefore if the $\theta^\mu$ are orthogonal, then so are rhe derivatives in question.


问:
Prove that $\sum_{n=1}^{\infty} \frac{\mu(n)}{10^n}$ is irrational
First of all, I'm aware that this question has been previously asked, (see: show that $\sum \frac {\mu(n)}{10^n}$ is irrational) however I did not find the solutions there particularly useful.

In general, I am comfortable with proving similar statements of this flavour, such as showing that $e$ is irrational, or $sin(1)$ (1 in radians). However this proof is a bit different, since the denominator of interest is not a factorial.

I know that I can think of this as an infinite sequence, and considering $\sum_{n=1}^{\infty} \frac{1+\mu(n)}{10^n}$, I can think of that sum as an infinite ternary sequence of $0,1,2$.

I've tried finding upper and lower bounds for this sum, splitting it into positive and negative sums, and I don't know how to proceed.

答:
The central concern about this problem is to determine whether $1+\mu(n)$ is periodic. If it is periodic, we can soon conclude that this number is rational. We will prove the irrationality by contradicting this statement.

If $1+\mu(n)$ is periodic, then within each period $T$, there should be only a fixed number of occurences of $1+\mu(n)=1$. This indicates that there should be a fixed number of non square-free integers within the period. However, if we were to let $p_k$ be $k$'th prime, we can show that for any $M$>0 the following system of congruences

$$
\begin{cases}
n & \equiv0\pmod{p_k^2} \\
n+1 & \equiv0\pmod{p_{k+1}^2} \\
n+2 & \equiv0\pmod{p_{k+2}^2} \\
\vdots & \vdots \\
n+M &\equiv0\pmod{p_{k+M}^2}
\end{cases}
$$

always has solution by the Chinese remainder theorem, which means that it is possible to have arbitrarily large consecutive non square-free sequence of integers, contradicting our assumption. Therefore, the series

$$
\sum_{n=1}^\infty{\mu(n)\over10^n}
$$

converges to an irrational number.


问:
How to prove a hard inverse trigonometry mathematical induction
$\arctan\left(\frac{1}{2\left(1\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(2\right)^{2}}\right)+\arctan\left(\frac{1}{2\left(3\right)^{2}}\right)+...+\arctan\left(\frac{1}{2\left(n\right)^{2}}\right)=\frac{\pi}{4}-\arctan\left(\frac{1}{2n+1}\right)$

I assume you need to do something like "take tan of both sides" but that doesn't really work for me given the summation on the LHS. Any hints?

答:
I'll assume you were able to prove the base case $n = 1$.

The inductive hypothesis is to assume $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+1}. \quad (1)$$

Then, you need to prove $$\arctan \dfrac{1}{2\cdot1^2}+\cdots+\arctan \dfrac{1}{2\cdot n^2} +\arctan \dfrac{1}{2\cdot (n+1)^2} = \dfrac{\pi}{4} - \arctan \dfrac{1}{2n+3}. \quad (2)$$

If you subtract $(1)$ from $(2)$, you get the equation $$\arctan \dfrac{1}{2\cdot (n+1)^2} = \arctan \dfrac{1}{2n+1} - \arctan \dfrac{1}{2n+3}. \quad (3)$$

So, after assuming the inductive hypothesis $(1)$, you need to prove equation $(3)$. Then you can add equation $(3)$ to equation $(1)$ to prove equation $(2)$, which completes the inductive step.


问:
If two vectors $A=ai-j-4k$ and $B=2i-bj+3k$ and $3A=2B$, what are the values of $a$ and $b$?
I have tried to solve it by first calculating $3A$ and $2B$. As these two vectors are equal their magnitudes should be equal too which gives me the equation

$ 9a^2+153=4b^2+52
\implies9a^2-4b^2=-101 .......(1)$

the answer in the book is given as $a=4/3$ and $b=3/2$. I think these two values should satisfy equation $(1)$. But it doesn't.
Can anyone verify if I am doing the math in the wrong way or the answers are wrong.

答:
Two vectors are equal if and only if each of their components is equal.

The condition $3A=2B$ is saying $$3a\vec i-3\vec j -12\vec k = 4\vec i-2b\vec j+6\vec k.$$ See that the $\vec k$-component is not equal. A contradiction. Thus, there is no such $a$ and $b$.


问:
Probability -- Proving Cantelli’s inequality by setting up a $E[φ(X)]$ and using Markov's inequality
I'm working on proving cantellis inequality. We're told to use this setup:
$P(X >= a) ≤ E[φ(X)]/ φ(α)$ where $var[X] = σ^2$ and $E[X] = 0$ and $φ(x) = (x+c)^2$

An intermediate step was proving that $P(X ≥ α ) ≤ (σ^2+c^2)/(α +c)^2$ which I did successfully.

The next part was finding a value of c that minimizes the right hand side of the equation, and therefore gets the best bound. We're to find the c that gets this bound:

$P(X ≥ α ) ≤ (σ^2)/(α^2 +σ^2)$

The answer is supposed to be $c = σ^2/α$ but I'm getting, by partial differentiation on c and setting that RHS to 0, $c =  ασ^2/(α- σ^2) $

$d/dc ((σ^2 + c^2)/(α + c)^2) = (2c(α+c)^2 - 2(α + c)(σ^2+c^2))/(α + c)^4 = 0$

$2c(α+c)^2 - 2(α + c)(σ^2+c^2) = 0$

$2c(α+2αc+c^2) - 2(ασ^2 + αc^2+ cσ^2+c^3 = 0 $

$2αc+2αc^2+2c^3 - 2ασ^2 - 2αc^2-2cσ^2-2c^3 = 0$

$2αc - 2ασ^2 -2cσ^2 = 0$

$αc - ασ^2 -cσ^2 = 0$

$c(α- σ^2) =  ασ^2 $

$c =  ασ^2/(α- σ^2) $

What did I do wrong?

答:
Given that $$\mathbb P(X\geqslant\alpha)\leqslant\frac{\sigma^2+c^2}{(\alpha+c)^2},$$
we consider the map $f:[0,\infty)\to\mathbb R$ defined by $f(c) = \frac{\sigma^2+c^2}{(\alpha+c)^2}$. Now, $f(0) = \left(\frac\sigma\alpha\right)^2$, and $\lim_{c\to\infty}f(c)=+\infty$. It is clear that $f$ is differentiable on $(0,\infty)$, as a rational function whose denominator is always positive, and so differentiating we find that
$$
f'(c) = \frac{2 \left(\alpha  c-\sigma ^2\right)}{(\alpha +c)^3}.
$$
It follows that $f$ is decreasing on $\left(0, \frac{\sigma^2}\alpha\right)$ and increasing on $\left(\frac{\sigma^2}\alpha,\infty\right)$, and thus attains a global minimum at $c^* = \frac{\sigma^2}\alpha$. We conclude that
$$
\mathbb P(X\geqslant\alpha)\leqslant\inf_{c\geqslant 0} \frac{\sigma^2+c^2}{(\alpha+c)^2} = \frac{\sigma ^2}{\alpha ^2+\sigma ^2}.
$$
(Note that


问:
supremum and infimum for A who is be a nonempty and bounded subset
Let A  be a nonempty and bounded subset on $\mathbb{R}$ and $\lambda \in \mathbb{R}$ whit  $\lambda\leqslant 0$. We difined a set $\lambda$A := { $\lambda$A : $a$ $\in$ A } Prove that sup($\lambda$A) = $\lambda$ ínf A and ínf($\lambda$A) = $\lambda$ sup A

we got a set  $\lambda$A = { $\lambda$A : $a$ $\in$ A }, which is bounded because A is bounded, and how is it bounded the set got sup and inf.
Because A is nonempty, A got a $a \in$ A, so  $\lambda$A its nonempty
Because A is bounded and got a inf, exist a $a' \in \mathbb{R}$ such that $a' \leqslant a $ si $\lambda a'\geqslant \lambda a$ for all $a \in$ A. For that reason $\lambda$A got a sup.

Let $\alpha = sup(\lambda A)$, so $\alpha$ its a upper bound of $\lambda A$ and if we do $\lambda \alpha = \lambda sup(\lambda A)$ so $\lambda \alpha$ its a lower bound of A.
Besides $\beta$ its any lower bound of $\lambda A$ for $\lambda \beta \geqslant \alpha$ and $\lambda \alpha \geqslant \beta$. For that $ \lambda \alpha$ its a inf of A

***Hi I got this problem and my solution but my teacher told me that's wrong. But I don´t know why. I hope you can help me

答:
If $\lambda=0$, $\sup{(\lambda A)}=\lambda\inf A=0$.

Now, if $\lambda<0$, define $\gamma=-\lambda$. So, $\gamma>0$. Now we find $\sup{(\lambda A)}=\sup{(-\gamma A)}=\sup\{-\gamma A: a\in A\}=-\inf\{\gamma A: a\in A\}=-\gamma\inf\{A: a\in A\}=\lambda \inf A$.

We can do a similar thing for the infimum.


问:
Identifying the distribution of the least squares estimate with linear regression
The model is set up as follows:
Random variables $Y_1,Y_2,...,Y_n$ are given by $Y_i=\alpha+\beta x_i+\epsilon_i$ where $\alpha,\beta$ and $x_1,...,x_n$ are constants with $\sum_{i=1}^n x_i=0$ and $\epsilon_1,...,\epsilon_n$ are independent $N(0,\sigma^2)$ random variables.
We define $$\hat{\alpha}=\frac{1}{n}\sum_{i=1}^n Y_i$$ and $$\hat{\beta}=\Big(\sum_{i=1}^n x_iY_i\Big)/\Big(\sum_{i=1}^nx_i^2\big)$$

The question is then to choose the first and second rows of an $n\times n$ orthogonal matrix $A$ appropriately so that $Z=AY$ satisfies $$R^2=\sum_{i=1}^n (Y_i-\hat{\alpha}-\hat{\beta}x_i)^2=\sum_{i=3}^n Z_i^2$$ and hence deduce the joint distribution of $\hat{\alpha},\hat{\beta}$ and $R^2/\sigma^2$.

So far, I have found the joint distribution of $\hat{\alpha}$ and $\hat{\beta}$, and computed their covariance matrix.

I then chose the first two rows of $A$ to be $\big(\frac{1}{\sqrt{n}},..., \frac{1}{\sqrt{n}}\big)$ and $\Big(\frac{x_1}{\sqrt{\sum x_i^2}},..., \frac{x_n}{\sqrt{\sum x_i^2}}\Big)$, respectively, which gave me the desired relation $R^2=\sum_{i=1}^n (Y_i-\hat{\alpha}-\hat{\beta}x_i)^2=\sum_{i=3}^n Z_i^2$.

This is when I get confused. Because $A$ is orthogonal, $Z_1,...,Z_n$ are independent and identically distributed $N(0,\sigma^2)$ random variables. Then $R^2/\sigma^2$ is $\chi_{n-2}^2$, and as this is independent from the $\hat{\alpha},\hat{\beta}$ (since the $Z_i$ are independent) we're done. But multiplying out the first row of $A$ with $Y$ in fact gives $Z_1=\sqrt{n}\hat{\alpha}$, which has mean $\alpha$, not $0$, which obviously does not make sense.

Should I choose my rows of $A$ slightly differently? The only thing is that choosing them in the way I have makes it clear that $A$ is in fact orthogonal, and it is really straightforward to deduce the relation $R^2=\sum_{i=1}^n (Y_i-\hat{\alpha}-\hat{\beta}x_i)^2=\sum_{i=3}^n Z_i^2$.

答:
A year late, but: Your error is in concluding that

$Z_1,\ldots,Z_n$ are independent and identically distributed $N(0,\sigma^2)$ random variables.

Given that $A$ is orthogonal and that $\operatorname{Cov}(Y)=\sigma^2I$ you can assert that $Z:=AY$ has covariance matrix
$$\operatorname{Cov}(Z)=\operatorname{Cov}(AY)=A\operatorname{Cov}(Y)A^T=A(\sigma^2I)A^T=\sigma^2I.$$
Since the joint distribution of the $Y$'s is multivariate normal, it follows that the $Z$'s are independent normal with common variance $\sigma^2$. However, this doesn't say anything about their means.


问:
Find a convergent subsequence in the sequence of rational primes in $\mathbb{Q}_p$
In this post, my main question is how to solve the following problem:

Consider the sequence of primes $2, 3, 5, 7, 11, \ldots$ in the $p$-adic field $\mathbb{Q}_p$,  find a convergent subsequence.

My attempts: Since $\mathbb{Q}_p$ is complete, it is enough for us to construct a subsequence which is a Cauchy sequence. So after all, I have to obtain an estimate of the gaps between primes (in the $p$-adic sense). Namely, arbitrarily picking two primes $p_{m}$ and $p_{n}$, we need to estimate the gap
$$\mathrm{Gap}(m,n) :=  \vert p_m - p_n \vert_p.$$
By the non-archimedian-ness of the absolute value $\vert\cdot \vert_p$, we have
$$
\mathrm{Gap}(m,n) :=  \vert p_m - p_n \vert_p \leq \max\{ \vert p_m \vert_p, \vert p_n \vert_p \} = 1
$$
if one of the primes $p_m$ is not $p$ itself. But this wild estimate is far from enough. I'm been stucking here and making no progress besides this.

Thank you all for commenting and answering this question! :)

答:
Define $a_1 :=$ the smallest prime $\neq p$, and if we have $a_m$, look at the set $\lbrace a_m + k p^{m+1} : k \in \mathbb N\rbrace$. By Dirichlet, it contains a prime (actually infinitely many); choose one of them which is $\neq p$, call it $a_{m+1}$.

The sequence $(a_i)_i$ is a subsequence of the sequence of all primes, and by definition, $\lvert a_m -a_{m+1}\rvert_p \le p^{-m-1}$.


问:
Is the contrapositive of a true statement always provable?
I had a very difficult time proving the contrapositive of a simple elementary number theory statement, and the more I think about, the more I believe it might not be provable that way using any proof techniques (that we accept as valid). I know there are true statements that aren't provable, but I'm not sure how much it relates to this. I have heard of something about models, or say, a model of Number Theory: NT. Are there generalizations of types of true statements we can prove in number theory in certain ways but not in others?</p>

答:
As others have noted in the comments, using the adjective true opens a can of worms: truth has a technical definition in semantics, and it almost certainly clashes with the question you want to ask. In particular, true things need not be provable, period.

Instead, let me state a proof-theoretic question that (I think) cuts close to the one you intend to ask:

I have managed to prove an implication $A \rightarrow B$, but I could do it only by taking the contrapositive $\neg B \rightarrow \neg A$, and proving that. In such a situation, can I always find a a more direct proof of $A \rightarrow B$ that does not involve taking contrapositives?

Based on your question, your gut feeling tells you that this need not be the case, and your gut feeling is correct. Sometimes, every proof of a statement requires you to take contrapositives.

Before I explain why, I have to explain some things regarding proof-theoretic terminology. To investigate proofs rigorously, we first have to define what we mean by proof. There is a large variety of such definitions, called proof systems (deductive systems).

Some proof systems define obscure or extremely constrained forms of reasoning, such as pedagogical reasoning, where you have to give explicit examples before introducing abstract concepts; or ethical reasoning, where you have to distinguish between permissible and forbidden statements. Different proof systems may prove different theorems. But even two proof systems which prove the same theorems can differ substantially from each other: a theorem can have multiple proofs in one system, yet only one proof in some other system. Proof theorists may whip up new deductive systems on demand, similarly to the way a group theorist would construct new groups to illustrate technical situations or to provide (counter)examples to mathematical conjectures and questions.

With these differences in mind, it should be clear that we'll have to fix a proof system up front to say anything sensible. From here on, I will focus on one specific proof system: the natural deduction of Gentzen and Prawitz. Natural deduction provides a rigorous substrate to the kind of proof you're likely to be interested in: mathematical proof of the kind that would be accepted in an ordinary mathematics textbook or journal article.

II. Writing a thorough introduction to natural deduction, let alone one that fits in a Math.SE answer, is a considerable challenge, so much so that I won't even try. If you can read and write mathematical proofs, you are equipped to understand natural deduction. You might be able to pick up how it works based purely on what's written below, but if you wish to learn all the rules and understand it more thoroughly, there are many tutorials, YouTube videos, and myriads of textbooks on the subject. I personally recommend the first three chapters of Jan von Plato's Elements of Logical Reasoning.

Natural deduction has many inference rules, all of them familiar to the working mathematician. These inference rules tell you how to construct new proofs from existing proofs. Each connective (conjunction $\wedge$, disjunction $\vee$, implication $\rightarrow$, negation $\neg$) and quantifier (universal $\forall$, existential $\exists$) comes equipped with one or more so-called introduction rules and one elimination rule. Examples:

If you have a proof of $A$ and you have a proof of $A \rightarrow B$, then you have managed to prove that $B$ holds (implication elimination).

If you have a proof that $A \wedge B$ ($A$ and $B$) holds, then you have managed to prove that $B$ holds (conjunction elimination, right).

If you have a proof that $A \wedge B$ holds, then you have managed to prove that $A$ holds (conjunction elimination, left).

If you have a proof that $A$ holds, and you have a proof that $B$ holds, then you have managed to prove that $A \wedge B$ holds (conjunction introduction).

If you have a proof that starts with "Suppose $A$..." and ends with "... therefore $B$", then you have managed to prove the conditional statement $A \rightarrow B$ (implication introduction).

If you have a proof that starts with "Suppose $A$..." and ends with "which is a contradiction", then you have managed to prove the negation $\neg A$ (negation introduction).

and so on...

Beyond these ordinary rules, there are three special rules as well:

If you have a proof that $\neg\neg A$ holds, then you have managed to prove that $A$ holds (double negation elimination).

You always have a proof of $A \vee \neg A$ (law of excluded middle).

If you have a proof that $\neg B \rightarrow \neg A$, then you have managed to prove that $A \rightarrow B$ (proof by contrapositive).

If you add any one of these rules to the ordinary rules of natural deduction, you can prove the other two. For example, if you add the law of excluded middle as a rule of inference, you can prove every instance of double negation elimination. Most importantly for us, you can use proof by contrapositive to prove every instance of double negation elimination. We can use this auxiliary proof:

Suppose that $\neg A$ holds.

Suppose that $\neg\neg A$ holds.

From 1 and 2 we have a contradiction.

From 2-3 we have $\neg\neg\neg A$ by negation introduction.

From 1-4 we have $\neg A \rightarrow \neg\neg\neg A$ by implication introduction.

From 5 we have $\neg\neg A \rightarrow A$ by contrapositive.

Now, imagine that we have a proof of $\neg\neg A$. Our auxiliary proof gives us $\neg\neg A \rightarrow A$, so implication elimination gives us a proof of $A$. So, as I claimed, you can use the proof by contrapositive inference rule, along with the ordinary rules, to prove every instance of double negation elimination.

The natural deduction proof system that has all the ordinary introduction and elimination rules, but none of the three special rules, is called Intuitionistic or Constructive Natural Deduction. We sometimes call the system that includes one of the special rules as well (most commonly double negation elimination, for technical reasons) Classical Natural Deduction. It is a well-known fact that $\neg\neg A \rightarrow A$ is not provable in Intuitionistic Natural Deduction. So in a very real sense, $\neg\neg A \rightarrow A$ is not provable without taking contrapositives, or one of the other two rules of inference equivalent to it.

III. If you put the axioms of elementary number theory into an Intuitionistic Natural Deduction proof system, you end up with a mathematical theory called Heyting Arithmetic. Keep in mind: just because a general logical principle is not provable without taking contrapositives, it does not follow that no instance is provable without taking contrapositives! E.g. Heyting arithmetic proves $\neg\neg t = 0 \rightarrow t = 0$ for any number $t$.

However, Heyting Arithmetic differs from usual (Peano) elementary arithmetic in a number of ways. In particular, Heyting Arithmetic does not prove the following:

For every polynomial $P$ with integer coefficients, there are integers $n$ such that for all $x$, $|P(n)| \leq |P(x)|$. [2]

This provides an answer to one of the weaker formulations of your question: every proof of the theorem above requires you to take contrapositives (or use an equivalent principle) somewhere in its proof.

I won't go into details, but a proof-theoretic result of De Jongh [1] allows us to answer a stronger formulation: we can also obtain explicit number-theoretic statements $A,B$ such that Heyting arithmetic proves the implication $\neg B \rightarrow \neg A$, but not the implication $A \rightarrow B$.

You could ask an even stronger question: find explicit number-theoretic statements $A,B$ such that every (sufficiently normalized) proof of $A \rightarrow B$ contains a proof of $\neg B \rightarrow \neg A$ itself as a subproof. Good luck with that: it sounds very very difficult :)

[1] De Jongh, D.H.J.: The maximality of the intuitionistic predicate calculus with respect to Heyting's arithmetic, tech. rep., Meeting of the Association for Symbolic Logic, Manchester UK, 1969

[2] Friedman, H.: Classical/Constructive Arithmetic, FOM mailing list, 18 Mar 2006


答:
HINT: Let $P_0$ be the set of primes. Given an infinite $P_k\subseteq P_0$ for some $k\ge 0$, there is an $r_{k+1}\in\{1,\ldots,p^{k+1}-1\}$ such that

$$P_{k+1}=\{n\in P_k:p_n\equiv r_{k+1}\pmod{p^{k+1}}\}$$

is infinite. Now use the sets $P_k$ to construct your Cauchy sequence.

Title: Find a convergent subsequence in the sequence of rational primes in $\mathbb{Q}_p$

Body:
In this post, my main question is how to solve the following problem:

Consider the sequence of primes $2, 3, 5, 7, 11, \ldots$ in the $p$-adic field $\mathbb{Q}_p$,  find a convergent subsequence.

My attempts: Since $\mathbb{Q}_p$ is complete, it is enough for us to construct a subsequence which is a Cauchy sequence. So after all, I have to obtain an estimate of the gaps between primes (in the $p$-adic sense). Namely, arbitrarily picking two primes $p_{m}$ and $p_{n}$, we need to estimate the gap
$$\mathrm{Gap}(m,n) :=  \vert p_m - p_n \vert_p.$$
By the non-archimedian-ness of the absolute value $\vert\cdot \vert_p$, we have
$$
\mathrm{Gap}(m,n) :=  \vert p_m - p_n \vert_p \leq \max\{ \vert p_m \vert_p, \vert p_n \vert_p \} = 1
$$
if one of the primes $p_m$ is not $p$ itself. But this wild estimate is far from enough. I'm been stucking here and making no progress besides this.

Thank you all for commenting and answering this question! :)


问:
Keyhole contour for $\frac{(\log{x})^2}{1+x^3}$
I am trying to evaluate the integral
$\displaystyle \int_0^{\infty}\frac{\log x}{x^3+1}dx$
and I was trying the keyhole contour with $\displaystyle f(z)=\frac{(\log z)^2}{z^3+1}$

The large circular contour $\gamma_R:t\mapsto Re^{it}$ and the small circular contour $\gamma_{\epsilon}:t\to \epsilon e^{it}$ around $0$ both have integrals which tend to $0$ as we let $R\to \infty$ and $\epsilon \to 0$.

The three residues at $z=-1, e^{i\pi/3}, e^{-i\pi/3}$ are

$\displaystyle \frac{(\log i)^2}{3i^2}=-\frac{\pi^2}{3}, \frac{(\log e^{i\pi/3})^2}{3(e^{i\pi/3})^2}=-\frac{\pi^2}{27}e^{-2i\pi/3}$ and $\displaystyle -\frac{\pi^2}{27}e^{2i\pi/3}$

The path above the branch (i.e. the positive real axes) converges to $\displaystyle \int_0^{\infty}\frac{(\log x)^2}{x^3+1}dx$ and the path below the branch $\displaystyle -\int_0^{\infty}\frac{(\log x+2i\pi)^2}{x^3+1}dx$

So when we add everything up we get
$\displaystyle -\int_0^{\infty}\frac{4i\pi\log x-4\pi^2}{x^3+1}dx=2i\pi[-\pi^2/3-\pi^2/27(e^{-2\pi i/3}+e^{2\pi i/3})]=2i\pi\left(-\frac{\pi^2}{3}+\frac{\pi^2}{27}\right)$

By taking the imaginary part, this would (incorrectly) seem to imply that
$\displaystyle 4\pi\int_0^{\infty}\frac{\log x}{x^3+1}dx=-2\pi\frac{8\pi^2}{27}$

$\displaystyle \int_0^{\infty}\frac{\log x}{x^3+1}dx=\frac{4\pi^2}{27}$

(The real answer is $\frac{2\pi^2}{27}$)

This also (again incorrectly) implies that

$\displaystyle \int_0^{\infty}\frac{1}{x^3+1}dx=0$

(The real answer is $\frac{2\pi}{3\sqrt{3}}$)

as the real part is $0$ on the right hand side.

I have verified that these answers are in fact wrong.

I appreciate that there are alternative contours to evaluate this integral but I would like to know what I missed in this particular computation.

答:
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\on}[1]{\operatorname{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, $\ds{\ln}$ is the logarithm principal branch and $\ds{\cal C}$ is a key-hole contour which "takes care" of the above mentioned branch-cut.

Contributions from the big arc and the smal arc around $\ds{z = 0}$ vanish out in the,respectively, radius $\ds{\to \infty}$ and radius $\ds{\to 0^{+}}$.

Poles of the integrand are, $\underline{according}$ to the above mentioned branch-cut, are
$\ds{r = \exp\pars{\pm{2\pi \over 3}\,\ic}}$.

Namely,
\begin{align}
&\bbox[5mm,#ffd]{\oint_{\cal C}{\ln^{2}\pars{z} \over
z^{3} - 1}\,\dd z}
\\ = &\
2\pi\ic\sum_{r = \exp\pars{\pm 2\pi\ic/3}}\,\,\,\,
\lim_{z \to r}\bracks{\pars{z - r}\,{\ln^{2}\pars{z} \over z^{3} - 1}}
\\[5mm] = &\
2\pi\ic\sum_{r = \exp\pars{\pm 2\pi\ic/3}}\,\,\,\,
\bracks{-4\pi^{2}/9 \over 3r^{2}} =
-\,{8\pi^{3} \over 27}\,\ic
\sum_{r = \expo{\pm 2\pi\ic/3}}\,\,\,\,r
\\[5mm] = &\
-\,{16\pi^{3} \over 27}\,\ic\cos\pars{2\pi \over 3} =
{8\pi^{3} \over 27}\,\ic\label{1}\tag{1}
\end{align}

\begin{align}
&\bbox[5mm,#ffd]{\oint_{\cal C}{\ln^{2}\pars{z} \over
z^{3} - 1}\,\dd z}
\\ = &\
\int_{-\infty}^{0}{\bracks{\ln\pars{-x} + \ic\pi}^{2} \over
x^{3} - 1}\,\dd x +
\int_{0}^{-\infty}{\bracks{\ln\pars{-x} - \ic\pi}^{2} \over
x^{3} - 1}\,\dd x
\\[3mm] = &\
-\int_{0}^{\infty}{\bracks{\ln\pars{x} + \ic\pi}^{2} \over
x^{3} + 1}\,\dd x +
\int_{0}^{\infty}{\bracks{\ln\pars{x} - \ic\pi}^{2} \over
x^{3} + 1}\,\dd x
\\[5mm] = &
-4\pi\ic\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1}
\,\dd x\label{2}\tag{2}
\end{align}

(\ref{1}) and (\ref{2}) yield:
\begin{align}
&\int_{0}^{\infty}{\ln\pars{x} \over x^{3} + 1}
\,\dd x =
{8\pi^{3}\,\ic\,/27 \over -4\pi\ic} =
\bbx{-\,{2\pi^{2} \over 27}} \approx -0.7311 \\ &
\end{align}


问:
Kronecker Delta help
Find the value of $\delta_{ii}\delta_{jk}\delta_{jk}$.

Apparently this is $9$, but I can't see why:

$\delta_{ii} = 3$ (as in this case this is defined in $\mathbb{R}^3$)

$\delta_{jk}\delta_{jk} = \delta_{jk}$

$\delta_{ii}\delta_{jk} = 3\delta_{jk}$

but is this not just 3 x the identity matrix? Why is this a direct number?

答:
Note that in the summation, $i,j,k$ each appear twice. According to the Einstein summation convention, we take a sum over every variable. That is, the expression should be interpreted as
$$
\sum_{i = 1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \delta_{ii} \delta_{jk}\delta_{jk}
=
\sum_{i = 1}^3 \sum_{j=1}^3 \sum_{k=1}^3 \delta_{ii} \delta_{jk}.
$$
We can see that this sum is indeed equal to $9$.

Title: Kronecker Delta help


问:
converting con-convex region to convex
I'm trying to model a problem using Linear Programming theory, though the feasible region of the problem is non-convex. Yet, I think using Big-M and some binary variables this region can be converted to a convex feasible region.
Is there any way to convert the following non-convex region to a convex region (with higher dimension) using integer programming "if-then" constraints?

答:
You can introduce a binary variable $z$, where $z=1$ indicates that the point $(x,y)$ is in the $1 \times 3$ rectangle and $z=0$ indicates that $(x,y)$ is in the trapezoid.  Now impose the following linear big-M constraints to enforce that:
\begin{align}
0 \le x &\le 2 + z \\
x - y &\le 3z \\
1-z \le y &\le 2-z \\
\end{align}
Checking the two cases, we see that
\begin{align}
z=1 &\implies (x,y) \in [0,3] \times [0,1] \\
z=0 &\implies x \in [0,y] \land y \in [1,2] \\
\end{align}
More generally, you can represent a union of polyhedra $P_i$ by introducing a binary variable $z_i$ for each one, along with linear constraints $\sum_i z_i=1$ and linear big-M constraints that enforce $z_i=1 \implies x \in P_i$.

Title: converting con-convex region to convex


问:
Show that there exists $N\in \mathbb N$ such that $A_N \subseteq U$.
Let $A_n$ be a sequence of nonempty compact subset of $\mathbb R$ with $A_{n+1}\subseteq A_n$. Let $A=\cap_{n\in \mathbb N}A_n$ Let $U $ be an open subset of $\mathbb R$ such that $A\subseteq U.$ Show that there exists $N\in \mathbb N$ such that $A_N \subseteq U$.

My attempt

$A=\cap_{n\in \mathbb N}A_n \subseteq A_n, \forall n\in \mathbb N.$

Since each $A_n$ is closed and bounded by Heine Borel Theorem. Hence, $A$ is also bounded. Arbitrary intersection closed subset of $\mathbb R$ is closed. Then, $A$ is closed.
Hence, By Heine Borel Theorem $A$ is compact. I don't know how to proceed from here. May I get help?

答: HINT: If $A_n\nsubseteq U$ for each $n\in\Bbb N$, we can pick a point $x_n\in A_n\setminus U$ for each $n\in\Bbb N$. Now use the sequence $\langle x_n:n\in\Bbb N\rangle$ to get a contradiction. You’ll need to use the fact that $A_0$ is compact. (My $\Bbb N$ includes $0$; if yours does not, replace $A_0$ by $A_1$.)


问:
Proving the quadratic formula (for dummies)
I have looked at this question, and also at this one, but I don't understand how the quadratic formula can change from $ax^2+bx+c=0$ to $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$. I am not particularly good at maths, so can someone prove the quadratic formula in a simple way, with no complicated words? All help appreciated.</p>

答:
You should work this way to prove it:
$ax^2+bx+c=0 \quad \iff \quad (\sqrt{a}x + \frac{1}{2}b/\sqrt{a})^2 + (c-\frac{1}{4}b^2/a) = 0$

$\iff \quad (\sqrt{a}x + \frac{1}{2}b/\sqrt{a})= + - \sqrt{\frac{1}{4}b^2/a -c}$

$\iff \quad \sqrt{a}x = -\frac{1}{2}b/\sqrt{a} + - \sqrt{\frac{1}{4}b^2/a -c} $

$\iff \quad x=-\frac{1}{2}b/a +- \sqrt{\frac{1}{4}b^2/a -c}/\sqrt{a}$

$\iff \quad x=-\frac{1}{2}b/a +- \sqrt{\frac{1}{4}b^2/a^2 -c/a}$

$\iff \quad \dots$

After some manipulations you will end up with the desired result.


问:
Measure that agrees with Lebesgue measure
Suppose $\mu $ is a measure defined on $(\mathbb{R} , \mathcal{B}( \mathbb{R}) ) $ such that $\mu (I) =\lambda (I) $ for every interval $I$ of $\mathbb{R} $ (where $\lambda $ is the Lebesgue measure) then how can I show $\mu =\lambda $.
While looking for this I came across the monotone class theorem. This does the trick but I was wondering whether there was quite a simple way to extend the result from intervals to any borel set.
I also thought about proving it using the fact that the smallest dynkin class generated a $\pi - $ system is precisely the smallest sigma algebra generated by the pi system.
But can we say that {$A \in \mathcal{B} (\mathbb{R}) : \mu(A) =\lambda (A) \
 $} is a dynkin class - I'm unsure as the measures are not finite.

Any help would be appreciated.

答:
(1).  For $E\subset \Bbb R$ let $O_E$ be the set of all open $O$ such that $E\subset O ,$  and let $C_E$ be the set of all compact $C$ such that $C\subset E.$ For any measure $m$ with $dom(m) \supset \mathcal B(\Bbb R),$ let $m^o(E)=\inf \{m(O): O\in O_E\}$ and $m^i(E)=\sup\{m(C):C\in C_E\}.$ Obviously, if $E\in dom(m)$ then $m^i(E)\le m(E)\le m^o(E).$

(2). $\mu$ agrees with $\lambda$ on bounded intervals, so $\mu$ agrees with $\lambda$ on any convex set $L$ because $L=\cup M$ where $M=\{L\cap [z,z+1):z\in \Bbb Z\}$ is a countable pairwise-disjoint family of bounded intervals.

(3). For any open $O$ there is a countable pairwise-disjoint family $G$ of convex open sets with $O=\cup G,$ so by (2) we have $\lambda(O)=\sum_{g\in G}\lambda(g)=\sum_{g\in G}\mu(g)=\mu(O).$

(4). By (3), therefore $\lambda^o(E)=\mu^o(E)$ for any $E\subset \Bbb R. $

(5). For any compact $C\subset \Bbb R$ take  a bounded open interval $J$ with $C\subset J. $ and let $D=J \setminus C.$ Then $D$ is also open so by (3) we have  $\mu(C)+\lambda(D)=\mu(C)+\mu(D)=\mu(J)=\lambda(J)=\lambda(C)+\lambda(D).$

So $\mu(C)=\lambda(C).$

(6). Let $E\subset \Bbb R$ be bounded.  Let $J$ be a bounded open interval with $\overline E\subset J.$ Let $C'_E=\{C\in C_E:C\subset J\}.$ Then by (5) we have $\mu^i(E)=\sup \{\mu(C): C\in C'_E\} =\sup  \{\lambda(C):C\in C'_E\}=\lambda^i(E).$

(7). $\lambda$ is outer-regular and inner-Radon-regular. That is, if $E\in dom(\lambda)$ then $\lambda(E)=\lambda^o(E)=\lambda^i(E).$ So if $E$ is a bounded Borel set then by (6) and (4):$$\lambda(E)=\lambda^i(E)=\mu^i(E)\le \mu(E)\le \mu^o(E)=\lambda^o(E)=\lambda(E)$$ and therefore $\mu(E)=\lambda(E).$ So for any Borel set $E$ we have $$\mu(E)=\sum_{z \in \Bbb Z}\mu(E\cap [z,z+1)\,)=\sum_{z\in \Bbb Z}\lambda(E\cap [z,z+1)\,)=\lambda(E).$$


问:
Relation between characteristic polynomials of $A$ and the linear map $\mathscr{L}(X)=AX$
For a fixed $A \in M_n (\mathbb{R})$, define $\mathscr{L}(X)=AX$ on the set $M_n (\mathbb{R})$. We have to find how characteristic polynomials of $\mathscr{L}$ and $A$ are related. I need to get my proof verified.

Here's what I have: I have proved(?) they both have the same set of eigenvalues. For if $\alpha$ is an eigenvalue of $A$ with corresponding eigenvector $v$, then the matrix $X$ with each of it's columns as $v$ will be an eigenvector for $\mathscr{L}$ with eigenvalue $\alpha$. If $\beta$ is an eigenvalue of $\mathscr{L}$ then $AX=\beta X$ for some $X \in M_n (\mathbb{R})$, or $(A-\beta I)X=0$. I think the first column of $X$ is an eigenvector for $A$ with eigenvalue $\beta$.
If this much is true, their characteristic polynomials  will have the same roots, only one is of degree $n$ and the other is of degree $n^2$. Can we say something more about them from here? Is it possible that the characteristic polynomial of $\mathscr{L}$ is just that of $A$ but raised to power $n$? How do we prove that?

答: Hint: Let $U_j$ denote the subspace of $M_{n}(\Bbb R)$ consisting of matrices whose entries outside of the $j$th column are all zero. Verify that $U_j$ is an invariant subspace of $\mathscr L$. What is the characteristic polynomial of $\mathscr L|_{U_j}$?


问:
Derivation of mutual information's closed-form analytical solution
$$I(X;Y) = -\frac{1}{2} \ln(1-\rho^2)$$ is the mutual information between two Gaussian random variables. What source derived this formula? Could we have the full derivation here as an answer.

First Attempt

Given $f(x)$ is the Gaussian p.d.f. of variable $X$ and $f(y)$ is the Gaussian p.d.f. of variable $Y$, and
\begin{align}
f(x,y)&=\frac{1}{\left( (2\pi)^{n}\det{(\boldsymbol
\Sigma)}\right)^\frac{1}{2}  }\exp\left(-\frac{1}{2}(\boldsymbol{x}-\boldsymbol{\mu})^\top\boldsymbol{\Sigma}^{-1}(\boldsymbol{x}-\boldsymbol{\mu})%
\right)\\
& = \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1-\rho^2}}
\exp \left\{ -\frac{1}{2\left(1-\rho^2\right)} \left[ \left(\frac{x-\mu_{X}}{\sigma_{X}}\right)^2
+ \left(\frac{y-\mu_{Y}}{\sigma_{Y}}\right)^2 -2\rho \left(\frac{x-\mu_{X}}{\sigma_{X}}\right) \left(\frac{y-\mu_{X}}{\sigma_{Y}}\right) \right] \right\}
\end{align}
is the joint distribution where $\boldsymbol{x}, \boldsymbol{\mu}$ and $\boldsymbol{\Sigma}$ are the data observations, means and covariance matrix of the joint distribution,

\begin{align}
I(X;Y) &= \int \int f(x,y) \ln \frac{f(x,y)}{f(x)f(y)} dx dy\\
&= \int \int f(x,y) \ln \frac{f(x,y)}{\left(2\pi \sigma_X^2\right)^{-\frac{1}{2}} e^{-(x-\mu_X)^2 / 2\sigma_X^2} \cdot \left(2\pi \sigma_Y^2\right)^{-\frac{1}{2}} e^{-(y-\mu_Y)^2 / 2\sigma_Y^2}} dx dy\\
&= ?
\end{align}

答:
Let $(X, Y) \sim \mathcal{N}(0, K),$ where
$$
K=\left[\begin{array}{cc}
\sigma^{2} & \rho \sigma^{2} \\
\rho \sigma^{2} & \sigma^{2}
\end{array}\right]
$$
Then $$h(X)=h(Y)=\frac{1}{2} \log (2 \pi e) \sigma^{2}$$ and $$h(X, Y)=\frac{1}{2} \log (2 \pi e)^{2}|K|=
\frac{1}{2} \log (2 \pi e)^{2} \sigma^{4}\left(1-\rho^{2}\right),$$ and therefore
$$
I(X ; Y)=h(X)+h(Y)-h(X, Y)=-\frac{1}{2} \log \left(1-\rho^{2}\right)
$$
If $\rho=0, X$ and $Y$ are independent and the mutual information is 0 . If $\rho=\pm 1, X$ and $Y$ are perfectly correlated and the mutual information is infinite.


问:
Write the expressoin in terms of $\log x$ and $\log y \log(\frac{x^3}{10y})$
What is the answer for this? Write the expression in terms of $\log x$ and $\log y$ $$\log\left(\dfrac{x^3}{10y}\right)$$

This is what I got out of the equation so far. the alternate form assuming $x$ and $y$ are positive $$3\log(x)-\log(10y)$$ or maybe $$\log\left(\dfrac{x^3}{y}\right)-\log(5)=\log(2)$$?

I would appreciate a solution or an input thanks so much.

答:
\begin{align*}
 \log (\frac{x^3}{10y}) & = \log (x^3) - \log (10y) \\
& = 3 \log x - (\log 10 +\log y) \\
& = 3 \log x - \log 10 - \log y
\end{align*}
using
$$
\log(ab) = \log a + \log b
,$$
$$
\log(\frac{a}{b}) = \log a - \log b
$$
and
$$
\log(a^n) = n \log a
$$</p>


问:
How to construct a nontrigonal curve?
$C$ is a smooth projective curve over an algebraically  closed field, we say $C$ is trigonal if $C$ admits a degree 3 cover of $\mathbb{P}^{1}$. I could give example which is trigonal. Now I want to construct a curve which is not trigonal. My idea is following:

If $C$ is trigonal, then there will be a linear system $|D|$ in $C$ such that $\mathrm{deg} D=3$ and $\mathrm{dim}|D|=1$. So my ideal is to construct a curve such that for all effective divsor of degree 3 in $C$, $\mathrm{dim} |D|=0$. However it seems not so easy to get such a curve.

Could you give an example?

答: If the canonical curve is an intersection of quadrics, it is not trigonal. So, the simplest example of a non-trigonal (and not hyperelliptic) curve is a complete intersection of three quadrics in $\mathbb{P}^4$.

Title: How to construct a nontrigonal curve?


问:
Joint Entropy closed-form analytical solution
Differential entropy of a single Gaussian random variable is

$$H(X) = \frac{1}{2} \ln (2\pi e \sigma^2)$$

What then is the closed-form analytical solution for joint entropy, $H(X,Y)$?

答:
Let $(X, Y) \sim \mathcal{N}(0, K),$ where
$$
K=\left[\begin{array}{cc}
\sigma^{2} & \rho \sigma^{2} \\
\rho \sigma^{2} & \sigma^{2}
\end{array}\right]
$$
Then differential entropy is $$h(X)=h(Y)=\frac{1}{2} \log (2 \pi e) \sigma^{2}$$ and joint entropy is

\begin{align}
h(X, Y)&=\frac{1}{2} \log (2 \pi e)^{2}|K|\\
&= \frac{1}{2} \log (2 \pi e)^{2} \sigma^{4}\left(1-\rho^{2}\right)
\end{align}


问:
Real and imaginary parts for $F(\varphi,m)$ with complex argument
I have this case: $F(\arcsin u,m)$, where $u>1$, so it ends up as $F(\frac{\pi}{2}-i\text{arccosh} u,m)$.

and I need to decompose it into real and imaginary parts. I have seen this question, but it has no answer. After some searches, looking through Abramowitz & Stegun, at 17.4.11 it describes how to do that:

$$F(\phi+i\psi,m)=F(\lambda,m)+iF(\mu,m)$$

where $\cot^2\lambda$ is the positive root of

$$x^2-\left[\cot^2\phi+m\sinh^2\psi\csc^2\phi-m_1\right]x-m_1\cot^2\phi=0\tag{1}$$

and

$$m\tan^2\mu=\tan^2\phi\cot^2\lambda-1\tag{2}$$

It turns out that the real part ends up as the complete elliptic integral of the first kind, $F(\lambda,m)=K(m)$, which suits me just fine, so all I have to do is determine the imaginary part, which means I need $\mu$, which derives from $\lambda$. For my case $\phi=\frac{\pi}{2}\Rightarrow \cot^2\phi=0,\space\csc^2\phi=1$, so $(1)$ becomes:

$$x^2-\left[m\sinh^2(\text{arccosh}u)-m_1\right]x-m_1=0$$

so, considering that $m<0.1$ (practical values are $0.01$ or less), gives the positive root (with the help of wxMaxima, I'm not versed in math):

$$\lambda_+=\frac{\sqrt{m^2(u-1)^2(u+1)^2+m_1\left[2m(1-u^2)+4\right]+m_1^2}+m(u^2-1)-m_1}{2}$$

So far, so good, but now I have problems in expressing $\mu$ from $(2)$:

$$\mu=\arctan\left(\frac{\sqrt{\tan^2\phi\cot^2\lambda-1}}{m}\right)$$

Did I do it right so far? If yes, what do I do with $\tan\phi=\tan\left(\frac{\pi}{2}\right)$? If not, how should I proceed?

Looking in the same A&S, section 17.4.8 describes the transforming of the completely imaginary argument:

$$F(i\phi,m)=iF(\theta,1-m),\space\theta=\arctan(\sinh\phi)$$

Knowing that the $\lambda$ calculated earlier is different than the $\mu=\text{arccosh}u$, I still tried plotting $F(\arctan(\sinh(\text{arccosh}u)),1-m)$ (got curious), where the argument reduces to (thanks to wxMaxima) $\arctan\sqrt{u^2-1}$. By mistake, I wrote $1+m$ in the modulus, instead of $1-m$, and this came up for $m=0.1$:

and for $m=0.01$, which is about the maximum value $m$ can take in my application:

Not only it gets closer, but for $m=0.1$ I used the hammer to tweak the modulus and the exponent to be something like $F(\arctan\sqrt{u^{2.2}-1},0.98625+m)$, and the result is this (the difference is in orange, multiplied by 100):

Is this a known "shortcut"? Is this worth following?

答:
There are in fact two sets of formulas for $F(\sin^{-1}u,m)$ with $u>1$, depending on whether $u>\frac1{\sqrt m}$ or not. Both are very simple and don't have the problem of the tangent blowing up at $\pi/2$.

If $1\le u\le\frac1{\sqrt m}$, Byrd and Friedman 115.02 gives
$$F(\sin^{-1}u,m)=K(m)+iF\left(\sin^{-1}\frac{\sqrt{u^2-1}}{u\sqrt{1-m}},1-m\right)$$
If $\frac1{\sqrt m}\le u$, Byrd and Friedman 115.03 gives
$$F(\sin^{-1}u,m)=F\left(\sin^{-1}\frac1{u\sqrt m},m\right)+iK(1-m)$$


答:
Suppose for contradiction $A_n \not \subseteq U$ for every $n \in \mathbb{N}$. Then pick $x_n \in A_n \setminus U$ for every $n \in \mathbb{N}$. Since $A_n \subseteq A_0$ for every $n$, $\{x_n\}_{n\in \mathbb{N}} \subseteq A_0$, and by the compactness of $A_0$, there is a convergent subsequence of $(x_n)$ that converges. For convenience let $\{x_n\}$ now refer to the convergent subsequence instead, and let $x$ be the limit point. Notice that for every $N \in \mathbb{N}$, then for all $n \geq \mathbb{N}$, $x_n \in A_N$. So $\{x_n\}_{n \geq N}$ is a convergent subsequence of each $A_N$, and hence the limit $x$ is in each $A_N$. Thus, since $A = \cap_{N \in \mathbb{N}}A_N$, $x \in A$ as well. Since $A \subseteq U$, $x \in U$. But then by the definition of convergence, there must be some $M \in \mathbb{N}$ such that for all $n \geq M$, $x_n \in U$. But this directly contradicts that we chose each $x_n$ to be in $A_n \setminus U$. By contradiction, we're done.

Note: By passing to the convergent subsequence and technically abusing a bit of notation, I sidestepped some finnicky indexing management that you'll have to do. I'll leave that to you. The general idea remains as above.

Body:
Let $A_n$ be a sequence of nonempty compact subset of $\mathbb R$ with $A_{n+1}\subseteq A_n$. Let $A=\cap_{n\in \mathbb N}A_n$ Let $U $ be an open subset of $\mathbb R$ such that $A\subseteq U.$ Show that there exists $N\in \mathbb N$ such that $A_N \subseteq U$.

My attempt

$A=\cap_{n\in \mathbb N}A_n \subseteq A_n, \forall n\in \mathbb N.$

Since each $A_n$ is closed and bounded by Heine Borel Theorem. Hence, $A$ is also bounded. Arbitrary intersection closed subset of $\mathbb R$ is closed. Then, $A$ is closed.
Hence, By Heine Borel Theorem $A$ is compact. I don't know how to proceed from here. May I get help?


问:
Find $\displaystyle\sum_{j=1}^{M-1} \sin\left(\frac{j\pi}{M}\right)\frac{j}{M}\left(1-\frac{j}{M}\right)$
I have been tasked to show $$\boxed{\displaystyle\sum_{j=1}^{M-1} \sin\left(\frac{j\pi}{M}\right)\frac{j}{M}\left(1-\frac{j}{M}\right)=\frac{1}{4M^2}\sin\left(\frac{\pi}{M}\right)\csc^4\left(\frac{\pi}{2M}\right)}$$

I think I could do this with a mess of complex exponentials ($\displaystyle\sum_{j=1}^{M-1}\exp\left(\frac{i\pi xj}{M}\right)$ evaluated at 1) and differentiating the geometric series sum twice but that seems very messy. I was wondering if there was an easier way.

If not, is there an easy way to instead find this sum to leading order in M?

答:
It is still manageable. Note
$$f(x)= \sum_{j=1}^{M-1}e^{i\frac{\pi xj}{M}}
=\frac{e^{i\frac{\pi x}{M}}-e^{i\pi x}}{1-e^{i\frac{\pi x}{M}} }
$$
$$f’(x)= i \sum_{j=1}^{M-1}e^{i\frac{\pi xj}{M}}\frac{\pi j}M,\>\>\>
f’’(x)= -\sum_{j=1}^{M-1}e^{i\frac{\pi xj}{M}}\frac{\pi^2j^2}{M^2}
$$
Then
\begin{align}
&\sum_{j=1}^{M-1} \sin\frac{j\pi x}{M}\left(1-\frac{j}{M}\right)\frac{j}{M}\\
= &Im \left(\frac1{i\pi}f’(x)+\frac1{\pi^2}f’’(x)\right)\\
= &\frac{d}{dx}Im \left(\frac1{i\pi}f(x)+\frac1{\pi^2}f’(x)\right)\\
=& \frac{d}{dx}\left(\frac1{2\pi}-\frac1{2\pi M}\sin^2\frac{\pi x}2 \csc^2\frac{\pi x}{2M} \right)\\
= &\frac1{4M^2}\left(2\sin^2\frac{\pi x}2\cot\frac{\pi x}{2M}-M\sin\pi x\right) \csc^2\frac{\pi x}{2M}
\end{align}
Set $x=1$ to obtain
\begin{align}
\sum_{j=1}^{M-1} \sin\frac{j\pi}{M}\left(1-\frac{j}{M}\right)\frac{j}{M}= \frac1{2M^2}\cot\frac{\pi x}{2M} \csc^2\frac{\pi x}{2M}
= \frac{1}{4M^2}\sin\frac{\pi}{M}\csc^4\frac{\pi}{2M}
\end{align}


答: $A_1 \cap U^{c} \subseteq \bigcup_{n \geq 2} A_n^{c}$. Since $A_1 \cap U^{c}$ is a closed subset of $A_1$ it is compact. This open cover therefore has a finite subcover. If $A_1 \cap U^{c} \subseteq \bigcup_{2\leq n \leq N} A_n^{c}$ the $A_N \subseteq U$.

Title: Show that there exists $N\in \mathbb N$ such that $A_N \subseteq U$.


问:
How to find the sum of the following series
How can I find the sum of the following series?
$$
\sum_{n=0}^{+\infty}\frac{n^2}{2^n}
$$
I know that it converges, and Wolfram Alpha tells me that its sum is 6 .

Which technique should I use to prove that the sum is 6?

答:
For $x$ in a neighborhood of $1$, let
$$
f(x) = \sum\limits_{n = 0}^\infty  {\frac{{x^n }}{{2^n }}} = \sum\limits_{n = 0}^\infty  {\bigg(\frac{x}{2}\bigg)^n }  = \frac{1}{{1 - x/2}} = \frac{2}{{2 - x}}.
$$
Thus, on the one hand,
$$
f'(x) = \sum\limits_{n = 0}^\infty  {\frac{{nx^{n - 1} }}{{2^n }}}  \;\; {\rm and} \;\; f''(x) = \sum\limits_{n = 0}^\infty  {\frac{{n(n - 1)x^{n - 2} }}{{2^n }}} ,
$$
and, on the other hand,
$$
f'(x) = \frac{2}{{(2 - x)^2 }} \;\; {\rm and} \;\; f''(x) = \frac{4}{{(2 - x)^3 }}.
$$
Hence,
$$
f'(1) = \sum\limits_{n = 0}^\infty  {\frac{n}{{2^n }}}  = 2  \;\; {\rm and} \;\; f''(1) = \sum\limits_{n = 0}^\infty  {\frac{{n(n - 1)}}{{2^n }}}  = 4.
$$
Finally,
$$
\sum\limits_{n = 0}^\infty  {\frac{{n^2 }}{{2^n }}}  = f'(1) + f''(1) = 6.
$$
The idea here was to consider the Probability-generating function of the geometric$(1/2)$ distribution.</p>


答: I think you are close to the expected answer with your first answer. That is, $$\log\left(\frac{x^3}{10y} \right) = \log(x^3) - \log(10y) = 3 \log(x) - \log(10) - \log(y).$$</p>

Title: Write the expressoin in terms of $\log x$ and $\log y \log(\frac{x^3}{10y})$

Body:
What is the answer for this? Write the expression in terms of $\log x$ and $\log y$ $$\log\left(\dfrac{x^3}{10y}\right)$$

This is what I got out of the equation so far. the alternate form assuming $x$ and $y$ are positive $$3\log(x)-\log(10y)$$ or maybe $$\log\left(\dfrac{x^3}{y}\right)-\log(5)=\log(2)$$?

I would appreciate a solution or an input thanks so much.


问:
Prove that if $\int_a^b f(x)^p dx = 0$, then $\int_a^b f(x) dx = 0$ for $p > 1$.
Assume that $f$ is Riemann integrable on $[a,b]$ with $f(x) \geq 0$ for all $x \in [a,b]$. Prove that if $\int_a^b f(x)^p dx = 0$, then $\int_a^b f(x) dx = 0$ for $p > 1$. I was thinking of bounding $f$ by $f(x)^p$ with a constant, but I'm not sure how to deal with the x-values when $f(x)$ approach $0$.

答: Let $\epsilon >0$. if $f(x) \geq \epsilon$ then $f(x)\leq \epsilon^{1-p} f(x)^{p}$. In any case $f(x) \leq \epsilon + \epsilon^{1-p} f(x)^{p}$. Hence $\int f(x)dx \leq \epsilon^{1-p}\int f^{p}(x)dx+\epsilon$ this gives $\int f(x)dx \leq \epsilon$. Since $\epsilon$ is arbitrary this gives $\int f(x)dx=0$.


问:
Find a constant $c_0 > 0$ that satisfies the following properties:
Find a constant $c_0 > 0$ that satisfies the following properties:

(i) The convergence of every series $\sum a_n$ with $a_n \geq 0$ implies the convergence of $\sum\limits_{n=1}^\infty b_n := \sum\limits_{n=1}^\infty \frac{\sqrt{a_n}}{n^c}$ for all $c > c_0$

(ii) There exists a convergent series $\sum a_n$ with $a_n \geq 0$ such that the series $\sum b_n$ diverges if $c = c_0$

My attempt:

$\sum\limits_{n=1}^\infty \frac{\sqrt{a_n}}{n^c} \leq \sum\limits_{n=1}^\infty \frac{\sqrt{M}}{n^c}$ since $a_n \geq 0$ is a convergent sequence and every convergent sequence is bounded. Note that this sum converges only when $c > 1$.
So, we get that $c_0 = 1$. But from Rudin Chapter 3 problem 7, we know that
the convergence of $\sum a_n$ automatically implies the convergence of $\sum \frac{\sqrt{a_n}}{n}$ for $a_n \geq 0$, so the bound of $c_0 = 1$ is not tight enough. Can anyone tell me what $c_0$ should be?

Any help would be appreciated!

答: Let $c_0=1/2$. To show $\sum_n b_n$ converges for $c=1/2+\epsilon,$ use AM-GM: $$\sqrt{\frac{a_n}{n^{1+2\epsilon}}} \le \frac{a_n}{2} +\frac{1}{2n^{1+2\epsilon}}.$$ For the case where $c=1/2,$ consider $a_n=\frac{1}{n\log^2(n+1)}.$


问:
Integral, set and parametric representation
I am to compute the following: $\displaystyle\iiint\limits_V 1\, dx\, dy\, dz$,

where $V= \{{(x,y,z) \in \mathbb R^3 : (x-z)^2 +4y^2 < (1-z)^2} \text{ and } 0<z<1\}.$

Does anyone have idea what parametric representation should I take? I think there will be something elliptic, for instance $y=\frac{1}{2}r\sin(\alpha)$ and $z=z$, but I don't know what about $x$. Or maybe another parametric representation? Could you help me?

答:
Try
$$
u=x-z\quad
v=2y\quad
w=1-z
$$
New domain will be $\{(u,v,w)\in\mathbb{R}^3:u^2+v^2<w^2, 0<w<1\}$. Jacobian is also easy to compute. </p>


问:
Let $f: \mathbb{R} \to \{0, 1\}$ be a non-decreasing, right-continuous function. Then $\sup\{x: f(x) = 0\} = c$, where for all $x \geq c$, $f(x) = 1$.

Let $f: \mathbb{R} \to \{0, 1\}$ be a non-decreasing, right-continuous function with $\lim_{x \to -\infty}f(x) = 0$ and $\lim_{x \to +\infty}f(x) = 1$. Then $\sup\{x: f(x) = 0\}$ is equal to the value $c$ such that for all $x \geq c$, $f(x) = 1$.

We observe that $S = \{x: f(x) = 0\}$ is bounded from above by a value $c \in \mathbb{R}$ such that for all $x \geq c$, $f(x) = 1$ ($c$ exists due to the limit at $+\infty$ and right continuity of $f$). Thus $s:= \sup S$ exists. Of course, $s \leq c$ by definition.

Could someone give me a hint on how to show $s \geq c$?

答: Just define $c$ to the supremum of $S$. If $x>c$ then $x \notin S$ so $f(x)=1$. By right continuity it follows that $f(x)=1$ for $x=c$ also.


问:
Show that $Y$ is not first-countable at $[0]$
(a) Show that $X := \{0, 1\}^{\mathbb R}$ is not first-countable.

(b) Let $X := \bigcup_{n=0}^∞ [2n, 2n + 1]$ be a union of countably many disjoint closed intervals. Let $Y$ be the quotient of $X$ obtained by gluing all the points $0, 2, 4, 6, \ldots$ together.

Show that $Y$ is not first-countable at $[0]$ (the equivalence class of $0$). Use a diagonalization argument.

My attempt:

$(a)$ I proceeded indirectly by showing that there exists a real-valued function on some subspace of $[0,1]^{\mathbb R}$ that is sequentially continuous but not continuous. It is the case of $F(f) = \int_0^1 f(t) \, dt$ where $f$ is measurable.

$(b)$ But I don't know how to prove this part. Any help please?

答: For (b): If $V=\{U_j:j\in \Bbb N_0\}$ is a countable set of nbhds of $[0]$ in $Y$ then for each $j$ there is a function $f_j:\Bbb N_0\to (0,1)$ such that $ U_j\supset (2n,2n+f_j(n))$ for every $n.$ Then $[0]\cup(\cup_{n\in \Bbb N_0}(2n, f_n(n)/2)\,)\,)$ is a nbhd of $[0]$ which is not a subset of any member of $V.$


问:
Schnitt/Dedekind Cuts addition to zero?
I'm being asked to prove that if T is a Schnitt, and (-T) is the negative of that Schnitt, then (T) + (-T) = 0.
Both the addition operation and the o are on the Rationals.
I think I need to use cases, and maybe further subcases, but I'm really stuck.</p>

答:
I am assuming that you define $-T$ to be $\{-q:q\in\Bbb Q\setminus T\}$, so that

$$T+(-T)=\{r+s:r\in T\text{ and }-s\in\Bbb Q\setminus T\}\,.$$

Let $r\in T$ and $-s\in\Bbb Q\setminus T$; then by the definition of a cut we know that $r<-s$, and hence that $r+s<0$.

Now let $q$ be any negative rational; we want to show that $q\in T+(-T)$. Let $t\in T$.

Show that there is an $n\in\Bbb Z^+$ such that $t-nq\notin T$.

Let $n_0=\min\{n\in\Bbb Z^+:t-nq\notin T\}$, $r=t-(n_0-1)q$, and $s=t-n_0q$; $r\in T$, and $s\notin T$. That is, $s\in\Bbb Q\setminus T$, so $-s\in -T$, and $r+(-s)\in T+(-T)$.

Verify that $r+(-s)=q$, and conclude that $T+(-T)=\{q\in\Bbb Q:q<0\}$, the cut corresponding to the rational number $0$.

Title: Schnitt/Dedekind Cuts addition to zero?


问:
How to prove this using Taylor's Theorem
If $g(x)= \sin x$, prove that the remainder of the Taylor's Theorem converges to zero when $n \to \infty$, for each $x_0$ and $x$ fixed.

My thoughts so long:

By the Taylor's Theorem and given fixed $x_0$ and $x$ we have:

$P_n(c)=\sin(x_0)+\cos(x_0)(x-x_0)-\frac{\sin(x_0)}{2!}(x-x_0)^2-\frac{\cos(x_0)}{3!}(x-x_0)^3...$

The remainder is $R_n(c)= \frac{\sin^{n+1} (c)}{(n+1)!}(x-x_0)^{n+1}$

But I'm stuck at this expression, does it make sense to take the limit as n approaches infinity of the $R_n$ expression?

答: You're not really limiting in a continuous way but rather as discrete sequence of terms. See here</p>


问:
Parametric equations for given line
How would you find the parametric equations for:

1) a line through $(3,1)$ and $(-5,4)$.

2) a segment joining $(1,1)$ and $(2,3)$.

Can anyone show me the steps of doing it cause the way my textbook shows it, it has gotten me in a loop.

答:
Let $p,q \in \mathbb{R}^2$ be points on the plane. The vector $v$ with starting point $p$ and ending point $q$ can be understood as a transport of points, in other words, the vector $v$ will be the subtraction of the points to give a kind of machine that when applied via the sum to the starting point gives the endinging point. Multiplying a vector by a number $k$ is equivalent to get a transport of point in the same direction if $k>0$ that is $k$ times the original transport. It'll reverse orientation if $k <0$.

Now, what's a line? A line can be understood as a curve that never changes it's direction. In other words, you take a point and transport it forever in the same direction forward and backward and you'll get a line in euclidean plane. Now one can describe a line by the following function $\alpha : \mathbb{R} \to \mathbb{R}$

$$\alpha(t) = p + tv$$

So this function says that you take any point on the line and transport it always on the same direction (that of $v$) all the way forever (this aims to allowing $t$ to run all the reals). Of course $p$ can be any point on the line and $v$ can be any transport from one point on the line to any other point on the line. A segment of $\alpha$ is obtained by restricting the domain to some closed subinterval of the reals (so that the transport gets bounded).

Now with this theory let's tackle the exercise. Since we want a line through $(3,1)$ and $(-5,4)$ we know that both points are on the line. So $(-5, 4) - (3,1) = (-8, 3)$ is the vector that transports one point of the line to the other one. So by the preceding discussion the line is parametrized by $\alpha : \mathbb{R} \to \mathbb{R}^2$ given by the following:

$$\alpha(t) = (-5,4)+t(-8,3)=(-5-8t, 4-3t)$$

Now the segment joining $(1,1)$ and $(2,3)$ can be obtained in analogous way. However the domain must be a closed subinterval. The natural choice is $[0,1]$, so that for $0$ we get $(1,1)$ and for $1$ we get $(2,3)$. Again the vector responsible for transporting the initial to the final point is $(2,3) - (1,1) = (1,2)$. So the parametrization is $\alpha : [0,1] \to \mathbb{R}^2$ given by:

$$\alpha(t) = (1,1) + t(1,2) = (1+t, 1+2t)$$

Title: Parametric equations for given line


问:
Checking whether set is closed or open
Let $X=\mathbb N × \mathbb Q$ with the subspace topology of the usual topology of $\mathbb R^2$. Let $P=\{(n,\frac {1}{n}): n\in \mathbb N\}$. Then check whether $P$ is closed or open or both.

I think P is closed but not open. Since P does not have a limit point outside P it is closed. If P is open then we must have some open set U in $\mathbb R^2$ such that $P=U\cap X$. Which is not possible as $U\cap X $ will contain infinitely many points other than $(n,\frac{1}{n})$. Am I right?

答: Your answer for closedness is correct. To show that it is  not open consider $(1,1)$. Let us show that this is not an interior point. Suppose some open neighborhood of this point is contained in $P$. Consider the points $(1,1+\frac 1 N)$. These points do not belong to $P$ but they will lie in the neighborhood for $n$ sufficiently large.

Title: Checking whether set is closed or open


答:
$$0\leq\bigg|\frac{\sin^{n+1}c(x-x_0)^{n+1}}{(n+1)!}\bigg|\leq \bigg|\frac{(x-x_0)^{n+1}}{(n+1)!}\bigg|.\tag{1}$$

Let $a_n= \frac{(x-x_0)^{n+1}}{(n+1)!}$,then $$\lim\frac{a_{n+1}}{a_n}=\frac{(x-x_0)^{n+2}\cdot(n+1)!}{(n+2)!\cdot(x-x_0)^{n+1}}=0$$
So $\sum a_n$ converges by ratio test, then $\lim a_n=0$.

Therefore the rightmost side of $(1)$ tends to zero.
Then by SQUEEZE theorem middle part of $(1)$ also tends to $0$.

Title: How to prove this using Taylor's Theorem

Body:
If $g(x)= \sin x$, prove that the remainder of the Taylor's Theorem converges to zero when $n \to \infty$, for each $x_0$ and $x$ fixed.

My thoughts so long:

By the Taylor's Theorem and given fixed $x_0$ and $x$ we have:

$P_n(c)=\sin(x_0)+\cos(x_0)(x-x_0)-\frac{\sin(x_0)}{2!}(x-x_0)^2-\frac{\cos(x_0)}{3!}(x-x_0)^3...$

The remainder is $R_n(c)= \frac{\sin^{n+1} (c)}{(n+1)!}(x-x_0)^{n+1}$

But I'm stuck at this expression, does it make sense to take the limit as n approaches infinity of the $R_n$ expression?


问:
Logic - Reducing a first-order logic involving a material conditional
Suppose that
$$
\forall x \forall y: P(x,y) \implies Q(x)
$$
and
$$
\forall x \exists y: P(x,y)
$$

Then, can I conclude that
$$
\forall x: Q(x)
$$

If it is true, what is the rationale for it?

What I tried:

$$
\begin{align}
&\forall x \forall y: P(x,y) \implies Q(x)\\
&\iff
\forall x \forall y:\lnot P(x,y) \lor Q(x)\\
&\iff
\forall x: (Q(x) \lor \forall y:\lnot P(x,y))\\
&\iff
\forall x: (Q(x) \lor \lnot (\exists y:P(x,y)))
\end{align}
$$
Combining the result with $\forall x \exists y: P(x,y)$, I concluded that $Q(x)$ should be true for all $x$ since $\lnot (\exists y:P(x,y))$ is always false.

$$
\begin{align}
&\forall x: (Q(x) \lor \lnot (\exists y:P(x,y)))
\land \forall x \exists y: P(x,y)\\
&\iff \forall x: ((Q(x) \lor \lnot (\exists y:P(x,y))) \land \exists y: P(x,y))\\
&\iff \forall x: (Q(x) \land \exists y: P(x,y))\\
&\implies \forall x: Q(x)
\end{align}
$$

答:
If it is true, what is the rationale for it?

$\def\boxit#1{\bbox[lemonchiffon,0.5ex]{#1}}$We have premises of $\boxit{\forall x~\forall y:(P(x,y)\to Q(x))}$ and $\boxit{\forall x~\exists y:P(x,y)}$.   Should we take an arbitrary variable, $\boxit a$, then we infer from the second premise that there is a witness variable, call it $\boxit b$, which satisfies $\boxit{P(a,b)}$.   For these variables, we also infer from the first premise that $\boxit{P(a,b)\to Q(a)}$ will be satisfied.   Thus, by modus ponens, we infer that $\boxit{Q(a)}$ is satisfied.   Since $\boxit b$ does not occur in this statement, and $\boxit a$ is arbitrary, we have therefore shown that $\boxit{\forall x:Q(x)}$ is entailed by these premises.

$$\def\fitch#1#2{~~~~{\begin{array}{|l}#1\\\hline#2\end{array}}}\fitch{~~1.~\forall x\,\forall y:(P(x,y)\to Q(x))\hspace{3.5ex}\textsf{Premise}\\~~2.~\forall x\,\exists y:P(x,y)\hspace{14ex}\textsf{Premise}}{\fitch{~~3.~\boxed a\hspace{23.5ex}\textsf{Assumption (Arbitrary)}}{~~4.~\forall y:(P(a,y)\to Q(a))\hspace{4ex}\textsf{Universal Elimination, 1}\\~~5.~\exists y:P(a,y)\hspace{14.5ex}\textsf{Universal Elimination, 2}\\\fitch{~~6.~\boxed b~P(a,b)\hspace{13.5ex}\textsf{Assumption (Witness)}}{~~7.~P(a,b)\to Q(a)\hspace{8ex}\textsf{Universal Elimination, 4}\\~~8.~Q(a)\hspace{18.5ex}\textsf{Conditional Elimination, 6, 7}}\\~~9.~Q(a)\hspace{21.5ex}\textsf{Existential Elimination 5, 6-8}}\\10.~\forall x:Q(x)\hspace{19.75ex}\textsf{Universal Introduction, 3-9}}$$