数学分析 - Eufisky - The lost book

几个恒等式证明

1.证明:

\[\sum_{j=1}^{n-1}{\frac{1}{1-\exp \left\{ \frac{2\pi ij}{n} \right\}}}=\frac{n-1}{2}.\]
 

2.设$N$为自然数, $\{x\}$表示$x$的小数部分.证明\[\sum_{n=1}^N{\left( \left\{ x+\frac{n}{N} \right\} -\frac{1}{2} \right)}=\left\{ Nx \right\} -\frac{1}{2}.\]

若$a,b$是互质的正整数,证明
\[\int_0^1{\left( \left\{ ax \right\} -\frac{1}{2} \right) \left( \left\{ bx \right\} -\frac{1}{2} \right) dx}=\frac{1}{12ab}.\]
 
3.求极限
\[\lim_{m\rightarrow \infty}\lim_{n\rightarrow \infty}\int_0^1{\int_0^1{\cdots \int_0^1{\sum_{i=0}^m{\exp \left\{ -\frac{n}{\sum_{k=1}^n{x_{k}^{m-1}}} \right\} \frac{\prod_{j=1}^i{\sum_{k=1}^n{x_{k}^{j-1}}}}{\left( \sum_{k=1}^n{x_{k}^{m-1}} \right) ^i}}dx_1dx_2\cdots dx_n}}}.\]
4.证明
\begin{align*}f\left( x \right) &=\frac{1}{a}+\frac{x}{a\left( a+d \right)}+\cdots +\frac{x^n}{a\left( a+d \right) \cdots \left( a+nd \right)}+\cdots\\&=\frac{e^{x/d}}{dx^{a/d}}\int_0^x{e^{-t/d}t^{a/d-1}dt}.\end{align*}
提示:考虑关于$f(x)$的微分方程.
 
5.证明Ramanujan的恒等式
\[\int_0^{\infty}{e^{-3\pi x^2}\frac{\sin\text{h}\pi x}{\sin\text{h}3\pi x}dx}=\frac{1}{\sqrt{3}e^{2\pi /3}}\sum_{n=0}^{\infty}{\frac{e^{-2n\left( n+1 \right) \pi}}{\left( 1+e^{-\pi} \right) ^2\left( 1+e^{-3\pi} \right) ^2\cdots \left( 1+e^{-\left( 2n+1 \right) \pi} \right) ^2}}.\]
参考:G. N. Watson 1936:The Final Problem : An Account of the Mock Theta Functions
6.Compute 
$$\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}.$$
By the software Mathematica, I find
$$\sum_{n=1}^{\infty}{\left( \frac{n^n}{n!e^n}-\frac{1}{\sqrt{2\pi n}} \right)}=-\frac{2}{3}-\frac{\zeta \left( 1/2 \right)}{\sqrt{2\pi}}.$$

Well, $-\frac{1}{\sqrt{2\pi}}\zeta\left(\tfrac{1}{2}\right)$ is the $\zeta$-regularization of the divergent series $\sum_{n\geq 1}\frac{1}{\sqrt{2\pi n}}$, hence the problem boils down to finding the $\zeta$-regularization of the divergent series $\sum_{n\geq 1}\frac{n^n}{n!e^n}$. As pointed out in the comments,
 
$$ W(x) = \sum_{n\geq 1}\frac{n^{n-1}(-1)^{n-1}}{n!}x^n $$
holds for any $x\in\left(-\frac{1}{e},\frac{1}{e}\right)$ by Lagrange inversion theorem, hence
$$ ze^{-z} W'(-ze^{-z})=\sum_{n\geq 1}\frac{n^n}{n!e^{nz}}z^{n}=\frac{z}{1-z} =\sum_{n\geq 1}z^n\tag{1}$$
holds for any $z\in(-W(e^{-1}),1)$. Pretty strange identity, I can give you that.<br>
Similarly, over the same interval
$$ -W(-z e^{-z})=\sum_{n\geq 1}\frac{n^{n-1}}{n!e^{nz}}z^n = z \tag{2}$$
$$ 1=\sum_{n\geq 1}\frac{n^{n}}{n!e^{nz}}z^{n-1}-\sum_{n\geq 1}\frac{n^{n}}{n!e^{nz}}z^{n}=\frac{1}{1-z}-\frac{z}{1-z}.\tag{3}$$
Since $\zeta(0)=-\frac{1}{2}$, it should not be difficult to prove from $(1)$ and $(2)$ that the $\zeta$-regularization of $\sum_{n\geq 1}\frac{n^n}{n!e^n}$ equals $-\frac{2}{3}$ as wanted, for instance by computing $\sum_{n\geq 1}\frac{n^{n-1-k}}{n!e^n}$ for any $k\in\mathbb{N}$:
$$ \sum_{n\geq 1}\frac{n^{n-2}}{n!e^n}=\int_{-1/e}^{1}\frac{W(x)}{x}\,dx = \frac{1}{2},\qquad \sum_{n\geq 1}\frac{n^{n-3}}{n!e^n}=-\int_{-1/e}^{1}\frac{W(x)}{x}(1+\log(-x))\,dx=\frac{5}{12} $$
$$ \sum_{n\geq 1}\frac{n^{n-4}}{n!e^n}=\frac{7}{18},\qquad \sum_{n\geq 1}\frac{n^{n-4}}{n!e^n}=\frac{1631}{4320},$$
$$ \sum_{n\geq 1}\frac{n^{n-1-k}}{n!e^n}= \frac{1}{\Gamma(k)}\int_{0}^{1}(1-x)(x-1-\log x)^{k-1}\,dx.\tag{4} $$
Indeed the substitution $x=e^{-s}$ in $(4)$ and the integral representation for the $\zeta$ function complete the proof.
Taking $$F\left(x\right)=\sum_{n\geq1}\frac{n^{n-1}}{n!e^{n}}x^{n}-\frac{1}{\sqrt{2\pi}}\sum_{n\geq1}\frac{x^{n}}{n^{3/2}}=-W\left(-\frac{x}{e}\right)-\frac{\mathrm{Li}_{3/2}\left(x\right)}{\sqrt{2\pi}},\,\left|x\right|<1$$ where $W\left(x\right)$ is the [Lambert $W$ function][1] and $\mathrm{Li}_{3/2}\left(x\right)$ is the [Polylogarithm function][2], we obtain, differentiating both sides,that $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)x^{n-1}=-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}$$ so $$\sum_{n\geq1}\left(\frac{n^{n}}{n!e^{n}}-\frac{1}{\sqrt{2\pi n}}\right)=\lim_{x\rightarrow1^{-}}\left(-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}-\frac{\mathrm{Li}_{1/2}\left(x\right)}{x\sqrt{2\pi}}\right).$$ Now, [we know][3] that $$\mathrm{Li}_{v}\left(z\right)=\left(\Gamma\left(1-v\right)\left(1-z\right)^{v-1}+\zeta\left(v\right)\right)\left(1+O\left(\left|1-z\right|\right)\right),v\neq1,\,z\rightarrow1$$  and now we claim $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1}{\sqrt{2\left(1-x\right)}}-\frac{2}{3}$$ as $x\rightarrow1^{-}$. This is true because, [since][5] $$W\left(z\right)\sim-1+\sqrt{2ze+2}-\frac{2}{3}e\left(z+\frac{1}{e}\right)$$ as $z\rightarrow-1/e$, we have $$-\frac{W\left(-\frac{x}{e}\right)}{x\left(W\left(-\frac{x}{e}\right)+1\right)}\sim\frac{1-\sqrt{2\left(1-x\right)}+\frac{2}{3}\left(1-x\right)}{x\sqrt{2\left(1-x\right)}-\frac{2}{3}\left(1-x\right)x}$$ $$=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\left(\frac{1}{1-\sqrt{2-2x}/3}\right)\right)=\frac{1}{x}\left(-1+\frac{1}{\sqrt{2\left(1-x\right)}}\sum_{k\geq0}\left(\frac{\sqrt{2-2x}}{3}\right)^{k}\right)$$ $$=\frac{1}{x}\left(-\frac{2}{3}+\frac{1}{\sqrt{2\left(1-x\right)}}+O\left(\sqrt{1-x}\right)\right)$$ then the claim.
 
[1]:https://en.wikipedia.org/wiki/Lambert_W_function
[2]:https://en.wikipedia.org/wiki/Polylogarithm
[3]:http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/06/01/02/01/01/
[4]:http://mathworld.wolfram.com/StirlingsSeries.html
[5]:http://functions.wolfram.com/ElementaryFunctions/ProductLog/06/01/02/
This is a general answer to the followup question by Jack D'Aurizio. 
 
**Proposition**
>Let $n\in\mathbb{N}$. We have the asymptotic expansion
$$n!\sim \sqrt{2\pi n} \frac{n^n}{e^n} \left[ 1+ \frac1{12n} +\frac1{288n^2}-\frac{139}{51840n^3}-\frac{157}{2488320n^4}+\cdots \right]$$
 
Note that this is not a convergent series, but an asymptotic expansion. The error in the truncated series is asymptotically equal to the first omitted term. Regard the series on the right as an element of the ring of power series over rational numbers $\mathbb{Q}[[T]]$. 
$$S(T)=1+ \frac1{12}T+\frac1{288}T^2-\frac{139}{51840}T^3-\frac{157}{2488320}T^4 + \cdots. $$
Consider the multiplicative inverse of $S(T)$ in $\mathbb{Q}[[T]]$.
$$S^{-1}(T)=1-\frac1{12}T+ g_2 T^2 + g_3 T^3 + g_4 T^4 + \cdots. $$
Let $Y_s(T)=\sum_{n=0}^{\infty} h_n(s) T^n \in\mathbb{Q}[s][[T]]$ be defined by
$$\left(\frac12 T^2\right)^{s-1}\sum_{n=0}^{\infty} h_n(s) T^n = \left[ \frac12 T^2 + \frac13 T^3 + \frac14 T^4+\cdots \right]^{s-1}.$$
Then we have
 
**Theorem**
>$$\sum_{n=1}^{\infty} n^p\left[ \frac{n^n}{n!e^n}- \frac1{\sqrt{2\pi n}} \sum_{k=0}^p \frac{g_k}{n^k}\right]=(-2)^p p!h_{2p+1}(-p)  - \frac1{\sqrt{2\pi}}\sum_{k=0}^p g_k \zeta\left(k+\frac12-p\right).$$
 
With $p=0$, it is the original series
$$\sum_{n=1}^{\infty} \left[\frac{n^n}{n!e^n} - \frac1{\sqrt{2\pi n}}\right]=-\frac23 - \frac{\zeta\left(\frac12\right)}{\sqrt{2\pi}}$$
 
With $p=1$, it gives the value of 
$$\sum_{n=1}^{\infty} \left[ \frac{n^{n+1}}{n!e^n} - \sqrt{\frac{n}{2\pi}} + \frac{1}{12\sqrt{2\pi n}}\right] = -\frac 4{135} - \frac{\zeta\left(-\frac12\right)}{\sqrt{2\pi}} + \frac{\zeta\left(\frac12\right)}{12\sqrt{2\pi}}.$$

7.Prove that $$\sum_{m\leqslant x}\sum_{n\leqslant x}\Big\{\frac{x}{m+n}\Big\}=\Big(2\log2-\frac{\pi^2}{12}\Big)x^2+O(x\log x),$$

where $\{x\}$ is the fractional part of the real number $x$.
 
I know
$$\sum_{n\leqslant x} \Big\{\frac{x}{n} \Big\}=(1-\gamma)x+O\big(x^{1/2}\big),$$

where $\gamma$ is Euler's constant. But I don't know whether it is useful. Can you help me?

参考:这里


Write $\{x\} = x - \lfloor x\rfloor$,s so that
\[\sum_{m \leq x} \sum_{n \leq x} \left\{ \frac{x}{m + n}\right\} = \sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} - \sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor.\]
Then note that
\[\sum_{m \leq x} \sum_{n \leq x} \left\lfloor \frac{x}{m + n}\right\rfloor = \sum_{m \leq x} \sum_{n \leq x} \sum_{\ell \leq \frac{x}{m + n}} 1 = \sum_{\ell \leq \frac{x}{2}} \sum_{n \leq \frac{x}{\ell} - 1} \sum_{m \leq \frac{x}{\ell} - n} 1.\]
The sum over $m$ is $\lfloor x/\ell\rfloor - n$. The ensuing sum over $n$ is
\[\left\lfloor \frac{x}{\ell}\right\rfloor \left(\left\lfloor \frac{x}{\ell}\right\rfloor - 1\right) - \sum_{n \leq \frac{x}{\ell} - 1} n.\]
Via partial summation,
\[\sum_{n \leq \frac{x}{\ell} - 1} n = \left(\frac{x}{\ell} - 1\right) \left(\left\lfloor \frac{x}{\ell} \right\rfloor - 1\right) - \int_{1}^{\frac{x}{\ell} - 1} \lfloor t\rfloor \\, dt,\]
and this integral is equal to $\frac{1}{2} \left(\frac{x}{\ell} - 1\right)^2 + O(\frac{x}{\ell})$. So the sum over $n$ and $m$ simplifies to
\[\frac{x^2}{2 \ell^2} + O\left(\frac{x}{\ell}\right).\]
The ensuing sum over $\ell$ is
\[\frac{x^2}{2} \sum_{\ell = 1}^{\infty} \frac{1}{\ell^2} - \frac{x^2}{2} \sum_{\ell > \frac{x}{2}} \frac{1}{\ell^2} + O(x \log x).\]
The first sum over $\ell$ is $\zeta(2) = \pi^2/6$. The second is $O(1/x)$. So this simplifies to
\[\frac{\pi^2 x^2}{12} + O(x \log x).\]
 
Now we deal with
\[\sum_{m \leq x} \sum_{n \leq x} \frac{x}{m + n} = x \sum_{m \leq x} \sum_{n \leq x} \frac{1}{m + n}.\]
We deal with the sum over $n$ via partial summation: it is equal to
\[\frac{\lfloor x\rfloor}{m + x} + \int_{1}^{x} \frac{\lfloor t \rfloor}{(m + t)^2} \\, dt = \log \frac{m + x}{m + 1} + O\left(\frac{1}{m}\right),\]
where we have used the fact that $\lfloor x \rfloor = x - \{x\} = x + O(1)$, the fact that the antiderivative of $t/(m + t)^2$ is $m/(m + t) + \log(m + t)$, and the fact that the antiderivative of $1/(m + t)^2$ is $-1/(m + t)$.
 
So it remains to evaluate
\[\sum_{m \leq x} \log \frac{m + x}{m + 1} = \lfloor x\rfloor \log \frac{2x}{x + 1} + (x + 1) \int_{1}^{x} \frac{\lfloor t\rfloor}{(t + x)(t + 1)} \\, dt.\]
The antiderivative of $\frac{t}{(t + x)(t + 1)}$ is $\frac{x}{x - 1} \log \frac{t + x}{t + 1}$, and so after some simplification, we arrive at $(2\log 2) x + O(\log x)$.
 

 

1-1+1-1+...=1/2?

来源:马明辉QQ空间
 
在一个寒风凛冽的晚上,我一手拿着烤红薯,一手玩着手机。
 
当时我正在看熊哥的微信公众号: Xionger的数学小屋,他在上面发了北大18年研究生考试题(侵删)。
 
当然了,我早已对这种考试题失去了兴趣,但唯独最后一题,却很是在意。我与向老师讨论了这个题目,向老师嘛,以前与之讨论过一些数学问题,感觉他特厉害,随手就写了份解答给我。
 
Suppose $f(x)>0$, $f''(x)\le 0$, and $f(+\infty)=+\infty$ on $[0,+\infty)$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}. \]
 
\begin{proof}
     Since $f(x)>0$, $f''(x)\le 0$, and $f(+\infty)=+\infty$, it is not hard to see $f'(x)>0$, then
\[ \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\sum\limits_{n=0}^{\infty} \left( \frac{1}{f^s(2n)}-\frac{1}{f^s(2n+1)} \right)=-\sum\limits_{n=0}^{\infty} g_s(\xi_n), \]
where $g_s(x)=(\frac{1}{f^s(x)})'=-\frac{sf'(x)}{f^{s+1}(x)}$ and $\xi_n \in (2n,2n+1)$. Since
\[ g_s'(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}>0, \]
we can get
\[ -\sum\limits_{n=0}^{\infty} g_s(\xi_n)\le -\sum\limits_{n=0}^{\infty} g_s(2n)\le -g_s(0)-\frac{1}{2}\sum\limits_{n=1}^{\infty} \int_{2n-2}^{2n} g_s(x)dx \]
\[ =-g_s(0)-\frac{1}{2}\int_{0}^{+\infty} g_s(x)dx=-g_s(0)+\frac{1}{2f^s(0)}\to \frac{1}{2} \]
as $s\to 0_+$. Similarly, we have
\[ -\sum\limits_{n=0}^{\infty} g_s(\xi_n)\ge -\sum\limits_{n=0}^{\infty} g_s(2n+1)\ge -\frac{1}{2}\sum\limits_{n=0}^{\infty} \int_{2n+1}^{2n+3} g_s(x)dx \]
\[ =-\frac{1}{2}\int_{1}^{+\infty} g_s(x)dx=\frac{1}{2f^s(1)}\to \frac{1}{2} \]
as $s\to 0_+$.
\end{proof}
 
想起前段时间刚好做过这样一道题,来源于Tenenbaum解析数论课后题230,证明基于Euler-Maclaurin公式(不了解的可自行wiki)。
 
Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sn^2}=\frac{1}{2}. $
 
\begin{proof}
     Let $f_s(x)=e^{-sx^2}$, by Euler-Maclaurin formula, we have
\[ \sum\limits_{n=0}^{N} f_s(n)=\int_{0}^{N} f_s(x)dx+\frac{f_s(0)+f_s(N)}{2}+\frac{f_s'(0)+f_s'(N)}{12}-\frac{1}{2}\int_{0}^{N} B_2(x)f_s''(x)dx. \]
Since $f_s'(x)=-2sxe^{-sx^2}$, $f_s''(x)=2se^{-sx^2}(2sx^2-1)$, we can get
\[ \int_{0}^{N} f_s(x)dx\to \int_{0}^{+\infty} f_s(x)dx=\int_{0}^{+\infty} e^{-sx^2}dx=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}} \]
\[ f_s(0)+f_s(N)=1+e^{-sN^2}\to 1 \]
\[ f_s'(0)+f_s'(N)=-2sNe^{-sN^2}\to 0 \]
\[ \int_{0}^{N} B_2(x)f_s''(x)dx\to \int_{0}^{\infty} B_2(x)f_s''(x)dx=\int_{0}^{\infty} B_2(x)2se^{-sx^2}(2sx^2-1)dx \]
\[ =2\sqrt{s}\int_{0}^{\infty} B_2(\frac{x}{\sqrt{s}})e^{-x^2}(2x^2-1)dx \]
as $N\to \infty$. Therefore, we have
\[ \sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}-2\sqrt{s}\int_{0}^{\infty} B_2(\frac{x}{\sqrt{s}})e^{-x^2}(2x^2-1)dx. \]
Since $B_2(x)$ is bounded, we can get $\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}+O(\sqrt{s})$ as $s\to 0_+$, then $\sum\limits_{n=0}^{\infty} (-1)^ne^{-sn^2}=2\sum\limits_{n=0}^{\infty} e^{-4sn^2}-\sum\limits_{n=0}^{\infty} e^{-sn^2}=\frac{1}{2}+O(\sqrt{s})$.\\
Another method: Let $\vartheta(t)=\sum\limits_{n\in\mathbb{Z}}e^{-\pi n^2t}$, $t>0$ be the Jocabi theta function. Since
\[ \sum\limits_{n=1}^{\infty} e^{-\pi n^2t}\le \sum\limits_{n=1}^{\infty} e^{-\pi nt}=\frac{e^{-\pi t}}{1-e^{-\pi t}}=O(e^{-\pi t}),  \]
we can get
\[ \vartheta(t)=1+2\sum\limits_{n=1}^{\infty} e^{-\pi n^2t}=1+O(e^{-\pi t}) \]
and
\[ \vartheta(\frac{1}{t})=\sqrt{t}\theta(t)=\sqrt{t}+O(\sqrt{t}e^{-\pi t}). \]
as $t\to \infty$. Then we have
\[ \sum\limits_{n=0}^{\infty} e^{-sn^2}= \frac{1}{2}+\frac{1}{2}\vartheta(\frac{s}{\pi})=\frac{\sqrt{\pi}}{2}\cdot \frac{1}{\sqrt{s}}+\frac{1}{2}+O(s^{-\frac{1}{2}}e^{-\frac{1}{s}}) \]
as $s\to 0_+$.
\end{proof}
 
可能利用Jacobi Theta function也能解决,我没试过。现在问题来了,把指数上的n²换成n³结论还对不对?我问了向老师,他也没什么简单的做法。
 
Suppose $p(x)=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ is a monic polynomial with degree $m\ge 1$, prove that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}. \]
 
这种结论一看感觉就是对的呀,但是为什么做不出来啊?这就尴尬了。
 
大概是大二那年十一假期,闲来无事写过一份文档,是讨论Grandi级数的某些意义下的收敛性的,有兴趣可以去看一下http://duodaa.com/blog/index.php/archives/351/。然后呢,我就想起了以前做过的一些题目。这个是借用Γ函数的性质做的,也可以在北大那道题中取f(x)=x。
 
Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{n^s}=-\frac{1}{2}. $
 
\begin{proof}
     Since
\[ \Gamma(s)=\int_0^{\infty}x^{s-1}e^{-x}dx=\int_0^{\infty}(at)^{s-1}e^{-at}dat=a^s\int_0^{\infty}t^{s-1}e^{-at}dt, \]
we have
\[ a^{-s}=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}e^{-at}dt. \]
Then
\[ \sum\limits_{n=1}^{\infty}\frac{(-1)^n}{n^s}=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}\sum\limits_{n=1}^{\infty}(-1)^ne^{-nt}dt=\frac{1}{\Gamma(s)}\int_0^{\infty}t^{s-1}\frac{-e^{-t}}{1+e^{-t}}dt \]
\[ =-\frac{1}{\Gamma(s)}\int_0^{\infty}\frac{t^{s-1}}{1+e^t}dt=-\frac{\int_0^{\infty}\frac{t^{s-1}}{1+e^t}dt}{\int_0^{\infty}t^{s-1}e^{-t}dt}\to -\frac{1}{2} \]
as $s\to 0_+$.
\end{proof}
 
上下极限讨论的细节可见http://www.duodaa.com/?/question/6738。小试身手之后,再看一道比较难的题,来源于biler数学分析问题集3.58。
 
Let $S(s)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{1+n^2s}}$, prove that $\lim\limits_{s\to 0+}S(s)=\frac{1}{2}$.
 
\begin{proof}
     It is generally known that $\sin z=z\mathop{\Pi}\limits_{n=1}^{\infty}(1-\frac{z^2}{n^2\pi^2})$, or
\[ \sin \pi z=\pi z\mathop{\Pi}\limits_{n=1}^{\infty}(1-\frac{z^2}{n^2}). \]
Take logarithmic derivative, we have
\[ \pi\cot \pi z=\frac{1}{z}+\sum\limits_{n=1}^{\infty}\frac{-2z}{n^2-z^2}=\frac{1}{z}+\sum\limits_{n=1}^{\infty}(\frac{1}{z+n}+\frac{1}{z-n})=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(\frac{1}{z+n}+\frac{1}{z-n}). \]
Since
\[ \frac{1}{\sin \pi z}-\cot \pi z=\frac{1-\cos \pi z}{\sin \pi z}=\tan \frac{\pi z}{2} \]
and
\[ \pi \tan \frac{\pi z}{2}=\pi\cot (\pi \frac{1-z}{2})=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(\frac{1}{\frac{1-z}{2}+n}+\frac{1}{\frac{1-z}{2}-n})  \]
\[ =-\sum\limits_{n\in \mathbb{Z}} \left(\frac{1}{z-(2n+1)}+\frac{1}{z+(2n-1)}\right)=-\sum\limits_{n\in \mathbb{Z}} \left(\frac{1}{z-(2n+1)}+\frac{1}{z+(2n+1)}\right) \]
we have
\[ \frac{\pi}{\sin \pi z}=\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}(-1)^n(\frac{1}{z+n}+\frac{1}{z-n})=z\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{z^2-n^2}, \]
then
\[ \frac{\pi}{\sinh \pi z}=\frac{i\pi}{\sin i\pi z}=z\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{z^2+n^2}. \]
It is easy to show
\[ \int_{0}^{\infty}\frac{\pi dt}{\sinh(\pi y \cosh t)}=\int_{0}^{\infty} y \cosh t\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{y^2\cosh^2t+n^2}dt\]
\[ =\int_{0}^{\infty} \sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{y^2\sinh^2t+(y^2+n^2)}d(y\sinh t) \]
\[ =\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}\arctan\frac{y\sinh t}{\sqrt{y^2+n^2}}|_{0}^{\infty} =\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}}\cdot\frac{\pi}{2}. \]
Let $s=\frac{1}{y^2}$, then we have $y\to+\infty$ as $s\to 0_+$, and
\[ S(s)=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{\sqrt{1+n^2s}}=\frac{1}{2}+\frac{1}{2}\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{1+n^2s}}=\frac{1}{2}+\frac{y}{2}\sum\limits_{n\in \mathbb{Z}}\frac{(-1)^n}{\sqrt{y^2+n^2}} \]
\[ =\frac{1}{2}+\frac{1}{\pi}\int_{0}^{\infty}\frac{\pi y dt}{\sinh(\pi y \cosh t)}. \]
\\
Since
\[ \frac{\pi y}{\sinh(\pi y \cosh t)}\to 0\]
as $y\to \infty$, and
\[ \frac{\pi y}{\sinh(\pi y \cosh t)}\le\frac{1}{\cosh t}\in L^1[0,\infty), \]
we have
\[ \int_{0}^{\infty}\frac{\pi y dt}{\sinh(\pi y \cosh t)}\to 0 \]
as $y\to\infty$ by dominated convergence theorem.
 
\end{proof}
 
下面这个问题又作何解呢?那汤松的实变函数论里面,证明傅里叶级数一些性质的时候,用到了这种级数。
 
Prove that $\lim\limits_{s\to 0+}\sum\limits_{n=1}^{\infty} (-1)^n(\frac{\sin ns}{ns})^2=-\frac{1}{2}$.\\ \\ \\ \\
 
 
好像没什么思路?算一下x²的傅里叶级数就行了。前面的证明几个例子基本上都依赖于一些恒等式,也就是说,每做一个这种类型的题目,就要找到一个恒等式。可万一找不到怎么办?比如多项式那个。
 
给一个题做一个题,效率太低。这就促使我们找到对这类问题一个统一的解法。
 
Suppose $a_s(x)=a(x,s)\in C([0,\infty)\times[0,1])$ and $a_s(x)\in C^1[0,\infty)$. Under the following conditions\\
$\text{(i)}a_0(x)=1,$\\
$\text{(ii)}a_s(\infty)=0 \text{ for } s>0,$\\
$\text{(iii)}\int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx \to 0 \text{ as } s\to 0_+,$\\
we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\frac{1}{2}. \]
 
\begin{proof}
     Since
\[ a_s(0)=\int_{0}^{\infty}a_s'(x)dx=\sum\limits_{n=0}^{\infty}\int_{n}^{n+1}a_s'(x)dx, \]
we have
\[ \sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\sum\limits_{n=0}^{\infty} \left( a_s(2n)-a_s(2n+1) \right)=\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1}a_s'(x)dx \]
\[ =\frac{1}{2}a_s(0)+\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1}a_s'(x)dx-\frac{1}{2}\sum\limits_{n=0}^{\infty}\int_{n}^{n+1}a_s'(x)dx \]
\[ =\frac{1}{2}a_s(0)+\frac{1}{2}\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} \left( a_s'(x)-a_s'(x+1) \right) dx. \]
Since $a_s(0)\to a_0(0)=1$ and
\[ |\sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} \left( a_s'(x)-a_s'(x+1) \right) dx|\le \sum\limits_{n=0}^{\infty}\int_{2n}^{2n+1} |a_s'(x)-a_s'(x+1)| dx\]
\[ \le \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx \to 0  \]
as $s\to 0_+$, we can get the conclusion.
 
\end{proof}
 
条件(iii)看起来有点奇奇怪怪的,那我们就把它换成一个弱一点但看起来很明了的条件。
 
Suppose $a_s(x)=a(x,s)\in C([0,\infty)\times[0,1])$ and $a_s(x)\in C^1[0,\infty)$. Under the following conditions\\
$\text{(i)}a_0(x)=1,$\\
$\text{(ii)}a_s(\infty)=0 \text{ for } s>0,$\\
$(\text{iii}')a_s(x) \text{ is a convex function },$\\
we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^na_s(n)=\frac{1}{2}. \]
 
\begin{proof}
     Since $a_s(x)$ is a convex function, $a_s'(x)$ is increasing, we have
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{\infty} a_s'(x+1)-a_s'(x)dx=a_s(0)-a_s(1)\to 1-1=0 \]
as $s\to 0_+$, it implys (iii).
\end{proof}
 
再看看北大的那道题,就成了上述结论的简单推论。
 
Suppose $f(x)>0$, $f''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}. \]
 
\begin{proof}
     If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}>0 \]
means $a_s(x)$ is convex.
\end{proof}
 
当然了,我们还可以稍微加强一点北大那题的结论,可自行检验这个要更强一点。
 
Suppose $f(x)>0$, $f''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, we have the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}. \]
 
\begin{proof}
     If we take $a_s(x)=e^{-sf(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}\ge 0 \]
means $a_s(x)$ is convex.
\end{proof}
 
现在我们就有能力解决下面的问题了。
 
Suppose $f\in C^{m+1}[0,\infty)$ with $f>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. If $f(\infty)=\infty$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}. \]
 
\begin{proof}
We have solved the case $m=1$ before, and we only consider $m\ge 2$. Without loss of generality, assume $f, f',\cdots, f^{(m)}>0$. Otherwise, we can replace $m$ by a smaller number. If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}. \]
By lemma, we have
\[ (\frac{f'^2}{f''})'\ge (1+\frac{1}{m-1}-\varepsilon)f'>0 \]
and
\[ \frac{f'^2}{f''}\ge (1+\frac{1}{m-1}-\varepsilon)f-C, \]
then $\frac{f''}{f'^2}$ is decreasing to $0$, so the number of roots of $a_s''(x)$ is no more than $1$. If $s$ is small enough, then there is $x_s$ s.t. $a_s''(x_s)=0$, i.e. $s=\frac{f''(x_s)}{f'^2(x_s)}$, and
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{x_s} a_s'(x)-a_s'(x+1)dx+\int_{x_s}^{\infty} a_s'(x+1)-a_s'(x)dx\]
\[ =-a_s(0)+a_s(1)+2(a_s(x_s)-a_s(x_s+1)). \]
Now, we only need to prove that $a_s(x_s)-a_s(x_s+1)\to 0$ as $s\to 0_+$. Since $a_s(x_s+1)-a_s(x_s)=a_s'(\xi_s)$, where $\xi_s\in(x_s, x_s+1)$, we have
\[ |a_s(x_s+1)-a_s(x_s)|\le sf'(\xi_s). \]
Since
\[ f'(\xi_s)=f'(x_s)+\cdots+\frac{f^{(m)}(x_s)}{(m-1)!}(\xi_s-x_s)^{m-1}+\frac{f^{m+1}(\eta_s)}{m!}(\xi_s-x_s)^m \]
\[ \le f'(x_s)+\cdots+\frac{f^{(m)}(x_s)}{(m-1)!}<<f'(x_s) \]
and
\[ s=\frac{f''(x_s)}{f'^2(x_s)}<<\frac{1}{f(x_s)}, \]
we have $sf'(\xi_s)<<\frac{f'(x_s)}{f(x_s)}\to 0$ as $s\to 0_+$.
\end{proof}
 
这里才用了Vinogradov记号,$f<<g$来表示$|f|≤C|g|$对某个正常数$C$成立。如果把上述首一多项式改成指数函数,结论还成立吗?
 
Suppose $f(x)>0$, $f'''(x)\le 0$, and $f(\infty)=\infty$ on $[0,\infty)$, we have the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} (-1)^ne^{-sf(n)}=\frac{1}{2}. \]
 
\begin{proof}
If there is $x_0\ge 0$ s.t. $f''(x_0)\le 0$, then $f''(x)\le 0$ for $x\ge x_0$, we have solved this case. So we can assume $f''(x)>0$, and $f'(x)>0$. If we take $a_s(x)=e^{-sf(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s(sf'^2(x)-f''(x))e^{-sf(x)}. \]
Since $f''(x)$ is decreasing, and $f'(x)$ is strictly increasing, then $\frac{f''(x)}{f'^2(x)}$ is decreasing to $0$, so the number of roots of $a_s''(x)$ is no more than $1$.\\
If $a_s''(x)>0$ for all $x\ge 0$, then $a_s(x)$ is convex.\\
If there is $x_s$ s.t. $a_s''(x_s)=0$, i.e. $s=\frac{f''(x_s)}{f'^2(x_s)}$, then
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=\int_{0}^{x_s} a_s'(x)-a_s'(x+1)dx+\int_{x_s}^{\infty} a_s'(x+1)-a_s'(x)dx\]
\[ =-a_s(0)+a_s(1)+2(a_s(x_s)-a_s(x_s+1)). \]
Since $a_s(x_s+1)-a_s(x_s)=a_s'(\xi_s)$, where $\xi_s\in(x_s, x_s+1)$, we have
\[ |a_s(x_s+1)-a_s(x_s)|\le sf'(\xi_s). \]
Since $f'(\xi_s)-f'(x_s)=f''(\eta_s)(\xi_s-x_s)\le f''(0)$, where $\eta_s \in(x_s, \xi_s)$, and
\[ f'(x_s)=\sqrt{\frac{f''(x_s)}{s}}\le \sqrt{\frac{f''(0)}{s}}, \]
we have
\[ |a_s(x_s+1)-a_s(x_s)|\le s(f''(0)+\sqrt{\frac{f''(0)}{s}})=sf''(0)+\sqrt{sf''(0)}\to 0\]
as $s\to 0_+$.
\end{proof}
 
也许你觉得这个应该也是成立的,但可惜的是,Hardy有个结论却说极限一定不存在。
 
Prove that
$$\lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!^s}=\frac{1}{2}. $$
 
 
Lemma: Suppose $f\in C^{m+1}[0,\infty)$ and $f, f',\cdots, f^{(m)}>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. Prove that
$$\limsup\limits_{x\to +\infty} \frac{f(x)f''(x)}{f'^2(x)}\le 1-\frac{1}{m}.$$
 
\begin{proof}
We can deal with this problem by induction. The case $m=1$ is trivial, so we can assume $m\ge 2$. Since
\[ f(x)=f(0)+f'(0)x+\frac{f''(\xi)}{2}x^2\ge f'(0)x, \]
we can get $f(x)\to\infty$ as $x\to \infty$. Fixed $\varepsilon>0$, we have
\[ \frac{f'f'''}{f''^2}\le 1-\frac{1}{m-1}+\varepsilon \]
for $x\ge x_0$ by induction. Since
\[ (\frac{f'^2}{f''})'=\frac{2f'f''^2-f'^2f'''}{f''^2}=(2-\frac{f'f'''}{f''^2})f'\ge (1+\frac{1}{m-1}-\varepsilon)f', \]
we have
\[ \frac{f'^2}{f''}\ge (1+\frac{1}{m-1}-\varepsilon)f-C, \]
that means
\[ \frac{ff''}{f'^2}\le \frac{f}{(1+\frac{1}{m-1}-\varepsilon)f-C}\to \frac{1}{1+\frac{1}{m-1}-\varepsilon} \]
as $x\to\infty$. Let $\varepsilon\to 0$, we can obtain
$$\limsup\limits_{x\to +\infty} \frac{ff''}{f'^2}\le 1-\frac{1}{m}.$$
\end{proof}
 
这里才用了Vinogradov记号,$f<<g$来表示$|f|≤C|g|$对某个正常数$C$成立。如果把上述首一多项式改成指数函数,结论还成立吗?
 
Suppose $f\in C^{m+1}[0,\infty)$ with $f>0$, $f^{(m+1)}\le 0$, where $m\ge 1$. If $f(\infty)=\infty$, prove the following conclusion
\[ \lim\limits_{s\to 0+} \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{f^s(n)}=\frac{1}{2}. \]
 
\begin{proof}
Without loss of generality, assume $f, f',\cdots, f^{(m)}>0$. Otherwise, we can replace $m$ by a smaller number. If we take $a_s(x)=\frac{1}{f^s(x)}$, then (i), (ii) are trivial, and
\[ a_s''(x)=s\cdot \frac{(s+1)f'^2(x)-f''(x)f(x)}{f^{s+2}(x)}. \]
By lemma, we can get $a_s''(x)>0$ for $x\ge x_0$, that means $a_s(x)$ is convex.
\end{proof}
 
现在我们就有能力解决有关首一多项式的那道题了。
 
Suppose $p(x)=x^m+a_{m-1}x^{m-1}+...+a_1x+a_0$ is a monic polynomial with degree $m\ge 1$, prove that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2}. \]
 
\begin{proof}
     Let $a_s(x)=e^{-sp(x)}$, then (i), (ii) are trivial. We only need to show that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=2N}^{\infty} (-1)^ne^{-sp(n)}=\frac{1}{2} \]
for some positive integer $N$, or equivalent
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-sp(n+2N)}=\frac{1}{2}, \]
so we can consider $p(x+2N)$ instead of $p(x)$. Without loss of generality, we can assume the coefficients of $p(x)$ are positive.
 
Since $a_s'(x)=-sp'(x)e^{-sp(x)}<<sx^{m-1}e^{-sx^m}$, we have $a_s(x)-a_s(x+1)=-a_s'(\xi)<<sx^{m-1}e^{-sx^m}$, where $\xi\in (x,x+1)$. Then
\[ \int_{0}^{\infty} |sp'(x)e^{-sp(x)}-sp'(x)e^{-sp(x+1)}|dx<<s^2\int_{0}^{\infty} x^{2(m-1)}e^{-sx^m}dx\]
\[ =s^{\frac{1}{m}}\int_{0}^{\infty} x^{2(m-1)}e^{-x^m}dx \to 0 \]
as $s\to 0_+$. Since $p'(x)-p'(x+1)<<x^{m-2}$, we have
\[ \int_{0}^{\infty} |sp'(x)e^{-sp(x+1)}-sp'(x+1)e^{-sp(x+1)}|dx<<s\int_{0}^{\infty} x^{m-2}e^{-sx^m}dx\]
\[ =s^{\frac{1}{m}}\int_{0}^{\infty} x^{m-2}e^{-x^m}dx \to 0 \]
as $s\to 0_+$. That means (iii).
\end{proof}
 
 
Suppose $f\in C^1[0,\infty)$, $f'\in L^1[0,\infty)$, and $f(0)=1$, $f(\infty)=0$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
 
\begin{proof}
     Let $a_s(x)=f(sx)$, then (i), (ii) are trivial. Since $a_s'(x)=sf'(sx)$, we have
\[ \int_{0}^{\infty} |a_s'(x)-a_s'(x+1)|dx=s\int_{0}^{\infty} |f'(sx)-f'(s(x+1))|dx\]
\[ =\int_{0}^{\infty} |f'(x)-f'(s+x)|dx=||f'-f_s'||_{L^1} \to 0 \]
as $s\to 0_+$.
\end{proof}
 
这个结论简洁又好用。随便取一个函数就可以出一道题,拿出来坑人真是不亦乐乎。
 
Let $f(x)=\frac{1}{\sqrt{1+x^2}}, (\frac{\sin x}{x})^2, 2(\frac{1}{x}-\frac{1}{e^x-1})$ and so on, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]\\
 
下面说明在某种意义下,可导的条件是不能必不可少的。这里的条件要求导数可积,如果说函数是单调递减的,那么导数自然是可积的,所以对于此类函数而言,连续可微似乎是多于的条件?然而并不是这样。
 
Suppose $F(x)$ is the Cantor-Lebesgue function, let $f(x)=F(1-x)$, then it is obvious that
\[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=1\]
for $s=\frac{1}{3^n}$, and
\[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2} \]
for $s=\frac{2}{3^n}$, where $n\ge 1$.\\
 
然后就猜测会不会对于Cantor-Lebesgue函数而言,对任意s,无穷级数的值都不小于1/2?感觉很有可能是对的,但是费了半天劲没证出来,拿给小罗他也没证出来。最后发现8/45的地方,级数的值是小于1/2的。行吧,竟然有反例。其实我当时的第一想法没想到Cantor-Lebesgue函数,而是强行构造了另一个函数。
 
Suppose $g(x), 0\le x\le 1$ is the unique continuous solution of the following equation
 
$ g(x)=\frac{1}{4}f(2x), 0\le x\le\frac{1}{2},$\\
 
$g(x)=\frac{1}{4}+\frac{3}{4}f(2x-1), \frac{1}{2}\le x\le 1.$\\
or equivalent,
 
\[ g: \sum\limits_{n=1}^{\infty} \frac{a_n}{2^n}\mapsto  \sum\limits_{n=1}^{\infty} \frac{a_n 3^{S_{n-1}}}{4^n} \]
where $a_n=0, 1$, and $S_n=\sum\limits_{i=1}^{n} a_i$, $S_0=0$. Let $f(x)=g(1-x)$, then it is obvious that
 \[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{3}{4} \]
for $s=\frac{1}{2^n}$, where $n\ge 1$.\\
 
对于这个函数而言,会不会极限刚好等于3/4呢?肯定不可能,猜也能猜到极限存在肯定就是1/2。
 
Suppose $f\in C[0,\infty)$ is an decreasing function with $f(0)=1$ and $f(\infty)=0$. If the limit
\[  \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\lambda \]
exists, then $\lambda=\frac{1}{2}$.
 
其实呢,我们有下述更一般些的结论。
 
Suppose $f\in C[0,\infty)$ is an increasing function with $f(0)=1$ and $f(\infty)=0$, then
\[  \liminf\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)\le\frac{1}{2}\le \limsup\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns).\]
 
\begin{proof}
     Let $f_1(x)=\int_{0}^{1}f(xt)dt=\frac{1}{x}\int_{0}^{x} f(t)dt$, then we can get the conclusion.
\end{proof}
 
再积分一次,还可以把连续弱化为在0处连续,结论也是一样的。槊神说我的做法跟他不谋而合。
 
如果函数导数不可积但是二阶导可积呢?比如下面这个例子?
 
Let $f(x)=\frac{\sin\sqrt{x}}{\sqrt{x}}$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
 
幸好,我们也有类似的结论,其证明也都是很类似的。
 
Suppose $f\in C^2[0,\infty)$, $f''\in L^1[0,\infty)$, and $f(0)=1$, $f(\infty)=0$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
\begin{proof}
     Since $f''\in L^1[0,\infty)$, $f'(\infty)$ exists, then we have $f'(\infty)=0$ by $f(\infty)=0$. Let $f_s(x)=f(sx)$, by Euler-Maclaurin formula, we have
\[ \sum\limits_{n=0}^{N} f_s(n)=\int_{0}^{N} f_s(x)dx+\frac{f_s(0)+f_s(N)}{2}+\frac{f_s'(0)+f_s'(N)}{12}-\frac{1}{2}\int_{0}^{N} B_2(x)f_s''(x)dx. \]
Since
$f_s(0)=f(0)=1$, $f_s'(x)=sf'(sx)$, $f_s''(x)=s^2f''(sx)$,
\[ \int_{0}^{N} f_s(x)dx=\frac{1}{s}\int_{0}^{sN}f(x)dx, \]
\[\int_{0}^{N} B_2(x)f_s''(x)dx=s^2\int_{0}^{N} B_2(x)f''(sx)dx=s\int_{0}^{sN} B_2(\frac{x}{s})f''(x)dx \]
we have
\[ \sum\limits_{n=0}^{N} f_{2s}(n)=\frac{1}{2s}\int_{0}^{2sN}f(x)dx+\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{6} \]
\[ -s\int_{0}^{2sN} B_2(\frac{x}{2s})f''(x)dx\]
and
\[ \sum\limits_{n=0}^{2N} f_{s}(n)=\frac{1}{s}\int_{0}^{2sN}f(x)dx+\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{12} \]
\[ -\frac{s}{2}\int_{0}^{2sN} B_2(\frac{x}{s})f''(x)dx\]
Since $f''\in L^1[0,\infty)$, we can get $f'$ is bounded.
Thus,
\[ \sum\limits_{n=0}^{2N} (-1)^nf(ns)=2\sum\limits_{n=0}^{N} f(2ns)-\sum\limits_{n=0}^{2N} f(ns)=2\sum\limits_{n=0}^{N} f_{2s}(n)-\sum\limits_{n=0}^{2N} f_s(n) \]
\[ =\frac{1+f(2sN)}{2}+\frac{sf'(0)+sf'(2sN)}{4}-2s\int_{0}^{2sN} B_2(\frac{x}{2s})f''(x)dx \]
\[ +\frac{s}{2}\int_{0}^{2sN} B_2(\frac{x}{s})f''(x)dx\]
Let $N\to \infty$, we have
\[ \sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}+s\frac{f'(0)}{4}-2s\int_{0}^{\infty} B_2(\frac{x}{2s})f''(x)dx+\frac{s}{2}\int_{0}^{\infty} B_2(\frac{x}{s})f''(x)dx \]
Since $B_2(x)$ is bounded, we can get $\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}+O(s)$.
\end{proof}
 
有兴趣还可以得到高阶导可积对应的情形。最后再考虑一个例子,它的任意阶导数不可积,所以我们只能截断一下。
 
Let $f(x)=\frac{\sin x}{x}$, then we have
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf(ns)=\frac{1}{2}. \]
 
\begin{proof}
     Let $f_N(x)=f(x)\chi_{[0,N\pi]}$, it is not difficult to show that
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^nf_N(ns)=\frac{1}{2}, \]
then we have
\[ \limsup\limits_{s\to 0+}|\sum\limits_{n=0}^{\infty} (-1)^nf(ns)-\frac{1}{2}|=\limsup\limits_{s\to 0+}|\sum\limits_{ns>N\pi}^{\infty} (-1)^nf(ns)|  \]
\[ =\limsup\limits_{s\to 0+}|\sum\limits_{ns>N\pi}^{\infty} \frac{(-1)^n\sin ns}{ns}| \]
By Abel formula, take $a_n=(-1)^n\sin ns=\sin n(\pi+s)$, $b_n=\frac{1}{ns}$, we have
\[ |\sum\limits_{ns>N\pi}^{\infty} \frac{(-1)^n\sin ns}{ns}|\le \frac{A}{N\pi}. \]
Let $N\to \infty$, we can get the conclusion.
\end{proof}
 
证明过程中用到了Abel分部求和公式。
 
\[ \lim\limits_{s\to 0+}\sum\limits_{n=0}^{\infty} (-1)^ne^{-se^n}=\frac{1}{2}? \]
 
也许你觉得这个应该也是成立的,但可惜的是,Hardy有个结论却说极限一定不存在。
 
 
For any $a>1$, the limit
\[ \lim\limits_{x\to 1-}\sum\limits_{n=0}^{\infty} (-1)^nx^{a^n} \]
does not exist.
 
也就是说,我们不能让指数上的部分增长的太快。其实我也忘了Hardy这个结论咋证的了,就不贴证明了。。。
 
当然了,也可以考虑x的傅里叶级数。所以做一些题目的时候千万别被套路了!就像Dirichlet有个定理是说(a,q)=1,那么等差数列a+nq中有无穷个素数一样,经常被拿来出题,比如4n+1,6n+5这种,要是不想想根本原因,不知道要被坑多少次。
 
这类例子很多,希望能起到抛砖引玉的作用,以后看问题能看到更深刻的地方,别老是被人家给套路了。如果想不到,更甚说不愿去想更深层的东西,庸俗的人生大多都相仿。
 
(2018年中科大考研题) Suppose $a_n>0$ and
\[ \left| \sum\limits_{n=1}^{\infty} \frac{\sin (a_nx)}{n^2} \right|\le |\tan x| \]
for $x\in (-1,1)$, prove that $a_n=o(n^2)$ as $n\to\infty$.
\begin{proof}
Since $\sum\limits_{n=1}^{\infty} \cfrac{\sin (a_nx)}{n^2}$ is uniform convergence, we have
\[ \int_{0}^{t}\sum\limits_{n=1}^{\infty} \frac{\sin (a_nx)}{n^2}dx=\sum\limits_{n=1}^{\infty} \frac{1}{n^2}\int_{0}^{t} \sin (a_nx)dx=\sum\limits_{n=1}^{\infty} \frac{1-\cos(a_nt)}{n^2a_n} \]
and
\[ \int_{0}^{t} \tan x dx=-\ln\cos t, \]
then we can get
\[ \sum\limits_{n=1}^{\infty} \frac{1-\cos(a_nt)}{n^2a_n}\le -\ln\cos t  \]
for $t\in (0,1)$, so that
\[ 2\sum\limits_{n=1}^{N} \frac{1-\cos(a_nt)}{n^2a_nt^2}\le -\frac{2\ln\cos t}{t^2}  \]
for any $N\in\mathbb{N}$. Let $t\to 0_+$, then
\[ \sum\limits_{n=1}^{N} \frac{a_n}{n^2}\le 1. \]
Let $N\to\infty$, it is obvious that
\[ \sum\limits_{n=1}^{\infty} \frac{a_n}{n^2}\le 1. \]
\end{proof}

第一届熊赛分析与方程部分试题

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%\title{中国科学技术大学\\
%2015年硕士学位研究生入学考试试题}
%\author{(线性代数与解析几何)}
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{\erhao \CJKfamily{kd}{第一届Xionger网络数学竞赛} }\\
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{\sanhao \textbf{分析与方程部分试题解答} }\\
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{\sihao 整理编辑: 马明辉 \qquad
2018年6月8日}
%{\sanhao \textbf{解答:Eufisky (Xiongge)}}\\
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\begin{enumerate}
%\renewcommand{\labelenumi}{\textbf{{\theenumi}.}}
% 重定义第一级计数显示
\renewcommand{\theenumi}{\textbf{\arabic{enumi}.}}
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\item Suppose $S_n=a_1+\cdots+a_n$, and $T_n=S_1+\cdots+S_n$. If $\lim\limits_{n\to\infty} \frac{1}{n}T_n=s$ and $na_n$ is bounded, then $\lim\limits_{n\to\infty} S_n=s.$
\begin{proof}
We may assume that $\lim\limits_{n\to\infty} \frac{1}{n}T_n=0$, or we can replace $a_1$ by $a_1-s$. For $\forall \varepsilon>0$, there is a large number $N$, $\forall n\ge N$, we have $|T_n|\le \varepsilon n.$ Since
$$T_{n+k}-T_n=S_{n+1}+\cdots+S_{n+k}$$
$$=kS_n+(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}),$$
we have
$$
kS_n=T_{n+k}-T_n-(ka_{n+1}+(k-1)a_{n+2}+\cdots+2a_{n+k-1}+a_{n+k}).
$$
Suppose $|na_n|\le C$, then
$$k|S_n|\le (2n+k)\varepsilon+C\left(\frac{k}{n+1}+\frac{k-1}{n+2}+\cdots+\frac{1}{n+k}\right)\le (2n+k)\varepsilon+C\frac{k^2}{n},$$
or equivalent,
$$|S_n|\le \left(\frac{2n}{k}+1\right)\varepsilon+C\frac{k}{n}.$$
Take $k=[\sqrt{\varepsilon}n]>\frac{1}{2}\sqrt{\varepsilon}n$, then we have
$$|S_n|\le \left(\frac{4}{\sqrt{\varepsilon}}+1\right)\varepsilon+C\sqrt{\varepsilon}=\left(4+\sqrt{\varepsilon}+C\right)\sqrt{\varepsilon}.$$
\end{proof}
 
%\begin{solution}
 
%\end{solution}
 
\item Suppose $f$ is a real value function on $\mathbb{R}$, and $f(x+y)=f(x)+f(y)$ for $\forall x, y\in \mathbb{R}$. If the set $\{(x,f(x)): x\in\mathbb{R}\}$ is not dense in $\mathbb{R}^2$, then $f$ is continuous.
\begin{proof}
In fact, we have $f(x)\equiv f(1)x$. It is not hard to see $f(rx)=rf(x)$ for $x\in\mathbb{R}, r\in \mathbb{Q}$. We denote $c=f(1)$, then $f(r)=cr$ for $r\in \mathbb{Q}$. If $f(x)\equiv cx$ is false, then there exists a number $x_0\in \mathbb{R}-\mathbb{Q}$, s.t. $y_0=f(x_0)\ne cx_0$. Then we have
$$f(x_0s+r)=y_0s+cr=c(x_0s+r)+(y_0-cx_0)s, \forall s, t\in \mathbb{Q}.$$
Fix $y\in\mathbb{R}$. For $\forall \varepsilon>0$, there is a number $s\in\mathbb{Q}$, s.t. $|s(y_0-cx_0)-y|<\varepsilon$, then $|s|< \frac{|y|+\varepsilon}{|y_0-cx_0|}\le \frac{|y|+1}{|y_0-cx_0|}=C$. Take $r\in\mathbb{Q}$ s.t. $|x_0+r|<\varepsilon$, we have
$$f(x_0+r)=c(x_0+r)+(y_0-cx_0),$$
and
$$f(s(x_0+r))=sc(x_0+r)+s(y_0-cx_0),$$
then $|s(x_0+r)|\le C\varepsilon$ and
$$|f(s(x_0+r))-y|\le |sc(x_0+r)|+|s(y_0-cx_0)-y|< (Cc+1)\varepsilon.$$
Fix $(x,y)\in\mathbb{R}^2$. For $\forall \varepsilon>0$, there exists a number $a\in \mathbb{R}$ s.t. $|a|<\varepsilon$, and $|f(a)-(y-cx)|< \varepsilon$. Take $r\in \mathbb{Q}$ s.t. $|x-r|<\varepsilon$, we have
$$f(a+r)=f(a)+cr=f(a)+cx+c(r-x)$$
then $|(a+r)-x|<2\varepsilon$ and
$$|f(a+r)-y|=|f(a)-(y-cx)|+|c(r-x)|< (1+c)\varepsilon.$$
It is a contradiction.
\end{proof}
 
 
 
\item Suppose $\Omega\subset \mathbb{C}$ is a domain, and $u_n=Re f_n$, where $f_n\in H(\Omega)$. If $\{u_n\}$ is uniform convergence on arbitrary compact subset of $\Omega$, and $\{f_n(z_0)\}$ is convergence for some $z_0\in \Omega$. Prove that $\{f_n\}$ is uniform convergence on arbitrary compact subset of $\Omega$.
\begin{proof}
Since the compact set has a finite open covering property, we only need to consider the case $\Omega=\mathbb{D}$. We can obtain the conclusion by the Borel-Carath$\acute{\text{e}}$odory lemma: Suppose $f\in H(\mathbb{D})$. Let $M(r)=\max\limits_{|z|=r} |f(z)|$, $A(r)=\max\limits_{|z|=r} Re f(z)$, then for $0<r<R<1$, we have
$$M(r)\le \frac{2r}{R-r}A(R)+\frac{R+r}{R-r}|f(0)|.$$
\end{proof}
 
 
\item  Suppose $\Omega\subset \mathbb{C}$ is a convex domain, and $f\in H(\Omega)$. If $Re f'(z)\ge 0$ for $\forall z\in\Omega$ and $f$ is not constant function, then $f$ is injective.
\begin{proof}
Consider $e^{-f'(z)}\in H(\Omega)$, then $|e^{-f'(z)}|=e^{-Re f'(z)}\le 1$. If $Re f'(z_0)=0$ for some $z_0\in\Omega$, then $e^{-f'(z)}=const$, or $f'(z)=const$ by the maximum principle. That is to say $f(z)=cz$ with $c\ne 0$, then $f$ is injective. If $Re f'(z)> 0$ for $\forall z\in\Omega$, then for $\forall z_1, z_2\in\Omega$ with $z_1\ne z_2$, we have
$$f(z_2)-f(z_1)=\int_{z_1}^{z_2}f'(z)dz=(z_2-z_1)\int_{0}^{1} f'(z_1+t(z_2-z_1))dt,$$
then
$$Re\frac{f(z_2)-f(z_1)}{z_2-z_1}=\int_{0}^{1} Re f'(z_1+t(z_2-z_1))dt>0.$$
\end{proof}
 
%\begin{enumerate}[itemsep=0pt,parsep=0pt,label=(\arabic*)]
%\item
%\end{enumerate}
 
%\begin{newproof}
 
%\end{newproof}
 
 
 
 
\item  Prove the linear span of $t^ne^{-t}, n=0,1,2,\cdots$ is dense in $L^2(0,\infty)$.
\begin{proof}
Let $M$ be the closed linear span of $t^ne^{-t}, n=0,1,2,\cdots$. Take any $\varphi\in M^{\bot}$, we have
$$\int_{0}^{\infty} t^ne^{-t}\varphi(t)dt=0, n=0,1,2,\cdots.$$
Let $z$ be a complex with $\Im\ z>-1$, and
$$f(z)=\int_{0}^{\infty} e^{izt}e^{-t}\varphi(t)dt.$$
Since $|\frac{e^z-1}{z}|\le C\max\{1,e^{|z|}\}$, we have
$$f'(z)=\lim\limits_{h\to 0}\frac{f(z+h)-f(z)}{h}=\lim\limits_{h\to 0} \int_{0}^{\infty} \frac{e^{iht}-1}{h}e^{izt}e^{-t}\varphi(t)dt$$
$$=\int_{0}^{\infty} ite^{izt}e^{-t}\varphi(t)dt$$
by the dominated convergence theorem. That means $f$ is analytic. Similarly, we have
$$f^{(n)}(z)=\int_{0}^{\infty} i^nt^ne^{izt}e^{-t}\varphi(t)dt.$$
Since $f^{(n)}(0)=i^n\int_{0}^{\infty} t^ne^{-t}\varphi(t)dt=0, n=0,1,2,\cdots$, we have $f(z)\equiv 0$, that means $e^{izt}e^{-t}\in M$. According to the Weierstrass approximation theorem, every continuous periodic function $h(t)$ is the uniform limit of trigonometric polynomials, we can get $h(t)e^{-t}\in M$. Let $g(t)$ be a continuous function with compact support, and $g_1(t)=g(t)e^t$. Denote by $h(t)$ a $T$ periodic function such that
$$h(t)\equiv g_1(t), t\in [0,T],$$
where $T$ is large enough so that the support of $g_1(t)$ is contained in the interval $[0,T]$. Then
$$|g_1(t)-h(t)|\le ||g_1||_{L^{\infty}}\chi_{(T,\infty)}(t),$$
so that
$$|g(t)-h(t)e^{-t}|\le ||g_1||_{L^{\infty}}e^{-t}\chi_{(T,\infty)}(t).$$
Let $T\to\infty$, we can get $g(t)\in M$. Since the set of all continuous functions with compact support is dense in $L^2(0,\infty)$, we have $M=L^2(0,\infty)$.
\end{proof}
 
 
\item Let  $A$ is a unital commutative Banach algebra that is generated by $\{1,x\}$ for some $x\in A$. Then the complement set of $\sigma(x)$ is connected.
\begin{proof}
Let us decompose $\sigma(x)^c$ into its connected components, obtaining an unbounded component $\Omega_{\infty}$ together with a sequence of holes $\Omega_1, \Omega_2, \cdots,$
$$\sigma(x)^c=\Omega_{\infty}\cup\Omega_1\cup\Omega_2\cup\cdots.$$
Let $\Omega=\Omega_1\cup\Omega_2\cup\cdots$. If $\sigma(x)^c$ is not connected, the $\Omega\ne \emptyset$. Suppose $\lambda\in\Omega$, then for arbitrary polynomial $p(z)$, since $p(z)$ is analytic, we have
$$|p(\lambda)|\le \max_{z\in \sigma(x)} |p(z)|=\max\limits_{\omega\in Sp(A)} |\omega(p(x))|\le ||p(x)||$$
by the maximum principle and Gelfand theorem. If we defind
$$\omega: p(x)\mapsto p(\lambda),$$
then $\omega$ is bounded on $\{p(x)\}$. Since $\{p(x)\}$ is dense in $A$, $\omega$ have unique extension on $A$, and $\omega(xy)=\omega(x)\omega(y)$, that means $\omega\in Sp(A)$. Then $\lambda=\omega(x)\in\sigma(x)$, it is a contradiction.
\end{proof}
 
 
\item Suppose $X$ is a compact  Hausdorff space. $\Omega$ is a family of colsed connected subset of $X$, and $\Omega$ is totally order with respect to inclusion relation. Then $Y=\cap\{A: A\in\Omega\}$ is connected.
\begin{proof}
If $Y$ is not connected, then are open set $B$ and $C$, with $B\cap C=\emptyset$, $B\cap Y\ne \emptyset$ and $C\cap Y\ne \emptyset$. Consider the set $Y_1=\cap\{A-(B\cup C): A\in\Omega\}$, then $Y_1=Y-(B\cup C)=\emptyset$. Since $A$ is connected, if $A-(B\cup C)=\emptyset$, or $A\subset B\cup C$, then $A\subset B$, or $A\subset C$, it is impossible. Thus $A-(B\cup C)\ne\emptyset$. Since $A-(B\cup C)$ is compact, and finite intersection is not empty, then $Y_1\ne\emptyset$. It is a contradiction.
\end{proof}
 
\item Suppose the measurable set $A\subset \mathbb{R}$ with $0<m(A)<\infty$. Let $f(x,r)=m(A\cap[x-r,x+r])/2r$, then there exists $x\in\mathbb{R}$ s.t.
$$0<\liminf\limits_{r\to 0_+}f(x,r)\le \limsup\limits_{r\to 0_+}f(x,r)<1.$$
\begin{proof}
Since $0<m(A)<\infty$, there are interval $I_1, I_2$ with $|I_1|=|l_2|=2r_0$ s.t. $m(A\cap I_1)>\frac{1}{2}|I_1|$, $m(A\cap I_2)<\frac{1}{2}|I_2|$.
Since $f(x,r)$ is continuous about $x$, there exists $x_0$ with $f(x_0,r_0)=\frac{1}{2}$. Since we have
$$m(A\cap [x_0-r_0,x_0+r_0])=m(A\cap [x_0-r_0, x_0])+m(A\cap [x_0, x_0+r_0])=r_0,$$
there exists $x_1\in [x_0-\frac{r_0}{2},x_0+\frac{r_0}{2}]$ with $f(x_1,\frac{r_0}{2})=\frac{1}{2}$. Or equivalently, there exists $x_1\in\mathbb{R}$ with
$$|x_1-x_0|\le\frac{r_0}{2}, f(x_1,\frac{r_0}{2})=\frac{1}{2}.$$
Similarly, there exists $x_n\in\mathbb{R}$ with
$$|x_n-x_{n-1}|\le\frac{r_0}{2^n},\quad f\left(x_n,\frac{r_0}{2^n}\right)=\frac{1}{2}.$$
Let $x=\lim\limits_{n\to\infty} x_n$, then $|x-x_n|=\left|\sum\limits_{k=n+1}^{\infty} (x_k-x_{k-1})\right|\le \sum\limits_{k=n+1}^{\infty} |x_k-x_{k-1}|\le\frac{r_0}{2^n}$. For any $r<r_0$, there exists unique $N\ge 1$ s.t. $\frac{r_0}{2^N}\le r<\frac{r_0}{2^{N-1}}$. Then we have
$$\left[x_{N+1}-\frac{r_0}{2^{N+1}}, x_{N+1}+\frac{r_0}{2^{N+1}}\right]\subset\left[x-\frac{r_0}{2^N},x+\frac{r_0}{2^N}\right]\subset[x-r,x+r],$$
and
\begin{align*}
f(x,r)&=\frac{m\left( A\cap \left[ x-r,x+r \right] \right)}{2r}\ge \frac{m\left( A\cap \left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)}{2\times\frac{r_0}{2^{N-1}}}\\
&=\frac{1}{4}f\left( x_{N+1},\frac{r_0}{2^{N+1}} \right) =\frac{1}{8}.
\end{align*}
 
On the other hand, we have
\begin{align*}
m\left( A\cap \left[ x-r,x+r \right] \right) &\le m\left( A\cap \left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)
\\
&+m\left( \left[ x-r,x+r \right] -\left[ x_{N+1}-\frac{r_0}{2^{N+1}},x_{N+1}+\frac{r_0}{2^{N+1}} \right] \right)
\\
&=\frac{r_0}{2^{N+1}}+2r-\frac{2r_0}{2^{N+1}}=2r-\frac{r_0}{2^{N+1}}\le 2r-\frac{1}{4}r=\frac{7}{4}r,
\end{align*}
then
$$f(x,r)=\frac{m(A\cap[x-r,x+r])}{2r}\le\frac{7}{8}.$$
That is to say
$$\frac{1}{8}\le\liminf\limits_{r\to 0_+}f(x,r)\le \limsup\limits_{r\to 0_+}f(x,r)\le\frac{7}{8}.$$
\end{proof}
 
\item Suppose $\{f_n\}_{n=1}^{\infty}$ is a bounded sequence in $L^p$ with $1\le p<\infty$. If $f_n\to f$ a.e., then $f\in L^p$ and
$$\lim\limits_{n\to\infty}\int |f_n|^p-|f_n-f|^p=\int |f|^p.$$
\begin{proof}
We denote $M=\mathop{\sup}_{n\ge 1}\int |f_n|^p<\infty$.Since $f_n\to f$ a.e., we have $|f_n|^p\to|f|^p$ a.e. and by Fatou Lemma
$$\int |f|^p\le\mathop{\underline{\lim}}\limits_{n\to\infty}\int |f_n|^p\le M<\infty,$$
that is to say $f\in L^p$. For $\forall a,b\ge 0$, we have
$$|a^p-b^p|=p\xi^{p-1}|a-b|\le p\max\{a,b\}^{p-1}|a-b|$$
by Lagrange Mean Value Theorem, where $\xi$ is a real number between $a$ and $b$. Then we obtain
$$||f_n|^p-|f_n-f|^p|\le p\max\{|f_n|,|f_n-f|\}^{p-1}|f|.$$
Fixed $\varepsilon>0$. Suppose $A$ is a measurable set with $m(A)<\infty$, and the follow inequality holds
$$\int_{A^c} |f|^p\le \varepsilon.$$
There is a $\delta>0$ such that
$$\int_B |f|^p<\varepsilon \text{ whenever } m(B)<\delta$$
by absolute continuity. Since $f_n\to f$ a.e. on $A$ and $m(A)<\infty$, we can find a measurable subset $a\subset A$ such $m(A\setminus a)<\delta$ and $f_n\to f$ uniformly on $a$ by Egorov Theorem. Then we have
$$\int_a |f_n|^p-|f_n-f|^p\to \int_a |f|^p,$$
as $n\to\infty$, and
$$\int_{a^c} |f|^p=\int_{A\setminus a} |f|^p+\int_{A^c} |f|^p< 2\varepsilon.$$
Since the function $\max\{|f_n|,|f_n-f|\}^{p-1}\in L^{p'}$, and $$||\max\{|f_n|,|f_n-f|\}^{p-1}||_{p'}=||\max\{|f_n|,|f_n-f|\}||_p^{p-1}\le (3M^p)^{\frac{1}{p'}}.$$
We have
\begin{align*}
\int_{a^c} ||f_n|^p-|f_n-f|^p|&\le p\int_{a^c}\max\{|f_n|,|f_n-f|\}^{p-1}|f|\\
&\le p(3M)^{\frac{1}{p'}}(\int_{a^c} |f|^p)^{\frac{1}{p}}
\le p(3M)^{\frac{1}{p'}}(2\varepsilon)^{\frac{1}{p}},
\end{align*}
by H\"{o}lder inequality. Thus we obtain
$$\mathop{\overline{\lim}}\limits_{n\to\infty}\left|\int |f_n|^p-|f_n-f|^p-\int |f|^p\right|\le (1+p(3M)^{\frac{1}{p'}})(2\varepsilon)^{\frac{1}{p}}.$$
\end{proof}
 
\item Suppose $D_n(t)$ are the Dirichlet kernels, and $F_N(t)$ is the $N$-th Fej$\acute{\text{e}}$r kernel given by
$$F_N(t)=\frac{D_0(t)+\cdots+D_{N-1}(t)}{N}.$$
Let $L_N(t)=\min\left(N,\frac{\pi^2}{Nt^2}\right)$. Prove
$$F_N(t)=\frac{1}{N}\frac{1-\cos Nt}{1-\cos t}\le L_N(t)$$
and $\int_{\mathbb{T}} L_N(t)dt\le 4\pi$. If $f\in L^1(\mathbb{T})$ and the $N$-th Ces$\grave{\text{a}}$ro mean of Fourier series is
$$\sigma_N(f)(x)=\frac{S_0(f)(x)+\cdots+S_{N-1}(f)(x)}{N},$$
then $\sigma_N(f)(x)\to f(x)$ for every $x$ in the Lebesgue set of $f$.
\begin{proof}
Since $D_N(t)=\sum\limits_{n=-N}^{N}e^{int}=\frac{\sin(N+\frac{1}{2})t}{\sin\frac{t}{2}}$, we have
$$F_N(t)=\frac{1}{N}\frac{\sin^2\frac{Nt}{2}}{\sin^2\frac{t}{2}}=\frac{1}{N}\frac{1-\cos Nt}{1-\cos t}.$$
Since $|D_N(t)|\le 2N+1$, we can get $F_N(t)\le N$. For $0<x<\frac{\pi}{2}$, we have $\sin x\ge\frac{2}{\pi}x$, then
$$F_N(t)=\frac{1}{N}\frac{\sin^2\frac{Nt}{2}}{\sin^2\frac{t}{2}}\le\frac{1}{N}\frac{1}{\sin^2\frac{t}{2}}\le\frac{\pi^2}{Nt^2}.$$
That mens $F_N(t)\le L_N(t)$. And
$$\int_{\mathbb{T}} L_N(t)dt=2\int_0^{\pi}L_N(t)dt=2\int_0^{\frac{\pi}{N}}Ndt+2\int_{\frac{\pi}{N}}^{\pi}\frac{\pi^2}{Nt^2}dt=4\pi-\frac{2\pi}{N} \le 4\pi.$$
Since $\int_{\mathbb{T}}F_N(t)=1$ and $F_N(t)\le L_N(t)$, we can get $\{F_N(t)\}$ is an approximation to the identity, then $\sigma_N(f)(x)=(f*F_N)(x)\to f(x)$ for every $x$ in the Lebesgue set of $f$.
\end{proof}
 
\item Suppose the sequence $\{a_n\}$ satisfying $a_{n+1}=(4n-2)a_n+a_{n-1}$. Prove that $\{a_n\}$ is convergence if and only if
$$(e-1)a_0+(e+1)a_1=0.$$
\begin{proof}
If $\{a_n\}$ is convengence and not vanishing, then it is obvious that $a_na_{n+1}<0$. We can assume $a_0>0$, $a_1<0$, then $a_{2n}>0$, $a_{2n+1}<0$. Let $b_n=4n-2$ and
$$a_n=p_{n-2}a_1+q_{n-2}a_0, n\ge 2.$$
Since $a_2=b_1a_1+a_0$, and $a_3=(1+b_1b_2)a_1+b_2a_0$, we can get
$$p_0=b_1, p_1=1+b_1b_2, q_0=1, q_1=b_2.$$
Since $a_{n+2}=b_{n+1}a_{n+1}+a_n=b_{n+1}(p_{n-1}a_1+q_{n-1}a_0)+(p_{n-2}a_1+q_{n-2}a_0)=(b_{n+1}p_{n-1}+p_{n-2})a_1+(b_{n+1}q_{n-1}+q_{n-2})a_0$, we can get
$$p_n=b_{n+1}p_{n-1}+p_{n-2}, q_n=b_{n+1}q_{n-1}+q_{n-2}.$$
That is to say
$$\frac{p_n}{q_n}=\left[b_1,b_2,\cdots,b_{n+1}\right]=b_1+\cfrac{1}{b_2+\cfrac{1}{\cdots+\cfrac{1}{b_{n+1}}}},$$
and we have $\frac{p_n}{q_n}\to [b_1,b_2,\cdots]$ as $n\to\infty$. Since
$$a_{2n+1}=p_{2n-1}a_1+q_{2n-1}a_0<0, a_{2n+2}=p_{2n}a_1+q_{2n}a_0>0,$$
we have
$$\frac{p_{2n}}{q_{2n}}(-a_1)<a_0<\frac{p_{2n-1}}{q_{2n-1}}(-a_1),$$
Let $n\to\infty$, we can get
$$a_0+[b_1,b_2,\cdots]a_1=0.$$
On the contrary, if $a_0+[b_1,b_2,\cdots]a_1=0$, since $\left|[b_1,b_2,\cdots]-\frac{p_n}{q_n}\right|<\frac{1}{q_nq_{n+1}}$, we can get
$$|a_n|=|a_1|\cdot\left|p_{n-2}-q_{n-2}[b_1,b_2,\cdots]\right|\le\frac{|a_1|}{q_{n-1}}\to 0.$$
Since
$$\frac{e-1}{e+1}=[0,2,6,10,\cdots]=\cfrac{1}{2+\cfrac{1}{6+\cfrac{1}{10+\cfrac{1}{\cdots}}}},$$
we can get $[b_1,b_2,\cdots]=\frac{e+1}{e-1}$, then $a_0+[b_1,b_2,\cdots]a_1=0$ is equivalent to
$$(e-1)a_0+(e+1)a_1=0.$$
\end{proof}
 
\item Suppose $\{x_n\}$ satisfying $x_1=1$, $x_{n+1}=x_n+\frac{1}{S_n}$, where $S_n=x_1+\cdots+x_n$. Prove that\\
(a) $x_n^2-2\ln S_n$ is increasing and $x_n^2-2\ln S_{n-1}$ is decreasing for $n\ge 2$.\\
(b) $x_n^2-2\ln n-\ln\ln n$ is convengence.\\
(c) $\lim\limits_{n\to\infty} \cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)=\frac{1}{4}$.
\begin{proof}
For $n\ge 2$, we have
\begin{align*}
&\left( x_{n+1}^{2}-\text{2}\ln S_n \right) -\left( x_{n}^{2}-\text{2}\ln S_{n-1} \right) =\left( x_n+\frac{1}{S_n} \right) ^2-x_{n}^{2}+\text{2}\ln \frac{S_{n-1}}{S_n}
\\
&=\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}+\text{2}\ln \left( 1-\frac{x_n}{S_n} \right) <\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-2\left( \frac{x_n}{S_n}+\frac{x_{n}^{2}}{2S_{n}^{2}} \right) =\frac{1-x_{n}^{2}}{2S_{n}^{2}}\le 0.
\end{align*}
 
Similarly,
\begin{align*}
&\left( x_{n+1}^{2}-\text{2}\ln S_{n+1} \right) -\left( x_{n}^{2}-\text{2}\ln S_n \right) =\left( x_n+\frac{1}{S_n} \right) ^2-x_{n}^{2}-\text{2}\ln \frac{S_{n+1}}{S_n}
\\
&=\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-\text{2}\ln \left( 1+\frac{x_{n+1}}{S_n} \right) >\frac{2x_n}{S_n}+\frac{1}{S_{n}^{2}}-2\left( \frac{x_{n+1}}{S_n}-\frac{x_{n+1}^{2}}{2S_{n}^{2}}+\frac{x_{n+1}^{3}}{3S_{n}^{3}} \right)
\\
&=\frac{x_{n+1}^{2}-1}{S_{n}^{2}}-\frac{2}{3}\frac{x_{n+1}^{3}}{S_{n}^{3}}=\frac{x_{n+1}^{2}}{S_{n}^{2}}\left( 1-\frac{1}{x_{n+1}^{2}}-\frac{2x_{n+1}}{3S_n} \right).
\end{align*}
Since $x_1=1, x_2=2$, and $S_1=1, S_2=3$, we have
$$x_{n+1}-S_n=x_n+\frac{1}{S_n}-S_n=\frac{1}{S_n}-S_{n-1}\le \frac{1}{S_2}-S_1=-\frac{1}{2}<0,$$
that means $\frac{2x_{n+1}}{3S_n}\le\frac{2}{3}$, then
$$1-\frac{1}{x_{n+1}^2}-\frac{2x_{n+1}}{3S_n}>\frac{1}{3}-\frac{1}{x_{n+1}^2}>0.$$
That implys (a).\\
Since $x_n^2-2\ln S_n$ is increasing, we can get $x_n^2-2\ln S_n\ge x_2^2-2\ln S_2=4-2\ln 3>1$. Since $x_n\ge 1$, we have $S_n\ge n$, then
$$x_n^2\ge 1+2\ln S_n\ge 1+2\ln n.$$
Since $x_n^2-2\ln S_{n-1}$ is decreasing, we can get $x_n^2-2\ln S_n\le x_{n+1}^2-2\ln S_n\le x_2^2-2\ln S_2=4$, then
$$x_n^2\le 4+2\ln S_n.$$
Since $S_n\ge n$, we can get $x_n\le 1+1+\frac{1}{2}+\cdots+\frac{1}{n-1}\le 2+\ln n$, then $S_n\le nx_n\le n(2+\ln n)$, and
$$x_n^2\le 4+2\ln n+2\ln(2+\ln n).$$
Thus, it is obvious that
$$\frac{x_n}{\sqrt{\ln n}}\to\sqrt{2},$$
and we have
$$\frac{S_n}{n\sqrt{\ln n}}\to \sqrt{2}$$
by Stolz formula. Since $x_n^2-2\ln S_n\le 4$, we can get $x_n^2-2\ln S_n$ is convengence, then
$$x_n^2-2\ln n-\ln\ln n=x_n^2-2\ln S_n+2\ln\frac{S_n}{n\sqrt{\ln n}}$$
is convengence. That implys (b).\\
Since $x_n^2=2\ln n+\ln\ln n+a+o(1)$, we can get
$$\frac{x_n^2}{2\ln n}-1=\frac{\ln \ln n}{2\ln n}+\frac{a+o(1)}{2\ln n},$$
then
\begin{align*}
\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)&=\left(\cfrac{x_n}{\sqrt{2\ln n}}+1\right)^{-1}\cfrac{\ln n}{\ln \ln n}\left(\frac{x_n^2}{2\ln n}-1\right)\\
&\to \frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}.
\end{align*}
Moreover, we have
$$\lim\limits_{n\to\infty} \ln \ln n\left(\cfrac{\ln n}{\ln \ln n}\left(\cfrac{x_n}{\sqrt{2\ln n}}-1\right)-\frac{1}{4}\right)=\frac{a}{4}.$$
\end{proof}
 
\item Suppose $f\in C[0,\infty)$ and for $\forall a\ge0$, we have
$$\lim\limits_{x\to\infty} f(x+a)-f(x)=0.$$
Then there exist $g\in C[0,\infty)$ and $h\in C^1[0,\infty)$ with $f=g+h$, such that
$$\lim\limits_{x\to\infty} g(x)=0, \text{ } \lim\limits_{x\to\infty} h'(x)=0.$$
\begin{proof}
In fact, $f$ is uniformly continuous. Otherwise, there are two sequences $\{x_n\}_{n=1}^{\infty}$, $\{y_n\}_{n=1}^{\infty}$ and a positive $\varepsilon$, such that
$$x_n, y_n\to\infty, |x_n-y_n|\to 0, |f(x_n)-f(y_n)|\ge\varepsilon_0$$
as $n\to\infty$. Consider the functions
$$\varphi_n(x)=f(x_n+x)-f(x_n)$$
and
$$\phi_n(x)=f(y_n+x)-f(y_n)$$
defined on the interval [0,1]. Then we have
$$\varphi(x), \phi(x)\to 0$$
as $n\to\infty$ due to $\lim\limits_{x\to\infty} f(x+a)-f(x)=0$. For $\forall 0<\varepsilon<\cfrac{1}{2}$, there is a set $A_{\varepsilon}\in[0,1]$ such that $m([0,1]\setminus A_{\varepsilon})<\varepsilon$ and $\varphi_n, \phi_n\to 0$ uniformly on $A_{\varepsilon}$ by Egorov Theorem. Take a integer $N$ such that $\forall n\ge N$ and $\forall x\in A_{\varepsilon}$, we have $|x_n-y_n|<1-2\varepsilon$ and
$$|\varphi_n(x)|\le\cfrac{\varepsilon_0}{3},\qquad |\phi_n(x)|\le\cfrac{\varepsilon_0}{3}.$$
Since $m((x_n+A_{\varepsilon})\cap(y_n+A_{\varepsilon}))=m(x_n+A_{\varepsilon})+m(y_n+A_{\varepsilon})-m((x_n+A_{\varepsilon})\cup(y_n+A_{\varepsilon}))\ge 2(1-A_{\varepsilon})-(1+|x_n-y_n|)=1-2\varepsilon-|x_n-y_n|>0$, there is a point $x\in(x_n+A_{\varepsilon})\cap(y_n+A_{\varepsilon})$. We have $x-x_n,x-y_n\in A_{\varepsilon}$, and then
$$|\varphi(x-x_n)|=|f(x)-f(x_n)|\le\cfrac{\varepsilon_0}{3}, \quad |\phi(x-y_n)|=|f(x)-f(y_n)|\le\cfrac{\varepsilon_0}{3},$$
thus $|f(x_n)-f(y_n)|\le \cfrac{2}{3}\varepsilon_0$, it is a contradiction.
 
Let $h(x)=\int_{x}^{x+1} f(t)dt$, and $g(x)=f(x)-h(x)$, then we have
$$h'(x)=f(x+1)-f(x)\to 0$$
as $x\to\infty$. Since $f$ is uniformly continous, there is positive $M$ such $\forall x,y\ge 0$ with $|x-y|\le 1$, we have $|f(x)-f(y)|\le M$. Since $f(x)-f(x+t)\to 0$ as $x\to\infty$ and $|f(x)-f(x+t)|\le M$ for $\forall t\in [0,1]$, by DCT (Dominated convergence theorem), we have
$$g(x)=\int_{0}^{1} f(x)-f(x+t)\to 0 \text{ as } n\to\infty.$$
\end{proof}
 
 
\end{enumerate}
 
\end{document}

向老师的题目

计算积分
\[\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}.\]

解.先作换元$x\to\tanh x$,可得

\begin{align*}&\hspace{0.5cm}\int_0^1{\frac{1}{1+a^2x^2}\left[\left(1-\frac{x}{2}\right)\ln\frac{1+x}{1-x}+\frac{\pi^2x^2}{4}\right]^{-1}\textrm{d}x}\\&=\int_0^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\\&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\left(\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2x^2}{4}}\right)\textrm{d}x}\end{align*}
于是我们考虑函数$$\displaystyle{f\left(z\right)=\frac{1}{1+a^2\coth ^2z}\cdot\frac{\tanh  ^2z-1}{\tanh  z-z}}$$ 的如下围道积分
注意到$\displaystyle{f\left(z\right)=\frac{1}{1+a^2\coth ^2z}\cdot\frac{\tanh  ^2z-1}{\tanh  z-z}}$ 的极点为$z=0$,$\displaystyle{z=\pm\frac{\pi}{2}\mathrm{i}}$ (这两个极点在围道边界上),以及$1+a^2\coth ^2z=0$的根$\displaystyle{z=\pm\mathrm{i}\cdot\mathrm{arccoth}\left(\frac{\mathrm{i}}{a}\right)=\pm\mathrm{i}\cdot\arctan(a)}$, 因此根据留数定理有
\begin{align*}&\int_{ - \infty  - \frac{\pi }{2}{\rm{i}}}^{\infty  - \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z}  - \int_{ - \infty  + \frac{\pi }{2}{\rm{i}}}^{\infty  + \frac{\pi }{2}{\rm{i}}} {f\left( z \right){\rm{d}}z} \\&= 2\pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = 0} \right] + \left( {{\rm{Res}}\left[ {f\left( z \right),{\rm{i}} \cdot \arctan \left( a \right)} \right] + {\rm{Res}}\left[ {f\left( z \right),z =  - {\rm{i}} \cdot \arctan \left( a \right)} \right]} \right)} \right)\\&+ \pi {\rm{i}}\left( {{\rm{Res}}\left[ {f\left( z \right),z = \frac{\pi }{2}{\rm{i}}} \right] + {\rm{Res}}\left[ {f\left( z \right),z =  - \frac{\pi }{2}{\rm{i}}} \right]} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}} - \frac{a}{{2\left( {a - \arctan \left( a \right)} \right)}}} \right) + \pi {\rm{i}}\left( {1 + 1} \right)\\&= 2\pi {\rm{i}}\left( {\frac{3}{{{a^2}}} - \frac{{\arctan \left( a \right)}}{{a - \arctan \left( a \right)}}} \right).\end{align*}
注意到$\displaystyle{\tanh  \left(z\pm\frac{\pi}{2}\textrm{i}\right)=\coth z,\coth \left(z\pm\frac{\pi}{2}\textrm{i}\right)=\tanh  z}$,于是
\begin{align*}&\int_{-\infty -\frac{\pi}{2}\textrm{i}}^{\infty -\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}-\int_{-\infty +\frac{\pi}{2}\textrm{i}}^{\infty +\frac{\pi}{2}\textrm{i}}{f\left(z\right)\textrm{d}z}=\int_{-\infty}^{\infty}{f\left(x-\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}-\int_{-\infty}^{\infty}{f\left(x+\frac{\pi}{2}\textrm{i}\right)\textrm{d}x}\\&=\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x-\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}-\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\coth x-\left(x+\frac{\pi}{2}\textrm{i}\right)^2}\textrm{d}x}\\&=-\pi\textrm{i}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}=2\pi\textrm{i}\left(\frac{3}{a^2}-\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}\right).\end{align*}
因此
\begin{align*}\int_0^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}&=\frac{1}{2}\int_{-\infty}^{\infty}{\frac{1}{1+a^2\tanh  ^2x}\cdot\frac{\coth ^2x-1}{\left(\coth x-x\right)^2+\frac{\pi^2}{4}}\textrm{d}x}\\&=\frac{\arctan\left(a\right)}{a-\arctan\left(a\right)}-\frac{3}{a^2}\end{align*}
 
在开始本文之前,积分镇楼
\[\int_0^1{x^{20}\left[ \left( 1-\frac{x}{2}\ln \frac{1+x}{1-x} \right) ^2+\frac{\pi ^2x^2}{4} \right] ^{-1}\text{d}x}=\frac{5588512716806912356}{374010621408251953125}\]
下面的这些问题其实最开始来自1998年的美国数学月刊征解题
结果1998年的期刊没有答案,1999年以后的期刊又无处可寻,于是我就自己想办法解决这几个题目.我想了好久,突然想到我在前面某一期公众号上解决过类似的问题,于是灵感来了,解决了这几个征解题的同时还导出了几个额外的式子.

设$\displaystyle{\mathrm{Si}(x)=\int_0^x\frac{\sin t}t\mathrm dt}$表示正弦积分函数,求和

\[\sum_{n=1}^\infty\frac{\mathrm{Si}(n\pi)}{n^3}\]

解.[原创]

首先利用分部积分得
\begin{align*}\text{Si}\left( n\pi \right) &=\int_0^{n\pi}{\frac{\sin t}{t}\text{d}t}=\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}=\int_0^{\pi}{\sin nx\text{d}\left( \ln x \right)}=-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\\&=-n\int_0^{\pi}{\cos nx\text{d}\left( x\ln x-x \right)}=n\left[ \left( -1 \right) ^{n-1}\left( \pi \ln \pi -\pi \right) -n\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x} \right]\end{align*}
于是我们可得
\[\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^3}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^2}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\]
而$\displaystyle{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}=\frac{\pi^2}{12}}$,把后一部分式子再分部积分得
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}.\end{align*}
现在考虑函数\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{3}{4}{x^2} - \frac{1}{2}{x^2}\ln x,}&{x \in \left( {0,\pi } \right]}\\{0,}&{x = 0}\end{array}} \right..\]作偶对称以后再作$2\pi$周期延拓,则$f(x)$的Fourier余弦级数为
\[\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}\]
其中$\displaystyle{a_n=\frac{2}{\pi}\int_0^{\pi}{f\left( x \right) \cos nx\text{d}x}=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}$,根据Fourier级数收敛定理可知
\[\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n}=f(0)=0\]
而$\displaystyle{a_0=\frac2\pi\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \text{d}x}=\frac{11}{18}\pi ^2-\frac{1}{3}\pi ^2\ln \pi}$,因此
\[\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \text{d}x}}=\frac{\pi}{2}\sum_{n=1}^{\infty}{a_n}=-\frac{\pi}{4}a_0=\frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3\]
于是最后得到
\[\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}=\frac{\pi ^2}{12}\left( \pi \ln \pi -\pi \right) -\left( \frac{\pi ^3}{12}\ln \pi -\frac{11}{72}\pi ^3 \right) =\frac{5\pi ^3}{72}\]
同样道理我们还能得到
\[\sum_{n=1}^{\infty}{\left( -1 \right) ^n\frac{\text{Si}\left( n\pi \right)}{n^3}}=-\frac{\pi ^2}{6}\left( \pi \ln \pi -\pi \right) -\left( \frac{2\pi ^3}{9}-\frac{\pi ^3}{6}\ln \pi \right) =-\frac{\pi ^3}{18}\]
只不过这时需要利用$\displaystyle{\sum_{n=1}^{\infty}{\frac{1}{n^2}}=\frac{\pi ^2}{6}}$和$\displaystyle{\frac{a_0}{2}+\sum_{n=1}^{\infty}{\left( -1 \right) ^na_n}=f\left( \pi \right) }$即可.在第一步分部积分中我们利用$\ln x$的Fourier展开可得$\displaystyle{\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}\text{Si}\left( n\pi \right)}{n}}=\frac{\pi}{2}}$.

设$\displaystyle{\mathrm{Si}(x)=\int_0^x\frac{\sin t}t\mathrm dt}$表示正弦积分函数,求和

\[\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}\]

解.同上,先分部积分得

\[\text{Si}\left( n\pi \right) =-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\]
于是得到
\[\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}\]
考虑函数$f(x)=\ln x,x\in(0,\pi)$,作偶函数延拓和$\pi$周期延拓得到的余弦级数为
\[\widetilde{f}\left( x \right) =\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n\cos nx}\]
其中$\displaystyle{a_n=\frac{2}{\pi}\int_0^{\pi}{\cos nx\ln x\text{d}x}},a_0=2\ln\pi-2$,由Parseval定理得
\[\frac{a_{0}^{2}}{2}+\sum_{n=1}^{\infty}{a_{n}^{2}}=\frac{2}{\pi}\int_0^{\pi}{f^2\left( x \right) \text{d}x}=\frac{2}{\pi}\int_0^{\pi}{\ln ^2x\text{d}x}=4-4\ln \pi +2\ln ^2\pi\]
因此我们最后得到
\[\sum_{n=1}^{\infty}{\left( \frac{\text{Si}\left( n\pi \right)}{n} \right) ^2}=\sum_{n=1}^{\infty}{\left( \int_0^{\pi}{\cos nx\ln x\text{d}x} \right) ^2}=\frac{\pi^2}2\]

好吧,算到这里,虽然解决了这三个美国数学月刊征解题,但是干脆再无聊点,下面我们算两个五次方的级数,其实套路都一样,但是计算量增加很多.

\[\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{269}{43200}\pi ^5,\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}\text{Si}\left( n\pi \right)}{n^5}}=\frac{4}{675}\pi ^5\]

从\begin{align*}\text{Si}\left( n\pi \right) &=\int_0^{n\pi}{\frac{\sin t}{t}\text{d}t}=\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}=\int_0^{\pi}{\sin nx\text{d}\left( \ln x \right)}=-n\int_0^{\pi}{\cos nx\ln x\text{d}x}\\&=-n\int_0^{\pi}{\cos nx\text{d}\left( x\ln x-x \right)}=n\left[ \left( -1 \right) ^{n-1}\left( \pi \ln \pi -\pi \right) -n\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x} \right]\end{align*}

出发,可得
\[\sum_{n=1}^{\infty}{\frac{\text{Si}(n\pi)}{n^5}}=\sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n^4}\left( \pi \ln \pi -\pi \right)}-\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\]
而$\displaystyle{\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4}=\frac{7\pi^4}{720}}$(利用$\zeta(4)$即可),把后一部分式子再分部积分得
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}\left( \frac{1}{2}x^2\ln x-\frac{3}{4}x^2 \right)}}\\&=\sum_{n=1}^{\infty}\frac1{n^2}{\int_0^{\pi}{\left( \frac{3}{4}x^2-\frac{1}{2}x^2\ln x \right) \cos nx\text{d}x}}\\&=\sum_{n=1}^{\infty}{\frac{1}{n^2}\int_0^{\pi}{\cos nx\text{d}\left( \frac{11}{36}x^3-\frac{1}{6}x^3\ln x \right)}}\\&=\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^n}{n^2}}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}\\&=-\left( \frac{11}{36}\pi ^3-\frac{\pi ^3}{6}\ln \pi \right) \frac{\pi ^2}{12}+\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}.\end{align*}
\begin{align*}\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\left( \frac{11}{36}x^3-\frac{x^3}{6}\ln x \right) \sin nx\text{d}x}}&=\sum_{n=1}^{\infty}{\frac{1}{n}\int_0^{\pi}{\sin nx\text{d}\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right)}}\\&=\sum_{n=1}^{\infty}{\int_0^{\pi}{\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right) \cos nx\text{d}x}}.\end{align*}
下面考虑函数\[f\left( x \right) = \left\{ {\begin{array}{*{20}{c}}{\frac{{25}}{{288}}{x^4} - \frac{1}{{24}}{x^4}\ln x,}&{x \in \left( {0,\pi } \right]}\\{0,}&{x = 0}\end{array}} \right.\]及其余弦级数可得\[\frac{a_0}{2}+\sum_{n=1}^{\infty}{a_n}=f\left( 0 \right) =0\]
其中$\displaystyle{a_n=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{25}{288}x^4-\frac{1}{24}x^4\ln x \right) \cos nx\text{d}x},a_0=\frac{\pi ^4\left( 137-60\ln \pi \right)}{3600}}$,于是$\displaystyle{\sum_{n=1}^{\infty}{a_n}=-\frac{a_0}{2}=-\frac{\pi ^4\left( 137-60\ln \pi \right)}{7200}}$,因此我们最后得到
\[\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^5}}=\frac{7\pi ^4}{720}\left( \pi \ln \pi -\pi \right) +\frac{\pi ^2}{12}\left( \frac{11\pi ^3}{36}-\frac{\pi ^3}{6}\ln \pi \right) -\frac{\pi}{2}\cdot \frac{\pi ^4\left( 137-60\ln \pi \right)}{7200}=\frac{269}{43200}\pi ^5\]
第二个级数同理.

最后再算一个级数

\[\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{\pi ^4}{27}\]

解.首先

\[\frac{\text{Si}\left( n\pi \right)}{n^2}=\frac{\left( -1 \right) ^{n-1}}{n}\left( \pi \ln \pi -\pi \right) -\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}\]
把上式两边乘以$\sin nx$求和,利用Fourier级数的收敛定理得
\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^2}\sin nx}&=\left( \pi \ln \pi -\pi \right) \sum_{n=1}^{\infty}{\frac{\left( -1 \right) ^{n-1}}{n}\sin nx}-\sum_{n=1}^{\infty}{\sin nx\int_0^{\pi}{\sin nx\left( x\ln x-x \right) \text{d}x}}\\&=\frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2}.\end{align*}
最后利用Paserval定理得
\[\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^4}}=\frac{2}{\pi}\int_0^{\pi}{\left( \frac{\left( \pi \ln \pi -\pi \right) x}{2}-\frac{\pi \left( x\ln x-x \right)}{2} \right) ^2\text{d}x}=\frac{\pi ^4}{27}\]

注:从这里,其实我们发现了对任意整数$k$,$\displaystyle{\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^{2k-1}}}(k>1),\sum_{n=1}^{\infty}{\left( -1 \right) ^{n-1}\frac{\text{Si}\left( n\pi \right)}{n^{2k-1}}},\sum_{n=1}^{\infty}{\frac{\text{Si}^2\left( n\pi \right)}{n^{2k}}}}$都可以算出来,只是当次数增加时,这个计算量会越来越大,所以再往上算的话估计只有欧拉和拉马努金这种人去了,不知道这个结果是否可以用类似伯努利数的方法给出通项呢?我也解决不了.

 
最后再做一点额外补充:我把这些题目放到了数学贴吧,某位吧友也提出了一个不错的方法,就是利用下面的Fourier级数
比如
\begin{align*}\sum_{n=1}^{\infty}{\frac{\text{Si}\left( n\pi \right)}{n^3}}&=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\frac{\sin nx}{x}\text{d}x}}=\sum_{n=1}^{\infty}{\frac{1}{n^3}\int_0^{\pi}{\sin nx\text{d}x}\int_0^{+\infty}{\text{e}^{-xy}\text{d}y}}\\&=\int_0^{+\infty}{\text{d}y}\int_0^{\pi}{\text{e}^{-xy}}\frac{x^3-3\pi x^2+2\pi ^2x}{12}\text{d}y=\frac{5}{72}\pi ^3.\end{align*}
类似的也可以处理,只不过这些Fourier级数光算出来计算量已经不得了了,如果提高到五次或者更高次,计算量更大了,不过对于有平方的式子,恐怕还只能利用Parseval定理了.好了,本期的问题就到这里了,下一期暂时没想好写什么QAQ.

美国数学月刊无穷乘积与西西新年祝福题

若$a_i$是超越方程\[\left( {\cos x} \right)\left( {\cosh x} \right) + 1 = 0\]的正实数根从小到大排成的数列, 求证:

$$\sum_{i=1}^{\infty}{a_{i}^{-6}\left(\frac{\sin a_i-\sinh a_i}{\cos a_i+\cosh a_i}\right)^2}=\frac{1}{80}.$$


首先可以肯定,这个方程里$a_i$肯定是解不出的.其次我们有\[{\left( {\sinh {a_i}} \right)^2} = {\left( {\cosh {a_i}} \right)^2} - 1 = \frac{1}{{{{\cos }^2}{a_i}}} - 1 = {\tan ^2}{a_i} \Rightarrow \sinh {a_i} = \left| {\tan {a_i}} \right|.\]
当$\sinh {a_i} = \tan {a_i}$时,我们有\[{\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} - \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\tan ^2}\frac{{{a_i}}}{2}.\]
当$\sinh {a_i} = -\tan {a_i}$时,我们有\[{\left( {\frac{{\sin {a_i} - \sinh {a_i}}}{{\cos {a_i} + \cosh {a_i}}}} \right)^2} = {\left( {\frac{{\sin {a_i} + \tan {a_i}}}{{\cos {a_i} - \frac{1}{{\cos {a_i}}}}}} \right)^2} = {\cot ^2}\frac{{{a_i}}}{2}.\]事实上,两种情况都会出现.接下来大家一起来思考下哈!

以下几个也是不同寻常的题,正是因为莫名其妙、不明觉厉才想一探究竟,希望大家一起来玩!


1、求无穷积分$$\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\int_{1/2}^{+\infty}{\frac{dxdydz}{\prod\limits_{cyc}{\left[242x^5-\left(y-1\right)^5-\left(z+1\right)^5\right]}}}}}.$$


2、设\[{\left( {1 + \frac{1}{x}} \right)^x} = e\left( {1 - \sum\limits_{k = 1}^\infty  {\frac{{{d_k}}}{{{{\left( {\frac{{11}}{{12}} + x} \right)}^k}}}} } \right),\]求
\[{\sum\limits_{k = 1}^\infty  {\frac{1}{{1 + d_k^2}}} }.\]


(2017年10月AMM征解题)求证

\[\prod\limits_{j \ge 1} {{e^{ - 1/j}}\left( {1 + \frac{1}{j} + \frac{1}{{2{j^2}}}} \right)} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.\]

\[{x_n} = \prod\limits_{k = 1}^n {\left( {1 + \frac{1}{k} + \frac{1}{{2{k^2}}}} \right)} = \prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} ,\]
\begin{align*}\frac{{\prod\limits_{k = 1}^{2n} {\left( {1 + \frac{1}{{{k^2}}}} \right)} }}{{{x_n}}} &= \frac{{\prod\limits_{k = 1}^{2n} {\frac{{{k^2} + 1}}{{{k^2}}}} }}{{\prod\limits_{k = 1}^n {\frac{{{{\left( {2k + 1} \right)}^2} + 1}}{{{{\left( {2k} \right)}^2}}}} }} = \frac{{\left( {{1^2} + 1} \right)\left( {{2^2} + 1} \right)\left( {{4^2} + 1} \right) \cdots \left[ {{{\left( {2n} \right)}^2} + 1} \right]}}{{{1^2}{3^2} \cdots {{\left( {2n - 1} \right)}^2}\left[ {{{\left( {2n + 1} \right)}^2} + 1} \right]}}\\&= 2\prod\limits_{k = 1}^n {\left( {1 + \frac{1}{{4{k^2}}}} \right) \cdot } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{4{n^2} + 4n + 2}},\end{align*}
由Wallis公式可知
\[\mathop {\lim }\limits_{n \to \infty } {\left[ {\frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}} \right]^2}\frac{1}{{2n + 1}} = \frac{\pi }{2}.\]
由$\mathrm{sinh} x$的无穷乘积
\[\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} = \prod\limits_{k = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{k^2}}}} \right)} \]
可知
\[\prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{{k^2}}}} \right)} = \frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }},\quad \prod\limits_{k = 1}^\infty {\left( {1 + \frac{1}{{4{k^2}}}} \right)} = \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi },\]
而调和数列
\[{H_n} = \sum\limits_{k = 1}^n {\frac{1}{k}} = \ln n + \gamma + o\left( 1 \right),\]
\[\mathop {\lim }\limits_{n \to \infty } n\prod\limits_{k = 1}^n {{e^{ - 1/k}}} = {e^{ - \gamma }}.\]
因此所求积分为
\[\frac{{\frac{{{e^\pi } - {e^{ - \pi }}}}{{2\pi }}}}{{{e^\gamma } \times \frac{\pi }{2} \times \frac{{{e^{\pi /2}} - {e^{ - \pi /2}}}}{\pi }}} = \frac{{{e^{\pi /2}} + {e^{ - \pi /2}}}}{{\pi {e^\gamma }}}.\]
 
事实上,我们还有
\begin{align*}\cosh \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{{\sinh \left( {\pi x} \right)}}{{\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 + \frac{{{x^2}}}{{{n^2}}}} \right)} ,\\\cos \left( {\pi x} \right) &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{4{x^2}}}{{{{\left( {2n - 1} \right)}^2}}}} \right)} ,&\frac{{\sin \left( {\pi x} \right)}}{{\pi x}} &= \prod\limits_{n = 1}^\infty {\left( {1 - \frac{{{x^2}}}{{{n^2}}}} \right)} ,\end{align*}
 
另外
\begin{align*}\sqrt 2 \sin \left( {\frac{{x + 1}}{4}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{{{{\left( { - 1} \right)}^n}x}}{{2n + 1}}} \right)} ,\\\sqrt {x + 1} \sin \left( {\frac{{\sqrt {x + 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{{4{n^2} - 1}}} \right)} ,\\- \sqrt {x - 1} \mathrm{csch}\left( {\frac{\pi }{2}} \right)\sin \left( {\frac{{\sqrt {x - 1} }}{2}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 - \frac{x}{{4{n^2} + 1}}} \right)} ,\\- \sqrt { - x - 1} \mathrm{csch}\left( {\frac{\pi }{{\sqrt a }}} \right)\sin \left( {\frac{{\sqrt { - x - 1} }}{{\sqrt a }}\pi } \right) &= \prod\limits_{n = 0}^\infty {\left( {1 + \frac{x}{{a{n^2} + 1}}} \right)} ,\\\frac{{{e^{ - \gamma x}}}}{{\Gamma \left( {1 + x} \right)}} &= \prod\limits_{n = 1}^\infty {\frac{{1 + x/n}}{{{e^{x/n}}}}},\end{align*}
 
对于求和,我们有
\begin{align*}\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} - {x^2}}}} &= \frac{1}{{2{x^2}}} - \frac{\pi }{{2x}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {{n^2} - {x^2}} \right)}^2}}}} &= - \frac{1}{{2{x^4}}} - \frac{{{\pi ^2}}}{{4{x^2}}}\mathrm{csc}^2\left( {\pi x} \right) + \frac{\pi }{{4{x^3}}}\cot \left( {\pi x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {2n - 1} \right)}^2} - {x^2}}}} &= \frac{\pi }{{4x}}\tan \left( {\frac{\pi }{2}x} \right),&& \left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left[ {{{\left( {2n - 1} \right)}^2} - {x^2}} \right]}^2}}}} &= \frac{{{\pi ^2}}}{{16{x^2}}}\sec \left( {\frac{\pi }{2}x} \right) - \frac{\pi }{{8{x^3}}}\tan \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty \\\sum\limits_{n = 1}^\infty {\frac{1}{{{n^2} + {x^2}}}} &= \frac{\pi }{{2x}}\coth \left( {\pi x} \right) - \frac{1}{{2{x^2}}}, &&\left| x \right| < \infty\\\sum\limits_{n = 1}^\infty {\frac{1}{{{{\left( {2n - 1} \right)}^2} + {x^2}}}} &= \frac{\pi }{{4x}}\tanh \left( {\frac{\pi }{2}x} \right),&&\left| x \right| < \infty\end{align*}
其中$\mathrm{sinh}x=\frac{e^x-e^{-x}}2,\mathrm{cosh}x=\frac{e^x+e^ {-x}}2,\mathrm{csch}x=\frac2{e^x-e^ {-x}},\mathrm{tanh}x=\frac{e^x-e^ {-x}}{e^x+e^ {-x}},\mathrm{coth}x=\frac{e^x+e^ {-x}}{e^x-e^ {-x}}$.
The Weierstrass factorization theorem. Sometimes called the Weierstrass product/factor theorem.
 
Let $f$ be an entire function, and let $\{a_n\}$ be the non-zero zeros of $ƒ$ repeated according to multiplicity; suppose also that $ƒ''$ has a zero at $z= 0$ of order $m\geq 0$ (a zero of order $m=0$ at $z=0$ means $f(0)\neq 0$.
Then there exists an entire function $g$ and a sequence of integers $\{p_n\}$ such that
 
\[f(z)=z^m e^{g(z)} \prod_{n=1}^\infty E_{p_n}\left(\frac{z}{a_n}\right).\]
 
====Examples of factorization====
 
\begin{align*}\sin \pi z &= \pi z \prod_{n\neq 0} \left(1-\frac{z}{n}\right)e^{z/n} = \pi z\prod_{n=1}^\infty \left(1-\left(\frac{z}{n}\right)^2\right)\\\cos \pi z &= \prod_{q \in \mathbb{Z}, \, q \; \text{odd} } \left(1-\frac{2z}{q}\right)e^{2z/q} = \prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+\tfrac{1}{2}} \right)^2 \right) \end{align*}
 
The cosine identity can be seen as special case of
\[\frac{1}{\Gamma(s-z)\Gamma(s+z)} = \frac{1}{\Gamma(s)^2}\prod_{n=0}^\infty \left( 1 - \left(\frac{z}{n+s} \right)^2 \right)\]
for $s=\tfrac{1}{2}$.
 
Mittag-Leffler's theorem.
 
== Pole expansions of meromorphic functions ==
Here are some examples of pole expansions of meromorphic functions:
 
\begin{align*}\frac{1}{\sin(z)}&= \sum_{n \in \mathbb{Z}} \frac{(-1)^n}{z-n\pi}= \frac{1}{z} + 2z\sum_{n=1}^\infty (-1)^n \frac{1}{z^2 - (n\,\pi)^2},\\\cot(z) &\equiv \frac{\cos (z)}{\sin (z)}= \sum_{n \in \mathbb{Z}} \frac{1}{z-n\pi}= \frac{1}{z} + 2z\sum_{k=1}^\infty \frac{1}{z^2 - (k\,\pi)^2},\\\frac{1}{\sin^2(z)} &= \sum_{n \in \mathbb{Z}} \frac{1}{(z-n\,\pi)^2},\\\frac{1}{z \sin(z)}&= \frac{1}{z^2} + \sum_{n \neq 0} \frac{(-1)^n}{\pi n(z-\pi n)}= \frac{1}{z^2} + \sum_{n=1}^\infty \frac{(-1)^n}{n\,\pi} \frac{2z}{z^2 - (n\,\pi)^2}.\end{align*}

 

好题

设$a,b>0$,而$\displaystyle \int_0^{\infty}{\frac{tdt}{\sqrt[3]{\left( a^3+t^3 \right) \left( b^3+t^3 \right) ^2}}}$.证明:
(1) $\displaystyle I\left( a,b \right) =I\left( \frac{a+2b}{3},\sqrt[3]{b\frac{a^2+ab+b^2}{3}} \right)$.
 
(2) 数列$\{a_n\},\{b_n\}$满足$a_0=a,b_0=b$,且满足$a_{n+1}=\frac{a+2b}{3},b_{n+1}=\sqrt[3]{b_n\frac{a_n^2+a_nb_n+b_n^2}{3}}$,求证$\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n=\frac{I(1,1)}{I(a,b)}$.
 
线性系统理论代数基础,韩京清:http://book.ixueshu.com/book/15b7a59545f6afe2.html
 
http://book.ixueshu.com/book/b9f03de828640dda.html
 
Ramanujan's golden ratio equation
\begin{align*}R\left( e^{-2\pi} \right) &=\frac{e^{-\frac{2\pi}{5}}}{1+\frac{e^{-2\pi}}{1+\frac{e^{-4\pi}}{1+\ddots}}}=\sqrt{\frac{5+\sqrt{5}}{2}}-\phi ,\\R\left( e^{-2\sqrt{5}\pi} \right) &=\frac{e^{-\frac{2\pi}{\sqrt{5}}}}{1+\frac{e^{-2\pi \sqrt{5}}}{1+\frac{e^{-4\pi \sqrt{5}}}{1+\ddots}}}=\frac{\sqrt{5}}{1+\left( 5^{3/4}\left( \phi -1 \right) ^{5/2}-1 \right) ^{1/5}}-\phi ,\end{align*}
其中$\phi=\frac{1+\sqrt5}{2}$为黄金分割数(golden ratio).
 
\begin{align*}R(q) & = q^{\frac{1}{5}}\prod_{n=1}^\infty \frac{(1-q^{5n-1})(1-q^{5n-4})}{(1-q^{5n-2})(1-q^{5n-3})} \\ &= \cfrac{q^{1/5}}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}.\end{align*}
 
Ramanujan–Sato series,https://en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series
\[\frac{1}{\pi} = \frac{2 \sqrt 2}{99^2} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{26390k+1103}{396^{4k}}.\]
Chudnovsky algorithm,https://en.wikipedia.org/wiki/Chudnovsky_algorithm
\[\frac{1}{\pi} = 12 \sum^\infty_{k=0} \frac{(-1)^k (6k)! (545140134k + 13591409)}{(3k)!(k!)^3 \left(640320\right)^{3k + 3/2}}.\]此公式对$\pi$有非常好的计算性能.
 
Ramanujan's Hypergeometric Identity,http://mathworld.wolfram.com/RamanujansHypergeometricIdentity.html
 
$$\sum_{k=-\infty}^\infty 2^k = 0.$$
 
Plot the graphs of the functions $$f(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}+\sqrt{16-x^2}$$ and $$g(x)=\dfrac{2(x^2+|x|-6)}{3(x^2+|x|+2)}-\sqrt{16-x^2}$$ in $x\in[-4,4]$ on the same plane. 
 
 
![enter image description here][1]
 
 
  [1]: http://i.stack.imgur.com/9PdtB.jpg
 
\[\sum_{n=1}^{\infty} \frac{n^{13}}{e^{2\pi n} - 1} = \frac{1}{24}.\]
 
美丽有两种,一种是全新的数学工具;一种是时刻浮现在脑海里你甜美的笑容。

sinx小于x小于tanx,Young不等式,等周问题
张筑生,谢惠民

https://math.stackexchange.com/a/411763/165013 


https://math.stackexchange.com/a/842310/165013 


https://math.stackexchange.com/a/2323155/165013 


https://math.stackexchange.com/a/718750/165013 


https://math.stackexchange.com/a/83952/165013 


https://math.stackexchange.com/a/61727/165013
 
 
 
乌班图下安装TeXLive2017:
 

曲面积分计算

(2012年中科院考研题)设$\rho (x,y,z)$是原点$O$到椭球面$\frac{x^2}2+\frac{y^2}2+z^2=1$的上半部分(即满足$z\geq 0$的部分) $\Sigma$的任一点$(x,y,z)$处的切面的距离,求积分\[\iint_\Sigma \frac z{\rho (x,y,z)}dS.\]
 
所求积分为
 
\[I=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}.\]
 
记$z=\varphi (x,y), (x,y)\in D$,其中$D$为$x^2+y^2=2$.首先有
 
\begin{align*}dS&=\sqrt{1+\left(\frac{\partial \varphi}{\partial x}\right)^2+\left(\frac{\partial \varphi}{\partial x}\right)^2}dxdy=\sqrt{1+\left(\frac{-x}{2z}\right)^2+\left(\frac{-y}{2z}\right)^2}dxdy\\&=\frac1{2z}\sqrt{x^2+y^2+4z^2}dxdy.\end{align*}
 
因此
 
\begin{align*}I&=\iint_{\Sigma}{\frac{z}{\rho \left( x,y,z \right)}dS}=\frac{1}{2}\iint_{\Sigma}{z\sqrt{x^2+y^2+z^2}dS}\\&=\frac{1}{4}\iint_D\sqrt{x^2+y^2+z^2}\sqrt{x^2+y^2+4z^2}dxdy=\frac{1}{4}\iint_D\sqrt{1+\frac{x^2}2+\frac{y^2}2}\sqrt{4-x^2-y^2}dxdy\\&=\frac14\int_0^{2\pi}d\theta\int_0^{\sqrt{2}}r\sqrt{1+\frac{r^2}2}\sqrt{4-r^2}dr=\frac\pi4\int_0^{\sqrt{2}}\sqrt{1+\frac{u}2}\sqrt{4-u}du\\&=\frac\pi{4\sqrt{2}}\int_0^{\sqrt{2}}\sqrt{8+2u-u^2}du=\frac{\sqrt{2}\pi}{16}\left(\sqrt{10-6\sqrt{2}}+9\arcsin \frac{\sqrt 2-1}3+2\sqrt{2}+9\arcsin \frac13\right).\end{align*}
这是因为
\begin{align*}&\int{\sqrt{8+2u-u^2}du}=u\sqrt{8+2u-u^2}-\int{\frac{u-u^2}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\frac{\left( 8+2u-u^2 \right) -8-u}{\sqrt{8+2u-u^2}}du}\\&=u\sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{8+u}{\sqrt{8+2u-u^2}}du}\\&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+\int{\frac{9}{\sqrt{9-\left( u-1 \right) ^2}}du}\\&=\left( u-1 \right) \sqrt{8+2u-u^2}-\int{\sqrt{8+2u-u^2}du}+9\arcsin \frac{u-1}{3}+C,\end{align*}
$$\int{\sqrt{8+2u-u^2}du}=\frac{u-1}{2}\sqrt{8+2u-u^2}+\frac{9}{2}\arcsin \frac{u-1}{3}+C.$$

 

国外论坛好题

Sylvester equation https://en.wikipedia.org/wiki/Sylvester_equation
To seek the maximum value of $S=x_1x_2+x_2x_3+\cdots+x_nx_1$ on this domain:
 
补充:这是Ky Fan-Tausski-Todd inequality,参考Converses of two inequalities of Ky Fan, O. Taussky, and J. Todd以及这里
 
$x_1+x_2+\cdots+x_n=0$ and 
$x_1^2+x_2^2+\cdots+x_n^2=1$.
 
I have made some trivial observations:
 
1) $S\in[-1,1)$ by the rearrangement inequality.
 
2) We can make $S$ arbitrarily close to $1$ by increasing $n$.
 
3) An equivalent problem is to minimise $(x_1-x_2)^2+\cdots+(x_n-x_1)^2$.
 
But does the maximum have a meaningful closed form  for each $n$?
I propose to do a discrete Fourier transform. To this end put $\omega:=e^{2\pi i/n}$. In the following  all sums are over ${\mathbb Z}_n$, unless indicated otherwise. Let
$$y_k:={1\over n}\sum_l x_l\omega^{-kl}\ .$$
Since the $x_l$ are real we have $$y_{-k}=\overline{y_k}\tag{1}$$ for all $k$, furthermore $y_0=\sum_lx_l=0$. One has Parseval's formula
$$n\sum_k y_k\overline{ y_k}=\sum_l x_l^2=1\tag{2}$$
and the inversion formula
$$x_j=\sum_k y_k\omega^{jk}\ .\tag{3}$$
Using $(3)$ one computes
$$S=\sum_j x_jx_{j+1}=\ldots=n\sum_k|y_k|^2\omega^k\ .\tag{4}$$
At this point we have to distinguish the cases (a) $n=2m$, and (b) $n=2m+1$.
 
(a) If $n=2m$ then $\omega^m=-1$, and $(4)$ gives
$$\eqalign{S&=n\left(\sum_{k=1}^{m-1}|y_k|^2(\omega^k+\omega^{-k}) \ +|y_m|^2\omega^m\right)\cr
&=n\left(\sum_{k=1}^{m-1}|y_k|^2\>2\cos{2k\pi\over n} \ -|y_m|^2\right)\ .\cr}$$
Given the conditions $(1)$ and $(2)$ it is easily seen that the optimal admissible choice of the $y_k$ is $$y_1=y_{-1}={1\over\sqrt{2n}}\>,\qquad y_k=0\quad(k\ne\pm1)\ .\tag{5}
$$  This leads to
$$S_{\rm opt}=\cos{2\pi\over n}\ .$$
In particular when $n=4$ one obtains $S_{\rm opt}=0$, as indicated in Zubzub's answer.
 
(b) If $n=2m+1$ then $(4)$ gives
$$S=2n\sum_{k=1}^m |y_k|^2\cos{2k\pi\over n}\ ,$$
and the choice $(5)$ leads again to
$$S_{\rm opt}=\cos{2\pi\over n}\ .$$

In particular when $n=3$ one obtains $S_{\rm opt}=-{1\over2}$, and when $n=5$ one obtains $S_{\rm opt}=\cos{2\pi\over5}\doteq0.309$, as indicated in Zubzub's answer.

见:https://math.stackexchange.com/questions/2213960/optimising-x-1x-2x-2x-3-cdotsx-nx-1-given-certain-constraints


Let $$
A=\left[ \begin{matrix}
A_1& A_3\\
0& A_2\\
\end{matrix} \right]\quad \text{and} \quad B=\left[ \begin{array}{c}
B_1\\
B_2\\
\end{array} \right],
$$and for any eigenvalue $s$ of $A$, we have
$$
\mathrm{rank}\,\left[ A-sI_n,B \right] =n.
$$
Prove there are a real matrix $K\in \mathbb{R}^{r\times n}$ and invertible matrix $T\in \mathbb{R}^{n\times n}$, such that
$$
T\left( A+BK \right) T^{-1}=\left[ \begin{matrix}
A_1& 0\\
0& \bar{A}_2\\
\end{matrix} \right] ,\qquad TB=\left[ \begin{array}{c}
\bar{B}_1\\
B_2\\
\end{array} \right],
$$
where $A\in \mathbb{R}^{n\times n},B\in \mathbb{R}^{n\times r}$. Meanwhile,  $\bar{A}_2$ and $\bar{B}_1$ is real matrix of the proper dimension, $\bar{A}_2$ and $A_1$ Have eigenvalues that are not identical to each other.
 
This problem is about controllability of linear system and pole assignment, so I think we may try controllability canonical form, or let $K=(K_1,K_2)$, then determine the $K_1,K_2$ and $T$, but it seems so difficult.

The statement about the rank of $\left[A - s\,I, B\right]$ implies that the pair $(A,B)$ is [controllable](https://en.wikipedia.org/wiki/Hautus_lemma). Therefore the poles of the resulting matrix should be able to be placed anywhere. The similarity transformation should allow us to separate the eigenvalues/modes and therefore achieve the stated goal
 
$$
T \left(A + B\,K\right) T^{-1} = 
\begin{bmatrix}
A_1 & 0 \\ 0 & \bar{A}_2
\end{bmatrix}, \quad 
T\,B = 
\begin{bmatrix}
\bar{B}_1 \\ B_2
\end{bmatrix},
$$
 
given that
 
$$
A = 
\begin{bmatrix}
A_1 & A_3 \\ 0 & A_2
\end{bmatrix}, \quad
B = 
\begin{bmatrix}
B_1 \\ B_2
\end{bmatrix}.
$$
 
For solving this it is easier to separate the problem in smaller problems. For this I will define $T$ and $K$ as
 
$$
T = 
\begin{bmatrix}
T_1 & T_2 \\ T_3 & T_4
\end{bmatrix}, \quad
K = 
\begin{bmatrix}
K_1 & K_2
\end{bmatrix}.
$$
 
The last goal requires
 
$$
T\,B = 
\begin{bmatrix}
T_1\,B_1 + T_2\,B_2 \\ T_3\,B_1 + T_4\,B_2
\end{bmatrix} = 
\begin{bmatrix}
\bar{B}_1 \\ B_2
\end{bmatrix}.
$$
 
The bottom half of this goal might have infinitely many solution if there is any overlap in the span of $B_1$ and $B_2$. But this problem should be solvable in general, in which case $T_3=0$ and $T_4=I$ should always solve it. Using this then the inverse of $T$ can shown to be
 
$$
T^{-1} = 
\begin{bmatrix}
T_1 & T_2 \\ 0 & I
\end{bmatrix}^{-1} = 
\begin{bmatrix}
T_1^{-1} & -T_1^{-1}\,T_2 \\ 0 & I
\end{bmatrix}.
$$
 
Since nothing is specified about $\bar{B}_1$ then $T_1$ and $T_2$ could be anything for now as long as $T_1$ is invertible. The left hand side of the first goal can now be written as
 
$$
T \left(A + B\,K\right) T^{-1} = 
\begin{bmatrix}
T_1\,A_1\,T_1^{-1} + \bar{B}_1\,K_1\,T_1^{-1} & T_1\,A_3 + T_2\,A_2 - T_1\,A_1\,T_1^{-1}\,T_2 + \bar{B}_1\left(K_2 - K_1\,T_1^{-1}\,T_2\right) \\
B_2\,K_1\,T_1^{-1} & A_2 + B_2\left(K_2 - K_1\,T_1^{-1}\,T_2\right)
\end{bmatrix}.
$$
 
It can be shown that the top and bottom left half can be set equal to the goal by using $T_1=I$ and $K_1=0$. This allows the top and bottom right of the first goal equation to be simplified to
 
$$
\begin{bmatrix}
A_3 + T_2\,A_2 - A_1\,T_2 + \bar{B}_1\,K_2 \\
A_2 + B_2\,K_2
\end{bmatrix} = 
\begin{bmatrix}
0 \\
\bar{A}_2
\end{bmatrix}.
$$
 
For the bottom half of this equation you could just use a pole placement algorithm to find $K_2$, such that none of the poles match those of $A_1$. This should be possible since the pair $(A_2,B_2)$ should be controllable. The top half can then be rewritten as
 
$$
A_3 + T_2\,\bar{A}_2 - A_1\,T_2 + B_1\,K_2 = 0,
$$
 
which can be transformed into a [Sylvester equation](https://en.wikipedia.org/wiki/Sylvester_equation)
 
$$
A\,X + X\,B = C,
$$
 
with $X = T_2$, $A = -A_1$, $B = \bar{A}_2$ and $C = -A_3 - B_1\,K_2$. This equation has an unique solution for $X$ when $A$ and $-B$ do not have a common eigenvalue. This is an identical constraint as mentioned by your problem statement.
 
So to solve this problem you can first do a pole placement with the pair $(A_2,B_2)$ to find $K_2$, avoiding the eigenvalues of $A_1$. And then solve a Sylvester equation after substituting in this obtained $K_2$ in order to find $T_2$. Using these values then the final solution can then be expressed using
 
$$
T = 
\begin{bmatrix}
I & T_2 \\ 0 & I
\end{bmatrix}, \quad
K = 
\begin{bmatrix}
0 & K_2
\end{bmatrix}.
$$

 

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一个偏微分方程求解

$$\frac{\partial J}{\partial t}=\frac{1}{4}\left( \frac{\partial J}{\partial x} \right) ^2-x^2-\frac{1}{2}x^4 \tag 1$$
$$\frac{\partial^2 J}{\partial t\partial x}=\frac{1}{2} \frac{\partial J}{\partial x}\frac{\partial^2 J}{\partial x^2} -2x-2x^3$$
Change of function :
$$\frac{\partial J}{\partial x}=u(x,t)\quad\to\quad \frac{\partial u}{\partial t} -\frac{1}{2} u\frac{\partial u}{\partial x}= -2x-2x^3 \tag 2$$
Characteristic system of equations :
$$\frac{dt}{1}=\frac{dx}{-\frac{1}{2} u}=\frac{du}{-2x-2x^3}$$
First family of characteristic curves, from $\quad -2\frac{dx}{ u}=\frac{du}{-2x-2x^3} :$
$$2udu-(8x+8x^3)dx=0 \quad\to\quad u^2-4x^2-2x^4=c_1$$
Second family of characteristic curves, from $\frac{dt}{1}=\frac{dx}{-\frac{1}{2} u} :$
$$dx+\frac{u}{2}dt=0=dx+\frac{\sqrt{c_1+4x^2+2x^4}}{2}dt$$
$$dt+\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=0 \quad\to\quad t+\int\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=c_2 $$
The integral can be expressed on closed form. The formula involves a special function, namely the Elliptic Integral of the first kind :
http://www.wolframalpha.com/input/?i=integrate+2%2Fsqrt(C%2B4+x%5E2%2B2+x%5E4)&x=0&y=0
 
[![enter image description here][1]][1]
 
In interest of space and in order to make easier the writing, this big formula will be symbolized as :
$$\int\frac{2dx}{\sqrt{c_1+4x^2+2x^4}}=\Psi\left(c_1,x\right)$$
Where $\Psi$ is the above known function. Thus the second family of characteristic curves is :
$$ t+\Psi\left(c_1,x\right)=c_2$$
 
The general solution of the PDE $(2)$ is expressed on the form of the implicit equation :
$$F\left(\left(u^2-4x^2-2x^4\right) \:,\: \left(t+\Psi\left(u^2-4x^2-2x^4\:,\:x\right)\right) \right)=0$$
where $F$ is any differentiable function of two variables.
 
The function $F$ might be determined according to a boundary condition which has to be derived from a given boundary condition of Eq.$(1)$. Nevertheless it appears doubtful to find a closed form for $F$ considering the complicated function $\Psi$.
 
Supposing that the function $F$ be determined, which is optimistic, a more difficult step comes after, to go from $u(x,t)$ to $J(x,y)$, which suppose possible to find a closed form for $\int u(x,t)dx$.
 
 
  [1]: https://i.stack.imgur.com/DLyqp.jpg
 
对于实矩阵$A=\left[ \begin{matrix}A_1& A_3\\0& A_2\\\end{matrix} \right] $