1-1+1-1+...=1/2? - Eufisky - The lost book
第一届熊赛分析与方程部分试题
几个恒等式证明

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

Eufisky posted @ 2018年7月22日 18:40 in 数学分析 with tags 逆逆 , 608 阅读
 
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}

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