正弦和除以项数的极限

hbghlyj

aops
\[\lim_{n\to\infty}\frac1n\sum_{r=1}^n|\cos r|\]
Theorem: if $f(x)$ is periodic (let's say of period $2\pi$ just for specificity) and continuous, and $t$ is not a rational multiple of $\pi,$ then
\[\lim_{n\to\infty}\frac1n\sum_{r=1}^nf(rt)=\frac1{2\pi}\int_0^{2\pi}f(x)\,dx\]
That theorem applies here. All you need to finish it is to compute the average value of $|\cos x|.$

Here's the plan for the proof.

First we prove this theorem is true for a trigonometric polynomial. That is, if $g(x)=\sum_{k=-m}^mc_ke^{ikx}$ and $T_n(g)=\frac1n\sum_{r=1}^ng(rt),$ then $\lim_{n\to\infty}T_n(g)=\frac1{2\pi}\int_0^{2\pi}g(x)\,dx=c_0.$

Then, we use the fact that the trigonometric polynomials are uniformly dense in $C(\mathbb{T}).$ Given an arbitrary continuous $f$ and a $\epsilon>0,$ write $f=g+b,$ where $g$ is a trig polynomial and $|b(x)|<\epsilon$ for all $x,$ then $\limsup_{n\to\infty}T_n(f)\le\frac1{2\pi}\int_0^{2\pi}g(x)\,dx+\epsilon$ and $\liminf_{n\to\infty}T_n(f)\ge\frac1{2\pi}\int_0^{2\pi}g(x)\,dx-\epsilon$, and $\frac1{2\pi}\int_0^{2\pi}g(x)\,dx$ is within $\epsilon$ of $\frac1{2\pi}\int_0^{2\pi}f(x)\,dx.$

Now, how to we prove this for trig polynomials? Suppose $k\ne 0.$ Then $T_n(e^{ikx})=\frac1n\sum_{r=1}^n e^{ikrt}=\frac1n\cdot\frac{e^{ikt}-e^{ik(n+1)t}}{1-e^{ikt}}.$ Therefore $|T_n(e^{ikx})|\le\frac{2}{n|1-e^{ikt}|},$ The condition that $t$ is not a rational multiple of $\pi$ guarantees that $1-e^{ikt}\ne 0,$ and hence $T_n(e^{ikx})\to 0$ as $n\to\infty.$ It should be clear what to do for $k=0.$

hbghlyj

\[\lim_{n\to \infty }\frac{\sum _{k=1}^n  |{\sin k}| }{n}=0\]

uk702

hbghlyj 发表于 2022-9-24 07:41
\[\lim_{n\to \infty }\frac{\sum _{k=1}^n  |{\sin k}| }{n}=0\]

真的？不太相信。

我们知道 |sin(k)| 位于 0~1 之间，单纯凭拍脑袋，假设 $\varepsilon$ 大于 0 但小于1， 感觉 |sin(k)| > $\varepsilon$ 的比例似应该大于某值。

uk702

uk702 发表于 2022-9-24 08:35
真的？不太相信。

我们知道 |sin(k)| 位于 0~1 之间，单纯凭拍脑袋，假设 $\varepsilon$ 大于 0 但小于1 ...

k (k=0,1,2,...) 在以 $\pi$ 为周期的数轴上致密，应该是可信的，这样  |sin(k)| 在 0~1 之间致密，似乎也是可信的。

uk702

用 gp/pari 测试了一下，
gp > n=1000;sum(x=1,n,abs(sin(x)))/n
0.636839697328780

gp > n=10000;sum(x=1,n,abs(sin(x)))/n
0.6366333618509846121

gp > n=100000;sum(x=1,n,abs(sin(x)))/n
0.6366202485919884060

gp > n=1000000;sum(x=1,n,abs(sin(x)))/n
0.63661994525437796572034054


gp > n=10000000;sum(x=1,n,abs(sin(x)))/n
0.6366197992563352204787

实现上，若假设 k（k=1,2,3,...）在以 $\pi$ 为周期的数轴上均匀致密，则应该有：
\begin{align*}
\lim_{n \to \infty} &\frac{\sum_{k=1}^n |sin(k)|}{n}   \\
&= \frac{1}{\pi}\int_0^{\pi} |sin(x)| dx  \\
&= \frac{2}{\pi}\int_0^\frac{{\pi}}{2} sin(x) dx  \\
&= \frac{2}{\pi} \\
&= 0.63661977236758134307553505349...
\end{align*}