Limit and Sum

kunny

The function $f(x)$ is defined as follows.

$f(x)=\left\{\begin{array}{cc}1\ \ \ \mbox{ if }\sin x\cdot\cos x \geqq0 \\0\ \ \ \mbox{ if }\sin x\cdot\cos x<0\end{array}...$

Evaluate

$\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \left|f(k+1)-f(k)\right|$

grobber

A first try (sorry if it's wrong: it's a quick mental computation): is it $\frac 2{\pi}$ ?

**Posted:** Fri Dec 17, 2004 12:30 am

limpet · New Member Offline Joined: 12 Dec 2004 Posts: 8

I get $1 - \frac 2{\pi}$ . Is it correct?

Edit: oops, $\frac 2{\pi}$

**Posted:** Fri Dec 17, 2004 12:57 am

kunny

To grober and limpet:

Yes,the answer is $\frac{2}{\pi}$ .How could you obtain it?

kunny

**Posted:** Fri Dec 17, 2004 4:06 pm

grobber

Well, it's very similar to other problems posted here, one or two by you, if I remember correctly.

Since $\pi$ is irrational, it means that the sequence $\{n\}$ is uniformly distributed in $(0,\pi)$ , where $\{\}$ is the fractional part " $\pmod \pi$ ", if I can say so Smile

. The sequence represents the fraction of $k$ 's in $\{1,2,\ldots,n\}$ for which $\{k\},\{k+1\}$ are in different regions of $(0,\pi)$ , where one of the regions is $(0,\frac \pi 2)$ , and the other region is $(\frac \pi 2,\pi)$ . In order for this to happen, $\{k\}$ must lie in $(\frac \pi 2-1,\frac \pi 2)\cup(\pi-1,\pi)$ . The length of this region is $2$ , and if we divide it by the length of the entire interval $(0,\pi)$ , we get the result.

**Posted:** Fri Dec 17, 2004 4:15 pm

kunny

Thank you for your reply, grobber!

I see.I will show you the answer just to make sure.But my idea is almost same as grobber's one.

From the definition in question $f(x)=1$ if $0 \leqq x \leqq\frac{\pi}{2},\pi \leqq x\leqq\frac{3\pi}{2}$ and $f(x)=0$ if $\frac{\pi}{2}<x<\pi,\frac{3\pi}{2}<x<{2\pi}$ .since $1<\frac{\pi}{2}$ , the condition such that $|f(k+1)-f(k)|=1\cdots$ [1]　is the existence of some natural numbers $l$ which satisfies $\frac{l\pi}{2}-1<k<\frac{l\pi}{2}\cdots$ [2].The width of this interval is 1,so $k$ which satisfy is unique and $k$ satifying [2] is one-to-one mapping to this interval. Now let $\frac{(m-1)\pi}{2}<n<\frac{m\pi}{2}\cdots$ [3],according to whether $n$ satisfies $\frac{m\pi}{2}-1 <n$ or not ,the value of $S_n=\sum_{k=1}^n |f(k+1)-f(k)|$ is $m$ or $m-1$ which yields $m-1\leqq S_n\leqq m$ .From [3], since we have $\frac{2n}{\pi}<m<\frac{2n}{\pi}+1$ ,we obtain $\frac{2n}{\pi}-1<S_n<\frac{2n}{\pi}+1$ i.e. $\frac{2}{\pi}-\frac{1}{n}<\frac{S_n}{n}<\frac{2}{\pi}+\frac{1}{n}$ .Consequently by Archimedes'axiom the desired limit is $\lim_{n\to\infty}\frac{S_n}{n}=\frac{2}{\pi}$ .

kunny

**Posted:** Fri Dec 17, 2004 5:58 pm

kunny

I could find the solution posted by Kent Merryfield on Thu Dec 02,2004.
This is copy which is printed out.
Here is the solution.

The answer is $1-\frac{2}{\pi}$ .

Choose $x_k \in\ [0,2\pi)$ such that $k\equiv x_k\mod 2\pi.$ Given the irrationality of $\pi$ , in the long run these $x_k$ may be considered instances a random variable uniformly distributed on $[0,2\pi).$ We have a random variable, $a_ka_{k+1}$ , which depends on $x$ and we are computing the expected value of this function.This function equals $-1$ if $x\in (\pi-1,\pi)$ or $x \in (2\pi-1,2\pi)$ and is $1$ otherwise.So expected value of this random variable is $\frac{1\cdot(2\pi-2)+(-1)\cdot 2}{2\pi}=1-\frac{2}{\pi}.$

**Posted:** Sat Dec 18, 2004 5:57 pm