Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

When I took measure theory with Frank Jones' books years ago, I did every problem in the book because I loved its teaching style.

There was one problem that took me 4-5 years to solve. It was problem 23 of Chapter 16 (I've rewritten it to simplify some obscure notation):

Let $F(x)=x^2 \cos(\frac{1}{x})$ away from 0 and 0 at 0. Define the function $V_F(x)=\int_0^x |F'(t)|dt$. Show that $V_F'(0)=\frac{2}{\pi}$.

I finally solved it by assuming part of it went to 0, doing a u-substitution, getting the answer as a series, estimating the series using an integral test, and taking the derivative. I'm not sure I could recreate it from memory.

How are you supposed to solve this problem? Is there an easier way? All the other exercises in this book are pretty self contained or have a warning. Can this be solved with just techniques from the book?

share|improve this question
add comment

1 Answer

up vote 3 down vote accepted

Since $F'(x)=2x\cos\frac{1}{x}+\sin\frac{1}{x}$ and since $V_F(x)=\int_0^x|F'(t)|dt$, $$V_F'(0)=\lim_{x\to 0}\frac{1}{x}\int_0^x|2t\cos\frac{1}{t}+\sin\frac{1}{t}|dt.$$ Noting that $$\frac{1}{x}\int_0^x|2t\cos\frac{1}{t}|dt\le \frac{2}{x}\int_0^xtdt=x,$$ we have $$V_F'(0)=\lim_{x\to 0}\frac{1}{x}\int_0^x|\sin\frac{1}{t}|dt\stackrel{s=\frac{1}{t}}{=}\lim_{x\to 0}\frac{1}{x}\int_{\frac{1}{x}}^\infty\frac{|\sin s|}{s^2}ds.$$

For every $k\in\mathbb{N}$, $\int_{k\pi}^{(k+1)\pi}|\sin s|ds=2$, so $$\frac{2}{\pi^2}(\frac{1}{k+1}-\frac{1}{k+2})<\frac{2}{(k+1)^2\pi^2}<\int_{k\pi}^{(k+1)\pi}\frac{|\sin s|}{s^2}ds<\frac{2}{k^2\pi^2}<\frac{2}{\pi^2}(\frac{1}{k-1}-\frac{1}{k}).$$ Therefore, when $n\pi\le\frac{1}{x}\le(n+1)\pi$, \begin{eqnarray*} \frac{2n}{(n+2)\pi}&=&\frac{2n}{\pi}\sum_{k=n+1}^\infty (\frac{1}{k+1}-\frac{1}{k+2})<n\pi\sum_{k=n+1}^\infty\int_{k\pi}^{(k+1)\pi}\frac{|\sin s|}{s^2}ds<\frac{1}{x}\int_{\frac{1}{x}}^\infty\frac{|\sin s|}{s^2}ds \\ &<&(n+1)\pi\sum_{k=n}^\infty\int_{k\pi}^{(k+1)\pi}\frac{|\sin s|}{s^2}ds<\frac{2(n+1)}{\pi}\sum_{k=n}^\infty(\frac{1}{k-1}-\frac{1}{k})=\frac{2(n+1)}{(n-1)\pi}. \end{eqnarray*} As $x\to 0$, $n\to\infty$, it follows that $V_F'(0)=\frac{2}{\pi}$.

share|improve this answer
    
Nicely done, +1. –  1015 Apr 27 '13 at 17:49
    
@julien: Thank you! –  23rd Apr 27 '13 at 17:52
1  
You're welcome. I've noticed you have produced a few nice answers to some nontrivial questions. And only one of them got the nice answer badge. While so many proofs that $\sqrt{2}$ is irrational get an unsane number of upvotes. I think it is a pity. But all I can legally do is upvote once nice answers like this one. –  1015 Apr 27 '13 at 17:57
1  
@julien: Sincerely thank you once again for your appreciation. To be honest, I also noticed many answers with high upvotes are solutions to uninteresting questions in my/our opinion, and it depressed me for a while. Now I have to accept this fact, and try to upvote nice answers as you do. –  23rd Apr 27 '13 at 18:15
1  
I'll read your answers more carefully from now on! Yeah, lots of votes are meaningless. Even on comments. I think it is a matter of sanity not to pay attention to that too much. –  1015 May 11 '13 at 18:22
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.