2
$\begingroup$

Let $f:[0,1]\rightarrow \mathbb R$ be continuous. Assume that $\displaystyle\int_{-1}^{1}f(t)\mathrm dt=1$. Evaluate

$$\lim_{n \to +\infty}\int _{-1}^1 f(t)\cos^2(nt) \mathrm dt$$

How to evaluate this? Integration by parts is not yielding anything.

$\endgroup$
  • $\begingroup$ First of all, what do you imagine the answer will be? $\endgroup$ – Simon S Dec 8 '14 at 13:18
  • $\begingroup$ See this. $\endgroup$ – David Mitra Dec 8 '14 at 13:19
  • $\begingroup$ I have no idea how to imagine an answer of a mathematical problem@SimonS $\endgroup$ – Learnmore Dec 8 '14 at 13:23
  • $\begingroup$ @learning maths, do you want me to detail more? Both links have complete proofs of the lemma, are they clear to you? It would be best that you precise what degree of knowledge you've reached, so that answers can be fit to your situation $\endgroup$ – mvggz Dec 8 '14 at 13:26
  • $\begingroup$ thanks @DavidMitra for the link $\endgroup$ – Learnmore Dec 8 '14 at 13:29
3
$\begingroup$

Hints:

1) $\cos^2(u) = \frac{1-cos(2u)}{2}$

2) Riemann's lemma, that is being proven here (you don't integrate on the same interval but the proof is exactly the same): lim$_{n\rightarrow \infty}\int _{-\pi}^\pi f(t)\cos nt\,dt$

If you combine these two you'll find what you want I think

$\endgroup$
  • $\begingroup$ Can u name some books where I can find this $\endgroup$ – Learnmore Dec 8 '14 at 13:31
  • $\begingroup$ @learning maths A book no, since I've learned that from my professor. This is a usual result of fourier analysis saying that Fourier coefficients tends to zero as n goes to infinity. That's also a classic exercice of integration. Here I'll give the wiki link (not very thorough ..) to help you get started on the subject: en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma $\endgroup$ – mvggz Dec 8 '14 at 13:34
  • $\begingroup$ @learning maths no pb :) . You should have a look at the link in my answer, you'll find two proofs of this result. Of course there are a bit of prerequisites but all is stated if I remember correctly. $\endgroup$ – mvggz Dec 8 '14 at 13:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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