8

I've come to a bit of a sticking point in my answer to problem 14A given here

http://www.maths.cam.ac.uk/undergrad/pastpapers/2011/ib/List_IB.pdf

(note that this is a past paper that I am trying for myself and not homework)

Anyway, things would maybe be ok if I could easily tell that

$\sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}=0$

Is this the case? If not what does the sum equal? Is it even convergent?

Also, if anyone has any ideas about the end of the question I posted, where it asks me to comment on my answer, that would be much appreciated.

Ewan Delanoy
  • 61,600
user27182
  • 2,124
  • 17
  • 30
  • if there is a way of me lifting the LaTeX from the link and then pasting it into here then I will happily do that, if someone tells me how. – user27182 Mar 19 '13 at 11:32
  • 1
    $.785398$ (approx.) by http://www.wolframalpha.com/input/?i=%5Csum_%7Bn%3D0%7D%5E%5Cinfty%5Cfrac%7Bsin(2n%2B1)%7D%7B2n%2B1%7D&t=crmtb01&f=rc – Sugata Adhya Mar 19 '13 at 11:43
  • I do not know the value, but it should converge. Use Abel's theorem. – Sungjin Kim Mar 19 '13 at 12:00
  • @Hyeder By the way, speaking about the second part of your question, I’d like to clarify a few things. Am I right, that you ask about the final phrase of the problem 14A? When (in this problem) you use Fourier expansion, do you suppose, that the function $2T$ periodic? And when $g(t)$ is introduced (on the interval $[-\pi,\pi]$) is it supposed, that it is periodic too? If yes, then its’ period is $2\pi$? Does it compare in some way to the period of expansion ($2T$) (they should be equal :))? – Caran-d'Ache Mar 19 '13 at 16:05
  • @Caran-d'Ache. Yes, we are to assume that the period of $g(t)$ is $2\pi$. I don't understand what you mean about its comparing with the expansion where out period is $2T$, can you explain please? And what has got me stuck is the phrase: "What is the sum of the Fourier series for $g(t)$ at $t = 0$? Comment on your answer." We should get $g(0) = \frac{1}{2}$, but the series seems to give me $\frac{1}{2} + \frac{\pi}{4}$ and I can't see why... the Gibbs phenomenon? – user27182 Mar 19 '13 at 16:48
  • 1
    @Hayeder Well, just the form of the Fourier series looks a bit strange to me. One can assume 2 options: 1) The period of $g(t)$ is $2\pi$ and is equal to $2T$, so one has the common "square wave" function. In this case integration over $[0,\pi]$ (where $g(t)\neq 0$) can be turned to integration over $[T, 2T]$, which will greatly simplify the answer (it will be cancel out in the argument of trig functions). 2) The period of $g(t)$ is $2\pi$ but is $2T\neq 2\pi $. So the duty cycle of the rectangular pulses wouldn’t be equal $2$. Then there will be more mess with algebra. – Caran-d'Ache Mar 19 '13 at 17:17
  • @Hayeder The question with Gibbs phenomenon is not that clear, because you do not approximate the function with finite series, but instead you sum up infinite series, where should not be a place for such a phenomenon. So I personally think this is not the point. But from the methodological point of view (as a lecturer) I think that’s what is being asked about. – Caran-d'Ache Mar 19 '13 at 17:18
  • Actually I should correct myself. The summing up of the Fourier series (even if it is infinite) will demonstrate Gibbs phenomenon if the initial function has discontinuity of the second kind. – Caran-d'Ache Mar 19 '13 at 17:27
  • @Caran-d'Ache. The interpretation which is intended, in my opinion, is that $g(t)$ is a square wave such that the shortest distance between any two distinct discontinuities of $g(t)$ is $\pi$. So yes, in the formula for the series at the top of the question, in the argument of our trig functions, we have a cancellation which leaves the argument as $\frac{nt\pi}{\pi} = nt$. So, what would you have written as the comment on your answer? – user27182 Mar 19 '13 at 18:08

2 Answers2

4

Using the result at this link and the references within, you can show that $$ \sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}={\pi\over4}. \tag1$$

To be more specific, use equation (17) on page 4 of Surprising Sinc Sums and Integrals. Plug in $N=0$ and $a_0=1$, then $N=0$ and $a_0=2$, subtract and divide by 2 to get the result (1).

Community
  • 1
3

The correct answer is: $\sum_{n=0}^\infty\frac{\sin(2n+1)}{2n+1}=\frac{\pi}{4}$

You can see for example here: Prudnikov, A. P.; Brychkov, Yu. A.; and Marichev, O. I. Integrals and Series, Vol. 1: Elementary Functions. New York: Gordon and Breach, 1986. Or use Wolfram Mathematica. :)

Caran-d'Ache
  • 3,564