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I know that this is on the boundaries of what's allowed, but hopefully someone'll answer before it gets closed!

What (periodic) function has Fourier series the harmonic series? I really want the even (cosine) terms to be the harmonic series and no odd terms.

Edit: so that the record is perfectly clear, what I wanted was a function with Fourier series

$\sum\sb {n \ge 1} \frac{1}{n} \cos(n \pi t)$

edited Nov 4 at 20:10

asked Nov 4 at 18:38

Andrew Stacey
10k●12●68

	I wouldn't vote to delete the questions. It's not one of your best, but it's perfectly reasonable. – Ben Webster♦ Nov 4 at 20:09
	I'm curious now, which is my best? You've only got 4 to choose from! Anyway, I wanted this for a lecture tomorrow so was hoping to save myself some time and I did. – Andrew Stacey Nov 4 at 20:12

Qiaochu Yuan · Accepted Answer · 2009-11-04 20:14:30Z UTC

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It's a standard series computation to show that $\sum_{n \ge 1} \frac{x^n}{n} = \log \frac{1}{1 - x}$ . Now substitute x = e^{it} and take the real part.

(As an aside, the reason I write the identity this way is that this is the version which is combinatorially significant.)

edited Nov 4 at 20:14

answered Nov 4 at 18:41

Qiaochu Yuan
18.6k●1●43●140

Andrew Stacey · Answer 2 · 2009-11-04 20:24:51Z UTC

Qiaochu Yuan's original answer was very helpful, but working through it left me with a question or two. I couldn't put the workings in a comment so I put them here. It turned out that there was a misunderstanding as to exactly what coefficients I wanted. Once that was resolved, his original answer sufficed. However, just in case someone else ponders this question, I'll leave my workings here to save them the effort.

Following the hint from Qiaochu, we start from

$\sum_{n=0} x^n = (1 - x)^{-1}$

and integrate term-by-term to get

$\sum_{n=1} \frac{1}{n} x^n = -\log(1 - x)$

now we substitute in $x = e^{\pi i t}$ to get

$\sum_{n=1} \frac{1}{n} e^{n \pi it} = -\log(1 - e^{\pi it})$

of which we then take the real part:

$\sum_{n=1} \frac{1}{n} \cos(n \pi it) = -\Re\log(1 - e^{\pi it})$

Excluding the case where t=0, we can use the standard branch of the logarithm and so use the identity $\log(re^{i\theta}) = \log(r) + i\theta$ to deduce that we want the logarithm of the absolute value of $1 - e^{\pi i t}$ . Since squaring is easily taken care of, we are lead to consider:

$|1 - e^{\pi i t}|^2 = (1 - e^{\pi i t})(1 - e^{-\pi it}) = 1 - e^{\pi i t} - e^{-\pi it} + 1 = 2 - 2 \cos(\pi t)$

and thus conclude that

$\sum_{n=1} \frac{1}{n} \cos(n \pi it) = - \frac{1}{2} \log 2(1 - \cos(\pi t))$

David Speyer · Answer 3 · 2009-11-04 18:46:09Z UTC

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To add to Qiaochu's answer: Of course, that sum does not converge for any real t. But I believe that it should Cesaro converge for all t not in 2piZ, because sum e^{nit}/n converges for all such t.

answered Nov 4 at 18:46

David Speyer
23.6k●1●26●89

For the record: This refers to Qiaochu's original answer. – Andrew Stacey Nov 4 at 21:46

What function has fourier series the harmonic series?

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