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.

The series $$\sum_{n=1}^\infty \frac{\sin nx}{n}$$ is the Fourier series of the odd $2\pi$-periodic extension of $(\pi-x)/2, 0<x<\pi$.

My question is : $$\sum_{n=1}^\infty \frac{\cos nx}{n}$$ is the Fourier series of the even $2\pi$-periodic extension of what function?

share|improve this question
This is Example 1.6.3 in Arne Broman, "Introduction to partial differential equations", 1970. –  TCL Apr 23 '13 at 23:30
add comment

2 Answers

We can ask Mathematica to sum the series and see what comes out.

Simplify[ComplexExpand[Re@Sum[Cos[n*x]/n, {n, 1, Infinity}]]]


Note that I've grabbed just the real part because numerical experiments show that the imaginary part is zero. Of course, we should always check these kinds of computations. Note for example, that this function has a logarithmic singularity at $x=0$, exactly as we'd expect, since the series boils down to the Harmonic series at $x=0$. We can also try to use the computer to compute the Fourier coefficients of the result. I can't quite seem to get the result fully symbolically, but here's a partial result:

(2/Pi) Table[
  Integrate[- Log[2 - 2 Cos[x]]*Cos[n*x]/2, {x, 0, Pi}],
{n, 1, 10}]
(* Out: {1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10} *)

Well, that looks good! I guess we could also look at a plot:

   -Log[2 - 2 Cos[x]]/2,
   Sum[Cos[n*x]/n, {n, 1, 15}]
 }, {x, -8, 8}]

enter image description here

share|improve this answer
Good use of computer! But can anyone prove this? –  TCL Mar 5 '13 at 20:56
add comment
up vote 3 down vote accepted

The function $-\frac{1}{2}\ln(2-2\cos x)$ simplifies to $f(x)=-\ln(2\sin(x/2))$. We will show that the Fourier cosine series of $f(x), 0<x<\pi,$ is $$\sum_{n=1}^\infty \frac{\cos nx}{n}.$$ Firstly, $$\frac{1}{\pi}\int_0^\pi f(x)\,dx=\frac{1}{\pi}\int_0^\pi \ln 2+\ln(\sin(x/2))\,dx=0$$ because $$\int_0^\pi \ln(\sin(x/2)\,dx=2\int_0^{\pi/2}\ln(\sin(u))\,du=\int_0^\pi \ln(\sin(u))\,du=-\pi\ln 2$$ where the last equality can be found in Ahlfors' Complex Analysis p. 159. Let $n\ge 1$ be an integer. Using integration by parts with $u=\ln(\sin(x/2)$ and $dv=\cos(nx)\,dx$, and the fact that $\lim_{x\to 0^+}\sin(nx)\ln(\sin(x/2)=0$, we get (note that $\int_0^\pi \ln2\cos(nx)\,dx=0$) $$\int_0^\pi f(x)\cos nx\,dx=\frac{1}{2n}\int_0^\pi \cot(x/2)\sin nx\,dx=\frac{\pi}{2n}$$ where the last equality is from here. This proves that $$\frac{2}{\pi}\int_0^\pi f(x)\cos nx\,dx=\frac{1}{n}.$$

Note that the even $2\pi$-periodic extension of $-\ln(2\sin(x/2)), 0<x<\pi$ is the $2\pi$-periodic function $$-\ln\left|2\sin\left(\frac{x}{2}\right)\right|=\ln\left| \frac{1}{2}\csc\left(\frac{x}{2}\right)\right|,$$ $-\infty<x<\infty$.

share|improve this answer
add comment

Your Answer


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.