I've been studying Fourier series and in trying to compute the Fourier series for the function $f: (-\pi,\pi)\to \mathbb{R}$ given by $f(x)=|\sin x|$ I've found something quite strange that I'm not being able to understand what I've done wrong. First, the Fourier series is

$$F_{(-\pi,\pi)}[f](x)=\dfrac{a_0}{2}+\sum_{n=1}^\infty a_n \cos nx+b_n \sin nx$$

first we see quite easily that $b_n = 0$ for all $n\in \mathbb{N}$ while $a_0 = 4/\pi$. On the other hand we have

$$a_n = \dfrac{1}{\pi} \int_{-\pi}^\pi f(x) \cos nx dx = \dfrac{2}{\pi}\int_0^\pi \sin x \cos nx dx$$

then I've used the fact that $\sin x = \frac{1}{2i}(e^{ix}-e^{-ix})$ and $\cos nx = \frac{1}{2}(e^{inx}+e^{-inx})$ to find out that

$$\sin x \cos nx = \dfrac{1}{4i}(e^{ix}-e^{-ix})(e^{inx}+e^{-inx})=\dfrac{1}{2}\left[\sin((n+1)x)-\sin((n-1)x)\right]$$

so that substituting this on the formula for $a_n$ gives

$$a_n = \dfrac{-2}{\pi(n^2-1)}[(-1)^n+1]$$

this certainly doesn't work for $n=1$ because we get a zero on the denominator. On the other hand, using the formula for $a_n$ with $n=1$ gives

$$a_1 = \dfrac{2}{\pi}\int_0^\pi \sin x \cos x dx = 0.$$

Why the calculation I did doesn't work for $n=1$? I can't find what I've done wrong. Is there something I've missed?

  • $\begingroup$ Well, you said "substituting this on he formula for $a_n$" but you did a lot more than that. Write it out and things will be more evident. $\endgroup$ – Hurkyl Aug 16 '15 at 2:18
  • $\begingroup$ You noticed that the denominator is 0, but have you looked at the numerator? $\endgroup$ – Paul Sinclair Aug 16 '15 at 2:26
  • $\begingroup$ @Hurkyl, now I saw it. On the integral, there is one $\sin ((n-1)x)$ which for $n=1$ is simply zero. When I integrate this I divide by $n-1$ so that for $n=1$ this can't be done. So I should treat the case $n = 1$ with more care and the case $n > 1$ as I've done. Thanks. $\endgroup$ – user1620696 Aug 16 '15 at 2:28

It does work in a (very poor) sense, because $(-1)^n = -1$ for $n=1$, so in $$ a_n = \frac{-2}{\pi(n^2-1)}[(-1)^n+1] $$ there are terms in the numerator and denominator that are zero. So if you prioritize the $(-1)^n+1 = 0$ part then we could interpret $a_1$ to be $0$. But it's more rigorous to say that $a_1$ is ill-defined (since $0/0$ is ill-defined), and derive a separate formula. (And the reasoning with competing denominators is bound to lead you to errors eventually; it works coincidentally here.)

To see more clearly why this is necessary, note that your formula for $\sin x\cos nx$ for $n=1$ gives $$ \sin x\cos x = \frac{1}{2}\sin 2x $$ since $\sin(n-1)x = \sin 0 = 0$. So you can't expect to use the formula for the integral of sine: $$ \int_{0}^\pi \sin((n-1)x)~dx = -\frac{1}{n-1}\cos((n-1)x)\bigg|_{0}^\pi = -\frac{1}{n-1}[(-1)^{n-1} - 1], $$ but the equality between the first to the second quantity only holds if $-\frac{1}{n-1}\cos((n-1)x)$ is a bona fide antiderivative of $\sin((n-1)x)$. But for $n=1$, $\sin((n-1)x) = \sin 0 = 0$ and $-\frac{1}{n-1}\cos((n-1)x) = -\frac{1}{0}$, which hardly makes for a convincing antiderivative of $0$. Therefore this case must be handled separately.


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.