fourier series of absolute value of function

Question

I am trying to find the Fourier series of

$$ |\cos(x)| \text{ from } -\pi \leq x<\pi$$

I know that the $$ b_n $$ terms go to 0 because we have the integrand as an odd function of x.

But how can I solve for

$$a_n? $$

I know that $$a_n $$ is even and that

$$ a_n = 1/\pi\int |\cos(x)|\cos(n\pi) $$

But I am not sure how to solve this.

Any ideas?

What is the variable $l$ here? Usually, $l$ is a bad letter to use as a variable, because it is hard to distinguish from the numeral $1$. If you must, use the script $\ell$, LaTeX \ell. In any event, I think there is no value of $l$ for which that is the correct formula... — Thomas Andrews, Nov 17 '14 at 00:04
if $f$ is even, or odd, then $|f|$ is even. $\cos$ is even, so your function is even. — MPW, Nov 17 '14 at 00:06
I know that my function is even...The b_n term was odd as it has sine with it — Jackson Hart, Nov 17 '14 at 00:06

dustin · Accepted Answer · 2014-11-17T00:35:16.620

3

Since $f$ is even, you can find your coefficients as \begin{align} a_0 &= \frac{1}{2\pi}\int_{-\pi}^{\pi}|\cos(x)|dx\\ &= \frac{1}{\pi}\int_{0}^{\pi/2}\cos(x)dx - \frac{1}{\pi}\int_{\pi/2}^{\pi}\cos(x)dx\\ a_n &= \frac{1}{\pi}\int_{-\pi}^{\pi}|\cos(x)|\cos(nx)dx\\ &= \frac{2}{\pi}\int_{0}^{\pi/2}\cos(x)\cos(nx)dx - \frac{2}{\pi}\int_{\pi/2}^{\pi}\cos(x)\cos(nx)dx \end{align} Useful identity to use $$ \cos[(n+1)x]+\cos[(n-1)x] = 2\cos(x)\cos(nx) $$

edited Nov 17 '14 at 00:35

answered Nov 17 '14 at 00:14

dustin

8,241

Can you explain how you go from step 3 to step 4? – Jackson Hart Nov 17 '14 at 00:17
@JacksonHart $x^2$ is even so $$\int_{-1}^1x^2dx = 2\int_0^1x^2dx$$ correct? – dustin Nov 17 '14 at 00:18
Are you sure we can say this though? I feel like this argument only is correct because a Fourier series is periodic, but would not be correct otherwise – Jackson Hart Nov 17 '14 at 00:19
@JacksonHart all even integrals can be written as 2 times the integral over half the original domain. – dustin Nov 17 '14 at 00:20
OK. And we solve this with integration by parts? – Jackson Hart Nov 17 '14 at 00:21
Are you sure, neither would reduce out? – Jackson Hart Nov 17 '14 at 00:23
but should it really effect the domain of the other intergral it is multiplied by? I mean I see what you are saying, it just seems weird that you can substitute like this and change domain of both integrals – Jackson Hart Nov 17 '14 at 00:30
@JacksonHart I dont understand your comment. – dustin Nov 17 '14 at 00:35
where did you find that identity? – Jackson Hart Nov 17 '14 at 00:39
Manipulation of $\cos(a+b)$ rules http://en.wikipedia.org/wiki/List_of_trigonometric_identities – dustin Nov 17 '14 at 00:41
so now 4 integrals? – Jackson Hart Nov 17 '14 at 00:42
Case $n=1$ is zero so only consider $n\neq 1$ and all easy integrals to handle. – dustin Nov 17 '14 at 00:45
@JacksonHart it can be done by int by parts as well. – dustin Nov 17 '14 at 00:52

fourier series of absolute value of function

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