# Thread: Fourier Series Expansion

1. ## Fourier Series Expansion

Hello everyone

I am considering the following equation:

$f(x) = - \sin( x)^2 - \cos( x)$

Which looks like this:

And need to expand it in terms of a Fourier series. I know the above function is continuous, differentiable and $2 \pi$-periodic, hence there should exist a convergent Fourier series of the form:

$\frac{1}{2} a_0 + \sum \limits_{n=1}^{\infty}[a_n \cos(nx) + b_n \sin(nx)]$

With coefficients given as:

$a_n = \frac{1}{\pi} \int \limits_{- \pi}^{\pi} f(x) \cos(n x) \mathrm{d}x$

$b_n = \frac{1}{\pi} \int \limits_{- \pi}^{\pi} f(x) \sin(n x) \mathrm{d}x$

However I get, in both cases, $a_n = 0$ and $b_n=0$. The only thing I find to be non-zero is $a_0 = -1$.
Can anybody help me realizing where I am going wrong?

Kind Regards

2. ## Re: Fourier Series Expansion

The $b_n$ will indeed be zero because $f(x)$ is an even function.

$\sin^2 x + \cos x = \tfrac12(1-\cos 2x) + \cos x$

So the Fourier series is $-\tfrac12 - \cos x + \tfrac12\cos 2x$

Your $a_0 = -1$ is correct though (because the series uses $\tfrac12a_0$).

3. ## Re: Fourier Series Expansion

Thank you very much! Is $\sin^2(x) = \frac{1}{2}(1-\cos(2x))$ a trigonometrical identity?

4. ## Re: Fourier Series Expansion

Yes, from $\cos 2A = \cos^2 A - \sin^2 A$.

Note that you can find the Fourier Series for all purely trigonometric functions simply by using trigonometric identities.

5. ## Re: Fourier Series Expansion

That is nice to know thank you