1
$\begingroup$

Given that $f(x)= \sin(x + \pi/4)$ is periodic with period $2\pi$. Find the complex Fourier series.

It's quite a moderate tough question. Can someone help me out. Thanks in advance

  • 0
    or expand $\sin(x+a)$ and rewrite $\sin(x)$ and $\cos(x)$ in exponential form...2012-09-01

2 Answers 2

2

Here is your Fourier series,

$ \sin(x+\frac{\pi}{4}) = \cos(\frac{\pi}{4})\sin(x)+ \cos(\frac{\pi}{4})\cos(x)=\frac{1}{\sqrt 2} \sin(x) + \frac{1}{\sqrt 2} \cos(x)\,.$

Note that, $\sin(x+\frac{\pi}{4})$ is orthogonal to $\{ 1, \sin(mx),\cos(mx)\} \,, m=2,3,\dots $ on the $[0,2\pi]$.

Use the identities $ \sin(x) = \frac{ {\rm e}^{ix} - {\rm e}^{-ix} }{2i} $ and $ \cos(x) = \frac{ {\rm e}^{ix} + {\rm e}^{-ix} }{2} $ to get it in the complex form

$ \sin(x+\frac{\pi}{4}) = \left( \frac{\sqrt {2}}{4}i+\frac{\sqrt {2}}{4} \right) {{\rm e}^{-ix}}+\left( -\frac{\sqrt {2}}{4}i+\frac{\sqrt {2}}{4} \right) {{\rm e}^{ix}} $

  • 0
    Thank you Mhenni Benghorbal. I think now I can understand something on it2012-09-03
3

Note: $\sin(x+\pi/4) = \frac{1}{2i}(e^{i(x+\pi/4)}-e^{-i(x+\pi/4)})$. Furthermore,

$ e^{i(x+\pi/4)} = e^{ix}e^{i\pi/4} $

Now, use $e^{i\theta} =\cos(\theta)+i\sin(\theta)$ to find the explicit value of the coeffient $e^{i\pi/4}$. I leave the rest to you.

  • 0
    Thanks James for your expalantion. I'm able to do understand now2012-09-03