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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

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hint: rewrite this as $u\,e^{ix}+v\,e^{-ix}$ –  Raymond Manzoni Sep 1 '12 at 18:13
    
$\sin(\theta) = \frac{1}{2i}(e^{i\theta}-e^{-i\theta})$ if you're looking for a sum of imaginary exponentials then this is probably a good way to look at it. –  James S. Cook Sep 1 '12 at 18:14
    
Could you please show a few step so that I able to continue –  Garett Sep 1 '12 at 18:33
    
use James' method with $\theta=x+\pi/4$ and put the $i\pi/4$ terms out of the $e^{\pm ix}$ exponentials... –  Raymond Manzoni Sep 1 '12 at 18:42
    
or expand $\sin(x+a)$ and rewrite $\sin(x)$ and $\cos(x)$ in exponential form... –  Raymond Manzoni Sep 1 '12 at 18:51

2 Answers 2

up vote 2 down vote accepted

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

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

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No this is not the fourier Series.What you do is the expansion only –  Garett Sep 1 '12 at 18:31
    
In a somewhat connected vein, see math.stackexchange.com/questions/71593/… –  James S. Cook Sep 1 '12 at 18:45
1  
@Garett: This is the Fourier series. You can use $ {\rm e}^{ix} =\cos(x)+i\sin(x) $ to change to the complex form. –  Mhenni Benghorbal Sep 1 '12 at 18:47
    
Thank you Mhenni Benghorbal. I think now I can understand something on it –  Garett Sep 3 '12 at 16:44

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.

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Thanks James for your expalantion. I'm able to do understand now –  Garett Sep 3 '12 at 16:44

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