# Given that f(x)= sin (x + pi/4 ) is periodic with period 2pi. Find the complex Fourier Series. Please someone really help me on this because this is my first time I'm posting a question in this...

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

Please someone really help me on this because this is my first time I'm posting a question in this forum. So If someone able to help me this question, then I definitely love to subscribe this forum. Thanks guys

### 2 Answers | Add Yours

If your answer needs to be in the form `sum_{n=-infty}^infty c_n e^{i nx}` , we can compare with the coefficients of the exponentials.

That is:

`c_1=(1+i)/{2isqrt2}`

and

`c_{-1}=-{1-i}/{2isqrt2}`

All other coefficients are zero since this is a finite Fourier series.

The complex Fourier series is a (potentially infinite) series of functions of the form `e^{iax}` where `a` is a number. Since `e^{iax}` is preiodic with period `2pi`, if we can represent `f(x)` in terms of the exponentials, we are done.

In this case, since `sinx` can be represented as `1/{2i}(e^{ix}-e^{-ix})`

we are very close to a solution already, and expect that we will end up with a finite Fourier series, instead of an infinite Fourier series.

Now we need to expand `f(x)` .

`f(x)=sin(x+pi/4)`

`=1/{2i}(e^{i(x+pi/4)}-e^{-i(x+pi/4)})`

`=1/{2i}(e^{ix}e^{{ipi}/4}-e^{-ix}e^{-{ipi}/4})`

Now from Euler's formula `e^{i theta}=cos theta+i sin theta` , we see that

`e^{+-{i pi}/4}=cos(pi/4)+-isin(pi/4)`

`=1/sqrt2+-i/sqrt2`

which means that the function becomes

`1/{2i}(e^{ix}1/sqrt2(1+i)-e^{-ix}1/sqrt2(1-i))`

`=1/{2isqrt2}(e^{ix}(1+i)-e^{-ix}(1-i))`

Although many Fourier series that you see in your classes will be infinite, there are many (such as this one) that are finite.

**The Fourier series is `1/{2isqrt2}(e^{ix}(1+i)-e^{-ix}(1-i))` .**

**Sources:**