Dirac Delta Function - Fourier Series

[BOAS]

1. Homework Statement

Find the Fourier series of

$f(x) = \delta (x) - \delta (x - \frac{1}{2})$ , $- \frac{1}{4} < x < \frac{3}{4}$
periodic outside.

Homework Equations

[/B]
$\int dx \delta (x) f(x) = f(0)$

$\int dx \delta (x - x_0) f(x) = f(x_0)$

The Attempt at a Solution

[/B]
I am struggling to visualise this function, which is making it hard to tell if my series makes sense. Plus I find the delta function a little confusing to reason about.

The function has a period of 1.

$a_0 = 2 \int^{3/4}_{-1/4} \delta (x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) dx = 0$

$a_n = 2 \int^{3/4}_{-1/4} \delta (x) \cos (n \pi x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) \cos (n \pi x)dx = 2 - 2 \cos (\frac{n \pi}{2})$

$b_n = 2 \int^{3/4}_{-1/4} \delta (x) \sin (n \pi x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) \sin (n \pi x) dx = -2 \sin (\frac{n \pi}{2})$

I think those are my Fourier coefficients, and I couldn't find a nice way to express them, so I think they're ok left as trig functions.

Does it look ok?

 

[HallsofIvy]

cos(\frac{n\pi}{2}) is 0 if n is odd, 1 if n is an even multiple of 2 and -1 if n is an odd multiple of 2. Sin(x) is 0 if n is even, 1 if n is of the form 4k+1 for some k and -1 if it is of the form 4k-1 for some k. So \sum_{n= 0}^\infty(2- 2cos(\frac{n\pi}{2})cos(n\pi x)+ 2 sin(\frac{n\pi}{2})sin(n\pi x)= \sum_{n=0}^\infty [(2- 2(-1)^{n})cos(2n\pi x)+ 2(-1)^n sin(((2n+1)/2]\pi x)]

 

[BOAS]

cos(\frac{n\pi}{2}) is 0 if n is odd, 1 if n is an even multiple of 2 and -1 if n is an odd multiple of 2. Sin(x) is 0 if n is even, 1 if n is of the form 4k+1 for some k and -1 if it is of the form 4k-1 for some k. So \sum_{n= 0}^\infty(2- 2cos(\frac{n\pi}{2})cos(n\pi x)+ 2 sin(\frac{n\pi}{2})sin(n\pi x)= \sum_{n=0}^\infty [(2- 2(-1)^{n})cos(2n\pi x)+ 2(-1)^n sin(((2n+1)/2]\pi x)]

Thanks a lot.

That's incredibly helpful and clear.

 

[vela]

$a_0 = 2 \int^{3/4}_{-1/4} \delta (x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) dx = 0$

$a_n = 2 \int^{3/4}_{-1/4} \delta (x) \cos (n \pi x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) \cos (n \pi x)dx = 2 - 2 \cos (\frac{n \pi}{2})$

$b_n = 2 \int^{3/4}_{-1/4} \delta (x) \sin (n \pi x) dx - 2 \int^{3/4}_{-1/4} \delta (x - \frac{1}{2}) \sin (n \pi x) dx = -2 \sin (\frac{n \pi}{2})$

I think those are my Fourier coefficients, and I couldn't find a nice way to express them, so I think they're ok left as trig functions.

You can also use some trig identities to simplify
$$a_n \cos n\pi x + b_n \sin n\pi x = 2\cos (n\pi x) - 2 \left[\cos \left( \frac{n \pi}{2}\right) \cos (n\pi x) + \sin \left(\frac{n \pi}{2}\right) \sin (n\pi x) \right]$$ to get
$$-4 \sin \frac{n \pi}{4} \sin\left[n\pi \left(x - \frac 14\right) \right]$$