MHB Find Fourier series of Dirac delta function

[ognik]

Hi - firstly should I be concerned that the dirac function is NOT periodic?

Either way the problem says expand $\delta(x-t)$ as a Fourier series...

I tried $\delta(x-t) = 1, x=t; \delta(x-t) =0, x \ne t , -\pi \le t \le \pi$ ... ('1' still delivers the value of a multiplied function at t)

from there I tried $a_0 = \lim_{{\in}\to{0}}\frac{1}{\pi}\int_{t-\in}^{t+\in}1 \,dx $, but that's not going to give the answer in the book ($\frac{1}{2\pi} +\frac{1}{\pi} \sum_{n=1}^{\infty} Cosn(x-t)$)

A hint on how to start this please?

 

[I like Serena]

Hi - firstly should I be concerned that the dirac function is NOT periodic?

Either way the problem says expand $\delta(x-t)$ as a Fourier series...

I tried $\delta(x-t) = 1, x=t; \delta(x-t) =0, x \ne t , -\pi \le t \le \pi$ ... ('1' still delivers the value of a multiplied function at t)

from there I tried $a_0 = \lim_{{\in}\to{0}}\frac{1}{\pi}\int_{t-\in}^{t+\in}1 \,dx $, but that's not going to give the answer in the book ($\frac{1}{2\pi} +\frac{1}{\pi} \sum_{n=1}^{\infty} Cosn(x-t)$)

A hint on how to start this please?

Hi ognik,

When we expand a function to a Fourier series, that expansion is only valid between $-\pi$ and $\pi$.
Outside of that interval, we'll get indeed repetitions that don't belong to the dirac function.

Btw, $\delta(x-t) = \infty$ when $x=t$.
Instead we should use the property that $\int_{-\pi}^{\pi} \delta(x-t) f(t)\,dt = f(x)$ if $-\pi < x < \pi$.

So we get:
$$a_0 = \frac 1 \pi \int_{-\pi}^{+\pi} \delta(x-t) \cdot 1 \,dt = \frac 1\pi \cdot 1$$

 

[ognik]

Thanks ILS, I hadn;t encountered that property - is it commonly used?

Dummy variables being something that can confuse me, please would you explain why it wouldn't be the usual $ a_0 = \frac1\pi \int \delta(x-t)f(x)\,dx = f(x)$ ?

Then is $ a_n = \frac 1\pi \int \delta(x-t)Cosnt\,dt = \frac 1\pi Cosnz|^\pi_-{\pi} $ ?

 

[I like Serena]

Thanks ILS, I hadn;t encountered that property - is it commonly used?

Yes. It is the fundamental property of the dirac function (although I modified it a bit to fit the problem at hand).
See for instance here.

Dummy variables being something that can confuse me, please would you explain why it wouldn't be the usual $ a_0 = \frac1\pi \int \delta(x-t)f(x)\,dx = f(x)$ ?

Then is $ a_n = \frac 1\pi \int \delta(x-t)Cosnt\,dt = \frac 1\pi Cosnz|^\pi_-{\pi} $ ?

When we take an integral, we integrate with respect to some (possibly dummy) variable.
After the integration that variable can no longer be present in the expression - it has been integrated.
In your example $x$ is being integrated, while $t$ is considered to be constant.
The resulting expression is not supposed to contain $x$ any more.

Note that when we write an indeterminate integral like $F(x) = \int f(x)\,dx$, there is an ambiguity in the usage of $x$.
It has 2 different meanings, which is something I recommend avoiding in Fourier analysis.
Properly it should be:
$$F(x) = \int^x f(\tilde x) \,d\tilde x$$

 

[ognik]

Thanks ILS.

Both x and t are variables, so why did you integrate w.r.t. t?

And please check my $a_n$ in my previous post - comparing with the book it looks wrong, but I can't see why?

 

[I like Serena]

Thanks ILS.

Both x and t are variables, so why did you integrate w.r.t. t?

And please check my $a_n$ in my previous post - comparing with the book it looks wrong, but I can't see why?

Your problem statement is actually rather confusing as to which variable we should integrate ($x$ or $t$).
Looking at the answer, I think it's neither $x$ nor $t$, but we should expand $\delta(y)$ instead, and afterwards substitute $y=x-t$.

 

[ognik]

Hi ILS, I believe I have it:
Your link shows everything we need, the property should be $ \int_{-\infty}^{\infty}\delta \left(x-t\right) \,dx = f(t)$

$a_0 =\frac{1}{\pi}$ is still correct

But $a_n= \frac{1}{\pi}\int_{-\infty}^{\infty} \delta \left(x-t\right) cos nx \,dx = \frac{1}{\pi}cos nt $ and similarly $b_n= \frac{1}{\pi}sin nt $

Then $ \int_{-\infty}^{\infty}\delta \left(x-t\right) \,dx = \frac{1}{2\pi} + \sum_{n=1}^{\infty} \left( Cosnt Cos nx +Sinnt Sin nx \right) = \frac{1}{2\pi} + \sum_{n=1}^{\infty} Cosn(x-t) $

 

[I like Serena]

Ah okay, it works like that as well.

 

[ognik]

Ah okay, it works like that as well.

Thanks again ILS - and I also got it using your suggested substitution :-)