It has been 20 years since failed to understand Fourier transforms during my formal education. I am trying to solve an optical problem, and it looks as though the Fourier transform should be helpful. I have been reading up, but I don't understand the transforms well enough to work through a simpler example. And in the end I really need to figure out how to deal with a more complicated one. A less complicated one, that I could feed through a solver program so that I know which answer I'm trying to reproduce is: $$\int_{-\infty}^\infty x^2 \frac 1{\sigma \sqrt{2\pi}}e^{-\frac 12 \frac {(x-\mu)^2}{\sigma^2}}dx=\mu^2+\sigma^2 $$ Which is a Gaussian distribution multiplied with $x^2$; $\sigma$ and $\mu$ have their usual meanings of standard deviation and mean. I have more complicated versions where I have all sorts of sin or cos terms instead of the $x^2$. But I'm focusing on this less complicated one, since I expect understanding this one will help me work my way up to the others. Taking the constant terms out from the integral is obvious: $$\int_{-\infty}^\infty x^2 \frac 1{\sigma \sqrt{2\pi}}e^{-\frac 12 \frac {(x-\mu)^2}{\sigma^2}}dx=\frac 1{\sigma \sqrt{2\pi}}\int_{-\infty}^\infty x^2 e^{-\frac 12 \frac {(x-\mu)^2}{\sigma^2}}dx$$ And I notice it vaguely resembles the formula for convolution where $f(x)=x^2$ and g is of the form $g(x-\mu)=e^{-ax^2}$. But I fail to understand the notation of the convolution theorem, and am inexperienced in the required maths (I do get the intuition: in the transform-domain we can add/subtract/use-algebra-to-simplify the transformed results and then leverage those results to solve our initial problem).
In the tutorials I found online, the $e^{-i\omega t}$ magically appears at this point; I guess if I can see behind this magic I am a big step forward.
So, to be specific about the questions:
- Am I right to suspect that we can relatively easily solve this using a Fourier transform?
- In the notation of convolution: what is f*g(x) as opposed to f(x)*g(x)? Am I right to understand that it is the same as f(g(x)), which is a very common notation to describe the chain rule for derivatives, which I know as (f(g(x))'=f'(g(x))*g'(x)?
- How do I move forward? In other words, can I use the convolution theorem to move forward, and how do I apply it?
thanks for your time and help.
UPDATE: All right, I have thought about it a little more. And I found a source that helps: $\hat F(0)=\int F(t)e^{-i*0*t}dt =\int F(t)dt$. Anyone who ever writes a tutorial on this aimed to inspire people to start using FT, without understanding the full proofs: you may want to lead with that. :)
doing this, I arrive at the correct answer! Could anyone confirm if I applied the time shifting and the derivative rules correctly?
I let $$F(t)=\frac {1}{\sigma \sqrt{2\pi}} e^{-\frac 1 2 \frac {t^2}{\sigma^2}} $$ And found in a table $$\mathcal{F}(F(t))=\hat F(\omega) = e^{-\frac 1 2 \sigma^2 \omega^2}$$ Time shift rule: $\mathcal{F}(f(t-t_0))=e^{-i\omega t_0} \hat{f}(\omega) $ so that:$$\mathcal{F}(F(t-\mu))=e^{-i\omega\mu} \hat{F}(\omega)=e^{-i\omega\mu-\frac 1 2 \sigma^2 \omega^2} $$ Derivative rule: $\mathcal{F}(t^2 f(t))= i^2 \frac{d^2}{d\omega^2}\hat{f}(\omega)=-\frac{d^2}{d\omega^2}\hat{f}(\omega)$, then I think that: $$\mathcal{F}(t^2 F(t-\mu))= \frac {d^2}{d\omega^2} e^{-i\omega\mu}\hat{F}(\omega)=\frac {d^2}{d\omega^2}e^{-i\omega\mu-\frac 1 2 \sigma^2 \omega^2}=(\mu^2+\sigma^2-2i\sigma^2\omega\mu-\sigma^4\omega^2)e^{-i\omega\mu-\frac 1 2 \sigma^2 \omega^2}$$ So that if I fill in $\omega = 0$, I arrive at the correct answer!
My specific remaining question: $\mathcal{F}(t^2 F(t-\mu))$, shouldn't that have been $\mathcal{F}((t-\mu)^2 F(t-\mu))$? (in other words, did I arrive at the correct answer despite making an error?)
Is it maybe also true that $\mathcal{F}(t^2 F(t-\mu))=\mathcal{F}((t-\mu)^2 F(t-\mu))$?
