Proof: Fourier Transform of f(x) = f(-x)

Chen
Messages
976
Reaction score
1
How can I prove that doing a Fourier transform on a function f(x) twice gives back f(-x)?

Thanks..
 
Physics news on Phys.org
I think you're asking about the double integral

\int_{-\infty}^{\infty} e^{i kx} dk \int_{-\infty}^{\infty}e^{ikx'} f(x')dx'

If so, then the outer integral

\int_{-\infty}^{\infty}e^{i k (x+x')} dk = \delta (x+x')

i.e. the Dirac delta function and you arrive at your result upon evaluating the inner integral. (That's ignoring factors of 2\pi which I am sure you can handle!)
 
Last edited:
I should hope so. Thanks!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top