MHB Proof of Fourier Series: F(ax) = (1/a)f(k/a) with F(x) as Fourier Transform

[Poirot1]

Let f(k) be the Fourier transform of F(x). Prove that the Fourier transorm of F(ax) is $\frac{1}{a}f(\frac{k}{a})$ where a>0 and the Fourier transform is defined to have a factor of 1/2pi.

 

[CaptainBlack]

Let f(k) be the Fourier transform of F(x). Prove that the Fourier transorm of F(ax) is $\frac{1}{a}f(\frac{k}{a})$ where a>0 and the Fourier transform is defined to have a factor of 1/2pi.

The particular form of the FT you are using is not that important, the basic idea is that:

\[\mathfrak{F} \big[ F(ax) \big] (k) = \int_{-\infty}^{\infty} F(ax) e^{-kx{\rm{i}}}\; dx\]

Now make the change of variable \(x^{*}=ax\) and the result drops out.

CB

 

[CaptainBlack]

Let f(k) be the Fourier transform of F(x). Prove that the Fourier transorm of F(ax) is $\frac{1}{a}f(\frac{k}{a})$ where a>0 and the Fourier transform is defined to have a factor of 1/2pi.

There is confusion here about the thread title (Fourier series) and content (Fourier transform). I have answered for the FT, is that what you intended?

CB