Inverse Fourier Transform of f(k): Yes

[RedX]

Suppose a function f(k) has a power series expansion:

$$f(k)=\Sigma a_i k^i$$

Is it possible to inverse Fourier transform any such function?

For example:

$$f(k)=\Sigma a_i k^{i+2}\frac{1}{k^2}$$

Since g(k)=1/k^2 should have a well-defined inverse Fourier transform, and the inverse Fourier transform of k*g(k) -> dg(x)/dx [where g(x) is the inverse Fourier transform of g(k)], then inverse Fourier transform of f(k) is an infinite sum of the inverse Fourier transforms of g(k)=1/k^2 and its derivatives d^n[g(x)]/dx^n.

 

[ObsessiveMathsFreak]

Since g(k)=1/k^2 should have a well-defined inverse Fourier transform,

Woah there. If you want to talk about the Fourier transforms (inverse or otherwise) of polynomials, etc, you'll have to move into distribution theory and all that that entails. In the process, you'll lose the ability to simply multiply "functions" and your second equation will raise hairs all over the place when you come to apply it.

As to your first question, if I remember correctly, the "answer" to your question exists only in the realm of distributions. Specifically, I think your power series has the inverse Fourier transform
$$F(x)= \sum_n (-i)^n a_n \delta^n (x)$$

Where $$\delta^n(x)$$ is the n^th derivative of the Dirac delta function. It's OK to write these as a function so long as you remember that they can never be evaluated; only integrated.

I'll finish by warning you to never, ever divide or multiply any dirac functions by other such. It's not a defined operation.

 

[vigvig]

As long as your function is square summable (or square integrable) you can always take the Fourier inverse.
Vignon S. Oussa