We can expand any analytic around a point function in a Taylor series around the same point (I consider real functions for now).

$$f(x)=\sum_{k=0}^\infty f^{(k)}(x_0) \frac{(x-x_0)^k}{k!}$$

By using fractional calculus, can we also represent non-analytic functions in a kind of a 'Taylor integral'?

$$g(x)=\int_0^\infty g^{(a)}(x_0) \frac{(x-x_0)^a}{\Gamma(a+1)}da$$

Here $g^{(a)}$ is a fractional derivative.

My reasoning is as follows. We have Fourier series for periodic functions and Fourier integral (Fourier transform) for non-periodic functions. Could it work the same way for Taylor series?

There is a very similar question even with almost the same notation (I've just found it though). However, it doesn't use the concept of fractional derivative.

  • 1
    $\begingroup$ Have you thought about writing your expansion on the Fourier side, where fractional differentiation is just pointwise multiplication? $\endgroup$ – Ian Jan 9 '17 at 11:45
  • $\begingroup$ See whether this helps you: epubs.siam.org/doi/abs/10.1137/0502004 $\endgroup$ – Rohan Jan 9 '17 at 11:49
  • $\begingroup$ And the Taylor expansion (with remainder) is obtained by integration by parts. You should obtain some similar expressions with the fractional derivative. $\endgroup$ – reuns Jan 9 '17 at 11:54

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