Given the definition of Taylor expansion: $$f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$ We can find the $m$'th derivative of $f(x)$ quite easily: $$\frac{d^m}{dx^m} f(x) = \sum_{n=0}^{\infty} \frac{d^m}{dx^m}\frac{f^{(n)}(a)}{n!}(x-a)^n$$

Guided by the above definition of $m$'th derivative of $f$ I propose that we can extend the derivative definition to all complex numbers $m$. By extending the use of factorial to gamma function. Applying a limit to get rid of the undefined negative integers at gamma function we are left with:

$$\frac{d^{m'}}{dx^{m'}} f(x) = \lim_{m\rightarrow m'}\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{\Gamma(n-m+1)}(x-a)^{n-m}$$

This definition seems to be relatively useless. Only the derivative of constant functions seems to be somewhat intuitive. Got a bunch of proofs showing convergence of sum under some function constraints which might be helpful but seem relatively basic.

My question, how to more rigorously define the domain of $m$ and $f$ it is applicable to? Is it possible, if so which avenues to use? Since the right hand side is very well defined and left hand side is undefined for $m\notin \mathbb{N}$.

EDIT: I am well aware of derivative extension Fractional Derivatives. Which has somewhat same derivation.

  • $\begingroup$ Riemann actually studied this when he was a student: part of his work on it was published in his Werke (XIX: translated title An attempt to generalise differentiation and integration). This was back when "algebraic analysis" was ruining mathematics, and Riemann's version of proof is not exactly up to modern standards. $\endgroup$ – Chappers Sep 29 '15 at 19:58

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