# I Taylor series to evaluate fractional-ordered derivatives

1. Feb 21, 2017

### Kumar8434

Can the Taylor series be used to evaluate fractional-ordered derivative of any function?
I got this from Wikipedia:
$$\frac{d^a}{dx^a}x^k=\frac{\Gamma({k+1})}{\Gamma({k-a+1})}x^{k-a}$$
From this, we can compute fractional-ordered derivatives of a function of the form $cx^k$, where $c$ and $k$ are constants (assuming that the constant term gets out of the derivative even in case of fractional derivatives, which must be true).
So, if the derivatives of fractional order have the property of distribution over addition, i.e. it distributes over the various terms of a function connected by the addition operator, then it should be easy to evaluate fractional ordered derivatives of any function $f(x)$ by applying the fractional ordered derivative to the Taylor series of the function.
But it wasn't mentioned anywhere on the Wikipedia article on fractional calculus about using Taylor series to evaluate fractional-ordered derivatives of any function. Does some problem come in evaluating fractional-ordered derivatives that way?

Last edited: Feb 21, 2017
2. Feb 21, 2017

### Stephen Tashi

Some functions don't have a Taylor series. Some functions don't have derivatives. If a function has derivatives of all orders and has a Taylor series, the Taylor series doesn't necessarily converge. So, to formulate a useful question, you need to ask specifically about functions that have a convergent Taylor series. That's worth investigating. I don't know the answer. There are several different types of fractional derivatives, so it isn't a precisely defined question till we pick a specific type of fractional derivative.

3. Feb 21, 2017

### Kumar8434

I've picked this gamma function definition of fractional derivatives. Can this be used to get fractional derivatives of $sinx$. $logx$, etc?

4. Feb 22, 2017

### hilbert2

Sometimes it also happens that a function has derivatives of all orders and has a convergent Taylor series, but it converges towards a wrong function... Think about a function defined by: $f(x) = 0$ if $x \leq 0$, $f(x)=e^{-1/x}$ if $x > 0$.