Hi. Two quick questions. Will be grateful for the answers!

If the fractional derivative, order a, of f(x) is written:

(d^a /dx^a )f(x)

then what does the above expression become when f(x)= xⁿ ?

And, again when f(x)=n, what happens when the above expression is differentiated with respect to a? I.e. what is

d/da ((d^a /dx^a )(xⁿ ))

Or for that matter for other functions.

Many thanks in advance.

The answer to your first question is

[n!/(n-a)!]x^n-a

You can see this by utilizing the formula,

d/dx(xⁿ )=nx^n-1 . After doing this a few times, both the pattern and the reason for the pattern should appear.

The answer to the second question is far more complex

It is

[n!/(n-a)!]*x^n-a *Psi(n-a+1)-[n!/(n-a)!]*x^n-a ln(x)

To be honest, that's only a result I've read off my computer screen, so I'll have to leave the explaining to someone else, but I think it has to do with $\Gamma x$ being very similar to $x!$. BTW,

$\mathrm{Psi}(x)=d/\mathrm{dx}(\ln (\Gamma (x)))$

and

$\Gamma (x)={\int }_0^{\infty }{t}^{x-1}{e}^{-t}\mathrm{dt}$

See here for a similar discussion on NRICH (now archived)

See here for Dr. Beardon's article "Fractional Calculus I"

See here for Dr. Beardon's article "Fractional Calculus"

See here for Dr. Beardon's article "Fractional Calculus III"

Kewin

Many thanks Brad. That's quite some expression, the answer to the second question. What does it work out to when a=0 and n=0, or (just in case the answer is trivial), when a=0 and n=1?

What I'd like is for there to be some nice constants around in this area somewhere, not 'only' e and Euler's gamma :-)

Brad, shouldn't the correct answer to the second question have psi(n-a) rather than psi(n-a+1) and an + rather than a - to combine the two expressions? I.e. [n!/(n-a)!]*x^(n-a)*Psi(n-a)+[n!/(n-a)!]*x^(n-a)*ln(x)

Anon, for your first question, keep in mind that
$x!=\Gamma (x+1)$
For the second question,

If you're evaluating

d/da( e^ln(x)(n-a) )

Try to make a sub of b=n-a, and then use the chain rule.

If you still have trouble, just post again (I very well could be in error and just have not realized it yet).

Peter, I don't know that for your questions we can get any closer than an expression in terms of Psi. We can put this in terms of a definite integral, and perhaps someone on this board could solve it, but I have no idea how...

Brad