# Extending derivatives and integrals into fractions :: 03 Jan 2009

Disclaimer: This is sloppy math. Don’t take it too seriously.

In the previous post, I wrote about a way to generalize derivatives and integrals into one function. What happens if a number other than an integer is passed to that function?

Here is the generalization from the last post:

$\frac{\partial^n}{\partial x^n} x^k = \frac{k!}{(k-n)!} x^{k-n}$

Set k = 2 and n = 1/2:

$\frac{\sqrt{\partial}}{\sqrt{\partial x}} x^2 = \frac{2!}{(2-\frac{1}{2})!} x^{2-\frac{1}{2}} = \frac{2!}{1.5!} x^{1.5}$

Now there’s a problem. Since a! is only defined for integers 0 or greater, a different way of calculating factorials is needed. Luckily, there exists a Gamma function defined for all real and complex numbers such that:

$\Gamma(n+1) = n!$

Substitute the Gamma function into the other equation and simplify:

$\frac{\sqrt{\partial}}{\sqrt{\partial x}} x^2 = \frac{2!}{1.5!} x^{1.5} = \frac{2}{\Gamma(1.5+1)} x^{1.5} = \frac{2}{\Gamma(2.5)} x^{1.5} = \frac{2}{(\frac{3 \sqrt{\pi}}{4})} x^{1.5}$ $= \frac{2}{1} (\frac{4}{3 \sqrt{\pi}}) x^{1.5} = \frac{8}{3 \sqrt{\pi}} x^{1.5} = \frac{8 \sqrt{\pi}}{3 \pi} x^{1.5} \approx 1.505 x^{1.5}$

So, the half derivative of $$x^2$$ is approximately equal to $$1.505 x^{1.5}$$.

This can be used for ‘fractional integration’ as well, if a negative number is used for n.