Is this derivative thing I found a defined mathematical concept? If so, what is it called?

Question

I'm sure you're aware that $\frac{d^n}{dx^n}\frac{1}{x}=\frac{(-1)^nn!}{x^{n+1}}$

Well, what if $n=\frac{1}{2}$?

$$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\frac{1}{x}=\frac{(-1)^\frac{1}{2}(\frac{1}{2})!}{x^(\frac{1}{2}+1)}$$ $$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\frac{1}{x}=\frac{\sqrt{-1}(\frac{1}{2})!}{x^\frac{3}{2}}$$ So would it be correct to say that: $$\frac{d^\frac{1}{2}}{dx^\frac{1}{2}}\frac{1}{x}=\frac{i\sqrt{\pi}}{x^\frac{3}{2}}$$ Or am I just an idiot? Please tell me. :)

Fractional calculus is a thing that addresses such questions. I don't really know enough about it to tell you anything other than what that page says (which does seem to agree with what you've written) — Milo Brandt, Sep 23 '15 at 02:59

score 2 · Accepted Answer · edited Apr 13 '17 at 12:19

2

You may be interested in https://en.wikipedia.org/wiki/Fractional_calculus; see also Half order derivative of $ {1 \over 1-x }$ and the answer to half-derivative of $x^2$.

edited Apr 13 '17 at 12:19

Community

1

answered Sep 23 '15 at 03:00

Noah Schweber

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Is this derivative thing I found a defined mathematical concept? If so, what is it called?

1 Answers1