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Could non integral degree derivative somehow be interpreted? What I mean:

  • $f^{(1)}(x) = \frac{df(x)}{dx}$
  • $f^{(2)}(x) = \frac{d^{2}f(x)}{dx^2}$

How could $f^{(1.5)}(x)$ be interpreted?

András Hummer
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2 Answers2

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This is a matter of fractional analysis, and you should use the definition of fractional derivative. Not only 1.5, you may find derivative and integration of real and complex valued function of any order.

There are several definition of fractional derivative having different uses. Let me consider Riemann's definitions.

$$D^{\mu} f(x) = D^n [D^{-\nu} f(x)]$$

Where $\nu = n - \mu$ and $0 \le \Re{(\nu)} < 1$. $[D^{-\nu} f(x)]$ is the fractional integration of $f(x)$ of order $\nu$.

$$[D^{-\nu} f(x)] = \frac{1}{\Gamma(\nu)} \int_0^x (x - t)^{\nu-1} f(t) dt$$

So you have all the definitions in your hand. Now you may calculate your required answer.

Supriyo
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It seems as though you are looking for fractional calculus.

torbonde
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  • Thank you, just what I was looking for. Seems like my definition is $\frac{f^{(1.5)}(x+h) - f^{(1.5)}(x)}{h}$ in this example (with h approaching +0 of course). – András Hummer Aug 14 '13 at 10:53
  • Can you please give me some examples for its practical usage, where it actually does make sense to use fractal derivative? – András Hummer Aug 14 '13 at 10:55
  • I'm afraid not. But you might want to add that to your question. I just remembered seeing this a while ago, so it was quite easy for me to search for right now. But I have no experience with it whatsoever.:( – torbonde Aug 14 '13 at 11:02