What is the meaning (if any) of $D^{i}f(x)$?

Question

Recently I came across fractional derivatives of functions of the type

$D^{\frac{1}{2}}f(x)$

and I was wondering if it made sense to take the $i$ derivative of a function, where $i = \sqrt{-1}$

$D^{i}f(x)$

If it does make sense, what does it mean? And how do you compute it? Thanks!

You should make it clear if you mean $i=\sqrt{-1}$, and not just a variable. — Bonnaduck, Nov 26 '21 at 20:21

GEdgar · Answer 1 · 2021-11-26T20:26:32.900

1

Derivative of order $k$, where $k$ is a nonnegative integer, is a standard notion in calculus. Derivative of fractional order ... or of complex order ... that is not a standard notion in calculus. If you use it you should define it somehow.

One way it can be defined uses the Laplace transform. Perhaps you will study that in the future.

See HERE

edited Nov 26 '21 at 20:26

answered Nov 26 '21 at 20:17

GEdgar

111,679

Also Wolframalpha has this to say. – Graham Kemp Nov 26 '21 at 20:23
Yes! I am aware of higher order derivatives, I forgot to include that I meant i as in the imaginary unit. Thanks anyway! – Francisco Maggi Nov 26 '21 at 20:29

What is the meaning (if any) of $D^{i}f(x)$?

1 Answers1