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        riddles >> putnam exam (pure math) >> square root of the derivative
        
(Message started by: Grimbal on Jul 2^nd, 2004, 1:14pm)

Title: square root of the derivative
Post by Grimbal on Jul 2^nd, 2004, 1:14pm

Since there are a number of people knowledgeable in math around here, I'd like to ask about an idea I had some time ago and never found out if it works.

If you take the derivation operator D, you can see that it operates very regularily on polynomials.
Dxⁿ = n x^n-1.
And if you apply it k times
D^kxⁿ = n!/(n-k)! x^n-k
You could want to generalize and replace k by a real a and replace the factorial by the gamma function.
D^axⁿ = [smiley=cgamma.gif](n)/[smiley=cgamma.gif](n-a) x^n-a
And then, you could extend D^a to any regular function that can be written as an infinite sum of xⁿ.

So, for instance, we could define an operator D^1/2 such that applied twice, it is equivalent to the derivation.

Does this kind of thing exist? If yes, how does D^1/2sin(x) look?

Title: Re: square root of the derivative
Post by THUDandBLUNDER on Jul 2^nd, 2004, 2:24pm

These are called fractional derivatives (http://mathworld.wolfram.com/FractionalDerivative.html).

This (http://www2.machinedesign.com/derivatives/fractionalderivatives_part1.doc) and this (http://www2.machinedesign.com/derivatives/fractionalderivatives_part2.doc) explain them well.

Title: Re: square root of the derivative
Post by towr on Jul 2^nd, 2004, 2:24pm

[e] damnit, Thud beat me by a fraction of a minute.. >:([/e]

on 07/02/04 at 13:14:30, Grimbal wrote:

Does this kind of thing exist? If yes, how does D^1/2sin(x) look?

Well, you've just defined it, if it wasn't so allready.
And as sin(x) = (e^{i x} - e^{-i x})/i (if I'm not mistaken)
and e^x = [sum]x^k/k!
So you can just apply the operator you just defined

([sum][smiley=cgamma.gif](k + 1)/[smiley=cgamma.gif](k-a + 1) (i x)^k-a/k! + [sum] ][smiley=cgamma.gif](k + 1)/[smiley=cgamma.gif](k-a + 1) (-i x)^k-a/k!)/i

(n! = [smiley=cgamma.gif](n+1), so I assume you meant D^axⁿ = [smiley=cgamma.gif](n+1)/[smiley=cgamma.gif](n-a +1) x^n-a )

Title: Re: square root of the derivative
Post by BNC on Jul 2^nd, 2004, 2:50pm

on 07/02/04 at 14:24:44, towr wrote:

And as sin(x) = (e^{i x} - e^{-i x})/i (if I'm not mistaken)

almost... sin(x) = (e^{i x} - e^{-i x})/2i

Title: Re: square root of the derivative
Post by Grimbal on Jul 2^nd, 2004, 2:54pm

I wonder what you can not find on Wolfram's MathWorld.

I have been looking at that site, they even give the expression of the fractional derivative of sin(z).
http://functions.wolfram.com/ElementaryFunctions/Sin/20/03/0001/
but it uses the "regularized generalized hypergeometric function", which I have no idea what it looks like. I don't even know if the half-derivative is real.