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Topic: square root of the derivative (Read 1820 times) |
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Grimbal
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square root of the derivative
« on: Jul 2nd, 2004, 1:14pm » |
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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 xn-1.
And if you apply it k times
Dkxn = n!/(n-k)! xn-k
You could want to generalize and replace k by a real a and replace the factorial by the gamma function.
Daxn = [smiley=cgamma.gif](n)/[smiley=cgamma.gif](n-a) xn-a
And then, you could extend Da to any regular function that can be written as an infinite sum of xn.
So, for instance, we could define an operator D1/2 such that applied twice, it is equivalent to the derivation.
Does this kind of thing exist? If yes, how does D1/2sin(x) look?
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| « Last Edit: Jul 2nd, 2004, 1:16pm by Grimbal » |
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ThudnBlunder
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Re: square root of the derivative
« Reply #1 on: Jul 2nd, 2004, 2:24pm » |
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These are called fractional derivatives.
This and this explain them well.
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| « Last Edit: Jul 2nd, 2004, 3:41pm by ThudnBlunder » |
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towr
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Re: square root of the derivative
« Reply #2 on: Jul 2nd, 2004, 2:24pm » |
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[e] damnit, Thud beat me by a fraction of a minute..  [/e]
on Jul 2nd, 2004, 1:14pm, Grimbal wrote:| Does this kind of thing exist? If yes, how does D1/2sin(x) look? |
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Well, you've just defined it, if it wasn't so allready.
And as sin(x) = (ei x - e-i x)/i (if I'm not mistaken)
and ex = [sum]xk/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 Daxn = [smiley=cgamma.gif](n+1)/[smiley=cgamma.gif](n-a +1) xn-a )
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| « Last Edit: Jul 2nd, 2004, 2:25pm by towr » |
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BNC
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Re: square root of the derivative
« Reply #3 on: Jul 2nd, 2004, 2:50pm » |
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on Jul 2nd, 2004, 2:24pm, towr wrote:
And as sin(x) = (ei x - e-i x)/i (if I'm not mistaken)
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almost... sin(x) = (ei x - e-i x)/2i
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Grimbal
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Re: square root of the derivative
« Reply #4 on: Jul 2nd, 2004, 2:54pm » |
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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.
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