Author |
Topic: square root of the derivative (Read 1809 times) |
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender: 
Posts: 7527
|
 |
square root of the derivative
« on: Jul 2nd, 2004, 1:14pm » |
 Quote  Modify
|
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?
|
| « Last Edit: Jul 2nd, 2004, 1:16pm by Grimbal » |
 IP Logged |
|
|
|
ThudnBlunder
Uberpuzzler
The dewdrop slides into the shining Sea
Gender:
Posts: 4489
|
|
Re: square root of the derivative
« Reply #1 on: Jul 2nd, 2004, 2:24pm » |
Quote Modify
|
These are called fractional derivatives.
This and this explain them well.
|
| « Last Edit: Jul 2nd, 2004, 3:41pm by ThudnBlunder » |
IP Logged |
THE MEEK SHALL INHERIT THE EARTH.....................................................................er, if that's all right with the rest of you.
|
|
|
towr
wu::riddles Moderator
Uberpuzzler
Some people are average, some are just mean.
Gender:
Posts: 13730
|
|
Re: square root of the derivative
« Reply #2 on: Jul 2nd, 2004, 2:24pm » |
Quote Modify
|
[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? |
|
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 )
|
| « Last Edit: Jul 2nd, 2004, 2:25pm by towr » |
IP Logged |
Wikipedia, Google, Mathworld, Integer sequence DB
|
|
|
BNC
Uberpuzzler
Gender:
Posts: 1732
|
|
Re: square root of the derivative
« Reply #3 on: Jul 2nd, 2004, 2:50pm » |
Quote Modify
|
on Jul 2nd, 2004, 2:24pm, towr wrote:
And as sin(x) = (ei x - e-i x)/i (if I'm not mistaken)
|
|
almost... sin(x) = (ei x - e-i x)/2i
|
|
IP Logged |
How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
|
|
|
Grimbal
wu::riddles Moderator
Uberpuzzler
Gender:
Posts: 7527
|
|
Re: square root of the derivative
« Reply #4 on: Jul 2nd, 2004, 2:54pm » |
Quote Modify
|
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.
|
|
IP Logged |
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print