Taking Fractional Derivatives In the Frequency Domain
Bret Cahill
transform is easy: Just multiply each amplitude by it's frequency.
Supposing the order of the derivative is 1/2?
How would that derivative be taken in the frequency domain?
Bret Cahill
Pfs...@aol.com
Just how bdo you define derivative of order 1/2?
What is D^(1/2)[sin(x)] ?
Sami Physics
news:325abe79-8368-487a...@l22g2000pre.googlegroups.com...
> Taking the first derivative of a function after taking the Fourier
> transform is easy: Just multiply each amplitude by it's frequency.
Apostrophe's and they're use's, you fucking imbecile. Everything you post
is crap.
alainv...@gmail.com
> >Taking the first derivative of a function after taking the Fourier
> >transform is easy: Just multiply each amplitude by it's frequency.
>
> >Supposing the order of the derivative is 1/2?
>
> >How would that derivative be taken in the frequency domain?
>
> >Bret Cahill
>
> Just how bdo you define derivative of order 1/2?
> What is D^(1/2)[sin(x)] ?
Good night,
Since sin(x)= (exp(ix)-exp(-ix))/(2*i)
D^1/2 => {sqrt(i)*exp(ix)-sqrt(-i)*exp(-ix)}/(2i)
...
Alain
Bret Cahill
> >transform is easy: Just multiply each amplitude by it's frequency.
>
> >Supposing the order of the derivative is 1/2?
>
> >How would that derivative be taken in the frequency domain?
>
> >Bret Cahill
>
> Just how bdo you define derivative of order 1/2?
> What is D^(1/2)[sin(x)] ?
http://www.maplesoft.com/support/help/Maple/view.aspx?path=fracdiff
Bret Cahill
Bret Cahill
> >transform is easy: Just multiply each amplitude by it's frequency.
>
> >Supposing the order of the derivative is 1/2?
>
> >How would that derivative be taken in the frequency domain?
>
> >Bret Cahill
>
> Just how bdo you define derivative of order 1/2?
> What is D^(1/2)[sin(x)] ?
William Elliot
> On 12 f�v, 17:53, Pfss...@aol.com wrote:
>> On Sat, 12 Feb 2011 08:42:55 -0800 (PST), Bret Cahill
>> <BretCah...@peoplepc.com> wrote:
>>> Taking the first derivative of a function after taking the Fourier
>>> transform is easy: �Just multiply each amplitude by it's frequency.
>>
>>> Supposing the order of the derivative is 1/2?
>>> How would that derivative be taken in the frequency domain?
>>
>> Just how do you define derivative of order 1/2?
>> What is D^(1/2)[sin(x)] ?
>
> Since sin(x)= (exp(ix)-exp(-ix))/(2*i)
> D^1/2 => {sqrt(i)*exp(ix)-sqrt(-i)*exp(-ix)}/(2i)
> ...
Since for all n in N, D^n e^ax = a^n e^ax,
by analogy for all r in R, D^r e^ax = a^r e^ax.
Thus D^(1/2) e^ix = (sqr i).e^ix.
Pfs...@aol.com
<ma...@rdrop.remove.com> wrote:
>Since for all n in N, D^n e^ax = a^n e^ax,
>by analogy for all r in R, D^r e^ax = a^r e^ax.
>
>Thus D^(1/2) e^ix = (sqr i).e^ix.
How does that then translate to D^(1/2) {x^n} ??
Bret Cahill
Going to make me do all the heavy lifting, are you?
Proof:
1st derivative:
multiplying each amplitude by its frequency
2nd derivative:
multiplying each amplitude by its frequency^2
Since there shouldn't be anything quirky going on:
3/2 derivative:
multiplying each amplitude by its frequency^(3/2)
So just raise each frequency to the order of the fractional
derivative.
Incredibly easy!
A Stanford youtube math prof said that people think in the time domain
but nature operates in the frequency domain.
You can really see it here.
Bret Cahill
alainv...@gmail.com
For partial derivation (d/dx)^a o x^n
We just have to use gamma fonction.
For a=1 (d/dx)^1 o x^n = n!/(n-1)!x^(n-1)
general case, a , r real:
(d/dx)^a o x^r Gamma(r+1)/Gamma(r-a+1)*x^(r-a)
try
(d/dx)^a and (d/dx)^(1-a) upon x^n , n integer ...
Alain
whit3rd
> > Taking the first derivative of a function after taking the Fourier
> > transform is easy: Just multiply each amplitude by it's frequency.
> 3/2 derivative:
>
> multiplying each amplitude by its frequency^(3/2)
>
> So just raise each frequency to the order of the fractional
> derivative.
This is the launchpoint for the book _Calculus in a New Key_
by D. L. Orth. It sure SOUNDS interesting, but the noninteger
derivative doesn't simplify anything I've ever worked on.
Bret Cahill
> > > transform is easy: Just multiply each amplitude by it's frequency.
> > 3/2 derivative:
>
> > multiplying each amplitude by its frequency^(3/2)
>
> > So just raise each frequency to the order of the fractional
> > derivative.
>
> This is the launchpoint for the book _Calculus in a New Key_
> by D. L. Orth. It sure SOUNDS interesting, but the noninteger
> derivative doesn't simplify anything I've ever worked on.
It's good to check your work.
If you want to do it with the Excel Fourier transform tool:
1. get the new phase angle from the derivative order, nu
2. get the real and imaginary parts from the new phase angle and the
absolute value of the transform and the sign of the original real and
imaginary parts.
3. multiply the phase adjusted real and imaginary by the frequency^nu
4. complex
5. inverse transform.
Bret Cahill
alainv...@gmail.com
Bonjour,
I do not know the purpose and use of 'your' fractional
derivative.
to sum up (d/dy)^r o exp(ay) = a^r*exp(ay)
(d/dy)^r o y^n = Gamma(n+1)/Gamma(n-r+1)*y^(n-r)
We may also directly build a function g(x,y) such as
g(x+r,y) = d/dy)^r o g(x,y)
Example g(x,y) = d/dy)^x o (exp(2y)+exp(3y)
g(x,y) = 2^x*exp(2y)+3^x*exp(3y) ,
Alain
Bret Cahill
If a function cannot be represented analytically, a spreadsheet is a
convenient way to quickly look at a lot of different fractional
derivatives.
I'm still having some divide by zero issues. Excel can sometimes be
tricked into working just by using very small numbers for zero.
Bret Cahill