/sci/ - Science & Math » Thread #12344720

4KiB, 299x53, fractional_derivative.jpg

View Same Google iqdb SauceNAO

How do you conceptualize this?

Anonymous Sun 15 Nov 16:50:07 2020 No.12344720 View Reply Original Report

Quoted By: >>12345091 >>12345104

You definitely can't think of growth in the conventional sense, so what's the intuitive approach here?

Anonymous

Anonymous Sun 15 Nov 2020 16:55:09 No.12344733 Report

Quoted By: >>12345079

I just take the derivative. Why do you math nerds have to make easy things difficult?

Anonymous

Anonymous Sun 15 Nov 2020 18:32:12 No.12345079 Report

Quoted By:

>>12344733
this. I have a PhD in operator theory and all the fucking algebraic geometry majors are spergs who obfuscate simple concepts for no reason. If they were asked to explain what a polynomial was to some middle schoolers they'd bring up rings and fields for no fucking reason but to dickwave.

Anonymous

Anonymous Sun 15 Nov 2020 18:36:11 No.12345091 Report

Quoted By: >>12345096

>>12344720
Fourier transform.

Anonymous

Anonymous Sun 15 Nov 2020 18:39:23 No.12345096 Report

Quoted By: >>12345132

>>12345091
It's a convolution brainlet

Anonymous

Anonymous Sun 15 Nov 2020 18:41:52 No.12345104 Report

Quoted By:

>>12344720
take function f(t), now multiply it by the fractional kernel, take the convolution up to x. This operator defines the rate of change of the resulting convolution. Simple as

Anonymous

Anonymous Sun 15 Nov 2020 18:50:04 No.12345132 Report

Quoted By: >>12345155

>>12345096
That's the point you retard. If you assume that f is from the Schwartz class, consider D^\alpha as a tempered distribution, then take the Fourier transform of that distribution, you will get a distribution of the form t^\alpha times the Fourier transform of f. Compere that to the transform of the derivative of f.

Anonymous

Anonymous Sun 15 Nov 2020 18:56:45 No.12345155 Report

Quoted By: >>12345207

>>12345132
https://hal.archives-ouvertes.fr/hal-01740424/document

I dont think you helped OP in any way

Anonymous

Anonymous Sun 15 Nov 2020 19:17:35 No.12345207 Report

Quoted By:

>>12345155
holy shit, I forgot about the i in the formula, s-sorry. To make it more clear, if you have differentiable function and you take the derivative of it you get
$(\mathcal{F}f')(t)=it (\mathcal{F})(t)$
so, whatever is the fractional derivative of f, when you act on it with the Fourier transform you should get
$(\mathcal{F}D^\alpha f)(t)=C(t)^\alpha (\mathcal{F})(t),$
for some constant C. If you use the definition from OP image you will get that formula with C=i^\alpha.

Capcode	All Only User Posts Only Moderator Posts Only Admin Posts Only Developer Posts
Show Posts	All Only With Images Only Without Images
Deleted Posts	All Only Deleted Posts Only Non-Deleted Posts
Ghost Posts	All Only Ghost Posts Only Non-Ghost Posts
Post Type	All Only Sticky Threads Only Opening Posts Only Reply Posts
Results	All Grouped By Threads
Order	Latest Posts First Oldest Posts First

Your latest searches