Fractal Derivative

[@NoNTr1v1aL]

That's the best clickbait I've ever seen!

[@drpeyam]

Haha, the funny thing is that I didn’t even intend this to be clickbait 😂

[@foreachepsilon]

We do this in statistics with the derivative of variance. It's the fractal derivative of the standard deviation with alpha = 2.

[@PeterBarnes2]

I've been obsessing over this obscure derivative stuff, and I wanted to find a continuous form of the general Leibniz Rule (higher-order product rule), so that we can do fractional derivatives of products of functions. A while ago I made a comment on what I've been calling the 'Taylor Transform,' defined: T[f(x)](z) = [d^z/dx^z](x=0) / Gamma(z+1), Which gives that, for 'friendly' functions f: f(x)= {series, n=0 to inf. of} ( T[f(x)](n) ) * x^n Preface: I don't know much about convolutions, but it'd be nice if the continuous convolution definitely converged to the discrete convolution, which seems plausible, though here I assume it doesn't, so I use discrete convolution. Continuous convolution would simplify the problem completely. If we could describe the Taylor Transform of f(x)*g(x), we'd have the continuous Leibniz Rule. Conveniently, you can use the Cauchy Product to get: f(x)*g(x) = {series, n=0 to inf. of} c_n * x^n where c_n = {sum, k=0 to n of} ( T[f(x)](k) * T[g(x)](n-k) ) And, clearly c_n = T[f(x)*g(x)](n), as well. This is equivalent to the discrete convolution of T[f(x)] and T[g(x)] . The problem is the n. So far, we have a mutilated form of the general Leibniz Rule. It isn't continuous because n is a non-negative integer. What you want to do is find a function X which interpolates the discrete convolution of T[f(x)] and T[g(x)] : X(n) = (T[f(x)] ∗ T[g(x)])[n] for non-trivial X(n). You then know that, for some functions M(n) = 1 and B(n) = 0 for non-negative integers 'n,' T[f(x)*g(x)](z) = M(z)X(z) + B(z) succinctly, T[f(x)*g(x)] = MX + B As for the interpolation, I recommend using Newton Series, which can be formulated pretty simply here. X(z) = Gamma(z+1) * {series, n=0 to inf. of} (a_n) / (n! * Gamma(z-n+1)) where a_n is defined with forward divided differences: a_n = [c_0, ... , c_n] = {sum, j=0 to n of} nCr(n, j) * (-1)^(n-j) * c_j The problem with this interpolation is the convergence. It diverges for real z

[@q44444q]

You should post this on the r/math subreddit!

[@wiskifrac]

Or on math stack exchange

[@numoru]

Yo Update?

[@tarushsharma1404]

That awkward moment when you are watching Dr. Peyam's videos instead of studying for your class 12th final examinations

[@mohammedfarhaan9410]

literally me i got bitsat tmrw morning 💀

[@jaikumar848]

Dr payam! I like the content but board visibility was not good. :(

[@drpeyam]

Yeah, not sure what happened 😕

[@jaikumar848]

Dr Peyam please try to use dark markers and different camera angle like u did in previous video.

[@nickpeterfy8150]

Great video Dr. P! Couple follow-up questions for you: 1.) What does alpha represent? The fractal dimension itself? Or is it related to the Hurst exponent? Or some other interpretation? 2.) Do you have any videos explaining Hausdorff measure? 3.) Where did you get the picture for the video thumbnail? I’ve seen it on Twitter recently so wondering if you found it elsewhere or someone on Twitter took it from you possibly if you made it. Thanks!

[@Koisheep]

1:50 Wouldn't this be like some sort of derivative for \alpha-Hölder continuous maps? which CASUALLY I only heard about in a subject about Sobolev spaces applied to PDEs lmao

[@leonardromano1491]

Isn't this trivial? Just apply the chainrule... Since x^a is differentiable you simply get lim(x-x0) {[f(x)-f(x0)]/[x-x0]} * {[x-x0]/[x^a-x0^a]} = f'(x0)/(dx^a/dx) = f'(x0)/[a*x0^(a-1)] so for f = x, a = 1/2 you get 1/(1/2 * x0^-(1/2)) = 2*sqrt(x0) which is the same as from the limit in the video but using standard definitions...

[@leonardromano1491]

You can generalise this to arbitrary differentiable functions really. Wanna calculate df/d(cos(x)) why not? This is really just trivial. For any differentiable functions f and g (g' not 0) you can calculate df/dg as df/dx * (dg/dx)⁻¹ using the chain rule really^^

[@drpeyam]

Yep, that’s the point!

[@firemaniac100]

if f is differentiable you are right, however there is the possibility of f not being differentiable and the limit for the fractal derivative exists, i.e. f is fractal differentiable but not differentiable.

[@drpeyam]

Yeah, like sqrt(x), not differentiable at 0, but half-differentiable at 0

[@אוריקאופמן-ו8ו]

How is fractional derivative different from fractal derivative?

[@md2perpe]

I first thought that dt^\alpha should be interpreted as (dt)^\alpha and was therefore surprised to see t^\alpha - t_0^\alpha in the denominator. Then, when you wrote d\sqrt{t} it was obvious that dt^\alpha should be interpreted as d(t^\alpha). Then t^\alpha - t_0^\alpha is natural.

[@elfourier1569]

Does exist the fractal integral?

[@newtonnewtonnewton1587]

Very nice i am fond of calculus

[@schlega2]

This got me thinking about a Mandelbrot derivative: f_[n+1]=(f_n)'' + f_0 That didn't seem to lead anywhere interesting, so I tried antiderivatives instead (integrate from 0 to x twice then add the original function). Using Mf for the Mandelbrot antiderivative of f: M1 = cosh(x) Mx = sinh(x) Mexp(x) gives an infinite sum that converges everywhere but I can't find a closed form for it.

[@rainbowbloom575]

Matrix derivative of a function! (d^A)(f(x))/(dx^A)

[@8dolev]

2:02 Like a boss!

[@aliwaqas2396]

I i want you to suggest two books one for ODE'S and one for PDE'S , i am interested in the theory , the ( concept ) . I hope you will suggest a master piece , a book for the life time , thanks ! and i have some questions which i wanted to be solved so how i can send you those .

[@drpeyam]

ODE: Hirsch Smale Devaney PDE: Evans

[@aliwaqas2396]

Dr Peyam many thanks e^+infinity 🙂

[@frolomaskor]

This video made my day! 👍

[@helloitsme7553]

df/dx^α = (df/dx) / (dx^α/dx). This is the way to calculate it without the limit things, basically the Chen lu

[@drpeyam]

Except that sometimes df/dx might not exist whereas df/dx^alpha might exist

[@telotawa]

it feels like you're doing a substitution, like df / d t^a is like you're doing x = t^a and then df/dx? then reversing the substitution to get back to t?

[@drpeyam]

Yeah, basically

[@tylershepard4269]

Could this also be done using the chain rule?

[@drpeyam]

Yes IF f is differentiable! But in general this is more general

[@soar8nalra]

Why dont you just use chain rule?

[@shiina_mahiru_9067]

Ummm... how about fractal integral

[@drpeyam]

Interesting idea! Maybe it’s the Stieltjes integral?

[@polaris_babylon]

The derivation is such a wide subject Btw, does all the derivatives have integrals?

[@willnewman9783]

Nope, not really. You can define derivatives in many many more contexts, most of them not having integrals

[@polaris_babylon]

@@willnewman9783 Ok

[@gamedepths4792]

The Factual Derivative

[@shkotariq6138]

please make the camera angle better

[@drpeyam]

Yeah

[@alinajmaldin]

Ur the best!

[@Flanlaina]

Fractal calculus?

[@aatifali845]

sir upload more video regarding fractal fractional

[@yangchen1466]

Thank you. Really interesting.

[@johnsalkeld1088]

Could be interesting to have an alpha and beta as a function of x and see what this derivation looks like

[@SlipperyTeeth]

So, f'(x)/(ax^(a-1))?

[@Patrickoliveirajf]

indeed

[@drpeyam]

Unless f is not differentiable

[@rafaellisboa8493]

I don't get what this has to do with fractals like mandelbrot set

[@drpeyam]

Fractals have a noninteger dimension and those derivatives are noninteger derivatives. It kind of has to do with Hausdorff measure

[@marouaniAymen]

So I guess we can have fractal anti derivative.

[@harikishan5690]

nicee ,poor marker tho😂

[@ericthegreat7805]

So in other words df(t)/dg(t) = dg(f(t))/df(t). This is where the fractal comes in :o

[@wooyoungkim2925]

프랙셔날 데리~버~띠~브.. !!

[@jamesshelton3827]

oh shit indeed dr peyam o.o

[@jamesshelton3827]

you must offer the math demon a donut if you wish forbidden lore this exchange will require sprinkles o.o

[@jamesshelton3827]

because thinking of the torus is always giving someone a donut dr peyam o.o

[@jamesshelton3827]

you wanted that torus to have sprinkles dr peyam o.O

[@jamesshelton3827]

everyone does o.o down and to the left now

[@ashsharif2453]

Dr please don’t be aggressive with the markers😅 it’s your weapon

[@tszhanglau5747]

You changed your profile pic!

[@drpeyam]

I did! :)

[@mauricepanero]

Completely lost me. Could you enlarge on this a bit?

[@drpeyam]

It’s just like a derivative, except instead of doing your usual x-a think on the denominator, you replace it with x^alpha - a^alpha

[@alexanderseton]

I can barely see what you draw!! Cool video, though

[@sandorszabo2470]

Wikipedia isn't crazy 😊

[@dgrandlapinblanc]

Not easy but thanks.

[@alinajmaldin]

Idgi🤨

[@ammaeaar]

WHY ARE YOU SCREAMING?

[@romajimamulo]

Probably because his microphone is off to the side