Half derivative and Fourier transform

[@drpeyam]

Some applications of fractional derivatives: There are surprisingly many applications of this, because it turns out that some differential equations in physics are written in terms of fractional derivatives, see en.wikipedia.org/wiki/Fractional_calculus#Applications There are two other ones I can think of: 1) In functional analysis, it's an important problem to find a square root of an operator (I don't really know why, maybe to decompose that operator?), and what we really did is to find a square root of the derivative operator, because if you apply D1/2 twice, you get D, so (D1/2)^2 = D, so D1/2 = sqrt(D) in some sense. 2) Fractional derivatives allow us to define nice spaces of functions (for example, those whose fractional derivatives exist and are square integrable), and sometimes in differential equations you have a solution that is not defined in the classical sense (i.e. continuously differentiable), but might belong to this nice space, which allows us to study those equations.

[@MathsatBondiBeach]

The functional analysis Mafia are steeped in such issues, a classic of which is Kato's square root problem which my late professor from the early 1970s, Alan McIntosh, solved over a long period in collaboration with with Yves Meyer, Ron Coifman, Pascal Auscher and some others. A research level paper on the subject is here: arxiv.org/pdf/math/0108029.pdf

[@codingphysics695]

Very interesting idea! But I think the Laplace transform would be more practical for the actual calculation of the half-derivative. I did some back-on-the-evelope calculation for the half-derivative of the simple power function f(x) = x^a and I got the result: D^(1/2) x^a = (Γ(a + 1)/Γ(a + 1/2)) x^(a - 1/2) where Γ(x) is the gamma function and the half-derivative is defined with the help of the Laplace transform by L[ D^(1/2) f(x) ] != s^(1/2) F(s)

[@drpeyam]

Wow, that’s pretty!!! ❤️

[@valeriobertoncello1809]

Well Fourier is just a special case of Laplace, afterall!

[@leonardromano1491]

It would be so cool if you could make a video about the fractional derivatives of our favourite distributions like the delta-distribution using this definition of the fractional derivative.

[@MathsatBondiBeach]

Schwartz space and Bochner's theorem sneakily worked into this short video....I'd love to bring Fourier back and show him how his theory has been used. He could then say to Legendre (or was it Lagrange?) and the others on the French Academy who doubted his theory - "See I told you!!"

[@hallfiry]

A thing to ponder: does the fractional gradient of a function have a fractal dimension? (I have no idea)

[@ffggddss]

Good question; actually, just what sort of mathematical object will that be? When you have a scalar function, ƒ:ℝⁿ→ℝ in n-space, ℝⁿ, its gradient, ∇ƒ, is an n-vector. If you wanted a "2nd gradient," you could take a divergence: ∇•∇ƒ = ∇²ƒ which has the del operator in a dot-product with itself, and results in a scalar function (the Laplacian of ƒ). But the most "raw" form would be the tensor product: ∇⨂∇ƒ which is a 2nd-rank tensor in ℝⁿ, whose trace is that Laplacian, ∇²ƒ; and whose dual is an (n-2)-rank tensor, the cross product. And similarly for higher-order "gradients," order k being a tensor of rank k. So would a fractional gradient give you a tensor of fractional rank? Is there such a thing? I suppose it's worth a try to see whether Dr. Peyam's Fourier-transform method can be adapted to this case... Fred

[@hallfiry]

@@ffggddss I'll propose a bold idea and suggest the output dimension depends on the input. So at (1,1) the gradient of f(x,y) might be 2D, while at (1,2) it might be 1D, leaving us with a sharply mixed set.

[@ffggddss]

@@angelmendez-rivera351 Agreed. I certainly can't imagine what that would be! Fred

[@valeriobertoncello1809]

@@angelmendez-rivera351 This is extremely thought-provoking to me. Especially when I try to imagine what the potential applications of these concepts would be. Basically all of fundamental modern physics is built not on fractional calculus (or fractal dimensions), but merely on finite integer calculus (and dimensions). And by finite integer I mean "mostly 0, 1, and 2", since most of fundamental physics is built on second-order differential equations. Such a mesmerizing topic, fractal multivariable calculus!

[@jbtechcon7434]

Okay, but what’s the physical meaning? If I find the transform of f(z) multiply it by (2*pi*i)^(1/2), then reverse-transform that, I have the half-derivate of f(z) wrt z. But what does it actually mean? If I use some fractional exponent alpha instead of 1/2, what does it mean? What is the meaning of the alpha-th derivate of f(z) wrt z in the limit alpha -> zero?

[@MusicEngineeer]

haha! nice! as an audio processing guy, i may think of taking the derivative as applying a first order highpass filter with 6 dB/oct slope ....yeah...i know - actually 6.02...blabla. so, the half derivative is just a 3 dB/oct highpass. a 6 dB/oct lowpass is an integral and a 3 dB/oct lowpass is a half-integral. and so on. this simplifies thinking about these ideas intuitively a lot hmmm...ok - strictly speaking, a filter has a cutoff frequency and these things would be a sort of limit as the cutoff goes to zero, but anyway

[@MusicEngineeer]

@@angelmendez-rivera351 well, it's intuitive for me personally because i have to deal a lot with such filters

[@61rmd1]

So clear, and so powerful property! Thank you Dr Peyam! I have a question: What about the fractional derivative on Fourier trasform of Generalised Functions? I guess that it can be possible...what is you r opinion? Greetings from Italy...

[@CalculusPhysics]

Fantastic video! the only thing that i’m still a little confused about is automatically assuming that it works for fractions. like when we do integration by parts for normal derivatives, it’s pretty clear that the Fourier transform of f’’(x) will be (2πiζ)²*f^(ζ) but how do we know this will always work for fractional values of α [alpha]? i guess what i’m asking is that is there a more rigorous way to show that relationship?

[@CalculusPhysics]

sorta like the Gamma Function. it’s a pretty similar process using integration by parts for positive integers and getting n!, but is there a more rigorous way to show that the definition holds for fractions and complex numbers?

[@tylershepard4269]

The Fourier transform is useful for developing a spectral decomposition of a finite signal. It comes out of the everlasting exponential, and you get the Fourier transform by plugging in w (omega) and convolving with the impulse response of a system. Pretty neat stuff!

[@NAMEhzj]

So what i haven't really understood (I don't know if you mentioned it somewhere) is, whether this is THE way to define half derivative. It certainly seems very natural but can there be a different definition of half derivative that still satisfies D^1/2 D^1/2 f = Df ? Would be very interesting to know/see a proof of that or the converse!

[@drpeyam]

I’m not sure if it’s unique, if that’s what you’re asking! There’s another definition of the half derivative which you can find on my playlist

[@katherinewallace8815]

Have you figured it out yet? Asking for a friend

[@NAMEhzj]

@@katherinewallace8815 Unfortunately not. I know that for positive, self-adjoint operators on Hilbert spaces one can find a square root with the spectral theorem. When you apply this to the Laplacian on R^n, you find that the Laplace is given by U^{-1}MU, where M is multiplication by |x|^2 and in this case U is given by the fourier transform, so it is like in this video. Then you can take the square root of the multiplication operator and get your awnser. In this case (Laplacian is positive, self-adjoint) you have uniqueness of a positive square root. (Adding a minus you can always get another solution, which is excluded by requiring positivity). However, since the derivative itself is neither positive nor self-adjoint this doesn't seem to work here. However, viewing it from this perspective suggests that this approach is somehow the right one, and not arbitrary.

[@jaikumar848]

Dr payam ! are you going to start series on FRACTIONAL CALCULUS?

[@drpeyam]

I’ve already done videos on that!

[@drpeyam]

Fractional Derivatives kzbin.info/aero/PLJb1qAQIrmmB_ma3YrfuOXTPOQawokYV_

[@PeterBarnes2]

With this definition of real derivatives, we get to do the whole story of function-derivatives, without all of the Maclaurin series stuff. Define { fourier of }[ {g(D)} [f(x)] ]= g( 2(pi)i*z ) * { fourier of }[f(z)] or {g(D)} [f(x)] = { inverse fourier of }[ g( 2(pi)i*z ) * { fourier of }[f(z)] ] You get back your 'fractional calculus' as 'powers of derivatives,' and this should give the same result as our Maclaurin-series defined function-derivative: Take g(x) = [series 'g' of (x^n)] so { g(D) } [f(x)] = [series 'g' of ( {D^n}[f(x)] ) ] We start with: { fourier of } [ { g(D) } [f(x)] ] By our Maclaurin deinition: = { fourier of } [series 'g' of ( {D^n}[f(x)] ) ] By linearity of fourier transform: = [series 'g' of ({ fourier of } [{D^n}[f(x)] ]) ] Fourier transform of a derivative: = [series 'g' of ( [(2(pi)i*x)^n] * { fourier of }[f(x)] ) ] Because "{ fourier of }[f(x)]" doesn't depend on 'n,' =[ { fourier of }[f(x)] ] * [series 'g' of ( (2(pi)i*x)^n ) ] Because of definition of g(x), and rearrange: = g(2(pi)i*x) * [ { fourier of }[f(x)] ] QED Well-behaved functions, g defined on C, yadda yadda. Very cool, pi-m.

[@PeterBarnes2]

Well, you'd want to prove that this 'Maclaurin Integral' would actually 1) converge and 2) converge to the function f. I wouldn't be surprised if Fourier transforms would be useful, in that. The tough thing is about where integration happens. I'd presume the entire complex plane. Conveniently, the Pi function approaches infinity on negative integers, meaning nth integrals of 'f' are excluded from the process.

[@ClevelandLemur]

I'm half impressed

[@ffggddss]

Next video: The 1½-derivative, by which you will be 150% impressed! Fred

[@RieMUisthegoaT]

Hey Dr. Peyam, What's is your research field? today i looked up the math department of my uni and i found out what do they research exactly, and i found some nice words lol like how my calc 1 dr researchs in integro-differential equations and fractional calculus. what means half derivatives aren't only useful but they are being developed right now...i think

[@drpeyam]

Partial Differential Equations, although I don’t do research on fractional derivatives

[@mehdisi9194]

Thank you dr peyam. I have a question about euler formula. Despite the fact that I have seen several proofs about this formula but I can not be convinced about the meaning of this formula at all and I see it artificially and not real because I do not have any idea of ​​the meaning of the power of imaginary numbers can you help me with introduction of any idea about it or any book where meaning of this formula is explained very well.thank you so much.

[@drpeyam]

I like the book by Brown and Churchill, it explains complex numbers quite well!

[@mehdisi9194]

@@drpeyam Thank you dr peyam. I have read this book thoroughly but in this book, too, I think the basic relationships are accepted by default And there is no discussion of the meaning of these relationships in this book. I'm looking for a book that at least discusses these relationships historically.

[@mehdisi9194]

@@angelmendez-rivera351 thank you angel. Yes, so I can not get along with this system of numbers. In my opinion, this system of numbers is simply a tool for launching, my field of study is more physical than mathematics and I think that should be why I can not accept this And that is why I'm looking for the historical evolution of this system of numbers and if you know a book in this field please introduce me too

[@ivansavchuk7956]

Can u do a video about Jordan decomposition in linear algebra? Previous year in Ukraine we had studied that but I didn’t understand what is that and the sense of this forms... Explain please...

[@drpeyam]

There’s already a video on that!

[@diegogambaro3823]

all this works because e^x is an autofunction of the derivate?

[@anthonyaportela217]

At first I didn't like your stuff, but i've grown to change my mind. You're amazing

[@drpeyam]

Thanks so much, I really appreciate it!

[@dominicellis1867]

Would it be possible to have a transcendental derivative like the order is phi or pi or e and would there be any application from this like in quantum mechanics or general relativity?

[@drpeyam]

Of course, just let alpha = pi

[@dominicellis1867]

@@drpeyam lol well that was easy do you know of any applications of alpha = phi derivative like maybe it models the rate of change of self similarity of the Mandelbrot set or maybe there’s some link between the product derivative and alpha = e?

[@drpeyam]

It appears apparently in fluid mechanics where a surface is cracked, or I guess in Brownian motion as well

[@peterwaksman9179]

Hi. How about working out some geodesics?

[@williamdavis2505]

Those are some big hats!

[@ffggddss]

Yup! Some expressions need a 10-gallon hat, podner! Fred

[@nathanisbored]

cool, but would it still give you the same half derivative of x as the other definition did? the one with pi in it?

[@nathanisbored]

@@angelmendez-rivera351 sick

[@funkycude57]

Dr.πm

[@nootums]

Bprp and the e^iπm

[@ffggddss]

Splendid! And very neat! Fred

[@drpeyam]

Thanks so much 🙂

[@GAoctavio]

Now do fractional FT :P

[@himanshumallick2269]

0:48 electrocuted!!

[@drpeyam]

himanshu mallick Hahahaha

[@remlatzargonix1329]

Cool!

[@xCorvus7x]

You have put on a hat. How do you undo that? You put on a reversed hat.

[@JeremyGluckStuff]

Dr. Peyam please use delimiters for your transforms those hats were ubsurd

[@drpeyam]

What do you mean?

[@emanuelmartinez3585]

First comment :)