Half derivative of 1

[@jdaman1989]

Applying the half derivative formula again, D^(1/2)[1/sqrt(pi*x)], does give you 0! Using the fractional derivative formula, the integral will give you a constant (independent of x), and differentiating that gives zero. P.S. Former Berkeley grad here. Keep up the great work!

[@drpeyam]

Go Bears! ❤️

[@helloitsme7553]

Oh so the equation O(O(f(x)))= f'(x) where O is an operation has two solutions when f(x)=1 and this is where this debate is coming from?

[@jdaman1989]

I think the issue is finding the functions f that satisfy the equation D^(1/2)[D^(1/2) f ] = 0. There are two solutions: f = 1, and f = 1/sqrt(x) (and any multiple of them). However, the "actual" derivative of 1/sqrt(x) is... not zero.

[@smiley_1000]

Julian Landaw no, square root of x isn't a solution. Where did you get this from?

[@jdaman1989]

Not sqrt(x), but 1/sqrt(x), i.e. x^(-1/2)

[@stydras3380]

What about a differential equation involving fractional derivatives! c:

[@pacolibre5411]

Its definately 1/sqrt(pi*x) because that satisfies both definitions, while 0 only satisfies one of them.

[@michalbotor]

while playing with the a-derivative myself i've noticed, that it's sometimes useful to remember that since 0

[@benjaminfox6206]

you should make a video deriving the fractional derivative formula

[@drpeyam]

It’s a definition, so there’s no derivation

[@benjaminfox6206]

Dr. Peyam's Show oh i see, my bad

[@mikewagner2299]

What about the intuition behind the definition? For example, the definition of the derivative is a way to formalize the slope between two infinitely close points on a curve

[@123pok456ey]

Dr. Peyam's Show ls it like how gamma function fit the definition of factorial? This definition fits some results of half derivative so we define it.

[@mikewagner2299]

wc k well the gamma function is the extension of the recurrence relation (n+1)! = (n+1)*n! to the similar continuous relation f(x+1)=x*f(x). Which, along with an initial value of f(1)=1 and a limit constraint, uniquely give f=gamma. There are infinitly many continuous functions that pass through all of the points (n, n!) But the gamma function does so (shifted by 1) while also possessing the fundamental behavior of the factorial function

[@Hexolero]

What is the motivation for defining the half-derivative using the complicated formula, instead of the (more intuitive) definition that (D^1/2)^2=D? It feels like any reasonable definition of the nth derivative is that (D^n)^1/n=D.

[@drpeyam]

I agree, but then it becomes basically impossible to calculate half derivatives. Try doing that for example with ln(x).

[@willnewman9783]

@@angelmendez-rivera351 You say that "the definition is still (D^1/2)^2=D" but this is by no means a definition. First, if such a D^1/2 exists, it is not unique, as -D^1/2 will also satisfy the same property. But more importantly, there is no linear function D^1/2 defined on the set of infinitely differentiable functions with (D^1/2)^2=D. So a definition like the one in the video is the best one can get.

[@willnewman9783]

@@angelmendez-rivera351 Well, I do not know why someone would want to consider such an operator if it is not linear. But regardless, (D^1/2)^2=D is not a definition, it is a property. And any D^1/2 must satisfy undesirable properties (if such a D^1/2 even exists)

[@kyleneilson1457]

@@angelmendez-rivera351 But D^1/2 as presented in the video does not satisfy these properties. In the video, he shows that D^1/2(1)=1/sqrt(pix). But then this gives that (1/sqrt(pix))'=D(1/sqrt(pix))=D^1/2(D^1/2(1/sqrt(pix))=D^1/2D^1/2D^1/2(1)=D^1/2(D(1))=0 But it is not true that the derivative of 1/sqrt(pix) is zero. Thus, (D^1/2)^2 is not the same as D.

[@kyleneilson1457]

@@angelmendez-rivera351 I am sorry Angel. I am not too good at this kind of stuff. I think I am just confused: If D^(1/2)(1/sqrt(πx))=0, and D^(1/2)D^(1/2)=D, then when we apply D^(1/2) to both sides of D^(1/2)(1/sqrt(πx))=0 we get D(1/sqrt(πx))=0, right? So where am I going wrong? Thank you for your patience with me

[@tom_szcz_org]

Love the half derivative videos

[@purim_sakamoto]

ガンマ関数おもしろいですねえ あと微分階数順にグラフを並べると、整数間で思わぬ変化をしたりする式があるのが衝撃です へやーー　本当はそんなんやったんやー！って

[@skylardeslypere9909]

Hi. So as we all know, you can define the derivative by saying it's the slope of the tangent of a function in a point. But can we define the fractional derivative as something similar? Can we define is to be the slope or positive/negative or...?

[@johnstotko6733]

I feel like it might be a useful exercise to get generalized expressions for the alpha-derivative. Similar to how one would use the definition of the derivative to get the product rule, power rule, chain rule, etc. How might one go about finding the null space of the half-derivative? How about the eigenfunctions of the half derivative? Are there any functions who's derivative is defined at some value but who's half derivative isn't? Likewise, the other way around? What's the laplace transform of the half derivative? So many questions...

[@tofu8676]

chill bro :D but i get the curiosity.

[@donfox1036]

John Stotko, looks to me that it's a scam. But then I can't even spell brachistocrone...

[@lightspd714]

Hi Dr.Peyam I believe the half derivative of 1 is 1/sqrt(pi x) given the Cauchy formula for repeated integrals derivation method for half derivatives. In this example however, I am left wondering what the interpretation would be. 1 is a constant so what would the significance of the x be in this result? If you have a line y=1 for example, half derivatives changing as you traverse through different x values doesn’t seem very intuitive.

[@helloitsme7553]

Fun question: find the solution to D^[3/2]f+D^[1/2]f=0 Update : too easy as you can just solve r^(3/2)+r^(1/2)=0. So maybe xD^[1/2]f=Df also still simple cause it's separable using a subtitution , at least it seems that but when you gotta solve D^[1/2]f=g it's a bit harder since the fractional derivative formula is not defined for α

[@mathunt1130]

What's in the kernel of sqrt(D)?

[@stevenwilson5556]

I want to love anything in life as much as Dr Peyam loves half derivatives and writing math on a white board (or chalk board) in general.

[@MrLordCoder]

take the i-th derivative of something. (complex unit i)

[@helloitsme7553]

Strichcoder i think he did the ith derivative of x in a video

[@drpeyam]

Already done ✅

[@yuvalpaz3752]

Hi, I just thought, in the definition video you said that alpha is in (0,1), so how do you deal with complex derivative? or did you meant |alpha|? Also, do you think that lim(a to 1-)D^af=Df?? and what about lim(a to 0+)D^af=f?

[@michaelriberdy475]

How can you derive a product lu so that the half derivative of x is equal to the half derivative of 1*x?

[@zeldasama]

He's back at it again with magical derivatives. Sorcery

[@federicopagano6590]

(-1)! is not defined but it goes to infinity so that constant would be c=0 so it all works good D (1)=0 even using both definitions

[@adj6006]

Could you have an nth derivative of 1, where n is any real number?

[@whozz]

Take the (n - floor(n)) derivative of 1 using the formula then derivate it the normal way floor(n) times

[@jameschen5775]

d^α/dx^α (k x^n) = kΓ(n + 1)/Γ(n - α + 1) x^(n - α) where {[k, n, α] ∈ ℂ}

[@Koisheep]

So, I've been looking for the motivation for Dr. Peyam's formula and it seems to be related to Laplace transform: en.wikipedia.org/wiki/Fractional_calculus#Laplace_transform

[@mcmage5250]

I am really confused now, ok so i used this definition to find fractional derivative of x^n. Answer i got was (x^n*(2n+1))/sqrt(pi*x) and yes i got the same answer when i plugged in 0 but here is the problem, you should remember the video you did on Half derivative of X or x^1 and answer you got was 2sqrt(x)/sqrt(pi). Now plug in n=1 into first equation which was for n greater or equal to 0. We get answer 3sqrt(x)/sqrt(pi) WHICH is not the same as your previous answer. It bothers me...

[@bashdaromerhussenmustaffa8861]

what is the half derivative (-x) Dr peyam

[@drpeyam]

- half derivative x

[@bashdaromerhussenmustaffa8861]

@@drpeyam thank you dr its very good . but can you tell me how it is happen ?its rule ( half derivative (-x)=-half derivative (x) ) again thank you for help me .

[@copperfield42]

so if I half derivative that again I should get 0, right? ... now I got stuck at the integral of 1/sqrt(t(x-t)) how I proceed from there?

[@wolffang21burgers]

Try u=x/2 - t :)

[@copperfield42]

Wolf Fang thanks :) I indeed got zero

[@dominikstepien2000]

So with this formula we have rational derivatives, but what about irrational derivatives?

[@drpeyam]

The same formula works for irrational alpha

[@povilasdapsys7765]

the 2/3'th derivative?

[@MF-lg8mt]

The fractional derivative of a constant, using the definition of the video is D^a K=Kt^(-a)/gamma(1-a). Now, substitute a=2/3 and K=1.

[@helloitsme7553]

I think I prefer the half derivative to be 0 , cause I think the formula isn't perfect, it is made up out of the definition that half differentiating twice should give you the derivative, but there is definitely some flaw in the formula. About the power rule part, the constant is 0 so the half derivative becomes 0. I believe any positive derivative of 1 is 0

[@milos_radovanovic]

so can this define half integral also?

[@drpeyam]

The half integral is the half derivative of the integral :)

[@boffo25]

Why do you like the half derivate so much? Why is it matematicaly so special?

[@stydras3380]

boffo25 Does it have to be special? Afterall math isn't about utility but about beauty.

[@boffo25]

Stydras Sure but form what branch of matematiche dies it come?

[@zeeek3348]

Awesome as always

[@stephenfreel2892]

I know that Cauchy’s integral formula has an application for nth derivatives. Couldn’t you use Cauchy’s integral formula to solve for the half derivative of 1?

[@drpeyam]

Yes, in fact that sort of explains why the formula in this video is as such

[@snejpu2508]

Hello, dr. Peyam! If you are interested in some another proposal for your channel's content, I propose derivations of formulas with pi. For example, derivation of Wallis formula for pi/2 comes from a sine formula, by setting x=pi/2. However, using x=pi/4 you can get a formula for sqrt(2) with pi. The same thing with x=pi/3 gives us a formula for sqrt(3) with pi. : )

[@drpeyam]

Ooooh, there will actually be a video on Wallis’ formula, but I haven’t thought about that way!

[@snejpu2508]

All right, it's your channel, so you decide. : )

[@medpop459]

I like your video sir ... it's awesome

[@SuperMtheory]

Very nicely done! Thank you.

[@gnikola2013]

What if the order of the derivative is also a function? I think it's an interesting idea, but I definitely lack the knowledge to make it possible

[@MF-lg8mt]

Here you can read more information about the variable-order fractional derivatives. www.sciencedirect.com/science/article/pii/S0165168410001404 If you don't have access to the article, I can share you the PDF file.

[@Schlaousilein67]

Nice video 👍

[@daaa2299]

Do 1/2 derivative of e^x

[@drpeyam]

Already done ✅

[@123pok456ey]

Dr. Peyam's Show How about by using the definition ?

[@iheb404-notfound3]

Gamma of half is sqrt(pi)/2 Dear Dr. Peyam, tell me if i am wrong but my calculator gives me something like 0.8862...

[@RalphDratman]

This is disturbing because it seems that the half derivative is not a single well-defined operator. Maybe these two definitions should be called two different things?

[@UnforsakenXII]

I see you everywhere, Ralph. : )

[@RalphDratman]

Andres Franco Valiente -- Hallucinations, maybe?

[@smiley_1000]

It actually is. If you differentiate again, you get 0.

[@RalphDratman]

I know that, but still there seem to be at least two definitions that sometimes give different answers. Why should we imagine they both in some sense describe the same operator?

[@smiley_1000]

@@RalphDratman I agree, we should think of them as being different. One Problem with the intuitive one is, that we can't give a formal definition.

[@pacolibre5411]

So do you mean to tell us that fractional derivatives aren’t linear? (Edit, they are)

[@drpeyam]

They’re still linear!

[@pacolibre5411]

Dr. Peyam's Show 2 things. 1) I wasnt thinking of linearity properly. 2) I just did out the half derivative of 1/sqrt(pi*x) and found out that it is in fact 0, which I did not expect to be true, so yeah, there is nothing wrong with that formula. I expected because of the split in definition that you mentioned at the end that the general formula did not obey the “half of half is normal derivative” rule

[@carlosvargas2907]

No querés iniciar una guerra pero voy a darte batalla!!

[@alvaroperezrivera5069]

And the product rule for the half derivative? First I"ll attemp that

[@MF-lg8mt]

The Leibniz rule is different for fractional operators. For classical derivative, Leibniz rule is a finite summation, however, for the fractional derivative is an infinite series.

[@theoleblanc9761]

Let f defined by for all x, f(x)=1. Let g=f(id-1) (id(x)=x for all x and 1 means the constant function, ie: f) g(x)=f(x-1)=1=f(x) Therefore g=f so D^0.5 f = D^0.5 g so (D^0.5 g)(x)=1/√(πx) But g is just f translated but the half derivative is not, or from an another point of view, g is f it self but just the origin has been moved (without moving f) and the half derivative of f (equal D^0.5 g) has moved with the origin but f don't. That is why I understand that can arbitrarily right D^0.5 f = 1/√(π(x+β)) with β a random real number... More, 1/√(πx) is only defined for x>0 but no matter x>0 or x