Won't somebody stop this madman!

Next video: "The derivativeth-derivative of e^(-x^2)"

Next zeta functionth derivative of gamma function pleeease.

This series in a few years will be like: Solve the differential equation sin(D)y+exp(D)y+arctan(D)y+xy'=0

There is no stopping this mad man.

These kind of objects are actually used in quantum mechanics. For example, e^(-2pi*i/h*p*a) is on operator which shifts the wave function of the system in space, expressed by the momentum of the syste.. p, the momentum operator is h/2pii*grad. Ultimately this is connected to continuous symmetries being expressed vy lie groups, and the elements of the lie algebra being the operators of conserved quantities.

This idea is very clever. However, before I saw your videos about exponential derivative, I had thought that it is generalization to fractional derivative. Since I had seen your thumbnail, I thought it should be written as ((d^(e^x))/(dx^(e^x))) f(x) and calculated like fractional derivative. For example, like D^(e-2) (f'' (1)) (where D stands for fractional derivative). Although this isn't what I expected, I find your videos about differential operators very interesting. And I know that my idea might not be even possible to calculate.

Hmmm, I feel as though saying this is the ‘sin’th derivative is wrong because we don't have D^sin(x) instead we have sin(D). The notation in the thumbnail is misleading in this sense. Now that I think about it, D^f(x) sounds really interesting and could be a good video idea.

I never knew I needed to know this, but boy I love it!

next: arithmetic differential equations lol

Love you channel! Thanks for sharing the stuff! %-) The math gets more fun with your vídeos

Next Xth derivative of X

Hey, Peyam! This was a great video! One thing I noticed was: sinD(sin x)=D(sin x)*sinh(1) and sinD(cosx)=D(cosx)*sinh(1), where D(sin x) and D(cos x) are the regular derivatives of sin x and cos x, respectively. And I thought that was pretty cool and interesting!

Awesome!! (As always)

Next do ζ(D) and determine the eigenspaces and eigenvalues of it. Then also maybe Γ(D).

The more natural way to write the derivative is to note that it be the imaginary part of exp(id/dx)f(x) = f(x+i) and the cos derivative will be the real part.

Dr Peyam, i think you could use the exponential derivative exp(D)f(x)=f(x+1) to redefine the cosine and the sine derivative as cos(D)f(x):= (f(x+i)+f(x-i))/2 and sin(D)f(x):= (f(x+i)-f(x-i))/2i, it gives the same results above in your video

There might be confusion around the notation sin(D)x. sin(D) is not sin of D, it’s more like another function based on sin(x).

Could you talk about dual numbers? Since the topic is unusual ways to take derivatives, I feel that it would be relevant.

DeaR Peyami Bro Best regard from Turkey U'r legend 😍❤

Hey! I've been messing w/ similar ideas in my freetime using residue; not to sound presumptuous, but can you find a closed form for e^(ad^n/dx^n)f(x)?

If you do exponential series but with the integral operator you get the shifting but with -1, which is kinda expected but still really cool.

Hi Dr Peyam. I don't know if anyone else has commented this yet but I think I have figured out how to simply express sine and cosine derivatives of functions f(x). The sine derivative of f(x) = (f(x+i)- f(x-i))/2i, whilst the cosine derivative of f(x) = (f(x+i)+ f(x-i))/2. Also, the sinh derivative of f(x) = (f(x+1)- f(x-1))/2 whilst the cosh derivative of f(x) = (f(x+1)+ f(x-1))/2. The exponential derivative of f(x) indeed equals the sum of the cosh derivative of f(x) and the sinh derivative of f(x). The sine and cosine derivative formulae look very similar to the Euler formulae for sine and cosine, where the top of the fraction is a combination of ith and -ith exponential derivatives. You can also easily prove that the sinh and cosh derivative formulae hold for x^m.

Hey, Peyam! One thing I noticed was that taking the sin derivatives of sine and cosine multiplied their respective regular derivatives by sinh(1), which made it very easy for me to remember. And I thought it was pretty cool!

Last time you said that it might be useful for the lie groups. In the last year I've been very reading about quaternions and dual quaternions. what where you tinking about the usefulness of this derivatives in the lie groups?

Is there any other way to go about this type of differentiation if the function is not analytic?

I wonder if you could prove that this definition would hold on branch cuts if the functions in the differential operator weren't entire.

Enjoyed it as always! Hmm, I've never seen any antiderivative versions. Something to explore? By the way, in vector calculus, you skipped showing that the integral definition of divergence (given here: en.wikipedia.org/wiki/Divergence) is equivalent to the differential version. The divergence theorem plays an obvious central role in proving this result. If you mentioned it already, and I missed it, then I apologize XD. But I think it would make for a nice video on its own. It would also give you a chance to show off your skills in (mathematical) analysis. ;) I hope this video wasn't as glitchy for everyone else as it was for me.

- sin(1)*sin(x) is equal to (cos(x + 1) - cos(x - 1))/2 via Werner's formulas.

Using Leibniz notation, you might write g(d) f(x) / g(dx) for the function-derivative by g of f with respect to x. This makes linear differential equations feel truly linear. You can say g(y) = y^2 + 1, g(d) f(x) / g(dx) = 0, which gives that f(x) = sin(x) {not 0-function, yadda yadda}. A trivial example, but a nifty one. Also, g(e^x) = g(d) (e^x) / g(dx). It just pops right out. g(d) (e^u) / g(dx) for any u(x) is a tougher problem, but not really. An interesting challenge would be using a parameter function g_n(y), then asking for a function derivative by g of f(x) in terms of x and n. I want to see what this kind of idea would mean if we wanted to talk about Fourier transforms instead of Maclaurin Series.

Are there cases where the sum diverges into infinity? Can those be represented algebraically?

You should do the sine of a fractional derivative next

That's odd cause if you divide the last 2 equations then you get: cot(x)=-tan(x) but it should be: arccot(x)=-arctan(x)??

What is the "overall name" for these g(x)-th derivatives of f(x) so that I could look it up on Wikipedia or sth?

What about the Gamma function of D?

Is there something like a half integral or something like that?

I found a general form for sin(D) Notice that exp(D)f(x) = f(x+1) exp(nD)f(x) = f(x+n) exp(iD)f(x) = f(x+i) Since sin(t) = Im[exp(it)] Re[exp(iD)f(x)] = Im[f(x+i)] So... sin(D)f(x) = Im[f(x+i)] , and cos(D)f(x) = Re[f(x+i)]

Can you make video series about derivations of e ? That would be very very interesting. Perhaps 2-3 episodes. /just kidding ;)

8:18 why it's linear ???

Infinite derivative?

Dr peyam can u wait a little bit lol when u finish a board before erasing and step a bit aside lol because a lot of us are taking screens hots of the final board lol thanks

I have a general formula for the sin derivative, sin(D)f(x)=(f(x+i)-f(x-i))/2 for any function admiting a Taylor serie.

Okay, there’re some cool suggestions in the comments but an exciting one is missed. Bow down before the imaginary derivative of f(x) :D

Make the inverse (Sine integral)

Why is such derivative useful? Is it a mathematical trick or is there an application worth mentioning?

What about i-th derivatives? (And next: complex ones)

ンアーッ! そんな危険なところにDを入れて･･･ と思ったら無限級数の方をいじるのか いやぁー拡張するの楽しいです

How about like arctan(d/dx) lnx?

So, I know this is more of a comment on my lack of knowledge, but what precisely makes this valid?

Haha this is crazy. I love it.

9:45 THE MEANING OF IT ALL!

do people commonly call sinh "sinch"? I never heard that before :D

For the thumbnail: wouldn't it be "sin(d/dx) x" ?

The derivativeth derivative of a function

I think I could do a power series of: x.

5:07 you said you wanted to find the sin-th derivative of cos... But why did you use i?

I want Gaussian derivative for Christmas please.

So sin(d/dx) [e^(pi*x)] is just 0. 🤯

sin(1)=Im(exp(i))

Happy Yalda Peyam agha 🍉🍎

this is me... what .... ohhh ... cool... the world just got bigger!

Tangent derivative for sin . Its look funny..

Does this have any practical use?

I went and looked at the g[D] of f(x), the general case for these function-derivatives, and I wasn't satisfied with Maclaurin series. It's pretty easy to find that g[D] (f(x)) = sum n=0 to inf. { [(D^n)[g] (0)] * [(D^n)[f] (x)] / n! To get a taylor series, you would write [(D^n)[g] (h)] * [((D-h)^n)[f] (x)] That means we've got a binomial expansion to do, which is pretty bad because we've already got a summation. The double summation will not go away cleanly in the end, either. Starting with the trouble spot (also, it makes more sense to use 'y' instead of 'h' for the end result, for a nifty 'symmetry'): ((D-y)^n) [f] (x) =" {binomial expansion, (D^k)[f(x)], (-y)^(n-k)} " and we eventually get ((D-y)^n) [f] (x) = (n!) * { sum k=0 to n (-1)^(n-k) * (y)^(n-k) * [(D^k)[f(x)]] } Putting it all together would get terribly cumbersome in a youtube comment (it's much cleaner when written properly), but the end result: - is a double summation that isn't clearly easy to clean up - contains the product: [(D^k)[f(x)]] * [(D^n)[g(y)]] -has an almost mystical 'symmetry,' where switching from g[D] @y of (f(x)) to f[D] @x of (g(y)) only switches [(D^k)[ f (x)]] * [(D^n)[ g (y)]] to [(D^k)[ g (y)]] * [(D^n)[ f (x)]] and changes y^(n-k) to x^(n-k) -definitely includes the solution where y=0 gives the Maclaurin expansion of g. (There's a tiny trick because 0^0 = 1 for binomial expansions 'like' this.)

Could you please write when you are deriving the function? Because you just wrote something like " function" = "derivative of the function", and it really complicates the comprehension of the video.

But why??? Why would you say that (d/dx)^(sin) =sin(d/dx) in the thumbnail (again!)? That is so wrong it even fails for reals: 2^3=/=3^2. I love you but this is really frustrating. You don't even acknowledge it. Is it just clickbait? It's frustrating because I would love you to explore the real deal, the actual (d/dx)^(sin).