That SquareSpace advertisement is simply genius. Did you and your team make it yourselves???

I've been using Laplace transforms to compute complex order derivatives for fun, it's super interesting and a really easy way of conceptualizing what's actually going on.

Any suggestions of bibliography about it?

@@matematicacommarcospaulo There are many books available about laplace transforms but not many about their applications to half derivatives

I convert the function into its representation as a hypergeometric function, add alpha to all the coefficients, then go to lunch.

@@plutothetutor1660 I'm looking for those with half derivatives

Agreed! He is a devoted father, engages in dangerous hobbies, teaches full time face to face AND produces at least one exemplary video a day!! Conclusion: He doesn’t get much sleep.

The f^-1(g(f(x))) pattern is super common throughout math and is generally referred to as conjugation. It plays a very central role in group theory.

Great connection! The laplace and fourier transforms are in fact very similar to diagonalizations. Basically, what you're doing is transforming from one domain (set of basis vectors) into a more convenient set of basis vectors. The choice of basis vectors is informed by the *choice of linear operators you wish to apply to it*. Let's start discrete. Say you're doing linear algebra and you have some symmetric matrix S. Any symmetric matrix can be decomposed into orthogonal eigenvectors: S = Q Λ QT. Now say you have some vector x, and you want to repeatedly apply S to it. S^n(x) = Q Λ^n QT x, since diagonalization provides a convenient way to take powers of a matrix. Look at the structure: (QT x) = Q transpose times x, this line transforms x from its original basis to the basis of eigenvectors of S. Λ^n is some simple operation applied to x in the eigenvector basis Q transforms back into the original basis. So Q Λ^n QT x transforms into eigenvector basis, does some simple operation in eigenspace, and then transforms back into the original basis. Now let's see the parallel with the Fourier transform of a function. S, the symmetric linear operator, will be the *second derivative of x*. The eigenfunctions of the second derivative are sin(kt) and cos(kt) or generally e^(ikt). As you may recall from math, sines and cosines are orthogonal functions. Inner products in function space (what we wrote as QT x for the discrete analogy) are the same as integrating the product of functions across their range: prod(f, g) = ∫ fg dt ∫ e^(ikt) e^(ipt) dt = δ_k,p (= 1 if k==p, 0 otherwise, you can prove this yourself) The Fourier transform then transforms some function f(t) from its original, cartesian basis into a basis of orthogonal eigenfunctions, just like QT x in the discrete case. You take the inner product of f(t) with every different frequency e^(ikt), and you get the frequency content F(k). The power of the Fourier and Laplace transforms is that when you transform into the eigenbasis of some operator, it becomes *very easy* to compose that operator with itself on the original function. For our discrete case, it makes it possible to take some arbitrary power of the matrix S. In the continuous case, the transformation makes it possible to take some arbitrary power of the derivative operator. And that's what this video is showing. For more information, check out Gil Strang's Computational Science and Engineering, section about Fourier.

@aidanhennessey5586 , And it's also an idea accessible to kids, who use it all the time rotating objects in Zelda TotK (where you can rotate a 3d shape in 2 axis, and need this trick to effectively do it in the third one)

can you use analytic continuation on annuli to express a Laurent series as a Taylor series on a punctured domain to obtain the fractional derivative there?

Now I'm curious what happens when you use this approach for the fractional derivatives of the exponential. We'd expect it to still be an identity, but it might break down in an interesting way since one of the hidden assumptions is that f(t) is sub-exponential. The offending expression is still valid and equals f(0) = e^0 equals one as long as s ≥ 1, so who knows 🤷‍♀️

@@ZieWeaver I'm pretty sure that Laplace transforms only require that the function to be transformed be of exponential order or less, not of sub-exponential order or less, so this method should still work for exponentials. Also, the exponential function's series expansion converges absolutely, so doing the fractional derivative term-by-term ought to work just fine. I'm not 100% sure on that though.

@@ChefSalad I'm referring to 16:05 when we make the sub-exponential assumption to evaluate e^{-st}f(t) on (0, inf). We can substitute e^{-st}e^t=e^{(1-s)t} which always converges at t=0 but conditionally diverges at t=inf as discussed earlier.

t-->infinity is a concept too, not a number.

the integral over the arc goes to zero in the limit of large radius

It's hard because Laplace transforms are only 'nice' for certain specific functions: en.wikipedia.org/wiki/List_of_Laplace_transforms

I can see the incels shrieking "WOKE MATH!" already.

Idea of "integrodifferential" combines integration and differentiation into one operation, where antiderivatives are treated as derivatives applied negative number of times.

Dr Peyam made a video about complex derivatives: kzfaq.info/get/bejne/qrORn9yexNScqoE.html

I've seen a bunch of videos where they plug in i into the fractional derivative and I guess it works

​@@frfr1022First I've seen that term; it piqued my interest because I have an operation (in my research) that combines integration and differentiation (but after looking at your term, in another way). Do you know of any other terms that describe combining these two operations?

sure! you can even put a matrix as the derivative, since you can define algebraic functions on matrices

The Matrix is resetting itself.

Yes the definition of the "fractional" derivative given in this video works for α=pi and α=e - "fractional" here is a bit of a misnomer like in "fractional part"; it doesn't actually mean a fraction, just a non-integer.

By definition it turns into antiderivatives since D^a D^b is the same as D^(a+b)

I think I can prove that this would work, although I'm not sure if I'm doing anything wrong: Given the regular D^a, show that D^-a D^a f(t) = f(t), i.e. if D^a forms the a-th derivative of f, then D^-a forms the a-th antiderivative. D^-a D^a f(t) = L^-1(s^-a L(L^-1(s^a L(f(t))))) = L^-1(s^-a s^a L(f(t))) = L^-1(L(f(t))) = f(t) If I'm going wrong with this anywhere, then I'd assume that this is because L probably isn't injective (and therefore L^-1 would actually be multi-valued or lose some possible results) but I don't really know anything about that because Michael Penn kind of glossed over how L^-1 may or may not be well-defined in this video.

“And bye”

I don't think so. The d/dx serves purely as notation in this case (but of course there are times where it's useful to think about it as an actual fraction). Most of the times half derivatives are written as f^(1/2) or D^1/2 f

You can write derivatives in an implicit way, eg: y=x^2 dy = 2 x dx (d^2)y = 2 dx^2 and then conclude (d^2/dx) y = 2 dx So yes, it does make sense. But there is no use for it - the only part that matters is the exponent in the numerator.

@@proxagonal5954 dy/(dx dx) = N, where N is greater than all natural numbers

@@pyropulseIXXI I don't think that's a well defined proposition

As long as a is a natural number, one could define this quite easily by generalizing the usual limit definition of the derivative. But I think in most cases, that limit would not exist.

L(f(t)) will not be defined then bc the Integral has its bounds at 0 and infinity. That being said he later assumes that t>0 which is probably the only case where f(0) not being defined works bc you can do workarounds with limits.

@@kju-uu8me is it possible that the limit as t approaches 0 does not exist?

You can make the same assumption Michael does for the case Df(t), that is L^(-1)(sf(0)) = sf(0) δ(t) which is 0 as δ(t)=0 for t>0.

​@@r.maelstrom4810 Thanks; I hadn't realized that the inverse Laplace transform of 's' is also a δ(t). I think this should have been explicitly pointed out, though, because following the previously shown transforms, s = s^1 = 1 / s^(-2+1), so the inverse should be t^(-2) / Γ(-1), which is ill-defined due to Γ(-1). Apparently, one needs to do a complex contour integration to obtain δ(t). *Update:* I mistakenly stated that L(-1)(s) = δ(t), when in fact it should be δ'(t) (the derivative of δ(t), which can be defined by an integration by parts). In that case, it is not at all clear to me how we end up with a zero term, and in any case the issue is involved enough that it should merit a brief explanation in the video.

I think i have seen complex derivatives, but not on KZfaq. Matrix, or quaternion derivatives i haven't heard of, but maybe Cauchy formula wouldn't fail in this context

Complex derivatives are already a thing look up "ith derivative" you'll probably find a few videos about it Matrix derivatives might work since we did find way to evaluate e to the power of a matrix but I'm not too sure about that yet

It probably has no practical value but you could define matrix derivatives as D^M = exp(M log D) and then use the series definition to get "something". At this point we are just making stuff up though so we can do whatever we want ;)