Exponential derivative

In this video, I define the exponential derivative of a function using power series, and then show something really neat: For “most” functions (those that have a power series expansion), the exponential derivative is just shifting the function by 1!

I also derive the product rule for exponential derivatives, which is much more elegant than the one for derivatives.

Enjoy!

[compphysgeek]

"In case you're wondering what are the Applications of this? I have absolutely no idea!" .. quintessential mathematician :D

[luisduarte]

After watching this I immediately thought "this is an operator! And it translates a function by an amaunt! Omg I can apply this to QM and discorver some new stuff!"
So I spent 2 days experimenting some stuff with this, had a lot of work and was even thinking about showing this to my teacher.
About 30minutes ago I found out all the stuff I though I was discovering were already discovered and that this is a well known operator, and there's so much more to it than what I had done...
Really though I was making a breakthrough in physics. Happy I could find it out on my own tho.

[morbidmanatee]

This is the well known shift/evolution operator in quantum mechanics.

[PeterBarnes]

Part 2: Fourier Derivative.
Part 3: Laplace Derivative
Part N: Types and Properties of Transformation-Derivatives

[NeolithicFellow]

I always joke about Math motto “Generalize all of them to death!” when I see how mathematicians derive any kind of patterns, but this one’s just blown my mind. Astonishing.

[ryanoftinellb]

Peano goes insane: Give each natural number _x_ a successor of the form e^D (x)

[curtiswfranks]

Hey, good job! You just rediscovered a cornerstone of quantum mechanics! (I am seriously congratulating you. You are creative!)

[GritliAhmed]

The fact that you got to find it out by your own is just awesome!

[pyrotas]

Actually, that would be nice to mention the action of exp(aD) with a being a scalar. Then one could make connection to the generator of translations in quantum mechanics. And perhaps even mention that given a pair of canonically conjugated operators A and B with eigenfunctions f(a) and g(B) are such that exp(lambda A) translates g(b) by lambda, and exp(lambda B) translates f(a) by lambda. This is of paramount interest in QM.

[michalbotor]

neat! acutally i checked, that exp(aD)f(x) = f(x+a), for whatever a. ;)

[billkerry]

Dr. Peyam. You are awesome and I am definitely learning more from you. Please keep sharing the knowledge. Its great to encourage others by thinking outside the box.

[Небудьбараном-км]

This man is so much fun to watch! Thanks for explanation!

[jordanweir]

That trick with the sum interchange is one I've somehow missed in the past, nice to be exposed to it as its pretty nifty imo

[ekueh]

Wow! Thank you.
My wish list is using negative derivatives rules to solve integration. This will add one more tool to the bag of tricks

[kobethebeefinmathworld]

7:00 "So there are no BlackPenRedPen magic here" - Dr. Peyam
A new, useful quote in 2018 lol

[LwLiPp]

This would be quite a neat introduction into the concept of operator (semi-) groups. In that context, the content of the video translates to the differential operator being the generator of the shift group (even in the weak sense!).

[Demki]

Well, not sure if there are applications of it when looking at it as "the exponential of the derivative operator", but there are definitely applications for a linear operator L which has the property L(f)(x)=f(x+1), linear recursion relations is one such application.

And the operator "exp(D)" coincides with L for analytic functions, so that might help get some insight on problems.
It might even be used to obtain the analytic solutions to something like (L-2)f=0, (if we don't constrain f to be analytic, that recurrence relation has infinitely many linearly independent solutions, if we look at f as a function from reals to reals)

[manuelgnucci]

I independently derived the exponential derivative like 5 years ago when I was just a studend, I used the Laplace transform to justify exp(D)f=f(x+1).
Then a guy with a bacherlor of mathematics degree came to me to say it was garbage and I let it go.

[GreenMeansGOF]

It’s an awesome feeling to discover something new in mathematics. I have also played with curiosities but im not sure if it is new.

[tryphonunzouave]

Finally I see a room that's full of pens lol, in my classes none of them work and everyone ends up having to bring their own