Exponential derivative
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- Опубликовано: 25 авг 2024
- 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!
"In case you're wondering what are the Applications of this? I have absolutely no idea!" .. quintessential mathematician :D
That was the best!
They're used for numerically solving differential equations :p
Functional calculus in infinite dimensional Hilbert space. In particular, Quantum Mechanics
@@AbrahamLozadaabe Where EXACTLY in quantum mechanics? My field of profession is physics & I've never come across with it! (Just because infinite dimensional Hilbert space comes in qm, it doesn't mean all the mathematics a mathematician can talk about it is used in qm - As how we can't claim that 'just because' we use calculus in physics, we r using ALL the mathematics that mathematicians are talking about in calculus!)
In my opinion, what Dr. Peyam is developing is very interesting only (as of now). It is esoteric to the mathematician.
@@TheAzwxecrv It's very elementary. Consider, for instance, a translation in the position representation. Remember that the generator of translations is the momentum operator (and the form of this operator in position representation).
This is the well known shift/evolution operator in quantum mechanics.
Any links to further reading?
It's vaguely related, but this isn't the time evolution operator.
@@Kalumbatsch it's the displacement operator: shifts a position eigenstate | x > to |x+x'> if you consider e^{-ix'P}
@@giannisniper96 That still isn't the same expression.
i am deeply convinced that +1 is a well known operator in many places of science
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.
That's still very magical :)
It feels good when u discover something on Ur own, I too was playing with infinite series and discovered this thing,refer ruclips.net/video/L4apIOi68jo/видео.html
still, it was a break througphysics in general ,but for you. You had an idea, tested it and had bad luck that someone had it before you. Know what, you still had the right idea, and the fact ,it´s already known only made it easier to verify it. Not gonna bring you a dime, but it still is worth a lot. hypothesis and experiment are part of all parts of science including mathematics
every time I discover something I assume I'm not the first one, but I guess if we keep going maybe we would eventually get to the edge of human knowledge :D
@@Metalhammer1993 agree!
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.
Peano goes insane: Give each natural number _x_ a successor of the form e^D (x)
circular reasoning since the peano axioms are part of the definition of e^D (sort of)
@@thedoublehelix5661 how about we start here at e(D) and then prove a had a successor a+1 for all a
Part 2: Fourier Derivative.
Part 3: Laplace Derivative
Part N: Types and Properties of Transformation-Derivatives
Lie derivatives
Log differentiation seems obvious and should at least be mentioned in Part 1. I used to use log differencing to make Economic indices from data.
I don't know what log differentiating means in other contexts, but you can't use ln(D) in the same way as exponential (e^D) or sine (sin(D)) derivatives with a Maclaurin Series.
To be able to use a function g(x) to create a g(D) operator, you need to define g(x) in terms of other functions f(x) for which the f(D) operator is already defined. As well, g(x) must be defined for all x in terms of those functions f(x). You can't restrict the domain of functions because the f(D) operator almost certainly won't behave like the f(x) function does.
For example, a Maclaurin Series consists of functions f_n(x) = x^n. f_n(D) = d^n () / dx^n. Changing the 'n' in f_n(D) is not like changing the 'n' in f_n(x). The second derivative is not necessarily the square of the first derivative, even though the 'powers' in the derivative operator are doubled.
Because the behavior is unpredictable, we can't establish what it would mean for a particular domain to evaluate correctly, even if any limits do converge. (A series can converge on a domain larger than the domain they approximate a particular function.) We can't restrict the domain, because we can't establish correct evaluation. If the series converges everywhere to a function, that's a guarantee that the operator will behave as a 'g(D)' operator.
The natural logarithm is defined on (0,2] with a Taylor Series. A Taylor Series could work if the domain wasn't restricted. It isn't, so you can't really do an ln(D) operator like that. Short of inventing some mystically non-arbitrary way of establishing how to obtain an inverse function of an operator from the original function on an operator, ln(D) will remain implausible.
@@PeterBarnes2 log derivative is f'/f, which of course is the same as (log(f))' when that logarithm exists, hence the name. This has nice properties, such as (fg)'/(fg)=f'/f+g'/g. I use these tricks frequently to make log derivatives much simpler to do
Don't most functions not have a Taylor series? Infinitely differentiable is a very strong condition
Hey, good job! You just rediscovered a cornerstone of quantum mechanics! (I am seriously congratulating you. You are creative!)
Thank you 😄
neat! acutally i checked, that exp(aD)f(x) = f(x+a), for whatever a. ;)
Yeah! That's the shift operator as used in quantum mechanics. Sooo cool to be able to deduce it in a completely different way.
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.
product rule and make one of the two functions the scalar
@@plislegalineu3005 this is clear, I'm just saying that it would be nice to mention some connection to physics.
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
The fact that you got to find it out by your own is just awesome!
7:00 "So there are no BlackPenRedPen magic here" - Dr. Peyam
A new, useful quote in 2018 lol
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
I'm actually not sure if this works well unless you're just exploring math for fun, as there are a lot of cases where negative derivatives aren't the same as integration
+Joshua Cohen I thought derivatives and integrals are inverses, and the effect of an integral reduces the derivative by 1. So how would there be examples of the negative derivative being not the same as an integral?
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
Except none of them work 😭
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.
Legend says Dr learned to write left handed so that it’s easier to see for the audience
Quantum Mechanics I assume?
Aaaaaaahahaha ! Like the reference to black pen/red pen.
Though it is amazing how the exponential's seem to follow the same rules as the bases.
This man is so much fun to watch! Thanks for explanation!
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!).
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.
Never give up on ideas, even if they’re silly :)
@@drpeyam thank you
Aha this is very interesting indeed, the vector a*d/dx is a Killing vector on the manifold R^1 which generates isometries on the manifold, which are really push forwards corresponding to diffeomorphisms that translates a given vector by "a" on the real axis. These also form a Lie algebra that when exponentiated to give exp(a*d/dx) gives group elements that generates translations of type x \to x+a and thus f(x) \to f(x+a) for any a and f \in C^(\infinity)(R)!. This is so deeply rooted in differential geometry, group theory and analysis!. Thanks Dr Peyam love your videos!
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.
Fubini is that guy who can get into any club without any questions even though he looks super shady
OMG MIND BLOWN!...you always come up with an interesting result!
Great explanation! Thanks a lot! This video helped me solving my HW in quantum chemistry.
That was a very interesting subject of calculus! Really creative! I enjoyed this video a lot.
One use case in the momentum operator in quantum mechanics can be written as the idx and then exp(- iadx ) is a unitary translation operator.
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)
It sounded absolutely ridiculous at first, but after learning linear algebra it makes perfect sense. Differentiation is a linear transformation, which means compatible with matrices, and exp of a matrix is well defined.
That’s a nice way of putting it wow
today (e)s a great day!!thanks for this video dr peyam
Omg, you are a genius, dr
I would love to take class in your classroom
4 years later ive come back for this and I have an application for this:
t(x) = a*(x+1) + a*t(x)^2 + a*(e^D t(x))
expand that as a series in terms of a
and then, the coefficient of a^n gives you the number of lambda expressions with n symbols
Nice video. Note that the first 15 minutes can be compressed as follows:
e^Df(x) = Sigma D^nf(x)/n! = Sigma f^(n)(x)*1^n/n! = f(1+x), provided that f has a Taylor series around x with radius > 1.
The exponential function can be thought of as a way of shifting multiplication to addition. That explains why the exponential derivative handles the product rule the same way that the regular derivative handles the chain rule. By implication, I would expect that a logarithmic derivative would handle the chain rule the same way that the regular derivative handles the product rule. Likewise, where the exponential derivative of a function so often just adds one to the variable, I would expect that a logarithmic derivative would subtract one.
Operator theory.(functional analysis).was developed to understand these objects, and it has applications in partial differential equations, quantum mechanics and Malliavin calculus.
We can calculate sums like the sum for n=1 to infinity of ((-1)^(n-1))/(n*x^n) (which is ln((x+1)/x)) with the exponential derivative
Wow! How?
We start by noticing that lnx having a Taylor serie de^x(lnx)/dx=ln(x+1) then by using the definition of exponential derivative we end up with a nice expression
That is beautiful, thank you!
I m very happy to learn such stuff and especially from u ..plzzz bring more videos on such new topics
25:21 definitely the very best part of the video XD I had to start laughing loudly XD Great!
Also, whenever Taylor's expansions exist, the last 10 minutes can be compressed and strengthened as follows:
e^D[g(f_1(x),...,f_n(x))] = g(f_1(x+1),...,f_n(x+1)) = g(e^D[f_1(x)],...,e^D[f_n(x)])
I can see how this ends: Dr Peyam ends up in the hospital and witnesses will he say he was integrating infinite ordinals laughing like a maniac until he collapsed.
Actually, the product rule can be proved in a easier ( and trivial) way for analytic functions... 😁
From the thumbnail I assumed this was a black pen red pen lol. It’s cool to see another calculus channel on here, awesome concept btw! Def a new sub!
Yay! 😄
This is a crazy-good video!! Thanks so much for sharing!
Aw man I loved k^nekx as a kid.
but srsly, beautiful stuff as ever. keep up the good work
He is really nice and excellent proff.😊
Thumb up, Nice work DR. . I came across these properties while during research on Hirota Bilinear equation. I actually wanted to expand this ln[cosh(aDx)f(x).f(x)];
@Dr Peyam, Sir, if you have any idea about the Taylor expansion of it pls help me.
This just the action of the translation operator known in quantum mechanics. Just expand the exponential operator and identify the resulting expression as a Taylor expansion.
In any case nice video
Best
I don't really use that notation (on the preview image) but what you seem to be describing I would call e^D and is quite common in physics.
For example, the time propagator for a Schrödinger wave function is written e^H/i where H is a differential operator:
H= -a*Laplacian + f(x,y,z)
An even more general result is that the Exp[k*D]f(x) = f(x+k), which is awsome because there are no restrictions on k so for k = i you can find Sin[D] and Cos[D] of any function by taking the real and imaginary parts of the functions
Have you ever heard of Umbral Calculus?
Supware has a video on it named "The Shadowy World of Umbral Calculus".
I'm asking this because his sequel video is "The Abstract World of Operational Calculus" which uses something really similar to what you've discovered.
I think these two videos will be very interesting to you
now let's take the derivative of the differential operator :O
omg thank you. I finally understand how to interchange the sigma notations
So could one expand a general linear operator as a power series of the derivative operator then?
Yep :)
Would be good to see the logarithmic derivative being applied and no doubt to wider classes of functions....Great tutorial thanks 🙂
I made one on sin derivative of cos or something :)
This is the exponential map of a traslations vector field representation of a Lie Algebra that give the elemento of the Lie Group of traslations.
Thank you Dr. Peyam
Was ab to go to bed but yeah.. screw that.. brotherman uploaded another vid so exponential derivative it is🤟🏼
This video is very nice, unfortunately is hard to see the calculations on the blackboard. exp(D) f(x) = f(x+1) is a proposed exercise in the 'modern algebra' of Birkhoff and MacLane . My guess is that such formula is a special case of exp(tD)f(x)=f(x+t) which is a semigroup generated by D and solves the first order linear wave equation .
This just got a fun tease in a 3Blue1Brown video! :)
I've remembered Peyam on all the video xD
Reminds me of making a unitary time operator by slamming the Hamiltonion into the exponential and summoning Daddy Taylor
Now I‘m curious about those half derivatives
Check out my playlist :)
Awesome!
Thanks!
Could you use this to turn any delay-differential equation into an ODE?
Exactly my thought... It becomes a nonlinear eigenproblem with an infinte number of eigenvalues.
(e^D -1)f(x) = f(x+1) - f(x)
let S = f(0) + f(1) + f(2) + ... + f(n-1)
(e^D-1) S = f(n) - f(0)
S = (1/D) ( D/(e^D-1) ) ( f(n) - f(0) )
This leads to Euler-Maclaurin summation fomula.
Just to clarify: Time 17:59 - Since u r replacing only the numerator by that second sum, the n factorial in the denominator should come out of the second summation, isn't it? (But it is still inside the first summation.) Am I correct?
Why am I seeing groups? Why are adders being turned into multipliers?
So cool Dr Peyam, l love all your derivative extravaganzas! What about antiderivative/intégral extravaganza? How would that look?
Usually if the shift is to the right, the antiderivative shifts to the left :)
The product rule proof
(e^D)(f(x) g(x)) = f(x+1) g(x+1) = (e^D)(f(x)) (e^D)(g(x)), or more compact (e^D)(f g) = (e^D)(f) (e^D)(g)
Isn't it trivial?
how about d^d operator? because i think, i ve seen something like ∫ p⁽⁾/(2π)^d d^d(p)=0
i forgot what the power of p was
Application: In quantum mechanics (QM) h/(2pi) id/dt is the Energy operator H = ih/(2pi)D (also called Schrödingers equation, there D= d/dt)). Thus the operator
exp(T 2pi i)H/h)
is shifting the quantum mechanical psi-function psi(t,x) by the time-difference T to psi(t+T,x), so it's describing the propagation of the quantum state in time (operator of time translation; quite a deep insight in QM: The energy Operator H is "ruling" the time development; that's, why you might say, "physics" is the science of energy :-). Nice, isn't it? Quantum physicist are quite aware of this. The same way you can get half groups of a shift in space using the momentum operator h/(2pi) id/dx.
In Quantum Mechanics thus the proof of Noether's theorem is trivial - if energy is conserved, tim translation is a symmetry, by which the laws of physics do not change.
That’s really cool, thanks for sharing!!
@@drpeyam Thank you for your really nice take on the exp(D)-operator. I really enjoyed that ... especially, because it is so fundamental in QM (well, in a mathematical well defined model for QM; an example, how a functional analystic viewpoint can give us a quite nice background on what's going on in nature).
For the product rule, before changing order of sumation it is just a "Cauchy product"
let's do sin derivative and cosine derivative
You can read my mind 👌 Will be posted in a week!
I guess it should follow using
cos(Df) = ( e^iDf + e^-iDf ) / 2
and using the result e^D(kf) for any constant k we are done.
Easy Mathematics,
cos(Df) is just cos(f') which entirely depends on f
I think you mean cos(D)f
= (e^iD + e^-iD)/2 * f
= e^iD * f/2 + e^-iD * f/2
= f(x+i)/2 + f(x-i)/2
sin(D)*f
= (e^iD - e^-iD)/2i * f
= f(x+i)/2i - f(x-i)/2i
= if(x-i)/2 - if(x+i)/2
@@MagicGonads Thanks for your correction.
Nice lecture thx
What I really like about this is how you can get derivation to define itself.
f'(x)=lim_(k→0) (f(x+k)-f(x))/k
e^(k d/dx) f(x) = f(x+k)
Thus: f'(x)=lim_(k→0) (e^(k d/dx) -1 ) f(x)/k
This is amazing, thank you!
@Dr Peyam, Sir, if you have any idea about the Taylor expansion of ln[cosh(aDx)f(x).f(x)]; pls help me.
it just came to me: can this be used to solve delayed ordinary differential equations like this one: f'(x) = f(x+a)?
if we write f'(x) =: Df(x) and f(x+a) = exp(aD)f(x), we get
0 = f'(x) - f(x+a) = Df(x) - exp(aD)f(x) = [D - exp(aD)]f(x).
this looks similar to the eigenvalue problem to me, so maybe there is an efficient way to find an 'eigenfunction' f that solves this?
Now the next questions Is there a logarithmic operator(LN) of exp(D) such that LN(exp(D)) = D, then this logarithmic operator maps f(x+1) -> f'(x)? That would be real crazy.
That was super interesting, thank you!
Maybe we can extend this result (being the shift operator) to the class of (piecewise) continuous functions instead of smooth functions using a limit of a Fourier expansion on compact subsets of R?
very intriguing...I'm thinking about the possibility to extend the series from -inf to +inf. Clearly, we should get other than exp derivative.
But curiosity brings me to imagine:
1) multivariable operators, so to have exponential partial derivative;
2) the possibility to compose series of fractional operators...
But i'm not a matematician, so i can't see so much ahead. Please, tell me if it is possible...many thanks!
I don’t see why it wouldn’t be possible 😊
Amazing video but wait a second...
Didn't you use Taylor expansion for the product lu?! 22:53
Wow!
this is very interesting... did you try to relate this operator with something already known?
It appears in quantum mechanics
Meu Deus, que loucura. Ótimo vídeo.
So you made that definition? That's really cool
Yeah but apparently it already exists!
It is interesting to think of the application of Maclaurin and Laurent series in D to polynomial operator identities. E.g., {P(n)a_n}n -> P(D)A(x) but when does the same hold for holomorphic or meromorphic functions of n?
I suppose the answer is most easily found by (modern) algebraic means or Lie theory.
Very cool
that video would have been super useful when i had my field theory course. oh well, 10 years too late, i guess.
Since you did Fractional Derivatives, could you cover Fractal Derivatives?
Also, regarding the video, is e^D(f) supposed to be interpreted as exp(f'(x)), or exp(d/dx)*f(x)?
Ooooh, never heard of it, but it sounds interesting! And it’s the second one
Coming back after a discussion to my abstac algebra teacher..
Wonder about ln (1-D)f(x) since you can expant it with taylor series, or it can't be done because interval of convergence?
And also, when i try to evaluated e^D^a (e^x) some properties is lost.
Buts still interesting to know
You can do 1/1-D as well, I believe, but there are definitely convergence issues
this is a very cool product rule :D
Really nice job and interesting properties.
One application for this shifting operator is in deriving Euler -Maclurain summation formula.
In this respect, could you please do the summation formula for two independent variables?
I need a simplified relation to obtain a sum for two variables.
5:50 oh no... the board!
So Excellent
does e^e^D converge to anything nice?
Did you come up with this idea all by yourself?
Yeah!
@@drpeyam You are awesome!
Some might say I’m Peyamazing 😄
Do you really really state that you invented this standard technique, which actually is the continuum limit of the exponential of a matrix? Are you really really sure Mr. Peyam?!?
Or, to state it even more properly, the exponential of a (linear) operator?!?
The exponential derivative is the pull-back transformation in differential geometry
I vaguely remember that, haha
@@drpeyam Actually, the exponential derivative is the Taylor extension in your case.
Halfway through.... i don’t understand why you haven’t mentioned that the shift of x to x +1 means that the exponential derivative is a translation operator, moving the scale backwards by1 or increasing all values of x by one..... am I “jumping the gun”? (You did mention Lié groups @ the end). I’ll have to review the dense summations done past the middle of the lecture & take notes!!
How is this (and or the half-derivative) tool used in fields other than pure mathematics? Physics seems to love positive integer derivatives for modeling Newtonian movement. Is there a model where this can be applied, or maybe for financial analysis?
This is used apparently in quantum mechanics, with the shift operator