The exponential of the derivative?

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[physicsatroeper]

This kind of thing is also very important in Quantum Mechanics, as this exponential operator is playing the role of the "Translation" operator acting on a wavefunction, for example. This is manifesting the fact that the operator "e^{ipx}" where "p" is the "momentum operator" is doing a translation, which in turn gives you things like "Momentum-conservation <=> translation-invariance" via Noether's Theorem. Similarly gives you a tool to prove Heisenberg's Uncertainty Principle in wave mechanics. Nice video!

[cd-zwtt]

michael again with the crazy math objects! wild stuff

[jeffbarrett]

I noticed this neat connection the other day. A linear time-invariant system can be written:

dx/dt = A x

Where x : R -> R^n, and A is an nxn matrix. The solution to this system is well known in terms of the matrix exponential: x(t) = exp(tA)x0, where x0 is the initial condition.

Now consider the PDE:
∂u/∂t = v ∂u/∂x

Its solution is u(x, t) = u0(x+vt), but from the result in this video, that is exactly u(x, t) = exp(vt ∂/∂x) u0(x). It's very interesting to see the similarities between the matrix A for the system of equations and the operator v∂/∂x for the partial differential equation. It's sort of like a continuous generalization.

[aproductions]

Oh my god, this is the shift operator from QM! I never understood its definition until now -- it was kind of skipped over in the lecture. This makes so much sense, thank you!

Of course, I still don't understand the subsequent result that the momentum operator must be the derivative term in the exponential (the generator of translation or whatever), but that's very much a physics result and outside the scope of this channel.

[manucitomx]

I loved this. Clear, straightforward, well explained.
Thank you, professor.

[kkanden]

i love how clear your explanation of each step is! what's important tho is that it's not overexplained either. perfect, just perfect!!

[christophdietrich]

This is an example of a strongly continuous one parameter operator semi group. What we did by hand is that the derivative operator generates the shift (semi-)group. There is a whole field of mathematics to explore if you are interested.

[zachbills]

This gives a very slick derivation of of the Euler-Maclurin summation formula. We get sum f(x+n)=sum If you then Taylor expand the final parentheses and interpret D^-1 as an integral you get your formula. The parentheses expression is the exponential generating function for the Bernoulli numbers, explaining their presence in the formula.

[danielmilyutin]

exp(t d/dx) is exactly shift operator
f(x) |-> f(x+t)

upd: but it works only on analytic functions

[elzilcho]

What's neat with that convergence requirement of x>1 is that, with a little fiddling, it is always satisfied. Because when x=1, the original question just becomes the trivial e^(d/dx)0=0, and when x<1 you can just pull the negative exponent out of the log function and then you're just doing the same solution but with a negative out the front.

[chester]

reminds me of the operator algebra from quantum mechanics

[maximilianarold]

17:30 I never felt so unvalued in my entire life

[squeezy]

The exercise suggested is excellent - using the cyclic nature of the derivative of sin(x) you can do the following:
- Write out the expansion, in which you will get something like: exp(a d/dx) sin(x) = sin(x) + acos(x) - a²/2! * sin(x) - a³/3! * cos(x) + ...
- Factor by grouping to get sin(x)[1 - a²/2! + a⁴/4! - ...] + cos(x)[a - a³/3! + a⁵/5! - ...]
- Recognise the series in brackets to be the Maclaurin/Taylor Series expansions of cos(x) and sin(x) respectively evaluated at a:
- Thus exp(a d/dx) sin(x) = sin(x)cos(a) + cos(x)sin(a)
- Using the addition formula, exp(a d/dx) sin(x) = sin(x+a) as required. [I don't believe there are any convergence issues here as this example only used the series expansions of cos(x) and sin(x) which are valid for all x, but do correct me if I'm wrong.]

Very interesting video, I'd love to see more operational calculus on this channel :)

[Leonardo-G]

Maybe e^(a d/dx) e^x would be interesting?

All derivatives of e^x are itself so we can factor it out of the summation:

e^(a d/dx) e^x = e^x (1 + a + a^2/2! + a^3/3! + a^4/4! ...)

The series on the right looks familiar. In fact, it's the maclauren series for e^a! So we replace it like so:

e^x (e^a)
e^(a d/dx) e^x = e^(x + a)

And we arrive at the same result. Amazing!

[iabervon]

Applying it to e^x is really easy, since the derivatives don't change anything, giving you the Taylor series for e^x evaluated at 1, with an e^x in each term. Then factor the e^x out, and you get e^xe^1, which is e^(x+1).

[Balequalm]

Isn't convergence used a step before what Michael claims? I'm pretty sure you need uniform convergence to interchange the integral and summation to begin with, which gives the same condition in the end.

[howardthompson]

The x^{n-m} should be x^{m-n} instead.

[euanthomas]

Control engineers use this a lot. In Laplace transforms the s in e^-st plays the role of d/dt and the e^-st is physically a phase shift ( s= jw), as you anticipated in your exercise suggestion for operating on the sine function. As other commentors below state, this is also a translation in quantum mechanics, or a phase shift for the momentum operator. Amazing how all this stuff from completely different fields fits together.

[eliosedrata]

I'll show that exp(aD(sin(x)))=sin(x+a) :
First you can show easily by induction that the nth derivative of sin is sin(x + n*pi/2).
Then exp(aD(sinx))=sum from 0 to infinity of (a^n)/n! *sin(x+npi/2)
At that point you split the sum between even and odd terms, so that cos(a) and sin(a) appear.
You get :
Sum of a^2n/(2n)! * sin(x+npi)
+ sum of
And sin(x+npi)=(-1)^n sin(x) and
sin(x+pi/2)=cos(x)
Finally you recognize the power series of cos(a) and sin(a), so you get
exp(aD(sinx))
=cos(a)sin(x)+sin(a)cos(x)
=sin(x+a)

And of course it works for every real x and a since the radius of convergence of sin and cos is infinity.

[samosamo]

Very nice explanation. We could start the course of one parameter semigroup by this! In fact the derivation operator is the infinitisimal generator of the shift semigroup. One should precise the domain.