Thread: linear differential operators

Hello, I am puzzled about this operation (1) with polynomials in a general proof of Exponential input theorem and I don't understand the the concept of a being s-fold zero.
p(D) = q(D)(D - a)^s
q(a) = nonzero
The above implies

p^(s)(a) = q(a)s!

The proof lets k be the degree of q(D) and writes it in powers of (D-a)

(1) q(D) = q(a) + c₁(D - a) + ... c_k(D - a)^k

(2) q(D) = q(a)(D−a)^s+c₁(D−a)^s+1 +...+c_k(D−a)^s+k

then substitute D with a

How did they figure (1) out?
if k = 2, I get

D² + c₁D +c₂ = A² + c₁A + c₂ + c₁D - c₁A + c₂D² - c₂2DA + c₂A²

I feel totally confused and as if I missed a lot
Could you please point out what I do not understand?
Thank you

Hey likelylad.

Can you clarify whether D is just a variable or whether it is a differential operator please?

I think it is a differential operator..

the proof is on page 5 of and is very short
http://ocw.mit.edu/courses/mathemati...18_03S10_o.pdf

Then the middle terms of (D-a)^k disappear becease Da is zero, right? Still do not see what they did there :/

If you are wondering how they got q(D) out then you should think about a Taylor expansion type result around the centre of a: In other words, they are expanding something around a instead of 0.

hey I can see it !

I am still wondering about a few things.

Can you rewrite
p^(s)(D) = q(a)s! + powers of (D-a)
D=a
p^(s)(a) = q(a)s!

into
D^sp(D) = q(a)s! + powers of (D-a)
D=a
p(a) = q(a)s!/a^s
?

It really got me confused:
Is there a proof that you can treat d/dx as a variable in some instances? It obviously works but..
Do people think about this (like in topology or some Hilbert spaces, I had a look into operator theory and it looks years
-away hard - I am noob at math) ? Or is it clear how to use it to everyone except for me?

Thank you for the enlightenment