# Thread: linear differential operators

1. ## Exponential input theorem

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) + ... ck(D - a)k

(2) q(D) = q(a)(D−a)s+c1(D−a)s+1 +...+ck(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

2. ## Re: linear differential operators

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

3. ## Re: linear differential operators

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 :/

4. ## Re: linear differential operators

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.

5. ## Re: linear differential operators

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
Dsp(D) = q(a)s! + powers of (D-a)
D=a
p(a) = q(a)s!/as
?

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