Hey likelylad.
Can you clarify whether D is just a variable or whether it is a differential operator please?
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) + c_{1}(D - a) + ... c_{k}(D - a)^{k }
(2) q(D) = q(a)(D−a)^{s}+c_{1}(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
D^{2} + c_{1}D +c_{2} = A^{2} + c_{1}A + c_{2} + c_{1}D - c_{1}A + c_{2}D^{2 }- c_{2}2DA + c_{2}A^{2 }
I feel totally confused and as if I missed a lot
Could you please point out what I do not understand?
Thank you
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 :/
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^{s}p(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