I'm looking over the section on differential operators, in particular this document about the proof of the generalized exponential response formula, ( http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf ). I just can't understand the step which says that q(D) can be written in powers of D-a. Can anyone explain the equation

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

?

Question

I'm looking over the section on differential operators, in particular this document about the proof of the generalized exponential response formula, ( http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf ).  I just can't understand the step which says that q(D) can be written in powers of D-a.  Can anyone explain the equation

q(D) = q(a) + c1 (D-a) + ... + ck (D-a)^k

?

anonymous · Answer

When are two polynomials equal. When they have equal coefficients. q(X) = q(a) + c1(x-a) + ... + ck(x-a)^k give you k equations in k unknowns (ck)

anonymous · Answer

try it for k=1 and you'll see how obvious it is.