Consequences of Cauchy's Formula (differential formula)


by StumpedPupil
Tags: application, cauchy's formula, complex, differentiation, power series
StumpedPupil
StumpedPupil is offline
#1
Oct24-08, 01:33 PM
P: 11
1. The problem statement, all variables and given/known data

How does Cauchy's Formula help find the power series of the complex function f(z) = e^z.

2. Relevant equations
e^z = ∑z^k/k! (sum from k = 0 to infinity)

Cauchy's Formula

Consequence of Cauchy's Formula: F(z) is analytic in a domain D and the point z1 is in D. If the disc |z - z1| < R is in D, then a power series expanded about the point z1 is valid in the disc. Furthermore, the coefficients of this power series is given by an application of Cauchy's Formula over a circle < R which is positively oriented.


3. The attempt at a solution
My book skips steps, but shows a solution. All it says is that e^z is entire, so it is analytic on the whole complex plane. e^z is its own derivative so if we take the derivative of the power series expanded about the center z1 = 0 we get ∑a*kz^(k-1), where a* are the coefficients of the series. This is again differentiable (infinitely differentiable in the disc < R). Some how, all this leads to the conclusion of what a* is by using Cauchy's Formula, but I don't understand why.
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Dick
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#2
Oct24-08, 02:23 PM
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If the original power series is sum a_k*x^k, then the coefficient of x^k in the derivative of the power series is (k+1)*a_k+1, If the derivative is equal to the original power series then the coefficients must be equal. So a_k=(k+1)*a_k+1, or a_k+1=a_k/(k+1). You know a_0=1, so a_1=a_0/(0+1)=1. a_2=a_1/(1+1)=1/2, a_3=a_2/(2+1)=1/(2*3) ... Can you see where this is going?


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