Taylor Series for Complex Variables

[eaglesmath15]

Homework Statement

Obtain the Taylor series e^z=e Ʃ(z-1)ⁿ/n! for 0[itex]\leq(n)[/itex]<[itex]\infty[/itex], (|z-1|<[itex]\infty[/itex]) for the function f(z)=e^z by (ii) writing e^z=e^z-1e.

Homework Equations

Taylor series:
f(z) = Ʃ(1/2\pi/i ∫(f(z)/(z-z₀)ⁿ⁺¹dz)(z-z₀)ⁿ

The Attempt at a Solution

The first part of this question called for the Taylor series to be found using the Cauchy Integral shortcut at fⁿ(1), so I assume that this is meant to be solved using the formula for Taylor series as is, but that's where I get stuck. I fill in the Taylor series, and reduce to:
f(z)=e Ʃ(z-1)ⁿ∫e^z-1/(z-1)ⁿ⁺¹}dz
and I have no idea how to get the integral to equal [itex]\stackrel{1}{n!}[/itex], which is what it must in order to equal the original condition.

 

[Dick]

Homework Statement

Obtain the Taylor series e^z=e Ʃ(z-1)ⁿ/n! for 0[itex]\leq(n)[/itex]<[itex]\infty[/itex], (|z-1|<[itex]\infty[/itex]) for the function f(z)=e^z by (ii) writing e^z=e^z-1e.

Homework Equations

Taylor series:
f(z) = Ʃ(1/2\pi/i ∫(f(z)/(z-z₀)ⁿ⁺¹dz)(z-z₀)ⁿ

The Attempt at a Solution

The first part of this question called for the Taylor series to be found using the Cauchy Integral shortcut at fⁿ(1), so I assume that this is meant to be solved using the formula for Taylor series as is, but that's where I get stuck. I fill in the Taylor series, and reduce to:
f(z)=e Ʃ(z-1)ⁿ∫e^z-1/(z-1)ⁿ⁺¹}dz
and I have no idea how to get the integral to equal [itex]\stackrel{1}{n!}[/itex], which is what it must in order to equal the original condition.

Well, what kind of expression does the Cauchy Integral formula give you for ∫e^z-1/(z-1)ⁿ⁺¹dz?

 

[brmath]

If you must use the Cauchy integral formula to keep your prof happy, you must, although calling it a shortcut strikes me as rather inaccurate. Maybe calling it the long way around?

Whenever I have to expand ##e^{f(z)}## for f(z) holomorphic, I use the expansion in z at 0 and shove f(z) in where previously I had a z. For example

##e^z = \sum z^n/n!## where the coefficients (all 1) can be computed with the elementary calculus derivative approach if you don't know them. You won't get anything different using Cauchy, so why go to all that trouble?

So for example ##e^{2z} = \sum (2z)^n/n!##. This works fine.

You can at least use this simpler approach to check your work.