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Taylor Series for Complex Variables

  1. Nov 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Obtain the Taylor series ez=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)=ez by (ii) writing ez=ez-1e.


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


    3. 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∫ez-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.
     
  2. jcsd
  3. Nov 13, 2013 #2

    Dick

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    Homework Helper

    Well, what kind of expression does the Cauchy Integral formula give you for ∫ez-1/(z-1)n+1dz?
     
  4. Nov 14, 2013 #3
    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.
     
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