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Finding a complex Taylor series

  1. Apr 20, 2012 #1
    1. The problem statement, all variables and given/known data
    Not much has gotten me in this class, and I almost want to say this has to be a typo, but I want someone else to check it out first.

    Homework question is that we need to show that

    cos(cos θ)*cosh(sin θ) = Ʃ(-1)ncos(nθ)/(2n)! for n>=0

    There is a similar one involving sin and sinh, but if the problem is sorted out here on this one, I can fix it for the other one.


    2. Relevant equations

    Since cosh(x) = cos(i*x), I can rewrite my original function as

    cos(cos θ)*cos(i*sin θ).

    3. The attempt at a solution

    Using a product-to-sum trig identity, this eventually gets me to

    1/2*[cos(exp(-iθ))+cos(exp(iθ))].

    But because of the Taylor series for cosine, I end up with an exp(-2inθ) and exp(2inθ).

    So I end up getting the series I expect, except I have a cos(2nθ) which does not match. Is there a division by 2 I'm missing somewhere or some quirky identity that makes it true that cos(2nθ)=cos(nθ)?
     
  2. jcsd
  3. Apr 20, 2012 #2

    Dick

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    Homework Helper

    No quirky identity can make cos(2nθ)=cos(nθ), because it's not true. I think you are right and there is a typo in the solution.
     
  4. Apr 21, 2012 #3
    That's what I thought. Thanks for clarifying. I'll point it out on Monday in class and hopefully I'm not the only one that noticed that.
     
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