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Taylor series for a complex function

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the 5 jet of the following function at z=0:

    [itex]f(z) = \frac{sinhz}{1+exp(z^3)}[/itex]

    2. Relevant equations
    [itex]\frac{1}{1-z}=\sum_{n=0}^\infty z^n[/itex] where [itex]z=-exp(z^3)[/itex]

    3. The attempt at a solution

    I have tried to multiply the series for sinhz by the series for [itex]\frac{1}{1-(-exp(z^3))}[/itex] but to no avail since I end up with an infinite string of z s to the power of 3. I've also tried to simply substitute the series for sinhz and exp(z^3) directly into the fraction but cannot separate the fraction to give separate terms. Any suggestions for another method?
     
    Last edited: Nov 16, 2013
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  3. Nov 16, 2013 #2

    LCKurtz

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    What is a 5 jet?
     
  4. Nov 16, 2013 #3
    Sorry I should have said the 5 jet is the expansion up to and including powers of z to the fifth power and no further.
     
  5. Nov 16, 2013 #4

    Dick

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    You can't apply the series to 1/1+exp(z^3). exp(z^3) isn't small near z=0. What does the taylor expansion of exp(z^3) look like?
     
  6. Nov 16, 2013 #5
    [itex]f(z)=exp(z^3)[/itex] = [itex]\sum_{n=0}^\infty\frac{z^{3n}}{n!}[/itex]
     
  7. Nov 16, 2013 #6
    [itex]\sum{0}^infty\z^n[/itex]
     
    Last edited: Nov 16, 2013
  8. Nov 16, 2013 #7
    edit: only just learning to use latex forget that last post
     
  9. Nov 16, 2013 #8

    Dick

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    Good. Now just write the terms of order <= 5 and substitute that into the denominator.
     
  10. Nov 16, 2013 #9
    [itex]\sum_{n=0}^\infty\frac{z^{3n}}{n!}[/itex] = [itex]1+z^3+\frac{z^6}{2}+\frac{z^9}{6}+\frac{z^{12}}{24}[/itex]
     
  11. Nov 16, 2013 #10

    Dick

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    Too many terms. z^6 has power greater than 5 already.
     
  12. Nov 16, 2013 #11
    [itex]1+z^3[/itex]
     
  13. Nov 16, 2013 #12
    Do I now expand [itex]*\sum(-1)^n(1+z^3)^n[/itex]
     
  14. Nov 16, 2013 #13

    Dick

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    You can't again. (1+z^3) isn't a small number either. You need to substitute (1+z^3) for e^(3z) into 1+e^(3z) and THEN think about expanding.
     
  15. Nov 16, 2013 #14
    I end up with [itex]\frac{1}{2}*\frac{1}{1-(-\frac{z^3}{2})}[/itex] . Now I presume I expand [itex]\frac{1}{2}\sum (\frac{-z^3}{2})^n[/itex] to get [itex]\frac{1}{2}\sum(-1)^n\frac{z^{3^n}}{2^n}[/itex] = [itex]\frac{1}{2}(1-\frac{z^3}{2})[/itex]. Now multiplying by the series for sinhz we get [itex](z+\frac{z^3}{6}+\frac{z^5}{120})(\frac{1}{2}-\frac{z^3}{4})[/itex] = [itex]\frac{z}{2}+\frac{z^3}{12}-\frac{z^4}{4}+\frac{z^5}{240}[/itex]
     
  16. Nov 16, 2013 #15

    Dick

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    And that's right. Good job! Your series formula for 1/(1-z) only converges if |z|<1. You can't apply it to things that aren't small.
     
  17. Nov 16, 2013 #16
    Thank you.
     
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