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Sum of a series

  1. Apr 14, 2010 #1
    1. The problem statement, all variables and given/known data
    [PLAIN]http://img263.imageshack.us/img263/9336/seriesgay.jpg [Broken]

    In the previous part of the question we had to show where the taylor expansion comes from, and calculated the maclaurin series for e^x, sin x and cos x. From that we had to prove De Moivre's theorem and so I would imagine that these things help in the last part of this question. I can see it looks like a Maclaurin series, just not sure where to start really.

    Any help would be appreciated, thanks.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Apr 14, 2010 #2
    What have you tried? Does that infinite series sort of look like any other infinite series you know?
     
  4. Apr 14, 2010 #3
    well it looks like a maclaurin series, but I don't really know how to work out what it's a Maclaurin series of.
     
  5. Apr 14, 2010 #4

    Mark44

    Staff: Mentor

    Not really. A Maclaurin series has powers of x (or whatever the variable happens to be).

    IOW, a Maclaurin series looks like this:
    [tex]\sum_{n = 0}^{\infty} a_n x^n[/tex]

    Your series is
    [tex]\sum_{n = 0}^{\infty} \frac{2^n~cos(n\theta)}{n!}[/tex]
     
  6. Apr 14, 2010 #5
    Yeah but I was thinking that it looked like something to do with e^(i[tex]\theta[/tex]) to the power of n which would give terms of cos(n[tex]\theta[/tex]). I'm not sure, if not how do I go about tackling the problem?
     
  7. Apr 14, 2010 #6
    Hint: The real part of exp(i x) = ???
     
  8. Apr 14, 2010 #7
    cos(x)... sorry it's been a long day! Still not sure how to use that to get any further.
     
  9. Apr 14, 2010 #8
    Replace all the cos(nx) by exp(i n x) in the summation and then take the real part of the summation.
     
  10. Apr 14, 2010 #9
    ok, I think I have it.

    Is it the Maclaurin series for

    e^(2*Re[e^(i[tex]\theta[/tex])])

    that seems to work I think :s, meaning that the sum is just what's written above right?
     
  11. Apr 14, 2010 #10
    or rather:

    Re[e^(2*e^(i0))]
     
  12. Apr 14, 2010 #11
    Yes, and now you can simply this using Euler's formula

    exp(ix) = cos(x) + i sin(x)
     
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