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The sum of series involving cosine

  1. Feb 1, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the sum of the series s(x) = 1 +cos(x)+ (cos2x)/2!+(cos3x)/3!...

    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 1, 2009 #2

    gabbagabbahey

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    Hi thanksie, what have you tried? What methods have you learned? Are you allowed to compare it to other known series?
     
  4. Feb 1, 2009 #3
    Hi, Yes we can compare it to known series. I've tried a comparison with the harmonic series and the expansion for e^x looks promising. I'm just not sure how to get started.
     
  5. Feb 1, 2009 #4

    gabbagabbahey

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    Comparing it to the series for e^x sounds good to me.....try writing [itex]\cos(nx)[/itex] in terms of complex exponentials....what does that give you?
     
  6. Feb 1, 2009 #5
    would x= cosnx? is there a property for cosine where (cos x)^n=cos nx ?
     
  7. Feb 1, 2009 #6

    gabbagabbahey

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    Not quite...try writing [itex]\cos(nx)[/itex] in terms of complex exponentials....what do you get?
     
  8. Feb 1, 2009 #7
    Do you mean:

    x^ni=cos nx + sin nx?
     
  9. Feb 1, 2009 #8

    gabbagabbahey

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    Huh?!:confused: Why on Earth would you think that were true?

    Have you not seen the formulas [tex]e^{i\theta}=\cos\theta+i\sin\theta[/tex] and [tex]\cos\theta=\frac{e^{i\theta}+e^{-i\theta}}{2}[/tex] before?
     
  10. Feb 1, 2009 #9
    sorry i was using that first one. i was confused as to what you meant with cos nx.

    using the second one makes a lot more sense. are you saying i should:

    cos n(theta) = (e^ni(theta) +e^-ni(theta))/2
     
  11. Feb 1, 2009 #10
    and how do i even find the sum of this series. i only know how to find a sum of a geometric series...i dont know how to deal with the factorial
     
  12. Feb 1, 2009 #11

    gabbagabbahey

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    Yes, so now you have

    [tex]1+\cos(x)+\frac{\cos(2x)}{2!}+\ldots=\sum_{n=0}^{\infty} \frac{\cos(nx)}{n!}=\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{i n x}+e^{-i n x}}{n!}[/tex]

    correct?

    Try breaking it into two separate sums:

    [tex]\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{i n x}+e^{-i n x}}{n!}=\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{i n x}}{n!}+\frac{1}{2}\sum_{n=0}^{\infty} \frac{e^{-i n x}}{n!}[/tex]


    Now use the fact that [tex]e^{n\theta}=\left(e^{\theta}\right)^n[/tex] and compare each sum to the series for e^u.
     
  13. Feb 1, 2009 #12
    That makes sense. Thanks for all your help.

    One more unrelated quick question:
    do the series n/(n+1) and n^2/(n^2+1)converge or diverge?
     
  14. Feb 1, 2009 #13

    gabbagabbahey

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    Well, try applying some of the convergence tests that you've learned....
     
  15. Feb 1, 2009 #14
    I looked at this and became interested but got stuck.

    [itex] e^{ix} = \cos x + i \sin x [/itex]

    [itex] e^{-ix} = \cos x - i \sin x [/itex]

    [itex] \cos x = \frac{e^{ix} + e^{-ix}}{2} [/itex]

    [itex] \displaystyle\sum_{n=0}^\infty \frac{\cos (nx)}{n!} = \displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} [/itex]

    [itex]\displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} = \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{2n!} + \displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{2n!}[/itex]

    [itex]\displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} = \frac{1}{2} \left[ \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{n!} + \displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{n!} \right][/itex]

    [itex] e^x = \displaystyle\sum_{n=0}^\infty \frac{x^n}{n!}[/itex]

    [itex] \displaystyle\sum_{n=0}^\infty \frac{e^{inx}}{n!} = e^{e^{ix}} [/itex]

    [itex] \displaystyle\sum_{n=0}^\infty \frac{e^{-inx}}{n!} = e^{e^{-ix}} [/itex]

    [itex]\displaystyle\sum_{n=0}^\infty \frac{e^{inx} + e^{-inx}}{2n!} = \frac{1}{2} \left[e^{e^{-ix}} + e^{e^{ix}} \right] [/itex]

    Edit: I think this is OK actually.
     
    Last edited: Feb 1, 2009
  16. Feb 1, 2009 #15
    yeah I was going to say that's what I got... lots of thanks to gab, I would have put my work in here but I don't know how to use this forum very well.
     
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