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Power series - finding the sum

  1. Jan 17, 2010 #1
  2. jcsd
  3. Jan 17, 2010 #2


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    Notice that (n+2)(n+1)xn is the second derivative of xn+2

    [tex]\sum_{n=0}^\infty x^{n+2}[/tex]

    is a geometric series of the form [itex]\sum ar^n[/itex] with a= x2 and r= xn. Use the usual formula for sum of a geometric series to write that in "closed form" and differentiate twice.
  4. Jan 17, 2010 #3
    Thank you very much (sun)
  5. Jan 17, 2010 #4
    Would you tell me if my solution to the problem is correct?
    Again, we are searching for the sum.




    [tex]1+C = 0 \rightarrow C=-1[/tex]
  6. Jan 17, 2010 #5
    Almost. Why do you introduce the constant C? You have shown:
    [tex]\int f(x) \text{ d}x = xe^{x^3}[/tex]
    so you get,
    [tex]f(x) = \left(xe^{x^3}\right)'[/tex]
    It's only when integrating you add the constant. Also you derive the wrong value for C because in the sum:
    [tex]\sum_{n=0}^\infty \frac{{x}^{3n}}{n!}[/tex]
    the first term actually doesn't make any sense for x=0 since it has 0^0, but we always interpret the first term to be 1 (this is just shorthand when dealing with Taylor series) so:
    [tex]\sum_{n=0}^\infty \frac{{0}^{3n}}{n!} = 1 + 0^3/1! + 0^6/2! + \cdots = 1[/tex]
    Thus C=0 and we get rid of the constant which shouldn't have been there in the first place.
  7. Jan 17, 2010 #6
    Thank you :)
  8. Jan 17, 2010 #7
    Ok, this is the last one, I promise :blushing:


    It's a little confusing for me because there is no x. I tried with [tex]1^n = x^n[/tex] and finding the sum
    but no result...
  9. Jan 17, 2010 #8
    Try working with (we use 2n+1 for exponent to cancel the denominator and get a geometric series when differentiating):
    [tex]f(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{2n+1}[/tex]
    Differentiate it, use the identities for geometric series and integrate again. It turns out f is actually a function you probably know very well.
  10. Jan 18, 2010 #9


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    And, of course, after you have found it as a function of x, set x= 1 to get your numerical sum.
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