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Taylor Series coefficient of x^n

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Find the Taylor series for 0.5x^2[e^x-e^(-x)] around x=0. What is the coefficient of x^n?


    2. Relevant equations
    e^x=∑x^n/n!


    3. The attempt at a solution
    I understand how to find the Taylor series for this equation (it being ∑[x^(2n+3)/n!]; x^3+x^5+x^7/2!+...) through manipulation of the Taylor series for e^x; however I'm not quite sure what it means by the "coefficient of x^n?" The answer is apparently 0 when n is 1 or even, and otherwise it is given by 1/[(n-3)/2]!, but how is this found? Thank you so much for any help!
     
  2. jcsd
  3. May 7, 2014 #2

    jbunniii

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    The coefficient of ##x^n## is simply the value by which ##x^n## is multiplied when you write out the series. So in your case, assuming your series is correct (which I haven't checked),
    $$x^3 + x^5 + \frac{x^7}{2!} + \cdots$$
    There is no constant term (##x^0##), so the coefficient of ##x^0## is zero. Similarly, the coefficient for ##x^1## is zero and likewise for all ##x^n## where ##n## is even, since these terms do not appear in the series.

    So look at what remains: the coefficient for ##x^3## is ##1##, the coefficient for ##x^5## is also ##1##, and the coefficient for ##x^7## is ##1/(2!) = 1/2##. If you plug ##n=3,5,7## into the formula given in the answer, you can see that it matches these values.

    To find the general formula, the question comes down to: what is the coefficient of ##x^n## in your series ##\sum x^{2n+3}/(n!)##?

    Hint: try using a different letter instead of ##n## to avoid confusion: ##\sum x^{2m+3}/(m!)##. What is the ##x^n## term of this series?
     
  4. May 7, 2014 #3
    The coefficient of ##x^n## is precisely what it means. If your given a value ##n## what is the coefficient for that specific term ##x^n##.
     
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