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Taylor Series Help

  1. Apr 4, 2007 #1
    Let ƒ be the function given by f (x) = e ^ (x / 2)

    (a) Write the first four nonzero terms and the general term for the Taylor series expansion of ƒ(x) about x = 0.
    (b) Use the result from part (a) to write the first three nonzero terms and the general term of the series expansion about x = 0 for g (x) = ((e^(x / 2)) – 1)/x

    For part a I got
    P4 = 1 + x/2 + ((x/2)^2)/2! + ((x/2)^3)/3!

    E ((x/2)^n)/n!
    E = summation sign n=0 to infinity

    For part b I got, which was marked wrong,
    x-1 + ((x/2)2 – 1)/(2!x) + ((x/2)3 – 1)/(3!x)

    E ((x/2)n - 1)/(n!x)

    On my paper my teacher marked only –1 once and I am not for sure what that means. Any help on part B would be greatly appreciated.
  2. jcsd
  3. Apr 4, 2007 #2


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    You were, of course, correct that the Taylor's series for e^(x2) is 1+ x/2+ ((x/2)^2)/2!+ ((x/2)^3)/3!+ ...
    But for (b) you seem to have just stuck a "-1" into the numerator, and divided some of the denominators by x. Do it step by step:
    e^(x/2)= 1+ (x/2)+ (x/2)^2/2!+ (x/2)^3/3!+ ... (which I would prefer to write as 1+ x/2+ x^2/((4)(2!))+ x^3/((8)(3!))+ ...) so
    e^(x/2)-1= x/2+ x^2/((4)(2!))+ x^3/((8)(3!))+ ... and finally
    (e^(x/2)-1)/x=(1/2)+ x/((4)(2!)+ x^2/(8)(3!)+...
    I think that is
    [tex]\sum_{n= 0}^\infty \frac{x^n}{2^{n+1}(n+1)!}[/tex]
  4. Apr 4, 2007 #3
    Thank you, it actually makes sense now.
  5. Feb 18, 2010 #4

    there was a part c for this too. it said:
    c) for the function g in part b, find g'(2) and use it to show that [tex]\sum_{n= 1}^\infty \frac{n}{4(n+1)!}[/tex] = 1/4
    i don't know how to start it, or what to do. help please?
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