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Maclaurin Series for Natural Log Function

  1. Aug 4, 2013 #1
    1. The problem statement, all variables and given/known data
    Use x=-1/2 in the MacLaurin series for e^x to approximate 1/sqrt(e) to four decimal places.


    2. Relevant equations



    3. The attempt at a solution
    [itex] \sum_{n=0}^\infty \frac{x^n}{n!} = 1 + x + x^2/2 + x^3/6 + ... [/itex]

    For this particular power series, I have:

    [itex] \sum_{n=0}^\infty \frac{(x+0.5)^n}{sqrt(e)(n!)} = 1/sqrt(e) + (x+0.5)/(sqrt(e)) + (x+0.5)^2/sqrt(e) + (x+0.5)^3/6*sqrt(e)... [/itex]

    (-1/2)^(n+1)/(n+1)! <= 0.00004
    n= 2

    (1/sqrt(e) + (x+0.5)/sqrt(e) + (x+0.5)^(2)/2*sqrt(e)) = (1/sqrt(e) + 0 + 0)= 1/sqrt(e)= 0.6065

    But this doesn't seem right because I should end up with the actual number instead of the 1/sqrt(e) itself. I thought I was doing the method correctly, but I guess not. I would appreciate any insight on this at all, as I don't know of any other method to apply here. Thanks!
     
  2. jcsd
  3. Aug 4, 2013 #2

    lurflurf

    User Avatar
    Homework Helper

    I don't know why you did that. From
    $$e^x \sim \sum_{k=0}^\infty \frac{x^k}{k!}$$
    you should simply have
    $$e^{-1/2} \sim \sum_{k=0}^\infty \frac{(-1/2)^k}{k!}$$.
    A partial sum will be a good estimate if enough terms are used. Do you know how to determine how many terms are needed?
     
  4. Aug 4, 2013 #3
    I think that I have to set it to be less than or equal to 0.00004 as I did above? So n=2?
     
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