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Limits: rate of convergence of Euler

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that the rate of convergence for Euler's method (in solving the problem y'=y) is comparable to 1/n by showing that [tex]lim_{n\rightarrow infinity} \frac{e-(1+\frac{1}{n})^n}{1/n}[/tex].


    2. Relevant equations
    Hint: Use L'Hopital's Rule
    If lim x-> infinity f(x)/g(x) is in indeterminate form, then it can be evaluated as lim x-> infinity f'(x)/g'(x).
    Hint: Use Maclaurin expansion for ln(1+t).
    ln(1+t) = t - t^2/2 + t^3/3 - t^4/4 + t^5/5 .......


    3. The attempt at a solution

    L'Hopital's Rule:
    Differentiating gives
    [tex]lim_{n\rightarrow infinity} \frac{\frac{1}{n+1} - ln(1 + \frac{1}{n})(\frac{n+1}{n})^n}{-1/n^2}[/tex]

    [tex] ln (1+1/n) = 1/n - 1/(2n^2) + 1/(3n^3) - 1/(4n^4) + 1/(5n^5) ... [/tex]

    Not sure where to go from here. When I plug in the series, what can I do? I tried multiplying the fraction by [tex]\frac{-n^2}{-n^2}[/tex] but I don't think that gets me anywhere. I could apply L'Hopital again, but that seems like a monstrous function to differentiate. Any ideas?
     
  2. jcsd
  3. Feb 12, 2009 #2

    Dick

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    I would say don't use l'Hopital if you are going to use series expansions anyway. (1+1/n)^n=exp(n*ln(1+1/n)). Use the series expansion on n*ln(1+1/n). Only keep 2 terms. Factor out the e. Now series expand exp. Only keep the terms that are going lead to terms with a power of n less than two.
     
  4. Feb 13, 2009 #3
    Great! that works out to e/2. Thanks for the help! I wonder why they hinted at L'hopital though.
     
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