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Homework Help: Series estimation/error

  1. Nov 27, 2012 #1
    The question is: estimate the sum, from 2 to infinity, of 1 / (n^2 + 4) to within 0.1 of exact value.

    I have the following: the integral from n to inf. of 1 / (x^2 + 4) is (pi/4 - 1/2 arctan(n/2)).

    Next, in order to find the number of partial sums to use, set Sn + the integral from n+1 to infinity < S < Sn + the integral from n to infinity.

    This gives (pi/4 - 1/2 arctan[(n+1)/2]) < (remainder) < (pi/4 - 1/2 arctan(n/2)).

    I can only assume from here that you are meant to subtract the lesser from the greater and set (1/2 arctan[(n+1)/2]) - 1/2 arctan(n/2) < 0.2. So arctan((n+1)/2) - arctan(n/2) < 0.4. What?
    Last edited: Nov 27, 2012
  2. jcsd
  3. Nov 27, 2012 #2
    The error of the integral approximation will be between the two integrals you have, so you want
    [tex]\int_{n+1}^\infty \frac{dx}{x^2 + 4} \leq 0.1 \leq \int_n^\infty \frac{dx}{x^2 + 4}[/tex]
    These two separate inequalities will tell you what range of values of n you should choose to approximate the series within an error bound of 0.1. For this problem, it turns out there is only one integer between these two integrals.
    [tex]\frac{\pi}{4} - \frac{1}{2}\arctan\left(\frac{n + 1}{2}\right) \leq 0.1[/tex]
    for example gives us the lower bound when solved of
    [tex]n \geq 2\tan\left(\frac{\pi}{2}-0.2\right)-1 \approx 8.86631[/tex]
    Once you have the value of n, you can use it in the inequality
    [tex]s_n + \int_{n+1}^\infty \frac{dx}{x^2 + 4} \leq s \leq s_n + \int_n^\infty \frac{dx}{x^2 + 4}[/tex]
    Adding the two values and dividing by 2 does not change the error of the value, so you can approximate the value s by taking the average of the two values outside.
    Last edited: Nov 27, 2012
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