1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook