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Calculus 2 - Infinite Series

  1. Nov 6, 2011 #1
    Determine how many terms of the convergent series must be summed to be sure that the remainder is less than 10^-4.

    Ʃ[n=1,∞] cos(k)/k^(3/2)

    I'm not sure how to solve this problem. I'm only aware of remainder formulas for the integral test and for alternating series. I'm not sure that this particular problem pertains to the remainder from the integral test sense we are not given the number of terms and it's what we are solving for and not exactly sure what I'm suppose to do.
     
  2. jcsd
  3. Nov 6, 2011 #2

    Mark44

    Staff: Mentor

    The summation index should match the one used in the function being summed.
    Your book should have a theorem about the remainder of a Taylor series when you approximate a Taylor series by its first n terms. That's what you need to use.
     
  4. Nov 6, 2011 #3
    Convergence of Taylor Series

    Let f have derivatives of all orders on an open interval I containing a. The Taylor series for f centered at a converges to f for all x in I if and only if lim n->inf R_n(x)=0 for all x in I, where

    R_n(x) = ( f(c)^(n+1) (x-a)^(n+1) )/(n+1)!

    is the remainder at x (with c between x and a)

    I'm not sure how to solve this problem because I don't know what function the series represents. My book has examples in which it finds the remainder term of the summation from k=0 to infinity of x^k/k! and such were the the function is actually known and proves that the interval of convergence is -infinity to positive infinity and such

    but none of this stuff were you have to find how many terms are needed to get to a certain range and I'm not sure how to do this type of problem with this type of series.
     
  5. Nov 6, 2011 #4

    Mark44

    Staff: Mentor

    Scratch that - I was thinking we were dealing with a Taylor series, but obviously we're not. Let me think about it a bit...
     
    Last edited: Nov 6, 2011
  6. Nov 6, 2011 #5

    Mark44

    Staff: Mentor

    I'm stumped. According to Wolframalpha, the series converges to about .210049. I calculated the partial sums up to 300 terms in Excel, and the partial sums still don't seem to be settling down very quickly.

    Checking in several of my calculus books, I don't see any error estimate theorems other than the ones you already cited.
     
  7. Nov 6, 2011 #6
    Ya I'm stumped to.
     
  8. Nov 6, 2011 #7
    Have you learned about the Lagrange error bound?
     
  9. Nov 6, 2011 #8
    Nope I have not.
     
  10. Nov 6, 2011 #9
  11. Nov 6, 2011 #10

    Mark44

    Staff: Mentor

    As far as I can tell, that won't be helpful here - the series isn't a Taylor series, which is what the thread in the link is working with.
     
  12. Nov 6, 2011 #11
    I thought you could use the error bound formula for power series, but I maybe wrong.

    EDIT:

    I'm wrong, it's for taylor only. My friend is about to post a technique he knows of.
     
    Last edited: Nov 6, 2011
  13. Nov 6, 2011 #12
    You could do this just by plugging in numbers like sooo, although this isn't very elegant:
    cos(1)/(3/2) = 0.3602
    cos(2)/(9/4) = -0.18495
    cos(5)/(3/2)^5 = 0.03735
    cos(10)/(3/2)^10 = -0.0145
    cos(20)/(3/2)^20 = 0.000122
    Were getting close..
    cos(21)/(3/2)^21 = -0.0001098
    cos(22)/(3/2)^22 = 0.000133
    cos(23)/(3/2)^23 = -0.0000475
    cos(24)/(3/2)^24 = 0.000025

    So you need 23 terms before you can be sure.
     
  14. Nov 7, 2011 #13

    Mark44

    Staff: Mentor

    This is incorrect - the general term in the series is cos(k)/k^(3/2), not cos(k)/(3/2)^(k).
     
  15. Nov 7, 2011 #14

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    |cos(k)/k^(3/2)|<=1/k^(3/2). It's bounded by a convergent p-series. Can't you use the integral test to get an error estimate as well?
     
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