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-Interval of Convergence

  1. Jul 8, 2013 #1
    1. The problem statement, all variables and given/known data
    [itex]\Sigma[/itex] from n=0 to infinity (x-4)^(2n)/((n+1)(11^n))
    Find the Radius of Convergence


    2. Relevant equations

    limit of (Cn/(Cn+1)
    Which I found it to be 11 (isn't this supposed to be the Radius of Convergence?)


    3. The attempt at a solution
     
  2. jcsd
  3. Jul 8, 2013 #2
    To find the radius of convergence of a geometric series [itex]\sum_{n=0}^{\infty}a_n[/itex] you take the following limit and solve for x, or an expression involving x.
    [tex]a_{n}=\frac{(x-4)^{2n}}{(n+1)11^{n}}[/tex]
    [tex]\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|<1[/tex]
    [tex]\lim_{n\rightarrow\infty}\left|\frac{(x-4)^{2(n+1)}}{(n+2)11^{n+1}}\cdot\frac{(n+1)11^{n}}{(x-4)^{2n}}\right|<1[/tex]
    Can you continue from here?

    Edit: I inserted the absolute value symbols that I originally forgot.
     
    Last edited: Jul 8, 2013
  4. Jul 8, 2013 #3

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    Let y = (x-4)^2, and re-write the series.
     
  5. Jul 8, 2013 #4
    From what u guys said I got ((x-4)^2)/11
     
  6. Jul 9, 2013 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Ok, so how does that translate into radius of convergence? Where is |(x-4)^2/11|<1?
     
  7. Jul 9, 2013 #6
    well i got sqr(-11) +4 < x < 15?
     
  8. Jul 9, 2013 #7

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's pretty wrong. Can you show how you got it??!!
     
  9. Jul 9, 2013 #8
    well

    -1< (x-4)^2/11 < 1 (is this right?)

    then i solved that and got to that soltuion. I had a feeling that its wrong since it came out with some weir dnumbers

    could it be -7<x<15
     
  10. Jul 9, 2013 #9

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Is there any way how (x-4)^2/11 could be negative? If not, what is the lower limit?

    Note that the upper limit (<1) gives two separate limits for x.

    What happens if you set x=14 in the equation above?
     
  11. Jul 9, 2013 #10
    O ok, ummm i stil hve to think about this but


    I guess it cant be negative, sorry I wasnt thinking. I solved it again and would it be 0< eqn < sqrt(11) +4?

    Im still kinda confused on how this works, sorry.
     
  12. Jul 9, 2013 #11

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    What is eqn?

    Please show all steps you did, otherwise it is hard to guess where you did something wrong.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Series-Interval of Convergence
Loading...